## 1st PUC Maths Question Bank Chapter 6 Linear Inequalities

Students can Download Maths Chapter 6 Linear Inequalities Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## karnataka 1st PUC Maths Question Bank Chapter 6 Linear Inequalities

Question 1.
Define inequality.
Two real numbers or two algebraic expressions related by the symbol ‘<’, V, ‘<’ or ‘>’ form an inequality.
For example: 3 < 5, 7 > 5 are numerical inequalities x < 5,y > 2, x > 3 are literal inequalities. ax + b < 0, ax + b > 0 are strick inequalities in one variable.
ax + b < 0, ax + b >0 are slack inequalities in one variable.
ax + by < c, ax + by > c are strick inequalities in two variables.
ax + by < c, ax + by > c are slack inequalities in two variables.
ax2 + bx + c < 0, ax2 + bx + c > 0 are quadratic inequalities in one variable.

Question 2.
Define solution set.
The set of all those values of x which satisfy the given inequation is called the solution set of inequation.

Rules for solving inequations:

• Rule 1: Adding the same number or expression to each side of an inequation docs not change the inequality.
• Rule 2: Subtracting the same number or expression from each side of an inequation does not change the inequality.
• Rule 3: Multiplying (or dividing) each side of an inequation by the same positive number does not change the inequality.
• Rule 4: Multiplying (or dividing) each side of an inequation by same negative number reverses the inequality.

Inequality solver is an application help you to solve linear inequality and tracing linear equations and root point.

Question 3.
Solve 30x < 200 when
(i) x is a natural number
(ii) x is an integer.
Given: 30x < 200
$$x<\frac{200}{30}=6.6$$

(i) x is a natural number:
Solution set = { 1, 2, 3,4, 5, 6}

(ii) x is a real number:
∴ Solution set = {… -3, -2, -1,0, 1,2, 3,4, 5, 6}

Question 4.
Solve 24x < 100, when
(i) x is a natural number
(ii) x is an integer.
Given: 24x < 100
$$x<\frac{100}{24}=4.16$$
(i) when x∈N, solution set ={ 1, 2, 3, 4}
(ii) when x∈Z,
∴ Solution set= {… -3, -2, -1, 0,1, 2, 3, 4}

Question 5.
Solve -12x > 30, when
(i) x∈N
(ii) x∈Z
Given: -12x > 30
$$\Rightarrow x<\frac{30}{-12}=-2.5$$
(i) When x∈N, solution set = φ
(ii) When x∈Z, solution set ={…, -5, -4, -3}

Question 6.
Solve: 5x – 3 < 7, when
(i) x is an integer
(ii) x is a real number.
Given: 5x – 3 < 7
⇒ 5x<7 + 3 = 10
∴ x< $$\frac{10}{5}=2$$
(i) If x is an integer,
solution set = {…, -3, -2, -1, 0, 1}.
(ii) If x is a real number, solution set = (-∞, 2).

Question 7.
Solve: 3x + 8 > 2, when
(i) x is an integer
(ii) x is a real number.
Given: 3x + 8 > 2
⇒ 3x > 2 – 8 = – 6
∴ x>-$$\frac{6}{3}=-2$$
(i) When x∈Z, solution set = {-1, 0, 1, 2,3,…}
(ii) When x∈M, solution set = (-2,∞).

Question 8.
Solve: 5x – 3 < 3x + 1 when
(i) x is an integer
(ii) x is a real number.
Given: 5x – 3 < 3x + 1
⇒ 5x – 3x <1 + 3
⇒ 2x<4
$$\Rightarrow x<\frac{4}{2}=2$$
(i) When x∈Z, solution set={…, -3,-2,-1, 0,1}
(ii) When x ∈M, solution set = (- ∞, 2).

Question 9.
Solve: 4x + 3 < 6x + 7.
Given: 4x + 3 < 6x + 7
⇒ -7 + 3 < 6x – 4x
⇒ -4 <2x
⇒ $$\Rightarrow x>\frac{-4}{2}=-2$$
∴ Solution set = (-2, ∞).

Question 10.
Solve: 4x + 3 < 5x + 7
Given: 4x + 3 < 5x + 7
⇒ -7 + 3 < 5x – 4x
⇒ -4 < x
∴ Solution set = (-4, ∞).

Question 11.
Solve: 3x – 7 > 5x -1
Given: 3JC – 7 > 5x – 1
⇒1 – 7 > 5x – 3x
⇒-6 > 2x
∴ ⇒ x>$$\frac{-6}{2}=-3$$
∴  Solution set = (-3, ∞)

Question 12.
Solve: 3(x – 1) < 2 (^ – 3)
Given: 3(x – 1) < 2 (x – 3)
⇒3x-3<2x-6
⇒ 3x – 2x < -6 + 3
⇒x<-3
∴  Solution set = (-∞,-3].

Question 13.
Solve: 3(2 – x) > 2(1 – x)
Given 6 – 3x > -2x + 3x
⇒ 6 – 2 > -2x + 3x
⇒ 4>x.
∴ Solution set = (-∞, 4]

Question 14.
Solve:
$$x+\frac{x}{2}+\frac{x}{3}<11$$

Question 15.
Solve:
$$\frac{x}{3}>\frac{x}{2}+1$$

Question 16.
Solve:
$$\frac{5-2 x}{3} \leq \frac{x}{6}-5$$

Question 17.
Solve:
$$\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}$$

Question 18.
Solve:
$$\frac{1}{2}\left(\frac{3 x}{5}+4\right) \geq \frac{1}{3}(x-6)$$

Question 19.
Solve: 2(2x+ 3) -10 < 6(x -2)
Given: 4x + 6-10<6x-12
⇒ 12-4<6x-4x
⇒ 8 < 2x
$$\Rightarrow \quad x>\frac{8}{2}=4$$
∴ Solution set = (4, ∞)

Question 20.
Solve: 37 – (3x+ 5) ≥ 9x – 8(x – 3)
Given: 37 – (3x+ 5) ≥ 9x – 8(x – 3)
⇒ 32 – 24 ≥ x + 3x
⇒ 8 >4x
⇒x<2
∴  Solution set = (-∞,2)

Question 21.
Solve:
$$\frac{x}{4}<\frac{5 x-2}{3}-\frac{7 x-3}{5}$$
$$\frac{x}{4}<\frac{5(5 x-2)-3(7 x-3)}{15}-\frac{4 x-1}{15}$$
⇒15x<16x-4
⇒ 4<16x-15x
⇒ x>4
∴ Solution set = (4, ∞).           .

Question 22.
Solve:
$$\frac{2 x-1}{3} \geq \frac{3 x-2}{4}-\frac{2-x}{5}$$

Question 23.
Solve the following inequalities and show the graph of the solution in each case on number line
(i) 3x – 2 < 2x +1
(ii) 5x – 3 > 3x- 5
(iii) 7x + 3 < 5x + 9
(iv) 3(1-x)<2(x + 4)
(v) $$\frac{3 x-4}{2} \geq \frac{x+1}{4}-1$$
(vii) $$\frac{x}{2} \geq \frac{5 x-2}{3}-\frac{7 x-3}{5}$$
(i) Given: 3x – 2 < 2x +1
⇒ 3x-2x<1 + 2
⇒ x < 3
∴ Solution set = (-<», 3)
The graph of the solution set is given below

Question 24.
The marks obtained by a student of class XI in first and second terminal examination are 62 and 48, respectively. Find the minimum marks he should get in the annual examination to have an average of at least 60 marks.
Let x be the marks obtained by student in the annual examination.
Then
\begin{aligned} &\frac{62+48+x}{3} \geq 60\\ &\Rightarrow 110+x \geq 180\\ &\Rightarrow x \geq 70 \end{aligned}
Thus, the student must obtain a minimum of 70 marks to get an average of at least 60 marks.

Question 25.
Ravi obtained 70 and 75 marks in first two unit test. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Let x be the marks obtained in the third test.
$$\frac{70+75+x}{3} \geq 60$$
⇒ 145+ x≥180
⇒ x> 180-145 = 35.
Thus, Ravi must obtain a minimum of 35 marks to get an average of atleast 60 marks.

Question 26.
To receive grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade ‘A’ in the course.
Let x be the marks in the fifth test obtained by Sunita
$$\frac{87+92+94+95+x}{5} \geq 90$$
⇒ 368 + x>450⇒x>82
So, Sunita must obtain a minimum of 82 marks to get grade ‘A’.

Question 27.
Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 40.
Let x be the smaller of the two consecutive odd natural number, so that the other one is x + 2.
Given x> 10 and x + (x + 2) < 40
⇒ 2x < 38
⇒ x<19
∴10 <x< 19
Since x is an odd number, x can take the values 11,13, 15 and 17.
∴ Required possible pairs are
(11,13), (13,15), (15, 17), (17,19).

Question 28.
Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.
Let x and x + 2 be the required pairs of consecutive odd positive integers.
Given: x + 2<10
⇒ x<8 and
⇒ x + x + 2>11
⇒ 2x>9
⇒ x > 4.5
∴ 4.5 < x < 8
x = 5, 7
∴ Required possible pairs are (5, 7), (7, 9).

Question 29.
Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.
Let x and x + 2 be the required pairs of consecutive even positive integers.
Given: x > 5
x + x + 2 <23
⇒ 2x< 21
⇒ x< 10.5
∴ 5 < x < 10.5
x = 6, 8, 10
∴ Required possible pairs are (6, 8), (8, 10), (10,12)

Question 30.
The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter then the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
Let x, 3x, 3x – 2 be the sides of a triangle
Given: x + 3x + 3x – 2 ≥ 61
7x ≥63
x ≥ 9.
∴ Minimum length of shortest side = 9 cm.
∴ Other sides are 27 and 25.

Question 31.
A man wants to cut three lengths from a single piece of board of length 91 cm. The second length is to be 3 cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5 cm longer than the second?
Let x be the length of the shortest board then
x + 3 and 2x are the lengths of the second and third piece, respectively.
⇒ x+(x+3) + 2x≤91 and 2x≥(x + 3) + 5
4x≤88 and x≥8
⇒ x≤22 and x ≥ 8
⇒ 8 ≤ x ≤22.
∴ The possible length of board is at least 8 cm long but not more than 22 cm long.

Question 32.
Solve the following inequalities graphically in two dimensional plane:
(i) x > -3
(ii) y < -2
(iii) x + y< 5
(iv) 2x – 3y > 6
(v) 3x + 2y > 6
(vi) 3y – 5x < 30
(vii) y>2
(viii) 3x – 6 > 0
(ix) 2x + y > 6
(x) 3x + 4y < 12
(xi) y + 8 > 2x
(xii) x-y <2
(xiii) -3x + 2y > -6
(i) Consider the equation x = -3.
Draw a graph of x = -3 by dotted line, which is parallel to y-axis, to the left of y-axis at distance
Put x = 0 in the given inequation x > -3, we get 0 > -3 which is true.
∴ Solution region consists of (0, 0).

∴ The shaded region is the solution region

(ii) y < -2 ………….(1)
Consider the equation y = -2.
Draw a graph of y = -2 by dotted line, which is parallel to x-axis, below x-axis and at distance 2. Put y = 0 in (1), 0 < -2 which is not true.
∴ Solution region is not containing (0, 0).

∴ Shaded region is the solution region.

(iii) x + y < 5  ………….(1)
Draw the graph of x + y = 5 by dotted line.
Points (5, 0) and (0, 5) lie on it. Join these points.
Put x = 0 and y = 0 in (1), we get 0 + 0 < 5 which is true.
∴ The solution region contains the origin.

∴ The shaded region is the solution region

(iv) 2x- 3y > 6  ………….(1)
Draw the graph of 2x – 3y = 6 by dotted line It passes through (3, 0) and (0, -2). Join these points.
Put x = 0 and y = 0 in (1), we get 0 – 0 > 6 which is not true.
∴ Solution region does not contain the origin

∴ Shaded region is the solution region

(v) 3x + 2y > 6 ………….(1)
Draw the graph of 3x + 2y = 6 by dotted line.
It passes through (2, 0) and (0, 3). Join these points.
Put x = 0 and y = 0 in (1), we get 0 + 0 > 6 which is not true.
∴ The solution region does not contain the origin.

∴ The shaded region is the solution region,

(vi) 3y- 5x< 30 …(1)
Draw the graph of 3y – 5x = 30 by dotted line.
It passes through (-6,0) and (0,10)
Join these two points put x = 0 and y = 0 in (1),
we get 0 – 0 < 30 which is true.
∴ The solution region contains the origin

The shaded region is the solution region

(vii) y > 2 ………………(1)
Draw the graph of y = 2 by dotted line Put y = 0 in (1)
we get 0 > 2 which is false.
∴ Solution region doesn’t contain the origin.

∴ The shaded region is the solution region.

(viii) 3x- 6 ≥ 0 ………………..(1)
Draw the graph of 3x -6 = 0 i.e., x = 2 by thick line
Put x = 0 in (1), we get -6 > 0 which is false.
∴ Solution region does not contains the origin

The shaded region is the solution region,

(ix) 2x + y >6 …………………..(1)
Draw the graph of 2x + y = 6 by thick line.
It passes through (3, 0) and (0, 6).
Join these points, put x = 0 and y = 0 in (1), we get 0 + 0 > 6 which is false.
∴ Solution region does not contain the origin

(x) 3x + 4y < 12 …………… (1)
Draw the graph of 3x + 4y = 12 by thick line.
It passes through (4, 0) and (0, 3).
Join these points, put x = 0 and y = 0 in (1), we get 0 + 0 ≤ 12 which is true.
∴ The solution region contains the origin.

∴ The shaded region is the solution region,

(xi) y + 8 ≥2x ………… (1)

Draw the graph of y+8 = 2 by thick line
It passes through (4, 0) and (0, -8)
Join these points, put x = 0 and y = 0 in (1), we get 0+8≤0 which is true.
∴ The solution region contains the origin
∴ The shaded region is the solution region,

(xii)  x – y ≤ 2 ………… (1)
Draw the graph of x – y ≤ 2 by thick line
It passes through (2, 0) and (0, -2)
Join these points, put x = 0 and y = 0 in (1), we get 0 – 0 < 2 which is true.
∴ The solution region contains the origin

∴ The shaded region is the solution region.

(xiii) -3x + 2y>-6 ………… (1)
Draw a graph of -3x + 2y = -6 by thick line.
It passes through (2, 0) and (0, -3)
Join these points, put x = 0 and y = 0 in (1), we get 0 + 0 > -6 which is true.
∴ The solution region contains the origin

∴ The shaded region is the solution region.

Question 33.
Solve the following system of inequalities graphically.
(j) x≥3,y≥2
(ii) x+y≥5,x-y≤3

(i) x≥3 ……………………… (1)
First draw the line x = 3 by thick line.
Put x = 0 in (1), we get 0 > 3 is not true.
∴ Solution set of (1) is not containing the origin.
y≥2 ………………….. (2)
Now draw the line y = 2 by thick line Put y = 0 in (2), we get, 0 > 2 is not true.
∴ Solution set of (2) is not containing the origin.
∴ Shaded region is the required solution region.

(ii) Consider x + 3 ≥ 5 ……………….. (1)
Draw the graph of x + y = 5 by thick line.
It passes through (5, 0) and (0, 5)
Join these points put x = 0 and y = 0 in (1), we get 0 + 0 ≥ 5 which is false
∴ Solution set of (1) is not containing the origin.

Consider x-y ≤3  ……………….. (2)
Draw the graph of x – y = 3 by thick line It passes through (3,0) and (0, -3)
Join these points, put x = 0 and y = 0 in (2), we get 0 – 0 < 3 which is true.
∴ Solution set of (2) is containing the origin.
Hence, the shaded region represents the solution set of the given system of inequalities

(iii) Consider 2x + y ≥ 6  ……………….(1)
Draw the graph of 2x + y = 6 i.e. $$\frac{x}{3}+\frac{y}{6}=1$$by thick line.
It passes through (3, 0) and (0, 6).
Join these points put x = 0 and y = 0 in (1), we get 0 + 0 ≥ 6 which is false.
∴ Solution set of (1) is not containing the origin.
Consider 3x+4y ≤12  ……………….. (2)
Draw the graph of 3x+4y ≤12 i.e
$$\frac{x}{4}+\frac{y}{3}=1$$by thick line.

It passes through (4, 0) and (0, 3).
Join these points put x = 0 and y = 0 in (2), we get 0 + 0 < 12 which is true.
∴ Solution set of (2) is containing the origin.
Hence, the shaded region represents the solution set of the given system of inequalities.

(iv) Consider x+y≥4 ………………(1)

Draw the graph of x+y=4 by thick line.
It passes through (4, 0) and (0,4)
Join these points put x = 0 and y = 0 in (1), we get 0 + 0 > 4 which is false.
∴ Solution set of (1) is not containing the origin.
Consider 2x – y > 0 ………………(2)
Draw the graph of 2x – y = 0 by dotted line.
It passes through (0, 0) and (1,2)
Join these points put x = 2 and y = 1 in (2),
we get
2(2) – 1 > 0 which is true.
∴ Solution set of (2) is containing the point (2, 1)
Hence, the shaded region represents the solution set of the given system of inequations.

(v) Consider 2x -y > 1 ………………(1)
Draw the graph of 2x – y = 1 by dotted line
It passes through $$\left(\frac{1}{2}, 0\right)$$[ and (0,-1).
Join these points, put x = 0 and y = 0 in (1), we get 0 – 0 > 1 which is false.
∴ Solution set of (1) is not containing the origin.
Consider x – 2y < -1 ……………… (2)
Draw the graph of x – 2y = -1 by dotted line.
It passes through (-1, 0) and $$(-1,0) \text { and }\left(0, \frac{1}{2}\right)$$
Join these points, put x = 0 and y = 0 in (2) we get 0 – 0 < -1 which is false.
∴ Solution set of (2) is not containing the origin.
Hence, the shaded region represents the solution set of the given system of inequations.

(vi) Consider x+y≤6 ………………(1)
Draw the graph of x + y = 6 by thick line. It passes through (6, 0) and (0, 6)
Join these points put x = 0 and y = 0 in (1), we get 0 + Q < 6 which is true.
∴ Solution set of (1) is containing the origin.
Consider x + y ≥ 4  ……………… (2)
Draw the graph of x + y = 4 by thick line It passes through (4, 0) and (0, 4).
Join these points, put x = 0 and y = 0 in (2), we get 0 + 0 ≥ 4 which is false
∴ Solution set of (2) is not containing the origin.
Hence, the shaded region represents the solution set of the given system of inequations.

(vii) Consider 2x+y≥8 ……………… (1)
Draw the graph 2x + y = 8 by thick line.
It passes through (4, 0) and (0, 8)
Join these points, put x = 0 and y = 0 in (1), we get 0 + 0 ≥ 8 which is false
∴ Solution set of (1) is not containing the origin.
Consider x + 2y > 10 ……………… (2)
Draw the graph x + 2y = 10 by thick line It passes through (10, 0) and (0, 5).
Join these points, put x = 0 and y = 0 in (2), we get 0 + 0 > 10 which is false
Solution set of (2) is not containing the origin.
Hence, the shaded region represents the solution set of the given system of inequations.

(viii) Consider 8x + 3y ≤100 ……………… (1)
Draw the graph of 8x + 3y = 100 by thick line
It passes through $$\left(\frac{25}{2}, 0\right) \text { and }\left(0, \frac{100}{3}\right)$$
Join these points, put x = 0 and y = 0 in (1),
we get 0+0 ≤ 100 which is true.
∴ Solution set of (1) contains the origin.
Since x ≥0, y ≥0, every point in the shaded region in the first quadrant, including the points on the line and the axes, represents the solution of the given system of inequations.

(ix) Consider 5x + 4y ≤ 40 ………………… (1)
Draw the graph of 5x + 4.y = 40 by thick line.
It passes through (8, 0) and (0, 10).
Join these points, put x = 0 and y = 0 in (1), we get 0 + 0 ≤ 40 which is true.
Solution set of (1) contains the origin.
Consider x ≥ 2 ………………… (2)
Draw the graph of x= 2.
Clearly solution set of (2) is not containing the origin.
Consider y ≥ 3  ………………… (3)
Draw the graph of y = 3
Clearly solution set of (3) is not containing the origin.
Hence, the shaded region represents the solution set of the given system of inequations.

(x) Consider 3x + 2y≤12 ………………… (1)
Draw the graph of 3x + 2y = 12 by thick line.
It passes through (4, 0) and (0, 6).
Join these points put x = 0 and y = 0 in (1), we get
0+0≤12 which is true.
∴ Solution set of (1) contains the origin
Consider x ≥1 …………….. (2)
Draw the graph of x= 1
Clearly solution set of (2) is not containing the origin.
Consider y ≥ 2 … (3)
Clearly solution set of (3) is not containing the origin.
Hence, the shaded region represents the solution set of the given system of inequations.

(xi) Consider 5x+4y≤20 …………(1)
Draw the graph of 5x+4y = 20 by thick line
It passes through (4, 0) and (0, 5).
Join these points, put x = 0 and y = 0 in (1), we get 0 + 0 < 20 which is true.
∴ Solution set of (1) contains the origin. Consider x > 1 …………….. (2)
Draw the graph of x = 1 clearly solution set of (2) is not containing the origin.
Consider y > 2 …………….. (3)

Draw the graph of y = 2 clearly solution set of (3) is not containing the origin.
Hence, the shaded region represents the solution set of the given system of inequations.

(xii) Consider x + y ≤ 9 …………….. (1)
Draw the graph of x + y = 9 by thick line.
It passes through (9, 0) and (0, 9).
Join these points, put x = 0 and y = 0 in (1), we get 0 + 0 < 9 which is true.
Consider y > x …………….. (2)
Draw the graph of y = x by dotted line.
Clearly solution set of (2) does not contain (1,0).
Since x ≥ 0, every point in the shaded region in the first quadrant, including the points on the line x + y = 9, excluding the points on the line y ≥ x and y-axis, represents the solution of the given system of inequations.

(xiii) Consider x + 2y≤8 …………….. (1)
Draw the graph of x + 2y = 8 by thick line. It passes through (8, 0) and (0,4).
Join these points, put x = 0 and y = 0 in (1), we get 0 + 0 < 8 which is true.
Solution set of (1) is containing the origin.
Consider 2x + y≤8 …………….. (2)
Draw the graph of 2x + y = 8 by thick line. It passes through (4, 0) and (0, 8)
Join these points, put x = 0 and y = 0 in (2), we get 0 + 0 < 8 is true.
Solution set of (2) is containing the origin. Since x > 0, y > 0, every point in the shaded region in the first quadrant, including the points on the lines, represents the solution of the given system of inequations.

(xiv) Consider 3x+4y≤60 …………. (1)
Draw the graph of 3x+4y=60 by thick line.
It passes through (20, 0) and (0, 15). Join these points, put x = 0 and y = 0 in (1), we get 0 + 0 < 60 is true.
∴ Solution set of (1) contains (0, 0). Consider x + 3y < 30 ………………… (2)
Draw the graph of x + 3y = 30.

It passes through (30,0) and (0,10)
Join these points, put x = 0 and y = 0 in (2), we get 0 + 0 ≤ 30 is true.
∴ Solution set of (2) contains (0, 0).
Since x ≥ 0, y ≥ 0 every point in the shaded region in the quadrant, including the points on the lines, represents the solution of the given solution of inequations.

(xv) Consider 2x + y > 4  …………. (1)
Draw the graph of 2x + y = 4 by thick line It passes through (2, 0) and (0, 4)
Join these points, put x = 0 and y = 0 in (1), we get 0 + 0 ≥ 4 is not true.
∴ Solution set of (1) does not contain (0, 0).

Consider x + y≤3 …………. (2)
Draw the graph of x + y = 3, by thick line.
It passes through (3, 0) and (0, 3).
Join these points, put x = 0 and y = 0 in (2), we get 0 + 0 < 3, which is true

Solution set of (2) contains (0, 0) Consider 2x-3y≤ 6 …………. (3)
Draw the graph of 2x – 3y = 6 by thick line It passes through (3, 0) and (0, -2)
Join these points, put x = 0 and y = 0 in (3), we get 0 – 0 < 6, which is true.
∴ Solution set of (3) contains (0, 0)
Hence, the shaded region represents the solution set of the given system of inequations.

(xvi) Consider x – 2y≤3 ………….. (1)
Draw the graph of x -2y = 3 by thick line.
It passes through $$(3,0) \text { and }\left(0,-\frac{3}{2}\right)$$
Join these points, put x = 0 and y = 0 in (1), we get 0 – 0 < 3, which is true.
∴Solution of (1) contains the origin.
Consider 3x + 4y > 12 …………. (2)
Draw the graph of 3x + 4y = 12 by thick line.
It passes through (4, 0) and (0, 3).
Join these points, put x = 0 and y = 0 in (2) we get 0 + 0 > 12 which is false.
∴ Solution set of (2) does not contain the origin.
Consider y>1   ………….. (3)
Draw the graph of y = 1 clearly solution set of (3) does not contain (0, 0).
Since x > 0, every point in the shaded region in the first quadrant, including the points on the lines, represents the solution of the given system of inequations.

(xvii) Consider 4x+3y≤60 ……….. (1)
Draw the graph of 4x+3y = 60 by the thick line.
It passes through (15, 0) and (0, 20)
Join these points, put x = 0 and y = 0 in (1), we get 0 – 0 ≤ 60 ,which is true.
∴ Solution set of (1) contains(0,0)
Consider y ≥2x ……………. (2)
Draw the graph of y = 2x by thick line
It passes through (0, 0) and (1,2)
Join these points, clearly solution set of (2) is not containing (1,1)
Consider x ≥ 3 ……………. (3)
Draw the graph of x = 3 clearly solution set of (3) does not contain the origin.
Since y > 0, every point in the shaded region in the first quadrant, including the points on the lines, represents the solution of the given system of inequations.

(xviii) Consider 3x+2y≤150 ……………. (1)
Draw the graph of 3x+2y = 150
It passes through (50, 0) and (0, 75) put x = 0 and y = 0 in (1), we get 0 +0 < 150, which is true.
∴ Solution set of (1) contains (0, 0)
Consider x + 4y < 80 ……………. (2)
Draw the graph of x + 4y = 80.
It passes through (80, 0) and (0, 20).
Join these points, put x = 0 and y = 0 in (2), we get 0 + 0 < 80, which is true.
∴ Solution set of (2) contains (0, 0) Consider x < 15 ……………. (3)
Clearly solution set of (3) does not contain the origin since x ≥ 0, y ≥ 0, every point in the shaded region in the first quadrant, including the points on the lines, represents the solution of the given system of inequations.

(xix) Consider x+2y≤10 …………(1)
Draw the graph of x+2y=10
It passes through (10, 0) and (0, 5) put x – 0 and y = 0 in (1), we get 0 + 0 < 10, which is true.
∴ Solution set of (1), contain (0, 0)
Consider x + y > 1  ………… (2)
Draw the graph of x + y = 1
It passes through (1,0) and (0, 1) put x = 0 and y = 0 in (2), we get 0 + 0 > 1, which is false.

∴ Solution set of (2) is not containing the origin
Consider x – y < 0 …………  (3)
Draw the graph of x – y = 0.
It passes through (0, 0) and (1, 1) put x = 2 and y = 0 in (3) we get
2 – 0 < 0 which is false
∴ Solution set of (3) does not contain (2, 0)
Since x > 0, y > 0, every point in the shaded region in the first quadrant, including the points on the lines, represents the solution of the given system of inequations.

Question 34.
Solve the Equations.
(i) 2≤3x-4≤5
(ii) -8≤ 5x – 3< 7
(iii) 6 < -3(2JC – 4) < 12
(iv) $$-3 \leq 4-\frac{7 x}{2} \leq 18$$
(v) $$-5 \leq \frac{5-3 x}{2} \leq 8$$
(vi) $$-15<\frac{3(x-2)}{5} \leq 0$$
(vii) $$7 \leq \frac{3 x+11}{2} \leq 11$$
(viii) $$-12<4-\frac{3 x}{-5} \leq 2$$

Question 35.
Solve the inequalities and represent the solution graphically on number line:
(i) 5x+ 1 > -24, 5x-1 < 24
(ii) 3x – 7 < 5 + x, 11 – 5x≤1
(iii) 3x – 7 > 2(x – 6), 6 – x > 11. – 2x
(iv) 2(x – 1) < x + 5,3(x + 2) > 2 – x
(v) 5(2x – 7) – 3(2x+ 3)≤0, 2x + 19≤6x + 47

Question 36.
In an experiment, a solution of hydrochloric acid is to be kept between 30° and 35° Celsius. What is the range of temperature in degree Fahrenheit if conversion formula is given by $$C=\frac{5}{9}(F-32)$$, where C and F represents temperature in degree Celsius and degree Fahrenheit, respectively.

Question 37.
A solution is to be kept between 68 °F and 77 °F. What is the range in temperature in degree Celsius (C) if the Celsius/Fahrenheit (F) conversion formula is given by
$$F=\frac{9}{5} C+32$$

Question 38.
A solution of 8% boric acid to be diluted by. adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 liters of the 8% solution, how many liters of the 2% solution will have to be added?
Let x be the number of liters of 2% boric acid solution.
∴ Total mixture = (640 + x ) liters.
$$\Rightarrow \frac{2 x}{100}+\frac{8}{100}(640)>\frac{4}{100}(640+x)$$
⇒ 2x+ 5120 >2560+ 4x
⇒ 5120-2560 >4x-2x
⇒ 2x< 2560
⇒ x< 1280 ………. (1)
Also, 2% of x+ 8% of 640 < 6% of (640 + x)
$$\Rightarrow \frac{2 x}{100}+\frac{8}{100}(640)<\frac{6}{100}(640+x)$$
⇒ 2x + 5120 < 3840 + 6x
⇒ 5120-3840 <6x-2x
⇒ 1280 < 4x
⇒ 320< x …………….. (2)
From (1) and (2), we get 320 < x< 1280

Question 39.
A manufacturer has 600 liters of 12% solution of acid. How many liters of a 30% acid solution must be added to it so that acid content in the resulting mixture will be more than 15% but less than 18%?
Let x liters of 30% acid solution is required to be added.
Then, total mixture = (x + 600) liters.
∴ 30% of x + 12% of 600 > 15% of (x + 600)
$$\Rightarrow \frac{30 x}{100}+\frac{12}{100}(600)>\frac{15}{100}(x+600)$$
⇒ 30x + 7200 > 15x + 9000
⇒ 15x> 9000 -7200= 1800
⇒ x > 120 …………………… (1)
Also 30% of x + 12% 600 < 18% of (x + 600)
$$\Rightarrow \frac{30 x}{100}+\frac{12}{100}(600)<\frac{18}{100}(x+600)$$
⇒ 30x + 7200 < 18x+ 10800
⇒ 12x < 10800 – 7200 = 3600
⇒ x <300 …………. (2)
From (1) and (2) we get 120 < x < 300
Thus, the number of liters of the 30% solution of acid will have to be more than 120 liters but less than 300 liters.

Question 40.
How many liters of water will have to be added to 1125 liters of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?

Question 41.
IQ of a person is given by the formula $$I Q=\frac{M A}{C A} \times 100$$ where MA is mental age and CA CA is a chronological age. If 80 < IQ < 140 for a group of 12 years old children, find the range of their mental age.

## Karnataka 1st PUC Hindi Textbook Answers Sahitya Vaibhav Chapter 4 बिन्दा

### बिन्दा Questions and Answers, Notes, Summary

I. एक शब्द या वाक्यांश या वाक्य में उत्तर लिखिएः

प्रश्न 1.
महादेवी वर्मा की बाल्य सखी का नाम लिखिए।
उत्तरः
महादेवी वर्मा की बाल्य सखी का नाम बिन्दा है।

प्रश्न 2.
महादेवी वर्मा को किसका अनुभव बहुत संक्षिप्त था?
उत्तरः
महादेवी वर्मा को संसार का अनुभव बहुत संक्षिप्त था।

प्रश्न 3.
पंडिताइन चाची के अलंकार उन्हें किसकी समानता दे देते थे?
उत्तरः
पंडिताइन चाची के अलंकार उन्हें गुड़िया की समानता दे देते थे।

प्रश्न 4.
बिन्दा का काम लेखिका को किसके तमाशे जैसा लगता था?
उत्तरः
बिन्दा का काम लेखिका को बाजीगर के तमाशे जैसा लगता था।

प्रश्न 5.
बिन्दा की आँखें लेखिका को किसकी याद दिलाती थीं?
उत्तरः
बिन्दा की आँखें लेखिका को पिंजड़े में बंद चिड़िया की याद दिलाती थीं।

प्रश्न 6.
बिन्दा ने तारे गिनते-गिनते एक चमकीले तारे की ओर उँगली उठाकर क्या कहा?
उत्तरः
बिन्दा ने तारे गिनते-गिनते एक चमकीले तारे की ओर उँगली उठाकर कहा कि – “वह रही मेरी अम्मा’।

प्रश्न 7.
महादेवी वर्मा ने रात को अपनी माँ से बहुत अनुनय पूर्वक क्या कहा?
उत्तरः
महादेवी वर्मा ने रात को अपनी माँ से बहुत अनुनयपूर्वक कहा कि माँ तुम कभी तारा न बनना।

प्रश्न 8.
महादेवी वर्मा किसके न्यायालय से मिलने वाले दण्ड से परिचित हो चुकी थीं?
उत्तरः
पंडिताइन चाची के न्यायालय से मिलने वाले दण्ड से महादेवी वर्मा परिचित हो चुकी थीं।

प्रश्न 9.
महादेवी वर्मा को किसकी पत्तियाँ चुभ रही थीं?
उत्तरः
महादेवी वर्मा को घास की पत्तियाँ चुभ रही थीं।

प्रश्न 10.
किसने बिन्दा के पैरों पर तिल का तेल लगाया?
उत्तरः
महादेवी वर्मा की माँ ने बिन्दा के पैरों पर तिल का तेल लगाया।

प्रश्न 11.
महादेवी वर्मा के लिए कौन त्रिकालदर्शी से कम न थी?
उत्तरः
महादेवी वर्मा के लिए रुकिया त्रिकालदर्शी से कम न थी।

अतिरिक्त प्रश्नः

प्रश्न 12.
‘बिन्दा’ रेखाचित्र की लेखिका कौन है?
उत्तरः
‘बिन्दा’ रेखाचित्र की लेखिका महादेवी वर्मा हैं।

प्रश्न 13.
पंडिताइन चाची को बिन्दा क्या कहकर पुकारती थी?
उत्तरः
पंडिताइन चाची को बिन्दा नयी अम्मा कहकर पुकारती थी।

प्रश्न 14.
नई अम्मा मोहन के साथ ऊपर के खण्ड में क्यों रहती थी?
उत्तरः
चेचक के डर से नई अम्मा मोहन के साथ ऊपर के खण्ड में रहती थी।

प्रश्न 15.
अंत में बालिका महादेवी वर्मा को बिन्दा के बारे में क्या पता चला?
उत्तरः
बालिका महादेवी वर्मा को पता चला कि वह तो अपनी आकाशवासिनी अम्मा के पास चली गई।

प्रश्न 16.
बहुत खुशामद करने पर रुकिया ने क्या बताया?
उत्तरः
बहुत खुशामद के बाद रुकिया ने बताया कि उसके घर में महारानी आई है।

II. निम्नलिखित प्रश्नों के उत्तर लिखिए:

प्रश्न 1.
महादेवी वर्मा को बिन्दा की याद क्यों आ गई?
उत्तरः
एक सज्जन बेटी को प्रवेश दिलाने स्कूल आये थे। उसकी कमजोर बेटी डरी हुई थी। उम्र ठीक न बताये जाने के कारण प्रवेश-पत्र लौटा दिया गया था। लेखिका ने सुना कि उन सज्जन की दूसरी पत्नी है, तो अचानक दो युगों से अधिक समय की धूल के नीचे दबी बिन्दा की याद आ जाती है।

प्रश्न 2.
महादेवी वर्मा को पंडिताइन चाची का कौन-सा रूप आकर्षित करता था?
उत्तरः
पंडिताइन चाची जब रंगीन साड़ी में सजकर, लाल स्याही की सिंदूर लगाकर, आँखों में काजल डालकर, चमकीले कर्णफूल, गले की माला, नगदार रंग-बिरंगी चूड़ियाँ और घुघरू वाले बिछुए पहनकर श्रृंगार करती, तो लेखिका महादेवी वर्मा को वह गुड़िया की तरह बहुत अच्छी लगती है।

प्रश्न 3.
महादेवी वर्मा के कभी-कभी छत पर जाकर देखने पर बिन्दा क्या-क्या करते दिखाई देती थी?
उत्तरः
जब कभी महादेवी वर्मा छत पर जाकर देखती, तो बिन्दा झाडू लगाती हुई, कभी आग जलाती हुई, कभी आँगन से पीने का पानी भरती हुई, तो कभी नयी अम्मा को दूध का कटोरा देती हुई दिखाई देती थी। ये सब बाजीगर के तमाशे की तरह लगता था।

प्रश्न 4.
बिन्दा अपनी नयी अम्मा से किस प्रकार डरती थी?
उत्तरः
बिन्दा नयी अम्मा से बहुत डरती थी। यहाँ तक कि थोड़ी बहुत ही आवाज आती अथवा आहट होती, तो वह काँपने लगती थी कि न जाने अब क्या होगा? नयी अम्मा की आवाज सुनते ही बिन्दा थर-थर काँपने लग जाती थी।

प्रश्न 5.
महादेवी वर्मा ने दोपहर के समय सबकी आँख बचाकर बिन्दा के घर पहुंचने पर क्या देखा?
उत्तरः
महादेवी वर्मा एक दिन दोपहर को सबकी आँख बचाकर बिन्दा के घर पहुंची, तो देखा कि नीचे के सुनसान खंड में बिंदा अकेली खाट पर पड़ी थी। आँखें धंस गई थीं। मुख दानों से भरकर न जाने कैसे हो गया था। लेखिका चकित चारों ओर देखती रह गई। बिन्दा ने कुछ संकेत और कुछ अस्पष्ट शब्दों में बताया कि नयी अम्मा मोहन के साथ ऊपर रहती है, शायद चेचक के डर से।

प्रश्न 6.
बिन्दा के घर के सामने भीड़ देखकर लेखिका के मन में क्या विचार आने लगे?
उत्तरः
बिन्दा के घर के सामने भीड़ देखकर लेखिका ने सोचा कि पंडितजी का विवाह तो दूसरी पंडिताइन चाची के मरने के बाद होगा। मोहन तो अभी छोटा है। शायद बिन्दा का विवाह हो रहा होगा और उसने मुझे बुलाया तक नहीं। अपमानित होकर बिन्दा को किसी भी शुभकार्य में न बुलाने की ठान ली।

अतिरिक्त प्रश्नः

प्रश्न 7.
बालिका महादेवी वर्मा ‘माँ’ नामधारी जीवों के कर्तव्यों के बारे में क्या समझा था?
उत्तरः
बालिका महादेवी ने यह समझा कि संसार का सारा कारोबार बच्चों को खिलाने-पिलाने और सुलाने का ही रहा है। इस महत्वपूर्ण कार्य में किसी भी प्रकार की चूक न होने देने का कार्य ‘माँ’ नामधारी जीवों को सौंपा गया है।

प्रश्न 8.
पंडिताइन चाची के स्वभाव के बारे में लिखिए।
उत्तरः
पंडिताइन चाची का स्वभाव विचित्र-सा था। उसके चिल्लाने की आवाजें बालिका महादेवी को सुनाई देती थीं। ‘उठती है या आऊँ, बैल के-से दीदे क्या निकाल रही है’, ‘मोहन का दूध कब गर्म होगा’, ‘अभागी मरती भी नहीं’ जैसे वाक्यों से वह बिन्दा को डाँटती रहती थी। वह बहुत ही कटु और उग्र स्वभाव की महिला थी।

प्रश्न 9.
बिन्दा के रूप-रंग का वर्णन कीजिए।
उत्तरः
बिन्दा नाटे कद की लड़की थी। उसके कद को देखकर ऐसा लगता था, मानो किसी ने ऊपर से दबाकर उसे कुछ छोटा कर दिया हो। उसके दुबले हाथ पाँव में से हरी-हरी नसें दीखती थी। उसके पाँव हरदम आने वाले भय के चलते थर-थर काँपते रहते थे।

प्रश्न 10.
बिन्दा और महादेवी वर्मा के घास की कोठरी में छिपने की कहानी लिखिए।
उत्तरः
एक दिन बिन्दा के हाथ से चूल्हे पर उफनते दूध की पतीली गिर पड़ी। वह नयी अम्मा के खौफ से भयभीत हो उठी। खौलते दूध से उसके पैर भी जल चुके थे। ऐसे में बिन्दा महादेवी का हाथ पकड़कर, उसे अपने घर में कहीं छिपा देने के लिए कहने लगी। दोनों गाय के लिए घास भरने वाली कोठरी में जा घुसी। महादेवी को घास की पत्तिया चुभ रही थी लेकिन बिन्दा अपने जले पैरों को घास में छिपाये महादेवी के हाथ पकड़े शांति से बैठी थी।

प्रश्न 11.
सबकी आँख बचाकर बिन्दा के घर पहुंचने पर महादेवी वर्मा ने वहाँ क्या देखा?
उत्तरः
महादेवी वर्मा ने देखा कि बिन्दा अकेली एक खाट पर पड़ी थी। उसकी आँखे गड्ढे में धंस गयी थी। मुख दानों से भर कर न जाने कैसा हो गया था। मैली सी चादर के नीचे छिपा उसका शरीर बिछौने से अलग नहीं जान पड़ता था। अर्थात् बिन्दा बहुत ही दुर्बल हो चुकी थी।

प्रश्न 12.
खिड़की से झाँक कर बिन्दा के दरवाजे पर जमा हुए आदमियों की भीड़ देखकर बालिका महादेवी ने क्या सोचा?
उत्तरः
बालिका महादेवी ने लोगों की भीड़ को देखकर सोचा कि शायद बिन्दा के घर में किसी का विवाह हो रहा हैं। किसका विवाह हो रहा है? इस तरह के प्रश्न उनके मन में तैरने लगे। वे इस नतीजे पर पहुँची कि शायद बिन्दा का विवाह हो रहा है और उसने मुझे बुलाया तक नहीं।

III. निम्नलिखित वाक्य किसने किससे कहे?

प्रश्न 1.
‘क्या पंडिताइन चाची तुम्हारी तरह नहीं है?’
उत्तरः
यह वाक्य महादेवी वर्मा ने अपनी माँ से कहा।

प्रश्न 2.
‘वह रही मेरी अम्मा’
उत्तरः
यह वाक्य बिन्दा ने महादेवी वर्मा से कहा।

प्रश्न 3.
‘तुम कभी तारा न बनना, चाहे भगवान कितना ही चमकीला तारा बनावें।’
उत्तरः
यह वाक्य महादेवी वर्मा ने अपनी माँ से कहा।

अतिरिक्त प्रश्नः

प्रश्न 4.
“आपने आयु ठीक नहीं भरी है।”
उत्तरः
महादेवी वर्मा ने सामने बैठे सज्जन से कहा।

प्रश्न 5.
“मेरी दूसरी पत्नी है और आप तो जानती ही होंगी ……..”
उत्तरः
सामने बैठे सज्जन ने लेखिका महादेवी वर्मा से कहा।

प्रश्न 6.
“अभागी मरती भी नहीं।’
उत्तरः
पंडिताइन चाची ने बिन्दा से कहा।

प्रश्न 7.
“क्या सवेरा हो गया?”
उत्तरः
महादेवी वर्मा ने अपनी सखी बिन्दा से कहा।

प्रश्न 8.
“क्या वे मुझसे नहीं मिल सकती?”
उत्तरः
महादेवी वर्मा ने अपनी सहेली रूकिया से कहा।

IV. ससंदर्भ स्पष्टीकरण कीजिएः

प्रश्न 1.
‘उठती है या आऊँ’, ‘बैल के-से दीदे क्या निकाल रही है’, ‘मोहन का दूध कब गर्म होगा’, ‘अभागी मरती भी नहीं’ आदि।
उत्तरः
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘बिन्दा’ नामक पाठ से लिया गया है जिसकी लेखिका महादेवी वर्मा हैं।
संदर्भ : पंडिताइन चाची यह शब्द बिन्दा को सुनाते हुए कहती हैं। इन वाक्यों में कठोरता की धारा बहती है।
स्पष्टीकरण : जब सर्दी के दिनों में लेखिका को देरी से उठाया जाता, गरम पानी से हाथ-मुँह धुलाये जाते, जूते और ऊनी कपड़े पहनाये जाते, जबरदस्ती गरम दूध पिलाया जाता, तब पड़ोस के घर में पंडिताइन चाची की आवाज कठोर शब्दों में सुनाई देती – अभागी तू अभी उठी नहीं? …… जल्दी से जाकर दूध ला ….. अभागी मर भी नहीं जाती।

प्रश्न 2.
‘तुम नयी अम्मा को पुरानी अम्मा क्यों नहीं कहती, फिर वे न नयी रहेगी और न डाँटेंगी।’
उत्तरः
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘बिन्दा’ नामक पाठ से लिया गया है जिसकी लेखिका महादेवी वर्मा हैं।
संदर्भ : जब बिन्दा महादेवी वर्मा को बताती है कि तारा बनकर मेरी माँ ऊपर से देखती रहती है और जो सजधज कर घर में आयी है वह नयी अम्मा है। तभी यह वाक्य महादेवी वर्मा अपनी सहेली बिन्दा से कहती है।
स्पष्टीकरण : बिन्दा की सौतेली माँ का उस पर अत्याचार होते देखकर लेखिका को बहुत दुख होता है। उनके बालसुलभ मन में यह बात बैठ जाती है कि जिस अम्मा को ईश्वर बुला लेता है, वह तारा बनकर ऊपर से बच्चों को देखती रहती है और जो बहुत सजधज से घर में आती है, वह बिन्दा की नयी अम्मा जैसी होती है। लेखिका की बुद्धि सहज ही पराजय स्वीकार करना नहीं जानती। इसी से उन्होंने सोचकर कहा कि यदि बिन्दा अपनी नयी अम्मा को पुरानी अम्मा कहेगी तो फिर वे नयी नहीं रहेगी और न डाँटेगी।

प्रश्न 3.
…. पंडिताइन चाची के न्यायविधान में न क्षमा का स्थान था, न अपील का अधिकार।
उत्तरः
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘बिन्दा’ नामक पाठ से लिया गया है जिसकी लेखिका महादेवी वर्मा हैं।
संदर्भ : नयी अम्मा अर्थात् पंडिताइन चाची के दिल में न दया थी और न क्षमा। वह बिन्दा को साधारण सी गलती पर भी उसे कठोर से कठोर दण्ड देती थी।
स्पष्टीकरण : एक दिन दूध गर्म करते समय दूध उफन कर बाहर निकलने लगा। तब बिन्दा उसे उतारने की कोशिश करने लगी पर वह दूध उसके पैरो पर गिर गया और पैर जल गया। लेकिन माँ के पास जाने की बजाय वह छुपना चाहती थी। वह अपनी सहेली के घर की कोठरी में छुप जाती है। तभी उन्हें पण्डिताइन चाची की उग्र आवाज सुनाई देती है। लेखिका की माँ बिन्दा के जले पैरों पर तिल का तेल लगाती है, और उसे उसके घर भिजवा देती है। पर चाची उसके प्रति सहानुभूति न दर्शाते हुए सेवा करने के बजाय कठोर शब्दों से डाटना, मारना शुरु करती है। इससे यही लगता है कि पंडिताइन चाची के न्यायविधान में न क्षमा का स्थान था, न अपील का अधिकार।

अतिरिक्त प्रश्नः

प्रश्न 4.
“नहीं, यह तो गत आषाढ़ में चौदह की हो चुकी।”
उत्तरः
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘बिन्दा’ नामक पाठ से लिया गया है। इसकी लेखिका महादेवी वर्मा जी हैं।
संदर्भ : महादेवी जी सामने बैठे सज्जन को प्रवेश पत्र लौटाते हुए जब कहती हैं कि इसमें उम्र सही नहीं भरी है, तब वह सज्जन यह जवाब देते हैं।
स्पष्टीकरण : एक सज्जन अपनी पुत्री का स्कूल में दाखिला करवाने आए हुए थे। महादेवी जी उस दुर्बल सी, छोटी सी लड़की पर नजर डालकर साथ आए सज्जन से पूछती है- शायद आपने इसमें इसकी उम्र सही नहीं भरी है। ठीक कर दीजिए, नहीं तो भविष्य कठिनाई आएगी। तब वे सज्जन कहते हैं, “नहीं, यह तो गत आषाढ़ में ही चौदह वर्ष की हो चुकी है।”

प्रश्न 5.
“आँखें गड्ढे में धंस गयी थी, मुख दानों से भर कर न जाने कैसा हो गया था।”
उत्तरः
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘बिन्दा’ नामक पाठ से लिया गया है। इसकी लेखिका महादेवी वर्मा जी हैं।
संदर्भ : एक दिन बालिका महादेवी बिन्दा को देखने चुपके से उसके घर में दाखिल हो जाती है। तब बिन्दा का ऐसा रूप देखती है।
स्पष्टीकरण : बहुत दिनों तक महादेवी ने जब बिन्दा को घर-आँगन में काम करते नहीं देखा तो एक दिन वह सबकी नजरें बचाकर बिन्दा के घर पहुंच गई। बिन्दा अकेली एक खाट पर पड़ी थी। उसकी आँखे गड्ढ़े में धंस गयी थीं। मुख दानों से भर कर न जाने कैसा हो गया था। बिन्दा बीमार हो गयी थी। उसको चेचक की बीमारी हो आयी थी। उसकी नयी अम्मा भी उसकी देखभाल नहीं करती थीं।

प्रश्न 6.
“तब क्या उस घर में विवाह हो रहा है, और हो रहा है तो किसका?”
उत्तरः
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘बिन्दा’ नामक पाठ से लिया गया है इसकी लेखिका महादेवी वर्मा जी हैं।
संदर्भ : बालिका महादेवी ने जब बिन्दा के घर के दरवाजे पर लोगों की भीड़ देखी तो उनके मन में यह प्रश्न पैदा हुआ।
स्पष्टीकरण : एक दिन रुकिया ने महादेवी की माँ को कुछ बताया और वह रामायण बन्द कर आँखे पोंछती बिन्दा के घर चल दी। ऐसे में महादेवी के लिए यह जानना आवश्यक हो गया कि आखिर हुआ क्या है। वे खिड़की से झाँककर देखती हैं कि बिन्दा के घर के दरवाजे पर बहुत आदमी जमा है। उन्होंने ऐसी भीड़ सिर्फ बारात-विवाह के समय ही देखी थी। तब उन्होंने सोचा ‘क्या उस घर में विवाह हो रहा है, और हो रहा है तो किसका?’ इस तरह के प्रश्न उनके मन में उठने लगते हैं।

V. कोष्ठक में दिए गए कारक चिन्हों से रिक्त स्थान भरिए:

(को, की, पर, से)

प्रश्न 1.
बिन्दा ………….. समस्या का समाधान न हो सका।
उत्तरः
की

प्रश्न 2.
बिन्दा ……….. मेरा उपाय कुछ ऊंचा नहीं।
उत्तरः
को

प्रश्न 3.
उसके घर जाने …………. माँ ने मुझे रोक दिया था।
उत्तरः
से

प्रश्न 4.
चूल्हे ………….. चढ़ाया दूध उफना जा रहा था।
उत्तरः
पर

प्रश्न 5.
बिन्दा ………….. सहेली महादेवी हैं।
उत्तरः
की।

VI. अन्य लिंग रूप लिखिए:

प्रश्न 1.
भिखारिन, पंडिताइन, लेखिका, बैल, चाची, नाना, दादी, विद्यार्थिनी, बालिका।
उत्तरः

• भिखारिन – भिखारी
• पंडिताइन – पंडित
• लेखिका – लेखक
• बैल – गाय
• चाची – चाचा
• नाना – नानी
• दादी – दादा
• विद्यार्थिनी – विद्यार्थी
• बालिका – बालक

VII. अन्य वचन रूप लिखिएः

प्रश्न 1.
बच्चा, कोठरी, सीढ़ी, मुद्रा, पंखा, दरवाजे, उँगली।
उत्तरः

• बच्चा – बच्चे
• कोठरी – कोठरियाँ
• सीढ़ी – सीढ़ियाँ
• मुद्रा – मुद्राएँ
• पंखा – पंखे
• दरवाजे – दरवाजा
• उँगली – उँगलियाँ

VIII. विलोम शब्द लिखिएः

प्रश्न 1.
दुर्बल, स्पष्ट, ज्ञात, मृत्यु, स्वर्ग, पुण्य, सुन्दर, न्याय, पराजय, स्वाभाविक, ऊपर।
उत्तरः

• दुर्बल × सबल
• स्पष्ट × अस्पष्ट
• ज्ञात × अज्ञात
• मृत्यु × जन्म
• स्वर्ग × नरक
• पुण्य × पाप
• सुन्दर × कुरुप
• न्याय × अन्याय
• पराजय × जय
• स्वाभाविक × अस्वाभाविक
• ऊपर × नीचे

बिन्दा लेखिका का परिचयः

श्रीमती महादेवी वर्मा हिन्दी साहित्य की प्रमुख छायावादी कवयित्री हैं। उनका जन्म 1907 ई. में फरुखाबाद में हुआ। प्रारंभिक शिक्षा इन्दौर में हुई। आपने प्रयाग विश्वविद्यालय से एम.ए. की उपाधि प्राप्त की। आप विक्रम विश्वविद्यालय, उज्जैन द्वारा मानद डॉक्टरेट की उपाधि से अलंकृत हुई|
मुख्य कृतियाँ : ‘अतीत के चलचित्र’, ‘स्मृति की रेखाएँ’, ‘पथ के साथी’ …. इत्यादि। आपको 1983 में अपने ‘यामा’ काव्य संग्रह पर ज्ञानपीठ पुरस्कार मिला है।
प्रस्तुत रेखाचित्र ‘बिंदा’ को ‘अतीत के चलचित्र’ से लिया गया है। यह एक हृदयस्पर्शी रेखाचित्र है। बिन्दा लेखिका के बचपन की सहेली है। सौतेली माँ द्वारा सताई गयी, वह खिलने से पहले ही मुरझा गई।
‘बिन्दा’ के माध्यम से मातृप्रेम से वंचित व सौतेली माँ के क्रूर व्यवहार को दर्शाने के उद्देश्य से इस रेखाचित्र का चयन किया गया है।

### बिन्दा Summary in Hindi

महादेवी वर्मा अच्छी कवयित्री ही नहीं, बल्कि विशिष्ट गद्य लेखिका भी हैं। वे रेखाचित्र और संस्मरण लिखने में प्रवीण हैं। इनके रेखाचित्रों में संस्मरण की शैली भी पाई जाती है। अतः इनके रेखाचित्र गंगा-जमुना का सुंदर संगम है। महादेवी वर्मा ने ‘बिन्दा’ रेखाचित्र में अपने बचपन की सखी की दर्दभरी कहानी की मार्मिक तस्वीर खींची है।

बिन्दा लेखिका के बचपन की सहेली थी। उसकी माँ उसके बचपन में ही चल बसी थी। उसके पिता ने दूसरी शादी कर ली। अतः बिन्दा का बचपन सौतेली माँ की कठोर निगरानी और कठिन दंड-संहिता में ही बीता।

बिन्दा की सौतेली माँ को सब लोग पंडिताइन चाची कहकर पुकारते हैं। बिन्दा की दिनचर्या घड़ी की सुई की तरह सदा चलती रहती है। उसके जीवन के शब्दकोश में ‘आराम’ या ‘चैन’ शब्द ही नहीं था। चाहे सर्दी का मौसम हो या गर्मी का – बिन्दा को दिन-रात घर का काम करना ही पड़ता था। फिर सौतेली माँ की सेवा भी करनी पड़ती थी।

बिन्दा पाँच-दस मिनट के लिए अदृश्य हो जाती, तो पंडिताइन चाची शेरनी की तरह दहाड़ती थी और बिन्दा हिरन की बच्ची की तरह थर-थर काँप उठती थी। पंडिताइन चाची बात-बात पर बिन्दा को डाँटा करती थी। इतना ही नहीं, वह बिन्दा को भद्दी से भद्दी गालियाँ भी देती थी।

लेखिका भी बिन्दा की उम्र की ही थी, पर बिन्दा की दुःख-भरी कहानी समझ नहीं पाती थी। वह घर में कोई भी काम नहीं करती थी, फिर भी उसकी माँ उसे नहीं डाँटती थी, लेकिन बिन्दा दिन-रात कोल्हू के बैल की तरह काम करती रहती थी, तब भी डाँट-फटकार सुनती थी। इसका कारण लेखिका का बाल-मन समझ नहीं पाया।

बिन्दा ने एक दिन चाँदनी रात में चमकते हुए एक तारे को दिखाकर अपनी बाल-सखी से कहा कि लो, वह देखो। मेरी अम्मा आकाश में तारा बनकर चमक रही है। लेखिका ने अपनी माँ से कहा कि माँ! तुम कभी भी तारा न बनना। तुम सदा मेरे पास ही रहो। लेखिका की माँ अपनी बेटी की बातें सुनकर सुन्न रह गई।

बिन्दा को चेचक की बीमारी लगी हुई थी। माँ-बाप उसे आँगन में खाट पर छोड़कर, घर के ऊपरी भाग में रहने लगते हैं, जब कि बिन्दा खाट पर अकेली पड़ी रहती थी।

बिन्दा कई दिन तक खाट पर कराहती रही। एक दिन अचानक लेखिका को पता चला कि बिन्दा अपनी माँ से मिलने आकाश में चली गई है। लेखिका के मन में अपनी बाल्य-सखी बिन्दा का दर्द-भरा चेहरा हमेशा बना रहा। वह उस दीन बिन्दा को नहीं भूल पाई।

### बिन्दा Summary in English

‘Binda’, written by Mahadevi Varma, presents a beautiful character sketch of her childhood friend.

Binda was the author’s childhood friend. Her mother had died during her childhood, and her father had remarried. Thus, her childhood was spent under the strict supervision of her stepmother and her harsh punishment.

Binda’s stepmother was called ‘Pandithayin Chachi’ by everyone. Binda’s daily routine and chores kept on ticking like the hands of a clock. Her vocabulary did not contain the words ‘rest’ or ‘leisure’. Whether it was summer or winter, Binda had to do the chores of the household, day and night. Then, she had to serve her stepmother.

. If Binda was out of sight even for a few minutes, her step-mother would roar as a lioness and Binda would shiver like a fawn. Binda’s step-mother would scold her for the smallest things. That was not all; she would shower Binda with very nasty swear-words and abuses.

The author was also of Binda’s age, but could not understand Binda’s painful story. The author would not do any of the household chores and yet her mother would never scold her. However, Binda would work day and night like an ox in the fields and still receive abuse at the end of the day. The author’s young mind could not understand her friend’s situation.

One moonlit night, Binda was looking up at the stars and told the author that her mother had become a star, up in the sky. The author, upon returning home, told her mother to never become a star. She told her mother to stay with her always. The author’s mother was shocked to hear this.

Binda was struck by small-pox. Her parents left her on a cot in the courtyard and began to live in the upper section of the house. Binda would lie alone on the cot.

Binda lay on the cot, groaning and crying in pain, for many days. One day, suddenly, the author found out that Binda had gone to meet her mother, up in the sky. The image of Binda’s sad face remained imprinted in the mind of the author. She could never forget her poor friend Binda.

कठिन शब्दार्थः

• सलज्ज – लाज से;
• बिछुए – पाँव की उँगलियों का एक गहना;
• फिरकनी – फिरकी/चकरी नामकं खिलौना;
• खजड़ी – खजरी, छोटी डफली;
• अवसन्न – खिन्न, उदास;
• पदच्युत – पद से हटाना;
• चेचक – शीतला नामक रोग (smallpox)

## Karnataka 1st PUC Chemistry Question Bank Chapter 5 States of Matter

### 1st PUC Chemistry States of Matter One Mark Questions and Answers

Question 1.
Write the ideal gas equation for one mole of a gas.
PV = RT

Question 2.
Write the Vander Waals1 equation for one mole of a real gas.

Question 3.
Write Vander Waals equation constants ‘a’ and ‘b’.
Units for ‘a’: L2 atm mol-2
Units for ‘b’: Litre mole-1

Question 4.
Gie the value of universal gas constant in SI units.
8.3124 J K-1 mol-1

Question 5.
Give the combined gas equation (or general equation for gases).
PV = nRT for n moles.

Question 6.
Write kinetic equation for gases.
PV = $$\frac { 1 }{ 3 }$$ mnc2

Question 7.
Write an equation for the pressure exerted by a gas in a container.
$$\mathrm{PV}=\frac{1}{3} \frac{\mathrm{mnc}^{2}}{\mathrm{V}}$$

Question 8.
Write an equation for the root mean square velocity of a gas.
$$V_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$$

Question 9.
Liquid ammonia bottle is cooled before opening the seal. Why?
NH3 is liquified at high temperature and pressure. It is cooled so as to reduce temperature so that it does not burst.

Question 10.
The size of weather balloon becomes larger and larger as it ascends up into higher altitudes. Why?
At higher altitudes, atmospheric condition is less, therefore, air inside balloon exerts less pressure released and it becomes larger and larger.

Question 11.
Which property of liquid is responsible for spherical shape of liquid drops?
Surface tension and viscosity.

Question 12.
What is the effect of temperature on viscosity and why?
Viscosity decrease in the temperature because intermolecular force of attraction decreases.

Question 13.
What is the effect of pressure on (i) viscosity, (ii) surface tension, (iii) density of liquid? –
Increase in pressure increases viscosity, surface tension and density because inter molecular force of attraction increases.

Question 14.
What is the boiling point of water at (i) higher altitudes, (ii) in pressure cooker?
(i) >100 °C (ii) < 100 °C.

Question 15.
Name four measurable properties of gases.

1. Volume
2. Pressure
3. Temperature
4. Mass

Question 16.
Name the SI unit of pressure and give its definition.
Pascal is SI unit of pressure. It is defined as pressure exerted when 1 newton force is acting per square metre area.

Question 17.
Define an ideal gas.
Ideal gas is gas which follows all the gas laws at all temperature and pressure.

Question 18.
Give various forms of ideal gas equation.

Question 19.
What are real gases?
Real gases are those which do not follow all the gas laws at all temperature and
pressure.

Question 20.
Under what conditions of T and P, most of gases deviate from ideal gas behavior?
At low temperature and high pressure most of gases deviate from ideal gas behavior.

the partial pressure formula of each gas is proportional to the number of molecules of the gas in the mixture.

Question 21.
State Dalton’s law of partial pressure.
It states that whenever two or more gases, which do not react chemically, are
enclosed in vessel, the total pressure is equal to sum of partial pressure of each gas.

Question 22.
What is meant by compressibility factor?
Compressibility factor is ratio of PV to nRT.
Z = $$\frac{\mathrm{PV}}{n \mathrm{RT}}b$$

Question 23.
Define boiling point of a liquid.
Boiling point is temperature at which vapour pressure of liquid becomes equal to atmospheric pressure.

Question 24.
How is density of gas related to its molar mass?
Density of gas is directly proportional to molar mass.

Question 25.
How is molar mass of gas related to rate of diffusion?
The rate of diffusion is inversely proportional to molecular mass
$$\frac{r_{1}}{r_{2}}=\sqrt{\frac{M_{2}}{M_{1}}}$$

Question 26.
What is meant by elastic collision?
Elastic collision is a collision in which there is no net loss energy rather there is transfer of energy.

Question 27.
Define aqueous tension.
Aqueous tension is pressure of water vapour at particular temperature.

Question 28.
State Graham’s law of diffusion.
It states that the rate of diffusion of gases is inversely proportional to square root of molecular masses at constant temperature and pressure.

Question 29.
Define “effusion”.
Effusion is the process of diffusion taking place from a tiny hole.

Question 30.
Arrange the following gases in the increasing orderof their van der waals1 constant ‘a’ values : CO2, H2> C6H6
H2 < CO2 < C6H6

Question 31.
What is SI unit of (i) viscosity, (ii) surface tension?
(i) Pascal second, (ii) Nm1

Question 32.
Define critical temperature of the gas. Give its expression.
Critical temperature is a temperature above which a gas cannot be liquefied.
$$\mathrm{T}_{\mathrm{c}}=\frac{8 \mathrm{a}}{27 b \mathrm{R}}$$ where ‘a’ and ‘b’ are Van der waal’s constants, R is gas constant.

Question 33.
What do you mean by Boyle temperature? Give its expression and its relation with inversion temperature.
Boyle temperature is a temperature at which most of real gases show ideal gas behavior over wide range of pressure. $$\mathrm{T}_{b}=\frac{\mathrm{a}}{b \mathrm{R}} \mathrm{T}_{\mathrm{i}}=2 \mathrm{T}_{b}$$

Question 34.
Define inversion temperature of a gas.
Inversion temperature is a temperature below which if gas is allowed to expand, it causes cooling effect.

Question 35.
Define (i) critical pressure, (ii) critical volume and give their expression.
(i) Critical pressure, it is pressure required to liquefy the gas at critical temperature.
$$\mathrm{P}_{\mathrm{c}}=\frac{\mathrm{a}}{27 \mathrm{b}^{2}}$$
(ii) Critical volume, it is volume occupied by 1 mole of gas at critical temperature and pressure. Vc = 3b

Question 36.
What type of graph will you get when PV is plotted against P at constant temperature.
A straight line parallel to pressure axis.

### 1st PUC Chemistry States of Matter Two Marks Questions and Answers

Question 1.
State Boyle’s law of gases. Give its mathematical expression.
Boyle’s law states volume of fixed mass of gas in inversely proportional to pressure at constant temperature. (P1V1 = P(2V2)T

Question 2.
State Charle’s law. Give the mathematical expression.
Charle’s law states that volume of fixed mass of gas is directly proportional to absolute temperature at constant pressure.
$$\left(\frac{\mathrm{V}_{1}}{\mathrm{T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{T}_{2}}\right)_{\mathrm{P}}$$

Question 3.
Give any two differences between ideal and real gas.

Question 4.
Mention the causes for the deviation of real gas from ideal behaviour.

• There is no intermolecular force of attraction within gaseous molecules (at high pressure and low temperature there is appreciable intermolecular force of attraction).
• The actual volume of the molecule is negligible as compared to total volume of the gas. (At high pressure the gaseous molecule occupy minimum volume, which is not negligible).

Note: Because of this deviation Van der waal modified the ideal gas pressure and ideal gas volume respectively.

Question 5.
What will be the density of CO2 in kg/m3 at 323 K and 101.3 kPa pressure?
T = 323 K, P = 101.3 kPa, M = 44 × 10-3kg mol-1, d = ?,
R= 8.314 × 10-3 kPa m3K-1mol-1

Question 6.
How many molecules of an ideal gas are there in 1 × 10-3 dm3 at STP?
22.4 dm3 of an ideal gas has 6.023 × 1023 molecules at STP
1 × l0-3dm3 of an ideal gas contains = $$\frac{6.023 \times 10^{23}}{22.4} \times 1 \times 10^{-3}=2.69 \times 10^{19}$$molecules.

grams to moles calculator and moles to grams calculator using a simply calculator.

Question 7.
Calculate the number of moles of hydrogen present in 500 cm3 of a gas under a pressure of 101.3 kPa at a temperature of 300 K. (R = 8.314 J K-1 mol-1).
n = ?

Question 8.
50 ml of oxygen were collected at 100 C under 750 mm pressure. Calculate volume at STP.
V1 = 50ml V2=?
P1= 750mm P2 = 760mm
T1 =10 + 273 = 283K T2 =273.15K

Question 9.
Why ethyl alcohol has lower boiling point than water?
Ethyl alcohol has higher vapour pressure than water due to less intermolecular forces of attraction.

Question 10.
A human adult breathes in approximately 0.50 L of air at 1 atm with each breath. If an air tank holds 10 L of air at 200 atm, how many breaths the tank will supply?
P1V1=P2V2
200 × 10 = 1 × V2
V2 = 2000 L
Number of Breaths = $$\frac{2000 \mathrm{L}}{0.5 \mathrm{L}}$$ = 4000

Question 11.
What will happen to volume of fixed amount of gas at a certain T and P if :
(a) T is kept constant but pressure is decreased to $$\frac { 1 }{ 4 }$$ th of original value?
(b) Pressure is halved and temperature in Kelvin is doubled?
(a) The volume will become 4 times
(b) The volume will remain same

Question 12.
10 g of O2 were introduced into an evacuated vessel of 5 L capacity at 27 °C,
Calculate pressure of gas in bar in the container [R = 0.083 bar L K-1 mol -1] [At. mass of O = 16u]
PV = raRT bar
P × 5 L = $$\frac { 10 }{ 32 }$$ × 0.083 × 300 K
P = $$\frac{24.9 \times 10}{160}=\frac{249}{160}$$  = 1.556 bar

Question 13.
Calculate the density of Rn gas at 298 K and 755 mm Hg pressure. [Atomic mass of Rn = 222u]

Question 14.
2NO (g) + O2 (g) → 2NO2 (g). A 1.98 L reaction vessel is filled with a mixture of NO and O2 to a pressure of 4.33 bar at 310 K. The reaction is allowed to continue until the total pressure decreases to 4.00 bar. What is the final pressure of NO2 ?
Reaction 2NO + O2 → 2NO2

Question 15.
Calculate the total pressure in a 10 L cylinder which contains 0.4 g of helium, 1.6 g of oxygen and 1.4 g of nitrogen at 27 °C. Also calculate the partial pressure of helium gas in the cylinder. Assume ideal behaviour for gases.
Ans:
Total number of moles = $$\frac{0.4}{4}+\frac{1.6}{32}+\frac{1.4}{28}$$ = 0.1 + 0.05 + 0.05 =0.2 moles.
pV = nRT
p × 10 L = 0.2 × 0.082 × 300 ⇒ p = $$\frac{24.6 \times 0.2}{10}$$
p = 0.492 atm
PHe = $$\frac{0.1}{0.2}$$ × 0.492 = 0.246 atm

Question 16.
Calculate the number of moles of hydrogen (H2) present in a 500 cm3 sample of hydrogen gas at a pressure of 760 mm Hg and 27 °C.
PV = nRT, P = 1 atm, V = 500 cm3, n = ?, R = 82.1 atm cm3K-1mol-1, T = 300 K.

Question 17.
34.05 ml of phosphorus vapour weighs 0.0625 g at 0.1 bar pressure. What is the molar mass of phosphorus?
pV = nRT

Question 18.
In terms of Charle’s law explain why -273 °C is the lowest possible temperature.
Charles Law plotted graphically with volume against temperature in °C. These plots when extended intersect the temperature axis at the same point -273 °C. Charles concluded that all gases at this temperature could have zero volume and below this temperature volume would be negative. It shows -273 °C is lowest temperature attainable.

Question 19.
(i) Why is Boyle’s law is obeyed by N2,O2 or CO2 only at low pressure and high temperature?
(ii) Compare the rate of diffusion of HCl and NH3. (Atomic masses of H = 1u, Cl = 35.5 u, N = 14u)
(i) It is because at low pressure and high temperature these gases follow ideal gas behavior due to negligible force of attraction and volume occupied by gas molecules can be ignored.
(ii)

Question 20.
Arrange solid, liquid and gas in order of energy giving reasons.
Solid<liquid

Question 21.
The Van der Waal’s constants for two gases are as follows:-

Which of them is more easily liquefiable and which has greater molecular size?
Greater the value of ‘a’, more easily the gas is liquefiable. Similarly, greater the value of V, greater is the molecular size. Hence, gas Y will be more easily liquefiable and will have greater molecular size.

Question 22.
Compare the rates of diffusion of 235UF6 and 238UF6
Molecular mass of 235UF6 = 235 + 6 × 19 = 349
Molecular mass of 238UF6 = 238 + 6 × 19 = 352

Question 23.
At a certain altitude, the density of air is 1/10 th of the density of the earth’s atmosphere and temperature is -10 °C. What is the pressure at that altitude? Assume that air behaves like an ideal gas, has uniform composition and is at STP at the earth’s surface.

Question 24.
What is the ratio of average kinetic energy of oxygen molecules to that of ozone molecules at 27 °C?
Average kinetic energy of any gas depends only on temperature and not upon the nature of the gas. Hence, both the gases will have same average kinetic energy at 27 °C, ie., the ratio will be 1: 1.

Question 25.
What is the difference between total kinetic energy and translational kinetic energy? For what type of molecules, the two are equal?
Total kinetic energy is the sum of translational, vibrational and rotational kinetic energies. The total kinetic energy is equal to the translational kinetic energy for monoatomic gases (He, Ne, etc.,) as they do not possess vibrational and rotational motion but have only translation motion.

Question 26.
Out of N2 and NH3, which one will have greater value of ‘a’ and which one will have greater value of ‘b’ ?

1. As NH3 is more easily liquefiable (due to hydrogen bonding), intermolecular forces of attraction are stronger then in N2. Hence, NH3 will have greater value of‘a’.
2. As NH3 molecule is larger in size than N2 , hence NH3 will have greater value for ‘b’.
(For NH3, a = 4.17 L2 atm mol-2 , b = 0.0371 L mol-1
For N2 , a = 1.39 L2 atm mol-2, b = 0.0319 L mol-1)

Question 27.
What would have happened to the gas if the molecular collisions were not elastic?
On every collision, there would have been loss of energy. As a result, the molecules would have slowed down and ultimately settle down’ in the vessel. Moreover, the pressure would have gradually reduced to zero.

Question 28.
CO2 is heavier than O2 and N2 gases present in the air but it does not form the lower layer of the atmosphere. Why?
Gases possess the property of diffusion which is independent of the force of gravitation. Due to diffusion, the gases mix into each other and remain almost uniformly distributed in the atmosphere.

Question 29.
N2O and CO2 have the same rate of diffusion under same conditions of temperature and pressure. Why?
Both have same molar mass (= 44 g mol-1). According to Graham’s law of diffusion, rates of diffusion of different gases are inversely proportional to the square root of their molar masses under same conditions of temperature and pressure.

Question 30.
Why liquids have a definite volume but no definite shape?
This is because the intermolecular forces of attraction are strong enough to hold the molecules together but not so strong as to fix them into definite positions (as in solids). Instead, they possess fluidity and hence no definite shape.

Question 31.
At a particular temperature, why vapour pressure of acetone is less than that of ether?
This is because the intermolecular forces of attraction in acetone are stronger than those present in ether.

Question 32.
A liquid is transferred from a smaller vessel to a bigger vessel at the same temperature. What will be the effect on the vapour pressure?
No effect, as it depends only on the nature of the liquid and temperature.

Question 33.
Why vegetables are cooked with difficulty at a hill station?
The atmospheric pressure is less and so the boiling point is lowered.

Question 34.
What is the approximate relationship between heat of vaporization and boiling point of a liquid?
∆Hvap / Tb =21 cal K-1 mol-1 (Trouton’s rule)

Question 35.
Explain Pressure coefficient of a gas.
Pressure coefficient (αp). At constant volume, the increase in the pressure of a gas per degree rise of temperature divided by its pressure at 0 °C is called pressure coefficient of the gas. Mathematically, $$\alpha_{\mathrm{P}}=\frac{\mathrm{P}_{\mathrm{t}}-\mathrm{P}_{0}}{\mathrm{P}_{0} \times \mathrm{t}}$$. For all gases, value of aP is αp is $$\frac { 1 }{ 273 }$$

Question 36.
Explain Atmolysis.
Atmolysis is the process of separation of two gases on the basis of their different rates of diffusion due to difference in their densities is called atmolysis. It has been applied with success for the separation of isotopes and other gaseous mixtures.

Question 37.
Write any two difference between evaporation and boiling.

Question 38.
Define surface tension.
Surface tension is defined as force per unit length acting perpendicular to the tangential line on the surface. It is due to imbalanced attractive forces acting downwards and tends to reduce the surface area of a liquid to minimum.

Question 39.
Define viscosity of a liquid.
The internal resistance to the flow of the liquids which one layer offers to another layer trying to pass over is called its viscosity. It depends upon the nature of the liquid and temperature.

Question 40.
Write an expression for kinetic gas equation.
PV = $$\frac { 1 }{ 3 }$$mnc2

Question 41.
Both N2O and CO2 have similar rates of diffusion under similar conditions of temperature and pressure. Explain.
Gram molar mass of both these gases is the same (44 g). According to Graham’s law of diffusion, the rates of diffusion of the gases are inversely proportional to the square root of their molar masses under similar conditions of temperature and pressure. Therefore, both these gases diffuse at the same rate.

Question 42.
If the density of a gas at the sea level at 0 °C is 1.29 kg mm-3. What is its molar mass? (Pressure = 1 bar)

Question 43.
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30 °C.
P1V1 = P2V2
1 × 500 = P2 × 200 P2 = 2.5 bar

Question 44.
Calculate the molar mass of an unknown gas which diffuses 1.117 times faster than oxygen gas through the same aperture under the same conditions of temperature and pressure.

### 1st PUC Chemistry States of Matter Three Marks Questions and Answers

Question 1.
How vapour pressure of liquid is related to (i) temperature, (ii) nature of liquid, (iii) boiling point, (iv) atmospheric pressure?
(i) Vapour pressure is directly proportional to temperature.
(ii) If intermolecular force of attraction is less in the liquid, its vapour pressure will be high.
(iii) Higher the vapour pressure, lower will be boiling point.
(iv) If atmospheric pressure is low, boiling point will be less and higher will be the vapour pressure.

Question 2.
Which type of intermolecular forces exist among the following molecules?
(i) H2S molecules (ii) H2O molecules (iii) Cl2 and CCl4 molecules (iv) SiH4 molecules (v) Helium atoms (vi) He atoms and HCl molecules.
(i) Dipole-dipole interactions (because H2S is polar)
(ii) Hydrogen bonding
(iii) London dispersion force (because both are non-polar)
(iv) London dispersion forces (because SiH4 is non-polar)
(v) London dispersion forces (because He atoms have symmetrical electron clouds)
(vi) Dipole-induced dipole forces (because HCl is polar while He atom has symmetrical r electron cloud)

Question 3.
Explain Relative humidity and percentage of humidity.
Relative humidity is a method of expressing the extent of moisture (water vapour) present in the air. It is defined as the ratio of the partial pressure of water vapour in the air to the vapour pressure of water at that particular temperature.

Thus, if the percent relative humidity of air is 60%, it means that partical pressure of water vapour in air is 0.6 times more than the vapour pressure of water at that temperature.

Question 4.
Give a reason for the following, (i) The size of weather balloon becomes larger and larger as it ascends into higher altitudes, (ii) Tyres of automobiles are inflated to lesser pressure in summer than in winter.
(i) As we go to higher altitudes, the atmospheric pressure decreases. Then the pressure outside the balloon decreases. To regasm equilibrium with the external pressure, the gas inside expands to decrease its pressure. Hence, the size of the balloon increases.

(ii) In summer, due to higher temperature, the average kinetic energy of the air molecules inside the tyre increases, ie., molecules start moving faster. Hence, the pressure on the walls of the tube increases. Its pressure inside is not kept low at the time of inflation, at higher temperature, the pressure may become so high that the tyre may burst.

Question 5.
The molecular speeds of gaseous molecules are analogous to those of rifle bullets. Then why is the odour of a gas not detected so fast?
Though the molecules of a gas travel at high speeds but they do not travel in straight lines in one direction like bullets. As they travel, they collide with the molecules of the gases present in the air. As a result, they are deflected. Thus, they follow a zig-zag path, i.e., the net distance travelled in a particular direction is quite small in a given time. That is why the odour is not detected so fast.

Question 6.
The average kinetic energy of a gas molecule at 0 °C is 5.621 × 10-27 J. Calculate Boltzmann constant. Also calculate the number of molecules present in one mole of the gas.

### 1st PUC Chemistry States of Matter Four / Five Marks Questions and Answers

Question 1.
Write the postulates of Kinetic theory of gases.

1. Gases are made up of large number of the minute particles.
2. Pressure is exerted by a gas
3. There is no loss of kinetic energy.
4. Molecules of gas attract on one another.
5. Kinetic energy of the molecule in directly proportional to absolute temperature.
6. Actual volume of the gaseous molecule very small.
7. Gaseous molecules are at always in motion.
8. There is more influence of gravity on movement of gaseous molecule.

Question 2.
Derive the expression for the pressure exerted by gas (Derive the Kinetic gas
equation PV = $$\frac { 1 }{ 3 }$$ mn2)

Step 1 :

Let the mass of each molecule of gas be ‘m’ kg and all are moving with root mean square velocity ‘c’.

Step 2 :
All gaseous molecules move in all direction randomly. Hence at a given
interval of time $$\frac { n }{ 3 }$$ molecules moves either along X or Y or Z axis.

Step 3 :
Consider n number of molecules of gas in a cube of length T within a 3 dimensional (X, Y and Z axis) network.
Then the area of the cube = l2 ……………..(1)
The volume of the cube = V = l3 ……………..(2)

Step 4 :
Calculation of change in momentum w.r.t. one axis.
Consider the side ‘A’ of the cube. Let one (n = 1) molecule of a gas moving along the axis towards the side ‘A’ with a velocity ‘c’ and strikes the wall ‘A’.
Initial momentum before striking = me ……………..(3)
Distance travelled = 1 units ……………..(4)
After striking the’wall ‘A’ molecule rebounds back with same speed in opposite direction.
∴ Final momentum after striking = -me ……………..(5)
Distance travelled = 1 units ……………..(6)
∴ Total change in momentum = (3) – (5) = mc – (- mc) = 2mc
Total distance travelled = (4) + (6) = 1 + 1 = 2l
Number of collisions per sec = C/21 ……………..(7)
∴ Total change in momentum per sec per molecule on face
‘A’ = collision per second × change in momentum $$\frac{\mathrm{C}}{2 \mathrm{l}} \times 2 \mathrm{m} \mathrm{C}=\frac{\mathrm{m} \mathrm{C}^{2}}{1}$$

Step 5 :
Total change in momentum per sec for one molecule strikes on the wall A = $$\frac{m c^{2}}{1}$$
∴The change in momentum per sec for $$\frac { n }{ 2 }$$ molecules strikes on wall A = $$\frac{n}{3} \times \frac{m c^{2}}{1}$$
= $$\frac{\mathrm{mnc}^{2}}{31}$$ …………(9)

Step 6 :
The change in momentum per sec = force. And the force in gaseous molecule is only due to pressure,

Question 3.
Derivation of Vander waal’s equation
Pressure correction (getting corrected ideal pressure)

Let us consider a container containing ‘n’ number of gaseous molecule. Now consider two gaseous molecule A and B inside it.

At ‘A’ : The gaseous molecule ‘A’ present at the interior of the container and is attracted (disturbed) by all neighboring gaseous directions. Hence the net force exerted by this molecule is zero.

At ‘B’ : The gaseous molecule ‘B’ is about to strike the walls of the container, it is influenced by other neighboring molecule as follows.
(a) It is attracted by large number of molecules inward and net effect is an inward pull. This inward force (pressure) is directly proportional to the number of molecules per unit volume (at the bulk of the container) ie., P ∝ $$\frac{1}{V}$$ …………(1)
(b) Within the container there are some molecules which are already striking the wall of the container. Thus there exist pulling of B inward by the molecules striking on the wall. This pressure (force) is directly proportional to the number of molecules striking on unit volume of the walls of the container. P ∝ $$\frac{1}{V}$$ …………(2)

From the above two points A and B it is clear that the actual pressure exerted on the wall due to gaseous molecular is less than the ideal pressure.
Actual Pressure (P) = Ideal Pressure (Pi) – Pressure Correction (Pc)
P = Pi – Pc
Pi = P + Pc …………….(3)

Volume correction : At high pressure the molecules are closer and covering/ occupying a considerable minimum area or volume. (Because the space available for the molecule to move is less than the actual volume of the container. Hence Vander waal accounted for this by subtracting volume correction ‘b’ from the observed volume V.
Ideal Volume (Vi) = V – b ………………(6)
We know that if Pi and Vi are pressure and volume of ideal gas then it satisfies PiVi = RT
But from (5) and (6) the ideal pressure and ideal Volumes are substituted.

Equation (7) is called Vander waal equation for one mole of gas. This equation for ‘n’ moles of gas is

Question 4.
Interpret PV – P curves based on Vander waals equation.
Consider Vander waals eqution for one mole of a gas $$\left(p+\frac{q}{V^{2}}\right)(V-b)=R T$$ ……………(1)

Case 1: At low pressure:-
When the pressure is low b = 0
∴ Equation becomes

PV > RT
Explains gradual increase in PV < RT. Hence the curve first descends at lower pressure.

Case 2 ; At high pressure : At high pressure, $$\frac{\mathbf{a}}{\mathbf{V}^{2}}$$ = 0 ∴ Equation becomes P(V – b) = RT or PV – Pb = RT PV > RT
Explains gradual increase in PV with simultaneous increase is pressure.

Case 3 : Change in temperature :
At low temperature the value of P and V are very small, hence deviate from ideal behavior.
At high temperature P and V are very large such that $$\frac{\mathbf{a}}{\mathbf{V}^{2}}$$ and b are negligible hence
Vander waals Eqn reduces to PV = RT.
∴ At high temperature almost all gases behaves like ideal gas.

Question 5.
Two flasks ‘A’ and <B’ have equal volumes. Flask ‘A’ contains H2 and is maintained at 300 E while ‘B’ contains equal mass of CH4 gas and is maintained at 600 E.
(i) Which flask contains greater number of molecules? How many times more?
(ii) In which flask pressure is greater? How many times more?
(iii) In which flask molecules are moving faster?
(iv) In which flask the number of collisions with walls are greater?
(i) Number of moles in flask A = 8 × number of moles in flask B.
Let the equal mass be x. Ratio of number of moles => $$\frac{x}{2}: \frac{x}{16}$$ = 8:1

(ii) Pressure α number of moles Pressure α temperature
If temperature had been equal, pressure in ‘A’ would have been 8 times but in ‘A’
temperature is $$\frac{1}{2}$$ of that in B
∴Pressure will be 4 times more

(iii) Flask B, because velocity is directly proportional to square root of temperature.
(iv) Flask A, because number of molecules of flask A is more.

Question 6.
Calculate the pressure exerted by 1.00 mol of CO2 (g) at 298 K that occupies 65.4 ml using Van der waal’s equation.
‘a’ for CO2 is 3.592 L2 bar/mol2, ‘6’ = 0.0427 L mol-1. Compare it with the pressure predicted by ideal gas equation for same conditions of T and P?

Question 7.
A gaseous mixture contains 2.2 bar He, 1.1 bar H2 and 4.2 bar N2 . What is mole fraction of N2 ?

Question 8.
Account for the following:
(a) The size of weather balloon becomes larger and larger as it ascends up into higher altitudes.
(b) Copper is malleable and ductile whiles sulphur is not.
(a) At higher altitudes, atmospheric pressure decreases, therefore, air inside the weather balloon exerts pressure and its size becomes larger and larger as it ascends higher and higher.
(b) Copper is metal and has strong metallic bond, that is why it is malleable and ductile. Sulphur is non-metal and has weak intermolecular force of attraction therefore it is brittle.

Question 9.
(a) Explain Boyle’s law with the help of kinetic theory of gases.
(b) An open beaker at 27 °C is heated to 477 °C. What fraction of air would have been expelled out?
(a) The kinetic theory of gases assumes that pressure of gas is due to collision of gas molecules with the walls of the container. The more will be frequency of collision, more will be pressure. The reduction in volume (v) of gas increases no. of molecules per unit volume to which pressure is directly proportional. Therefore, the volume of the gas is reduced if pressure is increased or we can say pressure is inversely propotionsl to volume P ∝ $$\frac{1}{V}$$

(b)

Question 10.
(a) What do you mean by ‘Surface Tension’ of a liquid?
(b) Equal volumes of two gases A and B diffuse through a porous pot in 20 and 10 seconds respectively. If the molar mass of A be 80, find the molar mass of B.
(a) Surface Tension of a liquid is the force by which surface molecules are attracted towards the bulk.

Question 11.
(a) Why air is dense at the sea level? Explain.
(b) Calculate the total pressure in a mixture of 4g of O2 and 2g of H2
confined to a total volume of 1L at 0 °C. (R = 0.821 L atm mol-1)
(a) Heavier gas will come down and lighter air goes up. Air at sea level is denser due to compression by mass of air above it.

Question 12.
Shown below is a volume vs temperature graph at different pressures P1>P2>P3 an<P4.
(a) Which gas law does this depict?
(b) What does each line of this graph called?
(c) What is the value of temperature at the point where the lines intercept at
temperature axis and what is the specific name given to this hypothetical temperature?
(d) Giving reason arrange the pressures, p1,p2,p3 and p4 in increasing order.
(a) Charle’s law.
(b) Isobar.
(c) – 273.15 °C, absolute zero is a lowest hypothetical temperature at which gases are supposed to occupy zero volume.
(d) p1 < p2 < p3 < p4
Reason : For a particular value of temperature : If volume is more it means pressure is less hence p1 is lesser than p2 and so on.

Question 13.
Density of a mixture of gases CO and CO2 is 1.5 g/L at 303 K temperature and 73 cm of Hg. What is the mole percentage of two gases in the mixture?

M = 38.85 (Molecular mass of the mixture)
Let total moles = 100
Moles of CO = a and Moles of CO2 = 100 – a
Mass of CO = 28a
Mass of CO2 =(100 – a)44
Total Mass = 28 a + (100 – a)44
Molecular mass of the mixture  = $$=\frac{28+(100-a) 44}{100}$$ = 38.85(evaluated)
a =32.19
Molecular percentage of CO = 32.19
Mole percentage of CO2 = (100 α) = 100 – 32.19 = 67.81

Question 14.
The van der waal’s equation of state for 1 mole of gas is $$\left(\mathbf{P}+\frac{a}{\mathbf{V}^{2}}\right)(\mathbf{V}-\mathbf{b})=\mathbf{R} \mathbf{T}$$
Arrange the following gases in order of increasing value of ‘a’. The gases are CO2, H2, He and N2. How does the value of V related to the ease of liquifaction?
H2 < He < N2 < CO2 is the order of increasing value of ‘a’. Measure of ‘a’ is proportional to the amount of force of intermolecular attraction. CO2 molecule being larger in size, has higher magnitude of van der Waal’s force of attraction and hence highest value of ‘a’.
Also higher is the intermolecular force of attraction, easier it is to liquefy a gas ie., higher the value of ‘a’, higher will be the case of liquefaction.

Question 15.
In an attempt to record critical temperature of some substances, the value of water has been missed out.

Will the value of x be more or less than (a) CO2 ? (b) NH3 ? Reason out your answers. Also arrange CO2, H2O and NH3 in the order of ease of liquefaction.
Value of x will be more than for both CO2 and NH3. This is because in CO2 only weak Van der waals forces exist but in H2O intermolecular H-bonding exist. Also magnitude of H-bonding in H2O is more than in NH3.

H2O has more intermolecular force of attraction therefore it should be easily liquefiable. It is already liquid at room temperature and atmospheric pressure.
Ease of liquefaction CO2 < NH3 < H2O.

Question 16.
An open vessel at the temperature 27 °C is heated until $$\frac { 3 }{ 5 }$$ parts of the air in it has been expelled. Assuming that the volume of the vessel remains constant, find the temperature to which the vessel has been heated.
For the open vessel, pressure and volume remains contant. If rej moles me present at T1 and n2 moles are present at T2 then
pV = n1RT1 and pV = n2RT2 => n1RT1 = n2RT2
n1T1 = nT2 => $$\frac{n_{1}}{\mathrm{T}_{1}}=\frac{n_{2}}{\mathrm{T}_{2}}$$

Let number of moles of air originally present = n
After heating, number of moles of air expelled = $$\frac { 1 }{ 2 }$$ n

Question 17.
In a hospital, an oxygen cylinder of capacity 10 L, oxygen is filled at 200 atmospheric pressure. If a patient breathes in 0.50 ml of oxygen at 1.0 atm with each breath, for how many breaths the cylinder will be sufficient? Assume temperature to be constant.
10 L at 200 atm pressure = V2L at 1 atm pressure
P1V1 = p2V2 => 200 × 10 = 1 × V2
V2 = 2000 L

Question 18.
Two flasks of equal volume are connected by a narrow tube of negligible volume and are filled with N2 gas when both the flasks are immersed in boiling water the gas pressure inside the system is 0.5 atm. Calculate the pressure of the system when one of the flasks is immersed in ice water while the other flask in boiling water. ‘
Temperature of the gas when flasks are in boiling water = 100 °C = 100 + 273 = 373 K
Pressure = 0.5 atm
Average temperature of the gas when one flask is in ice and other in boiling water. =50°C = 50 + 273 = 323 K

Question 19.
(i) Which of the liquids in each of the following pairs has higher vapour pressure : (a) Alcohol, glycerine ; (b) Petrol kerosene; (c) Mercury, water
(ii) Which one in each of the following pairs JU m ire viscous (a) coconut oil, castor oil ; (b) glycerine, kerosene ; (c) soft drinks, aerated water (soda water)?
(iii) Separate portions of chloroform and water at the same temperature are poured on your hands. The chloroform will feel colder. Account for this in terms of attractive forces.
(i) (a) Alcohol (b) Petrol (c) Water
(ii) (a) Castor oil (b) Glycerine (c) Soft drink
(iii) (a) Chloroform molecules have weak van der waal’s forces between them, while water molecules have hydrogen bonding between them. Since van der waal’s forces are weaker chloroform evaporates easily and causes cooling.

Question 20.
Deduce the relation PV = nRT, where R is a constant called universal gas constant.
According to Boyle’s law, V ∝ $$\frac { 1 }{ P }$$ (at constant T and n)
According to Charle’s law, V ∝ $$\frac { 1 }{ T }$$ (at constant P and n)
According to Avogadro’s law, V ∝ (at constant T and P)
Combining the three laws, V ∝ $$\frac { nT }{ P }$$ or PV ∝ nRT
Or PV = nRT where R is a constant called universal gas constant.

Question 21.
(i) What is the effect of temperature on (a) density, (b) surface tension, (c) viscosity and (d) vapour pressure of a liquid?
(ii) What is the effect of pressure on (a) volume, (b) boiling point and
(c) viscosity of a liquid?
(i) (a) Density decreases (b) Surface tension decreases (c) Viscosity decreases (d) Vapour pressure increases with rise in temperature.
(ii) (a) Volume decreases (b) B. pt increases (c) Viscosity increases with increase in pressure.

Question 22.
An LPG cylinder when full contains 14.2 kg gas and exerts pressure of 2.5 atm. If half of its gas is consumed what will be the pressure of the gas in the cylinder?
In a LPG cylinder, the gas is held as a liquid. There is an equilibrium between the LPG (liquid) and its vapour. The equilibrium vapour pressure does not depend upon the quantity of liquid present there as long as liquid ……….. vapour equilibrium is maintained. Therefore, even when half the gas has been used, the pressure excited by ’ the vapour of the gas remains equal to 2.5 atm.

Question 23.
Why at extremely low pressures, the real gases obey the ideal gas equation?
At low pressures, volume V is very large and hence the correction term b (a constant of small value) can be neglected in comparison to very large value of V. Thus the van der waal’s equation for 1 mole of a real gas.

Question 24.
Explain the following observations.
(a) Aerated water bottles are kept under water during summer.
(b) Liquid ammonia bottle is cooled before opening the seal.
(c) The tyre of an automobile is inflated at lesser pressure in summer than in winter.

(a) Aerated water contains C02 gas dissolved in aqueous solution under pressure and bottles are well stoppered. As in summer the temperature increases and we know that the solubility of the gases decreases with increase of temperature and as a result more of gas is expected to be generated in the bottle thereby pressure exerted by the gas in the bottle will increase. If the bottles are not kept under water, the gas generated may be large in quantity and hence pressure exerted by the gas may be very high and the bottle may explode. So, to decrease the temperature and hence to avoid explosion of the bottles.

(b) Liquid Ammonia bottle contains the gas under very high pressure. If the bottle is opened as such, then the sudden decrease in pressure will lead to a large increase in the volume of the gas. As a result, the gas will come out the bottle with force. This will lead to the breakage of the bottle. Cooling under tap water will result in the decrease of volume. It reduces the chances of accident.

(c) The pressure of the air is directly proportional to the temperature. During summer
due to high temperature the pressure in the tyre will be high as compared to that in water. The tube may burst under high pressure in summer. Therefore, it is advisable to inflate the tyres to lesser pressure in summer than in winter.

Question 25.
(i) State the law depicting the volume temperature relationship, (ii) Name the temperature at which the volume of the gas becomes equal to zero.
(i) The law is known as Charle’s law. Charle’s Law : Pressure remaining constant, the volume of a given mass of a gas increases or decreases by $$\frac { 1 }{ 273 }$$ of its volume at 0 °C for every one degree centigrade rise or fall in temperature.

Where Vt = Volume of the gas at t °C and V0 is its volume at 0 °C.
(ii) The temperature at which the volume of a gas becomes equal to zero is called absolute zero.

Absolute zero (0 K) = -273 °C
ie., the volume of gas = 0; because, before attaining this temperature, molecular motion of the gas is negligible, ie., they freeze. Gas no longer remains as gas.

Question 26.
Explain the following statements :
(i) Sodium chloride pieces are harder than sodium metal.
(ii) Copper is ductile and malleable but brass is not.
(iii) Latent heat of fusion of solid carbon dioxide is much less than that of silicon dioxide.
(iv) Water has its maximum density near 277 K.
(v) Ice floats on surface of water near the melting point.
(i) It has stronger electrostatic forces than metallic bond in sodium.
(ii) Copper is a pure element while brass is an alloy. Pure metal is more malleable.
(iii) SiO2 has covalent bonds while CO2 has only Van der waal’s forces (molecular solid). Hence SiO2 has higher heat of fusion.
(iv) Below 277 K and above 277 K, water expands hence its volume increases, density decreases and is maximum at 277 K
(v) At 0 °C or 273 K, its density is less than that of water hence floats.

Question 27.
What is the concept of Maxwell-Boltzmann distribution of molecular speeds, at a given temperature in a gas sample?
Concept s A gas consists of tiny particles (atom or molecules) separated from one another by large spaces. These particles are constantly moving in all directions. . During their motion, they collide with one another and also with the walls of the container. As a result of collisions, the speed and the direction of the molecules go on changing. Thus, all the molecules present in given sample of a gas do not possess the same speed. The speeds of individual molecules are different and are distributed over a wide range. The speeds of different molecules go on changing. However, the distribution of speed among different molecules remains the same at a particular temperature although the individual speeds of the molecules may change.

The distribution of molecules between different possible speeds was given by Maxwell and Boltzmann. He plotted the fraction of molecules, ie.,$$\frac{\Delta \mathrm{N}}{\mathrm{N}}$$ (along y-axis) having different speeds against the speeds of the molecules (along x-axis). The curve so obtained is shown in Fig. below and is known as Maxwell’s distribution curve.

The important features of Maxwell’s distribution curve can be summarized as follows:

• The fraction of molecules with very low or very high speeds is very small.
• The fraction of molecules possessing higher and higher speeds goes on increasing till it reaches the peak and then it starts decreasing.
• The maximum fraction of molecules possesses a speed, corresponding to the peak in the curve. The speed corresponding to the peak in the curve is referred to as most probable speed.

### 1st PUC Chemistry States of Matter Numerical Problems and Answers

Question 1.
Calculate the RMS velocity of oxygen at STP.
At STP P = 1.013 × 105 Pa
V = 22.4 × l0-3m3
Molecular weight of oxygen = 32 × 10-3 kg

= 461.23 metres per second, c = 461.23 ms-1

Question 2.
Calculate the RMS velocity of a molecule of nitrogen at STP.
At STP P = 1.013 × 105 Pa
v = 22.4 × 10-3m3
Molecular weight of nitrogen = 28 × 10-3 kg

c = 493.07ms-1

Question 3.
Calculate the RMS velocity of methane molecules at 298 K. R = 8.3 14 J K-1mol-1. Molecular weight of methane =16 × 10-3 kg.

Question 4.
The density of oxygen gas at STP is 1.429 gL-1. Find the RMS velocity.
At STP, P = 1.013 × 105Pa; d = 1.429gL-1 in mters it is

Question 5.
At what temperature will the rms velocity of hydrogen be the same as that of nitrogen at 350 K?

Question 6.
At what temperature will the RMS velocity of oxygen be twice at 200 K?
Let c1 be the rms velocity at 200 K (T1)
At temperature T2, c2 = 2c1

Question 7.
Calculate the total kinetic energy of 320 g of sulphur dioxide at 273 K.
(KE) n mole = $$\frac { 3 }{ 2 }$$ × nRT
T = 273K n = $$\frac { 320g }{ 64g }$$ = 5
$$\frac { 3 }{ 2 }$$ × 5 × 8.314 × 273 = 17020J

Question 8.
Calculate the total kinetic energy of 2 moles of chlorine at 300 K.
(KE)nmoies = $$\frac { 3 }{ 2 }$$ × nRT;
n = 2; T = 300 K
= $$\frac { 3 }{ 2 }$$ × 2 × 8.314 × 300 =7482.0 J

Question 9.
Calculate the kinetic energy of one mole of helium having RMS velocity K2ms-1.
Helium molecule is monoatomic, M = 4g = 4 × 10-3 kg
(KE)mole = $$\frac { 1 }{ 2 }$$ Mc2 = $$\frac { 3 }{ 2 }$$ x 4 x 10-3 x (102)2 = $$\frac { 1 }{ 2 }$$ × 4 × 10 = 20J

Question 10.
Calculate the kinetic energy of one mole of nitrogen at 400 K.
(KE)mol =$$\frac { 3 }{ 4 }$$ RT=$$\frac { 3 }{ 2 }$$ × 8.314 x JK-1mol-1 × 400K = 498.9 J mol-1.
Note: From problems 16 and 19, it can be observed that the kinetic energy of one mole of any two gases, at the same temperature will be the same.

Question 11.
Calculate the kinetic energy of 300 molecules of CO at 30 °C.

Question 12.
Calculate the total kinetic energy of the molecules of nitrogen per cm3 at STP and per m3 at STP.
1 mole of a gas occupies 22400 cc at STP
KE of 22,400 cc of a gas at STP = KE of 1 mole of a gas at STP
= $$\frac { 3 }{ 2 }$$ x RT = $$\frac { 3 }{ 2 }$$ x 8.314 × 273 =3404.58 J
KE of 1 cc of a gas = $$\frac { 3404.58 }{ 22400 }$$ = 0.152 J
KE of a m3 of a gas = 0.152 × 106 = 152 × 103 J.

Question 13.
Calculate the RMS speed of a molecule of hydrogen chloride at STP (The molar volume of any gas at STP = 22.4dm3
$$\mathrm{c}=\sqrt{\frac{3 \mathrm{PV}}{\mathrm{M}}}$$
where p = 1.01325 × 1015Nm-2
1.01325 × 1015kgm-1s-2
V = 22.4dm<sup<3mol-1 = 0.0224m3mol-1
M = 0.03645kgmol-1

This can be worked out using the relation $$\mathrm{c}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$$
R = 8.314 J K-1 mol-1
M = 0.03645 kg mol-1

Question 14.
Calculate the Root Mean Square velocity of a nitrogen molecule at 300 K and at a pressure of 2 × 10sNm2

Question 15.
Calcùlate the Root Mean Square velocity of a nitrogen molecule at 300 K and at a pressure of 2 × 105 Nm-2
1 mole of nitrogen at STP occupies a volume of 0.0224 m3 and weighs 0.028 kg.

Question 16.
Through the two ends of a glass tube of length 200 cm, hydrogen chloride gas and ammonia are allowed to enter. At what distance does ammonium chloride first appear?

Thus NH3 travel 1.465 times faster than HCl. In other words, NH3 will travel 1.465 cm in the same time in which HCl travels 1 cm.
Length of the tube = 200 cm

Thus, NH4Cl will first appear at a distance of 118.9 cm from NH3 end or 81.1 cm from HCl end.

Question 17.
Calculate the pressure exerted by 110 g of carbon dioxide in a vessel of 2 L capacity at 37 °C. Given that the van der waals constants are a = 3.59 L2atmmol-2 and b = 0.0427 L mol-1. Compare the value with the calculated value if the gas were considered as ideal.

Question 18.
One mole of SOa gas occupies a volume of 350 ml at 27 °C and 50 atm pressure. Calculate the compressibility factor of the gas. Comment on the type of deviation shown by the gas from ideal behavior.
Compressibility factor, Z = $$\frac{\mathrm{PV}}{\mathrm{nRT}}$$
Substituting, n = 1 mol, P = 50 atm, V = 350 × 10-3 L = 0.35 L
R = 0.0821 L atm K-1mol-1, T = 27 + 273 = 300 K, we get

For ideal gas, Z = 1. As for th given gas Z < 1, it shows negative deviation ie., it is more compressible than expected from ideal behavior.

Question 19.
The Van der waals constant ‘b’ for oxygen is 0.0318 L mol-1. Calculate the diameter of the oxygen molecule.

Question 20.
The boiling point of n – hexane is 68.9 °C. Calculate its approximate critical temperature.
According to Goldberg’s rule, for a liquid Tb = $$\frac { 2 }{ 3 }$$Tc
Hence Tc = $$\frac { 2 }{ 3 }$$Tb = $$\frac { 2 }{ 3 }$$ (38.9 + 273)K = 512.85 – 273°C

Question 21.
Assuming CO2 to be Van der waal’s gas, calculate its Boyle temperature. Given a = 3.59 L2atm mol-2 and b = 0.0427 Lmol-1.
Boyle temperature in terms of Van der waal’s constants is TB = $$\frac{\mathrm{a}}{\mathrm{Rb}}$$

Question 22.
A cylinder of 20.0 L capacity contains 160 g of oxygen gas at 25 °C. What mass of oxygen must be released to reduce the pressure of the cylinder at 1.2 atm?
Number of moles of oxygen gas present initially in the cylinder is
= $$\frac{160 \mathrm{g}}{32 \mathrm{g} \mathrm{mol}^{-1}}$$ = 5 moles
(Molar mass of 02 = 32 g mol-1)
To calculate the number of moles now present, we have P = 1.2 atm, T = 298 K, V = 20.0 L
Applying the relation, FV = nRT, we have,

∴ Number of moles of 02 required to be released = 5 – 0.98 = 4.02 mol
Or Mass of to be released = 4.02 × 32 g = 128.64 g.

Question 23.
50 litre of dry N2 is passed through 36 g H2O at 27 °C. After the passage of the gas, the mass of water was reduced to 34.80 g. Calculate the aqueous tension of water at 27 °C.
Water carried away by N2 gas = 36 – 34.80 = 1.20 g
As this water is carried away by 50 L of N2 gas, this means that volume occupied by water vapours = 50 L. Thus, now we have n = $$\frac { 1.20 }{ 18 }$$ = 0.0667mol, V = 50 L, T = 300
K, P = ?
Applying the gas equation, PV = nRT

= 0.03286 × 760 mm = 24.97 mm.

Question 24.
A 5 L vessel contains 1.4 g of nitrogen. When heated to 1800 K, 30% of molecules are dissociated into atoms. Calculate the pressure of the gas at 1800 K.

∴ Total number of moles = 0.035 + 0.030 = 0.065
i.e., n = 0.065 mol, V = 5L, T – 1800 K, P = ?
Applying PV = nRT, we get

Question 25.
The volume expansion of a gas under constant pressure is 0.037. Calculate its volume at – 100 °C if its volume at 100 °C is 685 cm3.
Volume expansion of a gas means increase or decrease in volume per degree rise or fall in temperature of its volume at 0 °C.
Vt = V0 + 0.0037 × V0 × t = V0(l + 0.0037t)
At 100 °C, V1oo°C = v0 (1 + 0.0037 × 100) = 685cm3 (Given)
Or V0= $$\frac{685}{1.37}$$ = 500cm3
At -100°C, V-1oo°C=500[1 + 0.0037 × (-100)] = 315cm3

## Karnataka 1st PUC Chemistry Question Bank Chapter 7 Equilibrium

### 1st PUC Chemistry Equilibrium One Mark Questions and Answers

Question 1.
Mention factors affecting rate (velocity) of reaction.

1. Nature of reactants
2. Catalyst
3. Temperature
4. Pressure
5. Concentration of reactants.

Question 2.
Define Law of mass action.
“At a given temperature the rate of chemical reaction is directly proportional to the product of active masses of the reactant.”

Question 3.
Define equilibrium constant of reaction.
“Is the ratio of velocity constant of forward reaction to velocity constant of backward reaction.” Also defined as, “It is the ratio of product to molar concentration of product to the product of molar concentration of the reactants”.

Question 4.
Write expression for Kc & KP for the reaction.

Question 5.
Write the relationship between Kp & Kc.
Kp = Kc [RT] Δn
Where Δn is = (number of gaseous product – number of gaseous reactant)

Question 6.
Define Le-Chatelier’s principle.
Definition : “when a constraint applied to a system at equilibrium in a reversible reaction, the equilibrium shifts so as to nullify the constraint”.
[Constraint is change in temperature or pressure, addition of reactant or product]

Question 7.
What is active mass ?
Active mass means molar concentration i.e., concentration expressed in mol/dm3.

Question 8.
What happens when the temperature of a reversible reaction at equilibrium is increased, if enthalpy change is positive?
Equilibrium shifts to the right.

Question 9.
Give an example for a reversible reaction in which Kp = KcRT.
PCI5(g) ⇌ PCI3(g)+CI2(g)

Question 10.
Write an expression for Kp for the following reaction: A + B ⇌ C + 2D.

Question 11.
How are KP and Kc related ? Mention the condition under which KP=KC.
Kp = Kc (RT)Δn when Δn = 0 i.e. no. of moles of gaseous products = no. of moles of
gaseous reactants, e.g., H2(g) + I2(g) ⇌ 2HI(g)Kp = Kc

Question 12.
Give an example of a reaction where Kp ≠ Kc.
PCI5(g) ⇌ PCI3(g) + CI2(g)

Question 13.
How does a catalyst influence the equilibrium constant of a reversible ‘ reaction ?
A catalyst does not influence the equilibrium constant of reversible reaction.

Question 14.
Which measurable property becomes constant in water , ~~ water vapour
equilibrium at constant temperature?
Vapour pressure.

Question 15.
Give one example of everyday life in which there is gas y..’ >. solution
equilibrium.
Soda-water bottle.

Question 16.
Under what condition, reversible process becomes irreversible?
If one of the products (gaseous) is allowed to escape out (i.e., in open vessel).

Question 17.
What is the effect on the value of equilibrium constant on adding catalyst?
No effect.

Question 18.
What is the effect of increasing pressure on the equilibrium?
N2 + 3H2 ⇌ 2NH3 ?
Equilibrium will shift in the forward direction forming more of ammonia.

Question 19.
What are the conditions for getting maximum yield of NHs by Haber’s process?
High concentrations of N2 and H2, low temperature, high pressure.

Question 20.
What happens to the dissociation of PCI5 in a closed vessel if helium gas is introduced into it at the same temperature?
No effect.

Question 21.
What is the law called which gives relationship between degree of dissociation of a weak electrolyte and its concentration in the solution?
Ostwald’s dilution law.

Question 22.
Write the expression for comparison of relative strengths of two weak acids in terms of their ionization constants.

Question 23.
What do you mean by reversible process?
Reversible process is a process in which products can react to form reactants back.

Question 24.
Define equilibrium.
Equilibrium is a stage at which rate of forward reaction is equal to rate of backward reaction.

Question 25.
What is physical equilibrium? Give one example.
Physical equilibrium is am equilibrium between two different physical states of same
substance, e.g., H2O(s) ⇌ H2O(I)

Buffer capacity can be also defined as quantity of strong acid or base that must be added to change the pH of one liter of solution by one pH unit.

Question 26.
Give two examples for Acedic buffer.
A solution containing a weak acid and its salt with a strong base.
(i) CH3COOH + CH3COONa
(ii) HCOOH + HCOOK

Question 27.
Write expression of Kc for reaction N2 (g) + 3H2 (g) ⇌ 2NH3 (g). Give units of K.

Question 28.
Name the factors that affect equilibrium position of a reversible reaction.
(i) Temperature (ii) Pressure, (iii) Concentration, (iv) Catalyst.

Question 29.
At what temperature the solid, and liquid are in equilibrium under 1 atm. pressure?
Melting point or freezing point.

Question 30.
Define solubility of substance.
Solubility of substance is maximum amount of solute that can be dissolved lOOg of solvent at a given temperature.

Question 31.
State Henry’s law.
Henry’s law states the solubility of gas is directly proportional to the partial pressure of the gas above the solution, i.e, mass of gas dissolved is directly proportional to pressure applied on gas.
m = k x p where ‘p’ is pressure applied on the gas.

Question 32.
What is the effect of temperature on solubility of gases in liquids?
Solubility of gases in liquids decreases with increase in temperature.

Question 33.
What is the effect of increasing pressure in the reactions?
(i) PCI5 (g) ⇌ PCI3 (g) + CI2 (g)
(ii) N2 (g)+ O2(g) ⇌ 2NO(g).
(i) The equilibrium will shift in backward reaction.
(ii) No effect.

Question 34.
What is the effect of temperature on the reactions?
(i) N2(g) + 3H2(g) ⇌ 2NH3(g) + Heat
(ii) N2(g) + O2(g) ⇌ 2NO3(g) – Heat
(i) The increase in temperature will favour backward reaction because the reaction is exothermic.
(ii) The increase in temperature favours forward reaction because the reaction is endothermic.

Question 35.
Write expression of KP for the reaction.
N2O(g) ⇌ 2NO2(g)

Question 36.
N2 (g) + 3H2 (g) ⇌ 2NH3 (g),K = 0.50 at 673K.
Write the equilibrium expression and equilibrium constant for reverse reaction.

Question 37.
What is difference between strong electrolyte, and weak electrolyte?
Strong electrolytes can dissociate into ions completely in aqueous solution whereas weak electrolytes do not dissociate into ions completely and there exist an equilibrium between ioins and unionized molecules, called Ionic Equilibrium.

Question 38.
Define degree of ionization or dissociation.
It is the ratio of number of ions or molecules produced to the total number of molecules.

Question 39.
State Ostwald’s dilution law.
Ostwald’s dilution law states that the degree of dissociation of weak electrolyte is inversely proportional to square root of its concentration.

Where Ka is acid dissociation constant and Kb is base dissociation constant.

Question 40.
Define ionic equilibrium.
The equilibrium between ions and unionized molecules is called ionic equilibrium.

Question 41.
Under what conditions a weak electrolyte can »have high degree of ionization?
On dilution, i.e., at a very low concentration, weak electrolyte has high degree of ionization.

Question 42.
What is the effect of temperature on degree of dissociation?
The degree of Dissociation increases with increase in temperature.

Question 43.
Define pH. Give its mathematical expression.
pH is defined as negative logarithm of [H2O+]concentration.

Question 44.
What is meant by ionic product of water (Ka)?
It is product of concentration of [H3O+] and [OH] at a specific temperature.
Kw = [H3O+ ][OH] = 1.0 × 10-14 at 298K

Question 45.
In qualitative analysis, on what basis cations are grouped?
Cations are grouped on the basis of their solubility product (Ksp).

Question 46.
What is the range of a pH indicator in terms of its dissociation constant (Kin)?
pH = pKin ±1.

Question 47.
What happens to the solubility of Agcl in water if NacI solution is added to it?
Solubility of AgCI decreases due to common ion effect.

Question 48.
What is the realationship between pKa and pKb values where Ka and Kb represent ionization constants of the acid and its conjugate base respectively?
pKa + pKb = pKw = 14.

Question 49.
What is the relationship between pH and pOH?
pH + pOH = pKw = 14.

Question 50.
Write the demerits of Brownsted-Lowry theory.
It does not explained why CaO, NH3 acts as base and BF3 and CO2 acts as acid.

Question 51.
Define acidolysis.
Acidolysis is the process where acidity of blood decreases from 7.3.

Question 52.
Define Alkolysis.
Alkolysis is the process where acidity of blood increases from 8.

Question 53.
What do you mean by buffer solution?
The solution which resist the pH or pOH when an acid or base is added to it. Buffer solution is a mixture of weak electrolyte and its salt.

Question 54.
Define buffer action.
The ability or capacity of a buffer solution to resist the pH by addition of acid or base.

Question 55.
Mention the types of buffer solution.
acidic buffer
basic buffer
neutral buffer.

Question 56.
Mention the uses of buffer solution.
It is used in chemical industries like paper, sugar, pharmaceuticals.
It is used in all biological fluids to maintain constant pH by buffer action.

### 1st PUC Chemistry Equilibrium Two Marks Questions and Answers

Question 1.
Explain rate of reaction with its mathematical forms.
Rate (Velocity) of reaction:
It is the ratio of change in concentration to unit interval of time.
Mathematically: Let us consider a general reaction, A (reactant) → B(products

Where dc = change in concentration : dt = change in time.
-ve sign = forward reaction where concentration of reactant decreases.
+ve sign = backward reaction where concentration of reactant increases.
Meaning of [A] = – concentration of reaction A
[B] = – concentration of Product B.

Question 2.
Define Rate equation with example.
An expression which relates the rate (velocity) of the reaction with change in concentration of reactant in a given interval of time.

Question 3.
Define Irreversible reaction with example.
A Chemical reaction in which product can not able to give back the reactant are called Irreversible reaction.
Example : C(s) + O2 (g) → CO2 (g)

Question 4.
Define Reversible reaction with example.
A Chemical reaction in which an infinitesimally small change in any one of the variable (Pressure, Temperature, Concentration of reactant or product) of the reaction is changed the direction (forward or backward) of the reaction changes.”
It is repreented by half arrow mark. An Example is H2 (g) +I2 (g) ⇌ 2HI(g)

Question 5.
Define Chemical Equilibrium with an example.
Is a state where the rate of forward reaction is equal to rate of backward reaction.” OR is also defined as “ A state where concentration of reactant and product remains
constant. ” Example : H2 (g) +I2 (g) ⇌ HI(g)

Question 6.
Chemical equilibrium is dyanamic. Justify.
Chemical equilibrium is dynamic : Because the velocities of forward and backward reactions are equal also the concentration of reactant and product unchanged at this equilibrium state, (but may not be equal). The equilibrium can be made to shift in either direction by changing the concentration. Pressure, Volume etc., Hence is dynamic in nature.

Question 7.
Write the characteristics of chemical equilibrium.

• Chemical equilibrium is dynamic.
• Equilibrium reaction will not come to an end.
• Equilibrium is attained only in a closed system.
• Catalyst does not alter equilibrium state.

Question 8.
The equilibrium constant for a reaction is 26.66. If the rate constant of the forward reaction is 3.2 × 10-4 s-1, what is the rate constant of the backward reaction?
Data given,
Equilibrium constant for a reaction Kc – 26.66
Rate constantof the forward reaction = k1= 3.2 × 10-4 /sec.
∴ Rate constant of the backward reaction = k2 = ?
Equilibrium constant of a reaction ‘Kc’ is given by the formula.

Question 9.
Define Arrhenius acid – base theory with one example.
An electrolyte which when dissolved in water, produces H+ ion is called acid.
An electrolyte which when dissolved in water, produces OH ion is called base.

Question 10.
Define Bronsted Lowry theory OR Protonic theory with one example.
A substance or molecule which donate proton is called acid [Protogenic]
A substance or molecule which accepts proton is called base [Protophillic]
Example:

Question 11.
What are amphoteric substance? Give one example.
A substance or molecule which acts both as acid as well as base are called amphoteric.
Example:

Question 12.
What are conjugate acid-base pair? Give one example.
Acid – base pairs which differed by one proton is called conjugate acid – base pair.
NH3+H2O → NH4+ + OH
Conjugate base of acid H2O is OH ; H2O + H2O → H3O+ + OH
Conjugate acid of base NH3 is NH3+ ; NH3 +NH3 → NH4+ +NH2.

Question 13.
Explain Lewi’s electron acid – base concept with an example.
According to Lewi’s theory, acid is one which accepts pair of electron and base is a – substance which donates pair of electron.

Question 14.
Calculate pH of 0.01 M HC1.
pH = -log10[H+] = -log10[0.01] = -log10[10-2] = 2

Question 15.
Calculate pH of 0.0001 M of HNO3.
pH = -log10[H+ ]-log10[0.0001] = -log10[10-4] = 4

Question 16.
Calculate pH of 0.00025 M HNO3.
pH = -log10[H+] = -Iog10[0.00025] – log10 [2.5 × 10-4] = 4 – log2.5
pH =4 – 0.3979 = 3.6021

Question 17.
Calculate pH of 0.1 M of H2SO4 (concentration 0.1 × 2 hydrogen = 0.2).
pH = -log10[H+] = -log10[0.1 × 2] = -log10[0.2] = -log10[2 × 10-1] = 1 – log2
pH = 1 – 0.3010 = 0.699

Question 18.
Calculate pH of 0.05 M HC1.
pH = -log10[H+] = -log10[0.05] – log10 [5 × 10-2] = 2 – log5 = 2 – 0.6990 =1.301

Question 19.
Calculate pH of 0.005 M H2SO4.
pH = -log10[H+] = -log10[0.005 × 2] = -log10[0.01] = -log10[ 1 × 10-2 ]
pH = 2 – log 1 = 2 – 0 = 2

Question 20.
Calculate pH of 3 × 10-9 M NaOH.
pOH = -log10[OH] = -log10[3 × 10-9] = 9 – log3 = 9 – 0.4771 = 8.5229

Question 21.
Calculate [OH] if pOH = 8.3.
pOH =-log10[OH]; 8.3 =-log10[OH]
Taking antilog on both the sides
[OH] = antilog (-8.3) = [OH] = antilog (-9 – 8.3 + 9) = antilog (0.7) × 10-9]
= 5.012 × 10-9 mol/dm3

Question 22.
Calculate [H+] if pOH = 9.23.
pH + pOH = 14 pH = 14 – pOH = 14 – 9.23 = 4.77
pH = -log10[H+] = -4.77 = log[H+]
Taking antilog on both the sides
[H+] = antilog (-4.77) = antilog [+5 -4.77 – 5] = antilog [0.23] × 10-5
= 1.698 × 10-5] mol/dm3]

Question 23.
Calculate [OH] if pH = 5.284
pH = -log10[H+]; [H+] = antilog [-5.284] = antilog [6 – 5.284 – 6]
= antilog (0.716 – 6) = 5.2 × 10-6 mol/dm3

Question 24.
What is the pH of a 0.05 M solution of formic acid ? (Ka = 1.8 × 10-4)

Question 25.
Calculate the [OH ] of a solution whose pH is 9.62.
pH + pOH = 14 pOH = 14 – pH = 14 – 9.62 = 4.38
pOH = 4.38 = -log10[OH]; 4.38 = -log10[OH”]
log10[OH] = -4.38; [OH-] = antilog (-4.38) = antilog (5 – 4.38 – 5)
= antilog (0.62 – 5) = antilog (0.62) × 10-5 = 4.169 × 10-5 mol dm-3.

Question 26.
The pOH of a solution is 5.725. Calculate the [H+].
pH + pOH = 14 pH = 14 – pOH = 14 – 5.725 = 8.275
pH = -log10[H+] ⇒ 8.275 = log10[H+] ⇒
[H+] = antilog (-8.275) = antilog(9 – 8.275 – 9).
[H+] = antilog (0.275 – 9) = 5.309 × 10-9 mol dm-3.

Question 27.
Calculate the pH of 0.125M H2SO4.
H2SO4 → 2H+ +SO42-
[H+] = 2[H2SO4] = 2 × 0.125 = 0.25
pH = -log10[H+] = -log10[0.25] = -log10 [2.5 × 10-1]
pH =-log2.5-log10-1 =-0.3979+1 = 0.6021.

Question 28.
Solubility product of barium sulphide (BaS04) is 2.4 × 10-9. Calculate it’s solubility.
BaSO4 is AB type of salt
Ksp = 2.4 × 10-9

Question 29.
If the solubility product of silver chloride is 1.8 × 10-10. What is the solubility of siliver ion if concentration of Cl is 0.01 molar.

Question 30.
Give any two differences between strong and weak electrolyte.

Question 31.
What is common ion effect ? Give example.
Supression in degree of dissociation of a weak electrolyte by addition of a common ion is called common ion effect.
Example : CH3COOH and CH3COONa

### 1st PUC Chemistry Equilibrium Three Marks Questions and Answers

Question 1.
Some processes are given below. What happens to the process if it subjected
to a change given in the brackets?
(i)

(ii) Dissolution of NaOH in water (Temperature is increased)
(iii) N2 (g) + O2 (g) ⇌ 2NO(g) -180.7kJ (Pressure is increased and temperature is decreased)
(i) Equilibrium will shift in the forward direction, i.e., more of ice will melt.
(ii) Solubility will decrease because it is an exothermic process.
(iii) Pressure has no effect. Decrease of temperature will shift the equilibrium in the backward direction.

Question 2.
What qualitative information can you obtain from the magnitude of equilibrium constant?
(i) Large value of equilibrium constant (>103) shows that forward reaction is favoured, i. e., concentration of products is much larger than that of the reactants at equilibrium.
(ii) Intermediate value of K(10-3 to 103) shows that the concentration of the reactants and products are comparable,
(iii) Low value of K(<10-3)shows that backward reaction- is favoured, i.e, concentration of reactants is much larger than that of the products.

Question 3.
How do you explain detection of I group basic radicals based on application and solubility product.
Group reagent: Dil. HCl → H+ + Cl
Redicals present: Pb2+, Hg2+, Cd2+
The group reagent HCl is a strong electrolyte under goes complete dissociation. The metal radicals forms it’s respective chlorides [PbCl2, HgCl2, CdCl2 ] Whose solubility product is less than ionic product. Hence, forms pricipitate.

Question 4.
How do you explain detection of H group basic radicals based on application of solubility product.
Group reagent: HCl → H+ + Cl; H2S ⇌ 2H+ + S2-

Here HCl is a strong electrolyte under goes complete dissociation where as H2S is a weak electrolyte under goes partial dissociation and attains equilibrium between it’s ions and molecules.

In this solution, because of common ion (H+) effect supression of degree of dissociation of H2S takes place. As a result [S2-] decreases. Hence the metal radical forms it’s respective sulphides [CuS, SnS, BiS] whose solubility product is less than ionic product. Hence forms pricipitate.

Question 5.
How do you explain the detection of HI group basic radicals based on application of solubility product ?
Group reagent: NH4OH ⇌ NH4+ + OH ; NH4Cl ⇌ NH4+ + Cl
Here NH4Cl is a strong electrolyte under goes complete dissociation where as NH4OH is a weak electrolyte under goes partial dissociation and attains equillibrium between it’s ions and molecules.

In this solution because of common ion (NH4+) effect supression of degree of dissociation of NH4OH takes place. As a result [OH] decreases. Hence the metal radical forms it’s respective hydroxides [Al(OH)3, Cr(OH)3, Fe(OH)3] Whose solubility product is less than ionic product. Hence forms precipitate.

Question 6.
Explain detection of IV group basic radicals using application of solubility product.
Group reagent: NH4Cl → NH4+ + Cl ; NH4OH ⇌ NH4+ + OH
H2S ⇌ 2H+ + S2- ; H+ + OH → H2O

Here, NH4Cl is a strong electrolyte undergoes complete dissociation, where as NH4OH – and H2S are weak electrolytes undergoes partial dissociation and attains equillibrium between their ions and molecules.

Along with common ion (NH4+) effect, the H+ and OH ion combines to form water molecules.
At the same time the metals forms their respective sulphides [ZnS, MnS] Whose solubility product is less than ionic product. Hence, forms precipitate.

Question 7.
Define Arrhenius acid – base theory with one example.
An electrolyte which when dissolved in water, produces H+ ion is called acid.
An electrolyte which when dissolved in water, produces OH- ion is called base.

Question 8.
Deduce equation for Ostwalds dilution law with respect to the weak acid.
Consider a weak acid HA with concentration C mol/dm3. After a time interval it reaches equillibrium with dissociation x mol/dm3 as follows.

Question 9.
Deduce equation for Ostwalds dilutions law with respect to weak base.
Consider a weak base BOH with concentration C mol/dm3. After a time internal it reaches equilibrium with dissociation x mol/dm3 as follows.

(ii) Where, Kb is dissociation constant of weak base.

Question 10.
Derive ionic product of water. Also find the value at 25 °C

Where, Kw is ionic product of water.
Value of Kw at 25°C :
It is found that at 25°C [H+] = [OH] = 10-7 mol/dm3
∴Kw = [H+][OH] = 10-7 × 10-7 = 10-14 (mol/dm3)2
(iii) At 25°C the value of Kw is 10-14( tmol/dm3)2

### 1st PUC Chemistry Equilibrium Five Marks Questions and Answers

Question 1.
How do you apply law of mass action to a gaseous reversible reaction.
If the reactants and products are in gaseous state, then for the reaction,

The total pressure P = PA + PB + PC + PD
If PA PB PC & PD are represents the partial pressure exerted by A, B, C and D respectively, then according to law of mass action,
Velocity of forward reaction Vf ∝ proportional PA PB Or
Vf=Kf PAPB………….(3)
Where Kf velocity (proportionality) constant of forward reaction.
Velocity of backward reaction Vb ∝ PC PD Or
Vb = Kb PCPD ……….(4)
Where Kb velocity (proportionality) constant of backward reaction.
At equilibrium, velocity of forward reaction is equal to velocity of backward reaction, from (3) & (4).
Vf = Vb Then

Where KP is equilibrium constant with respect topartial pressure of the reactant.

Question 2.
Write the characteristics of Equilibrium Constant (KP & KC).

• It is independent of initial concentrations of the reactants, the direction from (reactant or product) which equilibrium attained.
• The value of equilibrium constant independent of Volume, Pressure, and presence of catalyst or any inert material.
• Equilibrium constant gives relative idea about the extent to whick the reaction takes place.
• At a given temperature, the equilibrium constant of a particular reaction has definite value.
• If its (Kc or Kp) value is large, then the forward reaction takes place nearly to completion whereas backward reaction to a small extent Vice versa.
• If the forward reaction is exothermic, then the value of Kc or KP decreases. Vice versa.
• In a reversible reaction, the value of equilibrium constant of forward reaction is reciprocal of backward reaction, ieKf=1/Kb.
• The value of equilibrium constant depends on the stoichiometry of reactants, and products at equilibrium.

Question 3.
Write the conjugate acid for the following (a) Cl, (b) NO3, (c) HSO4, (d) HCO3, (e) SO42-, (f) CO32-, (g) NH2-, (h) N3-, (i) O2-, (j) H2O, (k) NH3.

Question 4.
Deduce Handersen equation for basic buffer.
Consider the basic buffer, a mixture of ammonium hydroxide and ammonium Chloride.

Question 5.
Deduce Hendersons equation for acidic buffer.
Consider the acidic buffer, a mixture of acetic acid and sodium acetate as.
CH3COOH ⇌ CH3COO + H+ → (1)
CH3COONa → CH3COO + Na+ →(2)

Where, Ka is dissociation constant of weak acid.
In this solution, [ CH3COO ] ≅ [Salt] → (4)
[CH3COOH] ≅ [Acid] → (5)

Question 6.
Explain Mechanism or working of acid buffer:
Consider acidic buffer mixture,
CH3COOH ⇌ CH3COO + H+ ; CH3COONa → CH3COO + Na+
Here, acetic acid is a weak electrolyte, in it’s solution there exists equillibrium between it’s ions and molecules. Where as sodium acetate completely dissociates into it’s ions.

Therefore, the buffer mixture contains large number of CH3COO ions followed by Na+, H+, CH3COOH.

Case i) When an acid added to this solution :
H+ ion of the acid combines with CH3COO ion in buffer solution and makes equillibrium with acetic acid.
CH3COO + H+ ⇌ CH3COOH
As a result pH remains constant.

Case ii) When a base is added to this solution :
OH ion of the base combines with H+ ion in buffer solution forms water molecule.
H++ OH → H2O.
As a result pH remains constant.

Question 7.
Explain the Mechanism or working of basic buffer.
Consider basic buffer solution,
NH4OH ⇌ NH4+ + OH ; NH4Cl → NH4 + Cl
Here, ammonium hydroxide is a weak electrolyte, in it’s solution there exist equillibrium between it’s ions and molecules.
Where ammonium chloride completely dissociates into it’s ions.
Therefore, buffer solution contains large number of NH4+ions followed by Cl, OH and NH4OH.

Case (i) When a base is added to this solution:
OH- ion of the base combines with NH4+ ion of the buffer solution and makes equillibrium with ammonium hydroxide.
NH4+ + OH⇌ NH4OH; As a result pOH remains constant.

Case (ii) When an acid is added to this solution :
H+ ion of acid combines with OH ion of buffer solution and forms water molecule. H+ + OH → H2O
As a result pOH remains constant.

Question 8.
The pKa value of acetic acid is 4.7447 at 25 °C. How would you obtain a buffer of acetic acid and sodium acetate with pH = 4?

So by mixing sodium acetate and acetic acid in the ratio 0.18 : 1 mole per dm3, the buffer of pH = 4 is obtained.

Question 9.
A buffer solution of pH = 4.7 is prepared from CH3COONa and CH3COOH. Dissociation constant of acetic acid is 1.75 × 10-5. Calculate the mole proportion of sodium acetate and acetic acid.

Question 10.
A buffer solution of pH 8.3 is prepared from nmnionium chloride and ammonium hydroxide. Dissociation constant of ammonium hydroxide is 1.8 × 10-5. What Is the mole proportion of ammonium chloride and ammonium hydroxide?

Question 11.
3.0 g of pure acetic acid and 4.1 g of anhydrous sodium acetate are dissolved together in water and the solution is made up to 500 ml. Calculate the pH of the solution. Given Ka of acetic acid is 1.75 × 10-5.
Calculation of concentration of acetic acid: mass/dm3 = N x g eq. mass

Molecular wt. of CH3COOH = 12 × 2 + 1 × 4 + 16 x 2 = 24 + 04 + 32 = 60.
[CH3COOH] $$\frac{3.0 \times 2}{60}=\frac{6.0}{60}$$ = 0.1N
Calculation of concentration of sodium acetate

## Karnataka 1st PUC Chemistry Question Bank Chapter 1 Some Basic Concepts of Chemistry

### 1st PUC Chemistry Some Basic Concepts of Chemistry One Mark Questions and Answers

Question 1.
What is the unit for equivalent weight?
Equivalent weight is a mere number, it has no units. When expressed in grams it is called gram equivalent weight.

Question 2.
Define Atomic weight.
The atomic weight of an element is the ratio of average mass of an atom of an element to $$\frac{1}{12}$$ of mass of an atom of C-12 Isotope.

Question 3.
Write the mathematical form of Atomic weights.

Question 4.
Mention the unit for equivalent weight.
There is no unit for atomic weight, when expressed in grams it is called gram atomic weight.

Question 5.
Write the Relationship between Atomic weight and Valency weight.

Question 6.
Define Valency.
Valency: The combining capacity of an atom (or the number of electrons present in the outermost orbital of an atom which involves in bond formation.

Question 7.
Mention the methods to determine Equivalent weight.

• Oxide Method
• Chloride Method
• Hydrogen Displacement method
• Interconversion method.

Question 8.
Define Molecular weight.
It is defined as “ The ratio of the mass of 1 molecule of the substance to $$\frac{1}{12}$$ mass of an atom of C-12 isotope”.

Question 9.
Write mathematical forms of Molecular weight.

Question 10.
Equal volume of all gases and vapours under the same condition of temperature and pressure contains an equal number of molecules.

Question 11.
Define Mole.
1 mole represents the amount of substance which contains an Avagadro number of particles.

Question 12.
What do you mean by Avagadro Number?
The number of molecules present in 1 gm molecular weight of any substance.
OR
The number of atoms present in 1 gm atomic weight of any element. It is represented by the symbol N = 6.022 × 1023

Question 13.
Define Molality.
The number of moles of solute dissolved in 1 kg of the solution.

Question 14.
Define Molarity.
A number of the molecular equivalent of solute dissolved in 1 dm3 of the solution.

Question 15.
What is mole fraction?
The ratio of the number of moles of a component to the number of moles present in the solution.

Question 16.
What do you mean by a Normal solution?
A solution containing 1 gm equivalent of the substance dissolved in 1 dm3 of the solution.

Question 17.
What do you mean by the decinormal solution?
A decinormal solution is the solution containing 0.1 g equivalent ($$\frac { 1 }{ 10 }$$ th) of solute / substance dissolved in 1 dm3 of solution.

Question 18.
What are indicators?
Indicators are the chemical substance used to indicate the endpoint of a volumetric reaction by the change in colour.

Question 19.
Mention the types of Indicators.

1. Internal indicator
2. External indicator
3. Self indicator

Formula of normality The Normality is defined as the number of gram equivalent of the solute dissolved per litre of the solution.

Question 20.
Define empirical formula.
It is a formula which gives the relative number of atoms of each element in the compound.

Question 21.
Define Molecular formula.
It is a formula which gives the actual number of atoms of each element present in the molecule of the given compound.

Question 22.
What is the relation between empirical and molecular formula?

Question 23.
State the law of definite proportion.
The law states that “a chemical compound, however, prepared, contains the same elements combined together in the same fixed proportion by mass”.

Question 24.
Define the law of conservation of mass.
It states that matter can neither be, created nor be destroyed.

Question 25.
State Gay Lussac’s law of combining volumes.
The law states that ’when gaseous reactants combine to give gaseous products, they do so in volumes which bear a simple whole number ratio to one another and to the volume of products, provided the volumes are measured under similar conditions of temperature and pressure”.

Question 26.
What is NTP?
(Normal temperature and pressure or standard temperature and pressure). The temp 273K (0 °C) and pressure 101.3Kpa (or N/m2) or 1 atm = 760 mm of Hg level are taken as STP or NTP.

Question 27.
What physical quantities are represented by the following units and what are their most common names?
(i) kg ms-2
(ii) kg m2s-2
(iii) dm3
(i) Force (newton)
(ii) Work (joule)
(iii) Volume

Question 28.
Rewrite the following after required corrections:
(i) The length of a rod is 10 cms
(ii) the work done by a system is 10 Joules.
(i) The length of a rod is 10 cm (s is not used)
(ii) The work done by a system is 10 joules (the small letter is used in place of the capital letter).

Question 29.
Classify the following substances into elements, compounds, and mixtures:
(i) Milk
(ii) 22-carat gold
(iii) Iodized table salt
(iv) Diamond
(v) Smoke
(vi) Steel
(vii) Brass
(viii) Dry ice
(ix) Mercury
(x) Air
(xi) Aerated drinks
(xii) Glucose
(xiii) Petrol/Diesel/Kerosene oil
(xiv) Steam
(xv) Cloud.
Element – (iv), (ix);
Compounds – (viii), (xii), (xiv), (xv);
Mixtures – (i), (ii), (iii), (v), (vi), (vii), (x), (xi), (xiii).

Question 30.
Why is air sometimes considered a heterogeneous mixture?
This is due to the presence of dust particles which change the phase, into heterogeneous.

Question 31.
1 L of a gas at STP weighs 1.97 g. What is the vapour density of the gas?
22.4 L of the gas at STP will weigh 1.97 × 22.4 = 44.1 g ie., molecular mass = 44.1.

Question 32.
Why atomic masses are the average values?
Most of the elements exist in different isotopes ie., atoms with different masses, eg. Cl has two isotopes with mass numbers 35 and 37 existing in the ratio 3:1. Hence, the average value is taken.

Question 33.
Determine the equivalent weight of each of the following compounds assuming the formula weights of these compounds are x, y, and z respectively, (i) Na2SO4 (ii) Na3P04.12H2O (iii) Ca3(PO4)2

Question 34.
What is the mass of mercury in grams and in kilograms if the density of liquid mercury is 13.6 g cm-3?
Mass = Volume x density = 1000 cm3 × 13.6 g cm-3 = 13,600 g = 13.6 kg

Question 35.
Vitamin C is known to contain 1.29 × 1024 hydrogen atoms. Calculate a number of moles of hydrogen atoms.

Question 36.
Define an organic compound. Give one example.
An organic compound contains essentially carbon, hydrogen along with certain elements like N, P, S, O, halogen, etc., eg: urea, sugar.

Question 37.
Define an inorganic compound. Give one example.
An inorganic compound is made up of two or more elements containing a rare percentage of carbon eg: NaCl, KC1, ZnS, etc.

Question 38.
Name the process used for refining petroleum.
Fractional distillation.

Question 39.
Calculate the percentage of nitrogen in NH4. (Atomic mass of N = 14, H = 1 amu)

Question 40.
State the law of multiple proportion.
It states ‘whenever two elements combine to form two or more compounds, the ratio between different weights of one of the elements which combines with fixed weight of other is always simple’.

Question 41.
Calculate the number of He atoms in (i) 52 u, (ii) 52 g, (iii) 52 moles of He. Atomic weight of He is 4 u.
(i) Atomic weight of Helium = 4, i.e. 4 u is mass of 1 atom,
52 u is mass of $$\frac { 1 }{ 4 }$$ × 52 = 13 atoms.

(ii) 4 g of He contains 6.022 × 1023 He atoms / 52 g of He contains ?
$$\frac{6.022 \times 10^{23} \times 52}{4}$$ = 7.8286 × 1023 atoms.

(iii) 1 mole of He contains 6.022 × 1023 atoms, 52 moles of He contains 52 × 6.022 × 1023 =3.131 × 1023 atoms.

Question 42.
How many electrons are present in 16 g of CH4 ?
1 molecule of CH4 =6 + 4 = 10 electrons, 16 g of CH4 contains 10 × 6.022 × 1023 = 6.022 × 1024

Question 43.
Boron occurs in nature in the form of two isotopes and $$\begin{array}{l}{11} \\ {5}\end{array}$$ B and $$\begin{array}{l}{10} \\ {5}\end{array}$$ in ratio of 81% and 19% respectively. Calculate its average atomic mass.

Question 44.
If 2 litres of N2 is mixed with 2 litres of H2 at a constant temperature and pressure. What will be the volume of NH3 formed?
N2(g) + 3H2(g) → 2NH3(g)

Question 45.
How many atoms are present in 4 ml of NH3 at STP?
22400 ml of NH3 contains = 4 × 6.022 × 1023 atoms [∵ NH3 contains 4 atoms]. 1ml of NH3 contains = $$\frac{4 \times 6.022 \times 10^{23}}{22400}$$ = 1.07 × 1020 atoms.

Question 46.
Which of these weighs most? (i) 32 g of oxygen, (ii) 2 g atom of hydrogen, (iii) 0.5 mole of Fe, (iv) 3.01 × 1023 atoms of carbon.
(i) 32 g of oxygen weighs most.
(ii) 2 g atom of H2 = 2g
(iii) 0.5 mole of Fe = mole × atomic weight = 0.5 × 56 = 28 g
(iv) 6.022 × 1023 = 1 atom of d 3.01 × 1023 atoms of C = $$\frac { 1 }{ 2 }$$ × 12 = 6 g.

Question 47.
An element has a specific heat of 0.113 cal/g °C. Calculate atomic weight of element.
According to Dulong Petitt’s law, Atomic weight × Specific heat = 6.4 cal
Atomic weight × 0.113 cal/g = 6.4 cal
Atomic weight = $$\frac { 6.4 }{ 0.113 }$$ = 56

Question 48.
Why are the atomic masses of most of the elements fractional?
This is because atomic masses are the relative masses of atoms as compared with an atom of C-12 isotope taken as 12.

Question 49.
Match the following:

A-(ii)
B-(iii)
C-(iv)
D-(i)

Question 50.
Match the following prefixes with their multiplies.

(i) micro 10-6
(ii) deca – 10
(ill) mega – 106
(iv) giga – 109
(v) femto – 10-15

Question 51.
Calculate the molar mass of the following:
(i) H2O (ii) CO2 (iii) CH4
(i) H2O = 2 × H+ 1 × O
2 × 1 + 1 × 16 = 18
(ii) CO2 = 12 × 1 + 2 × 16 = 44
(iii) CH4 = 1 × C + 4 × H
1 × 12 + 4 × 1 =16

### 1st PUC Chemistry Some Basic Concepts of Chemistry Two Marks Questions and Answers

Question 1.
Define the basicity of an acid with an example.
Basicity of an acid is the number of replaceable hydrogen atoms present on a molecule of acid is called the basicity of an acid.

Question 2.
Define acidity of a base with example.
The acidity of a base is the number of monobasic acids required to neutralize 1 molecule of a base. [The base which contains a number of replaceable – OH group].

Question 3.
Define equivalent weight of acid with example.
The number of parts by mass of the acid contains 1 part by mass of replaceable hydrogen.

Question 4.
Calculate the equivalent weight of the following acid, (a) H2C2O4.H2O (b) H2O2O4

Question 5.
Define equivalent weight of base with an example.
NaOH + HCl → NaCl + H2O
1 equivalent of HC1 = 36.45, here 1 equivalent of HCl reacts with NaOH.

Question 6.
Write the acidity and calculate the equivalent weight of the following base, a) Ca(OH)2 b) Fe(OH)3

Question 7.
Write the mathematical form for mole fraction.
Mathematically;
If n1 is the number of moles of solvent; If n2 is the number of moles of solute

Question 8.
What is ppm (parts per million)?
It is the mass of solute present in 1 million(104) of solution.

10-3 = 1 mg of solute /1 dm3 of solution.
∴ 1 ppm = 1 mg of solute /1 dm3 of solution.
n ppm = n mg solute / 1 dm3 of solution.

Question 9.
Define Normality.
Number of gram equivalent of the substance dissolved in 1 dm3 of solution.

Question 10.
Calculate molarity and normality of the following solution, a) HCl b) H2C2O4

Question 11.
Write a note on the internal indicator.
The internal indicator is the substance which is added to the titration flask at the beginning. Example: Methyl orange, starch, phenolphthalein, diphenylamine.
Internal indicators are further divided according to the type of reaction taking place during titration.

(a) Acid-base indicator: The choice of indicator for acid-base titration depends on the nature of the acid and base used in that reaction.
Acid-Base Indicator
Strong acid Strong base Methyl orange or phenolphthalein

(b) Redox indicator: Example: Diphenylamine. When titrating KMnO4 with oxalic acid.

Question 12.
What are external indicators? Give an example.
The external indicator is added in between the titration process by taking a drop of titrated
mixture.
Example: Potassium ferrocyanide is used when ferrous ammonium sulphate is titrated against K2Cr2O7.

Question 13.
What is the self indicator? Give an Example.
When in titration one of the solutions acts as a self indicator. Example: KMnO4.

Question 14.
What is gram molecular value? Explain.
The volume occupied by one gram mole of a substance in the gaseous or is the vapour state at STP is called the gram molecular volume. The gram molecular volume of all the substances is found to be 22.4 dm 3 or 22400 cm 3. The mass in gram of 22.4 dm 3 of a gas or vapour at STP is equal to its gram molecular weight.

1 g molecule of chlorine or 1 gram molecule of carbon-di-oxide or 1 gram of any other substance in the gaseous state occupies a volume of 22400 cm3 (22.4 dm3 ).

1 g molecule weight of hydrogen = 2.016 g, 2.016 g of hydrogen occupies 22.4 dm3 of volume at STP. But 1 g equivalent weight of hydrogen is 1.008 g occupies 11.2 dm3 of volume at STP or mass of 11200 cm 3 of hydrogen = its equivalent weight.

Question 15.
Define the following with examples, (i) Allotropy and allotropes, (ii) polymorphism, (iii) isomorphism.
(i) The existence of an element in two or more chemically similar but physically different forms is called allotropy and the different forms are called allotropes, e.g., diamond, graphite, wood charcoal, lamp black etc., are allotropes of carbon.
(ii) The existence of a compound in different crystalline forms is called polymorphism and the different forms are called polymorphs, e.g., ZnS has two polymorphs called zinc blende and wurtzite.
(iii) The existence of different compounds with similar chemical composition in the same crystalline form is called [M2SO4M2 (SO4)3 24H2O] are isomorphs.

Question 16.
Discuss smaller the quantity to be measured more precisely should be the instrument.
Quite often the uncertainty in measurement is expressed in terms of percentage by putting ± sign before it, e.g., 250 ± 1%, etc. If the same instrument is used for measuring different quantities, then smaller the quantity to equal to ± 1 mg, then if we weigh 100 g on it, the result can be reported as 100 ± 0.001%. If the same balance is used to weigh 10 g, the result reported will be 10 ± 0.01%, and if 1 g is weighed, the result reported will be 10 ± 0.1%. Hence, the smaller the quantity to be measured, the more precise should be the instrument.

Question 17.
Compute the mass of one molecule and the molecular mass of C6H6  (benzene). (At. mass of C = 12, H = 1 u).
Molecular weight of C6H6 = 6 × 12 + 6 × 1 = 78 g,
Mass of 1 molecule = $$\frac{78}{6.002 \times 10^{23}}$$ = 12.94 × 10-23 g = 1.294 × 10-22 g

Question 18.
An organometallic compound on analysis was found to contain, C = 64.4%, H = 5.5% and Fe = 29.9%. Determine its empirical formula. (At. mass of Fe = 56 u).

Question 19.
4g of copper chloride on analysis was founded to contain 1.890 g of copper (Cu) and 2.110 g of chlorine (Cl). What is the empirical formula of copper chloride? [At. mass of Cu = 63.5 u, Cl = 35.5 u].

Question 20.
Calculate the number of grams of oxygen in 0.10 mol of Na2CO3.10H2O.
1 mole of Na2CO3.10H2O contains 13 moles of oxygen atoms.
0.1 mole Na2CO3.10H2O contains 13 × 0.1 = 1.3 moles of oxygen atoms.
Mass of oxygen atoms = 1.3 × 16 = 30.8 g.

Question 21.
How many grams of Cl2 are required to completely react with 0.4 g of H2 to yield HC1? Also, calculate the number of HCl formed.
H2(g) + Cl2(g) → HCl(g) ,
2 × 1 = 2
2 × 35.5 = 71
2 g of H2 reacts with 71 g of Cl2 to give 73 g of HCl.
0.4 g of H2 reacts with $$\frac{71}{2}$$x 0.4 = 14.2 g of Cl2 to give $$\frac{73 \times 0.4}{2}$$ = 14.6 g of HCl.
14.2 g of Cl2 and 14.6 g of HCl.

Question 22.
Cone. HC1 is 38% HCl by mass. What is the molarity of this solution if d = 1.19g cm-3? What volume of conc. HCl is required to make 1.00 L of 0.10 M HCl?

Question 23.

Question 24.
Calculate the volume of O2 at STP liberated by heating 12.25 g of KClO3. (At. wt. of K = 39, Cl = 35.5, O = 16 u).

Question 25.
1 M solution of NaNO3 has a density of 1.25 g cm -3. Calculate its molality. (Mol. weight of NaNO3 = 85 g mol-1).
Mass of solution = Volume of solution × density of the solution
Mass of solution = 1000 cm3 × 1.25 g cm-3 = 1250 g
Mass of solute = 85 g; Mass of solvent = 1250 – 85 = 1165 g.

Question 26.
Explain how compounds differ from elements? Give two differences.
Elements consist of only one kind of atom.

Question 27.
Explain how mixture differs from pure substance? Give 2 differences.
Pure substances Mixtures

Question 28.
Classify each of the following as pure substance or mixture:

1. Ethyl alcohol
2. Oxygen
3. Blood
4. Carbon
5. Steel
6. Distilled water.

Pure substances:

1. Ethyl alcohol
2. Oxygen
3. Carbon
4. Distilled water

Mixtures:

1. Blood
2. Steel

Question 29.
Describe two factors that introduce uncertainty into measured figures.

1. Reliability of measuring instrument
2. Skill of the person making the measurement.

Question 30.
A 50 ml graduated cylinder has 1 ml graduation. What is the maximum number of significant figures of the volumes that can be reported from this graduated cylinder?
The maximum number of significant figures of the volumes that can be reported is 2.

Question 31.
What is the difference between the accuracy and precision of measurements?
Accuracy: It is related to the closeness of a signal measurement to its true value.
Precision: It refers to the closeness of the set of values obtained from identical measurements of a quantity.

Question 32.
Classify the following substances into elements, compounds, and mixture.
(1) Milk (2) Iodised table salt (3) Diamond (4) Steel (5) Dry Ice (6) Glucose (7) Petrol (8) Steam (9) Cloud (10) Aerated drinks
Elements (3), Compounds 5,6, 8,9, Mixture 1, 2,4, 7,10

### 1st PUC Chemistry Some Basic Concepts of Chemistry Five Marks Questions and Answers

Question 1.
Calculate the moles of NaOH required to neutralize the solution produced by dissolving 1.1 g P4O6 in water. Use the following reactions:
P4O6 + 6H2O → 4H3PO3;
2NaOH + H3PO3 → Na2HPO3 + 2H2O
(At. mass/g mol-1 P = 31, O = 16).
Molecular weight of P4O6 =4 × 31 + 6 × 16 =124 + 96 = 220 gmol-1

$$\frac { 1 }{ 200 }$$ moles of PiOfi produce 4 × $$\frac { 1 }{ 200 }$$moles of H3PO3
1 mole of H3PO3 requires 2 moles of NaOH
$$\frac { 1 }{ 50 }$$moles of H3PO3 requires 2 × $$\frac { 1 }{ 50 }$$ = $$\frac { 1 }{ 25 }$$ = 0.04 moles of NaOH.

Question 2.
(a) A sample of NaOH weighing 0.38 g is dissolved in water and the solution is made to 50.0 cm3 in a volumetric flask. What is the molarity of the solution?
(b) State and explain law of multiple proportion.

(b) Law of multiple proportion states whenever two elements react to form two or more compounds, the ratio between different weights of one of the elements which combine with fixed weight of another is always simple, e.g., nitrogen reacts with oxygen to form NO and NO2. In NO, 14 g of N reacts with 16 g of O. In NO2,14 g of N reacts with 32 g of O. The ratio between weights of oxygen which combine with fixed weight of nitrogen is 16 : 32. i.e., 1:2.

Question 3.
Calculate the amount of water (g) produced by the combustion of 16 g of methane.
CH4(g) + 2O2(g) >CO2(g) + 2H2O(1)
1 mole = 16 g of CH4 gives 2 moles of H2O i.e., 2 × 18 = 36 g of H2O.

Question 4.
How many moles of methane are required to produce 22 g of CO,2(g) after combustion?
44 g of CO2 will be produced by 1 mole of CH4.
22 g of CO2 will be formed by $$\frac{1}{44}$$ × 22 = 0.5 mole of CH4.

Question 5.
(a) How many significant figures are there in 1.00 × 106?
(b) One mole of sugar contains ……. oxygen atoms.
(c) Give an example of a molecule in which the empirical formula is CH2O and the ratio of molecular formula weight and empirical formula weight is 6.
(a) 3
(b) One mole of sugar C12H22O11 contains 11 × 6.023 × 1023 oxygen atoms.
= 66.253 × 10 23 = 6.6253 × 10 23 atoms of oxygen
(c) C6H12O6.

Question 6.
An organic monobasic acid was found to contain 39.5% carbon, 6.4% hydrogen and the rest oxygen. If the equivalent mass of the acid is 60, find out its molecular formula.
Percentage of oxygen = 100 – (39.5 + 6.4) = 100 – 45.9 – 54.1

∴Empirical formula = CH2O
Molecular mass = Eq. mass × basicity = 60 × 1 = 60
Empirical formula weight = CH20 = 12 + 2 + 16 = 30
Nows 30n = 60 ⇒ n = 2
∴Molecular formula = (Empirical formula) × n = (CH2O)2 = C2H4O2

Question 7.
Calculate the mass of 95% pure MnO2 to produce 35.5 g of Cl2 as per the following reaction.
MnO2 + 4HCl → MnCl2 + Cl2 + H2O.
(At. mass of Mn = 55)

71 g of Cl2 is produced by 87 g of MnO2
35.5 g of Cl2 is produced by $$\frac { 87 }{ 71 }$$ × 35.5 = 43.49 g of MnO2
Since MnO2 is 95% pure i.e., 95 g of MnO2 is present in 100 g of MnO2 sample.
43.49 of pure MnO2 will be present in $$\frac{43.49 \times 100}{95}$$= 45.8f of MnO2 sample

Question 8.
500 ml of Na2CO3 solution contains 2.65 g of Na2CO3 (mol. mass of Na2CO3 = 106). If 10 ml of this solution is diluted to 1 L, what is the concentration of the resultant solution?

For dilution M1V1 = M2V2
⇒ 0.05 × 10 = 1000 × M2
M2 = 0.0005 M (Molarity of the resultant solution).

Question 9.
2 L of a solution is prepared by dissolving 0.02 mole of NaBr and 0.02 mole of Na2SO4. 1 L of mixture X + excess of AgNO3 → Y (pale yellow ppt.).
1 L of mixture X + excess of BaCl2 → Z (white ppt.).
What is the ratio of the number of moles of Y and Z?
AgNO3 reacts with Br ions to form AgBr as pale yellow ppt. as per the following reaction

2 L of the solution will contain 0.02 mole of NaBr or 0.02 mole of Br.
⇒ 1 L of the solution will contain 0.01 mole of Br
∴ as per equation (i) number of moles of AgBr i.e., Y formed is 0.01 mole
BaCl2 reacts with SO42- ions present in the solution as per the following reaction

2L of solution contain 0.02 mole of Na2SO4 or 0.02 mole of SO42-.
⇒ 1 L of solution will contain 0.01 mole of SO42-.
∴ as per equation (ii) number of moles of BaSO4 i.e., Z formed is 0.01 mole
Ratio of Y and Z = 0.01: 0.01 = 1:1

Question 10.
How many grams of copper will get replaced in 2 L of CuSO4 solution having molarity 1.50 M, if it is made to react with 54 g of aluminium? (At. mass of Cu = 63.5 and A1 = 27.0)
The reaction with CuSO4 solution and aluminium may be represented as
3CUSO4 + 2Al → Al2(SO4)2 +3CU
(3 × 159.5)g = 478.5g (2 × 27)g = 54g
Calculation of wt. of CuSO4 in 2 L of CuSO4 solution having 1.50 molarity

478.5 g of CuSO4 reacts with 54 g of aluminium
Also, 159.5 g of CuSO4 has 63.5 g of Cu
1 g of CuSO4 has $$\frac{63.5}{159.5}$$ g of Cu
478.5 g of CuSO4 has $$\frac{63.5}{159.5}$$ × 478.5g of Cu = 190.5g of Cu

Question 11.
0.7 g of Na2CO3.xH2O dissolved to make 100 ml of solution. If 20 ml of this solution is required to completely neutralize 19.8 ml of 0.1 M HCl, then what is the value of x?
Calculation of molarity of sodium carbonate solution
n1M1V1 = n2M2V2 (n factor for sodium carbonate is 2)
2 × 20 × M1 =19.8 × 0.1
⇒ M1 = 0.0495

The value of x i.e., the number of water molecules of crystallization is 2.

Question 12.
Calculate the weight of CaO that can be obtained by heating 200 kg of limestone which is 95% pure. (At. mass of Ca = 40, C = 12 and O = 16).

100 kg of CaCO3 gives 56 kg of CaO.
Since lime stone is 95% pure; amount of CaCO3 present in 200 kg of impure sample = 190 kg CaCO3 100 kg of CaCO3 gives 56 kg of CaO
1 kg of CaCO3 gives $$\frac { 56 }{ 100 }$$kg of CaO
190 kg of CaCO3 give $$\frac { 56 }{ 100 }$$ × 190 = 106.4 kg of CaO.

Question 13.
20 g of sample containing Ba(OH)2 is dissolved in 10 ml of 0.5 M HC1 solution. The excess of HCl was then titrated against 0.2 M NaOH. The volume of NaOH used in the titration was 10 ml. Calculate the percentage of Ba(OH)2 in the sample. (Mol. wt. of Ba(OH)2 = 171)
Calculation of volume of HCl used in titration between NaOH and HCl.
VNaOH = 10 ml, MNaOH = 0.2M ,MHCl = 0.5M, VHCl = ?
MNaOH × VNaOH = MHCl × VHCl
VHCl = $$\frac{10 \times 0.2}{0.5}$$ = 4ml
This is the volume of HCl left unused when an excess of HC1 is added to Ba(OH)2 solution.
Total volume of HCl added = 10 ml.
Volume of HCl used to react with Ba(OH)2 = 10 – 4 = 6 ml
∴HCl reacts with Ba(OH)2 as per the equation
2HCl + Ba(OH)2 → BaCl2 + 2H2O
6 ml of 0.5 M HCl has reacted with Ba(OH)2 ⇒ Number of moles of HCl reacted,

Observing the molar ratio of HC1 and Ba(OH)2,
Moles of Ba(OH)2 reacted = $$\frac { 1 }{ 2 }$$ × moles of HCl reacted = $$\frac { 1 }{ 2 }$$ × O.003 = 0.0015 moles
Weight of Ba(OH)2 reacted = no. of moles × mol. wt. = 0.0015× 171 = 0.2565 g

### 1st PUC Chemistry Some Basic Concepts of Chemistry Numerical Problems and Answers

Question 1.
How many significant figures are there in each of the following numbers?
(i) 6.005, (ii) 6.002 × 1023, (iii) 8000, (iv) 0.0025, (v) π, (vi) the sum 18.5 + 0.4235, (vii) the product 14 × 6.345.
(i) Four because the zeros between the non-zero digits are significant figures.
(ii) Four because only the first term gives the significant figures and the exponential term is not considered.
(iii) Four. However, if expressed in scientific notation as 8 × 103, it will have only one significant figure, as 8.0 × 103, 8.00 × 103 or 8.000 x 103, it will have 2, 3, or 4 significant figures.
(iv) Two because the zeros on the left of the first non-zero digit are not significant.
(v) As π = $$\frac { 22 }{ 7 }$$ = 3.1428571…., hence it has infinite number of significant figures.
(vi) Three, because the reported sum will be only upto one decimal place i.e., 18.9.
(vii) Two, because the number with the least number of significant figures involved in the calculation (i.e., 14) has two significant figures.

Question 2.
Express the following to four significant figures: (i) 6.45372, (ii) 48.38250, (iii) 70000, (iv) 2.65986 × 103, (v) 0.004687.
(i) 6.454, (ii) 48.38, (iii) 7.000 × 104, (iv) 2.660 × 103, (v) 0.004687.

Question 3.
A sample of nickel weighs 6.5425 g and has a density of 8.8 g cm3 the volume? Report the answer to the correct decimal place.

The result should have two significant figures because the least precise term (8.8) has two significant figures.

Question 4.
Express the result of the following calculation to the appropriate number of $$\frac{3.24 \times 0.08666}{5.006}$$
$$\frac{3.24 \times 0.08666}{5.006}$$= 0.0560883 (Actual result)
As 3.24 has the least number of significant figures, viz., 3, the result should contain 3 significant figures only. Hence, the result will be reported as 0.0561 (after rounding off).

Question 5.
The mass of precious stones is pressed in terms of a carat. Given that 1 carat = 3.168 grains and 1 gram – 15.4 grains, calculate the total mass of the ring in grams and kilograms which contains 0.500 carat diamond and 7.00 gram gold.
The unit conversion factors to be used will be;

Question 6.
Two oxides of metal contain 27.6% and 30.0% of oxygen respectively. If the formula of the first oxide is M3O4, find that of the second.
In the first oxide, oxygen = 27.6, metal = 100 – 27.6 = 72.4 parts by mass.
As the formula of the oxide is M3O4, this means 72.4 parts by mass of metal = 3 atoms of metal and 4 atoms of oxygen = 27.6 parts by mass.
In the second oxide, oxygen = 30.0 parts by mass and metal = 100 – 30 = 70 parts by mass.
But 72.4 parts by mass of metal = 3 atoms of metal.
∴ 70 parts by mass of metal =$$\frac{3}{72.4}$$ × 70 atoms of metal = 2.90 atoms of metal.
Also, 27.6 parts by mass of oxygen = 4 atoms of oxygen.
∴ 4.35 atoms of oxygen.
Hence, ratio of M: O in the second oxide = 2.90 : 4.35 = 1:1.5 = 2:3.
∴ Formula of the metal oxide is M2O3.

Question 7.
What is the equivalent weight of KH(IO3)2 as an oxidant in presence of 4.0 (N) HCl when IC becomes the reduced form? (K = 39.0,1 = 127.0).
In KH(IO3)2, IO3 is present as IO3. Oxidation state of I will be

Question 8.
Calculate the percentage of the naturally occurring isotopes 35C1 and 37Cl that accounts for the atomic mass of chlorine taken as 35.45.
Suppose 35Cl present = x%. Then 37Cl present = (100 -x)%
∴ Average atomic mass = \frac{x \times 35+(100-x) \times 37}{100} = 35.45 (Given)
or 35x + 3700-37x = 3545 or 2x = 155 or x = 77.5%
This 35Cl = 77.5% and 37Cl = (100 – 77.5) = 22.5%.

Question 9.
Convert 22.4 L into cubic metres.

Question 10.
Calculate the molar mass of water if it contains 50% heavy water (D2O).
As water contains 50% D2O, this means that it contains $$\frac{1}{2} mole of H20 and \frac{1}{2}$$ mole
of D2O. Mass of $$\frac{1}{2}$$ mole of D2O = $$\frac{1}{2}$$ × (2 × 2 + 16) = 10 g.
Hence, molar mass of the given sample of water = 9 + 10 = 19 g mol-1.

Question 11.
60 cc of oxygen was added to 24 cc of carbon monoxide and the mixture ignited. Calculate the volume of oxygen used up and the volume of carbon dioxide formed.

∴ volume of oxygen used up is 12 cc. and the volume of carbon dioxide formed is 24 cc.

Question 12.
200 cm3 of carbon monoxide is mixed with 200 cm3 of oxygen at room temperature and ignited. Calculate the vol of CO2 formed on cooling to room temperature. What other gas if any may also be present.

Volume of carbon dioxide formed is 200 cm3 and the other gas present is 100 cm3 of oxygen.

Question 13.
Calculate the volume of oxygen required to burn completely a mixture of 22.4 dm3 of CH4 and 11.2 dm3 of H2 (all volumes measured at STP) [1 dm3 = 1 litre].

Total oxygen required = 44.8 dm3 +5.6 dm3 =50.4 dm3 Pits].
∴ the number of moles in 7 g of nitrogen is 0.25 moles.

Question 14.
Calculate the mass of 50 cc of CO at STP (C = 12, O = 16).
[1 mole = 1 g mol. wt. and occupies 22.4 lit. at STP]
g mol. wt. of carbon monoxide = 12 + 16 = 28 g
28 g of CO occupies 22400 cc at STP
How many grams of CO will occupy 50 cc at STP = $$\frac{28 \times 50}{22400}$$ = 0.0625 g

Question 15.
Calculate the volume at STP occupied by 6.023 × 1022 molecules of a gas X.
[1 mole of any substance contains 6.023 × 1023 number of molecules [Avogadro’s number]].
(a) 6.022 × 1023 molecules occupied 22.42 cm3 (Its) = $$\frac{22.4 \times 6.023 \times 10^{22}}{6.023 \times 10^{23}}$$ = 2.24 lits

Question 16.
Calculate the number of molecules in 1 kg of sodium chloride. [Na = 23, Cl = 35.5].
(a) 1 mol wt. of NaCl = 58.5 g contains 6.022 × 1023
(b) Contains molecules = $$\frac{1000}{58.5}$$ × 6.023 × 1023 = 102.96 × 1023 molecules.

Question 17.
Calculate the atomicity of a gas X if 1 g of X occupies 11,200 cc at STP [At. Wt. of X= 1].

Question 18.
0. 48 g of gas forms 100 cm3 of vapours at STP. Calculate the gram molecular wt. of the gas.
100 cm3 of the gas weighs 0.48 grams at STP
22,400 cm3 weight Mol. wt. at STP.
∴ $$\frac{22400}{100}$$ × 0.48 = 107.52

Question 19.
Calculate the weight of a substance X which in gaseous form occupies 10 litres at 27 °C and 700 mm pressure. The molecular Weight of X is 60.
Convert the volume to STP using the gas equation.
Initial conditions: P1 = 700 mm of Hg, V1 = 10 litres, T1 = 27 + 273 = 300 K
Final conditions (at STP): P0 = 760 mm of Hg, V0 = X litres, T0 = 273 K

Question 20.
Calculate the normality of oxalic acid solution containing 0.895 g crystals in 250 cm3 of its solution.
Molecular formula of oxalic acid crystals = H2C2O4.2H2O = 126

Question 21.
25 cm3 of ferrous ammonium sulphate solution require 20 cm3 of 0.1 N potassium dichromate solution. Calculate the amount of ferrous ammonium sulphate crystals dissolved in 250 cm3 of the solution. (Given equivalent)

Question 22.
What should be the normality of a solution prepared by diluting 250 ml of 0.4 NH2SO4 with 1000 ml of water?
Total volume of the diluted solution = 250 + 1000 = 1250 ml

Question 23.
20 cm3 of a solution of oxalic acid requires 25 cm3 of 0.2 N potassium permanganate to react completely (a) Calculate the normality of oxalic acid solution, (b) What volume of this oxalic acid solution when made up to 250cm3 gives 0.2 N solution?

Normality of oxalic acid solution = 200.
The volume of oxalic acid solution required = 200 cm3

Question 24.
200 cm3 of a solution of a dibasic acid contains 1.512 g of the acid and the normality of the solution is 0.12. Calculate (i) the equivalent mass and (ii) the molecular mass of the acid.
(i) Mass of the acid in one dm3 of the solution = 1.512 × 5 = 7.56 g.

(ii) Molecular mass of the acid = eq. mass × basicity = 63 × 2 = 126.

Question 25.
25 cm3 a of a solution of sodium hydroxide required 29 cm3 of $$\frac { n }{ 10 }$$ solution of oxalic acid for neutralization. Find the normality of sodium hydroxide solution and its amount dissolved in 500 cm3 of the solution.

Substituting we have 28 × 0.1 = 25 × N2
∴N2 = $$\frac{29 \times 0.1}{25}$$ = 0.116
∴ Normality of sodium hydroxide solution = 0.116
We know that W = N × E for 100 ml
i.e., Mass per dm3 = Normality × eq. nlass = 0.116 × 40 g
∴ Mass of sodium hydroxide in 500 cm 3 of the solution = $$\frac{0.116 \times 40}{2}$$ = 2.32 g.

Question 26.
Calculate the volume of concentrated nitric acid of normality 14 required to prepare 1 dm3 of \frac{N}{10} nitric acid.

Question 27.
Calculate the mass of hydrochloric acid in 200 cm3 of 0.2 N solution of it. What volume of this acid solution will react exactly with 25 cm3 of 0.14 N solution of sodium hydroxide?
Normality of hydrochloric acid = 0.2; Eq. mass of hydrochloric acid = 36.5
Mass of hydrochloric acid in one dm3 of the solution is given by W = N × E = 0.2 × 36.5 g

Question 28.
0. 99 g of acid was dissolved in water and the solution made up to 200 cm3, 20 cm3 of this solution required 15 cm3 of 0.105 N sodium hydroxide solution for complete neutralization. Find the equivalent mass of the acid.

Substituting we have 20 × N1 =15 × 0.105
∴ N1 = $$\frac{15 \times 0.105}{20}$$ = 0.07875
∴ Normality of acid solution = 0.07875
Mass of the acid in one dm 3 of the solution – 0.99 × 5 g = 4.95 g
Eq. mass of the acid is given by E = $$\frac{W}{N}=\frac{4.95}{0.07875}$$ = 62.85

Question 29.
2.15 g of impure oxalic acid crystals were dissolved in water and the solution made up to 250 cm3 of this solution required 20 cm3 of 0.16 N sodium hydroxide solution for complete neutalization. What is the percentage purity of oxalic acid crystals?

Substituting we have 25 × N1 =20 × 0.16
∴ N1 = $$\frac{20 \times 0.16}{25}$$ = 0.128
∴ Normality of oxalic acid solution = 0.128
Eq. mass of oxalic acid crystals H2C2O4.2H2O = 63.
Mass of pure oxalic acid in one dm 3 is given by W = N × E = 0.128 × 63
Mass of pure oxalic in 250 cm3 = $$\frac{0.128 \times 63}{4}$$ = 2.016 g
Mass of impure oxalic acid crystals dissolved in 250 cm3 = 2.15 g
% percentage purity of the given oxalic acid crystals = $$\frac{2.016 \times 100}{2.15}$$ = 93.76

Question 30.
A solution of sodium hydroxide contains 4.8 g of the substance per dm3, (a) What volume of the solution is required to prepare 500 cm3 of decinormal solution. (b) What volume of water has to be added to one dm3 of the solution in order to make it exactly decinormal?

Question 31.
Find the Molarity of hydrochloric acid containing 31.5% of hydrochloric acid. Its specific gravity is 1.16.
Mass of 1000 cm3 of the acid = 1000 × 1.16 = 1160 g
Mass of hydrochloric acid in 100 g of the solution = 3.15 g

Question 32.
100 cm3 of concentrated hydrochloric acid which is 11 M is diluted to 275 cm3. If 2.5 g of limestone are required to neutralize 10 cm3 of this diluted – acid, what is the percentage of calcium carbonate in the sample of limestone?

Substituting we get, 100 × 11 = 275 × M2

∴ Normality of the diluted acid = 4
1000 cm3 of the acid solution contain 4 gram
∴ 10 cm3 of the diluted acid contain = $$\frac{4 \times 10}{1000}$$ i.e., 0.4 gram
2.5 g of limestone contain 0.04 gram of calcium carbonate i.e., they contain 0.04 × 100 g = 4 g of calcium carbonate.
(Molar mass of calcium carbonate =100)
∴ percentage of calcium carbonate in the sample of limestone = $$\frac{4 \times 100}{2.5}$$ = 16.0

Question 33.
25.0 cm3 of acid required exactly 20.5 cm3 of deci molar base for complete neutralization. What is the normality of the acid?

Question 34.
Exactly 20.0 cm3 of nitric acid neutralized 28.4 cm3 of 0.25 M NaOH. What is the molarity of the acid?

Question 35.
18.5 cm3 of oxalic acid was completely neutralized by 20.0 cm3, 0.125 N base. Calculate the (a) normality (b) molarity and (c) mass of oxalic acid crystals in 1 dm3 of solution.

Mass of oxalic acid crystals in 1 dm3 of solution = normality × equivalent mass
= 0.1351 N × 63.0 (equivalent mass of oxalic acid crystals = 63.0)
= 8.513 g.

## Karnataka 1st PUC Physics Question Bank Chapter 5 Laws of Motion

### 1st PUC Physics Laws of Motion TextBook Questions and Answers

Question 1.
Give the magnitude and direction of the net force acting on

1. a drop of rain falling down at a constant speed.
2. a cork of mass 10 g floating on water
3. a kite skillfully held stationary in the sky.
4. a car moving with a constant velocity of 30 km/h on a rough road.
5. a high-speed electron in space far from ail material objects, and free of electric and magnetic fields.

1. In accordance with the first law of motion, there is no net force on the drop since it is moving with constant speed.
2. The weight of the cork is balanced by upthrust which is equal to the weight of water displaced. Hence no net force on the cork.
3. Since the kite is in the state of rest net force on it is zero.
4. From the first law of motion, since the velocity of the car is a constant net force on it is zero.
5. Since the electron is in free space no gravitational or electric or magnetic force is acting on it. Hence net force on it is zero.

Question 2.
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,

1. during its upward motion.
2. during its downward motion.
3. at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.

1. Force F = mg = 0.05 x 9.8 = 0.49 N, vertically downwards.
2. In this case also F = mg = 0.05 x 9.8 = 0.49 N, vertically downwards.
3. Again F = mg = 0.05 x 9.8 = 0.49 N, vertically downwards.
No, the horizontal velocity component remains the same.

Question 3.
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,

1. Just after it is dropped from the window of a stationary train.
2. Just after It is dropped from the window of a train running at a constant velocity of 36 km/h
3. Just after it is dropped from the window of a train accelerating with 1 ms-2
4. Lying on the floor of a train which is accelerating with 1 m s-2, the stone being at rest relative to the train. Neglect air resistance throughout.

1. Gravitational force is acting on the stone in downward direction F = mg = 0.1 × 10 = 1 N
2. Once the stone is dropped from the train, the only force acting on it is gravitation force =1 N.
3. Since there is no contact between train and stone the force acting on it is again gravitational force.
4. Since the stone is lying on the floor of train its acceleration in the same as that of the train. Hence the force excreted by train on the stone is F = ma = 0.1 × 1 = 0. 1 N in the direction of the train. The weight is balanced by the normal reaction of the floor of the train.

Question 4.
One end of a string of length L is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is:

1. T.
2. T – $$\frac{m v^{2}}{L}$$
3. T + $$\frac{m v^{2}}{L}$$
4. 0

T is the tension in the string. [Choose the correct alternative].
Net force = tension (T) in the string, which provides the necessary centripetal force to move the particle in the circle.

Question 5.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms-1. How long does the body take to stop?
Given F = – 50 N (retarding force)
m = 20 kg
u = 15 m/s. V = 0 m/s
t = ?
F = ma ⇒ a = $$\frac{\mathrm{F}}{\mathrm{m}}$$ = $$\frac{-50}{20}$$ = – 2.5m/s2
but we know that
V = u + at
0 = 15 + (- 2.5) t
⇒ t = 6 s.

Question 6.
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s-1 to 3.5 m s-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
Given m = 3 kg
u = 2 m/s
v = 3.5 m/s
t = 25 s.
F = ma
but we know that a = $$\frac{v-u}{t}$$
∴ F = m $$\left(\frac{v-u}{t}\right)$$ = 3 $$\left(\frac{3.5-2}{25}\right)$$
= 0.18 N
since direction acceleration ‘a’ is positive force is acting in the direction of motion.

Question 7.
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.

Given Fa = 8 N
Fb = 6 N
m = 5 kg
The result ant force F is given by,
F = $$\sqrt{\mathrm{Fa}^{2}+\mathrm{Fb}^{2}}$$
= $$\sqrt{64+36}$$ = 10 N
we know from the figure that
tan θ = $$\frac{F_{b}}{F_{a}}=\frac{6}{8}$$ = 0.75
⇒ θ = tan-1(0.75) = 37° with 8 N force

Question 8.
The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.
Given, u = 36 km/h
36 × $$\frac{1000}{3600}$$ m/s
=10 m/s
v = 0
t = 4 s
m = 400 + 65 = 465 kg
a = $$\frac{v-u}{t}$$ = $$\frac{-10}{4}$$ = – 2.5 m/s²
F = ma = 465 (- 2. 5)
= – 11625 N (retarding force)

Question 9.
A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms-2. Calculate the initial thrust (force) of the blast.
Initial thrust = upthrust needed to overcome gravitational pull of earth + upthrust needed to impart acceleration a
= mg + ma = m (a + g)
= 20,000 (5 + 9.8)
= 20,000 x 14.8
= 2.96 x 105 N.

Question 10.
A body of mass 0.40 kg moving initially with a constant speed of 10 m s-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t= 0, the position of the body at that time to be x= 0, and predict its position at t = – 5 s, 25 s, 100 s.
Given mass m = 0.4 kg
Retarding force F = – 8 N
∴ acceleration a = $$\frac{F}{m}$$ = $$\frac{-8}{0.4}$$ = – 20 m/s²
at t = – 5 s
a = 0 for t < 0
∴ s = u + $$\frac{1}{2}$$ at²
= 10 (-5) + $$\frac{1}{2}$$ (0)(-5)²
at t = 25 s
S = ut+$$\frac{1}{2}$$ at²
= 10 (25) + $$\frac{1}{2}$$ (- 20) (25)²
= – 6000 m
at t = 100 s
since there is a retarding force for 30 s
S1 = ut + $$\frac{1}{2}$$ at²
= 10 (30) + $$\frac{1}{2}$$ (-20) (30)²
= – 8700 m .
after 30 s it move with a constant velocity.
V = u + at
= 10 – 20 (30)
= – 590 m/s
for rest of 70 s.
S2 = – 590 (70) + $$\frac{1}{2}$$ (0) (70)² = -41300 m
∴ Total distance = S1 + S2 = 50000 m.

Question 11.
A truck starts from rest and accelerates uniformly at 2.0 m s-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the

1. velocity, and
2. acceleration of the stone at t = 11s? (Neglect air resistance.)

We have, Vt = u + at
i.e. Vt = 0 + 2 (10) = 20 ms-1
During the nextone second stone is under the effect of gravity.

Vg = u + gt
= 0 + 9.8 (1) = 9.8 m/s
∴ Net velocity of stone at t = 11 s
v = $$\sqrt{V_{t}^{2}+V_{g}^{2}}$$ = 22.27 m/s.
tan θ = $$\frac{V_{g}}{V_{t}}=\frac{9.8}{20}$$
⇒ θ = 26.1° with horizontal.

(b) The moment stone is dropped from truck only gravitational force is acting on it.
∴ acceleration = g = 9.8 m/s².

Question 12.
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s-1. What is the trajectory of the bob if the string is cut when the bob is

1. at one of its extreme positions
2. at its mean position.

1. As the bob has no velocity at the extreme position,(i.e. it is at rest), it will just fall vertically downwards when the string is cut at the extreme position.
2. At the mean position, if we cut the string the bob will follow a parabolic path under the effect of gravity.

Question 13.
A man of mass 70 kg stands on a weighing scale in a lift which is moving

1. upwards with a uniform speed of 10 m s-1
2. downwards with a uniform acceleration of 5 m s-2
3. upwards with a uniform acceleration of 5 ms-2. What would be the readings on the scale in each case?
4. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

The weighing scale will show the reaction of the floor of the lift on the man in each case. This is equal to the apparent weight of the person.

1. R= mg = 70 x 9-8 = 686 N
2. R’ = m(g – a) = 70(9-8 – 5) = 70 x 4.8 = 336 N
3. R”= m(g + a) = 70 (9-8 + 5) = 70 x 14.8 = 1036 N
4. R”‘= m(g – g) = 0. This is the condition of weightlessness.

Question 14.
Figure shows the position-time graph of a particle of mass 4 kg. What is the

1. force on the particle for t < 0, t > 4 s, 0 < t < 4 s?
2. impulse at t = 0 and t = 4 s? (Consider onedimensional motion only)

1. When t < 0. As this part of the graph is horizontal, so it can be concluded that the distance covered by the particle is zero i.e. particle is at rest, and hence force on the particle is zero.
When 0 < t < 4s. As OA has a constant slope, hence in this interval, particle moves with constant velocity
($$\frac { 3 }{ 4 }$$= 0.75ms1 ). Hence, net force on the particle is zero.
When t > 4s. As this portion shows the particle always remains at a distance of 3 m from the origin i.e., the particle is at rest. Hence, the net force on the particle is zero.

2. Impulse at t = 0. Here u = 0, -u = 0.75 ms-1, M = 4 kg
∴ Impulse = total change in momentum = Mu – Mu
= M(υ- u) = 4(0.75 – 0) = 3 kg ms-1
Impulse at t = 4s. Here u = 0.75 ms-1, v = 0
.’.   Impulse = M(υ – u) = 4 (0 – 0.75) = – 3 kg ms-1.

Question 15.
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string, a horizontal force F = 600 N is applied to

1. A
2. B along the direction of the string. What is the tension in the string in each case?

1. Acceleration of the whole system a = $$\frac{F}{M_{1}+M_{2}}=\frac{600}{10+20}$$ = 20ms-2
The net force, acting on A
= 600 – T = m1 (a)
∴ 600 – T = 10 × 20
⇒ T = 400 N.

2. similar to the above case.
The net force acting on B
= 600 – T = M2 a
∴ 600 – T = 20 × 20
⇒ T = 200 N.

Question 16.
Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses and the tension in the string when the masses are released.

Let ‘a’ be the acceleration of the masses. Then
for block m1, T – m1g = m1a → (1)
for block m2, m2g – T = m2 a → (2)
(1) + (2) (m2 – m1) g = (m1 + m2) a
⇒ a = $$\frac{12-8}{12+8}$$ g = 2m/s
substituting in (1)
T – 8 × 10 = 8 × 2
⇒ T = 96 N

Question 17.
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
Let m1 & m2 be the masses of smaller nuclei and let $$\vec{v}_{1}$$ & $$\vec{v}_{2}$$ be their velocities.
According to the law of conservation of momentum.
Initial momentum = final momentum
0 = m1$$\vec{v}_{1}$$ + $$\vec{v}_{2}$$
Or $$\vec{v}_{2}$$ = – $$-\frac{m_{1}}{m_{2}} \vec{v}_{1}$$
Hence v1 & v2 are in opposite direction.

Question 18.
Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s-1 collide and rebound at the same speed. What is the impulse imparted to each bail due to the other?
Impulse imparted to one ball by another ball = change in momentum.
Initial momentum of ball A, pi = mu1= 0.05 x 6 = 0.3 kg m s_1
Final momentum of ball A, pf = m(-u1) = 0.05 x (-6) = – 0.3 kg m s_1
∴  Impulse imparted to ball A by ball B
– pf – pi = – 0.3 – 0.3 = – 0.6 kg m s_1
Initial momentum of ball B, pi =m (-υ1) = -0 05 x 6 = – 0.3 kg m s_1
Final momentum of ball B, p= mυ2 = 0-05 x 6 = 0.3 kg m s_1
Impulse imparted to ball B by ball A
= pf- pi = 0.3 – (-0.3) = 0.6 kg m s-1

Question 19.
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s-1, what is the recoil speed of the gun?
Given
m1 = 0.02 kg, m2 = 100 kg
v1 = 80 m/s v2 = ?
According law of conservation of momentum
m1u1 + m2u2 = m1v1 + m2v2
0.02 (0) + 100 (0) = 0.02 × 80 + 100 × v2
v2 = – 1.6 × 10-2 m/s

Question 20.
A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)

given
m = 0.15 kg
u = 54 kmph
= 54 × $$\frac{1000}{3600}$$
= 15m/s
Along x -axis
Initial velocity = – u cos θ
= 15 cos (22.5°)
Final velocity = u cos θ
= 15 cos (22.5°)
∴ Impulse = change in momentum
= 0.15 [15 cos (22.5) – (-15 cos (22.5)°)]
= 4.16 kgm/s.
Along y-axis
Initial velocity = Final velocity = – u sin θ
∴ No impulse along the y-axis.

Question 21.
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev/min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Given m = 1.5m
r = 1.5 m
w = 40 rpm = $$40 \frac{\times 2 \pi}{60}$$ rad/s
= $$\frac{4}{3}$$π rad/s
Now Tension T = mrw²
= 0.25 × 1.5 × $$\left(\frac{4}{3} \pi\right)^{2}$$
= 0.58 N
Tmax = 200 N
Tmax = $$\frac{\mathrm{m} \mathrm{v}_{(\mathrm{max})}^{2}}{\mathrm{r}}$$
⇒ Vmax = $$\sqrt{\frac{200 \times 1.5}{0.25}}$$ = 34.6 m/s.

Question 22.
If, In Exercise 5.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks:

1. the stone moves radially outwards,
2. the stone flies off tangentially from the instant the string breaks,
3. the stoneflies off at an angle with the tangent whose magnitude depends on the speed of the particle?

The answer is 2. When a particle moves in a circular path, at each point the velocity is directed along the tangent of the circular path. Hence when a string, breaks, it moves along the tangent in accordance with Newton’s 1st law of motion.

Question 23.
Explain why

1. a horse cannot pull a cart and run in empty space.
2. passengers are thrown forward from their seats when a speeding bus stops suddenly.
3. it is easier to pull a lawnmower than to push it.
4. a cricketer moves his hands backward while holding a catch.

1. Noninertial, because the giant wheel has accelerated motion.
2. Inertial frame, because the speed is uniform in a straight line.
3. Non-inertial, because the aeroplane accelerates while taking off.
4. Non-inertial, because the cyclist has now accelerated motion.
5. Non-inertial, because the train is retarding.

### 1st PUC Physics Laws of Motion Additional Exercises Questions and Answers

Question 24.
The figure shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse?

The graph could be representing a ball rebounding between two walls separated by 2 cm with a constant velocity in free space. After receiving the impulse ball changes its direction. Hence time between two impulses is 2 seconds.
velocity = $$\frac{\text { displacement }}{\text { time }}$$ = $$\frac{2 \times 10^{-2}}{2}$$ = 0.01m/s
Initial momentum,
mu = 0.04 × 10-2kgm/s
Final momentum,
mv = – 0.04 × 10-2kgm/s
∴ Change in momentum = 0.08 × 10-2kgm/s

Question 25.
Figure 5.18 shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 m s-2. What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt?
(Mass of the man = 65 kg.)

Given acceleration of conveyor belt a = 1 m s-2
µs = 0.2
mass of man m = 65 kg
Then man experiences a pseudo force Fs = ma as he is in an accelerating frame as shown in the figure. Hence to maintain his equilibrium he exerts a force F = – Fs = ma = 65 × 1 = 65 N in direction of motion of belt.
∴ Net force acting on man = 65 N The man continue to be stationary with respect to belt if force of friction equal to force acting on man i.e.

µs N = mamax
µs . m .g = mamax
a(max) = µs × g
= 0.2 × 10
= 2m s-2

Question 26.
A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are : [Choose the correct alternative]

T1 and v1 denote the tension and speed at the lowest point. T2 and v2 denote corresponding values at the highest point.

The net force acting on stone at the lowest point directed vertically downward = mg – T1 & and the highest point = mg + T2. Hence option (a) is correct answer.

Question 27.
A helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms-2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the

1. force on the floor by the crew and passengers,
2. action of the rotor of the helicopter on the surrounding air,
3. force on the helicopter due to the surrounding air.

mass of helicopter = mh = 1000 kg mass of crew = mc = 300 kg
vertical acceleration, a =15 m/s².
1. force on the floor by crew & passenger = apparent weight of crew & passenger
= mc (g + a)
= 300(10 + 15)
= 7500 N.

2. The action of motor of helicopter on surrounding air is vertical downwards. The helicopter rises on account of reaction to this force
= (mn + mc) (g + a)
= (1000 + 300) (10 + 15)
= 32500 N.

3. force on helicopter due to the surrounding air is nothing but the reaction to the action of rolor = 32500 N in vertically upward direction.

Question 28.
A stream of water flowing horizontally with a speed of 15 m s-1 gushes out of a tube of cross-sectional area 10-2 m² and hits a vertical wall nearby. What is the force exerted on the wall by the Impact of water, assuming It does not rebound?
The volume of water hitting the wall per second
= (Area × Velocity) of a stream of water
= 10-2 × 15
= 0.15 m3s-1
density of water = 1000 kg/m3
∴ mass of water hitting the wall per second
= 0.15 × 1000= 150 kg/s
Initial momentum of water hitting the wall per second
= 150 × 15
= 250 kg m/s² or 2250 N
Final momentum per second = 0
∴ Force exerted on the wall: change in momentum per second
= 2250 N.

Question 29.
Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of

1. the force on the 7th coin (counted from the bottom) due to all the coins on its top,
2. the force on the 7th coin by the 8th coin,
3. the reaction of the 6th coin on the 7th coin.

1. There are 3 coins above the 7th coin
Hence force = (3m) g
= 3mg N

2. The 8th coin has two coins above it. Hence force exerted by 8th coin on 7th is, it’s weight plus the weight of two coins
= mg + 2 mg
= 3 mg N.

3. The 6th coin is under the weight of 4 coins above it
Reaction R = – F = – 4 mg N.

Question 30.
An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?
υ = 720 km/hr = 720 × $$\frac{1000}{3600}$$
= 200 m/s
θ = 15°

From the relation
tan θ = $$\frac{v^{2}}{r g}$$
⇒ r = $$\frac{v^{2}}{\tan \theta \times g}$$ = $$\frac{200 \times 200}{\tan 15^{\circ} \times 10}$$
= $$\frac{200 \times 200}{0.2679 \times 10}$$
= 14931 m

Question 31.
A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 106 kg. What provides the centripetal force required for this purpose The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?
velocity υ = 54 km/h = 54 × $$\frac{1000}{3600}$$= 15 m/s
mass m = 106 kg
The centripetal force F = $$\frac{m v^{2}}{r}$$ is provided by the lateral frictional force between rails and wheels of train.
The angle of banking required to prevent the wearing out of rail
tan θ = $$\frac{v^{2}}{r g}$$ = $$\frac{15 \times 15}{30 \times 10}$$ = 0.75
θ = tan-1 (0.75) ≈ 37°.

Normal Force formula Calculator will determine the force exerted by a surface to prevent the object from falling.

Question 32.
A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding?

In the first case man applies an upward force of 25 kg weight. Hence the action on the floor by man is
= 50 kg weight + 25 kg weight = 75 kg weight
= 75 × 10
= 750 N.

In the second case man applies a downward force of 25 kg weight. Hence the action on the floor by man is
= 50 kg weight – 25 kg weight = 25 × 10
= 250 N.
(other 500 N is applied on the ceiling) Hence man should adopt the second case.

Question 33.
A monkey of mass 40 kg climbs on a rope (Fig.) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey

1. climbs up with an acceleration of 6 m s-2
2. climbs down with an acceleration of 4 m s-2
3. climbs up with a uniform speed of 5 m s-1
4. falls down the rope nearly freely under gravity? (Ignore the mass of the rope).

1. When monkey climbs up with an acceleration ‘a’ then
T – mg = ma
Or T = m (g + a)
= 40 (10 + 6)
= 640 N
which exceeds the maximum tension which rope can withstand (600 N), hence rope breaks.

2. when monkey is climbing down with an acceleration a
mg – T = ma
or T = m (g – a)
= 40 (10-4)
= 240 N
The rope will not break.

3. when the monkey climbs up with uniform speed then
T = mg
= 40 × 10
= 400 N
The rope will not break.

4. when the monkey is falling freely, it would be a state of weightlessness. So, there won’t be any tension in the rope hence it will not break.

Question 34.
Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wail (Fig). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are

1. the reaction of the partition
2. the action-reaction forces between A and B?
3. What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? ignore the difference between µs and µk.

1. As the blocks are at rest against the rigid walls, reaction of the partition = – (force applied on A)
= 200 N towards left.

2. The action-reaction force between A & B are 200 N each.

3. when the wall is removed, the pushing force gives acceleration to the system. On taking the coefficient of friction into account,
200 – µ (m1 + m2) g = (m1 + m2) a
a = $$\frac{200-0.15(5+10) \times 10}{(5+10)}$$
= 11.8 ms-2
Let the force exerted by A on B be FBA. On considering the equilibrium of the only block
A, 200 – fk1 = m1 a + FBA
FBA = 200 – µ m1 g – m1 a
= 200 – 7.5 – 59
= 133.5 N towards left.

Question 35.
A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 m s-2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as
viewed by

1. a stationary observer on the ground,
2. an observer moving with the trolley.

1. Force experienced by block
F = ma = 15 × 0.5 = 7.5 N
Force of friction, Ff = µ mg = 0.18 × 15 × 10 = 27 N
Since the force experienced block is less than frictional force it wilt remain stationary with respect to trolley. For an observer on the ground block appears to move with same acceleration as trolley,

2. For an observer moving with trolley the block appears to he stationary as there is no relative motion between him and trolley and the block.

Question 36.
The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s-2. At what distance from the starting point does the box fail off the truck? (ignore the size of the box).

Force experienced by box F = ma = 40 × 2 = 80 N
frictional force Ffriction= µ mg = 0.15 × 40 × 10 = 60 N
∴ Net force on the box = F – Ffriction = 80 – 60 = 20 N .
∴ The backward acceleration experienced by box is given by,
a = $$\frac{\text { Net force }}{\text { mass }}$$ = $$\frac{20}{40}$$ = 0.5 m/s²
Let‘t’ be the time taken by box to move through 5m backwards
We have, S = ut + $$\frac{1}{2}$$ at²
∴ 5 = 0 × t + $$\frac{1}{2}$$ × 0.5 × t²
t = $$\sqrt{20} \approx$$ 4.47 s
The distance travelled by truck in t = 4.47s is
s = ut + $$\frac{1}{2}$$ at² (a = 2m /s2)
s = 0 × $$(\sqrt{20})$$ + $$\frac{1}{2}$$ × 2 $$(\sqrt{20})^{2}$$
s = 20 m
The box will off the truck after 20 m from starting point.

Question 37.
A disc revolves with a speed of $$33 \frac{1}{3}$$ rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?
If the coin is to revolve with the record then the force of friction must be enough to provide the necessary centripetal force.
i.e. mr ω² ≤ µs mg or r ≤ $$\frac{\mu_{\mathrm{s}} \mathrm{g}}{\omega^{2}}$$
Here, ω = $$33 \frac{1}{3}$$ rpm

$$\approx$$ 0.12 m
The coin placed within the radial distance of 0.12 m will revolve with the record. Hence coin at 4 cm will revolve with the record.

Question 38.
You may have seen in a circus a motorcyclist driving in vertical loops Inside a ‘deathwell’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m?
When the motor cyclist is at the highest point of death well, the normal readction R on the motor cycle by the ceiling of the chamber acts downwards. His weight mg also acts downwards. These two forces are balanced by the outward centrifugal acting on him.
∴ R + mg = $$\frac{m v^{2}}{r}$$ → (1)
The minimum speed required to perform vertical loop is given by equation (1) when
R = 0
∴ mg = $$\frac{m v^{2}_{(\min )}}{r}$$
υmin = $$\sqrt{\mathrm{rg}}$$ = $$\sqrt{25 \times 10}$$
≈ 15.8 m/s.

Question 39.
A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/mln. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?

The horizontal force N exerted by the wall on the man provides the necessary centrepetal force.
∴ N = m ω² r
The static frictional force (vertically upwards) balances the weight of the man mg.
The man remains stuck to the wall after the floor is removed if mg 〈 µ N
i.e., mg 〈 µ mRω²
∴ Minimum angular speed of rotation is
⇒ ωmin = $$\sqrt{\frac{g}{\mu_{s}} r}$$ = $$\sqrt{\frac{10}{0.15 \times 3}}$$ = 4.6 rad/s
Thus the minimum rotational speed of cylinder required to hold the man stuck to the wall is 4.6 rad/s

Question 40.
A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω

1. Show that a small bead on the wire loop remains at its lowermost point for ω ≤ $$\sqrt{g / R}$$
2. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω = $$\sqrt{2 \mathrm{g} / \mathrm{R}}$$? Neglect friction.

1. Let the radius vector joining the bead to the center of the wire make an angle θ with vertical downward direction. Let N be the normal reaction. From fig,
mg = N cos θ → (1)
mr ω² = N sin θ → (2)
or m (R sin θ ) ω² = N sin θ
mRω² = N
On substituting in (1)
mg = (m Rω²) cos θ
0r ω = $$\sqrt{\frac{9}{\mathrm{R} \cos \theta}}$$
For the bead to remain in lower most position θ = 0
⇒ cos θ = 1
⇒ ω ≤ $$\sqrt{\frac{\mathrm{g}}{\mathrm{R}}}$$

2. when ω = $$\sqrt{\frac{2 g}{R}}$$
cos θ = $$\sqrt{\frac{g}{R \omega^{2}}}$$ = $$\frac{g}{R\left(\frac{2 g}{R}\right)}$$
⇒ θ =60°
∴ apparent weight = m (g – a).

### 1st PUC Physics Laws of Motion One Mark Questions and Answers

Question 1.
Define force.
Force is defined as that external agent acting on a body changes its state of rest or uniform motion along a straight line.

Question 2.
Define inertia.
The tendency of a body to oppose any change in its state of rest or of uniform motion is called inertia.

Question 3.
State Newton’s first law of motion (or Law of inertia).
Everybody continues to be in its state of rest or uniform motion along a straight line unless compelled to change that state by an external force.

Question 4.
Define Linear momentum.
The momentum of a body to defined to be the product of mass and velocity and is denoted by P $$\overrightarrow{\mathrm{P}}$$ = $$m \bar{v}$$

Question 5.
State Newton’s Second law of motion.
The rate of change of momentum of the body is directly proportional to the applied force and takes place in the direction of the force.

Question 6.
Define newton.
One newton is defined as that force which acting on a body of mass 1kg produces an acceleration of 1 m/s2

Question 7.
Give the dominion formula for
a)Force
b) momentum
Force – MLT-2
Momentum – MLT-1

Question 8.
What quantity is conserved during rocket propulsion?
Linear Momentum.

Question 9.
Action and reaction forces do not cancel each other. Why?
Action and reaction forces do not cancel each other because they act on different objects.

Question 10.
What is the apparent weight measured? When a person of mass ‘m’ is standing in a lift accelerating with an acceleration of ‘a’
(i) Downwards
The weighting machine measures the reaction force given by the floor. So when lift is going
i) downwards the weight measured is lesser

Question 11.
ii) Upwards
ii) Upwards the weight measured is more
∴ apparent weight = m (g + a)

Question 12.
State Newton’s third law of motion.
For every action, there is an equal and opposite reaction.

Question 13.
Which is the weakest force in nature?
Gravitational force.

Question 14.
Is it possible for the weight of a body to be zero?
Yes, whenever a body is on a free fall its weight is zero, but its mass remains unaltered.

Question 15.
Two masses are In the Nation 1:2. What is the ratio of their Inertia?
Inertia of a body is directly proportional to its mass. Therefore the ratio of their inertia is Their inertia is also in the ration 1:2.

Question 16.
Passengers in buses tend to fall back as it accelerates. Why?
Due to inertia, the passengers tend to continue their state of rest, when the bus moves by accelerating.

Question 17.
A cricket player catches the ball by moving his hand along the direction of the motion of the ball. Why?
By moving his hand along the direction of motion of the ball, the player increases the time of contact, thus reduces the impulse felt.

Question 18.
A stone breaks the window glass, but a bullet make only a hole. Why?
Since the velocity of the bullet is much greater than that of the stone, the bullet is in contact with the glass for a very short time that the glass can’t give enough resistance.

Question 19.
Can a moving body be in equilibrium?
Yes, If a body is in a state of uniform motion in a straight line (net force acting on it is zero), its a moving body in equilibrium.

Question 20.
State law of conservation of momentum?
When the net external force on a system is zero, then there is no change in momentum of the system.

Question 21.
What is friction?
The property by virtue of which an opposing force is created between two bodies in contact, which opposes their relative motion is called friction.

Question 22.
What is frictional force?
The force, which opposes the motion of one body over the other in contact with it, is called frictional force

Question 23.
What is static friction?
Frictional force, which balances the applied force when the body is in the state of rest is called static friction.

Question 24.
What is limiting friction?
The maximum static friction that a body can exert on the other body in contact with it is called limiting friction.

Question 25.
What is sliding friction?
The frictional force that opposes the relative motion between the surfaces when one body slides over the other body is called sliding friction.

Question 26.
What is rolling friction?
Rolling friction is defined as the force of friction acting when a body rolls over the other body.

Question 27.
Define angle of friction.
Angle made by the resultant of normal reaction and limiting friction with the normal reaction is called angle of friction.

Question 28.
Define angle of repose.
Angle of repose is defined as the angle that an inclined plane makes with the horizontal when a body placed on it just starts sliding.

Question 29.
Define coefficient of static friction.
The coefficient of static friction is defined as the ratio of limiting friction to the normal reaction between the surfaces.

Question 30.
What are the units and dimensions of the coefficient of friction?
Co-efficient of friction is a ratio, so it is unitless.

Question 31.
When a wheel is rolling, what is the direction of the friction?
Friction is tangential to the wheel and in the direction opposite to motion.

Question 32.
Which of the following is a scalar quantity? Force, momentum & Inertia.
Inertia.

Question 33.
If the string rotating stone is cut. Which direction will the stone move?
The stone will move in the direction tangential from the point where it got cut.

Question 34.
Does a stone moving in a uniform circular motion (constant speed) has no net external force on it?
The speed is uniform but direction is changing, so, there is a change in velocity (acceleration is non zero) Hence stone is under the influence of a net external force.

Question 35.
A man of mass 60 kg is on a lift which is moving up with uniform speed, [g = 10ms 2]. Find apparent weight?
Since it is moving with uniform speed, there no additional force (a = 0) So, apparent weight = m(g + 0) = (60 kg) × (10 ms-2)
= 600 N.

Question 36.
What happens to the coefficient of friction if the weight of a body is doubled?
The coefficient of friction remain constant.

Question 37.
What provides the centripetal force for a car taking a turn on a level road.
Frictional force.

Question 38.
Find force on a body if change in momentum of a body is 20 kg ms-1 over 5 seconds.
Force is the rate of change of momentum
F = $$\frac{\Delta(\text { momentum })}{t}$$ = $$\frac{20 \mathrm{kg} \mathrm{ms}^{-1}}{5 \mathrm{s}}$$ 4 N

Question 39.
A 50 kg mass is subjected to a force of 5 N. What is acceleration of the body.
We know that, F = ma
⇒ α = $$\frac{F}{m}$$
= $$\frac{5 \mathrm{N}}{50 \mathrm{kg}}$$ = 0.1 ms-1

Question 40.
A 25 g body is moving with uniform velocity of 5ms-1. What is the force acting on the body?
Since there is no change in velocity the a = 0.
⇒ F = ma= (25 × 10-3) × (0) = 0 N

### 1st PUC Physics Laws of Motion Two Marks Questions and Answers

Question 1.
State the 4 basic forces of nature.

1. Gravitational force
2. Elector magnetic force
3. Strong nuclear force
4. Weak nuclear force.

Question 2.
Which is the strongest & weakest force in nature?
Strongest – strong nuclear force Weakest – gravitational force.

Question 3.
Explain why does a cyclist bends while riding a curved road?

On bending the cyclist part of his normal force to acts as centrepetal force that helps him staying within the circular path.
Fa = N sin θ = $$\frac{m v^{2}}{\alpha}$$

Question 4.
Show that impulse of force is equal to the change in momentum of a body.
Let a force F act on a body of mass ‘m’ for a short interval of time‘t’.
Then impulse of the force = F t
= mat = $$m\left(\frac{v-u}{t}\right) t$$ = m (v – u). Therefore impulse of force is equal to the change in momentum.

Question 5.
Define Impulse of a force and Impulsive force.
The product of the force & the time for which it acts on a body is called impulse of a force. The force acting on a body for short interval of time is called impulsive force.

Question 6.
A ball hits the ground with a momentum $$\overrightarrow{\mathbf{p}}$$ and bounce back with the same magnitude of momentum. Find the change in momentum.
Initial momentum = $$\overrightarrow{\mathbf{p}}$$
Final momentum = – $$\overrightarrow{\mathbf{p}}$$
change in momentum =
Δ p = – $$\overrightarrow{\mathbf{p}}$$ – $$\overrightarrow{\mathbf{p}}$$ =- 2$$\overrightarrow{\mathbf{p}}$$

Question 7.
While Jumping on a cement floor, we weigh less than what weighs on cement floor. Why?)
When we are on the floor, exerts reaction on us. When we jump the reaction on us is zero. Therefore while jumping on a cement Floor we weigh less than on cement floor.

Question 8.
Distinguish between conservative and nonconservative force.
Conservative forces are those forces against which the work done doesn’t depend on the path followed but depends only on the initial and final positions. Non-conservative forces are those in which the work done depends on the path taken

Question 9.
Calculate the Impulse of a force of 50 N acting for 0:1s.
F = 50N t = 0:1 s
impulse = 50 N × 0.1s= 5 Ns

Question 10.
What are the methods of reducing friction?

1. Friction between two surfaces can be reduced by polishing them.
2. Jets, aeroplanes, and cars are given streamlined shape to reduce friction due to air resistance.
3. The use of lubricants like oil, grease, etc. reduces the friction in machines.
4. By using ball bearings friction in wheels of a car or cycle can be minimised.

Question 11.
Static friction is a self-adjusting force comment.
The magnitude of static friction depends? on the magnitude of the applied force. As the applied force increases the magnitude of the static friction also increases. Thus static frictional force is a self-adjusting force.

Question 12.

1. Brakes of the vehicles work due to friction.
2. Friction helps in driving vehicles.
3. A match stick is lighted because of friction.

Question 13.
Is earth an inertial frame of reference?
No. earth can not be considered as an inertial frame of reference, because the earth is rotating and revolving, which means it is accelerating.

Question 14.
Derive an expression for recoil velocity of gun.
Let mg be mass of gun,
$$\vec{v}_{g}$$ = recoil velocity of gun,
mb be mass of Bullet $$\vec{v}_{b}$$
Initial momentum = 0,
Final momentum = mg $$\vec{v}_{g}$$ + mb $$\vec{v}_{b}$$
By law of conservation of momentum
0 = mg $$\vec{v}_{g}$$ + mb $$\vec{v}_{b}$$
⇒ $$\vec{v}_{b}$$ = – $$\left(\frac{m_{b} \vec{v}_{b}}{\vec{m}_{g}}\right)$$

Question 15.
How does lubricants help in reducing friction?
The lubricants spread over the irregularities on the surface that makes the contact. So, the contact between the lubricant and the moving objects reduces the friction.

Question 16.
A bubble generator is kept at the bottom of an aquarium which is on a free fall. Will the bubbles generated rise to the top?
No, the bubbles generated at the bottom will not rise to the surface, ‘this is because the water in the aquarium is in a state of weightlessness and does not give the bubbles a reactional upward force.

Question 17.
Why is it easier to pull a roller than push it?

When we pull, the force ‘F’ the normal force is reduced by a value F sin θ. So the friction experienced is lesser which makes it is easier to pull.

Question 18.
An object of mass m collides with a another object of mass ‘2 m’. if the initial velocity of the object of mass ‘m’ is v1 and of mass ‘2m’ is ‘0’. Then find the final velocity assuming they get stick to each other.
Initial momentum =
(m) (v1) + (2m) × (0)
= mv1
Final momentum= m(vx) + 2m (vx)
= 3m vx.
Where Vx is the velocitycombined mass By Law of conservation of momentum
mv1 = 3mvx
vx = $$\frac{m v_{1}}{3 m}$$
vx = $$\frac{v_{1}}{3}$$

Question 19.
If a boats sail is blown by air produced by a fan on the boat, can the boat move forward?
No, the boat can not be moved by a fan on the boat, this is because when the fan pushes the sail blowing air, the air pushes the fan backwards with the same force. Since there is no external force on the system – the net change in momentum is zero.

Question 20.
A retarding force Is applied to a motor car. If speed Is doubled how much more distance will it travel.
Let original force be F1, mass be m1 velocity be V and distance be S then,
F = ma, & V² = 2as
⇒ F = $$\frac{m v^{2}}{2 s}$$ ⇒ S = $$\frac{m v^{2}}{2 F}$$
If velocity is doubled, then the distance it will travel before coming to halt will be increased by 4 times.

### 1st PUC Physics Laws of Motion Three Marks Questions and Answers

Question 1.
Distinguish between mass and weight.

1. Mass is the amount of matter contained in a body while weight is the gravitational force acting on a body.
2. Mass of body remains same while weight of body varies from place to place.
3. Mass is a scalar but weight is a vector.
4. Unit of mass is kilogram and that of weight is newton.
5. Mass is measured using a physical balance and weight is measured using a spring balance.

Question 2.
Derive the equation F = ma.
Consider a body of mass ‘m’ moving with a velocity ‘u’. Let a constant force ‘F’ applied on a body changes its velocity to V in ‘t’ seconds.
Initial momentum of the body = mass × initial velocity = m u
Final momentum = mass × Final velocity = m v
Change of momentum in ‘t’ seconds = mv – mu.
= $$\frac{m v-m u}{t}$$ = m $$\left(\frac{v-u}{t}\right)$$
∴ α = ma
∵ $$\frac{v-u}{t}$$ = a, acceleration
According to Newton’s second law, the rate of change of momentum is directly proportional to the applied force or vice versa.
i. e. Force a rate of change of momentum
F α ma
F = kma
Where ‘k’ is a proportionality constant. In SI system k=1.
∴ F = ma

Question 3.
State and explain Newton’s third law of motion. Give illustrations for the same.
Newton’s third law states that for every action, there is an equal and opposite reaction.
Let F1 be the force exerted by the body A on body B, F1 is called action. Then force F2 exerted by B on A is called reaction. According to the third law F1 = – F2.
Illustrations:

1. When a book is placed on the table the weight of the book is acting vertically downwards (action). The table exerts an equal and opposite force vertically upwards (reaction).
2. A swimmer pushes the water in the backward direction with a certain force (action) and the water pushes him in the forward direction with equal and opposite force (reaction).
3. The sailing of a boat is due to the action of the boat on water and reaction from water on the boat.
4. When an object is suspended from the string, the weight of the object acts vertically downwards. The reaction in the string called the tension acts vertically upwards.
5. The earth attracts the moon with a force that constitutes action. In turn the moon attracts earth with equal and opposite force (reaction).

Question 4.
Derive a relation for the safe velocity of negotiating a curve by a body in a banked curve with fractional coefficient ‘µ’.
The net force along the x-direction inwards should provide the centripetal force
∴ Ffriction cos θ + N sin θ = $$\frac{m v^{2}}{2}$$ → (1)
∵ there is is no motion in y-direction
N cos θ Ffriction sin θ = mg → (2)
we know that
Ffriction = µ N.
divide (1) by (2)

$$\frac{\mu \mathrm{N} \cos \theta+\mathrm{N} \sin \theta}{\mathrm{N} \cos \theta-\mu \mathrm{N} \sin \theta}$$ = $$\frac{m v^{2}}{r(m g)}$$
⇒ $$\frac{N(\mu+\tan \theta)}{N(1-\mu \tan \theta)}$$ = $$\frac{\mathrm{v}^{2}}{\mathrm{rg}}$$
⇒ v² = rg$$\left(\frac{\mu+\tan \theta}{1-\mu \tan \theta}\right)$$
The max velocity to safely negotiate the turn is $$\sqrt{r g\left(\frac{\mu+\tan \theta}{1-\mu \tan \theta}\right)}$$

Question 5.
Write the equation corresponding to the ones given, for rotational motion about a fixed axis.
(i) x (t) = x (0) + v (0) t + 1a/2 t²
θ (t) = θ (0) + ω(0) + $$\frac{1}{2}$$ α t²

(ii) v² (t) = v² (0) + 2a [x t) – x (0)]
ω² (t) = ω² (0) + 2 a α [θ (t) – θ (0)]

(iii) $$\overline{\mathbf{v}}$$ = $$\frac{v(t)-v(0)}{2}$$
$$\bar{\omega}$$ = $$\frac{\omega(t)-\omega(0)}{2}$$

(iv) v(t) = v(0) + at
ω (t) = ω (0) + α t

Question 6.
Two masses m1 and m2 are connected to ends of string passing over a pulley. Find tension and acceleration associated.

Assuming mass mt moves down with an
acceleration ‘a’
m1g -T = m1a1 ………….. (1)
T – m2 g = m2 a1 ………. (2)
(1) + (2)
m1g – m2g = (m1 + m2) a1
⇒ a1 = $$\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) g$$
& T = m1g – m1 $$\left(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) g\right)$$
⇒ T = m1g $$\left[1-\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right]$$
T = $$\frac{2 m_{1} m_{2} g}{m_{1}+m_{2}}$$

Question 7.
Name a mass varying system. Derive an expression for the velocity of the rocket at any instant of time
A rocket-propelled into space is a mass varying system as it losses the weight of the fuel burnt.
Let the velocity of gas used for propelling be ‘vg’ & let the rate of decrease in mass of the body be $$\frac{\mathrm{d} m}{\mathrm{dt}}$$
Then, by law of conservation of momentum since initial momentum is zero, dp = 0
⇒ d (mv) = 0
⇒ (dm) v + m d v = 0
⇒ vdm = – mdv ⇒ dm = – $$\frac{\mathrm{d} \mathrm{v}}{\mathrm{v}}$$
Integrating on both sides
v = vg (Inm) + c
⇒ v = – vg logc m + c
where c is a constant.

Question 8.
Indicate the force acting on a block of mass ‘m’ at rest on an Inclined plane of angle θ.

Ffriction = μ N, N = mg cos θ & mg sin θ = Ffriction

Question 9.
Distinguish between static friction, limiting friction & kinetic friction. How do they vary with applied force? Explain.
The static friction is a friction that acts on a body at rest.
Limiting friction is the maximum value of static friction. It is the force that is required for the body to just start moving.
Kinetic friction is the frictional force that action a body which is in motion.
On increasing the applied force, static friction increase, until it reaches limiting friction which is fixed, and kinetic friction also remains constant.

Question 10.
Define Impulse. What graphical methods can be used to calculate impulse in the following cases

1. constant force
2. variable force acting on a body

Impulse is a force that acts a body for a very short duration of time. It is defined as the product of force and the time for which it acts.
1.

In case of a constant force, say F1, the impulse is simply product of force and duration
Impulse = F1 × t1
2.

In case of a variable force, the impulse will be the integral of F2 one of the interval [0,t2]
Impulse = $$\int_{0}^{t_{2}} F_{2}(t) d t$$

Question 11.
An object of mass ‘m’ is on a inclined plane (θ). Find

1. The effective resistant force on the body if it’s moving downwards.
2. The minimum force if it is being pushed upwards.

Assume a friction with a coefficient of μ

1. The net force along the slope is
⇒ Feq = F1 – Mg sin θ
= μ N – Mg sin θ
But, N = mg cos θ
⇒ Feq = mg (μ cos θ – sin θ)

2. Additional force required be Fa
Fa = mg sin θ + μ N
= mg sin θ + μ mg cos θ
⇒ Fa = mg (sin θ + μ cos θ)

### 1st PUC Physics Laws of Motion Five Marks Questions and Answers

Question 1.
State and prove the law of conservation of momentum.
In a closed system, the total linear momentum of the system remains constant or conserved.
Proof:
Consider two bodies A and B of masses m1 and m2 moving in the same direction with uniform velocities u1 and u2 respectively. After the collision let their uniform velocities be v1 and v2. Let‘t’ be the time of impact.
Change in momentum of A
= m1v1 – m1u1
Rate of change of momentum of A
= $$\frac{m_{1} v_{1}-m_{1} \mu_{1}}{t}$$
change in momentum of B
= m2v2 – m2u2
Rate of change of momentum of B
= $$\frac{m_{2} v_{2}-m_{2} u_{2}}{t}$$
If F1 is the force exerted by A on B then according to second law,
F1 = $$\frac{m_{2} v_{2}-m_{2} u_{2}}{t}$$ (action)
If F2 is the force exerted by B on A then
F2 = $$\frac{m_{1} v_{1}-m_{1} \mu_{1}}{t}$$ (reaction)
According to Newton’s third law, action and reaction are equal and opposite i.e.
F1 = – F2
$$\left[\frac{m_{2} v_{2}-m_{2} u_{2}}{t}\right]$$ = – $$\left[\frac{m_{1} v_{1}-m_{1} \mu_{1}}{t}\right]$$
m2v2 – m2u2 = – m1v1 + m1u1
OR
m1u1 + m2u2 = m1v1 + m2v2
i.e., Total momentum before collision = Total momentum after collision. Hence the momentum is conserved.

Question 2.
A stone weighing 5kg. falls from the top of a tower 100m high and buries itself 1 m deep in the sand. What is the average resistance offered by sand?

Mass of the stone, m = 5kg
Height of the tower h = s = 100m
Initial velocity u =0
Final velocity v = ?
From the relation, v² = u² + 2gs
v² = 0 + 2 × 9.8 × 100
v² = 1960
V = $$\sqrt{1960}$$
v = 44.27 ms-1
Then the stone penetrates through the sand with a initial velocity, u = 44.27 ms-1
Distance travelled, S = 1 m
Final velocity, v = 0
acceleration, a =?
From the equation, V² = u²+2as
0² = (44.27)² + 2 × a × 1
– 1960 = 2a
a = -980ms-2
∴ The average resistance offered by the sand is F = ma
= 5 × 980
F = 4900 N

Question 3.
State the Newton’s laws of motion. Write any two illustrations.
Newton’s first law of motion states that, everybody continues to be in its state of rest or uniform motion along a straight line unless compelled to change its state by an external force. Newton’s second law of motion states that the rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force.
Newton’s third law of motion states that for every action there is. an equal and opposite reaction.
Illustrations for Newton’s third law of motion are

• when a book is placed on the table the book exerts force on the table (action), in turn the table exerts an equal and opposite force (reaction) on the book in the upward direction.
• The sailing of a boat is due to the action of the boat on water and the reaction from water on boat.

Question 4.
Consider a body of mass ‘m’ attached to a string of length ‘L’. If the ring is forming a vertical circle, derive an expression for velocity and tension at any point.
Also, find the velocity that is required for mass to just reach the peak point of the circle.

Let ‘θ’ be the angle the string makes with the vertical at any instant of time. Let vx be the velocity of body at the lowermost point x. The distance traveled by the body from the given point ‘p’ to ‘X’ in ‘y’ direction is given by,
XY = L – Lcos θ = L(1 – cos θ)

T – mg cos θ = $$\frac{m v^{2}}{L}$$
Using the equation v² = u² + 2as, we can write
Vx² = v² + 2g (XY)
i.e. V² = Vx² – 2g L (1 – cos θ) ……….. (1)
Tension, T = mg cos θ + $$\frac{m v^{2}}{L}$$ using (1)
T = mg cos θ + $$\frac{m}{L}$$ (Vx² – 2gL (1 – cos θ)
= mg cos θ + $$\frac{m v_{x}^{2}}{L}$$ – 2 mg (1 – cos θ)
T = $$\frac{m v_{x}^{2}}{L}$$ + mg (3 cos θ – 2)
For the vertical circle to be reached v = 0 & vx = ? & θ = 180°
⇒ vx² = v² + 2g L (1 – cos θ)
vx² = 0 + 2g L (1 – (- 1))
vx = $$\sqrt{4g L}$$
vx = $$2 \sqrt{g L}$$

Question 5.
If the system Is on a frictionless surface. Find the ratio of tensions in the string.

The acceleration of the system is given
by a = $$\frac{\text { Force applied }}{\text { Total mass }}$$
= $$\frac{120 \mathrm{N}}{(10+20+30) \mathrm{kg}}$$ = 2 ms-2
For the last block
120 – T2 = ma
⇒ T2 = 120- (30) × (2)
= 60 N
For the middle block
T2 – T1 = ma
60 – T1 = (20) × (2)
T1 = 60 – 40 = 20 N
The ratio of tensions is,
T1 : T2 = 20 : 60 = 1 : 3

Question 6.
For the figure shown, find acceleration produced and the force of contact between the blocks. What is the effect of this force if it is applied to other blocks.

The acceleration of system is
a = $$\frac{F}{\left(m_{1}+m_{2}\right)}$$
When the force is applied on block m, we have,
F – Fc = m1 a
where Fc is force of contact
⇒ Fc = F – m1 $$\left(\frac{F}{m_{1}+m_{2}}\right)$$ = $$\left(\frac{F m_{2}}{m_{1}+m_{2}}\right)$$
If F is applied to other block ,

Fc = m1 a = $$\left(\frac{\mathrm{m}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)$$ F

Question 7.
In the system shown if μk (kinetic friction coefficient) is 0.04. Find acceleration of the trolley,
[g = 10ms-2]

Free body diagram of trolley

We know that Ff = μ N
= μk mg
= 0.04 × 15 × 10
= 6N
⇒ T – 6N = ma
= 15 × a
⇒ T – 5 a = 6 …………. (1)
Free body diagram of mass
⇒ 2g – T = ma
⇒ 20 – T = 20
⇒ 2a + T = 20 ……….. (2)
0n,(1) – (2)
– 17 a = – 14
a = $$\frac{14}{17}$$ = 0.82ms-2

Question 8.
Weights of 250 g & 200 g are connected by a string over a smooth pulley. I system is traveling 4.95 m in the first 3 second. Find the value of g.

For the pulley system,
T – (200 g × 10-3) g = (200 × 10-3kg) a
T – (0.29) = 0.2a …………….(1)
an (0.25g) – T = (0.25a) …………(2)
⇒ From (1) & (2)
0.45 a = (0.25 – 0.2) g
a = $$\frac{0.05}{0.45}$$g = $$\frac{g}{9}$$ ms-2
Now, S = 4.95 m, t = 3s u = 0
From S = ut + $$\frac{1}{2}$$ at²
4.95 = 0 + $$\frac{1}{2}$$ × a × (3)²
4.95 = $$\frac{1}{2}$$ × $$\frac{10}{9}$$ × g × 9²
g = $$\frac{4.95}{5}$$ = 9.9 ms-2

Question 9.
A force of 80N acting on a body at rest for 2 sec imparts it a velocity of 20ms-1 what is the mass of the body calculate the distance traveled by the body in 2 seconds?
Force, F = 80N
Initial velocity, u =0
Time for which force acts on the body, t = 25s,
Final velocity, v = 20ms-1
From the equation v = u + at
20 = 0 + a × 2
a = 10ms-2
∴ The mass of the body, from the equation F = ma
m = $$\frac{F}{a}$$ = $$\frac{80}{10}$$ = 8kg
∴ The distance travelled by the body in a time, t = 2s. From the equation
s = ut + $$\frac{1}{2}$$ at²
= 0 × 2 + $$\frac{1}{2}$$ × 10 × 2²
= 0 + 20 = 20m.

Question 10.
A man of 60 kg is standing on a weighing machine placed on the floor of a lift which reads the force in newtons. Find the reading of the weighing machine when the lift is

1. stationary
2. moving upwards with uniform speed of 10 ms-1
3. moving downwards with uniform acceleration of 5 ms-2
4. moving upwards with uniform acceleration of 5 ms-2.
5. What would be the reading of the weighing machine if the connecting rope of the lift suddenly breaks and lift begins to fall freely under gravity, g = 10 ms-2.

Weight of the man is due to the reaction from the floor of the lift. It is given by,
R = mg + ma*
where a* is the acceleration of the lift.
1. when the lift is at rest a* = 0
∴ Reading of the weighing machine
R = mg = 60 × 10 = 600 N.

2. when the lift is moving up or down with uniform velocity, a* = 0
∴ Reading of the weighing machine,
R = 600 N.

3. When the lift is moving downwards with an acceleration a*.
R = mg – ma* = m(g – a*)
= 60 × (10 – 5) = 300 N.

4. When the lift is moving upwards with acceleration a*,
R = mg + ma* = m(g + a*)
= 60 × (10 + 5)
= 900 N.

5. If the lift falls freely under gravity, a* = g
∴ R = m(g – a*) = 0.

Question 11.
A rubber ball of mass 0.1 Kg is dropped on the ground from a height of 2.5 m and it rises to a height of 0.4m. Assuming the time of contact with the ground to be 0.01 s, calculate the force exerted by the ground on the wall, g = 10ms-2
Mass of the ball m = 0.1 kg
Time in contact with the ground, t = 0.01s
1. When the ball is dropped on ground,
u =0, s = 2.5m, g = 10ms-2
From the equation, v² = u² + 2gs
v² = 0 + 2 × 10 × 2.5
v = 7.07ms-1, downwards.

2. When the ball rises up from the ground,
v = 0, s = 0.4m, g = 10ms-2
From the equation v² = u² + 2gs
0 = u² + 2(- 10)0.4
u² = 8
u =2.83ms-1, upwards.
Assuming the upward velocity +ve & downward velocity -ve,
change of velocity = 2.83 – (- 7.07)
i.e v – u = 9.9ms-1
∴ Force exerted by the ground on the ball is,
F = m $$\left(\frac{v-u}{t}\right)$$ = 0.1 $$\left(\frac{9.9}{0.01}\right)$$ = 99 N.

Question 12.
A Bullet flying with a velocity of 50ms-1 hits a block of wood and penetrates through a distance of 0.2 m before coming to rest. The Mass of the bullet is 0.03kg. Calculate the resistance offered by the block of wood.
Initial velocity, u = 50ms-1
Distance travelled, s = 0.2
Final velocity v = 0
Mass of the bullet, m = 0.03 kg
From the equation, v² = u² + 2as
0 = 50² + 2 × a × 0.2
a = – $$\frac{2500}{0.4}$$ = -6250ms-2
∴ The resistance offered by the wood is
F = ma = 0.03 × 6250 = 187.5 N.

Question 13.
A machine gun fires 200 bullets per minute with a velocity of 60ms-1. If the mass of each bullet Is 0.02kg, calculate the power of the gun.
Number of bullets fired in one minute = 200
work done by the gun in one minute is,
W = kinetic energy of 200 bullets
W = 200 ($$\frac{1}{2}$$mv²)
= 200 × $$\frac{1}{2}$$ × 0.02 × (60)²
= 7200J
∴ Power, P = $$\frac{W}{t}$$, t = 60 seconds
P = $$\frac{7200}{60}$$ = 120 watt.
∴ Power of the gun = 120 watts.

Question 14.
Two metal balls of masses 10kg and 8kg are moving In the same direction with velocities 10m/s and 4m/s respectively. They stick together after collision. Find their common velocity after collision. If they are moving

1. In the same direction,
2. In opposite direction before collision.

m1 = 10kg, m2 = 8kg
u1 = 10m/s and u2= 4m/s.
1. v1 = v2 = v, common velocity.
m1u1 +m2u2 = m1v1 + m2v2
10 × 10 + 8 × 4 = (l0 + 8)v
or v = 7.33m/s

2. m1u1 – m2u2 = (m1 + m2) v
10 × 10 – 8 × 4 = (10 + 8)v
v = 3.778m/s.

Question 15.
Derive the equation F = ma.
Consider a body of mass ‘m’ moving with a velocity ‘u’. Let a constant force ‘F’ applied on a body changes its velocity to ‘v’ in ‘t’ seconds.
Initial momentum of the body = mass × initial velocity = m u
Final momentum = mass × Final velocity = m v
Change of momentum in ‘t’ seconds = mv – mu.
Rate of change of momentum
= $$\frac{m v-m u}{t}$$ = m$$\left(\frac{v-u}{t}\right)$$ = ma
∵ $$\frac{v-u}{t}$$ = a, acceleration
According to Newton’s second law, the rate of change of momentum is directly proportional to the applied force or vice versa.
i.e. Force a rate of change of momentum
F α ma
F = kma
Where ‘k’ is a proportionality constant. In SI system k =1.
∴ F = ma.

Question 16.
Name the basic forces in nature.
Basic forces in nature are,

1. Gravitational force
2. Electromagnetic force
3. Nuclear force and
4. Weak force

Question 17.
A body is moving on a frictionless curved path of radius of 1.8 km with a speed of 30 ms-1. Find the banking angle required.
The centripetal force required to keep the body in circular motion is $$\frac{m v^{2}}{r}$$
Here, v = 30 ms-1
r = $$\frac{1.8 \times 10^{3} m}{2}$$ = 0.9 103 = 900m
N cos θ = mg
and $$\frac{m v^{2}}{r}$$ = N sin θ
⇒ $$\frac{m v^{2}}{r}$$ = $$\frac{m g}{\cos \theta}$$ sin θ
⇒ tan θ = $$\frac{v^{2}}{r g}$$
⇒ θ = tan-1 $$\left(\frac{30^{2}}{900 \times 10}\right)$$
⇒ θ = tan-1 (0.1) = 5.71°.

Question 18.
What is the acceleration of a body moving on a circular path of radius 400 m. If it has

1. constant speed of 40 ms-1
2. speed increases at 3 ms-2

A body on a circular path has two kinds of accelerations: radial & linear
1. If speed is constant linear acceleration is zero & radical acceleration is ar = $$\frac{v^{2}}{r}$$
v = 40ms-2
r = 400 m ⇒ ar = $$\frac{40 \times 40}{400}$$ = 4ms-2

2. If the speed increases at 3 m/s², it has a linear acceleration of 3ms-2
a = $$\sqrt{a_{r}^{2}+a_{1}^{2}}$$
= $$\sqrt{3^{2}+4^{2}}$$
= 5 ms-2.

Question 19.
An aeroplane at 360 km hr-1 has its wing banked at an angle 20°. Find the radius of the circle traversed by the plane, [g = 10ms-2]
Speed of the plane = 360 km/hr
= $$\frac{360 \times 10^{3}}{3600}$$ = 100 ms -1
we know that tan θ = $$\frac{v^{2}}{r g}$$
⇒ r = $$\frac{v^{2}}{\tan \theta \times g}$$ = $$\frac{100 \times 100}{\tan \left(20^{\circ}\right) \times 10}$$
= 2.747 km.

Question 20.
A uniform chain of length L is kept on a table of coefficient of static friction μ(limiting value). Find the maximum length of chain that can be outside the table, without it sliding away.

Let x be the length of the chain that can be outside the table.
Let ‘M’ be the total mass of the chain.
Mass on the table is $$\frac{M}{L}$$(L – x)
Mass of the chain outside = $$\frac{M}{L}$$ x
For, equilibrium,
Force of friction = weight of the hanging part.
i.e., μN = $$\left(\frac{M}{L} x\right)$$g × α
i.e., μ$$\left(\frac{M}{L}(L-x) g\right)$$ = $$\left(\frac{M}{L} x\right)$$g
μ(L – x) = x or x = $$\frac{\mu L}{1+\mu}$$

Question 21.
For the system shown in the figure, the coefficient of Kinetic friction between the mass and plane is 0.25.
Given that M2 = 5kg & M3 = 7kg. Find M1, such that the body M1, is moving with uniform velocity. Sin37° = $$\frac{3}{5}$$, cos 37° = $$\frac{4}{5}$$

For the mass m1 to have a uniform velocity the system should be in equilibrium.
⇒ T1 = m1 g
T1 = 10m1 …………… (1) (g =10 m/s²)
T1 =T2 + F11 + m2 g sin θ
= T2 + μ N + m2 g sin 37°
= T2 + μm2g cos 37° + m2g sin 37°
= T2 + m2g $$\left(\mu \frac{4}{5}+\frac{3}{5}\right)$$ (g =10 m/s²)
T1 = T2 + m2[8μ + 6] ………….. (2)
T2 = F12
T2 = μ N
= μ m3 g
T2 = (0.25) (7) (10)
T2 = 17.5 N ………….. (3)
Substituting (1) & (3) in (2)
10m1 = 17.5 + 5 [8(0.25) + 6]
10m1 = 57.5 N
m1 = 5.75 kg.

1st PUC Physics Laws of Motion Numerical Problems Questions and Answers

Question 1.
Two masses 4 kg & 2 kg are connected by a massless string and they are placed on a smooth surface. The 2 kg mass is pulled by a force of 12 N as shown

1. Find the acceleration of the system.
2. If the string is replaced by a spring then what change do you notice in the acceleration
3. In the string, system find the Tension

1. We know that from Newtons second Law,
F = ma
F Force on the system
⇒ a = $$\frac{F}{m}$$ = $$\frac{\text { Force on the system }}{\text { Total mass }}$$
= $$\frac{12 \mathrm{N}}{(4+2) \mathrm{kg}}$$
a = 2 ms-2

2. If the string is replaced by spring, there is no change in mass of system. So there is no change in the acceleration a = 2m s-2
c) We know that, F = m a
(12 – T) = (m a)
T = 12 – (2 × 2)
T = 8 N

a = 2 ms-2

Question 2.
A force of 98 N acts on a body of mass 10 kg which is at rest. Calculate

1. Velocity at the end of 5 seconds.
2. Distance traveled by the body in 5 seconds.

Solution:
1. To find the velocity at the end of 5 seconds.
We have, F = ma
Given, F = 98 N and
m = 10 kg
∴acceleration a = $$\frac{F}{m}$$ = $$\frac{98}{10}$$
= 9.8 ms-2
velocity v = u + at
Here a = 0,
a = 9.8 ms-2 and
t = 5 s
∴ v = 0 + 9.8 × 5
= 49.0 ms-1

2. To find the distance travelled
we have, s = ut + $$\frac{1}{2}$$ at²
Here, u = 0,
a = 9.8 ms-2
t = 5 seconds
∴ s = 0 × 5 + $$\frac{1}{2}$$ × 9.8 × (5)²
= 122.5 m.

Question 3.
A truck of mass 3000 kg is moving with a velocity of 10 m/s is accelerated by a force of 600N.

1. What is the rate at which its velocity increases?
2. How far will it travel In 10s?

Solution:
1. To find the rate at which velocity is increasing
Force F = 600 N
mass m = 3000 kg
Rate of increase in speed,
a = $$\frac{F}{m}$$
= $$\frac{600}{3000}$$
= 0.2 ms-2

2. To find the distance travelled in 10 s
We have, s = ut + $$\frac{1}{2}$$ at²
Here, u =10 ms-1
t = 10 s
a = 0.2 ms-2
∴ s = 10 × 10 + $$\frac{1}{2}$$ × 0.2 × (10)²
= 100 + 10
= 110 m.

Question 4.
A certain force acting on a body of mass 10 kg at rest moves it through 125 m in 5 seconds. If the same force acts on a body of mass 15 kg, what is the acceleration produced?
Solution:
in the case of the first body,
u = 0;
t = 5 s;
s = 125 m
Substituting these values in the equation,
s = ut + $$\frac{1}{2}$$ at
125 = 0 × 5 + $$\frac{1}{2}$$ × a × (5)²
$$\frac{1}{2}$$ × a × 25
∴ a = $$\frac{2 \times 125}{25}$$ = 10 ms-2
F = m × a = 10 × 10 = 100 N
If the same force acts on another body of mass 15 kg, the amount of acceleration produced is,
a = $$\frac{F}{m}$$ = $$\frac{100}{15}$$
= 6.67 ms-2

Question 5.
A cricket ball of mass 0.15 kg is moving with a velocity of 12 ms-1 and is hit by a bat so that the ball is turned back with a velocity of 20 ms-1. If the force of blow acts for 0.01 s, find the average force exerted on the ball by the bat.
Solution:
Initial velocity u = 12 ms-1
Final velocity v = 20 ms-1
Change in velocity =20 – (- 12)
= 20 + 12
= 32 ms-1
(-ve sign is taken because initial and final velocities are in opposite direction)
Time for which force is acting, t = 0.01 s
∴ acceleration a = $$\frac{\text { change in velocity }}{\text { time }}$$
= $$\frac{32}{0.01}$$
= 3200 ms-2
Force F = ma
= 0.15 × 3200
= 480 N.

Question 6.
A hammer of mass 1 kg moving with a speed of 6 ms-1 strikes a wall and comes to rest in 0.1 s. Find the

1. Impulse
2. Retarding force on the hammer
3. Retardation

Solution:
1. The initial momantum of the hammer is,
m × v = 1 kg × 6 m s-1
= 6 kg m s-1
= 6 Ns
Impulse = F . t = Δ P
= 0 – mv
= – 6 Ns.

2. The force on the hammer
F = $$\frac{\text { Impulse }}{\text { time }}$$ =$$\frac{6 \mathrm{Ns}}{0.1 \mathrm{s}}$$
60 N.

3. Retardation = a =$$\frac{F}{m}$$ = $$\frac{60 \mathrm{N}}{1 \mathrm{kg}}$$
= 60 m s-2

Question 7.
A disc of mass 200 g is kept floating horizontally by throwing 40 pebbles per second against it from below. If the mass of each pebble is 2g, calculate the velocity with which the pebbles are striking the disc. Assume the pebbles strike the disc normally and rebound with the same speed.
Solution:
Mass of the disc M = 200 g
= 0.2 kg
Total downward force
F = Mg
= 0.2 × 9.8 =1.96 N
Mass of one pebble m = 2 g = 2 × 10-3 kg Let v be the velocity with which the pebbles strike the disc. Momentum given by one pebble = mv The pebbles rebound downward and strike from below.
∴ net momentum given to the disc in the upward direction
= change in velocity of the pebble × m = (2v) m
Total momentum given in one second
=40 × m × 2v
= 80 mv
The disc remains horizontal if this is equal to the weight of the disc, Mg
∴ 80 mv = Mg
80 × 2 × 10-3 × v = 1.96
v = 12.25 ms-1

Question 8.
Water ejects with a speed of 0.2 ms-1 through a pipe of area of cross-section 1 × 10-2 m². If the water strikes a wall normally, calculate the force on the wall in newtons, assuming the velocity of the water normal to the wall is zero after the collision.
Solution:
Volume of water striking the wall per second = 0.2 × 10-2 = 2 × 10-3 m3
Mass of the water striking the wall in one second = volume × density = 2 × 10-3 × 1000
= 2 kg
Change in velocity of water on striking the wall in one second = 0.2 – 0 = 0.2 ms-1
Force acting on the wall
= change in momentum
=2 × 0.2
= 0.4N.

Question 9.
A gun of mass 5 tons fires a bullet of mass 20g with a velocity of 110.2ms-1. Find the velocity of the gun.
Solution:
Initially, both the gun and the bullet are at rest.
∴ The total initial momentum of the system is f zero.
If v1 and v2 are the final velocity of the gun and the bullet, final momentum is given by,
pf = m1v1 + m2v2
According to the law of conservation of momentum, pi = pf
m1v1 + m2v2 = 0
i.e., v1 = – $$\frac{m_{2} v_{2}}{m_{1}}$$
= $$\frac{-20 \times 110.2}{5 \times 1000}$$
= – 0.44 ms-1
∴ Recoil velocity of the gun is 0.44 ms-1.

Question 10.
A gun weighing 1000 kg recoils with a velocity of 3 × 10-2 m/s when a shell of mass 1 kg is shot from it. If the shell hits the target in 8 seconds, find the gun target distance.
Solution:
Initial momentum of the gun & that of shell is zero as they are at rest.
Recoil velocity of the gun v1 = – 3 × 10-2ms-1
Mass of the gun m1 = 1000kg
Mass of the shell m2 = 1 kg
velocity of the shell v2 =?
From the equation
m1 u1 + m2 u2 = m1v1 + m2 v2
0 = 1000(- 3 × 10-2) +1 .v2
∴ v2 = 30ms-1
The gun target distance,
s = v2 × t
= 30 × 8
= 240m.

Question 11.
A machine gun has a mass of 20 kg. The firing rate of 500 bullets per second and mass of each bullet Is 20 g. If the speed of the bullets 500 m s-1. Find the force required to keep the gun in its position.
Solution:
mgun = 20 kg, mb = 20 g
vgun = ? vb = 500 ms-1
From Law of conservation of momentum
Mgun Vgun + mb vb = 0
⇒ Vgun = –$$\frac{20 \times 10^{-3} \times 500}{20}$$
= – 0.5 -1
∴ Force required to hold its position
F = m$$\left(\frac{v-u}{t}\right)$$ = 20 × $$\frac{(0.5-0)}{\left(\frac{1}{500}\right) s}$$ = 5000 N

Question 12.
A body of mass 20 kg moving with a velocity of 10 ms-1 collides with another body of mass 40 kg moving In the same direction with a velocity of 5 ms-1. If both the bodies stick together after the collision, find the common velocity after collision.
Solution:
If u1 and u2 are the initial velocities of the two bodies before the collision, the total momentum before the collision is
pi = m1u1 + m2u2
Let v be the common velocity of the two bodies after collision. Then the final mo-mentum after collision is,
pf = (m1 + m2)v
From the law of conservation of momentum
Pi = Pf
m1u1 + m2u2 = ( m1 + m2) v

= 6.67 ms-1.

Question 13.
A shell of mass 10 kg flying horizontally with a velocity of 36 kmph explodes in air into two fragments. The larger fragment has a velocity of 25ms-1 & is directed In the same direction as the initial velocity of the shell. The smaller fragment has a velocity of 12.5 ms-1 in the opposite direction. Find the masses of the fragments.
Solution:
Let the mass of larger fragment be m1 = x Then the mass of smaller fragment is m2 = 10 – x
Initial velocity of larger fragment,
u1 = 36kmph = 10 ms-1.
Final velocity of larger fragment,
v1 = 25ms-1.
Initial velocity of smaller fragment,
u2 = 10ms-1.
Final velocity of smaller fragment,
v2 = – 12.5ms-1.
According to the law of conservation of momentum,
mu = m1v1 + m2v2
10 × 10 = x.25 + (10 – x) – 12.5
100 = 25x – 125 + 12.5x
225 = 37.5x
∴ x = $$\frac{225}{37.5}$$ = 6Kg
∴ Mass of larger fragment, m1 =. 6kg
Mass of smaller fragment m2 = (10 – x) = 4kg.

Question 14.
A neutron (mass = 1.67 × 1o-27kg) at a speed of 108m s-1. Collides with detron and gets sticked to it Find the velocity of the composite particle.
Solution:
Mass of neutron 1.67 × 10-27 kg
= Mn,
mass of detron = (mn) = 3.34 × 10-27 kg
= md,
velocity of neutron = 108m s-1 = Vn
velocity of detron = 0 m s-1 = Vd
On collision,
mass of composite particle = Mc = Mn + Md
= (1.67 + 3.34) × 10-27 kg
= 6.01 × 10-27 kg
velocity of composite particle = vc
From Law of conservation of momentum
Mn + Vn + McVc = McVc
1.67 × 10-27 × 10+8 + 0 = (5.01 × 10-27) Vc
⇒ Vc = $$\left(\frac{1.67}{5.01}\right)$$ × 10+8
Vc = 0.33 × 108m s-1

Question 15.
A projectile is fired a with velocity ‘V’ at an angle ‘θ’. If the projective breaks into 2 equal parts and one of them retraces the path then find the velocity of the other part
Solution:

At the highest point the projective will have only x-direction velocity and it is constant throughout the path.
S0, Vx = Vi cos θ
Let ‘M’ be the initial mass, $$\frac{M}{2}$$ be mass of the halves. Velocity of 1 half changes from vx to – vx. Let the velocity of other half be V0. From Law of conservation of momentum.

⇒ V0 = 3 Vx
⇒ v0 = 3 vi cos θ.

## Karnataka 1st PUC Physics Question Bank Chapter 12 Thermodynamics

### 1st PUC Physics Thermodynamics Textbook Questions and Answers

Question 1.
A geyser heats water flowing at a rate of 3.0 litres per minute from 27°C to 77°C. If geyser operates on a gas burner, what is the rate consumption of the fuel if the heat of combustion is 4. 0 × 104 J/g
Rate of flow of water = 3 litres/minute
∴ Mass of water flowing per minute,
M = 3000g / minute
Initial temperature, T1 = 27°C
Final temperature, T2=77°C
Rise in temperature of the flowing water,
∆ T = T2 – T1 = (77 – 27)° C
= 50° C
specific heat of water, c = 4200 J kg-1 °C-1
= 4.2 Jg-1 °C-1
∴ Total heat supplied by geyser,
∆ Q = mc ∆T
= 3000 × 4.2 × 50
= 630000 J/min
= 6.3 × 105 J/min
Heat of combustion of geyser
= 4 × 104Jg-1
∴ Rate of consumption of fuel,
r = $$\frac{6.3 \times 10^{5}}{4.0 \times 10^{4}}$$ = 15. 75 g/min.

Question 2.
What amount of heat must be supplied to 2.0 × 10-2 kg of Nitrogen (at room temperature) to raise its temperature by 45°C at constant pressure? (Molecular mass of N2 is 28; R = 8.3 J mol-1 k-1)
Molecular mass of N2 = 28
∴ 1 mole of N2 gas has a mass of 28g.
Given mass of Nitrogen = 2.0 × 10-2 kg
= 20g
∴ Amount of Nitrogen (in moles),
n = $$\frac{20}{28}$$ moles = 0. 714.
Molar specific heat capacity at constant
pressure of Nitrogen, Cp = $$\frac{7}{2}$$ R
Rise in temperature, ∆ T = 45° C
Total amount of heat to be supplied,
Q = nCp ∆T =0.714 × $$\frac{7}{2}$$ × 8.3 × 45
= 933.376 J
Therefore, amount of heat required is 933.376 J.

Question 3.
Explain why
1. Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1+T2)/2.
2. The coolant in chemical or nuclear plant (i.e, the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
3. Air pressure in car tyre increases during driving.
4. The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
1. When two bodies at different temperatures T1 and T2 are brought in thermal contact, heat energy flows from the body at higher temperature to the body at lower temperature till thermal equilibrium is attained.
Let m be the mass of both bodies and C1 and C2 be the specific heat capacities of bodies at temperatures T1 and T2 respectively. Also, assume that T1> T2.
Since the body at lower temperature absorbs all the energy supplied by a hot body,
∆ Q1 = – ∆ Q2
∴ mc1 ∆T1 = – mc2∆T2 ……..(1)
If ‘T’ is the equilibrium temperature
∴ mc1 (T1 – T) = – mc2(T2 – T) …….(2)
Re-arranging equation (2), we get
T = $$\frac{\mathrm{c}_{1} \mathrm{T}_{1}+\mathrm{C}_{2} \mathrm{T}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$$
The equilibrium temperature depends on the specific heat capacities of the two bodies and hence cannot be (T1 + T2)/2 in general.
(Note: The common temperature will be (T1 + T2)/2 only if the specific heat capacities are the same, i.e, if C1= C2, then (T1 + T2)/2.

2. The function of a coolant in a chemical or nuclear plant is to absorb as much heat as possible and prevent rise in temperature of the system.
The coolant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant.

Mathematical analysis:
The equilibrium temperature when two bodies come in thermal contact is given by
T = $$\frac{\mathrm{c}_{1} \mathrm{T}_{1}+\mathrm{c}_{2} \mathrm{T}_{2}}{\mathrm{c}_{1}+\mathrm{C}_{2}}$$ ……(1)
where C1, T1, C2, T2 represent specific heat capacity and temperature of first and that of second bodyis respectively.
Let T2<T1, ie the second body is the coolant. Re-writing (1), we have
T = $$\frac{\left(\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}\right) \mathrm{T}_{1}+\mathrm{T}_{2}}{\left(\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}\right)+1}$$ ……(2)
Since the function of coolant is to keep the temperature of the system constant, the term $$\left(\frac{c_{1}}{c_{2}}\right)$$ T1 in equation (2) should be as small as possible. This can be a achieved by having C2>> C1.
Therefore, the specific head capacity of the coolant should be very high.

3. When a car is in motion, the temperature of the tyre rises due to increases in the kinetic energy of the air molecules inside the tyre. Assuming air to be ideal gas, the equation of state is given by
PV = n RT
∴ Change in the state of the system is given by
∆ PV = ∆ n RT ……(1)
Since there is no change in volume or composition, we can rewrite equation (1) as V∆ p = nR ∆ T
Since temperature rises during motion, ∆T is positive. Therefore, ∆P is also positive implying that pressure inside tyre increases when the car is in motion.

4. The humidity in harbour town is generally much greater than humidity in a desert. Since humidity is a measure of water vapor content in the atmosphere and the specific heat of water vapor is very high (≃ 1.86 kJ kg-1 k-1 at 300k) the temperature fluctuations in harbour towns are generally lower than those in desert regions. Hence, the climate in harbour towns is more temperate than that of a town in a desert at the same latitude.

Question 4.
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston Is Insulated by having a pile of sand on it. By what factor does the pressure of the gas its original volume?
Since the cylinder wall and piston are heat insulators, there is no exchange of heat. Hence, the process is “Adiabatic”.
∴ The equation for the given system is
P1 V1γ = P2 V2γ
where P1 is the Initial pressure of hydrogen in cylinder, V1 isthe initial volume of hydrogen in the cylinder, P2 is the final pressure of hydrogen and V2 is final volume of hydrogen in cylinder, γ is ratio of water specific heat at constant pressure to that at constant volume of hydrogen.
$$\gamma=\left(\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}\right)_{+_{2}} \simeq 1.4$$
∴ $$P_{1} V_{1}^{1.4}=P_{2} V_{2}^{1.4}$$
Since V2 = $$\frac{v_{1}}{2}$$, P2=P1 $$\left(\frac{V_{1}}{V_{2}}\right)^{1.4}$$

Thus on decreasing the volume of hydrogen to one half its original value, the pressure increases by a factor of 2.639.

Question 5.
In changing the state of gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3J is done on the system. If the gas is taken from state A to B via a process In which the net heat absorbed by the system is 9.35 cal, how much is the net work done by . the system In the later case ? (Take 1 cal = 4.19 J)
Process 1 :
Since the process is adiabatic, net heat supplied to the system is 0. From the first law of thermodynamics, we have
∆ Q = ∆ U + ∆ W ……..(1)
where   ∆ Q is the net heat supplied to the system, ∆ U is the change in the internal energy of the gas and ∆ W is the net work done by the gas.
For adiabatic process, ∆ Q = o
∴ From equation (1), ‘
∆ U = – ∆ W
Net work done on the system is 22.3 J
i.e. ∆ w = – 22.3 J
∴ U = – (-22.3) J = 22.3 J.

Process 2 :
Since internal energy, U is a state variable it does not depend on the process but only on the initial and final state of the gas. As the initial and final states for both Process 1 and 2 are A and B respectively, change in internal energy is same as in Process 1.
∴ ∆ U = 22.3 J ………(2)
Net heat absorbed by the system,
∆ U = +9.35 cal =9. 35 × 4.19 J
∴ ∆ Q =39.176 J ……..(3)
From the first law of thermodynamics,
∆ Q = ∆ U + ∆ W
∴ ∆ W = ∆ Q – ∆ U
From equations (2) and (3),
∆ W = 39. 176 – 22. 3 J = 16. 876 J
∴ The network done by the system in the second process is 16. 876 J.

Question 6.
Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally Insulated. The stopcock is now opened. Answer the following.

1. What is the final pressure of the gas in A and B?
2. What is the change in Internal energy of the gas?
3. What is the change in temperature of the gas?
4. Do the intermediate states of the system (before setting to the final equilibrium state) lie on its P-V-T surface?

1. The above process is a case of free expansion of gas. As soon as the stopcock is removed the gas expands to a total volume of twice its original value (as VA = VB).
From Boyle’s law, we have $$P \alpha 1 / v$$
Since V doubles after the stopcock is removed. P reduces to one half the original value.
At STP, pressure of the gas is 100 kPa = 1 Bar
∴ Pressure after expansion = 50 kPa = 0.5 Bar.

2. From the first law of thermodynamics
∆ Q = ∆ U + ∆ W,
Since the process does not involve any work done by the gas such as moving a piston, and no heat is exchanged,
∆ Q = ∆ W = 0
∴ ∆ U = O
∴ There is no change in internal energy of the gas.

3. Since the internal energy of the gas is fixed for the given process, the temperature of the gas also does not change.
ie, ∆ T = ∆ U and if ∆ U = 0, then ∆ T = 0

4. In case of free expansion of gas, the gas does not go through states of thermodynamic equilibrium before reaching the final state. There fore, the thermodynamic parameters such as pressure, volume and temperature are not well defined for these intermediate states and hence, they do not lie on the P-V-T surface for the gas.

Question 7.
A steam engine delivers 5.4 × 108 J of work per minute and services 5.6 × 109 J of heat per minute from its boiler. What Is the efficiency pf the heat engine? How much heat is wasted per minute 7
Heat supplied from boiler per minute is 3.6 × 108 J/min
Work done by the steam engine per minute = 5.4 × 108 J/min
Efficiency of steam engine,
$$\eta=\frac{\text { work done }}{\text { heat supplied }}$$
$$\text { i.e. } \eta=\frac{5.4 \times 10^{8}}{3.6 \times 10^{9}}=\frac{5.4}{36}$$ = 0.15
∴ Efficiency of the steam engine is 15% Amount of heat wasted per minute
= Heat supplied to steam engine per minute – work done by the engine per minute.
= (3. 6 × 109 – 5.4 × 108) J/ min
= 3. 06 × 109 J/ min
∴ The steam engine wastes 3. 06 × 109 J of energy per minute.

Question 8.
An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 joules per second, at what rate is the internal energy increasing?
Heat supplied,ΔQ = 100 W = 100 J/s
Useful work done, ΔW = 75 J/s
Increase in internal energy/sec, ΔU =?
According to first law of thermodynamics, ΔQ = ΔU + ΔW
ΔU = ΔQ – ΔW = 100 – 75 = 25 J/s = 25 W

Question 9.
A thermodynamic system is taken from an original state D to an Intermediate state E by the linear process shown in Fig. Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F.

In a PV diagram,
1. Work done for a process = Area below the curve for the given process.
∴ For the given PV diagram, work done = Area of ∆ DEF
= $$\frac{1}{2}$$ × base × height = $$\frac{1}{2}$$ × FE × DF
= $$\frac{1}{2}$$ × (5 – 2) × (600 – 300) = $$\frac{1}{2}$$ × 3 × 300
= 450J
∴ The total work done by the gas is 450 J.

Question 10.
A refrigerator is to maintain eatables kept Inside at 9°C. If the room temperature is 36°c, calculate the coefficient of performance.
Temperature inside the refrigerator,
T1 = 9°C = (9+273) K = 282 K
Room temperature, T2 = 36°C
= (36 + 273) K
= 309 K
The coefficient of performance is,

∴ The coefficient of performance of the given refrigerator is 10.444.

### 1st PUC Physics Thermodynamics One Mark Questions and Answers

Question 1.
What is thermodynamics?
Thermodynamics is the branch of physics that deals with interconversion of heat and other forms of energy.

Question 2.
Define thermal equilibrium.
A system is said to be in thermal equilibrium if the macroscopic variables that characterise the system to not change whit time.

Question 3.
State zeroth law of thermodynamics.
Zeroth law of thermodynamics states that – two systems in thermal equilibrium with a third system separately by are in thermal equilibrium with each other.

Question 4.
What is the change in internal energy during Isothermal expansion of an ideal gas?
Zero, because ∆U α ∆T.

Question 5.
What is an Isotherm?
An isotherm is a P – V curve for a fixed temperature.

Question 6.
What is the network done by an ideal gas in an Isochoric process?
Zero. Because as the heat absorbed by the gas is entirely used to raise its internal energy.

Question 7.
State the relation between molar specific heat at constant pressure and that at constant volume for an ideal gas.
Relation between molar specific heat at constant pressure and that at constant volume for an ideal gas is,
Cp – Cv = R,
where Cp is the molar specific heat at constant pressure, Cv is the molar specific heat at constant volume and R is the Universal gas constant.

specific heat calculator helps to calculate the amount of heat needed to raise the temperature of 1 gram of a substance 1 degree Celsius (°C)

Question 8.
What is the network done in a reversible process?
Zero. Because,  the heat absorbed and change in internal energy are zero.

Question 9.
What is the temperature at which the value in the Celcius scale is equal to that in the Fahrenheit scale?
– 40°C

Question 10.
What is refrigeration?
The process of transfer of heat from a cold body to a hot body by doing work on the system is called refrigeration.

### 1st PUC Physics Thermodynamics Two Marks Questions and Answers

Question 1.
State and explain first, law of thermodynamics.
The first law of thermodynamics states that the net heat energy supplied to the system is equal to sum of change in internal energy of the system and work done by the system.
Let d Q amount of heat be supplied to the system resulting in a change of internal energy by dU. If the work done by the system is dW then, according to first law of thermodynamics,
dQ = dU + dW

Question 2.
Define molar specific heat at constant pressure, Cp and molar specific heat at constant volume, Cv .
Molar specific heat at constant pressure, Cp, is the amount of heat required to raise the temperature of 1 mol of a gas through 1K at constant pressure.
Molar specific heat at constant volume, Cv, is the amount of heat required to raise the temperature of 1 mol of a gas through 1K at constant volume.

Question 3.
A gas is found to obey PV2=Constant. The initial volume and temperature of the gas are Vi and Ti respectively. If the final volume of the gas is Vf, find its final temperature(Tf).
We have, P.V2 = constant
i.e. Pi V$$\mathrm{V}_{\mathrm{i}}^{2}$$ = Pf V$$\mathrm{V}_{\mathrm{f}}^{2}$$ …….(1)
For ideal gas , $$\frac{P V}{T}$$ = a constant
$$\therefore \frac{P_{i} V_{i}}{T_{i}}=\frac{P_{f} V_{f}}{T_{f}}$$ …….(2)
From equation (2), $$\frac{P_{i}}{f_{i}}=\frac{V_{f}}{T_{f}} \frac{T_{i}}{V_{i}}$$ …….(3)
Using equation (3) and (1),
$$\frac{V_{f}}{T_{f}} \cdot \frac{T_{i}}{V_{i}} V_{i}^{2}=V_{f}^{2}$$
⇒ Ti Vi = TfVf or Tf = $$\frac{T_{i} V_{i}}{V_{f}}$$

Question 4.
Distinguish between isothermal and adiabatic processes.
A thermodynamic process in which temperature remains constant is called isothermal process. The change in internal energy for this process is zero. A thermodynamic process in which there is no transfer of heat from system to surroundings is called adiabatic process. The change in internal energy in this process is non-zero.

Question 5.
1 kg of water initially at a temperature 27°C is heated by a heater of power 1KW. If the lid is opened, heat is lost at a constant rate of 200J/S. Find the time required for water to attain a temperature of 80°C with the lid open. (Specific heat of water = 4.2kJ/ kg/k)
Amount of water = 1 kg
Net heat supplied = (1000 – 200) = 800 W
Change in temperature = ∆ T
= (353 – 300) = 53K
∴ ∆ T = 53K
Time required, t = $$\frac{\mathrm{H}}{\mathrm{R}}$$
where H is the required energy and R is the rate of heat transfer
H = mC ∆T
= 1 × 4.2 × 103 × 53
= 2.226 × 105 J
R = 8 × 102 J/s
∴ t = $$\frac{2.226 \times 10^{5}}{8 \times 10^{2}}$$ = 2.783 × 102
∴ t ≈ 279 = 4 min 39 secs.

Question 6.
Obtain an expression for the coefficient of performance of a refrigerator.
A refrigerator transfers heat from a body at lower temperature to a body at higher temperature by doing work on it. If Q2 is the heat absorbed from body at temperature T2 (sink) and Q1 is the heat liberated by the refrigerator to a body at temperature T2 (source) then work done by there refrigerator,
W = Q1 – Q2
∴ The coefficient of performance,
α $$=\frac{Q_{2}}{W}$$
$$=\frac{Q_{2}}{Q_{1}-Q_{2}}$$
$$=\frac{T_{2}}{T_{1}-T_{2}}$$

Question 7.
The interior of the refrigerator is maintained at 5°C. If the surrounding temperature is 30°C, calculate the coefficient of performance of the refrigerator.
The co-efficient of performance of the refrigerator is given by
$$\alpha=\frac{Q_{1}}{w}=\frac{T_{2}}{T_{1}-T_{2}}$$
where T2 is the temperature inside the refrigerator and T1 is the temperature of the surroundings.
∴ $$\alpha=\frac{(30+273)}{\{(30+273)-(5+273)\}}$$
$$=\frac{303}{25}=12.12$$

Question 8.
What is a heat engine? Define efficiency of a heat engine.
Heat engine is a device that performs the conversion of heat energy to mechanical work through cyclic process. The efficiency of a heat engine is defined as the ratio of work done by the heat engine to heat absorbed per cycle. If a heat engine absorbs Q1 amount of energy from the source and dissipates Q2 amount of energy to sink, the efficiency η, is given by,
$$\eta=\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{Q}_{1}}=1-\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}$$.

Question 9.
The efficiency of a Carnot engine is 50% when the temperature of the sink is 300k, calculate the temperature of the source.
For a heat engine, efficiency η, is given by
$$\eta=1-\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}=1-\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}$$
where, T1 is the temperature of the source and T2 is temperature of the sink ∴

∴ T1 = (300) × (2)
T1 = 600 K
∴ Temperature of the source is 600K.

Question 10.
Prove that for an adiabatic process TVγ-1 constant.
We know that for an adiabatic process
PVγ = constant ……(1)
For an ideal gas, $$\frac{P V}{T}$$ = constant, C …….(2)
∴ $$P=\frac{C \cdot T}{V}=$$ …………(3)
using equation (3) in (1),
$$\left(\frac{C \cdot T}{V}\right) V^{\gamma}$$ = constant
i.e. T.Vγ-1 = constant.

### 1st PUC Physics Thermodynamics Five Marks Questions and Answers

Question 1
Derive the expression for work done by a gas in isothermal process.
For an ideal gas,
PV = n RT …….(1)
∴ For isothermal process. T = constant,
work done by an ideal gas,
dW = Pdv ……(2)
where dv is the change in volume of the gas.
using (1) in (2),
$$\mathrm{dW}=\frac{\mathrm{n} \mathrm{RT}}{\mathrm{V}} \mathrm{dv}$$
∴ Total work done,

or W = 2.303 n RT log10$$\left(\frac{v_{f}}{v_{i}}\right)$$

Question 2.
Derive an expression for work done by an ideal gas in adiabatic process.
Consider n moles of an ideal gas in container such that it is perfectly insulated (thermally) from the surroundings. Let the gas be compressed adiabatically by a small volume, dV.
For adiabatic process of ideal gas,
pvγ = constant ……..(1)
where P is the pressure of the gas, V is the volume of the gas and γ = $$\left(\frac{\mathrm{cp}}{\mathrm{cv}}\right)$$ is the ratio of specific heat capacities.
There fore, work done on the gas,
dW = P.dV ……..(2)
From equation (1),

where C is an arbitrary constant.
∴ dW =$$\frac{C}{V^{\gamma}}$$.dv
Total work done, W =

where, Vi is the initial volume of the gas and Vf is the final volume of the gas.

∴ From equation (3),
$$\therefore W=\frac{1}{(1-\gamma)}\left[P_{f} V_{f}-P_{i} V_{i}\right]$$ ………..(4)
For an ideal gas,
PV = nRT ….(5)
where n is the no of moles, R is the Universal gas constant and T is the Temperature of the gas.
Using (5) and (4),
$$W=\frac{1}{(1-\gamma)}\left[n R T_{f}-n R T_{i}\right]$$
where Tf is the final temperature and Ti Is the initial temperature
$$\therefore W=\frac{n R}{1-\gamma}\left(T_{t}-T_{1}\right)$$

Question 3.
What is Carnot engine? Describe the different parts of the Carnot heat engine.
Carnot engine is an ideal heat engine which has ideal gas as working substance.

The above diagram shows the working of a carnot heat engine. It consists of the following parts:
1. Source :
It is a source of heat maintained at constant temperature T1
2. Sink:
It is a sink for heat maintained at constant temperature T2.
3. Working substance:
Ideal gas contained in a cylinder with perfectly non-conducting walls and perfectly conducting bases.

Question 4.
Describe the working of Carnot engine and mention the expression for its efficiency.
A carnot engine is a reversible heat engine operating between two temperatures,

The steps involved in a carnot cycle are explained as follows.
Step 1:
Isothermal expansion of gas from state (P1, V1, T1) to state (P2, V2, T1) as shown in the figure.
Step 2:
Adiabatic expansion of gas from (P2, V2, T1) to (P3, V3, T2)
step 3:
Isothermal compression of gas from (P3, V3, T2) to (P4, V4, T2)
step 4:
Adiabatic compression of gas from (P4, V4, T2) to (P1, V1, T1)
These steps constitute one carnot cycle. The process is repeated thereafter.
The efficiency of carnot engine, η is given by
$$\eta=\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{Q}_{1}}=\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{T}_{1}}$$
$$\therefore \eta=1-\frac{T_{2}}{T_{1}}$$

1st PUC Physics Thermodynamics Numerical Problems Questions and Answers

Question 1.
Two moles of an ideal gas undergoes Isothermal expansion when heat is supplied to It. If the gas is at room temperature (298K) and the final volume is thrice the Initial volume, find the total heat supplied.
No of moles, n = 2
temperature, T = 298K
let initial volume = Vi
∴ final volume, Vf = 3Vi
The work done for isothermal process is given by,
W = 2.303 nRT log10 $$\left(\frac{v_{2}}{v_{1}}\right)$$
where V2 is the final volume and V1 is the initial volume.
Here $$\frac{v_{2}}{v_{1}}$$ = $$\frac{v_{f}}{v_{i}}=3$$
∴ W = 2.303 × 2 × 8.314 × 298 × $$\log _{10}^{3}$$
W = 5.44kJ
Since, for isothermal process, Heat supplied is equal to work done by the gas, the total heat supplied is 5.44 kJ.

Question 2.
An Ideal gas under goes adiabatic expansion to twice original volume. If the gas was initially at 40°C, calculate its temperature after the process.
[Take 1 = $$\frac{5}{3}$$]
Let the initial volume, V = V0
Final volume, V = 2.V0
Initial temperature, Ti = 40°C
= 40 + 273
= 313K
Let final temperature = Tf
TVγ-1 = constant
∴ Ti $$V_{i}^{\gamma-1}$$ = Tf $$V_{f}^{\gamma-1}$$
∴(3/3).$$\left(V_{0}\right)^{\frac{5}{3}-1}$$ =Tf $$\left(2 V_{0}\right)^{\frac{5}{3}-1}$$
∴ $$\mathrm{T}_{\mathrm{f}}=\frac{3 / 3}{2^{\frac{2}{3}}}$$
∴ Tf = 197.18 K
Therefore, on adiabatic expansion, the temperature of the gas drops to 197.18 K.

Question 3.
One mole of an ideal gas is supplied 2KJ of heat. If the temperature of the gas rises from 0°C to 200°C, calculate work done by the gas and change in its internal energy if

1. The process Is isobaric
2. The process is isochoric.

number of moles, n = 1
Initial temperature, Ti = (0 + 273)
= 273 K
Final temperature, Tf = (200 + 273)
= 473 K
Net heat supplied, H = 2000 J

1. Isobaric process :
work done, W = P – ∆ V
or W = n R∆T
∴ W= 1 × 8.314 × (473-273)
∴ W= 1662.8J = 1.66KJ
∴ Change in internal energy,
∆U = H – W= 2000 – 1662.8
= 337.2J

2. isochoric Process:
Work done in an isochoric process , W =0. Therefore change in internal energy
∆U = H = 2 kJ

Question 4.
One mole of an ideal gas undergoes a cyclic process as shown in the figure.

Calculate the work done by the gas in each of the process. A-B, B-C, and C-A. What is the network done by the gas at the end of one cycle?
1. The process A – B is isobaric. Therefore work done is given by,
WAB = P. ∆V
= 200 × 103 × (5-3) × 10-3
= 400J
∴ Also, work done is given by area under the curve, AB.
Area = (AB) × (200) × 103
= (2 × 10-3) × 200 × 103 =400
∴ Area = WAB.

2. Work done in the process BC,
WBC = Area under the curve BC = 0

3. Work done in process CA,
WCA= – (Area under the curve CA)
∴ WCA = – [ $$\frac{1}{2}$$ × (100 + 200) × 103 × (2 × 10-3)]
∴ WCA = – 300 J
∴ Total work done by the gas at the end of cycle,
Wtotal = (400 + 0 – 300) J = 100 J

Question 5.
Two samples A and B of same gas have same Initial pressure and volume. A undergoes an Isothermal expansion and B is subjected to adiabatic expansion. If the final volume in both cases is double the initial volume and work done In both the cases are same, then prove that

Let the initial pressure, pi = p1
initial volume, Vi = V1
final volume, Vf = V2
= 2V1
Work done in isothermal process,
Wiso = 2.303 RT log $$\left(\frac{V_{2}}{V_{1}}\right)$$
Since PV = RT (1 mol of gas),
Wiso = P1 V1 ln2 ……(1)
wad = $$\frac{1}{\gamma-1}$$ [P1 V1 – P2 V2]
$$=\frac{P_{1} V_{1}}{\gamma-1}\left[1-\frac{P_{2} V_{2}}{P_{1} V_{1}}\right]$$
P1 $$V_{1}^{\gamma}$$ = P1 $$V_{2}^{\gamma}$$ = K (constant)

Since work done is same for both processes,
∴ $$P_{1} V_{1} \ln 2=\frac{P_{1} V_{1}}{\gamma-1}\left(1-2^{1-\gamma}\right)$$ ……….(3)
Re-arranging and simplifying (3),
21-γ – 1 = (1 – γ) ln2.

Question 6.
Two moles of a monoatomic gas is taken through a cyclic process starting from A as shown in the figure. The
volume ratios are $$\frac{v_{B}}{V_{A}}$$ =2 and $$\frac{v_{D}}{V_{A}}$$= 4. If the temperature TA at A is 27°C, calculate,

1. Temperature of the gas at point B
2. heat absorbed or released by the gas in each process
3. the total work done by the gas during the complete cycle.

1. From the figure, for the process A – B, we have
V = k. T
where V = Volume of the gas
T = Temperature of the gas
k = proportionality constant
∴ $$\frac{\mathrm{v}}{\mathrm{T}}$$ = k
∴ $$\frac{V_{A}}{T_{A}}=\frac{V_{B}}{T_{B}}$$ and $$\frac{V_{B}}{V_{A}}=\frac{T_{B}}{T_{A}}$$
∴ TB = TA × 2
TB = (27 + 273) × 2 = 600K

2. Process A – B :
HAB = nCp∆T
HAB = 2 × [$$\frac{5}{2}$$ R] × (600 – 300)
HAB = 1500 R                 (Heat absorbed)

3. Process B – C:
HBC = 2.303nRTB log $$\frac{V_{D}}{V_{B}}$$
VD = 4VA
VB = 2VA
∴ $$\frac{V_{D}}{V_{B}}$$ = 2
∴ HBC = 2.303 × R × 600 × log 2
∴ HBC ≃ 831.6 R

4. Process C – P :
HCD = nCv ∆T
HCD = 2 × [$$\frac{3}{2}$$ R] × (-300)
∴ HCD = – 900 R

5. Process D – A :
HDA = 2.303 × nRTA × log $$\frac{V_{A}}{V_{D}}$$
∴ HDA = 2.303 × 2 × R × 300 × log 4
∴ HDA ≃ – 831.6 R
(3) From first law of thermodynamics,
∆H = ∆U + ∆W
where,
∆ H = net heat supplied to the system
∆ U = Net change in internal energy and
∆W = total work done by the system.
For a cyclic process, ∆U = 0
∴ ∆H = ∆W
∴ Total work done by the system,
∆W = HAB + HBC + HCD + HCA
= 1500 R + 831.6 R – 900 R – 831.6R
∴ ∆W = 600 R
Therefore, total work done by 2 mols of the given monoatomic gas is 600 R.

Question 7.
One mole of an Ideal gas Initially at 300K Js expanded Isothermally so that Its volume increases 5 times. It is then heated at constant volume to restore initial pressure. If the net heat supplied Is 29KJ find γ for the given gas.
Total heat supplied,
Htot = 2.303RT1log$$\frac{V_{2}}{V_{1}}$$ + Cv ∆T
= 2.303 × 8.314 × 300 × log5 + $$\frac{C_{v}}{R}$$ (R∆T)
Here, Htot = 29000 J
R∆T = ∆PV = P3 V3 – P2 V2
P3 = P1
V3 = 5.V1 = V2
P2 = $$\frac{P_{1}}{5}$$
∴ ∆PV = P1.5V1 – $$\frac{P_{1}}{5}$$ . 5V1
=4 P1 V1 = 4. R. T1
$$\therefore \frac{C_{v}}{R}=\left(\frac{29000-2.303 \times 8.314 \times 300 \times \log 5}{4 \times 8.314 \times 300}\right)$$
∴ $$\frac{C_{v}}{R}$$ = 2.505
∴ Cv ≃ 2.5R
∴ Cp ≃ 3.5R                   (Cp = Cv + R)
∴ γ = $$\frac{C_{p}}{C_{v}}=\frac{3.5}{2.5}=1.4$$

Question 8.
A Carnot engine operates between 900K and 300K.The efficiency of the engine has to be increased to 80%

1. By how much should the temperature of the source alone be increased?
2. By how much should the temperature of the sink alone be lowered?

Efficiency of a carnot engine is given by,
$$\eta=1-\frac{T_{2}}{T_{1}}$$
where T2 = Temperature of the sink
T1 = Temperature of the source
1.  0.8 = 1 – $$\frac{300}{T_{1}}$$
∴ $$T_{1}=\frac{300}{1-0.8}=\frac{300}{0.2}=1500 \mathrm{K}$$
∴ Temperature of the source alone has to be increased by 600K.

2. 0.8 = 1 – $$\frac{T_{2}}{900}$$
∴ T1 = 900 (1-0.8) = 180 K
∆T = 300-180 = 120K
Therfore temperature of the sink alone has to be lowered by 120K.

Question 9.
The efficiency of a Carnot’s engine is 20%. When the temperature of the sink Is reduced by 100°C, efficiency increases to 40%. Find the initial temperature of the source and sink.
Let the initial temperature of source, T1i = T1 and that of the sink be T2i = T2
∴Final temperature of sink, T2f = T2 – 100
Efficiency of the Carnots engine is given T2
by, $$\eta=1-\frac{T_{2}}{T_{1}}$$
where, T2 = Temperature of sink
T1 = Temperature of source
∴ 0.2 = 1- $$\frac{T_{2}}{T_{1}}$$ …….(1)
$$0.4=1-\frac{\left(T_{2}-100\right)}{T_{1}}$$ ………..(2)
Re – arranging equation (2),
$$0.4=1-\frac{T_{2}}{T_{1}}+\frac{100}{T_{1}}$$ …….(3)
Using equation (1) in equation (3),
$$0.4=0.2+\frac{100}{T_{1}}$$
∴ $$\mathrm{T}_{1}=\frac{100}{0.4-0.2}=500 \mathrm{K}$$
On substituting in (1) we have,
$$0.2=1-\frac{100}{T_{1}}$$
∴ T2 = 500 (1 – 0.2)
= 400 K
∴ Initial temperatures of source and sink are respectively 500K and 400 K.

Question 10.

1. Two Carnot engines are working between 400K and 300K, 300K and 200K respectively. Find the ratio of the efficiency of first system to the second system.
2. If 200 J of heat is supplied to a carnot engine at 400K and 50J of work Is done by the system, find the temperature of the sink and efficiency of the engine.

1. System 1:
Temperature of source, T1 = 400K
Temperature of sink, T2 = 300K
∴ efficiency of the engine is,
$$\eta_{1}=1-\frac{T_{2}}{T_{1}}=1-\frac{300}{400}=0.25$$

System 2:
Temperature of source, T1 = 300K
Temperature of sink, T2 = 200K
∴ efficiency of this engine is,
$$\eta_{2}=1-\frac{T_{2}}{T_{1}}=1-\frac{200}{300}=0.33$$
∴ Ratio of efficiencies is,
$$\frac{\eta_{1}}{\eta_{2}}=\frac{0.25}{0.33}=\frac{1 / 4}{1 / 3}=\frac{3}{4}=0.75$$

2. Temperature of source, T1 = 400 K
Hear supplied to engine, Q1 = 200 J
Work done by the engine, W = 50 J
∴ Heat rejected of the engine,
Q2 = Q1 – W = 150 J
∴ efficiency of the engine is,
$$\eta=1-\frac{Q_{2}}{Q_{1}}=1-\frac{150}{200}=1-\frac{3}{4}=0.25$$
Let temperature of sink, T2 = T
∴ $$\eta=1-\frac{T_{2}}{T_{1}}=1-\frac{T}{T_{1}}$$
∴ T = T1 (1 – η) = 400 (1 – 0.25)
= 300 K

Question 11.
A carnot engine has an efficiency of $$\frac{1}{2}$$. On increasing the temperature of the sink by 100°C, the efficiency drops to $$\frac{1}{3}$$. Find the temperature of the source and sink in the original state. By what amount should the source temperature be increased to restore the original efficiency?
Let temperature of source = T1
let temperature of sink = T2
∴ efficiency of the engine, $$\eta=1-\frac{T_{2}}{T_{1}}$$
$$\therefore 1-\frac{T_{2}}{T_{1}}=\frac{1}{2}$$ ……(1)
on increasing T2 by 100°C, efficiency,
$$\eta=1-\frac{\left(T_{2}+100\right)}{T_{1}}=\frac{1}{3}$$
$$\therefore\left(1-\frac{T_{2}}{T_{1}}\right)-\frac{100}{T_{1}}=\frac{1}{3}$$ …..(2)
substituting (1) in (2),

Thus initial temperature of source and sink are 600K and 300 K respectively After raising temperature of sink,
T2f = 300 + 100 =400 K
To restore original efficiency $$\left(\eta=\frac{1}{2}\right)$$ let the source temperature be raised by ‘T’ K.

Therefore, the temperature of the source should be raised by 200 K to restore original efficiency.

## Karnataka 1st PUC Political Science Question Bank Chapter 6 Legislature

### 1st PUC Political Science Legislature One Mark Questions and Answers

Question 1.
What is the root word of parliament?
The French word ‘parlor’ is the root word of parliament.

Question 2.
State the primary function of the legislature.
Framing of law is the primary function of the legislature.

Question 3.
Which are the two houses of the Indian Parliament?
Lok sabha and Rajyasabha.

Question 4.
Name the two houses of the United Kingdom.
House of Lords and House of Commons.

Question 5.
Who is an integral part of the Indian Parliament?
President is an integral part of the Indian parliament.

Question 6.
Who summons the parliament of India?
The President summons the Parliament of India.

Question 7.
Who can promulgate ordinance?
President can promulgate an ordinance.

Question 8.
What is the maximum gap between two sessions of parliament?
6 months is the maximum gap between two sessions of parliament.

Question 9.
Which lists are followed by the parliament to make laws?
Union lists are followed by the parliament to make laws.

Question 10.
Who presides over the joint sitting of the two houses?
President presides over the joint sitting of the two houses.

Question 11.
Which is popularly elected house of Indian parliament?
Lok Sabha is a popularly elected house of Indian parliament.

Question 12.
Who presides over the proceeding of the Loksabha?
Speaker presides over the proceeding of the Loksabha.

Question 13.
Who elects the speaker of Lok Sabha?
Members are Lok Sabha elects the speaker of Lok Sabha.

Question 14.
What is the strength of Lok Sabha?
The strength of Lok Sabha is 543 + 2.

Question 15.
How many seats are reserved for the Union Territories in Lok Sabha?
Under article 81, not more than 20 Lok Sabha members represent the Union Territories.

Question 16.
How many members of the Anglo-Indian community can be nominated to Lok sabha? or How many nominated members are there in Lok Sabha?
2 members represent by Anglo-Indian Community.

Question 17.
What is the term of a Loksabha member? or What is the normal tenure of the Lok Sabha?
5 years.

Question 18.
When the term of the Lok Sabha can be extended?
During the National emergency, the term of the Lok Sabha can be extended.

Question 19.
Where the money bill is introduced?
The money bill is introduced in Lok Sabha.

Question 20.
What is the total strength of the Rajya Sabha?
The total strength of the Rajya Sabha is 250.

Question 21.
How many members are nominated by the President to Rajya Sabha?
12 members are nominated by the President to Rajya Sabha.

Question 22.
Who presides over the Rajya Sabha?
The Vice president presides over the Rajya Sabha.

Question 23.
What is the tenure of the members of the Rajya Sabha?
The tenure of Rajya Sabha is 6 years.

Question 24.
Which union territory does not have representation to the Rajya Sabha?
Lakshadeep does not have representation to the Rajya Sabha.

Question 25.
Which house of the parliament has exclusive power to create new All India services?
Rajya Sabha can create new All India services.

Question 26.
Which state has a maximum strength in Lok Sabha?
Uttar Pradesh has maximum strength in Lok Sabha.

Question 27.
What is the term of members of the Legislative Assembly?
The term of the legislative assembly is 5 years.

Question 28.
What is the term of members of the legislative council?
6 years.

Question 29.
What is the maximum strength of the legislative council?
The maximum strength of the council shall not exceed and 1/3rd of the total members in the assembly.

Question 30.
Who dissolves Lok Sabha?
President dissolves Lok Sabha.

Question 31.
What are the three organs of the government?
Legislature, Executive and Judiciary.

Question 32.
What type of legislature is existing at the centre in India?
Bicameral legislature.

Question 33.
Give the Meaning of legislature.
The legislature is one that makes the laws. It expresses the will of the state.

Question 34.
What is the strength of Lok sabha?
545.

Question 35.
What is the term of Rajya sabha?
Rajyasabha is a permanent chamber in the term of office of the members in 6 years.

Question 36.
Who is the speaker of Lok sabha today?
Meera Kumar.

Question 37.
Who is the chairman of Rajya sabha?
The vice president of India Hamid Ansari.

Question 38.
Which state sends the largest number of representatives to Lok sabha?

Question 39.
What is the age limit to become a member of Lok sabha?
25 Years and above.

Question 40.
Who is the Custodian of the finance of the country?
Loksabha is the custodian of the finance of the country.

Question 41.
Explain the meaning of the theory of separation of powers.
In this, all the 3 branches of government are separate and independent.

Question 42.
Who wrote the book “The spirit of laws”?
French political thinker Montesquieu.

Question 43.
What are the functions of the legislature?
Legislature is the law-making organ of government.

Question 44.
What is the other name for the legislature?
Parliament.

Question 45.
Who elects the president of India?
Electoral College.

Question 46.
Which house is called a permanent house?
Rajya Sabha.

Question 47.
Who is the Presiding officer of the Rajya sabha?
Vice President of India.

Question 48.
Give any other two names of Lok sabha.
House of Representatives and lower house.

Question 49.
Who is the presiding officer of the Lok sabha?
Speaker.

Question 50.
Name the presiding officer of the VidhanaSabha.

Question 51.
What is an amendment?
It is a process of amending, altering or changing a Law.

Question 52.
Mention the stages involved in the legislative procedure.
There are five stages and three readings.

### 1st PUC Political Science Legislature Two Mark Questions and Answers

Question 1.
What is the legislature?
The legislature is the most important organ of government. It is the law-making organ of government. It expresses the will of the state.

Question 2.
Name the two houses of the American congress.
The two houses are the House of Representatives the lower house and the Senate the upper house.

Question 3.
Which part of the constitution deals with the Indian Parliament?
Part V of the constitution deals with the Indian parliament.

Question 4.
Write the qualifications to become a member of Loksabha.

1. He must be a citizen of India.
2. He must not be less than 25 years of age.
3. Must be eligible as voters.
4. Must process any other qualification as may be prescribed by parliament.

Question 5.
Write the two important functions of the speaker.

1. To preside over the session.
2. To maintain law and order in the session.

Question 6.
What are the functions of the Deputy speaker?
To preside over the house and perform all the functions in the absence of the speaker.

Question 7.
What is the quorum of the house?
To conduct the proceedings of the house 1/10th of the attendance is essential. It is called a quorum.

Question 8.
Who can create or abolish the legislative councils of a state?
Parliament can create or abolish the legislative councils by the recommendation of the concerned state Assembly.

Question 9.
What are the electoral powers of the Vidhan Sabha?
Electing the president and speaker of the assembly is the electoral power of the Vidhan Sabha.

Question 10.
Who dissolve the legislative assembly?
Governor dissolve the legislative assembly.

Question 11.
Mention the three types of legislative proceedings.

Question 12.
What are the responsibilities of the joint sitting of the house?
In case of disagreement between the two houses on a matter of legislation, it is resolved by a Joint Sitting of both the houses presided over by the Speaker. In a Joint Sitting, Loksabha would emerge triumphant because the decisions are taken by a majority of the total number of members of both the house present and voting in which the numerical superiority of Loksabha prevails.

Question 13.
Write the two function of the speaker of the Vidhana Sabha.

1. To preside over the session.
2. To maintain law and order in the session.

Question 14.
What are the qualifications of a member of Rajyasabha?
Qualifications of members of the Rajyasabha:

1. Must be a citizen of India.
2. “Should have attained the age of 30 years.
3. Owe allegiance to the constitution.
4. Must not hold any office of profit under the Government – National, Regional or local.
5. Should not be insolvent or man of unsound mind.
6. Must not have acquired the citizenship of a foreign state.

Question 15.
Write the significance of the legislature.
It is the most important organ of the government. The will of the state is expressed through the legislature. In modem representative democracy it acts as a mirror of the nation.

Question 16.
What type of legislature is existing in the centre in India?
The legislature with two houses is called bicameralism. There will be upper and lower house. Different nations call the two houses by different names and the composition of house differs.

Question 17.
Write a note on the speaker of Loksabha?
Speaker is the presiding officer of Loksabha the conducts all proceedings of Loksabha except the resolution which seeks his removal. He is the spokesman of the House.

Question 18.
Write a note on the Indian parliament.
The constitution of India lays down that there shall be a parliament and two houses to known as the house of people (The Loksabha) upper house is Rajyasabha. In other words, we have a bicameral legislature in India.

Question 19.
Who summons the parliament of India?
The president is empowered to summon each house of parliament to meet at such a time and place as he thinks fit. The time interval between two sessions of parliament should not be more than six months.

Question 20.
Mention the methods by which the legislature controls the executive.
The legislature controls the executive by way of discussions and debates during the Question Hour, Adjournment Motion, Zero Hour, Cut motion. Call-attention motion etc. The soundest way of controlling the executive is by way of moving the No-confidence motion. If the executive fails to win the support of Lok sabha, they must step down.

Question 21.
What is an electoral college?
The elected by the members of Loksabha and Rajyasabha along with the members of Legislative Assemblies of the states from the Electoral College.

Question 22.
Briefly explain the composition of Rajyasabha.
Composition of the Rajyasabha:
The Rajyasabha consist of 250 members, of which 238 members are elected from the states through indirect elections by a single transferable vote. 12 members are nominated by the president from among those with special knowledge and experience in the field of art, literature, science, social service, etc.

Question 24.
Name the three sessions of the Indian parliament.
The parliament meets in three sessions: the Budget session (February-May), the Monsoon session (July-September) and the winter session (November/December).

Question 25.
What is delegated legislation?
The rules and the regulations made by the executive for the formulation and implementation of law in the parliament are called delegated legislation.

Question 26.
What is No-confidence motion?
It is a resolution passed by the lower house of the legislature indicating loss or lack of faith in the Prime Minister and Council of Ministers. Hence the prime minister and his council will have to resign.

### 1st PUC Political Science Legislature Five Mark Questions and Answers

Question 1.
Write the meaning and significance of the legislature.
Legislature is the most important institution in a democratic system of government. It is the law-making organ of the government. It brings the will of the people into laws. The importance of the legislature is increasing because the executive and judiciary have to work on the basis of the laws made by the legislature.

The legislature is the primary and most powerful organ of the government. It controls the executive especially in the parliamentary form of government. It sanctions the budget and controls the national finances. In a parliamentary system, the real executive is chosen from and is also controlled by it.

Question 2.
Write a short note on the Indian parliament.
The constitution of India lays down that there shall be a parliament and two houses to be known as the house of people (The Loksabha) upper house is Rajyasabha. In other words, we have a bicameral legislature in India.

Question 3.
Write a briefly note on the composition of Lok Sabha.
The members of Lok Sabha are elected by the people. All adult citizens unless disqualified for other reasons have the right to select their representatives. Qualifications to become the members are must be a citizen of the country and must have attained the minimum age fixed by the constitution. The term of office is five years. Speaker is the presiding officer. He is elected from among the members of the house.

Question 4.
Describe the composition of Rajya Sabha.
In federal representation the state is important. In India, the members of the Rajyasabha are indirectly elected for 6 years but l/3rd of them will retire every 2 years. The Rajyasabha has 250 members out of whom 12 are nominated by the president for their contributions to science, literature, art and social sendee. The remaining 238 members are indirectly elected by the state legislative assemblies through a system of proportional representation.

Question 5.
Discuss the powers and functions of Loksabha.
1. Legislative functions:
The power of Loksabha extends to all subjects falling under the Union List and the Concurrent list. In case of emergency in operation, its power also extends to the State list as well. No bill can become a law without the consent of Loksabha. The Loksabha has equal powers of law-making with Rajyasabha except on financial matters where the supremacy of Loksabha is total.

In case of disagreement between the two houses on a matter of legislation, it is resolved by a Joint Sitting of both the houses presided over by the Speaker. In a Joint Sitting, Loksabha would emerge triumphant because tire decisions are taken by a majority of the total number of members of both the house present and voting in which the numerical superiority of Loksabha prevails.

2. Financial functions:
On financial matters, the supremacy of the Loksabha is total and complete. “One, who holds a purse, holds power,” said James Madison. By establishing its authority over the national purse, Loksabha establishes its authority over the Rajyasabha. It is expressly stated that the Money bill can originate only in the house of people. Regarding budget, Loksabha being a representative house enjoys total authority. Loksabha’s position on financial matters is such that the demands for grants are placed only before the Loksabha.

3. Control over the executive:
The Loksabha enjoys direct control over the executive because; the executive is directly responsible to the lower house and stays in office as long as it enjoys the confidence of the house. The Loksabha not only makes laws but also supervises the implementation. The lower house being a debating house, the members are free to seek information from the executive and raise questions and seek clarifications. The members can effectively seek information from the government by way of discussions and debates during the Question Hour (seek clarification), the Adjournment Motion (raises issues of national importance), the Zero Hour, the Cu motion, the Call-attention motion, etc. The soundest way of controlling the executive is by way of moving the No-confidence motion, if the executive fails to win the support of Lok sabha, they must step down.

4. Constituent functions:
The Loksabha shares equal powers in regard to amending provisions of the constitution. An amendment may be initiated either in the Rajyasabha or Loksabha and must be passed by a 2/3 majority in both the houses present and voting. The agreement of Rajyasabha is compulsory for the success of the constitutional amendment.

5. Electoral functions:
The Loksabha and Rajyasabha elect the highest constitutional- functionaries such as the President and the Vice-president. The President is elected by the members of Loksabha and Rajyasabha along with the members of Legislative Assemblies of the states. The Vice-president is elected by members of Loksabha and Rajyasabha.

6. Judicial functions:
The Loksabha acts as a judge in the impeachment of the President. Either house can prefer the charge of impeachment. If Rajyasabha prefers the charge, Loksabha investigates the charge and if it passes a resolution by a 2/3 majority of the total membership of the house. President stands impeached from the office. The Loksabha also sits in Judgement, along with the Rajyasabha, in removing high constitutional functionaries such as the Comptroller and Auditor General, The Chief Vigilance Commissioner, the Chief Election Commissioner, etc.

Question 6.
Explain the powers and functions of Rajyasabha.
1. Legislative functions:
On legislative matters, the Rajyasabha enjoys powers with the Loksabha except in case of a Money Bill or Financial Bill. Non-money Bill can originate in Rajyasabha and must get a 2/3 majority in the House and then proceeds to the Loksabha. The approval of both Houses is essential for a bill to become a law. In case of disagreement between the two Houses on a bill, both the houses sit oh a joint sitting presided by the speaker and the deadlock is resolved by a majority of the total number of members of both the houses present and voting.

2. Financial functions:
On the financial front, the Rajyasabha virtually has no powers. The procedure to deal with Money bills clearly states that a Money bill or financial bill cannot originate in the Rajyasabha. The Rajyasabha may discuss and suggest changes but have no right to reject or amend a Money Bill. It is left to the Loksabha to accept or reject its recommendations. In case, the Rajyasabha does not send back a Money bill back to the Loksabha within 14 days from the date of receipt of the bill, the bill is deemed passed, in the original form, by both the Houses.

3. Control over the executive:
The Rajyasabha’s hold over the executive is very minimal because the executive is not directly responsible to the upper house. However, it can seek information, and make clarifications on various policy matters. On issues of national and local importance, the members can grill the executive during debates and discussions during the Question Hour, the Adjournment motion, the Zero Hour, the Cut-Motion. Call- attention Motion etc.

4. Constituent functions:
The Rajyasabha enjoys full powers with Loksabha in executing constituent functions. An amendment to provisions of the constitution can be initiated in either House of the parliament and must be passed by a 2/3 majority in both the Houses present and voting. If Rajyasabha does not pass an amendment bill, the amendment Bill stands defeated. In some special provisions apart from the 2/3 majority in both house of parliament and ratification by not less than 1/2 of the states is necessary.

5. Electoral functions:
The Rajyasabha shares the privilege of electing the highest constitutional functionaries, the President and Vice-president. The President is elected by an electoral college consisting of the members of parliament along with the members of the State legislative assemblies. The members of both the houses of parliament elect the Vice-president.

6.Judicial functions:
The impeachment move against the President may be initiated in either house of the parliament. If Loksabha prefers the charge, Rajyasabha investigates the charge and passes a resolution by a 2/3 majority of the total membership of the house, then the President stands impeached. There is no need for an impeachment against the Vice-president who may be removed by a resolution of Rajyasabha passed by the majority of its members and consented to it by Loksabha. The Rajyasabha also participates in the removal of the highest constitutional functionaries such as the Chief Election Commissioner, the Vigilance Commissioner, etc.

7. Miscellaneous functions:
The Rajyasabha performs other functions as well

1. By a resolution, Rajyasabha can create one or more All India Services.
2. Continuation of emergency beyond the specified time must come before the Rajyasabha and Loksabha.
3. Orders made by the President suspending enforcement of fundamental rights is required to be laid before the Rajyasabha and Loksabha.
4. According to Article 249, the Rajyasabha by a resolution can ask the parliament to legislate on certain subjects in the State list.

Question 7.
Discuss the financial powers of both the houses of Parliament.
Financial functions are a very important function of the modem legislature. The national finances are controlled by the legislature. The legislature enacts the annual budget. The entire financial administration of the country comes under its general supervision direction and control.

Question 8.
Write a short note on the composition of the Vidhana Sabha.
There is a legislative assembly for every state. The number of members depends upon the population of the state. But it can not have less than 60 and more than 500 members. The members are chosen by direct election by people of the state. The governor has been given the power to nominate one or two members of the Anglo Indian community legislative assembly is five years.

Question 9.
Write a short note on the composition of the Vidhana Parishad.
The Composition of the legislative council is as follows:

• 1/3 – of the members are elected from the local bodies such as municipalities and district boards;
• 1/3 – of the members are elected from members of the Legislative Assembly;
• 1/12 – of the members are elected by the graduates from graduate constituencies.
• 1/12 – of the members are elected from teacher’s constituencies consisting of secondary- school, college and university- teachers; and
• 1/6- of the members are nominated by the Governor from the fields of science, art, social service, the co-operative movement, literature, etc.

Question 10.
Explain the powers and functions of Vidhana Sabha.
1. Legislative Functions:
The Legislative Assembly is entitled to pass laws on all subjects that fall under the state list such as police, public health, education, local-self governments, etc. Without the consent of the Vidlianasabha, no bill can become a law. Though the Vidhanasabha is competent enough to make laws on subjects listed in the concurrent list along with the central legislature, if parliament passes a law contained in the concurrent list, the legislative assembly is not competent to pass a law on the same subject.

However, some bills require the previous permission of the President before they are introduced in the state legislature. In case of breakdown of constitutional machinery in a state or when the proclamation of emergency is in operation, parliament has the power of making laws on matters falling under the state list. In case of a conflict between state law and the law of the parliament, the law of the parliament shall supreme.

2. Financial Functions:
The Vidhanasabha enjoys total control over the finance of the state. No new tax can be levied or collected without the consent of the Vidhanasabha. The authority of the Vidhanasabha over Vidhanaparishad is strengthened by the fact that a Money bill or Financial bill can only originate in the Vidhanasabha and the Vdhanaparishad can at the most delay it by 14 days but cannot reject or amend the Bill. The annual income-expenditure statement of the year the Budget must get the approval of the Vidhanasabha. Every- year during March-April, the beginning of the financial year, it is the responsibility of the government to place the budget before the house and seek its approval.

3. Control over the Executive (Administration):
The Vidhanasabha enjoys direct control over the administration, as the executive is directly, collectively, responsible to the Vidhanasabha and remains in office as long as they enjoy the confidence of the house. The members of the house can seek information from the government through questions and supplementary questions. It is the responsibility of the ministers to clarify points raised by members and give a satisfactory explanation.

Any attempt to lie or mislead the house is considered an offence against the house punishable under Contempt of the House. The debating occasions such as the Question Hour, Adjournment motion, the Emergency Adjournment motion, the Zero Hour, the Cut motion, the Call-attention motion keeps the executive under constant check and tire executive must be alert and ready with answers. However, ministers can ask for time to answer questions. The most effective weapon in the hands of the Vidhana sabha is the No-confidence motion, which can bring down a government.

4. Electoral Functions:
The members of the Vidhana sabha along with the members of the parliament constitute an electoral college to elect the President of India. They also take part in electing the members of Rajyasabha and also of the members of Legislative council.

5. Constituent Functions:
The state legislative assembly takes part in amending a few constitutional provisions. The Assembly does not initiate any amendment to constitution neither does it has such powers. But ratification of at least not less than half of the State legislative assemblies is necessary1 for amending certain provisions of the constitution. For instance, if there has to be an amendment made to electoral procedure of electing president of India then it has to be ratified by 1/2 of the states, which in turn is done by state legislative assemblies.

Question 11.
Discuss the powers and functions of Vidhana Parishad.
The powers and functions of the Vidhana Parishad are as follows:
1. Legislative functions:
No bill can become a law unless agreed upon by both the Legislative council and the Legislative assembly. Any bill other than a Money Bill can originate in the upper house. The lower house on passing a bill sends it to the upper house for consideration and recommendations.

If the upper house rejects the bill after keeping it with them for 120 days, the low er house can again send the bill. And if the Legislative council does not give approval the second time, the bill is deemed to have been passed by both the Houses of the state legislature.

2. Financial functions:
The Legislative council does not enjoy any position of power on financial matters. It is expressly stated in the constitution that a Money bill or financial bill cannot originate in the upper house. The Vidhanasabha on passing a Money bill sends it to the Vidhana Parishad for its recommendations, but it is to the lower House to accept or reject the recommendations. The Vidhana Parishad can neither correct nor amend a Money Bill.

If a Money Bill is not returned to the Vidhana sabha within 14 days of receipt of the bill, the bill is deemed passed by both the houses in its original form. On Financial matters, the Vidhana Parishad is sub-ordinate to the Vidhana Sabha. The Vidhana Parishad can only discuss the Budget but cannot make changes.

3. Control over the executive:
The influence of the Legislative council over the executive is minimal because the executive is directly responsible to the Legislative assembly. It has the right to seek necessary information and records for suggestions but cannot directly control the administration. The upper House can ask questions and supplementary questions during which ministers are duty-bound to provide satisfactory answers. The Vidhana Parishad can not initiate a motion of No-confidence to bring down the government as it is a nominated House.

Question 12.
Discuss the powers and functions of the speaker of the Loksabha.
The presiding officer of Loksabha is the Speaker who is elected from among the members along with the Deputy Speaker and stays in office till the life of the House i.e., 5 years. His primary task is to protect the dignity and decorum of the House and to see that the proceedings of the House are conducted in an orderly and a focused manner. He is the principal spokesperson of the House and must be impartial and even-handed in dealing as the custodian of the House.

In order to ensure impartiality, the speaker resigns his party membership in the election. The Deputy speaker discharges the duty even the office of the speaker falls vacant due to resignation, death or removal by a 2/3 majority of the total membership of the House or in the absence of the speaker. The salary of the speaker is determined by the parliament from time to time. The Speaker’s position in the House is one of dignity and authority.

• All orders of the house are executed through the Speaker
• Communication from the President is made known through the Speaker.
• It is the power of the speaker to declare whether a bill is a money bill or not.
• He enjoys the authority of interpreting the Rules of procedure and has the power to vote except in case of a tie.
• No member can speak in the House without the permission of the speaker and it is the speaker who fixes the time limit for speech.
• He presides over the Joint sittings of the parliament.
• During discussions, the members must address the Chair.
• In case of a tie, the speaker has the right to cast a vote.
• Speaker’s decisions cannot be questioned in a court of law.

Question 13.
Discuss the role of the ex-officio chairman of the Rajya Sabha.
The presiding officer of the Rajyasabha is the Vice-President who acts as the ex-officio chairman of the Rajyasabha. He holds office for a period of 5 years and salaries and allowances are as determined by the parliament. The position of the Chairman is that of dignity and honour and his task is to see that the transaction of business is conducted with dignity and purpose. Smooth transaction of business and maintenance of order is the responsibility of the Chairman.

Question 14.
Explain the law-making procedure.
A bill in order to become a law has to go through the law-making process. A bill, other than a Money bill, can originate in either house of the parliament before it goes to the President for assent. Non-money bill goes through three readings, involving five stages, in both the houses, before becoming a Law. A Bill may be moved by a private member called a Private Member bill or by a minister representing the Government called a Public bill. It is also called a Government bill.

A minister or a member can introduce a bill with the permission of the Speaker. No formal debate or no speeches relating to contents of the bill are made at this stage. After the bill has been introduced, it is immediately published in the Gazette of the Government of India. Now. the First reading of the bill is complete.

At this stage, the concerned minister provides details such as the purpose, objectives, and background of the bill in general as well as specifics relating to various clauses, schedules, and amendments to take place. No corrections or amendments can be made at this stage and after detailed discussion, the bill is put to vote.

3. The Committee Stage:
At this stage, the bill is submitted to a Committee and the names of. the members of the Committee are published. The committee headed by a chairman examines the Bill and its provisions and discusses it clause by clause. In the process, it may ask for relevant information and suggest its own changes and modifications. And now, the report and the bill are published in the Gazette.

4. The Report Stage:
Based on recommendations made by the Committee, the bill goes through detailed discussions. Changes can be suggested at this stage and the Report Stage is the last chance for the members to make any amendments to the bill.

This is the formalization stage of the bill where general discussions about the concerned bill take place. At this stage, no formal amendments can be made except informal changes. Then the bill is put to vote. Though the bill is open tor rejection, even at this stage, it is not normally resorted to.

After completing five stages in the House from which it originated, say Rajyasabha, it goes to the loksabha and has to undergo the same procedure. After getting passed in both the houses, it goes for the acceptance of the President. In case the other house rejects the bill altogether, the president may call for a joint sitting of the parliament to resolve the deadlock (Article 108).

Question 15.
Write a brief note on the Legislative powers of the president.
1. Legislative Functions:

The legislative functions are detailed below:
1. To summon, prorogue and dissolve the Parliament.

2. The President enjoys the power to address the Parliament. It is normally done after general elections or the first session of the year. It is generally called Presidential speech. This inaugural speech outlines the objectives and priorities of the government.

3. In passing the bills, if a deadlock arises due to non-agreement between two houses of the parliament, the President may call for a joint session of both the houses.

4. The President may address Lok sabha or Rajya sabha or both any time and also may send a message to both the houses of parliament to look into a bill.

5. In the considered view of the President, if he is satisfied that the Anglo-Indian com¬munity is not adequately represented, he may nominate 12 members to Rajya Sabha and 2 members to Lok sabha.

6. Prior permission of the President is essential while dealing with bills relating to the formation of new states, alteration of boundaries and some special bills like the finance bills.

7. No bill can become a law without the assent of the President. He enjoys the power to withhold a bill. This power is called ‘Veto power’. However, he cannot refuse his assent for finance bills. But he can withhold assent for a nonmoney bill. But if the same is resubmitted for signature even without changes, he cannot refuse to sign it.

8. The President enjoys the power of issuing Ordinance when the parliament is not in session. It will have the same power and effect similar to that of a law made by the Parliament provided the same is ratified by the Parliament within 6 weeks of its passage. Otherwise, it ceases to be a law and is considered null and void or zero.

Question 16.
Which is more powerful? State legislative assembly or state legislative council? Explain.
The legislative council is a weaker chamber. It is not only the second but also the secondary chamber of the state legislature. Money bills must originate only in the legislative assembly. The council has no control over the ministry in the state.

Question 17.
What are the miscellaneous functions of the Rajyasabha?
The miscellaneous functions of the Rajyasabha are:

1. By a resolution, Rajyasabha can create one or more All India Services.
2. Continuation of emergency beyond the specified time must come before the Rajyasabha and Loksabha.
3. Orders made by the President suspending enforcement of fundamental rights is required to be laid before the Rajyasabha and Loksabha.
4. According to Article 249, the Rajyasabha by a resolution can ask the parliament to legislate on certain subjects in the State list.

Question 18.
What are the functions of the legislature?
1. Lawmaking: The legislature enacts new laws, amends old laws and repeals unwanted laws.

2. Deliberative: The Legislature conducts extensive discussions on matters of public importance, bills and the budget.

3. Formulates policies: The legislature formulates domestic and foreign policy.

4. Controlling the executive: The legislature controls both the political and administrative executive, through questions, resolutions, and motions. A no-confidence motion is to express a lack of faith in and to oust the government.

5. Financial control: To enact money bills and budgets.

6. Judicial: The judiciary can impeach the heads of state and the judges of the higher courts.

7. Constitutional: the legislature amends the constitution.

8. Electoral: It is part of the Electoral College that elects the head of the state.

9. Ventilation of grievances: The legislature is a forum of complaints and formation of public opinion.

Question 19. Briefly explain the importance of the legislature.
The Importance of the legislature is:
The importance of Legislature has grown many-fold due to the ever-increasing role of government in the developmental activity. In present times, the role of the executive and judiciary cannot be imagined without legislature. That’s why Gilchrist has rightly observed thus: “The legislature is the preamble, executive the content and judiciary the conclusion”.

1. Lawmaking:
The Lawmaking the role of the legislature is so vital because even though the executive is strong, a wrong law passed by the legislature will have disastrous consequences. So, efficient lawmaking is essential and that task is handled by legislature. While making laws, it is important to evaluate the pros and cons of that legislation.

2. Determines the government:
In democracies, people elect a government of their choice from among competing political parties in elections, held periodically. Party or parties that secure majority support of the people form the government and others act as the opposition. The government is also called as “the ruling party or ruling coalition” continues as long as it enjoys the confidence of the legislature (lower house). If the government fails to secure a motion of no confidence it has to vacate the office. In India, parliamentary elections are held once in 5 years whereas in U.S.A, presidential elections are held once in 4 years.

3. Public platform:
The importance of legislature is reflected in its role serving as a public platform. The legislature consists of people’s representatives from different nooks and comers of the country representing various castes, religions, languages, customs, traditions, cultures and vivid socio-economic backgrounds. These representatives assemble in one place to discuss problems concerning people of the whole nation. So, legislature plays a key role in mirroring national sentiment.

Question 20.
Briefly explain the theory of separation of powers.
The theory of separation of powers deals with the relationship between the three organs of government. This theory was put forward by Montesquieu, a French political thinker in his book “The spirit of the laws”. He suggested that the governmental powers must be separated between i.e., legislature, executive and judiciary, Montesquieu theory had a great influence on the US constitution.

### 1st PUC Political Science Legislature Ten Mark Questions and Answers

Question 1.
Describe the composition, powers, and functions of Loksabha.
The members of Lok Sabha are elected by the people. All adult citizens unless disqualified for other reasons have the right to select their representatives. Qualifications to become the members are must be a citizen of the country and must have attained the minimum age fixed by the constitution. The term of office is five years. Speaker is the presiding officer. He is elected from among the members of the house.

The powers and functions of LokSabha are as follows
1. Legislative functions:
The power of Loksabha extends to all subjects falling under the Union List and the Concurrent List. In case of emergency in operation, its power also extends to the State list as well. No bill can become a law without the consent of Loksabha. The Loksabha has equal powers of law-making with Rajyasabha except on financial matters where the supremacy of Loksabha is total.

In case of disagreement between the two houses on a matter of legislation, it is resolved by a Joint Sitting of both the houses presided over by the Speaker. In a Joint Sitting, Loksabha would emerge triumphant because the decisions are taken by a majority of the total number of members of both the house present and voting in which the numerical superiority of Loksabha prevails.

2. Financial functions:
On financial matters, the supremacy of the Loksabha is total and complete. “One, who holds purse, holds power” said James Madison. By establishing its authority over the national purse, Loksabha establishes its authority over the Rajyasabha. It is expressly stated that the Money bill can originate only in the house of people. Regarding budget, Loksabha being a representative house enjoys total authority. Loksabha’s position on financial matters is such that the demands for grants are placed only before the Loksabha.

3. Control over the executive:
The Loksabha enjoys direct control over the executive because; the executive is directly responsible to the lower house and stays in office as long as it enjoys the confidence of the house. The Loksabha not only makes laws but also supervises the implementation. The lower house being a debating house, the members are free to seek information from the executive and raise questions and seek clarifications.

The members can effectively seek information from the government by way of discussions and debates during the Question Hour (seek clarification), the Adjournment Motion (raises issues of national importance), the Zero Hour, the Cu motion, the Call-attention motion, etc. The soundest way of controlling the executive is by way of moving the No-confidence motion, if the executive fails to win the support of Lok sabha, they must step down.

4. Constituent functions:
The Loksabha shares equal powers in regard to amending provisions of the constitution. An amendment may be initiated either in the Rajyasabha or Loksabha and must be passed by a 2/3 majority in both the houses present and voting. The agreement of Rajyasabha is compulsory for the success of the constitutional amendment.

5. Electoral functions:
The Loksabha and Rajyasabha elect the highest constitutional functionaries such as the President and the Vice-president. The President is elected by the members of Loksabha and Rajyasabha along with the members of Legislative Assemblies of the states. The Vice-president is elected by members of Loksabha and Rajyasabha.

6. Judicial functions:
The Loksabha acts as a judge in the impeachment of the President. Either house can prefer the charge of impeachment. If. Rajyasabha prefers the charge, Loksabha investigates the charge and if it passes a resolution by a 2/3 majority of the total membership of the house. President stands impeached from the office. The Loksabha also sits in Judgement along with the Rajyasabha, in removing high constitutional functionaries such as the Comptroller and Auditor General, The Chief Vigilance Commissioner, the Chief Election Commissioner, etc.

Question 2.
Describe the composition, powers, and functions of Rajyasabha.
In federal representation the state is important. In India, the members of the Rajyasabha are indirectly elected for 6 yrs but 1/3rd of them will retire even. 2 yrs. The Rajyasabha has 250 members out of whom 12 are nominated by the president for their contributions to science, literature, art and social service. The remaining 238 members are indirectly elected by the state legislative assemblies through a system of proportional representation.

The powers and functions of Rajyasabha are as follows:
1. Legislative functions:
Oh legislative matters, the Rajyasabha enjoys powers with the Loksabha except in case of a Money Bill or Financial Bill. Non-money Bill can originate in Rajyasabha and must get a 2/3 majority in the House and then proceeds to the Loksabha. The approval of both Houses is essential for a bill to become a law.

In case of disagreement between the two Houses on a bill, both the houses sit on a joint sitting presided by the speaker and the deadlock is resolved by a majority of the total number of members of both the houses present and voting.

2. Financial functions:
On the financial front, the Rajyasabha virtually has no powers. The procedure to deal with Money bills clearly states that a Money bill or financial bill cannot originate in the Rajyasabha. The Rajyasabha may discuss and suggest changes but have no right to reject or amend a Money Bill. It is left to the Loksabha to accept or reject its recommendations. In case, the Rajyasabha does not send back a Money bill back to the Loksabha within 14 days from the date of receipt of the bill, the bill is deemed passed, in the original form, by both the Houses.

3. Control over the executive:
The Rajyasabha’s hold over the executive is very minimal because the executive is not directly responsible to the upper house. However, it can seek information, and make clarifications on various policy matters. On issues of national and local importance, the members can grill the executive during debates and discussions during the Question Hour, the Adjournment motion, the Zero Hour, the Cut-Motion, Call- attention Motion, etc.

4. Constituent functions:
The Rajyasabha enjoys full powers with Loksabha in executing constituent functions. An amendment to provisions of the constitution can be initiated in either House of the parliament and must be passed by a 2/3 majority’ in both the Houses present and voting. If Rajyasabha does not pass an amendment bill, the amendment Bill stands defeated. In some special provisions apart from the 2/3 majority in both house of parliament and ratification by not less than 1/2 of the states is necessary.

5. Electoral functions:
The Rajyasabha shares the privilege of electing the highest constitutional functionaries, the President and Vice-president. The President is elected by an electoral college consisting of the members of parliament along with the members of the State legislative assemblies. The members of both the houses of parliament elect the Vice-president.

6. Judicial functions: The impeachment move against the President may be initiated in either house of the parliament. If Loksabha prefers the charge, Rajyasabha investigates the charge and passes a resolution by a 2/3 majority of the total membership of the house, then the President stands impeached.

There is no need for an impeachment against the Vice president who may be removed by a resolution of Rajyasabha passed by the majority of its members and consented to it by Loksabha. The Rajyasabha also participates in the removal of the highest constitutional functionaries such as the Chief Election Commissioner, the Vigilance Commissioner, etc.

7. Miscellaneous functions:
The Rajyasabha performs other functions as well as

1. By a resolution Rajyasabha can create one or more All India Services.
2. Continuation of emergency beyond the specified time must come before the Rajyasabha and Loksabha.
3. Orders made by the President suspending enforcement of fundamental rights is required to be laid before the Rajyasabha and Loksabha.
4. According to Article 249, the Rajyasabha by a resolution can ask the parliament to legislate on certain subjects in the State list.

Question 3.
Explain the composition, powers, and functions of Vidhana Sabha.
There is a legislative assembly for every state. The number of members depends upon the population of the state. But it can not have less than 60 and more than 500 members. The members are chosen by direct election by people of the state. The governor has been given the power to nominate one or two members of the Anglo Indian community legislative assembly is five years.

The powers and functions of Vidhanasabha are as follows:
1. Legislative Functions:
The Legislative Assembly is entitled to pass laws on all subjects that fall under the state list such as police, public health, education, local-self governments, etc. Without the consent of the Vidhanasabha, no bill can become a law. Though the Vidhanasabha is competent enough to make laws on subjects listed in the concurrent list along with the central legislature, if parliament passes a law contained in the concurrent list, the legislative assembly is not competent to pass a law on the same subject.

However, some bills require the previous permission of the President before they are introduced in the state legislature. In case of breakdown of constitutional machinery’ in a state or when the proclamation of emergency is in operation, parliament has the power of making laws on matters falling under the state list. In case of a conflict between state law and the law of the parliament, the law of the parliament shall supreme.

2. Financial Functions:
The Vidhanasabha enjoys total control over the finance of the state. No new tax can be levied or collected without the consent of the Vidhanasabha. The authority of the Vidhanasabha over Vidhanaparishad is strengthened by the fact that a Money bill or Financial bill can only originate in the Vidhanasabha and the Vidhanaparishad can at the most delay it by 14 days but cannot reject or amend the Bill. The annual income-expenditure statement of the year the Budget must get the approval of the Vidhanasabha. Even year during March-April, the beginning of the financial year, it is the responsibility of the government to place the budget before the house and seek its approval.

3. Control over the Executive (Administration):
The Vidhanasabha enjoys direct control over the administration, as the executive is directly, collectively, responsible to the Vidhanasabha and remains in office as long as they enjoy the confidence of the house. The members of the house can seek information from the government through questions and supplementary questions. It is the responsibility of the ministers to clarify points raised by members and give a satisfactory explanation.

Any attempt to lie or mislead the house is considered an offence against the house punishable under Contempt of the House. The debating occasions such as the Question Hour, Adjournment motion, the Emergency Adjournment motion, the Zero Hour, the Cut motion, the Call-attention motion keeps the executive under constant check and the executive must be alert and ready with answers. However, ministers can ask for time to answer questions. The most effective weapon in the hands of the Vidhanasabha is the No-confidence motion, which can bring down a government.

4. Electoral Functions:
The members of the Vidhanasabha along with the members of the parliament constitute an electoral college to elect the President of India. They also take part in electing the members of Rajyasabha and also of the members of Legislative council.

5. Constituent Functions:
The state legislative assembly takes part in amending a few constitutional provisions. The Assembly does not initiate any amendment to constitution neither does it has such powers. But ratification of at least not less than half of the State legislative assemblies is necessary for amending certain provisions of the constitution. For instance, if there has to be an amendment made to electoral procedure of electing president of India then it has to be ratified by 1/2 of the states, which in turn is done by state legislative assemblies.

Question 4.
Describe the composition, powers, and functions of Vidhana Parishad.
The Composition of the legislative council is as follows:

• 1/3 – of the members are elected from the local bodies such as municipalities and district boards.
• 1/3 – of the members are elected from members of the Legislative Assembly.
• 1/12 – of the members are elected by the graduates from graduate constituencies.
• 1/12 – of the members are elected from teacher’s constituencies consisting of secondary school, college, and university teachers; and
• 1/6- of the members are nominated by the Governor from the fields of science, art, social service, the co-operative movement, literature, etc.

The powers and functions of the Vidhana Parishad are as follows:
1. Legislative functions:
No bill can become a law unless agreed upon by both the Legislative council and the Legislative assembly. Any bill other than a Money Bill can originate in the upper house. The lower house on passing a bill sends it to the upper house for consideration and recommendations. If the upper house rejects the bill after keeping it with them for 120 days, the lower house can again send the bill. And if the Legislative council does not give approval the second time, the bill is deemed to have been passed by both the Houses of the state legislature.

2. Financial functions:
The Legislative council does not enjoy any position of power on financial matters. It is expressly stated in the constitution that a Money bill or financial bill cannot originate in the upper house. The Vidhanasabha on passing a Money bill sends it to the Vidhana Parishad for its recommendations, but it is to the lower House to accept or reject the recommendations. The Vidhana Parishad can. neither correct nor amend a Money Bill. If a Money Bill is not returned to the Vidhana sabha within 14 days of receipt of the bill, the bill is deemed passed by both the houses in its original form. On Financial matters, the Vidhana Parishad is sub-ordinate to the Vidhana Sabha. The Vidhana Parishad can only discuss the Budget but cannot make changes.

3. Control over the executive:
The influence of the Legislative council over the executive is minimal because the executive is directly responsible to the Legislative assembly. It has the right to seek necessary information and records for suggestions but cannot directly control the administration. The upper House can ask questions and supplementary questions during which ministers are duty-bound to provide satisfactory answers. The Vidhana Parishad cannot initiate a motion of No-confidence to bring down the government as it is a nominated House.

Question 5.
What are the special powers of both the Houses of the Parliament?
The Rajyasabha performs other functions as well as

1. By a resolution, Rajyasabha can create one or more All India Services.
2. Continuation of emergency beyond the specified time must come before the Rajyasabha and Loksabha.
3. Orders made by the President suspending enforcement of fundamental rights is required to be laid before the Rajyasabha and Loksabha.
4. According to Article 249, the Rajyasabha by a resolution can ask the parliament to legislate on certain subjects in State list. The Loksabha enjoys direct control over the executive because; executive is directly responsible to the lower house and stays in office as long as it enjoys the confidence of the house. The Loksabha not only makes laws but also supervises the implementation. The lower house being a debating house, the members are free to seek information from the executive and raise questions and seek clarifications.

The members can effectively seek information from the government by way of discussions and debates during the Question Hour (seek clarification), the Adjournment Motion(raises issues of national importance), the Zero Hour, the Cu motion, the Call-attention motion, etc. The – soundest way of controlling the executive is by way of moving the No-confidence motion, if the executive fails to win the support of Lok sabha, they must step down.

Question 6.
Explain the role, powers, and functions of the speaker of Loksabha.
The presiding officer of Loksabha is the Speaker who is elected from among the members along with the Deputy Speaker and stays in office till the life of the House i.e., 5 years. His primary task is to protect the dignity and decorum of the House and to see that the proceedings of the House are conducted in an orderly and a focused manner. He is the principal spokesperson of the House and must be impartial and even-handed in dealing as the custodian of the House.

In order to ensure impartiality, speaker resigns his party membership on election. The Deputy speaker discharges the duty when the office of the speaker falls vacant due to resignation, death or removal by a 2/3 majority of the total membership of the House or in the absence of the speaker. The salary of the speaker is determined by the parliament from time to time. The Speakers’ position in the House is one of dignity and authority.

• All orders of the house are executed through the Speaker
• Communication from the President is made known through the Speaker.
• It is the power of the speaker to declare whether a bill is a money bill or not.
• He enjoys the authority of interpreting the Rules of procedure and has the power to vote except in case of a tie.
• No member can speak in the House without the permission of the speaker and it is the speaker who fixes the time limit for speech.
• He presides over the Joint sittings of the parliament.
• During discussions, the members must address the Chair.
• In the case of a tie, the speaker has the right to cast a vote.
• Speaker’s decisions cannot be questioned in a court of law.

Question 7.
Explain the role, powers, and functions of the speaker of Vidhana Sabha.
The presiding officer of Vidhana sabha is the Speaker who is elected from among the members along with the Deputy Speaker and stays in office till the life of the House i.e., 5 years. His primary task is to protect the dignity and decorum of the House and to see that the proceedings of the House are conducted in an orderly and a focused manner. He is the principal spokesperson of the House and must be impartial and even-handed in dealing as the custodian of the House. In order to ensure impartiality, the speaker resigns his party membership in the election.

The Deputy speaker discharges the duty when the office of the speaker falls vacant due to resignation, death or removal by a 2/3 majority of the total membership of the House or in the absence of the speaker. The salary of the speaker is determined by the parliament from time to time. The Speaker’s position in the House is one of dignity and authority.

All orders of the house are executed through the Speaker

• Communication from the Governor is made known through the Speaker.
• It is the power of the speaker to declare whether a bill is money bill or not.
• He enjoys the authority of interpreting the Rules of procedure and has the power to vote except in case of a tie.
• No member can speak in the House without the permission of the speaker and it is the speaker who fixes the time limit for speech.
• During discussions, the members must address the Chair.
• In case of a tie, the speaker has the right to cast a vote.
• Speaker’s decisions cannot be questioned in a court of law.

## Karnataka 1st PUC Hindi Textbook Answers Sahitya Vaibhav Chapter 8 नालायक

### नालायक Questions and Answers, Notes, Summary

I. एक शब्द या वाक्यांश या वाक्य में उत्तर लिखिए :

प्रश्न 1.
कितने अनट्रेण्ड अध्यापकों का चुनाव था?
उत्तरः
तीन सौ अनट्रेण्ड अध्यापकों का चुनाव था।

प्रश्न 2.
कितने उम्मीदवार आये थे?
उत्तरः
दस हजार उम्मीदवार आये थे।

प्रश्न 3.
प्रति उम्मीदवार के पीछे कितनी सिफारिशें आयी थीं?
उत्तरः
प्रति उम्मीदवार के पीछे लगभग 5 – 6 सिफारिशें आयी थीं।

प्रश्न 4.
ज्योतिषी कहाँ पर बैठे थे?
उत्तरः
ज्योतिषी फुटपाथ पर बैठे थे।

प्रश्न 5.
पुलिस की कितनी गाड़ियाँ आयी थीं?
उत्तरः
पुलिस की दो गाड़ियाँ आई थीं।

प्रश्न 6.
किसका नाम सभापति के लिए प्रस्तावित किया गया?
उत्तरः
श्री रघुपति राघव का नाम सभापति के लिए प्रस्तावित किया गया।

प्रश्न 7.
‘नालायक’ पाठ के लेखक कौन हैं?
उत्तरः
‘नालायक’ पाठ के लेखक विवेकी राय हैं।

अतिरिक्त प्रश्नः

प्रश्न 8.
उम्मीदवारों में कौन अधिक संख्या में थे?
उत्तरः
उम्मीदवारों में थर्ड डिवीजनर अधिक संख्या में थे।

प्रश्न 9.
लेखक के पास पहुंचे हुए उम्मीदवार किस डिविजन में पास है?
उत्तरः
लेखक के पास पहुंचे हुए उम्मीदवार थर्ड डिविजन में पास है।

प्रश्न 10.
अध्यापकों के चुनाव की कार्रवाई शुरू होने में कितनी देरी थी?
उत्तरः
अध्यापकों के चुनाव की कार्रवाई शुरू होने में एक घण्टे की देरी थी।

प्रश्न 11.
नया स्टाक लाने के लिए कौन शहर की ओर दौड़े?
उत्तरः
दुकानदार नया स्टाक लाने के लिए शहर की ओर दौड़े।

प्रश्न 12.
किसकी सहायता से खाँचियों में शिफारिशी पत्र डालने का कार्य सम्पन्न हुआ?
उत्तरः
पुलिस की सहायता से खाँचियों में शिफारिशी पत्र डालने का कार्य सम्पन्न हुआ।

प्रश्न 13.
सभी थर्ड डिविजनर कहाँ पर एकत्र हुए?
उत्तरः
शहर के बाहर लगभग डेढ़ मील दूर नदी के किनारे एक बगीचे में सभी थर्ड डिविजनर एकत्र हुए।

प्रश्न 14.
अत्यंत गुप्त रूप से क्या ज्ञात हुआ?
उत्तरः
अत्यंत गुप्त रूप से ज्ञात हुआ कि शहर के बाहर जहाँ पुलिस और सी.आई.डी. वाले न पहुँच सके वहाँ कॉन्फ्रेन्स होगी।

प्रश्न 15.
कौन-सा नारा बहुत देर तक दुहराया गया?
उत्तरः
‘किसको मिले मास्टरी चाँस, थर्ड डिवीजन मैट्रिक पास’ – यह नारा बहुत देर तक दुहराया गया।

II. निम्नलिखित प्रश्नों के उत्तर लिखिएः

प्रश्न 1.
अनट्रेण्ड अध्यापकों के चुनाव के माहौल का वर्णन कीजिए।
उत्तरः
उस दिन जिला परिषद में प्राइमरी स्कूलों के लिए अनट्रेण्ड अध्यापकों का चुनाव था। उसमें तीन सौ अध्यापक चुने जानेवाले थे। दस हजार उम्मीद्वार आये थे। हर एक उम्मीद्वार के साथ लगभग 5-6 सिफारिश करनेवाले थे। दर्शक भी काफी थे और वह एक मेला जैसा लग रहा था। अनट्रेण्ड अध्यापकों के चुनाव के लिए शहर में सुबह से ही इतनी भीड़ लग गई थी कि मानो कुंभ का मेला लग गया हो। सड़कों पर लोगों की चहल-पहल और शोरगुल शुरू हो गया था। ऐसा लग रहा था कि किसी नेता को देखने भीड़ इकट्ठी हो रही है।

प्रश्न 2.
जिला परिषद के बाहर का दृश्य प्रस्तुत कीजिए।
उत्तरः
जिला परिषद में प्राइमरी स्कूलों के लिए अनट्रेण्ड अध्यापकों की नियुक्ति होनी थी। इसलिए जिला परिषद् के बाहर अपार जनसागर उमड़कर जमा हो गया था। प्रवेश-द्वार से लेकर बाहर आँगन, लान, मैदान, चौक, सड़क और लगभग दो-तीन फर्लाग तक ठसमठस आदमी भरे हैं, किसी को पता नहीं कि क्यों खड़े हैं, बस खड़े हैं। जो पहले आये वे पहले से फाटक की ओर खड़े होते गए। जो आगे थे उनका निकलना कठिन हो रहा था। भीतर परिषद कार्यालय था, जो बन्द था।

प्रश्न 3.
लाउड स्पीकर से क्या घोषणा की जा रही थी?
उत्तरः
चेयरमैन साहब के बंगले के बाहर भीड़ जमा होने के कारण, पुलिस की गाड़ियाँ भीड़ को नियंत्रित करने पहुंची। लाउड़ स्पीकर से घोषणा की जाने लगी कि “सज्जनों, ज्ञात हुआ है कि आप लोग थर्ड डिवीजनरों के सिफारिशी हैं। आप लोगों को बहुत समझाया गया कि सिफारिश से काम नहीं चलेगा, परन्तु आप लोग मानते ही नहीं हैं। अब स्कूल-मास्टर के लिए थर्ड डिवीजनर्स नहीं लिए जायेंगे। यदि आप लोग बिना उस सिफारिशी कागज को दिये टलनेवाले नहीं हैं तो दे दीजिये। पचास व्यक्ति खांची लेकर तैनात किये जाते हैं। इन्हीं खांचियों में डाल दें।”

प्रश्न 4.
दौड़ कर आते हुए उम्मीदवार ने क्या कहा?
उत्तरः
दौड़कर आते हुए उम्मीदवार ने गुस्से में कहा – “साहब लोग कहते हैं कि अध्यापक पद के लिए थर्ड डिविजनर नहीं लिये जाएँगे। क्यों नहीं लिये जायेंगे? जिसका ज्यादा वोट होगा, उसे लिया जायेगा। हम सभी फर्स्ट और सेकेंड डिवीजनर के खिलाफ़ आंदोलन करेंगे।”

प्रश्न 5.
थर्ड डिवीजनरों की व्यथा को प्रकट कीजिए।
उत्तरः
थर्ड डिवीजनरों का जीवन नरक के समान होता है। उनके लिए हर कहीं दरवाजा बंद रहता है जैसे वे आदमी नहीं, बैल हैं। उनकी उपेक्षा होती है। उन्हें नौकरी के लिए नाक रगड़नी पड़ती है। कदम, कदम पर निराशा होती है। अतः थर्ड डिविजनवालों को नालायक समझकर कोई नौकरी नहीं देता।

अतिरिक्त प्रश्नः

प्रश्न 6.
चबूतरे पर खड़े युवक ने क्या भाषण दिया?
उत्तरः
चबूतरे पर खड़े युवक ने ‘थर्ड डिवीजनर्स’ की कॉन्फ्रेन्स में ‘करो या मरो’ का नारा दिया। उसने कहा हमने स्कूल की फीस दी, परीक्षा दी। परीक्षा में थर्ड डिवीजन आया तो यह सरकार, स्कूल एवं स्कूल अधिकारियों का दोष है। हम लोगों को जिन्होंने थर्ड क्लास का बताया, वे अपराधी हैं, चोर है। इन्होंने हमारा जीवन नरक बना दिया है। जैसे हम आदमी नहीं बैल हैं। यह प्रजातंत्र का युग है। अगर निरक्षर एम.एल.ए. हो सकता है तो थर्ड डिवीजन अध्यापक क्यों नहीं हो सकता। हमें इसके लिए आन्दोलन करना होगा।

प्रश्न 7.
चैयरमैन साहब के बँगले पर का दृश्य प्रस्तुत कीजिए।
उत्तरः
चैयरमैन साहब के बंगले पर जिला परिषद के कार्यालय से भी दूनी भीड़ जमा थी। मैदान में लोग ज्वार बाजरे की तरह खड़े थे। जैसे चुनाव के समय पोलिंग स्टेशनों पर बातें होती है वैसी ही बातें यहाँ भी हो रही थी। सबके हाथों में सिफारिशी कागज थे। सभी लोगों की निगाहें चैयरमैन साहब के बंगले की ओर थी।

प्रश्न 8.
परिषद कार्यालय बंद रहने से उम्मीदवार क्या सोच रहे थे?
उत्तरः
सभी बेरोजगार परिषद कार्यालय के प्रवेश द्वार पर खड़े थे। परिषद कार्यालय बंद था। यह रहस्य का विषय था। सबको इसमें रहस्य लग रहा था। शायद अन्दर कुछ हो रहा हो। सब यही सोच रहे थे। सब लोग प्रवेश द्वार पर ही जमे हैं कि कहीं बाहर गये और इधर कहीं कुछ हो न जाए।

प्रश्न 9.
शिफारिश करने का तरीका क्या था?
उत्तरः
सिफारिश के लिए एक कागज पर उम्मीदवार का नाम, गांव का नाम एवं शिक्षा संबंधी जानकारी लिखी गयी। उस कागज के नीचे सिफारिश. करने वालों ने अपने-अपने हस्ताक्षर किया।

प्रश्न 10.
‘यह कैसा कुम्भ पर्व कुछ पहले ही यहाँ आ गया? लेखक किसके बारे में कह रहे हैं?
उत्तरः
जनवरी महीने की शीतलहर में शहर में अचानक असाधारण भीड़ बढ़ गयी थी। इस भीड़ को देखकर लेखक कहते हैं- ‘यह कैसा कुम्भ पर्व कुछ अरसे पहले ही यहाँ आ गया?’

III. निम्नलिखित वाक्य किसने किससे कहे?

प्रश्न 1.
आपका उम्मीदवार किस डिवीजन में पास है?
उत्तरः
यह वाक्य लेखक ने एक उम्मीदवार से कहा।

प्रश्न 2.
थर्ड डिवीजनर के लिए दुनिया में कोई जगह नहीं है।
उत्तरः
यह वाक्य लेखक ने सिफारिशी लोगों से कहा।

प्रश्न 3.
ये लोग हम लोगों का हक मार रहे हैं।
उत्तरः
यह वाक्य नौकरी के लिए आए एक थर्ड डिविजन उम्मीदवार ने लेखक से कहा।

प्रश्न 4.
हमारे लिए हर जगह दरवाजा बंद है।
उत्तरः
यह वाक्य एक थर्ड डिविजनर युवक ने बाकी थर्ड डिविजनर युवकों से कहा।

अतिरिक्त प्रश्नः

प्रश्न 5.
“हाँ, चलिए आपके साथ चल सकता हूँ।”
उत्तरः
लेखक विवेकी राय ने उम्मीदवार से कहा।

प्रश्न 6.
“हम लोग फर्ट और सेकंड डिवीजन वालों के खिलाफ आंदोलन करेंगे।
उत्तरः
थर्ड डिवीजन पास एक उम्मीदवार ने अन्य उम्मीदवार साथियों से कहा।

IV. ससंदर्भ स्पष्टीकरण कीजिए:

प्रश्न 1.
जीवन में पहली बार सिफारिश का यह तरीका जाना।
उत्तरः
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘नालायक’ पाठ से लिया गया है जिसके लेखक विवेकी राय हैं।
संदर्भ : स्कूलों के लिए अध्यापक के चुनाव के लिए आए उम्मीदवारों के साथ उनके शिफारिश करनेवाले तथा साथ में एक कागज पर उम्मीदवार का नाम, ग्राम, शिक्षा आदि लिखकर हस्ताक्षर किए थे।
स्पष्टीकरण : पता चला कि थर्ड डिविजनर्स को प्राइमरी अध्यापक की नौकरी नहीं दे रहे हैं, तो जो युवक उम्मीद लगाए बैठे थे, वे गुस्से में लाल-पीले होकर जवानों की तरह लड़ने-मारने को तैयार हो गए।

प्रश्न 2.
सुबह का सिकुड़ा वह बालक अब तनकर जवान की तरह लग रहा था।
उत्तरः
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘नालायक’ पाठ से लिया गया है जिसके लेखक विवेकी राय हैं। .
संदर्भ : एक उम्मीदवार दौड़ता हुआ लेखक के पास आया और उसने कहा – लोग कह रहें है कि थर्ड डिवीजनर्स को नहीं लेंगे। हम आन्दोलन करेंगे।
स्पष्टीकरण : सुबह जो उम्मीदवार लेखक के पास आया था, वही दौड़ता हुआ लेखक के पास आया और उसने कहा – कि लोग कहते हैं वे थर्ड डिवीजनर्स को नहीं लेंगे? हम फर्स्ट डिवीजन और सेकेण्ड डिवीजन वालों के खिलाफ आंदोलन करेंगे। वे हमारा हक मार रहे है, यही बालक जो सुबह से सिकुड़ा बैठा था अब वह तनकर जवान की तरह बातें कह रहा है।

प्रश्न 3.
हम हरगिज़ नहीं मानते कि कागज की तीन लकीरों के कारण हम लोग नालायक हैं।
उत्तरः
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘नालायक’ पाठ से लिया गया है जिसके लेखक विवेकी राय हैं।
संदर्भ : इस वाक्य को मंच पर खड़े युवा उम्मीदवार ने कहा। अंको के आधार पर नहीं बल्कि प्रतिभा के आधार पर युवकों को नौकरी मिलनी चाहिए।
स्पष्टीकरण : जब थर्ड डिविजनरों को खारिज कर दिया गया, तो वे लोग इसका दोष शिक्षाविभाग तथा अध्यापकों पर डालने लगे। सिर्फ कागज पर थर्ड डिविजन लिख देने से हम नालायक हो गए क्या? नहीं।

प्रश्न 4.
किसको मिले मास्टरी चाँस, थर्ड डिवीज़न मैट्रिक पास।
उत्तरः
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘नालायक’ पाठ से लिया गया है जिसके लेखक विवेकी राय हैं।
संदर्भ : थर्ड डिवीजन वालों को नालायक समझकर कोई नौकरी नहीं देता। फर्स्ट और सेकेण्ड डिवीजन वालों के खिलाफ उनका आंदोलन होता है। उसी में यह नारा लगाते हुए लेखक ने यह नारा सुना।
स्पष्टीकरण : थर्ड डिवीजनरों के अधिकार छिन जाने से वे उत्तेजित हो गए और सभी मिलकर ‘थर्ड डिविजनर्स जिंदाबाद’ के नारे लगाते हुए आन्दोलन पर उतर गए।

अतिरिक्त प्रश्नः

प्रश्न 5.
“शिक्षा का स्तर गिर रहा है। उसे अब और नहीं गिरने दिया जायेगा।”
उत्तरः
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘नालायक’ पाठ से लिया गया है। इसके लेखक विवेकी राय हैं।
संदर्भ : लेखक यह बात थर्ड डिवीजनर लोगों को कहते है।
स्पष्टीकरण : लेखक के यहाँ एक उम्मीदवार और उसके चार सिफारिशी पहुंच गए। थर्ड डिवीजनर ने लेखक से कहा – दुनिया में कोई जगह हो चाहे न हो, पर स्कूल मास्टर की जगह तो रिजर्व है। तब लेखक जवाब देते हैं कि आप लोग गलत समझते है। अब योग्य लोगों को अध्यापक के रूप में भरती किया जायेगा। शिक्षा का स्तर गिर रहा है। उसे अब और नहीं गिरने दिया जाएगा।

प्रश्न 6.
“ये लोग आज उल्टा चोर कोतवाल को डांटे’ की कहावत चरितार्थ कर रहे हैं।”
उत्तरः
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘नालायक’ पाठ से लिया गया है। इसके लेखक विवेकी राय हैं।
संदर्भ : एक युवक ने थर्ड डिवीजनर्स की मीटिंग में भाषण देते हुए कहा।
स्पष्टीकरण : काली माई के टूटे चबूतरे से थर्ड डिवीजनर बेरोजगारों को संबोधित करते हुए एक युवक कह रहा है- हमने फीस दी, परीक्षा दी और अगर हमारा थर्ड डिवीजन आया तो इसमें स्कूल का दोष है। हम लोग इन संस्थानों और लोगों पर मुकदमा कायम करेंगे। हम लोग फर्स्ट क्लास के आदमी हैं। हम लोगों को जिन्होंने थर्ड क्लास का बनाया, वे अपराधी हैं, चोर हैं। ये लोग आज ‘उल्टा चोर कोतवाल को डांटे’ की कहावत चरितार्थ कर रहे हैं।

प्रश्न 7.
“सो मैं पाँच वर्ष फेल होकर छठे साल के हाईस्कूल में थर्ड डिवीजनर भाई श्री रघुपतिराघव का नाम आज की सभा के लिए प्रस्तावित करता हूँ।”
उत्तरः
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘नालायक’ पाठ से लिया गया। है। इसके लेखक विवेकी राय हैं।
संदर्भ : थर्ड डिवीजनर्स की मीटिंग में बोलते हुए युवक ने यह घोषणा की।
स्पष्टीकरण : जब थर्ड डिवीजनर्स को अध्यापक की नौकरी के लिए अयोग्य घोषित कर दिया तब वे शहर में काली माई के टूटे चबूतरे के पास इकट्ठे हुए। उन युवकों में से एक युवक चबूतरे पर खड़े होकर भाषण देने लगा। उसने अपने भाषण के अंत में थर्ड डिवीजनर्स की सभा के लिए सभापति के रूप में पाँच वर्ष फेल होकर छठे साल के हाईस्कूल में थर्ड डिवीजनर भाई श्री रघुपति । राघव का नाम प्रस्तावित किया।

प्रश्न 8.
“आज हम थर्ड डिवीजनर्स कॉन्फ्रेन्स में ‘करो या मरो’ का नारा लगाएँगे और जीवन का संकल्प लेंगे।”
उत्तरः
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘नालायक’ पाठ से लिया गया है। इसके लेखक विवेकी राय हैं।
संदर्भ : काली माई के टूटे चबूतरे पर खड़े होकर भाषण देने वाले युवक ने कहा।
स्पष्टीकरण : युवक ने थर्ड डिवीजनर्स की कान्फ्रेंस में युवाओं को संबोधित करते हुए कहा – दोस्तों आज हम इस थर्ड डिवीजनर्स कान्फ्रेन्स में ‘करो या मरो’ का नारा लगाएँगे। अध्यापक की नौकरी के लिए हम थर्ड डिवीजनर्स के पास सबसे ज्यादा संख्या है। इस पर फर्स्ट-सेकिण्ड डिवीजन वालों ने धावा बोल दिया है। वे मास्टर की गद्दी छोड़े। इसके लिए हमें आन्दोलन करना होगा। इसके लिए वह युवक सभी बेरोजगार थर्ड डिवीजनर्स को ‘करो या मरो’ के लिए तैयार करता है।

V. निम्नलिखित वाक्यों को सूचना के अनुसार बदलिए:

प्रश्न 1.
तीन सौ अध्यापक चुने जायेंगे। (वर्तमानकाल में बदलिए)
उत्तरः
तीन सौ अध्यापक चुने जाते हैं।

प्रश्न 2.
सभी सिफारिशें तिरस्कृत की गईं। (भविष्यत्काल में बदलिए)
उत्तरः
सभी सिफारिशें तिरस्कृत की जाएंगी।

प्रश्न 3.
पुलिस लोगों की रक्षा कर रही है। (भूतकाल में बदलिए).
उत्तरः
पुलिस लोगों की रक्षा कर रही थी।

VI. अन्य लिंग रूप लिखिए:

प्रश्न 1.
शेर, अध्यापक, मास्टर, साहब, युवक।
उत्तरः

• शेर – शेरनी
• अध्यापक – अध्यापिका
• मास्टर – मास्टरनी
• साहब – साहिबा
• युवक – युवती

VII. अन्य वचन रूप लिखिएः

प्रश्न 1.
पुस्तक, गाड़ी, लड़का, बेटियाँ।
उत्तरः

• पुस्तक – पुस्तकें
• गाड़ी – गाड़ियाँ
• लड़का – लड़के
• बेटियाँ – बेटी

VIII. समानार्थक शब्द लिखिएः

प्रश्न 1.
सम्राट, अध्यापक, स्वतंत्र, प्रयत्न, सहायता।
उत्तरः

• सम्राट – राजा
• अध्यापक – शिक्षक
• स्वतंत्र – आजाद
• प्रयत्न – कोशिश
• सहायता – मदद

IX. मुहावरेः

• गोबर बारूद होना – नाकाम होना
• नाक रगड़ना – गिड़गिड़ाना
• हक मारना – अधिकार छीनना।

नालायक लेखक परिचयः

विवेकी राय का जन्म सन् 1923 नवंबर 20 को गाजीपुर जिले के सोनवानी नामक गाँव के किसान परिवार में हुआ। आप स्वयं को किसान साहित्यकार कहते हैं। अतीत के अभावग्रस्त जीवन, ग्रामीण शिक्षा और प्राइमरी स्कूल के शिक्षक से आरम्भ जीवन यात्रा को, उच्च शिक्षण संस्था’. में रीडर पद पर सेवा, बड़े लोगों से संपर्क-सान्निध्य तथा ख्याति प्राप्त रचनाकार के रूप में स्थापित हो जाने पर भी सादगी से परिचय में प्रस्तुति उनकी लेखकीय विश्वसनीयता का उत्कृष्ट साक्ष्य है।

प्रमुख कृतियाँ : काव्य – ‘अर्गला’, ‘रजनीगंधा’, ‘दीक्षा’। कहानी संग्रह – ‘जीवन परिधि’, ‘नयी कोयल’, ‘गूंगा जहाज’, ‘कालातीत’, ‘अतिथि’, ‘दलित विमर्श और विवेकी राय की कहानियाँ’ आदि।

प्रस्तुत ‘नालायक’ कहानी में मुख्य रूप से थर्ड डिविज़न में पास होनेवालों की मनोदशा को दर्शाया गया है। साथ ही इसमें बेरोजगारी जैसी व्यापक समस्याओं को उजागर किया गया है।

आज युवाओं में नौकरी के लिए होड़ लगी है। नौकरी पर ही निर्भर न रहकर, किसी न किसी काम पर या व्यवहार में जुट जाने की कला को अपनाना चाहिए। अंको के आधार पर ही नहीं बल्कि प्रतिभा के आधार पर भी युवकों को नौकरियों में अवसर मिले। इन विचारों को ध्यान में रखकर इस पाठ का चयन किया गया है।

### नालायक Summary in Hindi

डॉ. विवेकी राय हिन्दी के प्रसिद्ध कहानीकार, उपन्यासकार तथा निबन्धकार हैं। उन्होंने ‘नालायक, कहानी में शिक्षा के क्षेत्र में गिरते हुए मूल्यों तथा फैलते हुए भ्रष्टाचार का पर्दाफाश किया है। उन्होंने परोक्ष रूप से शिक्षा के क्षेत्र में अध्यापकों का बौद्धिक-स्तर कायम रखने की ओर संकेत दिया है।

जनवरी महीने में एक दिन शहर में युवाओं की भीड़ जमा हुई थी। जहाँ देखों वहाँ पढ़े-लिखे नौजवानों का जमावड़ा समुद्र जैसा फैला हुआ था। पता चला कि ये सब जिला-परिषद के स्कूलों के लिए अनट्रेण्ड अध्यापकों के पद के लिए आवेदन-पत्र देने आये बेरोजगार थे। अध्यापकों की जगहें केवल 300 थीं, जब कि आवेदक दस हजार से ज्यादा थे। उन उम्मीदवारों में कुछ उदास थे, कुछ मौन थे, तो कई जोशीले थे। उनकी आँखों से आग निकल रही थी। इन उम्मीदवारों की हालत बड़ी विचित्र थी। कई लोग सिफारिशी-पत्र लाये थे, तो कई थर्ड डिविजन में उत्तीर्ण थे।

जिला परिषद अध्यक्ष का कार्यालय बेरोजगार युवा-पीढ़ी से भरा हुआ था। पैर रखने की भी जगह नहीं थी। चारों ओर खलबली मची हुई थी। सब लोग परिषद के अध्यक्ष की राह देखने लगे थे। लम्बे समय के बाद पता चला कि अध्यक्ष अपने बंगले पर ही हैं। सभी उस ओर चल पड़े। परिषद के अध्यक्ष घबरा गए। पल भर में पुलिस की गाड़ियाँ आईं। कोई पुलिसवाला लाउडस्पीकर से घोषणा करने लगा कि अध्यापक पद के लिए थर्ड डिविजन प्राप्त आवेदक जा सकते हैं। युवाओं में दहशत फैल गई। क्योंकि दस हजार लोगों में थर्ड डिविजन वाले ही ज्यादा थे।

कुछ समय के बाद अध्यक्ष के बंगले से दूर नदी के किनारे, मैदान में सभी इकट्ठे हुए। उनमें से कुछ नेता गण आये। लाउडस्पीकर से ये उद्गार निकलने लगे – करो या मरो! थर्ड डिविजन वाले जिंदाबाद! कुछ युवा नेता गरजने लगे – क्या थर्ड डिविजनवाले शिक्षक पद के लिए नालायक हैं? ऐसा क्यों? सेकेन्ड या फस्ट डिविजन वालों ने कौनसा तीर मारा है? सरकार को झुकना पड़ेगा। “किसे मिले मास्टरी चान्स? थर्ड डिविजन मैट्रिक पास!” नारे आसमान में गूंजने लगे।

### नालायक Summary in English

Dr. Viveki Rai is a renowned writer of stories, novels and essays in Hindi. In this story, he sheds light on the falling standards and increasing corruption in the field of education. He says that in the field of education, teachers must maintain a steady state of mind.

One day in the month of January, there was a large congregation of youths in the town. As far as the eye could see, there were educated young people assembled like an unending ocean. It soon became clear that these youths were all unemployed, and hand-assembled in order to hand in their applications for the post of untrained teachers in the Zilla Parishad’s schools. There were only 300 posts while the number of applicants exceeded 10,000. Among the applicants, some were dejected and some were silent, while others were very enthusiastic and brimming with confidence. There was a burning fire in their eyes. It was a strange collection of applicants. While some of the applicants had come with glowing recommendation letters, others had barely managed to pass with a third division.

The office of the president of the Zilla Parishad was filled to the brim with unemployed youths. There was no place even to stand. The whole office was engulfed in commotion. Everyone was waiting for the arrival of the Zilla Parishad President. However, it became known that the President was at his bungalow. The whole crowd then began to proceed in that direction. Seeing such a large crowd coming towards his residence, the president was scared. The police was called in, and they soon arrived in police vans. One of the policemen made an announcement on the loudspeaker. He informed the crowd that all those applicants who had cleared their exams with a third division could leave, as they would not be considered for the post. Hearing this, panic gripped the entire crowd as most of them were third division pass.

After some time, most of the crowd had reassembled on the banks of the river, far away from the president’s bungalow. Some of the members in the crowd were involved in politics. Their cries were heard from the loudspeaker – “Do or Die!”, “Long live the third division!” Some youth leaders began to give speeches, asking whether third division applicants were unworthy of a teacher’s post. They wanted to know how people with second or even first division were better than them. They said that the government would have to bow to their demands. Slogans like “Who should get the teacher’s job? Third division metric pass!” began to resonate across the sky.

कठिन शब्दार्थः

• मवेशी – जानवर;
• बकलम – हस्ताक्षर;
• मुंड – सर;
• फतह – जीत;
• ढाठा – डंठल;
• खांची – टोकरी;
• भुसहुड़ – भूसे का ढेर;
• हरजाना – भरपाई;
• चरितार्थ – हकीकत;
• दुत्कार – तिरस्कार;
• मुर्रिसी – मास्टरी।

## Karnataka 1st PUC Hindi Textbook Answers Sahitya Vaibhav Chapter 6 दिल का दौरा और एनजाइना

### दिल का दौरा और एनजाइना Questions and Answers, Notes, Summary

I. एक शब्द या वाक्यांश या वाक्य में उत्तर लिखिएः

प्रश्न 1.
हृदय रोग के दो रूप कौन-से हैं?
उत्तरः
हृदय रोग के दो रूप हैं- दिल का दौरा तथा एनजाइना।

प्रश्न 2.
दिल का दौरा और एनज़ाइना आम तौर से कितने वर्ष से अधिक उम्र के व्यक्तियों में देखे जाते हैं?
उत्तरः
आम तौर से दिल का दौरा और एनजाइना 45 वर्ष से अधिक उम्र के व्यक्तियों में देखे जाते हैं।

प्रश्न 3.
‘मूगा टेस्ट’ किसे कहते हैं?
उत्तरः
न्यूक्लियर स्कैन से यह पता चलता है कि रोगी के दिल का कितना हिस्सा दौरे की चपेट से बेकार हुआ है। इसे ‘मूगा टेस्ट’ कहते हैं।

प्रश्न 4.
कोरोनरी धमनियों में सिकुड़न आने का एक बड़ा कारण क्या है?
उत्तरः
कोरोनरी धमनियों में सिकुड़न आने का सबसे बड़ा कारण उनमें वसा की परत का जम जाना है।

प्रश्न 5.
क्या बायपास सर्जरी भारत में संभव है?
उत्तरः
बाईपास सर्जरी भारत में भी संभव है।

प्रश्न 6.
हृदय रोगियों के लिए किस तरह का भोजन अच्छा नहीं है?
उत्तरः
हृदय रोगियों के लिए तले हुए एवं अधिक वसायुक्त भोजन अच्छा नहीं है।

अतिरिक्त प्रश्नः

प्रश्न 7.
“दिल का दौरा और एनजाइना’ लेख का लेखक कौन है?
उत्तरः
‘दिल का दौरा और एनजाइवा’ लेख का लेखक डॉ. यतीश अग्रवाल हैं।

प्रश्न 8.
ज्यादा तेज़ दौरा पड़ा तो रोगी की क्या दशा हो जाती है?
उत्तरः
ज्यादा तेज़ दौरा पड़ा तो रोगी बेहोश होकर गिर सकता है।

प्रश्न 9.
क्या मदिरा सेवन और धूम्रपान हृदय के लिए अच्छा है?
उत्तरः
मदिरा सेवन और धूम्रपान हृदय के लिए अच्छा नहीं है।

प्रश्न 10.
दिल के दौरे का प्रमुख लक्षण क्या है?
उत्तरः
दिल के दौरे का प्रमुख लक्षण सीने के बाई ओर प्राणलेवा दर्द उठना है।

प्रश्न 11.
दौरा पड़ने पर जीभ के नीचे रखकर चूसने के लिए कौन सी गोली देते हैं?
उत्तरः
दौरा पड़ने पर जीभ के नीचे रखकर चूसने के लिए सार्बिट्रट की गोली देते हैं।

प्रश्न 12.
ई.सी.जी. के बाद क्या करवाने के लिए सलाह देते हैं?
उत्तरः
ई.सी.जी. के बाद न्यूक्लियर स्कैन कराने की सलाह देते हैं।

प्रश्न 13.
हृदय रोगी को किस पर कड़ी नज़र रखना चाहिए?
उत्तरः
हृदय रोगी को अपने वज़न पर कड़ी नज़र रखना चाहिए।

प्रश्न 14.
कैफिन दिल की धड़कन को बढ़ाता है या कम करता है?
उत्तरः
कैफिन दिल की धड़कन को तेज करती है।

II. निम्नलिखित प्रश्नों के उत्तर लिखिएः

प्रश्न 1.
दिल का दौरा और एनजाइना किसे कहते हैं?
उत्तरः
दिल का दौरा और एनजाइना हृदय रोग के दो प्रकार है। एनजाइना में हृदय तक पहुंचने वाली ऊर्जा नहीं पहुंच पाती, जो कोरोनरी धमनी की शाखाओं में सिकुड़न की वजह से होता है। जब हृदय रक्त को पंप करने में असमर्थ हो जाता है, तो उसे ‘एनजाइना’ कहते हैं। इसी के भयंकर स्वरूप को ‘दिल का दौरा’ कहते हैं। किसी एक धमनी में रुकावट आ जाने में दिल का दौरा पड़ता है।

प्रश्न 2.
दिल का दौरा और एनजाइना किन कारणों से होता है?
उत्तरः
उच्च रक्तचाप, मोटापा, तनाव, धूम्रपान, वसा के खाद्य पदार्थों का सेवन, मदिरापन, कोलेस्ट्रोल का बढ़ना, परिवार में अन्य किसी को यह रोग होना, व्यायाम न करना, परिश्रम न करना ये सब दिल का दौरा और एनजाइना के मुख्य कारण माने गए हैं। उग्रता का स्वभाव, अकेलापन, शहरी जिन्दगी का तनाव, आधुनिकीकरण से जीवन के तौर तरीकों में बदलाव भी कारण हो सकते हैं।

प्रश्न 3.
दिल का दौरा और एनजाइना रोग के प्रमुख लक्षण क्या हैं?
उत्तरः
जब रोगी तनावपूर्ण स्थिति में होता है और उसके सीने में बाईं ओर दर्द होने लगता है, बेचैनी और भारीपन लगने लगता है, तो ये दिल के दौरे के लक्षण माने जाते हैं। खास तौर से सीने में असहनीय दर्द होता है और यह दर्द शरीर के विभिन्न भागों तक फैल जाता है। रोगी को ऐसा महसूस होता है जैसे उसकी छाती पर कोई बहुत भारी चीज रख दी गई हो। कभी-कभी असहनीय दर्द से रोगी बेहोश भी हो सकता है।

प्रश्न 4.
दिल का दौरा पड़ने पर प्राथमिक उपचार के लिए क्या-क्या कदम उठाने चाहिए?
उत्तरः
दिल का दौरा पड़ते ही डॉक्टर को बुलाना चाहिए। समाधान के लिए तब-तक कोई दर्द-निवारक गोली दी जा सकती है। पहले से ही यदि रोगी इसका शिकार है, तो उसे ‘सार्बिट्रट’ गोली दे देनी चाहिए। यदि रोगी की धड़कन रुक गई है, तो उसे पीठ के बल लिटाकर, उसके कपड़े ढीले करके, छाती पर मालिश करनी चाहिए। यथाशीघ्र अपनी साँस देने का प्रयास करें।

प्रश्न 5.
हार्ट अटैक के रोगी का इलाज़ किस तरह किया जाता है?
उत्तरः
रोगी को तुरन्त अस्पताल ले जाना जरूरी है। यदि धमनी का काम रुक गया हो, तो उसे ऐसी दवा दिलवानी चाहिए जिससे धमनी खुल जाये। स्ट्रेप्टोकाइनेज या यूरोकाइनेज जैसी दवा इंजेक्ट भी किया जा सकता है। आवश्यकता होने पर दवा और पेसमेकर की जरूरत भी हो सकती है।

प्रश्न 6.
‘बैलून एंजियोप्लास्टी’ तकनीक क्या है?
उत्तरः
जब एक-दो धमनियाँ अवरुद्ध हो जाती है, तो उन्हें ठीक करने के लिए बैलून एंजियोप्लास्टी’ की जाती है। एक पतली ट्यूब के द्वारा गुब्बारे को धमनी के सिकुड़े हुए भाग तक पहुँचाकर फुलाया जाता है। उससे वसा की परत दबकर धमनी खुल जाती है। फिर रक्त बहाव सामान्य होकर, रोग मिट जाता है।

प्रश्न 7.
हृदय रोग से बचने व काबू पाने के लिए क्या-क्या एहतियात बरतने चाहिए?
उत्तरः
रोजाना व्यायाम करें। इसके लिये सुबह-शाम की सैर अच्छी होती है। अपने वजन पर कड़ी नजर रखना चाहिए। उसे बिल्कुल न बढ़ने दें। मानसिक तनाव को भी दूर रखें। ब्लड प्रेशर बढ़ा हुआ हो या फिर शुगर या मधुमेह हो, तो ठीक समय दवा लेते रहें। इससे दिल को राहत मिलेगी। तले हुए, अधिक वसा वाले भोजन से बच कर रहें। घी-मक्खन, अंडे, चिकन, पूड़ी-कचौड़ी, समोसे, पराँठे, टिक्की, पकौड़ों आदि से दूर रहें। जहाँ तक हो सके, कॉफी और चाय भी कम से कम लें। धूम्रपान और मदिरा पर रोक लगा दें।

अतिरिक्त प्रश्नः

प्रश्न 8.
हृदय रोगी के खान-पान में क्या परहेज़ जरूरी हैं?
उत्तरः
हृदय रोगी को तले हुए, अधिक वसा वाले भोजन से बचना चाहिए। घी-मक्खन, अंडे, चिकन, पूड़ी-कचौड़ी, समोसे, पराँठे, टिक्की, पकौडो से दूर रहना चाहिए। कैफिन भी दिल की धड़कन तेज़ करती है। इसलिए जहाँ तक हो सके कॉफी और चाय भी कम से कम लेना चाहिए। धूम्रपान एवं शराब का सेवन भी वहीं करना चाहिए।

प्रश्न 9.
‘बायपास सर्जरी के बारे में लिखिए।
उत्तरः
बायपास सर्जरी में रूकी हुई कोरोनरी धमनी को बायपास करते हए, टाँग की शिरा का टुकड़ा इस तरह लगाया जाता है कि दिल के क्षतिग्रस्त भाग को फिर से खुराक मिलने लगती है।

प्रश्न 10.
दिल का दौरा और एनजाइना का निदान किस आधार पर किया जाता है?
उत्तरः
दिल के दौरे के तात्कालिक निदान में ई.सी.जी. और खास तरह का ब्लड टेस्ट किया जाता है। रोगी का न्यूक्लियर स्कैन से भी पता लगाया जाता है कि दिल का कितना हिस्सा दौरे की चपेट में बेकार हुआ है। इसे मूगा टेस्ट कहते हैं।
एनजाइना रोग में भी ई.सी.जी. महत्वपूर्ण है। ट्रेडमिल नामक टेस्ट भी होता है। कुछ रोगियों में ‘हाल्टर’ जाँच के अन्तर्गत रोगी का 24 से 72 घंटे तक ई.सी.जी. ली जाती है। ‘कोरोनरी एंजियोग्राफी’ से भी इस रोग का निदान किया जाता है।

प्रश्न 11.
हृदय के महत्व के बारे में लिखिए।
उत्तरः
हृदय शरीर का महत्वपूर्ण अंग है जो पूरे शरीर में जीवनदायी रक्त को पंप करता है। इस काम के लिए वह दिन रात कड़ी मेहनत करता है और उसे स्वयं भी उर्जा की जरूरत पड़ती है।

III. ससंदर्भ स्पष्टीकरण कीजिए :

प्रश्न 1.
“हृदय ही शरीर का वह महत्वपूर्ण अंग है जो शरीर के समस्त भागों को जीवनदायी रक्त को पंप करता है|”
उत्तरः
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘दिल का दौरा और एनजाइना’ नामक पाठ से लिया गया है। इसके लेखक डॉ. यतीश अग्रवाल हैं।
संदर्भ : लेखक शरीर में दिल के महत्वपूर्ण स्थिति के बारे में एवं उसके महत्व के बारे में बता रहे हैं।
स्पष्टीकरण : लेखक दिल के महत्त्व के बारे में बताते हुए कह रहे हैं कि यह हमारे संपूर्ण शरीर में जीवनदायी रक्त का संचरण करता है। पूरे शरीर में रक्त का प्रवाह सुचारू रूप से होता रहे इसके लिए दिल का स्वस्थ रहना निहायत जरूरी है। इसके लिए दिल को दिन-रात कड़ी मेहनत करनी पड़ती है।

प्रश्न 2.
“इस काम के लिए वह दिन-रात कड़ी मेहनत करता है और उसे स्वयं भी ऊर्जा की ज़रूरत पड़ती है|”
उत्तरः
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘दिल का दौरा और एनजाइना’ नामक पाठ से लिया गया है। इसके लेखक डॉ. यतीश अग्रवाल हैं।
संदर्भ : लेखक शरीर में दिल की कार्यक्षमता के बारे में बता रहे हैं।
स्पष्टीकरण : हृदय हमारे शरीर का सबसे महत्वपूर्ण अंग है। वह पूरे शरीर में जीवन प्रदायक रक्त को पंप करता है। इस कार्य को वह दिन रात करता है। जाहिर सी बात है कि इसके लिए दिल को भी उर्जा की जरूरत पड़ती है। दिल को यह उर्जा दो मुख्य कोरोनरी धमनियों से प्राप्त होती है।

प्रश्न 3.
“जिन्दगी में मानसिक तनाव को दूर रखें। उतार-चढ़ाव, लाभ-हानि – इसे जीवन का अंग मान लें|”
उत्तरः
प्रसंग : प्रस्तुत गद्यांश हमारी पाठ्य पुस्तक ‘साहित्य वैभव’ के ‘दिल का दौरा और एनजाइना’ नामक पाठ से लिया गया है। इसके लेखक डॉ. यतीश अग्रवाल हैं।
संदर्भ : लेखक हृदय रोग से बचाव के बारे में पाठकों को बता रहे हैं।
स्पष्टीकरण : हृदय रोग से बचने के लिए नियमित व्यायाम करना चाहिए। हृदय रोग जिनको है वे भी सुबह-शाम सैर करें। अपने वजन का ध्यान रखें। लेखक कहते हैं कि जिंदगी में मानसिक तनाव को दूर रखें। जीवन में उतार-चढ़ाव आते रहते हैं। लाभ-हानि यह सब जीवन का अभिन्न अंग है। इसे स्वीकार कर लेना चाहिए अर्थात् सकारात्मक सोच के साथ जीवन यापन करना चाहिए।

IV. कोष्टक में दिए गए कारक चिन्हों से रिक्त स्थान भरिएः

(का, में, के, पर, की)

प्रश्न 1.
उसके सीने ……….. दर्द उठने लगता है।
उत्तरः
में

प्रश्न 2.
इसके साथ ही जोरों ………. पसीना छूटने लगता है।
उत्तरः
का

प्रश्न 3.
मितली ……. शिकायत भी हो सकती है।
उत्तरः
की

प्रश्न 4.
पास ……….. डॉक्टर को बुला भेजें।
उत्तरः
के

प्रश्न 5.
कास्टेल उपस्थियों के संगम स्थल ……. जोरों का दर्द हो सकता है।
उत्तरः
पर

V. अन्य वचन रूप लिखिए:

प्रश्न 1.
पसली, तंत्रिका, दवा, सीमा, धमनी।
उत्तरः

• पसली – पसलियाँ
• तंत्रिका – तंत्रिकाएँ
• दवा – दवाइयाँ/दवाएँ
• सीमा – सीमाएँ
• धमनी – धमनियाँ

VI. विलोम शब्द लिखिएः

प्रश्न 1.
पास, नीचे, छोटा, चैन, ज्यादा, समर्थ।
उत्तरः

• पास × दूर
• नीचे × ऊपर
• छोटा × बड़ा
• चैन × बेचैन
• ज्यादा × कम
• समर्थ × असमर्थ

दिल का दौरा और एनजाइना लेखक परिचयः

डॉ. यतीश अग्रवाल वरिष्ट चिकित्सक एवं चिकित्सा विज्ञान के सुप्रसिद्ध लेखक हैं। आपने अनेक रोगों के लक्षण और रोग-निदान के उपायों पर कई लेख लिखे हैं। आप संप्रति दिल्ली के सफदरजंग अस्पताल में वरिष्ट आचार्य के पद पर सेवारत हैं।

आपके लेख अनेक पत्र-पत्रिकाओं में प्रकाशित हो चुके हैं। आपने औषध विज्ञान संबंधी विषयों पर संशोधन लेख भी प्रस्तुत किया है। आपकी कई कृतियाँ भारतीय भाषाओं के अतिरिक्त अंग्रेजी, जापानी एवं चीनी भाषाओं में भी अनूदित हैं।

डॉ. अग्रवाल की प्रमुख पुस्तकों में – “ए हंड्रेड्ज लाईफ्स’, ‘दग्दर बाबू’, ‘रिव्हर्सिंग बैक पैन’, ‘हाई ब्लड प्रेशर’, ‘हार्ट अटैक’ आदि प्रमुख हैं।

उद्देश्य : प्रस्तुत लेख को ‘स्वास्थ्य के २०० सवाल’ पुस्तक से लिया गया है। आज के सामाजिक संघर्ष एवं तनाव से भरे जीवन में दिल का दौरा एक सामान्य बीमारी है। इससे बचाव और मुक्ति पाने के उपाय इस लेख में दिये गए हैं।

विद्यार्थियों को ‘दिल का दौरा’ और ‘एनजाइना’ जैसी घातक बीमारी से अवगत कराने के उद्देश्य से इस लेख का चयन किया गया है।

### दिल का दौरा और एनजाइना Summary in Hindi

‘हम आज विज्ञान के युग में जी रहे हैं। वैज्ञानिकों ने अपनी प्रतिभा के द्वारा जीवन के हर क्षेत्र को प्रभावित किया है। चिकित्साशास्त्र के अनुसंधानों ने भयंकर बीमारियों का भी इलाज सुलभ बना दिया है। एक जमाने में दिल का दौरा और एनजाइना लाइलाज रोग थे। अब इनका उपचार सबके लिए आसान हो गया है। हृदय शरीर का अत्यन्त प्रधान अंग है। यह शरीर के प्रत्येक अंगों तक रक्त पहुंचाता है। यह काम दिल की दो कोरोनरी धमनियों और उनकी शाखाओं के द्वारा किया जाता है।

किसी कारण से दिल की धमनियों या उनकी शाखाओं में सिकुड़न आ जाती है, तो रक्त का प्रवाह ठीक तरह से नहीं हो पाता और दिल खून को पंप करने में असमर्थ हो जाता है। ऐसे समय हृदय को ऊर्जा पहुँचाने वाली किसी एक धमनी में एकाएक रुकावट आ जाने से दिल का दौरा पड़ता है। रक्त प्रवाह ठीक से न होने पर मनुष्य कुछ भी करने में असमर्थ हो जाता है। इसी को एनजाइना कहते हैं। दिल का दौरा उग्र रूप है, तो एनजाइना उसका हल्का रूप है।

उच्च रक्तचाप, मधुमेह, मोटापा, मानसिक तनाव, तंबाकू का उपयोग, मद्यपान आदि कारणों से दिल का दौरा पड़ता है। यह आनुवंशिक रोग भी है। नियमित रूप से व्यायाम न करना भी एक प्रमुख कारण है। दिल का दौरा किसी भी उम्र के व्यक्तियों को पड़ सकता है। सीने में असहनीय पीड़ा होती है, तो समझना चाहिए कि दिल का दौरा पड़ा है। यह दर्द बाएँ कंधे, गर्दन, बाँह और अंगुलियों के पौरों तक फैलता है। इसके साथ ही जोरों का पसीना छूटने लगता है।

एनजाइना रोग के निदान के कुछ उपाय ये हैं – ई.सी.जी., ट्रेड मिल नामक टेस्ट। इनसे रोग का पता नहीं लगता है, तो कोरोनरी एंजियोग्राफी करवाने से रोग का पता जरूर लगता है। हार्ट-अटैक की जानकारी ई.सी.जी. तथा खास तरह के ब्लड टेस्ट के द्वारा मिलती है। बाद में जरूरत होने पर दिल का न्यूक्लियर स्कैन कराने के द्वारा दिल की सही स्थिति का पता चलता है। बायपास सर्जरी से रुकी हुई कोरोनरी धमनी को बायपास करने पर दिल के उस भाग को फिर से पूरी खुराक मिलने लगती है।

दिल के दौरे का निवारण करने के लिए रोगियों को कुछ सतर्कता बरतनी पड़ती है। मितभोजन, तेल-मसाले से परहेज, धूम्रपान न करना, मद्यपान नहीं करना, मानसिक तनाव से दूर रहना, रक्तचाप और मधुमेह को नियंत्रण रखना आदि का पालन करने से हम दिल की बीमारी से बच सकते है|

दिल की बीमारी घातक है, लेकिन अब इसका इलाज होने लगा है। बड़े-बड़े शहरों में हृदयरोगों के विशेषज्ञ मिलते हैं। जब कभी दिल की बीमारी की भनक लगती है, तो तुरंत दिल-हृदय विशेषज्ञों के पास जाना चाहिए। अब हृदय-रोग से घबराने की आवश्यकता नहीं है।

### दिल का दौरा और एनजाइना Summary in English

This lesson, titled ‘Dil ka Daura aur Angina’, is taken from the book, ‘Svasthya ke 200 savaal’, written by Dr. Yathish Agarwal.
We live in an age of Science. Through their efforts, scientists have impacted every sphere of human life and activity. Practitioners of medical science have simplified the treatment of even the worst diseases that affect us. Once upon a time, heart attack and angina were untreatable. Now, getting treated for these conditions has become easy for anyone. The heart is one of the most important organs of our body. It pumps blood to every organ in our body. This is done through the two coronary arteries and their branches.

For some reason, if there is contortion or shrinkage in these arteries or their branches, then the blood does not flow properly and the heart is unable to pump blood. At such times, if any of the arteries providing energy and blood flow to the heart muscles have a blockage, or stop working, the person has a heart attack. If the blood doesn’t flow properly, a person is unable to do’any activity. This is called angina. If a heart attack is a severe form, then angina is the milder form.

High blood pressure, diabetes, obesity, mental stress, use of tobacco or alcohol can cause a heart attack. This is also a hereditary disease. One of the major causes of heart disease is a lack of regular exercise. A heart attack can happen to a person of any age. If one feels unbearable pain in the chest, then it might be a heart attack. The pain can spread to the right shoulder, the neck, arms and even the tips of fingers. Along with this, there is profuse sweating.

There are a few ways to diagnose angina – E.C.G. and treadmill tests. If these are inconclusive, then a coronary angiography must be conducted, which will certainly diagnose the problem. To know if one is at risk of heart attack, E.C.G. and some blood tests are conducted. Later, if required, a nuclear scan of the heart will show the exact state of the heart. Coronary arteries which have blockages can be bypassed through a heart bypass surgery, which provides an alternative route for the blood to reach the heart muscles.

To decrease the chances of a heart attack, patients must take care of and follow certain precautions. Restricted diet, avoiding too much oil or spices in the diet, not using tobacco products, not consuming alcohol, staying away from mental stress, keeping diabetes and blood pressure under control are some of the precautions which will keep diseases of the heart at bay.
The disease of the heart is very dangerous, but now there are treatments available for these conditions. In big cities, there are heart specialists. If we ever feel that we are at risk of heart disease, or we feel unwell, or feel pain in the chest, we must immediately visit a heart specialist. Now there is no need to be afraid of heart diseases.

कठिन शब्दार्थः

• ऊर्जा – शक्ति, बल;
• वसा – फैट, चरबी;
• गुब्बारा – हवा भरी रबर की थैली, ಬಲೂನ್;
• खाद्य – आहार;
• निदान – चिकित्सा;
• भाना – अच्छा लगना;
• बह्वाल – स्वस्थ;
• कूचकर जाना – मरजाना;
• गम – दुःख।
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