KSEEB Solutions for Class 7 Maths Chapter 6 Tribhuja Mattu Adara Gunagalu Ex 6.5

Students can Download Maths Chapter 6 Tribhuja Mattu Adara Gunagalu Ex 6.5 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths in Kannada helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

KSEEB Solutions for Class 7 Maths Chapter 6 Tribhuja Mattu Adara Gunagalu Ex 6.5

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2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3

Students can Download Basic Maths Exercise 18.3 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3

Part – A

2nd PUC Basic Maths Differential Calculus Ex 18.3 One or Two Marks Questions and Answers

Question 1.
3x2 + 4y2 = 10
Answer:
Given 3x2 + 4y2 = 10
Diff w.r.t x
6x + 8y \(\frac{d y}{d x}\) = 0
\(\Rightarrow \quad \frac{d y}{d x}=\frac{-8 y}{6 x}=\frac{-4 y}{3 x}\)

Question 2.
\(\sqrt{x}+\sqrt{y}=3\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 1

KSEEB Solutions

Question 3.
y2 = 4ax.
Answer:
Given y2 = 4ax.
Differentiate with respect to x
2y \(\frac{d y}{d x}\) = 4a.1 ⇒ \(\frac{d y}{d x}\) = \(\frac{4 a}{2 y}=\frac{2 a}{y}\)

Question 4.
\(x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 2

Question 5.
x2 = 4ay
Answer:
Given x2 = 4ay
Differentiate with respect to x, 2x = 4a \(\frac{d y}{d x}\) ⇒ \(\frac{d y}{d x}\) = \(\frac{2 x}{4 a}=\frac{x}{2 a}\)

Question 6.
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 3

Question 7.
x3 + y3 = 3axy
Answer:
Given x3 + y3 = 3axy
3x2 + 3y2 \(\frac{d y}{d x}\) = 3a \(\left(x \cdot \frac{d y}{d x}+y \cdot 1\right)\)
\(\frac{d y}{d x}\) (3y2 – 3ax) = 3ay – 3x2 = \(\frac{d y}{d x}=\frac{a y-x^{2}}{y^{2}-a x}\)

Question 8.
x – y = 0
Answer:
Given x – y = 0
Differentiate with respect to x,
1 – \(\frac{d y}{d x}\) = 0 ⇒ \(\frac{d y}{d x}\) = 1

KSEEB Solutions

Question 9.
x2 – y2 = a2
Answer:
Given x2 – y2 = a2
Differentiate with respect to x we get,
2x – 2y. \(\frac{d y}{d x}\) = 0 ⇒ \(\frac{d y}{d x}\) = \(\frac{2 x}{2 y}=\frac{x}{y}\)

Question 10.
x + \(\sqrt{x y}\) = x2.
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 4

Part-B

2nd PUC Basic Maths Differential Calculus Ex 18.3 Three marks Questions and Answers

Question 1.
log(xy) = x2 + y2
Answer:
Given log(xy) = x2 + y2
log x + log y = x2 + y2 differentiate w.r.t x
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 5

Question 2.
2x + 2y = 2x+y
Answer:
Given 2x + 2y = 2x+y
Differentiate w.r.t. x we get
2x log 2 + 2y log 2 \(\frac{d y}{d x}\)
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 6

KSEEB Solutions

Question 3.
xy = yx.
Answer:
Given xy = yx., taking logm both sides
y log x = x log y differentiate
Both sides w.r.t x
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 7

Question 4.
sin xy = cos(x + y).
Answer:
Given sin xy = cos(x + y), diff w.r.t x.
cos(xy) \(\left[\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}\right]\)
\(\frac{d y}{d x}\) [sin(x + y) + xcos (xy)]
= -sin(x + y) – y cos (xy)
\(\frac{d y}{d x}=\frac{-[\sin (x+y)+\cos x y]}{(\sin (x+y)+x \cos x y)}\)

Question 5.
y = 4x+y
Answer:
Given y = 4x+y, diff. w r.t. x
\(\frac{d y}{d x}\) = 4x+y log 4(1 + \(\frac{d y}{d x}\)) = 4x+y
\(\frac{d y}{d x}\)(1 – 4x+y log 4) = 4x+y log 4
∴ \(\frac{d y}{d x}=\frac{4^{x+y} \cdot \log 4}{1-4^{x+y} \cdot \log 4}\)

KSEEB Solutions

Part-C

2nd PUC Basic Maths Differential Calculus Ex 18.3 Five Marks Questions and Answers.

Question 1.
If \(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}\) = a, Prove that x . \(\frac{d y}{d x}\) = y.
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 8

Question 2.
If xy = ey – x, show that \(\frac{d y}{d x}\) = \(\frac{2-\log x}{(1-\log x)^{2}}\)
Answer:
Given xy = ey – x . Taking log both sides
y log x = (y – x)log ee
x = y (1 – log x) ∵ log ee = 1
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 9

KSEEB Solutions

Question 3.
If cos y = x cos(a + y). show that \(\frac{d y}{d x}\) = \(\frac{\cos ^{2}(a+y)}{\sin a}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 10

Question 4.
If ex = yx show that \(\frac{d y}{d x}\) = \(\frac{(\log y)^{2}}{\log y-1}\)
Answer:
Given ex = yx Taking logm both sides
y log ee = x log y
y = x log y differentiate w.r.t x
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 11

Question 5.
If ex+y = xy show that \(\frac{d y}{d x}\) = \(\frac{y(1-x)}{x(y-1)}\)
Answer:
Given yex+y = xy
Taking log m both sides
(x + y) loge = log(xy)
x+y = log x + log y diff w.r.t x
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 12

KSEEB Solutions

Question 6.
If yx = xy show that \(\frac{d y}{d x}\) = \(\frac{y(y=x \log y)}{x(x-y \log x)}\)
Answer:
Given yx = xy, Taking logm both sides
x log y = y log x, diff w.r.t x
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.3 - 13

KSEEB Solutions for Class 7 Maths Chapter 1 Purnamsagalu Ex 1.4

Students can Download Maths Chapter 1 Purnamsagalu Ex 1.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths in Kannada helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

KSEEB Solutions for Class 7 Maths Chapter 1 Purnamsagalu Ex 1.4

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KSEEB Solutions for Class 7 Maths Chapter 10 Prayogika Rekhaganita Ex 10.1

Students can Download Maths Chapter 10 Prayogika Rekhaganita Ex 10.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths in Kannada helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

KSEEB Solutions for Class 7 Maths Chapter 10 Prayogika Rekhaganita Ex 10.1

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KSEEB Solutions for Class 7 Maths Chapter 8 Parimanagala Holike Ex 8.3

Students can Download Maths Chapter 8 Parimanagala Holike Ex 8.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths in Kannada helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

KSEEB Solutions for Class 7 Maths Chapter 8 Parimanagala Holike Ex 8.3

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KSEEB Solutions for Class 7 Maths Chapter 8 Parimanagala Holike Ex 8.2

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KSEEB Solutions for Class 7 Maths Chapter 8 Parimanagala Holike Ex 8.2

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2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3

Students can Download Maths Chapter 6 Application of Derivatives Ex 6.3 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3

2nd PUC Maths Application of Derivatives NCERT Text Book Questions and Answers Ex 6.3

Question 1.
Find the slope of the tangent to the curve y = 3x4 – 4x at x = 4.
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.1

Question 2.
Find the slope of the tangent to the curve
\(y=\frac{x-1}{x-2}, x \neq 2 \text { at } x=10 \)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.2

Question 3.
Find the slope of the tangent to curve y = x3 – x + 1 at the point whose x- coordinate is 2.
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.3

Question 4.
Find the slope of the tangent to the curve y = x3 – 3x + 2 at the point whose x-coordinate is 3.
Answer:
\(\frac{d y}{d x}\) = 3x2 – 3 dx
slope at x = 3 is 3 (9) – 3 = 24.

KSEEB Solutions

Question 5.
Find the slope of the normal to the curve
\(x=a \cos ^{3} \theta, y=a \sin ^{3} \theta \text { at } \theta=\frac{\pi}{4}\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.5
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.6

Question 6.
Find the slope of the normal to the curve
\(x=1-a \sin \theta, y=b \cos ^{2} \theta \text { at } \theta=\frac{\pi}{2}\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.7

Question 7.
Find points at which the tangent to the curve y = x3 – 3x2 – 9x + 7 is parallel to
the x-axis.
Answer:
\(\frac{d y}{d x}\) = 3x2 – 6x – 9, slope of the tangent
since the tangent is parallel to x-axis \(\frac{d y}{d x}\) = 0
3x2 – 6x – 9 = 0
3 (x + 1) (x – 3) = 0, x = 3, x = -1
when x = 3, y = 27 – 27 – 27 4- 7 = -20
when x = -1, y = -1 -3 + 9 + 7 = 12
The points at which the tangent parallel to x – axis are (3, -20) and (-1, 12).

Question 8.
Find a point on the curve y = (x – 2)2 at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.8
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.9

Question 9.
Find the point on the curve y = x3 – 11x + 5 at which the tangent is y = x – 11.
Answer:
Slope of the line y = x – 11 is 1
slope of the curve \(\frac{d y}{d x}\) = 3x2 – 11
∴ 3x2 – 11 = 1                   ‘
3x2 = 12 ⇒ x2 = 4, x = +2
when x = 2, y = (2)3 – 11 (2) + 5 = -9
when x = -2, y = -8 + 22 + 5 = 19
points are (2, -9) and (-2, 19)
equation of tangent at (2, -9) and slope is 1
y + 9 = 1 (x – 2)
y = x- 11 equation of tangent at (-2, 19)
y – 19 = 1 (x + 2) ⇒ y = x + 211
∴ (2, -9) is the only point at which the tangent is y = x – 11.10.

KSEEB Solutions

Question 10.
Find the equation of all lines having slope – 1 that are tangents to the curve
\(y=\frac{1}{x-1}, x \neq 3\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.10
x = 2, x = 0
when x = 2, y = 1
when x = 0, y = -1
The points are (2, 1) and (0,-1)
equation of line though (2,1) having slope = -1
is y – 1 = -1 x (x – 2) ⇒ y + x = 3
or x + y – 3 = 0 and equation of line through (0,- 1)y + 1 = -1 (x – 0)
∴  y + x + 1 = 0.

Question 11.
Find the equation of all lines having slope 2 which are tangents to the curve
\(y=\frac{1}{x-3}, x \neq 3\)
Answer:
slope of the line = 2
slope of the curve = \(\frac{-1}{(x-3)^{2}}\)
\(\frac{-1}{(x-3)^{2}}\) =2
⇒ 2 (x -3)2 = -1
⇒ 2 (x2 – 6x + 9) = -1
⇒ 2x2 – 12x + 19 = 0
which has no real roots b2 – 4ac < 0
hence there is no point on the curve.

Question 12.
Find the equations of all lines having slope 0 which are tangent to the curve
\(y=\frac{1}{x^{2}-2 x+3}\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.11

Question 13.
Find the points of curve \(\frac{x^{2}}{9}+\frac{y^{2}}{16}=1\) which the tangents are
(i) parallel to x-axis
(ii) parallel to y-axis.
Answer:
16x2 + 9y2 = 144
16 x 2 x + 9 x 2y x \(\frac{d y}{d x}\) = 0
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.12
(0,4) and (0, -4) are the points on the curve at which tangent is parallel to x – axis.

KSEEB Solutions

(ii) For tangents parallel to y – axis \(\frac{d y}{d x}\) = \(\frac{1}{0}\)
\(\frac{-16 x}{9 y}=\frac{1}{0} \Rightarrow y=0\)
when y = 0, x2 = 9x = ± 3 point is (3,0) and (-3,0)
(3, 0) and (-3, 0) are the points at which the curve is parallel to y – axis.

Question 14.
Find the equations of the tangent and normal to the given curves at the indicated points:
(i) y = x4 – 6x3 + 13x2 – 10x + 5 at (0, 5)
Answer:
y = x4 – 6x3 + 13x2 – 10x + 5 at (0, 5) dy
\(\frac{d y}{d x}\) = 4x3 – 18x2 + 26x – 10 dx
slope at (0,5) = -10
∴ Equation of tangent at (0, 5) is
y – 5 = -10 (x – 0)
y – 5 = -10 x 10 x + y – 5 = 0
slope of the normal at (0,5)
\((0,5)=\frac{-1}{-10}=\frac{1}{10}\)
∴ equation of normal is
y – 5 =\(\frac{1}{10}\) (x – 0) = 10y – 50 = x
x – 10y + 50 = 0

(ii) y = x4 – 6x3 + 13x2 – 10x + 5 at (1, 3)
Answer:
\(\frac{d y}{d x}\) = 4x3 – 18x2 + 26x – 10 dx
slope of the tangent at x = 1

(iii) y = x3 at (1,1)
Answer:
\(\frac{d y}{d x}=3 x^{2}\) 
∴ slope of the tangent at x = 1 = 3
∴ Equation of tangent is y – 1 = 3(x – 1)
3x – y – 2 = 0
∴ slope of normal = \(\frac{-1}{3} \)
∴ equation of normal is y – 1 = \(\frac{-1}{3} \)(x – 1)
3y – 3 = – (x – 1)
x + 3y – 4 = 0.

KSEEB Solutions

(iv) y = x2 at (0, 0)
Answer:
\(\frac{d y}{d x}=2 x\)
∴ slope at x = 0 =0
∴ Equation of tangent is y – 0 = 0 (x – 0) ⇒ y = 0
slope of the normal = -1/(0)
∴ equation of normal \(y – 0=\frac{-1}{0}(x-0) \Rightarrow x=0\)

(v) x = cos t, y = sin t at \(\frac{\pi}{4}\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.13
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.14

Question 15.
Find the equation of the tangent line to the curve y = x2 – 2x +7 which is
(a) parallel to the line 2x – y + 9 = 0
Answer:
\(\frac{d y}{d x}\) = 2x – 2 also slope = 2
2 x – 2 = 2 ⇒2x = 4 ⇒ x = 2
when x = 2, y = (2)2 – 2 (2) + 7 = 7
∴ equation of tangent is y – 7 = 2 (x – 2)
2x – y + 3 = 0 ⇒ 2x – y + 3 = 0.

(b) perpendicular to the line 5y – 15x = 13.
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.15
12x + 36y – 227 = 0

Question 16.
Show that the tangents to the curve y = 7x3 + 11 at the points where x = 2 and x = – 2 are parallel.
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.16
KSEEB Solutions

Question 17.
Find the points on the curve y = x3 at which the slope of the tangent is equal to the y-coordinate of the point.
Answer:
y = x3
\(\frac{d y}{d x}\) = 3x2,
Given that \(\frac{d y}{d x}\) = y = x3
∴ 3x2 = x3 ⇒ x2 (3 – x) = 0 ⇒ x = 0 or x = 3
when x = 3, y = 33 = 27, when x = 0, y = 0
∴ The required points are (0, 0), (3, 27).

Question 18.
For the curve y = 4x3 – 2x5, find all the points at which the tangent passes through the origin.
Answer:
y = 4x3 – 2x5 dy
\(\frac{d y}{d x}\) = 12x2 – 10x4 dx
Let (a, b) be the point on the curve at which the tangent passes through the origin.
∴ Equation of tangent is
y – b = (12a2 – 10a2) (x – a)
but this passes through the origin
∴ 0 – b = (12a2 – 10a4) (-a) b = 12a3 – 10a5 ….(1)
Also from the equation b = 4a3 – 2a5 …. (2)
from (1) and (2) 12a3 – 10a5 = 4a3 – 2a5
8a3 = 8a5
a3 (1 – a2) = 0 ⇒ a = 0, a = + 1
when a = 0, b = 0, (0, 0)
a = 1,b= 12(1)- 10(1) = 2, (1,2)
a = -1, b = 12 (-1) -10 (-1) = -2, (-1, -2)
Hence the required points are (0,0), (1,2), (-1,-2).

Question 19.
Find the points on the curve x2 + y2 – 2x – 3 = 0 at which the tangents are parallel to the x-axis.
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.17

KSEEB Solutions

Question 20.
Find the equation of the normal at the point (am2,am3) for the curve ay2 = x3.
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.18

Question 21.
Find the equation of the normal to the curve y = x3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0.
Answer:
y = x3 + 2x + 6
\(\frac{d y}{d x}\) = 3x2+ 2
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.19

Question 22.
Find the equations of the tangent and normal to the parabola y2 = 4ax at the
point (at2, 2at).
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.20
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.21

Question 23.
Prove that the curves x = y2 and xy = k cut at right angles* if 8k2 = 1.
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.22

Question 24.
Find the equations of the tangent and normal to the hyperbola
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) at the point (x0, y0)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.23
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.24

Question 25.
Find the equation of the tangent to the curve
\(y=\sqrt{3 x-2}\) which is parallel to the line 4x – 2y +5 = 0
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.25

2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.26

KSEEB Solutions

Choose the correct answer in Exercises 26 and 27.

Question 26.
The slope of the normal to the curve y = 2x2 + 3 sin x at x = 0 is
(A) 3
(B) 1/3
(C) -3
(D) -1/3
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.27

Question 27.
The line y = x + 1 is a tangent to the curve y2 = 4x at the point
(A) (1, 2)
(B) (2,1)
(C) (1, – 2)
(D) (- 1, 2)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.28
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3.29

1st PUC Chemistry Model Question Paper 3 with Answers

Students can Download 1st PUC Chemistry Model Question Paper 3 with Answers, Karnataka 1st PUC Chemistry Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Chemistry Model Question Paper 3 with Answers

Time: 3.15 Hours
Max Marks: 70

Instruction:

  1. The questions paper has five parts A, B, C, D and E. All parts are compulsory.
  2. Write balanced chemical equations and draw labeled diagram wherever allowed.
  3. Use log tables and simple calculations f necessary (use of scientific calculations is not allowed).

Part – A

I. Answer ALL of the following (each question carries one mark): ( 10 × 1 = 10 )

Question 1.
‘Cis platin’ a medicine used in the treatment of which disease?
Answer:
Cancer/Cancer Tumour.

Question 2.
Write the mathematical expression for Boyle’s law.
Answer:
V ∝ \(\frac{1}{P}\) at constant temperature.

Question 3.
Give the example for gaseous reversible reaction for which Kp = Kc
Answer:
H2(g) + I2(g) ⇌ 2HI(g)

Question 4.
Which group elements in the periodic table are called Noble gases?
Answer:
18th group.

KSEEB Solutions

Question 5.
What is the oxidation number of the element in its free state?
Answer:
0 or Zero.

Question 6.
Write the general electronic configuration of First group elements.
Answer:
ns1

Question 7.
Why is boric acid considered as weak acid?
Answer:
Due to small size of B and presence of only six electrons in the valence shell i.e. protonic acid.

Question 8.
Mention the structure of SiO44-.
Ans.
Tetrahedral structure.

KSEEB Solutions

Question 9.
Give an example for non-benzenoid compound.
Answer:
Pyridine

Question 10.
Write the expanded form of ‘CNG
Answer:

Part – B

II. Answer any FIVE of the following questions carrying TWO marks ( 5 X 2 = 10 )

Question 11.
Convert 37°C to °F.
Answer:
°F = \(\frac{9}{5}\) + °C +32 = \(\frac{9}{5}\) X 37 + 32 = 98.6°

Question 12.
Under what conditions of temperature and pressure real gases tend to behave ideally?
Answer:
Low pressure and high temperature.

KSEEB Solutions

Question 13.
What is dipole moment? What is its SI unit.
Answer:
It is the product of +ve or negative charge and the distance between the centre of them, μ = e x d
SI unit coulomb meter or cm.

Question 14.
Write any two diagonal relationship between Beryllium and Aluminium.
Answer:
(a) Both forms covalent compounds and soluble in organic compounds.
(b) Both BeCl2 and AlCl3 act as Lewis acids.

Question 15.
Give the reaction for the synthesis of water gas and producer gas.
Answer:
1st PUC Chemistry Model Question Paper 3 with Answers - 1

Question 16.
How is chloromethane converted to methane?
Answer:
Chloromethane heated with zinc and NaOH to methane.
1st PUC Chemistry Model Question Paper 3 with Answers - 2

Question 17.
Illustrate Markovnikov’s rule with an example.
Answer:
1st PUC Chemistry Model Question Paper 3 with Answers - 3

The Br of HBr is added to double bonded carbon as containing less of OH atoms to give 2–bromopropanol.

Question 18.
What is BOD? What is its significance?
Answer:
The amount of oxygen consumed by the microorganisms in decomposing the organic matter present in water is called BOD. It is a water quality parameter to know the amount of organic matter available for bacteria.

KSEEB Solutions

Part – C

III. Answer any FIVE of the following questions; carrying THREE marks: ( 5 x 3 = 15 )

Question 19.
(a) Give reason : Ionic radius of Fis more than atomic radius of F.
Answer:
Due to more number of electrons in F than F.

(b) How does ionization enthalpy varies down the group?
Answer:
Ionisation enthalpy decreases down the group due to increase in atomic size of atoms.

(c) Ionization enthalpy of nitrogen is more than that of oxygen. Give reason.
Answer:
Due to completely half filled p-orbitals (1s2 2s2 2p3) in nitrogen but in oxygen atom p-orbital is more than half-filled.

Question 20.
With the help of MOT write the energy level diagram of hydrogen molecule. What is its bond order and predict magnetic property.
Answer:
BO = \(\frac{1}{2}\) (NBMO NABMO)
= \(\frac{1}{2}\)(2 – 0 ) = 1
It is diamagnetic due to absence of unpaired electrons in the molecule.
1st PUC Chemistry Model Question Paper 3 with Answers - 4

Question 21.
Calculate the formal charge of each oxygen atom of ozone molecule
Answer:
1st PUC Chemistry Model Question Paper 3 with Answers - 5

Question 22.
(a) Give any two differences between sigma and pi bonds.
Answer:
Sigma Bond

  1. Head to Head overlapping of AO.
  2. It is strong bond due to large overlapping.
  3. Free rotation is possible.

Pi Bond

  1. Lateral (sideways) overlapping of AO.
  2. It is weak bond due to small overlapping.
  3. Free rotation about n bond is not possible.

(b) What is the magnetic nature of oxygen molecule?
Answer:
It is paramagnetic due to presence of unpaired electrons in πpx πpy antibonding molecular orbitals.

Question 23.
Balance the following redox reaction using oxidation number method.
MnO2 + Br→ Mn2+ +Br2 + H2O (In Acidic medium)
Answer:
1st PUC Chemistry Model Question Paper 3 with Answers - 6
Step II – Oxidation Reaction x 2
Reduction Reaction x 1
1st PUC Chemistry Model Question Paper 3 with Answers - 7
Balance the O atoms (add to RHS) and balance the H+ ions (To LHS)
MnO2 + 2B + 4H+ → Mn+2 +2Br2 +2H2O

Question 24.
(i) Explain the laboratory preparation of dihydrogen.
Answer:
In the laboratory, dihydrogen is prepared by adding dil. H2S04 to granulated zinc.
i.e. Zn + H2SO4 (dil) → ZnSO4 + H2
The H2 gas liberated is collected by downward displacement of water.

(ii) Give an example of ionic hydride.
Answer:
LiH/NaH/CaH2/BaH2 etc.

KSEEB Solutions

Question 25.
Give any three points of differences between lithium and other alkali metals.
Answer:

  1. Small size of lithium and its ion.
  2. High charge to size ratio (polarising power).
  3. High ionization enthalpy and low electropositive character of lithium.
  4. Absence of d-orbitals.

Question 26.
(i) Write the molecular formula of silica.
Answer:
SiO2

(ii) Write the partial structure of silicone.
Answer:
1st PUC Chemistry Model Question Paper 3 with Answers - 8

(iii) Name the catalyst used in gasoline production.
Answer:
ZSM-5.

Part – D

IV. Answer any FIV E of the following questions’carrying FIVE marks: ( 5 x 5 = 25 )

Question 27.
(a) An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 gram of this substance is subjected to complete combustion.
Answer:
Given % carbon = 69%, H = 4.8%, w = 0.20g
1st PUC Chemistry Model Question Paper 3 with Answers - 9

(b) Give an example each for element and compound.
Answer:
Element = sodium/Na.
Compound = NaOH/sodium hydroxide.

Question 28.
(a) Write the significance of quantum numbers n, I and m.
Answer:
(i) Principal quantum number (n): Determines energy and size of the orbits (electrons).
(ii) Azimuthal Q.N. (1): Determines the shape of the suborbitals (s,p,d,f).
(iii) Magnetic Q.N. (m): Determines the orientation of the orbitals.

(b) State Pauli’s exclusion principle.
Answer:
For two electrons of an atom the values of n, 1 and m but s is different.
OR
No two electrons of an atom can have the same values of n, 1 and m are same 3 is different.

(c) Write the electronic configuration of copper (Z = 29).
Ans. Cu29= 1s22s22p63s23p63d104s1

KSEEB Solutions

Question 29.
(a) The FM station of All India Radio, Hassan, broadcast on a frequency of 1020 kilohertz. Calculate the wavelength of the electromagnetic radiation emitted by the transmitter.
Answer:
Given γ = 1020 kilohertz = 1020 x 1000 hertz
C = 3 x108 ms-11 λ = ?
1st PUC Chemistry Model Question Paper 3 with Answers - 10

(b) What is the maximum number of electrons present in third main energy level?
Answer:
Third Main energy level = M r /.
∴ No. of electrons in M shell = 18e.

Question 30.
(a) Give any three postulates of kinetic molecular theory of gases.
Answer:

  1. ll gases are made up of with many number of tiny particles called molecules.
  2. Gas molecules are separated from each other by large distance. Hence the volume of the gas molecules is negligible compared to total volume of the gas.
  3. There is no force of attraction between the gas molecules.
  4. The average kinetic energy of gas molecules is directly proportional to Kelvin temperature.

(b) On a ship sailing in Pacific Ocean where temperature is 23.4°C, a balloon is filled with 2L air. What will be the volume of the balloon when the ship reaches 1 the Indian Ocean where temperature is 26.1°C?
Answer:
V1= 2L V2 = ?
T1= 23.4°C + 273 = 296.4 K
T2 = 26.1°C + 273K = 299.1K
1st PUC Chemistry Model Question Paper 3 with Answers - 11

Question 31.
(a) The combustion of one mole benzene takes place at 298K and 1 atm. After combustion, CO2(g) and H2O (l) are produced and 3267.0 kJ of beat is liberated. Calculate the standard enthaipy of formation of benzene.
Given : Standard enthalpy of formation of CO2(g) and H2O (1) are – 393.5 kJ mol-1 and -285.0 kJ mol-1 respectively.
Answer:
Required equation : 6C (s) + 3H2(g) → C6H6 (l); ∆Hf = ?
(i) C6H6(l) + \(\frac{15}{2}\)O2(g) → 6CO2(g) + 3H2O(l); ∆Hc = -3267KJ
(ii) C(s) + O2(g) → CO2 (g); ∆Hf = -393.5 KJ
(iii) H2(s) + \(\frac{1}{2}\)O2(g) → H2O(l); ∆Hf= -285 KJ
reverse the equation (i) + (ii) 6 + (iii) * 3
1st PUC Chemistry Model Question Paper 3 with Answers - 12

(b) Write the mathematical expression for First law of thermodynamics.
Answer:
∆U = q +w or ∆U = q – P∆V

(c) Give an example of isolated system.
Answer:
Tea placed in a thermos flask.

KSEEB Solutions

Question 32.
(a) What are the values for ArH° and ArS° for reaction to be spontaneous at all
temperatures?
Answer:
When ∆rH° = -ve and ∆rS° = +ve the process is spontaneous at all temperture.

(b) Write the relationship between
(i) Enthalpy (H) and infernal energy (U) (ii) Cp and Cv.
(iii) Free energy (G) Enthalpy (H) and Entropy (S).
Answer:
(i) ∆H = ∆U + P∆V
(ii) Cp – Cv= R for 1 mole of an ideal gas
Cp – Cv =nR for n moles of an ideal gas
(iii) ∆G = ∆H – T∆S

Question 33.
(a) The following concentrations were obtained for the formation of NH3 from N2 and H2 at equilibrium 500K. [Ni] = 1.5 x 10-2 M, [H2] = 3.0 x 10-2 M and [NHJ = 1.2 x 10-2 m. Calculate the equilibrium constant.
Answer:
Given [N2] = 1.5 x 10-2 M, [H2] = 3.0 x 10-2 M and [NH3] = 1.2 x 10-2m
For N2(g) + 3H2 (g) → 2NH3 (g)
1st PUC Chemistry Model Question Paper 3 with Answers - 13

b) Define common ion effect
Answer:
It is the process of decreasing (or suppression) the dissociation of a weak electrolyte by adding a strong electrolyte having a common ion is called common ion effect.
1st PUC Chemistry Model Question Paper 3 with Answers - 14

(c) What is the relationship between [HsO+] and [OH-] for neutral solution?
Answer:
[H3O+] = [OH].

Question 34.
(a) What is acid and bases according to Arrhenius concept?
Answer:
Arrhenius Acid : Substances which give hydrogen ion in water are called Arrhenius Acid. Example: HCl. HNO3
Arrhenius base : Substances which gives hydroxyl ion (OH-) in water are called Arrhenius bases.
Example: NaOH, KQH, NH4OH, etc.

(b) Give an example for (i) so.M-vapoisr equilibrium (ii) liquid-vapour equilibrium,
(i) I2(s) → I2 (vapour) / NH4Cl(s)⇌ NH4Cl(vapour) (ii) H2O(l) → H2O(v).

(c) Write the equilibrium K. expression for H2 + I2
Answer:
Kc = \(\frac{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}{[\mathrm{HI}]^{2}}\)

KSEEB Solutions

Part – E

V. Answer any TWO of the following questions carrying FIVE marks: ( 2 x 5 = 10 )

Question 35.
(a) For the molecule CH3CH2CH2OH
(i) Identify functional group
Answer:
Functional group = – OH

(ii) Write the bond line forarulbu
Answer:
Bond line formula =
1st PUC Chemistry Model Question Paper 3 with Answers - 15
(iii) Write the succeeding homologue.
Succeeding homologue = CH3CH2CH2CH2OH

(b) Explain inductive effect with a suitable example.
Answer:
The permanent displacement of a (sigma) electrons along the saturated carbon chain away/ towards the group/atom attached at the end of the chain.
1st PUC Chemistry Model Question Paper 3 with Answers - 16

Question 36.
(a) An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.
Answer:
Given % carbon = 69%, H = 4.8%, w = 0.20g
1st PUC Chemistry Model Question Paper 3 with Answers - 9

(b) What are nucleophiles? Give an example.
Answer:
Chemical species having -ve charge reacts with nucleus of an atom/molecule is called nucleophile, e.g: H, CN, H2O, NH3, CP, Br etc.

KSEEB Solutions

Question 37.
(a) Explain the mechanism of chlorination of methane.
Answer:
Mechanism of chlorination of methane :
Step I : Chain initiation: Methane and Cl2 heated in UV light, Cl2 undergoes homolytic fission giving chlorine free radicals,
1st PUC Chemistry Model Question Paper 3 with Answers - 18

Step II: Chain propagation: Cl reacts with methane gives methyl free radical,
1st PUC Chemistry Model Question Paper 3 with Answers - 19
Methyl free radical reacts with Cl2 molecule gives C1‘ radical, i.e. CH3 + Cl2 → CH3Cl + Cl’ These are repeated several times.

Step III: Chain termination:
The reaction goes to the end when the free radicals reacts with each other.
1st PUC Chemistry Model Question Paper 3 with Answers - 20

(b) Write the Newman’s projections of ethane.
Answer:
1st PUC Chemistry Model Question Paper 3 with Answers - 21

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