1st PUC Physics Question Bank Chapter 4 Motion in a Plane

You can Download Chapter 4 Motion in a Plane Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Physics Question Bank Chapter 4 Motion in a Plane

1st PUC Physics Motion in a Plane TextBook Questions and Answers

Question 1.
State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.
Answer:
Scalar: Volume, mass, speed, density, number of moles, angular frequency.
Vector: Acceleration, velocity, displacement, angular velocity.

Question 2.
Pick cut the two scalar quantities in the following list:
force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.
Answer:
Scalar quantities: Work, current.

Question 3.
Pick out the only vector quantity in the following list:
Temperature, pressure, impulse, time, power, total, path length, energy, gravitational potential, coefficient of friction, charge.
Answer:
Vector quantity: Impulse.

KSEEB Solutions

question 4.
State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful:

  1. adding any two scalars,
  2. adding a scalar to a vector of the same dimensions,
  3. multiplying any vector by any scalar,
  4. multiplying any two scalars,
  5. adding any two vectors,
  6. adding a component of a vector to the same vector.

Answer:

  1. No, scalars must represent same physical quantity.
  2. No, vector can be added only to another vector.
  3. YesYes
  4. The two vectors must represent the same physical quantity.
  5. Adding a component of a vector to same vector is meaningless.

Question 5.
Read each statement below carefully false:

  1. The magnitude of a vector is always a scalar,
  2. each component of a vector is always a scalar,
  3. the total path length is always equal to the magnitude of the displacement vector of a particle,
  4. the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of the average velocity of the particle over the same interval of time,
  5. Three vectors not lying In a plane can never add up to give a null vector.

Answer:

  1. True. The magnitude of a vector gives its length, which is always a scalar.
  2. False. Each component of a vector is a vector by itself.
  3. False. The total path length is always greater than or equal to the magnitude of the displacement vector of the particle.
  4. True. Since the path length is always greater than or equal to the magnitude of the displacement vector of the particle, the average speed is always greater than or equal to the magnitude of the average velocity.
  5. True. If the vectors are not coplanar, their addition will always yield some nonzero component in at least one direction.

Question 6.
Establish the following vector inequalities geometrically or otherwise:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 1
When does the equality sign above apply?
Answer:
a) Geometrical solution:-
Consider the triangle shown in the figure.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 2
which is true by the triangle inequality (Sum of the lengths of any two sides of a triangle is always greater than the length of the third side.)
Equality holds in the case of a degenerate triangle (collinear vectors).
Hence proved.
Analytical solution:-
We must prove that
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 65 sqaure both sides,
To prove that
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 4
i.e., to prove that
a² + b² + 2ab cos θ ≤ a² + b² + 2ab,
( here \(|\overrightarrow{\mathrm{a}}|\) = a , \(|\overrightarrow{\mathrm{b}}|\) = b)
i.e., to prove that
2ab cos θ ≤ 2ab i.e., TPT cosθ ≤ 1
Since the range of cose is [-1, 1], the above inequality is true. The equality holds when θ = 0° (collinear vectors).
Hence Proved.

b) Geometrical solution:
Consider the triangle shown in the figure.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 5
Which is true by the triangle inequality. (Difference in the lengths of any two sides of a triangle is always lesser than the length of the third side.) The equality holds for a degenerate triangle (collinear vectors).
Proved.
Analytical solution:-
We are to prove that
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 6
Square both sides we must prove that
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 7
i.e., To prove that
a² + b² + 2ab cos θ ≥ a² + b² – 2ab
i.e., to prove that 2ab cos θ ≥ – 2 ab
to prove that cos θ ≥ -1
Since the range of cos θ is [-1, 1], the above inequality is true.
The equality holds when θ = 180° (collinear vectors)
Hence Proved.

c) geometrical solution:-
Consider the triangle shown in the figure.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 8
we must prove that \(|\overrightarrow{\mathrm{c}}| \leq|\overrightarrow{\mathrm{a}}|+|\overrightarrow{\mathrm{b}}|\)
or to prove that \(|\overrightarrow{\mathrm{c}}| \leq|\overrightarrow{\mathrm{a}}|+|\overrightarrow{\mathrm{- b}}|\)
(Since \(|-\vec{b}|=|\vec{b}|\)) which is true by the triangle inequality. (Sum of the lengths of any two sides of a triangle is always greater than the length of the third side.)
The equality holds when the triangle is degenerate (collinear vectors).
Hence proved.
Analytical solution:-
We must prove that \(|\vec{a}-\vec{b}| \leq|\vec{a}|+|\vec{b}|\)
Square both sides.
To prove that \((|\vec{a}-\vec{b}|)^{2} \leq(|\vec{a}|+|\vec{b}| |)^{2}\)
i.e., To prove that a² + b² – 2ab cos θ ≤ a² + b² + 2ab
i.e., to prove that -cos θ ≤ 1
Since the range of cos θ is [-1, 1], the range of -cos θ is also [-1, 1].
∴ The above inequality is true. Equality holds when θ = 180° (collinear vectors).
Hence proved.

d) Geometrical solution:-
Consider the triangle shown in the figure.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 9
i.e., To prove that \(|\vec{c}| \geq|| \vec{a}|-|-\vec{b}||\)
(Since \(|-\vec{b}|=|\vec{b}|\))
Which is true by the triangle inequality. (Difference in the lengths of any two sides of a triangle is always lesser than the length of the third side.)
The equality holds in the case of a degenerate triangle (collinear vectors).
Hence proved.
Analytical solution:-
We must prove that \(|\vec{a}-\vec{b}| \geq|| \vec{a}|-| \vec{b}||\)
Square both sides.
To prove that \((|\vec{a}-\vec{b}|)^{2} \geq(|| \vec{a}|-| \vec{b}||)^{2}\)
i.e., To prove that a² + b² – 2ab cose ≥ a² + b² – 2ab
i.e., to prove that cos θ ≤ 1
Since the range of cos θ is [-1, 1],
∴ The above inequality is true.
Equality holds when θ= 0 (collinear vectors).
Hence proved.

KSEEB Solutions

Question 7.
Given \(\vec{a}+\vec{b}+\vec{c}+\vec{d}=\vec{0}\). which of the
following statements are correct:
(a) \(\vec{a}, \vec{b}, \vec{c}\) and \(\vec{d}\) must each be a null vector,

(b) The magnitude of \((\vec{a}+\vec{c})\) equals the magnitude of \((\vec{b}+\vec{d})\),

(c) The magnitude of \(\vec{a}\) can never be greater than the sum of the magnitudes of \(\vec{b}, \vec{c}\) and \(\vec{d}\)

(d) \(\vec{b} + \vec{c}\) must lie in the plane of \(\vec{a}\) and \(\vec{d}\) if \(\vec{a}\) and \(\vec{d}\) if they are not collinear, and in the line of \(\vec{a}\) and \(\vec{d}\), if they are collinear?
Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 10
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 11
c) Correct. By an extension of the triangle inequality (sum of the lengths of any (n-1) sides of an n – sided closed polygon is always greater than the length of the remaining side.)

d) Correct. Given \(\vec{a}+\vec{b}+\vec{c}+\vec{d}=\vec{0}\)
⇒ \((\vec{b}+\vec{c}) = -(\vec{a}+\vec{d})\)
If \(\vec{a}\) and \(\vec{d}\) are not collinear, then \((\vec{a}+\vec{d})\) defines a plane. Clearly \((\vec{b}+\vec{c})\) has to lie in this plane for the given condition to hold good. Now if \(\vec{a}\) and \(\vec{d}\) are collinear, \(\vec{a}\) + \(\vec{d}\) is a line and clearly \((\vec{b}+\vec{c})\) must be collinear with this line for the given condition to hold good.

Question 8.
Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 4.20. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate?
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 12
Answer:
The magnitude of the displacement vector for each girl = Shortest distance between points P and Q = Diameter PQ
= 2 × 200 m = 400 m
Since girl B skates along the diameter PQ, for her, the magnitude of the displacement vector is equal to the actual length of path skated.

Question 9.
A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 4.21. If the round trip takes 10 min, what is the
(a) net displacement
(b) average velocity, and
(c) average speed of the cyclist?
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 13
Answer:
a) Since the cyclist returns to the starting point, his net displacement is zero.
b) The average velocity = \(\frac{\text { Net displacement }}{\text { Time taken }}\) = 0
c) Average speed = \(\frac{\text { length of the path }}{\text { Time taken }}\)
= \(\frac{\mathrm{OP}+\mathrm{arc} \mathrm{PQ}+\mathrm{QO}}{10 \mathrm{min}}\)
= \frac{1 \mathrm{km}+1/{4} \times 2 \pi \times 1 \mathrm{km}+1 \mathrm{km}}{1 / 6 \mathrm{hour}}[/latex] = 21.4 km / hour.

Question 10.
On open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 14
Let the origin be O. The path of the motions is as shown in the figure.
a) After 3 rd turn:-
In triangle ABC, AC = AB cos (∠ACB) + BC cos (∠BAC)
∠ACB = ∠BAC = 30° (Since AB = BC and ∠ABC = 120°).
∴ AC = 500 cos 30° + 500 cos 30°
AC = 866 m
In triangle OAC,
OC = \(\sqrt{\mathrm{OA}^{2}+\mathrm{AC}^{2}}\) (Since ∠OAC – 90°)
= \(\sqrt{500^{2}+866^{2}}\) = 1000 m
∴ Magnitude of the displacement = 1 km
Angle with the original direction
= tan-1 \(\left(\frac{\mathrm{AC}}{\mathrm{OA}}\right)\) = tan-1 \(\left(\frac{866}{500}\right)\) = 60°
Path length = 3 × 500 m = 1500 m Clearly magnitude of displacement < path length b) After sixth turn:- The net displacement here = 0 (because the motionist returns to his starting point) Angle with the original direction = 0° Path length = 6 × 500 m = 3 km Clearly, path length > magnitude of the displacement.

c) After the eighth turn:-
In triangle OAB,
OB = \(\sqrt{\mathrm{OA}^{2}+\mathrm{AB}^{2}+2 . \mathrm{OA} . \mathrm{AB} . \cos \theta}\)
= \(\sqrt{500^{2}+500^{2}+2 \times 500 \times 500 \times \cos 60^{\circ}}\)
= 866 m
Net displacement = 866 m
Angle with the original direction
= \(\tan ^{-1}\left(\frac{A B \sin \theta}{O A+A B \cos \theta}\right)\)
= \(\tan ^{-1}\left(\frac{500 \sin 60^{\circ}}{500+500 \cos 60^{\circ}}\right)\)
= \(\tan ^{-1}\left(1/{\sqrt{3}}\right)\)
= 30°
Path length = 8 × 500 m = 4 km
Clearly, path length > Magnitude of net displacement.

Question 11.
A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cab man takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is

  1. the average speed of the taxi,
  2. the magnitude of average velocity?
  3. Are the two equal?

Answer:
1. Average speed of the taxi
= \(\frac{\text { Path length }}{\text { Time taken }}\)
= \(\frac{23 \mathrm{km}}{28 \mathrm{min}}\) = \(\frac{23 \mathrm{km}}{28 \mathrm{28}_{60} \mathrm{h}}\) = 49.3 km h-1

2. Magnitude of Average Velocity
= \(\frac{\text { Net displacement }}{\text { Time taken }}\)
= \(\frac{10 \mathrm{km}}{28 \mathrm{min}}\) = \(\frac{10 \mathrm{km}}{28 \mathrm{28}_{60} \mathrm{h}}\) = 21.4 km h-1

3. Clearly, average speed is greater than magnitude of average velocity. The two are not equal.

KSEEB Solutions

Question 12.
Rain is falling vertically with a speed of 30 ms-1. A woman rides a bicycle with a speed of 10 ms-1 in the north to south direction. What is the direction in which she should hold her umbrella?
Answer:
The direction of θ is given by
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 15
= 18° with the vertical.

Question 13.
A man can swim with a speed of 4.0 km/h in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at 3.0 km/h and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?
Answer:
Time taken to cross the river
= \(\frac{\text { Width of the river }}{\text { Speed in this direction }}\)
= \(\frac{1 \mathrm{km}}{4 \mathrm{km} \mathrm{h}^{-1}}\) = \(1 / 4\) h = 15 min
Distance along the river covered in this time
= Speed of the river x Time taken to cross the river
= 3 km h-1 × 15 min
= 0.75 km.

Question 14.
In a harbour, wind Is blowing at the speed of 72 km/h, and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/ h to the north, what is the direction of the flag on the mast of the boat?
Answer: When the boat moves, the flag experiences wind blowing at 51 km h-1 in the south direction. Let the flag be at an angle 0 with the East direction.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 16
∴ The flag blows along the east direction.

Question 15.
The ceiling of a long hall is 25 m high. What Is the maximum horizontal distance, that a ball thrown with a speed of 40 m s-1 can go without hitting the ceiling of the hall?
Answer:
Given hm = 25 m , g = 9.8 ms-2 v0 = 40 ms-1
hm = \(\frac{\left(v_{0} \sin \theta_{0}\right)^{2}}{2 g}\)
25 = \(\frac{\left(40 \sin \theta_{0}\right)^{2}}{2 \times 9.8}\)
i.e. (40 sin θ0)² = 490
⇒ 40 sin θ0 = \(7 \sqrt{10}\)
⇒ sin θ0 = \(\frac{7 \sqrt{10}}{40}\)
⇒ θ0 = 33.6°
Maximum horizontal distance
R = \(\frac{v_{0}^{2}}{g}\) Sin 2 θ
= \(\frac{40^{2}}{9.8}\) sin(2 × 33.6°)
= 150.5 m

Question 16.
A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?
Answer:
Given R = \(\frac{v_{0}^{2}}{g}\) sin 2 θ = 100 m
Since R is maximum, θ = 45°
R = \(\frac{v_{0}^{2}}{g}\) = 100 m
Maximum height is achieved when throwing angle is 90°.
hmax = \(\frac{\left(v_{0} \sin 90^{\circ}\right)^{2}}{29}\)
= \(\frac{v_{0}^{2}}{g}\) = \(\frac{1}{2} \times \frac{v_{0}^{2}}{g}\)
1/2 × 100 m = 50 m.
∴ Maximum height hmax = 50m.

Question 17.
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?
Answer:
Radius of the circle r = 80 cm = 0.8m Angular speed
υ = \(\frac{14 \times 2 \pi r}{t}\)
υ = \(\frac{14 \times 2 \times 22 / 7 \times 0.8}{25}\)
υ = 2.816 ms-1
Acceleration = \(\frac{v^{2}}{r}\) = \(\frac{2.816^{2}}{0.8}\)
= 9.9 ms-2
Direction of acceleration: Towards the centre of the circle and along the radius.

Question 18.
An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.
Answer:
Radius r = 1.00 km = 1000 m
Speed v = 900 km h-1 = 250 ms-1
Centripetal acceleration
a = \(\frac{v^{2}}{r}=\frac{250^{2}}{1000}\)
a = 62.5 ms-2
Acceleration due to gravity g = 9.8 ms-2.
a = \(\frac{62.5}{9.8}\) = 6.4 g

question 19.
Read each statement below carefully and state, with reasons, if it is true or false:

  1. The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre
  2. The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point
  3. The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector

Answer:

  1. False. The statement is true only if the particle is executing circular motion at a uniform speed (no tangential acceleration).
  2. True. The instantaneous velocity is in the direction tangential to the instantaneous path of the particle.
  3. True. Consider the two points and on the circle as shown in the figure.

1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 17
At both these points, acceleration is towards the centre P. The net acceleration is O (equal magnitude but of opposite directions). Similarly, over one cycle, the net acceleration is a null vector.

Question 20.
The position of a particle is given by
\(r=3.0 t \hat{i}-2.0 t^{2} \hat{j}+4.0 \hat{k} m\)
where t is in seconds and the coefficients have the proper units for r to be in meters.

  1. Find the v and an of the particle?
  2. What is the magnitude and direction of velocity of the particle at t = 2.0 s?

Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 18
= 8.54 ms-1
Direction θ = \(\tan ^{-1}\left(\frac{v_{y}}{v_{x}}\right)\)
= \(\tan ^{-1}\left(\frac{-8}{3}\right)\)
= – 69.44° with the horizontal (x – axis)

KSEEB Solutions

Question 21.
A particle starts from the origin at t = 0 s with a velocity of 10.0 J m/s and moves in the x-y plane with a constant acceleration of (8.0 I + 2.0 J) m s-2.

  1. At what time is the x – coordinate of the particle 16 m? What is the y – coordinate of the particle at that time?
  2. What is the speed of the particle at the time?

Answer:
1. Along x – direction,
vx(0) = 0, ax = 8 ms-22
rx(t) = rx(0) + vx(o) t + 1/2 ax
= 0 + 0 + 1/2 × 8 × t² = 4 t²
Given rx(t1) = 16 m
4 t² = 16m
⇒ t = 2 s
∴ At time t = 2.0 s, x – coordinate of the particle = 16 m
Along y – direction,
vy(o) = 10 ms-1, ay = 2 ms-2
ry(t) = ry(0) + vy(0)t +1/2 ay
0 + 10 t + 1/2 × 2 × t²
= 10 t + t²
ry(2) = 10(2) + (2)² = 24 m
∴ y – coordinate at time t = 2.0 s is 24 m

2. Along x – direction,
Vx(t) = vx (0) + axt
= 0 + 8t = 8t
vx(2) = 16 ms-1
Along y – direction,
vy(t) = vy(0) + ayt
= 10 + 2t
vy (2) = 14 ms-1
Speed at time t a 2.0 s is \(\sqrt{\left(v_{x}(2)\right)^{2}+\left(v_{y}(2)\right)^{2}}\)
= \(\sqrt{16^{2}+14^{2}}\) = 21.26 ms-1

Question 22.
I and J are unit vectors along the x- and y-axis respectively. What is the magnitude and direction of the vectors I+ J, and I – J? What are the components of a vector A = 2I+3J along with the directions of i + j and i – j? [You may use graphical method]
Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 19
Question 23.
For any arbitrary motion in space, which of the following relations are true:

  1. vaverage = (1/2) (v (t1 + v (t2)))
  2. vaverage = [r(t2) – r(t1)] /(t2 – t2)
  3. v (t) = v (0) + a t
  4. r (t) = r (0) + v (0) t + (1/2) a t²
  5. aaverage = [v(t2) – v(t1)]/(t2 – t1)

(The ‘average’ stands for an average of the quantity over the time interval t1 to t2)
Answer:

  1. False. Average velocity is not the arithmetic mean of the individual velocities.
  2. True. By definition.
  3. False. This is true only if the acceleration \(\overrightarrow{\mathrm{a}}\) is constant.
  4. False. This is true only if the acceleration \(\overrightarrow{\mathrm{a}}\) is constant.
  5. True. By definition.

Question 24.
Read each statement below carefully and state, with reasons and examples, if it is true or false:
A scalar quantity is one that

  1. is conserved in a process
  2. can never take negative values
  3. must be dimensionless
  4. does not vary from one point to another in space
  5. has the same value for observers with different orientations of axes.

Answer:

  1. False. Mass is a scalar quantity and it is not conserved in a nuclear reaction.
  2. False. Angle is a scalar quantity and it can take negative values.
  3. False. Mass, current, etc. are scalar quantities that have dimensions.
  4. False, Temperature, electric potential, etc. are scalar quantities that vary from point to point in a medium.
  5. True. Change of axis does not change a scalar quantity. For example, mass is a scalar quantity that is independent of the axis chosen.

Question 25.
An aircraft is flying at a height of 3400 m above the ground if the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft?
Answer:
Let O be the observation point. A and B are the initial and final positions of the aircraft.
The angle subtended θ = 30°
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 20
= \(\frac{\text { Distance coverd }}{\text { Time taken }}=\frac{1960 \mathrm{m}}{10 \mathrm{s}}\) = 196 ms-1.

1st PUC Physics Motion in a Plane Additional Exercises Questions and Answers

Question 26.
A vector has magnitude and direction. Does it have a location in space? Can it vary with time Will two equal vectors a and b at different locations in space necessarily have Identical physical effects? Give examples in support of your answer.
Answer:
Yes, a vector has a defined location in space. It may or may not vary with time. For example, \(3 \hat{i}+4 \hat{j}\) is time independent whereas \(4 t \hat{i}-7 \sqrt{t} \hat{k}\) is time dependent.
Two vectors at different locations need not have the same physical effects. Consider the weight of a person in newtons, ‘k’ newtons of weight have different meanings on earth and on the moon, and so they have different physical effects.

Question 27.
A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector?
Answer:
A quantity that has magnitude and direction need not necessarily be a vector. Consider finite rotations of a particle about an axis. 2 complete rotations bring back the particle to its initial position but these rotations do not add up as per. Parallelogram law of addition. Hence, rotation is not a vector.

Question 28.
Can you associate vectors with

  1. the length of a wire bent into a loop,
  2. a plane area,
  3. a sphere? Explain.

Answer:

  1. No.
  2. Yes. The area \(\overrightarrow{\mathrm{A}}\) can be defined as \(\overrightarrow{\mathrm{A}}\) = \(\overrightarrow{I} \times \overrightarrow{\mathbf{b}}\) where \(\overrightarrow{i}\) and \(\overrightarrow{b}\) are the length and breadth vbctors.
  3. No.

KSEEB Solutions

Question 29.
A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to the fixed, and neglect air resistance.
Answer:
Let ‘u’ be the muzzle speed.
0=30°
Range R = \(\frac{\mathfrak{u}^{2} \sin 2 \mathfrak{\theta}}{\mathfrak{g}}\)
3000 m = \(\frac{\mathrm{u}^{2}}{\mathrm{g}}\) × sin (2 × 30°)
⇒ \(\frac{\mathrm{u}^{2}}{\mathrm{g}}\) = \(\frac{3000}{\sin 60^{\circ}}\) = 3464 m.
Maximum range possible Rmax = \(\frac{\mathrm{u}^{2}}{\mathrm{g}}\)
= 3.46 km
∴ It is not possible to hit a target 5 km away.

Question 30.
A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes directly overhead an antiaircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m s-1 to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit?
(Take g = 10 m s-2).
Answer:
Plane’s altitude
h = 1.5 km = 1500 m
Plane’s speed
Vp = 720 km/h = 200 ms-1
Shell’s speed
Vs = 600 ms-1
Let the shell be fired at an angle θ with the vertical.
Horizontal distance covered by the shell in time t = Horizontal distance covered by the plane in time t
⇒ Vs sin θ .t = Vpt
⇒ 600 sin θ = 200
⇒ θ = sin-1(1/3) = 19.5°
Minimum altitude to ensure no hit = Maximum height achieved by the shell
= \(\frac{\mathrm{u}^{2} \cos ^{2} \theta}{2 \mathrm{g}}\)
= \(\frac{600^{2} \cos ^{2}\left(19.5^{\circ}\right)}{2 \times 10}\)
= 1600 m
= 16 km.

Question 31.
A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
Answer:
Speed of the cyclist v = 27 km/h
= 7.5 ms-1.
Radius of the turn r = 80 m
Centripetal acceleration ac = \(\frac{v^{2}}{r}=\frac{7.5^{2}}{80}\)
= 0.70 ms -2
Tangential retardation ar = \(\frac{0.5}{1}\) = 0.5 ms-2
Net acceleration a = \(\sqrt{a_{c}^{2}+a_{r}^{2}}\)
= \(\sqrt{0.70^{2}+0.5^{2}}\)
= 0.86 ms-2
Angle θ with the tangent is given by
θ = tan-1 \(\left(\frac{\mathrm{a}_{\mathrm{r}}}{\mathrm{a}_{\mathrm{t}}}\right)\)
θ = tan-1\(\left(\frac{0.70}{0.50}\right)\)
θ = 54.5°.

An online projectile motion calculator allows you to compute the velocity, maximum height, and flight parameters at a given time in a fraction of a second.

Question 32.
(a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 21
(b) Shows that the projection angle θ for a projectile launched from the origin is given by
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 22
where the symbols have their usual meaning.
Answer:
(a) Let the horizontal component of the initial velocity be vox and the vertical component be voy.
During motion, the horizontal component is still vox but the vertical component
= Voy – gt
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 23
(b) For the projectile
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 24

1st PUC Physics Motion in a Plane One Mark Questions and Answers

Question 1.
What is the work done by a body in circular motion?
Answer:
Zero.

Question 2.
What is the trajectory of a projectile?
Answer:
Parabola.

Question 3.
What is velocity of projection?
Answer:
The velocity with which the body is projected is called velocity of projection.

Question 4.
What is angle of projection?
Answer:
The angle between the direction of projection and the horizontal drawn at that point is called the angle of projection.

Question 5.
What is range of projectile?
Answer:
The maximum horizontal distance covered by the projectile is called the range.

Question 6.
What is time of flight?
Answer:
The time taken to describe the range is called the time of flight.

Question 7.
When is the range of a projectile is maximum?
Answer:
When the angle of projection is 45° the range of a projectile is maximum.

Question 8.
Why should an athlete throw the javelin or shot put approximately at an angle of 45°?
Answer:
An athlete must throw shot – put or a javelin approximately at an angle 45°to achieve the maximum range.

Question 9.
What is uniform circular motion?
Answer:
Motion of a body along a circular path with constant speed is called uniform circular motion.

KSEEB Solutions

Question 10.
What is the direction of motion (or velocity) of a body in uniform circular motion?
Answer:
The direction of motion of the body at any instant will be along the tangent to the circular path.

Question 11.
Define angular displacement.
Answer:
The angle through which the radius vector rotates is called angular displacement.

Question 12.
Define angular velocity.
Answer:
The rate of angular displacement is called angular velocity.

Question 13.
Define period of revolution of a body.
Answer:
It is the time taken by the body to complete one revolution.

Question 14.
Define frequency of revolution of a body.
Answer:
Number of revolutions completed by the body in one second is called frequency of revolution.

Question 15.
Define centripetal acceleration.
Answer:
In uniform circular motion, the acceleration produced on the body acts along the radius, towards the centre of the circular path and is known as centripetal acceleration or radial acceleration.

Question 16.
What is the source of centripetal force for the planets revolving round the sun?
Answer:
The gravitational force of attraction between the sun and the planets provides the centripetal force for the revolution of the planets round the sun.

Question 17.
What Is banking of roads?
Answer:
Raising the outer edge of the road to get the required centripetal force for a vehicle is known as banking of roads (or track).

Question 18.
What is the work done by centripetal force?
Answer: Zero.

Question 19.
A body moves along a circle of radius 3m. What is the distance travelled. When it completes one circle?
Answer:
Given: radius, r = 3m.
Distance travelled by. the body when it completes one circle = circumference of the circle
= 2 π r = 2 × 3.4 × 3
= 18.84 m

Question 20.
Give an example for three dimen-sional motion.
Answer:
Flight of Aeroplane or movement of gas molecule in space.

Question 21.
Give an example of centripetal force.
Answer:
Electrostatic force between an electron and nucleus provides centripetal force for the revolution of the electron in its orbit.

Question 22.
Which vector quantity becomes zero at highest point of motion of projectile?
Answer:
Vertical component of velocity becomes zero at the highest point.

Question 23.
State the law of parallelogram of two forces.
Answer:
If two forces acting at a point are represented both in magnitude and direction by the two adjacent sides of a parallelogram then their resultant is represented by the diagonal of the parallelogram drawn from the same point.

Question 24.
What will be the effect on the horizontal range of a projectile when its initial speed is doubled keeping the angle the same?
Answer:
Since R α v0², doubling the initial speed will increase the range by a factor of 2² = 4 times.

KSEEB Solutions

Question 25.
What is the minimum number of forces acting on a object in a plane that can produce a zero resultant force?
Answer:
Two. (The forces should have equal magnitude but opposite direction).

Question 26.
A unit vector Is represented by \(\mathbf{a} \hat{\mathbf{i}}+\mathbf{b} \hat{\mathbf{j}}+\mathbf{c} \hat{\mathbf{k}}\). If the values of a and b are
0.6 and 0.8 respectively, find the value of C.
Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 25
⇒ c = 0

1st PUC Physics Motion in a Plane Two Mark Questions and Answers

Question 1.
When is a body said to have two-dimensional motion? Give an example.
Answer:
Motion of a particle in a plane is known as two-dimensional motion. Example:

  • A car moving along a zig-zag path on a road.
  • Motion of the planet around the sun in its orbit.

Question 2.
What is a projectile? Give an example.
Answer:
Anybody was thrown into space, that moves under the effect of gravity is called a projectile.
E.g.:

  • A bullet fired from the gun.
  • A javelin thrown by an athlete.

Question 3.
Obtain the relation between angular velocity and linear velocity.
Answer:
Consider a particle moving in a circle of radius ‘r’ with uniform angular velocity ‘ ω ’ and linear speed ‘V’. Let the particle moves from A to B in time ‘t’ seconds through a small distance ‘s’ on the circumference. Let θ be the angular displacement.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 26
Angular velocity ω = \(\frac{\theta}{\mathfrak{t}}\)
Angular displacement θ = \(\frac{s}{\mathfrak{r}}\)
∴ ω = \(\frac{\mathrm{s}}{\mathrm{tr}}\)
But \(\frac{s}{\mathfrak{t}}\) = v, linear velocity
∴ ω = \(\frac{\mathrm{v}}{\mathrm{r}}\) or
v = r ω
linear velocity = radius × angular velocity.

Question 4.
What is angle of banking? Give the expression for it.
Answer:
The angle through which the outer edge of the roads are raised is called the angle of banking. The angle of banking is given by,
θ = tan-1\(\left(\frac{v^{2}}{r g}\right)\)
where m is the mass, v is the velocity and r is the radius.

Question 5.
A body travels one round in a circle of radius ‘R’- What is the

  • displacement and.
  • distance traveled.

Answer:

  • The displacement of a body = 0
  • The distance travelled = 2nR

Question 6.
Write the expressions for the maximum height reached by a projectile, and explain the terms.
Answer:
The expression for the maximum height reached by the projectile is
\(\mathrm{H}=\frac{u^{2} \sin ^{2} \theta}{2 \mathrm{g}}\)
Where u, is the initial velocity of the projectile, θ is the angle between the horizontal & the initial velocity, g is the acceleration due to gravity.

Question 7.
What is banking of roads? Give an example of it.
Answer:
Vehicles while taking a turn cannot incline themselves to one side to get the required centripetal force. To provide necessary centripetal force without slipping. The outer edge of the road is raised over the inner edge.

Question 8.
Why a cyclist bends inward when to Is crossing a curve in a road?
Answer:
He bends inward to get sufficient centripetal force.

KSEEB Solutions

Question 9.
Two equal forces have their resultant equal to either. What is the inclination between them?
Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 27
= A
⇒ 2A² (1 + cos θ) = A²
⇒ 1 + cos θ = 1/2
⇒ cos θ = -1/2
⇒ θ = 120°

Question 10.
A cyclist has to bend a little Inwards from his vertical position while turning. Why?
Answer:
Bending provides a component of the normal reaction force from the ground to provide the cyclist the necessary centripetal force for turning. Hence, bending is necessary.

Question 11.
Suppose you have two forces \(\overrightarrow{\mathrm{F}}\) and \(\overrightarrow{\mathrm{F}}\). How would you combine them in order to have resultant forces of magnitudes

  1. zero
  2. 2 \(\overrightarrow{\mathrm{F}}\), and
  3. \(\overrightarrow{\mathrm{F}}\)?

Answer:

  1. If the forces act in opposite directions, the resultant is zero.
  2. If the forces act in the same direction, resultant \(\overrightarrow{\mathrm{R}}\) = 2 \(\overrightarrow{\mathrm{F}}\).
  3. For \(\overrightarrow{\mathrm{R}}\) = 2 \(\overrightarrow{\mathrm{F}}\)
    \(\sqrt{\mathrm{F}^{2}+\mathrm{F}^{2}+2 \mathrm{F} \cdot \mathrm{F} \cdot \cos \theta}\) = F
    ⇒ 2F² (1 + cos θ) = F²
    ⇒ 1 + cos θ = 1/2
    ⇒ θ = 120°

Question 12.
Prove the following statement: “For elevation which exceeds or falls short of 45° by equal amounts, the range Is equal”.
Answer:
At 45° – θ, the range is
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 28
At 45° + θ, the range is
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 29
⇒ R2 = R1 Proved.

Question 13.
What is the distance traveled by a point during time ‘t’, if it moves in x – y plane according to the relation x = a sin cot and y = a(1 – cos cot ωt)?
Answer:
Distance travelled d(t) =
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 30

Question 14.
A lady walking due east on a road with velocity 10 ms-1 encounters rain falling vertically with a velocity of 30ms-1. At what angle should she hold her umbrella to protect herself from the rain?
Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 31
⇒ θ = 18°16′ with the vertical direction.

Question 15.
If the vectors of equal magnitude add to either of them by magnitude, what is the angle between them?
Answer:

  1. If the forces act in opposite directions, the resultant is zero.
  2. If the forces act in the same direction, resultant \(\overrightarrow{\mathrm{R}}\) = 2 \(\overrightarrow{\mathrm{F}}\).
  3. For \(\overrightarrow{\mathrm{R}}\) = 2 \(\overrightarrow{\mathrm{F}}\)
    \(\sqrt{\mathrm{F}^{2}+\mathrm{F}^{2}+2 \mathrm{F} \cdot \mathrm{F} \cdot \cos \theta}\) = F
    ⇒ 2F² (1 + cos θ) = F²
    ⇒ 1 + cos θ = 1/2
    ⇒ θ = 120°

Question 16.
A swimmer can swim with velocity of 10km h-1 w.r.t. the water flowing in a river with velocity of 5 km h-1. In what direction should he swim to reach the
point on the other bank just opposite to his starting point?
Answer:
Let the angle of the swimmer be e w.r.t the horizontal.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 66
⇒ cos θ = 1/2
⇒ θ = 60°

Question 17.
The angle between vector \(\overrightarrow{\mathbf{A}}\) and \overrightarrow{\mathbf{B}} is 60°. What is the ratio of \(\overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}}[latex] and [latex]|\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}|[latex]?
Answer:
Required ratio [latex]=\frac{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}}{\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}}\)
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 33
= cot θ
= cot (60°)
= \(1 / \sqrt{3}\)

Question 18.
Two forces P=10 N and Q = 15 N are acting at a point making an angle 30° with each other. What is the cross product of P
and Q?
Answer:
P= 10N, Q = 15N, θ =30°
From the equation P × Q = P.QSine
= 10.15 Sin 30° = 75N.

1st PUC Physics Motion in a Plane Three Mark Questions and Answers

Question 1.
Find a unit vector parallel to the vector \(3 \hat{i}+7 \hat{j}+4 \hat{k}\)
Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 34
Unit vector parallel to \(\overrightarrow{\mathrm{a}}=\frac{\overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{a}}|}\)
\(=\frac{3 \hat{i}+7 \hat{j}+4 \hat{k}}{\sqrt{74}}\)

Question 2.
What is a projectile? Show that the path followed by an oblique projectile Is a parabola.
Answer:
Anybody was thrown into space, that moves under the effect of gravity is called a projectile.
E.g.:
1. A bullet fired from the gun.
2. A javelin thrown by an athlete., Consider a particle thrown up at an angle θ to the horizontal with a velocity u. The velocity of projection can be resolved into two component u cos θ and usinq along OX and OY respectively.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 35
The horizontal component remains constant while the vertical component is affected by gravity.
Let P(x, y) be the position of the projectile after ‘t’ seconds.
Then distaince travelled ‘x’ along OX is given by,
x = horizontal component of the velocity time
x = u cos θ × t ………… (1)
The distance travelled along the vertical direction in same time‘t’ is,
y = (u sin θ) t – 1/2 gt² ……………… (2)
Substituting, t = \(\frac{x}{u \cos \theta}\) from (1) in (2),
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 67
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 36…………………..(3)
where a = tan θ, \(\mathrm{b}=\frac{\mathrm{y}=\mathrm{gax}-\mathrm{bx}^{2}}{2 \mathrm{u}^{2} \cos ^{2} \mathrm{\theta}}\) are constant for given values of θ, u and g.
The equation (3) represents a parabola. Therefore trajectory of a projectile is a parabola.

Question 3.
Establish a relation between linear velocity and angular velocity in a uniform circular motion and explain the direction of the velocity.
Answer:
The distance ‘s’ covered by a body travelling in an arc of radius Y and turning its radial line by ‘θ’ is given by
s = r θ
Differentiating both sides w.r.t. time, we have
\(\frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{r} \frac{\mathrm{d} \theta}{\mathrm{dt}}\)
i.e., v = rw
or Linear velocity = radius × angular velocity.
At each point the body moves along the tangent.
The presence of centripetal force
F =\(\frac{m v^{2}}{r}\) makes the body to travel in the circular path.
Thus, the direction of velocity is always along the tangent at any point in the circular path.

Question 4.
Find the magnitude and direction of the resultant of two forces p and Q in terms of their magnitudes and angle Q between them.
Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 37
P and Q are two forces acting at a point O; making an angle θ. They are represented by OA and OB. the diagonal OC of the parallelogram OBCA represents the resultant R.
Draw CD perpendicular to extended line of OA. In the right angle triangle ODC.
OC² = OD² + CD²
= (OA + AD)² + CD²
= OA² + 2OA AD + AD² + CD²
= OA² + 2OA. AC cos θ + AC²
R² = P² + 2PQcos θ + Q² v
From triangle ADC,
AC² = AD² + CD² & \(\frac{\mathrm{AD}}{\mathrm{AC}}\) = cosq
Magnitude of the resultant
R = \(\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ} \cdot \cos \theta}\)
To find the direction:
If a is the angle made by R with P, then from the right angled triangle ODC
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 38

Question 5.
In long jump, does it matter how high you jump? What factors determine the span of the jump?
Answer:
In order to increase the span of the jump, it is necessary for the horizontal component of the velocity to be more than the vertical component. The height of the jump does not matter.
The factors that determine the span of the jump are:

  1. angle of elevation at the time of jumping.
  2. Speed of the runner at the time of jumping.

Question 6.
Determine a unit vector which Is per-pendicular to both \(\vec{A}=2 \hat{i}+\hat{j}+\hat{k}\) and \(\overrightarrow{\mathbf{B}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\)
Answer:
The unit vector \(\hat{n}\) is given by
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 39

Question 7.
An accelerating train is passing over a high bridge. A stone is dropped from the train at an instant when Its speed is 10 ms-1 and acceleration is 1ms-2. Find the horizontal and vertical components of the velocity and acceleration of the stone one second after it is dropped.Take g = 10 ms-2.
Answer:
Horizontal acceleration = 0 (because there is no force acting on the stone in the horizontal direction to provide any acceleration)
Horizontal velocity = 10 ms-1.
Vertical acceleration = g = 10ms-2.
Vertical velocity = ut + 1/2 gt²
= 0(1) + 1/2 (10)(1)²
= 5 ms-1.

Question 8.
A projectile is projected with velocity ‘u’ making an angle θ with the horizontal direction. Find:

  1. Time of flight
  2. Horizontal range

Answer:
1. Let H be the maximum height reached by the projectile in time t1
For vertical motion,
The initial velocity = usin θ
The final velocity = 0
Acceleration = – g
using, v² = u² + 2as
0 = u² sin2 θ – 2gH
2gH =u²sin2 θ
u² sin² θ
H = \(\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{g}}\)

2. Let t, be the time taken by the projectile to reach the maximum height H.
For vertical motion,
initial velocity =usin0
Final velocity at the maximum height = 0
Acceleration a = – g
Using the equation v = u + at1
0 = usin θ – gt1
gt1 = usin θ
t1 = \(\frac{\mathrm{usin} \theta}{\mathrm{g}}\)
Let t2 be the time of descent.
But t1 = t2
i.e. time of ascent = time of descent.
Time of flight T = t1 + t2 = 2t1
∴ T = \(\frac{2 u \sin \theta}{g}\)

Question 9.

  1. Show that for two complementary angles of projection of a projectile with the same velocity, the horizontal ranges are equal.
  2. For what angle of projectile is the rangle maximum?
  3. For what angle of projection of a projectile are the horizontal range and height attained by the projectile equal?

Answer:
1. For projection angle θ,
R1 = \(\frac{u^{2} \sin 2 \theta}{g}\)
For projection angle 90° – θ,
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 40
But sin (180° – 2θ) = sin2θ
∴ R2 = \(\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) = R1

2. Range is maximum when projection angle = 45°.

3. Given R = H
⇒ \(\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) = \(\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{g}}\)
⇒ 2 sin θ cos θ = \(\frac{\sin ^{2} \theta}{2}\)
⇒ cot θ = 1/4
θ = cot-1(1/4)
⇒ θ = 75.96°.

KSEEB Solutions

Question 10.
What is meant by centripetal acceleration? Derive the formula for centripetal acceleration.
Answer:
In uniform circular motion, the acceleration produced on the body acts along the radius, towards the centre of the circular path. This acceleration is known as centripetal acceleration or radial acceleration.
Let a particle of mass m be moving around the circle of radius r with a uniform speed ‘v’.
Let the particle moves from A to B in a time ‘t’ seconds covering a small angle θ.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 41
At the point A the velocity of the particle is along the tangent AC and therefore there is no component along AO (i.e. along radius).
At the point B the velocity of the particle is along BD. This can be resolved into two components v cos θ and v sin θ. Since θ is small sin θ ≅
θ and cos θ ≅ 1.
∴ change in velocity parallel to AC
= v cos θ – v = v -v = 0. ∵ cos θ ≅ 1.
∴ There is no acceleration along the tangent,
Change in velocity parallel to OA is
= vsin θ – 0 = v θ ( sinθ ≅ θ)
∴ Acceleration along AO = \(\frac{\mathrm{v} \theta}{\mathrm{t}}\)
i.e., centripetal acceleration = \(\frac{\mathrm{v} \theta}{\mathrm{t}}\) = v ω
∴ centripetal force = mass × centripetal acceleration
F=m v ω
F = mw²r = \(\frac{m v^{2}}{r}\) ∵v = ω r

Question 11.
From the top of a tower of 100m height, a ball is dropped and at the same time another ball is projected vertically upwards from the ground with a velocity of 25ms-1. Find when and where the two balls will meet. Take g = 9.8 ms-2
Answer:
Let AC = x
Then BC = 100 – x
x = ut + 1/2 g t²
x = 4.9 t² …………… (1)
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 42
BC = ut – 1/2 g t²
= 25 t – 1/2 × 9.8 t²
100 – x = 25 t – 4.9 t² ………………. (2)
Substituting 4.9 t² = x from (1) in (2)
100 – 2 = 25 t – x
⇒ t = 4 seconds
x = 4.9 t²
= 4.9 × 4²
= 78.4 m from the top or 21.6 m from the ground.

1st PUC Physics Motion in a Plane FourFive Marks Questions and Answers

Question 1.
Show that the trajectory of a projectile is a parabola.
Answer:
Consider a particle thrown up at an angle q to the horizontal with a velocity u. The velocity of projection can be resolved into two component u cos θ and u sin θ along OX and OY respectively.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 43
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 44
The horizontal component remains constant while the vertical component is affected by gravity.
Let P(x, y) be the position of the projectile after ‘t’ seconds.
Then distance travelled ‘x’ along OX is given by,
x = horizontal component of the velocity × time
x = ucos θ × t …………………. (1)
The distance travelled along the vertical direction in same time ‘t’ is,
y = (usin θ) t – 1/2gt² ………… (2)
Substituting, t = \(\frac{x}{u \cos \theta}\) from (1) in (2),
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 45
y = ax – bx² …………………… (3) where
a = tan θ, \(\mathrm{b}=\frac{\mathrm{y}=\mathrm{gax}-\mathrm{bx}^{2}}{2 \mathrm{u}^{2} \cos ^{2} \mathrm{\theta}}\), are constants
for given values of θ, u and g.
Equation (3) represents a parabola. Therefore trajectory of a projectile is a parabola.

Question 2.
Derive an expression for the maximum height reached, time of flight and range of a projectile.
Answer:
(i) Let H be the maximum height reached by the projectile in time t1 For vertical motion,
The initial velocity = usin θ
The final velocity = 0
Acceleration = – g
∴ using, v² = u² + 2as
0 = u²sin² θ – 2gH
2gH = u²sin² θ
H = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)

(ii) Let t, be the time taken by the projectile to reach the maximum height H.
For vertical motion,
initial velocity = usin θ
Final velocity at the maximum height = 0
Acceleration a = – g
Using the equation v = u + at1
0 = usin θ – gt1
gt1 = usin θ
t1 = \(\frac{\mathrm{usin} \theta}{\mathrm{g}}\)
Let t2 be the time of descent.
But t1 = t2
i.e. time of ascent = time of descent.
∴ Time of flight T = t1 + t2 = 2t1
∴ T = \(\frac{2 u \sin \theta}{\dot{g}}\)

(iii) Let R be the range of the projectile in a time T. This is covered by the projectile with a constant velocity ucos θ.
Range = horizontal component of
velocity × Time of flight
i.e, R = ucos θ. T
R = ucos θ. \(\frac{2 u \sin \theta}{g}\)
R= \(\frac{u^{2} \sin 2 \theta}{g}\)
∵ 2 sin θ.cos θ = sin2 θ.

Question 3.
Derive an expression for the centripetal force.
Answer:
Let a particle of mass m be moving around the circle of radius r with a uniform speed ‘V’.
Let the particle moves from A to B in a time ‘t’ seconds covering a small angle θ.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 46
At the point A the velocity of the particle is along the tangent AC and therefore there is no component along AO (i.e. along radius).
At the point B the velocity of the particle is along BD. This can be resolved into two components v cos θ and v sin θ. Since θ is small sin θ ≅
θ and cos θ ≅ 1.
∴ change in velocity parallel to AC
= v cos θ – v = v -v = 0. ∵ cos θ ≅ 1.
∴ There is no acceleration along the tangent,
Change in velocity parallel to OA is
= vsin θ – 0 = v θ ( sinθ ≅ θ)
∴ Acceleration along AO = \(\frac{\mathrm{v} \theta}{\mathrm{t}}\)
i.e., centripetal acceleration = \(\frac{\mathrm{v} \theta}{\mathrm{t}}\) = v ω
∴ centripetal force = mass × centripetal acceleration
F=m v ω
F = mw²r = \(\frac{m v^{2}}{r}\) ∵v = ω r.

KSEEB Solutions

Question 4.
Define centripetal force and give example.
Answer:
In uniform circular motion the force on the rotating body acts along the radius towards the centre of the circular path and is known as centripetal force.
Eg:

  1. In the case of a stone rotated round the circle by means of a string, the centripetal force is provided by the tension in the string.
  2. The force of gravitational attraction towards the sun is the centripetal force keeping the planets orbiting round the sun.
  3. In an atom the electrons revolve round the nucleus in the circular orbit the centripetal force is provided by the electrostatic force of attraction between the nucleus and the electrons.

Question 5.
What is the angle of banking for a curve road radius 180m suitable for a maximum speed of 30m? Calculate the maximum speed to be maintained when the angle of banking is
θ = 30° (g = 9.8ms1) (Rural 2005)
Answer:
Radius of the curve road, r =180m, Maximum speed v=30ms-1
Angle of banking 0=? Acceleration due to gravity, g =9.8ms-2
From the equation,
tan θ = \(\frac{v^{2}}{r g}\) = \(\frac{30^{2}}{180 \times 9.8}\) = \(=\frac{900}{1764}\)
= 0.5102
θ = tan-1(0.5102)
= 27° 1°
When the angle of banking θ = 30°,
From the equation tan θ = \(\frac{v^{2}}{r g}\)
tan θ 30° = \(\frac{v^{2}}{180 \times 9.8}\)
v2 = 180 × 9.8 × tan30°
= 1764 × 0.5774 = 1018.5
v = 32ms-1

Question 6.
A bullet fired from rifle attains a maximum height of 50m and crosses a range of 200m. Find the angle of projection.
Answer:
Maximum height attained by the bullet, H=50m,
Horizontal range R=200m,
Angle of projection 0 =?
From the equation H = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
50 = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\) ……………….(1)
From the equation R = \(\frac{u^{2} 2 \sin \theta \cos \theta}{g}\)
200 = \(\frac{u^{2} 2 \sin \theta \cos \theta}{g}\) ………………. (2)
Divide equ (1) by equ (2), we have
\(\frac{50}{200}=\frac{\sin \theta}{4 \cos \theta}\)
i.e \(\frac{1}{4}\) = \(\frac{1}{4}\) tan θ Thus tan θ = 1
θ = tan-1(1.0000) = 45°.

Question 7.
A train of mass 10,000 kg moving at 72km ph rounds a curve whose radius of curvature is 200m. What is its acceleration? What is the centripetal force?
Answer:
m = 10,000kg, v = 72 km/h
= \(\frac{72 \times 1000}{60 \times 60}\) = 20m/s
r = 200 m.
Acceleration a = v ω
= \(v\left(\frac{v}{r}\right)\)
= \(\frac{v^{2}}{r}=\frac{(20)^{2}}{200}\) = 2m/s2
Centripetal force, F = \(\frac{m v^{2}}{r}\)
= \(\frac{10000 \times(20)^{2}}{200}\) = 20000 N.

KSEEB Solutions

Question 8.
An object is projected with a velocity of 60ms-1 in a direction making an angle of 60° with the horizontal. Find

  1. the maximum height
  2. the time taken to reach maximum height
  3. the horizontal range

Answer:
u = 60m/s, θ = 60°.
1. Maximum height
H = \(\frac{u^{2} \sin ^{2}(\theta)}{2 g}\) = \(\frac{60^{2} \sin ^{2}(60)}{2 \times 9.8}\) = 137.74 m.

2. Time to reach maximum height
t = \(\frac{u \sin (\theta)}{g}\) = \(\frac{60 \times \sin (60)}{9.8}\) = 5.30s.

3. Range
R = \(\frac{u^{2} \sin (2 \theta)}{g}\) = \(\frac{60^{2} \sin (2 \times 60)}{9.8}\) = 318.12 m

Question 9.
Drive an expression for magnitude and direction of the resultant of two forces acting at a point.
Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 47
P and Q are two forces acting at a point O, making an angle θ. They are represented by OA and OB. the diagonal OC of the parallelogram OBCA represents the resultant R.
Draw CD perpendicular to extended line of OA. In the right angle triangle ODC.
OC² = QD² + CD²
= (OA + AD)² + CD²
= OA² + 2.0A AD + AD² + CD²
= OA² + 20A. AC cosq + AC²
R² = P² + 2PQcos θ + Q²
∵ From triangle ADC,
AC² = AD²+ CD² & \(\frac{A D}{A C}\) = cos θ
Magnitude of the resultant
R = \(\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ} \cdot \cos \theta}\)
To find the direction:
If a is the angle made by R with P, then from the right angled triangle ODC
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 68

Question 10.
A projectile shot at an angle of 60° above the horizontal ground strikes a vertical wall 30 m away at a point 15m above the ground. Find the speed with which the projectile was launched and the speed with which it strikes the wall. Take g = 10 ms-2
Answer:
Let the initial velocity of the projectile be u
Horizontal component = ux
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 49
Horizontal distance, d = ux t
i.e., 30 = \(\frac{\mathrm{u}}{2}\) t
⇒ t = \(\frac{60}{\mathrm{u}}\) ………… (1)
Vertical component uy = usin60°
= \(\frac{\mathrm{u} \sqrt{3}}{2}\)
Vertical displacement s = uyt – 1/2 g t²
15= \(\frac{\mathrm{u} \sqrt{3}}{2}\) t – 1/2 × 10 × t²
Substitute t = 60/u from (1),
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 69
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 50
⇒ u = 22.07 ms-1
∴Speed of launch = 22.07 ms-1
At the time of striking the wall,
vx = ux = 22.07 cos 60°
= 11.04 ms-1
vy = uy – gt
= 22.07 sin 60° – 10 × \(\left(\frac{60}{22.07}\right)\)
vy = -8.07 ms-1
Resultant speed v = \(\sqrt{v_{x}^{2}+v_{y}^{2}}\)
= \(\sqrt{11.04^{2}+(-8.07)^{2}}\)
= 13.68 ms-1</sup

Question 11.
Define projectile. Show that the path of projectile is a parabola. Find the angle of projection at which the horizontal range and maximum height of the projectile are equal.
Answer:
Consider a particle thrown up at an angle q to the horizontal with a velocity u. The velocity of projection can be resolved into two component u cos θ and u sin θ along OX and OY respectively.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 51
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 52
The horizontal component remains constant while the vertical component is affected by gravity.
Let P(x, y) be the position of the projectile after ‘t’ seconds.
Then distance travelled ‘x’ along OX is given by,
x = horizontal component of the velocity × time
x = ucos θ × t …………………. (1)
The distance travelled along the vertical direction in same time ‘t’ is,
y = (usin θ) t – 1/2gt² ………… (2)
Substituting, t = \(\frac{x}{u \cos \theta}\) from (1) in (2),
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 53
y = ax – bx² …………………… (3) where
a = tan θ, b = \(\frac{\mathrm{g}}{2 \mathrm{u}^{2} \cos ^{2} \theta}\), are constants
for given values of θ, u and g.
Equation (3) represents a parabola. Therefore trajectory of a projectile is a parabola.

1st PUC Physics Motion in a Plane Numerical Problems Questions and Answers

Question 1.
A ball is thrown Into air with a speed of 62 ms-1 at an angle of 45° with the horizontal. Calculate

  1. the maximum height attained
  2. the horizontal range
  3. the time of flight and
  4. the velocity of the ball after 4 seconds.

Solution:
1. Maximum height, H = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\) Here,
u = 62 ms-1, θ = 45°, g = 9.8 ms-2
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 54
= \(\frac{62 \times 62}{4 \times 9.8}\) = 98.06 m.

2. Horizontal range. R is given by,
R = \(\frac{u^{2} \sin 2 \theta}{g}\)
= \(\frac{62 \times 62 \times \sin 90}{9.8}\)
= \(\frac{62 \times 62}{9.8}\) = 392.2m.

3. The time of flight,
T = \(\frac{2 u \sin \theta}{g}\)
= \(\frac{2 \times 62 \times \sin 45}{9.8}\)
= \(\frac{2 \times 62}{\sqrt{2} \times 9.8}\) = 8.95 s.

4. The horizontal component of the velocity remains constant throughout.
∴ Horizontal component of the velocity after 4 seconds vx = ucos θ
Here u = 62 ms-1 and θ = 45°
∴ vx = 62 cos45° = \(\frac{62}{\sqrt{2}}\) = 44.3 s.
The vertical component of the velocity changes due to the acceleration due to gravity.
We have vy = uy + at,
Here, uy = usin θ; a = -g; t = 4 s;
From vy = usin θ – 4g
∴ vy = 62 sin45 – 9.8 × 4
= \(\frac{62}{\sqrt{2}}\) -39.2 = 4.64 ms-1
Magnitude of the resultant velocity after 4 seconds is,
v = \(\sqrt{v_{x}^{2}+v_{y}^{2}}\)
= \(\sqrt{(44.3)^{2}+(4.64)^{2}}\) = 44.5 ms-1
Direction of the velocity is given by,
tan a = \(\frac{v_{y}}{v_{x}}\) = \(\frac{4.64}{44.3}\) = 0.104
∴ a = tan-1(0.104) = 5°5’

Question 2.
A bullet is fired at an angle of 60° with the vertical with certain velocity hits the ground 3 km away is it possible to hit the target 5 km away by adjusting the angle of projection? If not, what must be the velocity of projection for the same angle of projections?
Solution:
Let ‘u’ be the velocity of projection.
Angle of projection ‘θ’ = 90 – 60° = 30°
Horizontal Range,
R = \(\frac{u^{2} \sin 2 \theta}{9}\) = \(\frac{u^{2} \sin 60}{g}\)
∴ u2 = \(\frac{\mathrm{Rg}}{\sin 60}\) …………… (1)
For a given velocity, the maximum range attained is given by making sin2 θ =1, and
Rmax \(\frac{u^{2}}{g}\).
Substituting for u² from (1), we have
Rmax = \(\frac{\mathrm{R}}{\sin 60}\)
= \(\frac{3}{0.866}\) = 3.46km
Since the maximum range is less than 5 km, it is not possible to hit the target 5km away.
From equation (1), we have,
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 55
= 56582
∴ u = 237.9 ms-1
Thus the bullet will hit the target if it is projected with a velocity of 237.9 ms-1

Question 3.
A stone of mass 0.5 kg is tied to one end of a string and whirled along the horizontal circle of 1.2 m radius, If the period of rotation is 5 s, calculate the tension in the string. If breaking tension is 1.5 N, calculate the maximum speed with which it is rotated and also calculate the corresponding period of rotation.
Solution:
When the mass is rotated along a horizontal circle, the centripetal force is equal to the tension of the string.
Hence, T= mw²r,
Here, m = 0.5 kg,
w = \(\frac{2 \pi}{T}\) = \(\frac{2 \pi}{5}\)
r = 1.2 m.
∴ T = 0.5 × \(\left(\frac{2 \pi}{5}\right)^{2}\) × 1.2
= \(\frac{0.5 \times 4 \pi^{2} \times 1.2}{25}\)
= 0.9475 N
The maximum tension which the string can withstand without breaking is 1.5 N and the corresponding velocity ‘v’ is given by, 1.5 = \(\frac{m v^{2}}{r}\)
∴ v2 = \(\frac{1.5 \times r}{m}\) or
v = \(\sqrt{\frac{1.5 \times 1.2}{0.5}}\)
= 1.89 ms-1
The corresponding angular velocity is given by,
w = \(\frac{\mathrm{v}}{\mathrm{r}}\) = \(\frac{1.89}{1.2}\)
= 1.575 rads-1
∴ The period of rotation,
T = \(\frac{2 \pi}{1.575}\) = 3.99 s.

Question 4.
At what angle should a cyclist lean over from the vertical while negotiating a curve of radius 58m with a speed of 15 km/hour?
Solution:
The angle of banking is given by,
tan θ = \(\frac{v^{2}}{r g}\)
Here, v = \(\frac{15000}{60 \times 60}\) = 4.17 ms-1
∴ tan θ = \(\frac{(4.17)^{2}}{58 \times 9.8}\) = 0.03059
θ = tan-1(0.03059)
= 1°45’.

KSEEB Solutions

Question 5.
A train is moving round a curve of 52m radius and the distance between the rails is 1.2 m by how much should the outer rails be raised above the inner one so that a train running at the rate of 45 kmhr-1 may not skid.
Solution:
The angle of banking is given by,
tan θ = \(\frac{v^{2}}{r g}\)
Here, v = \(\frac{45000}{60 \times 60}\) = 12.5 ms-1 and
r = 52 m
∴ tan θ = \(\frac{(12.5)^{2}}{52 \times 9.8}\) = 0.3066
θ = tan-1(0.3066)
= 17°2’
If ‘h’ is the height of the outer rail above the inner one, then
h = 1.2’sin 17.05°
= 0.352 m.

Question 6.
A golf ball leaves the tee with a velocity of 50ms-1 at an angle of 300 with the horizontal. Find its

  1. time of flight
  2. Range of projectile and
  3. The velocity with which it hits the ground at the end of its flight.

Answer:
u = 50m/s, θ = 30°.
1. Time of flight
T = \(\frac{2 u \sin (\theta)}{g}\) = \(\frac{2 \times 50 \times \sin (30)}{9.8}\) = 5.10 s

2. Range R = \(\frac{u^{2} \sin (2 \theta)}{g}\) = \(\frac{50^{2} \sin \left(2 \times 30^{\circ}\right)}{9.8}\) = 220.92 m

3. Velocity at the end = velocity of projection = 50 m/s

Question 7.
The greatest and the least resultants of two forces acting at a point are 5 N and1N respectively. Find the forces. What is the resultant of these two forces when they act at an angle of 50°?
Solution:
Let P and Q be the two forces. The greatest resultant is
P + Q = 5 …………… (1)
The minimum resultant is,
P – Q = 1…………….. (2)
Adding (1) and (2)
2P = 6
P =3N
∴ Q = 2 N
When they are acting at an angle,
θ =50°
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 56
= 4.55 N
The direction of the resultant is,
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 57
= 0.3575
∴ a = 19.67° = 19° 40′
Therefore the resultant is making an angle of 19°40′ with 3N.

Question 8.
Four forces of magnitudes 2N, 3N, 4N, and 5N are acting on a body at a point are inclined at 30°, 60°, 90°, and 120° respectively with the horizontal. Find their resultant.
Solution:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 58
Given, P1 =2N, P2 = 3N
P3 = 4N and P4 = 5N
These forces can be resolved into components along the X and Y directions.
Rx= P1 cos θ1 + P2 cos θ2 + P3 cos θ3 + P4 cos θ4
= 2cos 30° + 3cos 60 + 4cos 90 + 5 cos 120
= 2\(\frac{\sqrt{3}}{2}\) + 3 × \(\frac{1}{2}\) + 4 × 0 + 5 × \(\frac{1}{2}\) = 0.732 N

Ry= P1 sin θ1 + P2 sin θ2 + P3 sin θ3 + P4 sin θ4
= 2 sin 30 + 3 sin 60 + 4 sin 90 + 5 cos 120
= 2 × \(\frac{1}{2}\) + 3 × \(\frac{\sqrt{3}}{2}\) + 4 × 1 + 5 × \(\frac{\sqrt{3}}{2}\)
= 1 + 1.5\(\sqrt{3}\) + 4 + 2.5\(\sqrt{3}\)
= 1 + 2.59 + 4 + 4.33
= 11.92 N
∴ Resultant force R = \(\sqrt{\mathrm{R}_{\mathrm{x}}^{2}+\mathrm{R}_{\mathrm{Y}}^{2}}\)
= Vo.7322 +11.92 2
= 11.96 N.

Question 9.
The resultant of two forces acting at 60° is \(\sqrt{13}\) N. When the same forces act at 90° the resultant is \(\sqrt{10}\) N. Find the magnitude of the two forces.
Solution:
Let P and Q be the two forces. When the angle between them is 60°, resultant force is \(\sqrt{13}\) N. Thus
\(\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ} \cos 60^{\circ}}\) = \(\sqrt{13}\)
i.e.\(\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+\mathrm{PQ}}\) = \(\sqrt{13}\)
P + Q + PQ = 13 ……………(1)
When the angle is 90° resultant force is
\(\sqrt{10}\) N.
\(\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ} \cos 90^{\circ}}\) = \(\sqrt{10}\)
or \(\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}}\) = \(\sqrt{10}\)
P2+ Q2 =10 ……………….. (2)
Substituting (2) in (1)
10 + P = 13
PQ = 3 ……………………(3)
(P + Q)² = P²+ Q² + 2PQ
= 10 + 2 × 3 = 16
P + Q =4 ………………. (4)
(P – Q)² = P² + Q² – 2PQ
= 10 – 6 = 4
P – Q = 2 ………………. (5)
2P =6; 2Q = 2
P = 3N and Q = 1N.

KSEEB Solutions

Question 10.
The sum of the two forces is 8 kg wt and the magnitude of the resultant which is at right angles to the smaller force is 4 kgwt. Find the two forces.
Solution:
Let P be the smaller force.
P + Q = 8 kg wt.,
a = 90°, R = 4 kg wt.,
tan α = \(\frac{\mathrm{Q} \sin \theta}{\mathrm{P}+\mathrm{Q} \cos \theta}\)
tan 90° = \(\frac{\mathrm{Q} \sin \theta}{\mathrm{P}+\mathrm{Q} \cos \theta}\)
∞ = \(\frac{\mathrm{Q} \sin \theta}{\mathrm{P}+\mathrm{Q} \cos \theta}\)
⇒ P + Q cos θ = 0
cos θ = \(-\frac{P}{Q}\)
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 59
Q² – P²= 16
(Q-P)(Q+P) = 16
(Q-P) × 8 = 16
Q – P = 2
P + Q = 8
∴ 2Q = 10 ⇒ Q = 5 kg wt. and P = 3 kg wt.

Question 11.
The resultant of two forces acting at an angle of 150° is perpendicular to the smaller force. If the smaller force is 12\(\sqrt{3}\) N, find the greater force and its resultant.
Solution:
θ = 150°,
Let P be the smaller force P = 12\(\sqrt{3}\) N
a = 90°, Q = ?, R = ?
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 60

Question 12.
Two equal, non parallel forces acting at a point on a body. The square of the resultant is found to be three times the product of the forces, what is the angle between them.
Answer:
Given, P = Q & R² = 3PQ = 3P²
we have R² = P² + Q² + 2PQ cos θ
i.e. 3P²= P² + P² + 2P² cos θ
i.e. 3P² = 2P² + 2P² cos θ
1P² = 2P²cos θ
\(\frac{1}{2}\) = cose
⇒ θ = cos-1(\(\frac{1}{2}\)) = 60°.

Question 13.
The greatest and least resultant of two forces acting at a point are 17N and 7N respectively. Find the two forces. If these forces act at right angles, find the magnitude of their resultant.
Answer:
For maximum resultant
P + Q=17N ……………….. (1)
For minimum resultant
P – Q = 7N ……………….. (2)
Eq(1) + (2) gives 2P = 24 Or P = 24/2 = 12N.
From (1), Q = 17 – 12 = 5N
When P and Q are in perpendicular directions,
R = \(\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}}\)
= \(\sqrt{12^{5}+5^{2}}\) = 13N(∵ cos 90° = 0).

Question 14.
Two equal forces are acting at a point and angle between them in 60°. Calculate magnitude of Resultant of these forces.
Answer:
Q = P and θ = 60°.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 61

1st PUC Physics Motion in a Plane Hard Questions and Answers

Question 1.
The total speed v1 of a projectile at its greatest height is \(\sqrt{\frac{6}{7}}\) of its speed v2 at half its greatest height. Find the angle of projection.
Answer:
Velocity at the highest point = ucos θ = v1
Hmax = \(\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{g}}\)
Using v = u – 2gh,
At h = \(\frac{\mathrm{H}_{\mathrm{max}}}{2}\), vertical component of v2 is
given by,
v2y2 = u2sin2θ – 2g × \(\frac{\mathrm{H}_{\mathrm{max}}}{2}\)
v2y2 = u2 sin2θ – \(\mathrm{g}\left(\frac{\mathrm{u}^{2} \sin 2 \theta}{2 \mathrm{g}}\right)\)
⇒ v = \(\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2}\)
Horizontal component v2 is
v2x = v1 = ucosθ
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 62
Squaring both sides,
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 70
⇒ 7 cos²θ = 6 cos²θ + 3 sin² θ
⇒ cos²θ = 3 sin²θ
⇒ tan²θ = 1/3
⇒ tan θ = ± \(1 / \sqrt{3}\)
⇒ θ = ± 30°
θ = 30°
(because negative angle is not possible)

Question 2.
Prove that the path of one projectile as seen from another projectile is a straight line.
Answer:
The coordinates of first projectile as seen from the second projectile are
X = x1 – x1
= u1cos θ1 t – u2cos θ2 t
X = (u1cos θ1 – u2cos θ2)t
Y = y1 – y2
= u1sin θ1 t – 1/2 gt² – (u2sin θ2 t – 1/2 gt²)
Y = (u1 sin θ1 – u2sin θ2)t
∴ \(\frac{Y}{X}=\frac{\left(u_{1} \sin \theta_{1}-u_{2} \sin \theta_{2}\right) t}{\left(u_{1} \cos \theta_{1}-u_{2} \cos \theta_{2}\right) t}\)
= a constant, say m
This equation is of the form Y = mX, which is the equation of a straight line. Thus the path of a projectile as seen from another projectile is a straight line.

Question 3.
A particle is projected with some speed at an angle a to the horizontal from the end B of the horizontal base BC of a triangle ABC. It rises to the vertex A and after just grazing it, falls down to reach point C of the base BC. If the base angles of the triangle are β and γ, show that 4 cot α = cot β + cot γ.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 63
Answer:
From the figure,
cot β = \(\frac{\mathrm{BD}}{\mathrm{h}_{\max }}\)
and cot γ = \(\frac{\mathrm{DC}}{\mathrm{h}_{\max }}\)
cot β + cot γ = \(\frac{\mathrm{BD}+\mathrm{DC}}{\mathrm{h}_{\max }}\)
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 64
\(\frac{\text { Range }}{h_{\max }}\)
⇒ cot β + cot γ = \(\frac{\left(u^{2} \sin 2 \alpha\right) / g}{\left(u^{2} \sin ^{2} \alpha\right) / 2 g}\)
⇒ cot β + cot γ = 2 × \(\frac{2 \sin \alpha \cos \alpha}{\sin ^{2} \alpha}\)
cot β + cot γ = 4 cot α.

1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation

You can Download Chapter 3 Classification and Tabulation Questions and Answers, Notes, 1st PUC Statistics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation

1st PUC Statistics Classification and Tabulation Two Marks Questions and Answers

Question 1.
What is classification of the data?
Solution:
Classification is the process of arranging the data in to groups or classes according to common

Question 2.
What are the objectives of classification?
Solution:

  • To reduce the size of the data
  • To bring the similarities together.
  • To enable further statistical analysis

KSEEB Solutions

Question 3.
What are the basis/types of classification?
Solution:

  1. Chronological classification
  2. Geographical classification
  3. Qualitative classification
  4. Quantitative classification.

Question 4.
Give the formula of Sturge’s to find the number of classes. Or Give the formula used to determine the number of classes.
Solution:
The numbers of classes are obtained using Sturges’s Rule: k = 1 + 3.22 log N;
N-number of items.

Question 5.
For what purpose is correction factor used in frequency distribution?
Solution:
To get a better continuity between the class interval of a frequency distribution exclusive class intervals are used, so, if the frequency distribution is in inclusive class intervals isconverted into exclusive class intervals using correction factor.

Question 6.
What are the guidelines of classification?
Solution:
following are some guidelines following while classification:

  • The number of classes should generally between 4 and 15.
  • Exclusive classes should be formed for better continuity between the class intervals.
  • The width of the classes should be usually kept constant throughout the distribution.
  • Avoid open-end classes.
  • The classes should be arranged in ascending or descending order.
  • The lower limit of the first class should be either ‘o’ or multiple of 5.

Question 7.
Define the term tabulation
Solution:
Tabulation is a systematic arrangement of the classified data in to columns and rows of a table

Question 8.
Mention the parts of a Table
Solution:
Table No., Title of the table, Headnote/Sub-title, Captions, Stubs, Body of the table, Footnote & Source.

KSEEB Solutions

Question 9.
What are the objects of Tabulation?
Solution:
The object/Purpose of tabulation are:-

  • It simplifies the complex data
  • To facilitate for comparison
  • To give an identity to the data
  • To reveals trend and patterns of the data

Question 10.
How the number of classes using Prof. H.A.Sturge’s?
Solution:
The number of classes are obtained using Sturges’s Rule: k = 1 + 3.22 Log N

Question 11.
What are inclusive & exclusive class intervals?
Solution:
If in a class, both lower and upper limits are included in the same class are inclusive class intervals, eg. 0-9, 10-19, 20-29…
If in a class, the lower limit is included in the same class but the upper limit is included in the next class are exclusive class intervals, eg. 0-10, 10-20, 20-30…

Question 12.
Define frequency distribution
Solution:
A systematic presentation of the values taken by a variable and the corresponding frequencies is called Frequency distribution.

KSEEB Solutions

Question 13.
What are Marginal and Conditional frequency distributions?
Solution:
If in a bivariate frequency distribution, if the distribution of only one variable is considered, the distribution is called marginal frequency distribution.
If in a bivariate frequency distribution, if the distribution of only one variable is formed subject to the condition of the other variable it is called conditional frequency distribution.

Question 14.
What is the tabulation of the data?
Solution:
Tabulation is a process of systematic arrangement of the classified data in rows and columns, in the form of a table.

Question 15.
What are the parts of a table?
Solution:
The parts of a table are: Table No., Title of the table, Headnote/Sub-title, Captions, Stubs, Body of the table, Footnote & Source

Question 16.
What is the purpose of ‘table number’ in tabulation?
Solution:
A number should be given to each and every table, in order to distinguish and also for easy reference.

Question 17.
What are captions and stubs of a table?
Solution:
Captions: Column headings are called captions. They explain what the column represents. Captions are always written in one or two words on the top of each column.
Stubs: Row headings are called Stubs. They explain what the row represents. Stubs are usually written in one or two words at the left extreme sid/of each row.

KSEEB Solutions

Question 18.
What is headnote of table?
Solution:
It is a brief explanatory note or statement given just below the heading of the table put in a bracket. The statistical units of measurements, such as in ‘000s, Rs., Million tones, crores, Kgs., etc., are usually put in bracket.

Question 19.
What is indicated by source note of a table?
Solution:
Below foot note or below the table, source of the data may be mentioned for verification to the reader, regarding publications, organizations, pages, Journals etc,

1st PUC Statistics Classification and Tabulation Five Marks Questions and Answers

Question 1.
Explain chronological classification and geographical classification of data with examples.
Solution:
Temporal/Chronological classification: when the data enumerated over a different period of time, the type of classification is called chronological classification. The above type of classification is called as time series, e.g. Time series of the population, is listed in chronological order starting with the earliest period.
Table showing the population of India
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 1
Geographical classification :
In this type of classification the data are classified on the basis of geographical or locational or area wise differences between various items. Such as cities, districts, states etc. e.g., production of sugar in India may be presented state wise in the following manner:-
Table showing the production of sugar in India
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 2

KSEEB Solutions

Question 2.
Explain qualitative and quantitative classification with examples.
Solution:
Qualitative classification: Classification of the different units on the basis of qualitative characteristics (Called Attributes). Such as sex, literacy, employment etc.
e. g., the members of a club can be classified on the basis of sex wise distribution as follows:-
Table showing the sex distribution of members of a club
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 3
Quantitative classification: classification of the number of units on the basis of quantitative data, such as according to Height, weight, Wages, Age (years), Number of children, etc, Thus the groups of a student may be classified on their heights as follows:-
Table showing the heights of students
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 4

Online midpoint calculator and find the midpoint of aline segment joining two points using the midpoint calcultor in just one click.

Question 3.
Define the following terms :
i. Frequency, class frequency:
Solution:
Frequency refers to the number of times an observation repeated (f). The number of observations corresponding to a particular class is known as the class Frequency Class frequency is a positive integer including zero

ii. Class limits:
Solution:
Class limits- Lowest and the highest values that are taken to define the boundaries of the class are called class limits

iii. Range of the class: It is the difference between highest and lowest value in the data, i.e., Range = H.V. – L.V.

iv. Width of the class: The difference between the upper and lower limits of class called width of the class. It is denoted by c or i.
i/c = Upper limit(UL) – Lower limit(LL)

v. Class mid point
Solution:
The central value of a class called mid value/point; \(\mathrm{m} / \mathrm{x}=\frac{\mathrm{LL}+\mathrm{UL}}{2}\)

vi. Define Frequency Density:
Solution:
The frequency per unit of class interval is the frequency density.
i.e. frequency density = \(\frac{\text { Frequency the class }}{\text { width of the class }}=f / w\)

vii. Relative frequency:
Solution:
Relative frequency.. is the ratio of frequency of class to the total frequency of the distribution
Relative Frequency \(\frac{\mathrm{f}}{\mathrm{N}}\)

viii. Class interval-inclusive, exclusive and open-end classes:
Solution:
If in a class, lower as well as upper limits are included in the same class are called Inclusive class
e. g. 30-39,40-49….
If in a class, the lower limit is included in the same class and upper limit is included in the next class, such a class is called Exclusive class, eg. 30-40, 40-50…
If in a class, the lower and upper limits of the class is not specified are called open end classes” e.g. less than/below, or more than/ above 100

ix. Cumulative frequency- less than and more than cumulative frequency:
Solution:
The added up frequencies are called cumulative frequencies.
The number of observations with values less than upper limit is less than cumulative frequency. (l.c.f) i.e. Frequencies added from the top.
The number of observations with values more than lower class limit is more than cumulative frequency (m.c.f)

x. Correction factor:
Solution:
It is half of the difference between lower limit of a class and upper limit of the preceding class. Thus,
Solution:
Correction facctor (C.F) = \(\frac{\text { Lower limit of a class-Upper limit of the precending class }}{2}\)

xi. A Frequency distribution Discrete, Continuous, Bi-variate, Marginal
Solution:
A systematic presentation of the values taken by a variable and the corresponding frequencies is called frequency is called Frequency distribution.
While framing a frequency distribution, if class intervals are not considered, is called discrete frequency distribution.
1. Example:
The number of families according to number of children .
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 5
While framing a frequency distribution, if class intervals are considered, is called continuous frequency distribution.

2. Example:
The following table showing the weight (kgs.) of persons
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 6
A frequency distribution formed on the basis of two related variables is called bi-variate frequency distribution.
For example, we want to classify data relating to the Height and Weights of a group of individuals, Income and Expenditure of a group of individuals, Ages of Husbands and Wives, Ages of mothers and Number of children, etc .
In a Bivariate frequency distribution, the frequency distribution of only one of the variables is considered, it is marginal frequency distribution.

KSEEB Solutions

Question 4.
Mention/what are the rules/principles of formation of Frequency of distribution?
Solution:

  1. The lower limit of the first class should be either 0 or a multiple of 5
  2. Exclusive classes should be formed for better continuity
  3. The number of classes should be generally between 4 & 15
  4. The width of the classes should be kept constant throughout the distribution
  5. Avoid open-end classes
  6. The classes should be arranged in ascending or descending order.

Question 5.
Mention the rules/principles of the tabulation
Solution:

  1. The size of the table should be according to the size of the paper
  2. The stubs and captions should be arranged logically according to the alphabetical, Chronological, Geographical order and items to be arranged according to the size.
  3. If desired to locate the particular quantity in the body of the table that should be showed by thick colored inks or shaded off.
  4. The table should not be overloaded with number of characteristics, rather can be prepare another table.
  5. The table should be complete in all respects by captions and stubs, titles heading and no cell is left blank if left do not put ‘o’ but give (—) dash marks or write N.A
  6. Miscellaneous column can be provided for the data which do not fit to the data, such as Ratio, percentages.
  7. Ditto ( “ ) marks should not be used, as they may confuse with the no. 11
  8. Footnote may contain about errors, omissions, remarks about the data.
  9. Sources if provided regarding publications, organizations, pages, Journals etc.

Question 6.
The employees of a college can be classified on the basis of sex wise distribution as follows:-
Solution:
Table showing the sex distribution of employess of a college.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 7

Question 7.
The Employees of a college can be classified according to their occupations as :
Solution:
Table-1
The blank table given below represents the number of employees with different occupation in a college

Occupations Number of employees
Teaching staff
clerks
Attenders
Security men
Total

Question 8.
The employees of a commercial Bank can be classified according to their occupations and sex is :
Solution:
Table-2
The blank table given below represents the number of employees with different occupation and their sex in a commercial Bank
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 8

Question 9.
The employees of a college can be classified according to their occupations , sex and their marital status is :
Solution:
Table-3 : The blank table given below represents the number of employees with different occupation, sex and their marital status in a college
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 9

Question 10.
In a survey of 40 families in a certain locality, the number of children per family was recorded and the following data were obtained.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 10
Represent the data in the form of a discrete frequency distribution.
Solution:
Frequency distribution of the number of children.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 11

Question 11.
Prepare a frequency table from the following table regarding the number fatal accidents occurred in a day in Bangalore in June 2010.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 12
Solution:
Frequency distribution of the number of children.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 13
Question 12.
From the following paragraph prepare a discrete frequency table with the number of letters present in the words .
“Success in the examination confers no right to appointment unless government is satisfied, after such enquiry as may be deemed necessary that the candidate is suitable for appointment to the public service”
Solution:
The number of digits in the above statement: Highest digit = 11 and Lowest digit = 2
Frequency distribution of the number of letters in the words present in the statement
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 14

KSEEB Solutions

Question 13.
The following are the marks obtained by 50 college students in a certain test.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 15
Take suitable width of the class interval marks using struge’s rule.
Solution:
Here, N= 50; Range = H.V.-L. V. = 49-12 = 37
The number classes as per Sturge’s rule are obtained as follows:
Number of class intervals (K) = 1 + 3.322 logN= 1 + 3.22 log 50 = 1 + (3.22 × 1.6990) = 6.47=7
classes (Approx.) Size/width of class intervals – e = \(\frac{\text { Range }}{\text { Number of class intervals }}=\frac{37}{7}\)
5.28 = 6 (Approx.)
The size/width of each class is 6 and there are 7 classes. Thus, the required continuous frequency distribution with exclusive class intervals width is prepared as :
Frequency distribution of marks of students
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 16

Question 14.
The following data gives ages of 32 individuals in a locality. Using Sturge’s rule form a frequency table with exclusive type of class intervals.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 17
Solution:
Here N= 32; Range = H .V.- L.V. = 59- 01 = 58
The number classes as per Sturge’s rule are:
Number of class intervals (K) =1 + 3.22 logN = 1 + 3.22 log (32) = 1+(3.22 × 1.5051) = 5.85=6
classes (Approx.) Size/width of class intervals – e = \(\frac{\text { Range }}{\text { Number of class intervals }}=\frac{58}{6}\) = 10
(approx). Teh size / width of each class is 10 and there are 6 classes.
Frequency distribution of marks of students
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 118

KSEEB Solutions

Question 15.
The following are the marks obtained by 50 students in statistics; prepare a frequency table with class intervals of 10 marks.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 19
Solution:
Range: H.V-L.V = 93 – 23 = 70; take width as 10 marks, and then the number of classes will be: 70/10=7.
Frequency distribution of marks of students in statistics test
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 20

Question 16.
Prepare a bivariate frequency distribution of the marks in Accountancy & Statistics:
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 21
Solution:
Here the both variables are discrete in nature no need to prepare class interval.
Bi-variate frequency table showing Ages (years) of Mothers and Number of children
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 22

Question 17.
Below are the ages of husbands and wives prepare a bivariate frequency distribution with suitable width :
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 23
Here both are continuous variables form the class intervals as below :
Solution:
Ages of Husbands: Highest age = 47, Lowest age = 25, Difference = 22/(i)5width =5. classes. Ages of wife: Highest age = 47, Lowest age = 21, Difference = 26/(i)5width = 6 (Approx.) classes.
Let X and Y be the ages of Husbands and ages of Wives.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 24

KSEEB Solutions

Question 18.
Below are given the marks obtained by a batch of 20 students in mathematics and statistics:
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 25
Solution:
Marks in Mathematics: Highest Marks = 72, Lowest marks = 25, Difference = 47/(i)l Owidth 5 = 5 (Approx.) classes.
Marks in statistics: Highest marks = 85, Lowest marks = 20, Difference = 65/(i)10width = 7 (Approx.) classes.
Let x and y be marks inmathematics and marks in statistics.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 26

TABULATION

Question 19.
Give a general format of a table.
Solution:
General format of a table
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 27

Question 20.
What are the requisites of a good table? Or What are the General rules to the tabulation?
Solution:

  • Size:- the size of the table should be according to the size of the paper with more rows than columns. Exchange of the data can be done by altering the column and rows. A sufficient space should be provided in a particular to enter any new or to alter any affected figures.
  • Logical order:- The stubs and captions should be arranged logically according to the alphabetical, Chronological, Geographical order and items to be arranged according to the size.
  • Identity:- If desired to locate the particular quantity in the body of the table that should be showed by thick colored inks or shaded off.
  • The table should not be overloaded with number of characteristics, rather can be prepare another table.
  • The table should be complete in all respects by captions and stubs, titles heading and no cell is left blank if left do not put ’o’ but give (—) dash marks or write N.A
  • Miscellaneous column can be provided for the data which do not fit to the data, such as Radius, percentages.
  • Ditto ( “ ) marks should not be used,as they may confuse with the number 11
  • Footnote may contain about errors, omissions, remarks
  • Sources if provided regarding publications, organizations pages Journals etc.

KSEEB Solutions

Question 21.
Elucidate the difference between classification and tabulation.
Solution:
Comparison between Classification and Tabulation:-
The following points may be given as comparison:

  • Classification and Tabulation are not two distinct processes. Before tabulation, data are classified and then displayed under different columns and rows of a table.
  • Classification is the process of arranging the data in to groups or classes according to common characteristics possessed by the items of the data;
    Whereas Tabulation is a process of systematic arrangement of the classified data in rows and columns, in the form of a table.
  • Table contains precise and accurate information, where as classification gives only a classified groups of data.
  • Classification reduces the size of the data and brings the similarities together and tabulation facilitates comparison, reveal the trend and tendencies of the data.

Question 22.
In the college out of total of 1200 applications received for I puc admission, 450
were applied for science and 580 applied for commerce and remaining applied for
arts faculty. Tabulate the above information.
Solution:
Table 1
Table showing the distribution of applicants for I year puc to different faculties
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 28

Question 23.
In the college out of total of 1200 applications received for I puc admission 780 were boys. 450 were applied for science of which 300 were boys and 580 applied for commerce 250 were girls and remaining applied for arts faculty. Tabulate the above information.
Solution
Table 2
Table showing the sex-wise distribution of applicants for I year puc to different faculties
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 29

KSEEB Solutions

Question 24.
In a Sigma multinational accountant consultants there are 180 were accountants, 210 were article helpers, 300 were practitioner trainees. Of all the members 30% were women among accountants, 20% in Articles helpers and 15% among trainees. Tabulate the data.
Solution:
Table 3
Table represents the members of Sigma accountants according to cadre and sex
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 30
Footnote: * 180 × 30%= 54, 210 × 20% = 42

Question 25.
In a trip organized by a college there were 80 persons each of whom paid Rs.15.50 on an average? There were 60 students each of whom paid Rs.16. members of the teaching staff were charged at higher rate. The number of ser ants was 6 (all males) and they were not charged anything. The number of females was 20 percent of the total of which one was a lady staff member. Tabulate the information.
Solution:
Table 4
Table showing the contribution (in Rs.) details made by members of a college trip.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 31

Question 26.
In a state, there are 30 Medical colleges, 10 Dental colleges and 50 Engineering colleges. Among the Medical colleges, 5 are government colleges, 10 are aided private colleges and remaining are the unaided private colleges. Of the unaided colleges, 5 colleges are run by minority institutions.
Among the Engineering colleges, 10 are government colleges, 20 are aided private colleges and the rest are unaided private colleges. Of the unaided colleges, 10 colleges are run by minority institutions.
Among the dental colleges, 2 are aided private colleges and the rest are the unaided private colleges of which one is run by a minority institutions.
Tabulate the above information.
Solution:
Table showing the Medical, Engineering and Dental colleges run by Government, Private Aided and Private Unaided colleges in a State
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 32

Question 27.
The number of cases filed, hearing made and disposed by different bench judges in a day at High court of Karnataka are as given:
(i) Criminal cases filed 12, hearings made in 8 cases and disposed 3,
(ii) Land dispute cases filed 18, hearings made in 12cases and disposed 5,
(iii) Government service cases filed 6, hearings made in 4 cases and disposed 3,
(iv) Cheating cases filed 15; hearing made in 12 cases and disposed 8.
Tabulate the above information.
Solution:
Table showing the different types of cases filed, heard and disposed in a day at High court of Karnataka
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 33

KSEEB Solutions

1st PUC Maths Question Bank Chapter 8 Binomial Theorem

Students can Download Maths Chapter 8 Binomial Theorem 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 8 Binomial Theorem

Question 1.
State and prove Binomial theorem.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 1
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 2

Some observations in a binomial theorem:
(1) The expansion of {a + b)n has (n + 1) terms
(2) The coefficients nCr occurring in the binomial theorem are known as binomial coefficients.
(3) The indices of V go on decreasing and that of ‘a’ go on increasing by 1 at each stage.
i.e., for each term: index of a + index of b-n.
(4) Since nCr=nCn_r we have
nC0 =nCn, nCx =nCn_r and so on.
Thus the coefficients of the terms equidistant from the beginning and the end in a binomial theorem are equal.
(5) General term in (a + b)n: Tr+1 – nCr anrbr
(6) Middle terms in (a + b)n

  • When ‘n’ is even, the middle term
    \(=\left(\frac{n}{2}+1\right)^{t h} \text { term }\)
  • When ‘n’ is odd ,the middle term are
    \(\frac{1}{2}(n+1)^{n} \text { term an } \frac{1}{2}(n+3)^{n} \text { term }\)

(7) Taking a = x and b = -y in the expansion, we get (x-y)n =[x + (-y)]n

1st PUC Maths Question Bank Chapter 8 Binomial Theorem 3

KSEEB Solutions

Question 2.
Expand each of the expression
(i)\(\left(x^{2}+\frac{3}{x}\right)^{4}\)
(ii)\((1-2 x)^{5}\)
(iii)\(\left(\frac{2}{x}-\frac{x}{2}\right)^{5}\)
(iv)\((2 x-3)^{6}\)
(v)\(\left(\frac{x}{3}+\frac{1}{x}\right)^{5}\)
(vi)\(\left(x+\frac{1}{x}\right)^{6}\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 4
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 5
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 6

Question 3.
Using binomial theorem, evaluate each of the following:
(i) (98)5
(ii) (96)3
(iii) (102)5
(iv) (101)4
(v) (99)5
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 7
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 8

KSEEB Solutions

Question 4.
Which is longer \((1.01)^{1000000} \text { or } 10,000 ?\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 9

Question 5.
Using Binomial theorem, indicate which number is larger \( (1.1)^{10000} \text { or } 1000 ?\)
Answer:
Spilitting 1.01 and using Binomial theorem write the first few terms, we have.
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 10

Question 6.
Find (a + b)4 – (a -b)4. Hence evaluate
\((\sqrt{3}+\sqrt{2})^{4}-(\sqrt{3}-\sqrt{2})^{4}\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 11

Question 7.
Find (x +1)6 +(x -1)6. Hence or otherwise evaluate
\((\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 12
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 13

Question 8.
Evaluate:
\((\sqrt{3}+\sqrt{2})^{6}+(\sqrt{3}-\sqrt{2})^{6}\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 14

Question 9.
Find the value of
\(\left(a^{2}+\sqrt{a^{2}-1}\right)^{4}+\left(a^{2}-\sqrt{a^{2}-1}\right)^{4}\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 15
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 16

KSEEB Solutions

Question 10.
Show that 9n+1 – 8n – 9 is divisible by 64, whenever ‘n’ is a positive integer.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 17

Question 11.
Using binomial theorem, prove that 6n -5n always leaves remainder 1 when
divided by 25.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 18

Question 12.
Prove that
\(\sum_{n=0}^{n} y \cdot c_{n}=4\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 19
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 20

Question 13.
Find the 4th term in the expansion of (x-2 y)12.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 21

Question 14.
Find the 13th term in the expansion of
\(\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}, x \neq 0 \)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 22

Question 15.
Write the general term in the expression of
(i) (x2 -y)6
(ii) (x2-yx)12,x≠0
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 23

KSEEB Solutions

Question 16.
Find the coefficient of
(i) x5 in (x + 3)8
(ii) a5b7 in (a – 2b)12
(iii) x6y3 in (x + 2y)9
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 24

Question 17.
Find a, if the 17th and 18th terms of the expansion (2 +a)50 are equal.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 25

Question 18.
In the expansion of (1+ a)m+n, prove that coefficients of am and an are equal.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 26

Question 19.
Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1+x)2n-1
Answer:
In (1+x)2n we have
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 27
From (1) and (2) we get, the coefficient of xn in (1 + x)2n is twice the coefficient of xn in (1 + x)2n-1.

KSEEB Solutions

Question 20.
Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 28

Question 21.
Find the middle terms in the expansions of
(i) \(\left(3-\frac{x^{3}}{6}\right)^{7}\)
(ii)\(\left(\frac{x}{3}+9 y\right)^{10}\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 29
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 30

Question 22.
Show that the middle term in the expansion of \((1+x)^{2 n} \text { is } \frac{1 \cdot 3 \cdot 5 \ldots(2 n-1)}{\lfloor n} 2^{n} x^{n} \)where ‘n’ is a positive integer.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 31

Binomial Expansion Calculator to make your lengthy solutions a bit easier. Use this and save your time. Binomial Theorem & Series Calculator.

Question 23.
The second, third and fourth terms in the binomial expansion (x + a)n are 240, 720 and 1080, respectively. Find x, a and n
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 32
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 33

Question 24.
The coefficients of three consecutive terms in the expansion of (1 + a)n are in the ratio 1:7: 42. Find n
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 34
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 35

KSEEB Solutions

Question 25.
The coefficients (r-1)th, rth,and (r + 1)th, terms in the expansion of (x + 1)th, are in the ratio 1:3:5. Find n and r.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 36
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 37

Question 26.
Find the term independent of x in the expansion of \(\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{6}\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 38

Question 27.
Find the term independent of x in the expansion of \(\left(\sqrt[3]{x}+\frac{1}{2 \sqrt[3]{x}}\right)^{18}, x>0\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 39

Question 28.
Find a, 6 and n in the expansion of (a + b)n if the first three terms of the expansion are 729,7290 and 30375, respectively.
Answer:
Given: Tx = 729, T2 = 7290 and T3 = 30375
∴ an=729……………….(1)
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 40

KSEEB Solutions

Question 29.
Find a if the coefficients of x2 and x3 in the expansion of (3 +ax)9 are equal.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 41

Question 30.
If the coefficients of (r – 5)th and (2r -1)th terms in the expansion of (1 + x)34 are equal, find r.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 42
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 43

Question 31.
If the coefficients of ar-1, ar and ar+1 in the expansion of (1 + a)n are in arithmetic progression, prove that n2 – n(4r +1) + 4r2 -2 = 0.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 44

Question 32.
Show that the coefficient of the middle term in the expansion of (1 + x)2n is equal to the sum of the coefficients of two middle terms in the expansion of (1 + x)2n-1.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 45
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 46

Question 33.
Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of
\(\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{n} \text { is } \sqrt{6}: 1\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 47

Question 34.
Find the rth term from the end in the expansion of (x + a)n.
Answer:
rth term from the end in (x + a)n
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 48

Question 35.
If a and b are distinct integers, prove that a-b is a factor of an -bn, whenever n is a positive integer.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 49

Question 36.
The sum of the coefficients of the first three terms in the expansion of
\(\left(x-\frac{3}{x^{2}}\right)^{m}, x \neq 0, m \)being a natural numbers, is 559. Find the term of the expansion containing x3.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 50

Question 37.
Find the coefficient of x5 in the product (1 + 2x)6(1 – x)7 using binomial theorem.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 51

Question 38.
Find the coefficient of a4 in the product (1 + 2a)4(2-a)s using binomial theorem.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 52

Question 39.
Expand using Binomial theorem
\(\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}, x \neq 0\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 53

KSEEB Solutions

Question 40.
Find the expansion of (3x2 – 1ax + 3a2)3 using binomial theorem.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 54

1st PUC Business Studies Question Bank with Answers Karnataka

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Karnataka 1st PUC Business Studies Question Bank with Answers

1st PUC Business Studies Question Bank with Answers

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Karnataka 1st PUC Business Studies Question Bank with Answers in Kannada

Karnataka 1st PUC Business Studies Blue Print of Model Question Paper

1st PUC Business Studies Blue Print of Model Question Paper

For practical oriented questions, questions are to be selected from each of the 3 groups (Group 1: First 4 chapters, Group 2: Next four chapters and Group 3: Last three chapters. Please follow the chapters list given in this table)
Note: While selecting the Practical Oriented questions in Part E, care should be taken to avoid duplication of questions in Section A, B, C and D).

Total number of questions asked is 43 for 139 marks. Students have to answer 33 for 100 marks.
Section Wise Distribution of Marks

1st PUC Business Studies Blue Print of Model Question Paper 1

We hope the given Karnataka 1st PUC Class 11 Business Studies Question Bank with Answers Solutions, Notes, Guide Pdf Free Download of 1st PUC Business Studies Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2020-2021 in English Medium and Kannada Medium will help you.

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1st PUC History Question Bank with Answers Karnataka

Expert Teachers at KSEEBSolutions.com has created Karnataka 1st PUC History Question Bank with Answers Solutions, Notes, Guide Pdf Free Download of 1st PUC History Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2020-21 in English Medium and Kannada Medium are part of 1st PUC Question Bank with Answers. Here KSEEBSolutions.com has given the Department of Pre University Education (PUE) Karnataka State Board NCERT Syllabus 1st Year PUC History Question Bank with Answers Pdf.

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Karnataka 1st PUC History Question Bank with Answers

1st PUC History Question Bank with Answers

Karnataka 1st PUC History Question Bank with Answers in English

Karnataka 1st PUC History Question Bank with Answers in Kannada

Karnataka 1st PUC History Syllabus

  1. Introduction to History
    1.1 Meaning and definition of History – Herodotus, Augustine, Karl Marx, J.B-Bury, Nehru and Toynbee
    1.2 Importance of the study of History
  2. The Story of Human Evolution
  3. History of Ancient Civilization
    3.1 Introduction
    3.2 Egyptian Civilization
    3.3 Mesopotamian Civilization
    3.4 Chinese Civilization
  4. Establishment of Greek and Roman Empires-Contributions
    4.1 Greek City – States – Cultural Contributions
    4.2. Roman Republics – Legacy of Romans
  5. Rise and Spread of Christianity and Islam
    5.1. Life and Teachings of Jesus Christ Spread of Christianity
    5.2 Life and Teachings of Prophet Mohammed
  6. Medieval Period Towards Change
    Chruch, Society and State – Feudalism
  7. Beginning of Modern Age
    7.1 Geographical Discoveries
    7.2 Renaissance
    7.3 Reformation-Martin Luther-Counter Reformation
  8. World Revolutions
    8.1 Industrial Revolution
    8.2. The American War of Independence
    8.3. The French Revolution (1789-1795 CE)
    8.4. Russian Revolution
  9. Napoleon and Rise of Nationalism
    9.1 Napoleon Bonaparte
    9.2. Unification of Italy
    9.3 Unification of Germany
  10. World Wars and International Organizations
    10.1 World War I-Treaty of Versailles
    10.2 Rise of dictatorships
    10.3 World Warn
    10.4 UNO-Organs-Achievements
  11. Contemporary World
    11.1 Cold War
    11.2 Disintegration of USSR
    11.3 Formation of CIS
  12. Non-Aligned Movement-Emergence of the Third world
  13. Map Work: Historical Places of World Importance

Karnataka 1st PUC History Blue Print of Model Question Paper

Karnataka 1st PUC History Blue Print of Model Question Paper

We hope the given Karnataka 1st PUC Class 11 History Question Bank with Answers Solutions, Notes, Guide Pdf Free Download of 1st PUC History Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2020-2021 in English Medium and Kannada Medium will help you.

If you have any queries regarding Karnataka State Board NCERT Syllabus 1st Year PUC Class 11 History Question Bank with Answers Pdf, drop a comment below and we will get back to you at the earliest.

1st PUC Political Science Question Bank with Answers Karnataka

Expert Teachers at KSEEBSolutions.com has created Karnataka 1st PUC Political Science Question Bank with Answers Solutions, Notes, Guide Pdf Free Download of 1st PUC Political Science Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2020-21 in English Medium and Kannada Medium are part of 1st PUC Question Bank with Answers. Here KSEEBSolutions.com has given the Department of Pre University Education (PUE) Karnataka State Board NCERT Syllabus 1st Year PUC Political Science Question Bank with Answers Pdf.

Students can also read 1st PUC Political Science Model Question Papers with Answers hope will definitely help for your board exams.

Karnataka 1st PUC Political Science Question Bank with Answers

1st PUC Political Science Question Bank with Answers

Karnataka 1st PUC Political Science Question Bank with Answers in English

Karnataka 1st PUC Political Science Question Bank with Answers in Kannada

Karnataka 1st PUC Political Science Blue Print of Model Question Paper

1st PUC Political Science Blue Print of Model Question Paper 1 1st PUC Political Science Blue Print of Model Question Paper 2

We hope the given Karnataka 1st PUC Class 11 Political Science Question Bank with Answers Solutions, Notes, Guide Pdf Free Download of 1st PUC Political Science Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2020-2021 in English Medium and Kannada Medium will help you.

If you have any queries regarding Karnataka State Board NCERT Syllabus 1st Year Class 11 PUC Political Science Question Bank with Answers Pdf, drop a comment below and we will get back to you at the earliest.

1st PUC Sociology Question Bank with Answers Karnataka

Expert Teachers at KSEEBSolutions.com has created Karnataka 1st PUC Sociology Question Bank with Answers Solutions, Notes, Guide Pdf Free Download of 1st PUC Sociology Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2020-21 in English Medium and Kannada Medium are part of 1st PUC Question Bank with Answers. Here KSEEBSolutions.com has given the Department of Pre University Education (PUE) Karnataka State Board NCERT Syllabus 1st Year PUC Sociology Question Bank with Answers Pdf.

Students can also read 1st PUC Sociology Model Question Papers with Answers hope will definitely help for your board exams.

Karnataka 1st PUC Sociology Question Bank with Answers

1st PUC Sociology Question Bank with Answers

Karnataka 1st PUC Sociology Question Bank with Answers in English

Karnataka 1st PUC Sociology Question Bank with Answers in Kannada

Karnataka 1st PUC Sociology Blue Print of Model Question Paper

1st PUC Sociology Blue Print of Model Question Paper

Karnataka 1st PUC Sociology Syllabus

Chapter 1 Nature of Sociology (Teaching 26 Hours, 18 Marks)

  • 1.1 Meaning, Definition, Characteristics and types.
  • 1.2 Meaning, Definition, Elements and types.
  • 1.3 Meaning, Definition and characteristics.
  • 1.4 Meaning, Definition and Characteristics.
  • 1.5 Importance of studying Sociology
  • 1.6 Contributions of Western and Indian sociologists to the development of Sociology.
    (a) Auguste Comte (b) Herbert Spencer (c) Emile Durkheim (d) Karl Mark (e) Dr. G.S. Ghurye (f) Dr. A. R. Desai

Chapter 2 Basic Concepts (Teaching 20 Hours, 24 Marks)

  • 2.1 Basic concepts: Meaning
  • 2.2 Society: Meaning, Definition, Characteristics and types
  • 2.3 Community: Meaning, definition, Elements and types.
  • 2.5 Institution: Meaning, definition and characteristics.
  • 2.6 Association: Meaning, definition and characteristics.
  • 2.7 Social Groups: Meaning, definition, characteristics and classification of social groups. (Primary – secondary, unorganized – organized groups).
  • 2.8 Social Control: Meaning and definition, nature, Purpose and types. (Folkways, Mores and Laws)

Chapter 3 Social Process (Teaching 16 Hours, 18 Marks)

  • 3.1 Social Process: Meaning, Definition and types.
  • 3.2 Co-operation: Meaning, Definition, Characteristics, Types and Importance.
  • 3.4 Conflict: Meaning, Definition, Characteristics and Types.
  • 3.5 Accommodation: Meaning, Definition, Characteristics, Methods and Importance.
  • 3.6 Assimilation: Meaning, Definition, Characteristics and Importance.

Chapter 4 Culture and Socialization (Teaching 16 Hours, 18 Marks)

  • 4.1 Culture: Meaning, Definition and Characteristics.
  • 4.2 Types of Culture: Material and Non-material Culture.
  • 4.3 Socialization: Meaning and Definition.
  • 4.4 Stages of Socialization: Oral, anal, oedipal, adolescence and adulthood.
  • 4.5 Agencies of Socialization: (Family, peer groups, schools, mass media and state)
  • 4.6 The Role of culture in socialization.

Chapter 5 Social Institutions (Teaching 14 Hours, 16 Marks)

  • 5.1 Marriage: Meaning and definition, characteristics, functions and types. Monogamy, polygamy, (with examples).
  • 5.2 Family: Meaning and Definition, Characteristics, primary and secondary functions of family, Types (nuclear and joint family)
  • 5.3 Education: Meaning and Definition, Characteristics and importance.
  • 5.4 Education: Meaning and definition, Characteristics functions and types.

Chapter 6 Social Change (Teaching Hours: 12, 13 Marks)

  • 6.1 Social change: meaning, definition and characteristics of social change.
  • 6.2 (a) Evolution (b) Progress (c) Development
  • 6.3 Factors for social change
    • 6.3.1 Natural factors or geographical factor.
    • 6.3.2 Biological factors
    • 6.3.3 Cultural factors
    • 6.3.4 Technological factors.
  • 6.4 Resistance to social change.
  • 6.5 Consequences of social change.

Chapter 7 Social Research (Teaching 12 Hours, 13 Marks)

  • 7.1 Social Research: Meaning, Definition
  • 7.2 Sources of Data
  • 7.3 Tools and techniques of data collection
    • 7.3.1 Observation: Meaning and Types
    • 7.3.2 Interview: Meaning and Types
    • 7.3.3 Questionnaire: Meaning and Types
    • 7.3.4 Report writing
  • 7.5 Importance of Social Research

Chapter 8 Environment and Study (Teaching 12 Hours, 13 Marks)

  • 8.1 Society and Environment: Meaning and definition
  • 8.2 Environmental pollution: Meaning types and factors.
    • 8.2.1 Air pollution: Meaning and causes
    • 8.2.2 Water pollution: Meaning and causes
    • 8.2.3 Noise pollution: Meaning and causes
    • 8.2.4 Soil pollution: Meaning and causes
    • 8.2.5 Sources of pollution
    • 8.2.6 Nuclear Radioactive pollution
    • 8.2.7 Thermal pollution
    • 8.2.8 Pollution by pesticides
    • 8.2.9 Pollution from the solid waste
    • 8.2.10 Sea pollution
    • 8.2.11 Some environmental issues:
      a. Global warming
      b. Destruction of the ozone layer
      c. Acid rain
      d. Greenhouse effect
  • 8.3 Bio-diversity
  • 8.4 Environmental protection: Procedures and responsibility of society

We hope the given Karnataka 1st PUC Class 11 Sociology Question Bank with Answers Solutions, Notes, Guide Pdf Free Download of 1st PUC Sociology Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2020-2021 in English Medium and Kannada Medium will help you.

If you have any queries regarding Karnataka State Board NCERT Syllabus 1st Year Class 11 PUC Sociology Question Bank with Answers Pdf, drop a comment below and we will get back to you at the earliest.

1st PUC Sanskrit Model Question Papers with Answers 2020-21 Karnataka

Expert Teachers at KSEEBSolutions.com has created New Syllabus Karnataka 1st PUC Sanskrit Model Question Papers with Answers 2020-21 Pdf Free Download of 1st PUC Sanskrit Previous Year Board Model Question Papers with Answers are part of 1st PUC Model Question Papers with Answers. Here We have given the Department of Pre University Education (PUE) Karnataka State Board Syllabus Second Year Model Question Papers for 1st PUC Sanskrit Model Question Papers with Answers 2020-2021 Pdf.

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Karnataka 1st PUC Sanskrit Model Question Papers with Answers 2020-2021

Karnataka 1st PUC Sanskrit Blue Print of Model Question Paper

1st PUC Sanskrit Blue Print of Model Question Paper 2

1st PUC Sanskrit Blue Print of Model Question Paper 3

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1st PUC Statistics Question Bank with Answers Karnataka

Expert Teachers at KSEEBSolutions.com has created Karnataka 1st PUC Statistics Question Bank with Answers Solutions, Notes, Guide Pdf Free Download of 1st PUC Statistics Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2020-21 in English Medium and Kannada Medium are part of 1st PUC Question Bank with Answers. Here KSEEBSolutions.com has given the Department of Pre University Education (PUE) Karnataka State Board NCERT Syllabus 1st Year PUC Statistics Question Bank with Answers Pdf.

Students can also read 1st PUC Statistics Model Question Papers with Answers hope will definitely help for your board exams.

1st PUC Statistics Question Bank with Answers

Karnataka 1st PUC Statistics Question Bank with Answers

1st PUC Statistics Features of the Question Bank

  • For the first time Pre-University Department has been released the Question Bank for the First Year PUC Statistics.
  • First Year PUC Statistics Text Book contains 10 units
  • The questions in the Question Bank are framed for all the units on the basis of the text book.
  • Following is the pattern of the Question Bank.
    Section A-each question carries one mark.
    Section B – each question carries two marks.
    Section C – each question carries five marks.
    Section D- each question carries ten marks.
    Section E- each question carries five marks (Practical – oriented questions).
  • Tests, Mid-term and Annual Examination Question Papers should be based on this Question Bank.
  • Model Question Papers are given at the end of the question bank.

Karnataka 1st PUC Statistics Question Paper Pattern

Karnataka 1st PUC Statistics Question Paper Pattern

Karnataka 1st PUC Statistics Blue Print of Model Question Paper

1st PUC Statistics Blue Print of Model Question Paper 1

Karnataka 1st PUC Statistics Syllabus and Marking Scheme

Unit I Introduction to Statistics and Some Basic Concepts:

Meaning – Origin – Scope of Statistics. Definitions – in singular and plural sense. Characteristics, Branches, Functions and limitations of Statistics. Statistical applications in other subjects. Distrust of Statistics – causes and remedies. Some basic concepts – units, population, sample, qualitative characteristic, quantitative characteristic, attribute, variables (discrete and continuous), nominal scale and ordinal scale.

Unit II Organization of Data:

Statistical enquiry and its stages. Primary and Secondary data. Methods of collection of primary data, with merits and demerits. Essentials of a good questionnaire. Questionnaire and schedule with respect to their relative merits and demerits. Sources of secondary data. Census Enumeration and Sample Survey with respect to their relative merits and demerits. Pilot survey. Sampling – Methods of sampling – simple random sampling, stratified sampling and systematic sampling. Errors of sampling.

Unit III Classification and Tabulation:

Classification – Introduction, Meaning and objectives of classification, Types – (i) Chronological (Temporal) (ii) Geographical (Spatial) (iii) Qualitative (Simple and manifold) (iv) Quantitative, Explanation with examples.

Formation discrete and continuous frequency distributions: Explanation of range, class, class limits, class intervals inclusive, exclusive open-end classes, correction factors conversion of inclusive classes to exclusive classes, class mid-point, width of the class interval, formation of less than and more than cumulative frequency distributions, frequency density, relative frequency. Rules of classification. Formation of uni-variate and bi-variate frequency distributions.

Tabulation : Meaning, difference between classification and tabulation. Parts of a table- Format of a table and brief explanation of parts of table, Rules of tabulation, Types of tables-
(i) simple and complex (many-fold) tables (ii) General and special purpose tables; preparation of blank tables, tables with numerical information.

Unit IV Diagrammatic and Graphical Representation of Data:

Diagrams – Meaning, needs of diagrams, General rules of construction of diagrams – simple, multiple, component, percentage bars – problems and constructions. Component (sub divided) Pie diagram.

Graphs – Explanation, types – Histogram (Equal and Unequal width of class intervals), examples for obtaining mode from Histogram. Frequency Polygon and frequency curve – meaning, method of construction with and without histogram. Ogives – Meaning method of construction of both less than and more than Ogives. Examples. Obtaining the values of Median and Quartiles from less than Ogive. Comparison of tables and diagrams, difference between diagrams and graphs.

Unit V Analysis of Univariate Data:

(a) Measures of central tendency (Averages) – Meaning, objectives, definition of an average, definition of central tendency, essentials of a good average. Types of averages,

  1. Arithmetic mean – Definition, problems for both ungrouped and grouped data, one missing frequency problems, properties of mean with numerical applications, corrected mean in case of one wrong observation, merits and demerits.
  2. Median – Definition, problems for ungrouped and grouped data, one missing frequency problems, graphical solution of median, merits and demerits.
  3. Mode – Definition, problems for ungrouped and grouped data (excluding grouping method). Graphical solution of mode, merits and demerits, Empirical relationship among mean, median and mode.
  4. Geometric mean – Definition, problems for ungrouped and grouped data, problems on growth rates and interest rates, merits and demerits.
  5. Harmonic mean – Definition, problems for ungrouped and grouped data, problems on average speed, average number of days required for the completion of a given work, merits and demerits. Relationship between AM, GM and HM.

(b) Measures of position – Meaning, definitions of quartiles, deciles and percentiles. Problem on ungrouped and grouped data.

(c) Measures of dispersion – Meaning, objectives, definition, essentials of a good measure
of dispersion. Absolute and relative measures of dispersion.
Types – absolute and relative measures.

  1. Range – Definition, absolute and relative measures formulae, problems on ungrouped data.
  2. Quartile deviation – Definition, absolute and relative measures formulae, problems on ungrouped and grouped data.
  3. Mean deviation – Definition, absolute and relative measures formulae, problems on ungrouped and grouped data based on mean, median and mode.
  4. Standard deviation – Definition, problems on ungrouped and grouped data, properties of standard deviation with numerical applications and variance. Coefficient of variation for ungrouped and grouped data.

(d) Moments, Skewness and kurtosis-

  1. Moments – Meaning, definition of centrla moments, description of first four central moments. Formulae for β1, γ1, β2 and γ2.
  2. Skewness – Definition, types with diagram measures of skewness – Karl Pearson’s and Bowley’s coefficient of skewness for grouped data. Skewness based on moments (for given values of µ2 and µ3 finding β2).
  3. Kurtosis – Definition, explanation of kurtosis with neat diagram, measure of kurtosis based on moments (given the values of µ4 and µ2 finding β2).

Unit VI Analysis of Bivariate Data:

(a) Correlation – Definition, Types – Simple, multiple, partial. Causation – Spurious, positive, negative, perfect and no correlation explanation with examples. Significance of study of correlation analysis.
Measurement of correlation – scatter diagram explanation with charts, merits and demerits. Problems regarding construction of scatter diagram.

Karl Pearson’s coefficient of correlation – definition, formulae for ungrouped and grouped data. Properties of coefficient of correlation, interpretation. Problems – ungrouped and grouped data.
Spearman’s coefficient of rank correlation – without ties and tie (one or two), interpretation. Problems – with ties (One or two repeated ranks) and without ties.

(b) Regression – Definition, regression lines/equations of x and y and y on x. Properties of regression coefficient and regression lines/equations, Problems on ungrouped and grouped data, uses of regression analysis. Comparison between correlation and regression.

Unit VII Association and attributes:
Introduction, definition, notations – meaning and methods of association. Yule’s coefficient of association and its applications.

Unit VIII Interpolation and Extrapolation:
Meaning and utilities of interpolation and extrapolation. Binomial expansion method of interpolation (with one missing value) and Extrapolation for next successive value. Merits and demerits, applications.

Unit IX Theory of probability:

Introduction to probability, Definition of Experiment, Outcomes, Deterministic experiment, Random experiment, Sample space, Null event, Simple event, Compound event, Sure event with examples. Meaning of Favourable and Exhaustive outcomes, Equally likely events, Union and intersection of events, Mutually exclusive events, Complement of an event with examples. Meaning of Classical and Empirical/statistical methods of assigning probabilities. Classical/Mathematical, Empirical/Statistical and axiomatic definitions of probability. Statement and proofs (on basis of Mathematical definition) of P(Φ) = 0 P(S) = 1, 0 ≤ P(A) ≤ 1, P(A) + P(A’) = 1, Statement and proofs (on basis of Mathematical definition) of addition theorem of probability – for two non-mutually and mutually exclusive events. Definition and examples of independent, dependent events and conditional probability. Statement and proofs (on basis of Mathematical definition) of multiplication theorem of probability – for dependent and independent events and applications.

Unit X Random Variable and Mathematical expectation of a discrete random variable:

Definition with examples of discrete and continuous random variables. Definition of probability mass function and probability density function. Bivariate and marginal probability distributions definitions with examples.

Definition of Expected value/mean, variance and standard deviation of a discrete random variable. Related functions defined on a discrete random variable, applications (solving numerical and verbal problems), including the case of missing probabilities. Expectation and variance of following functions with proofs – a, aX, aX ± b where ‘a’ and ‘b’ are any two constants and related examples.

Statement and proofs of addition and multiplication theorem of Expectation. Covariance and correlation coefficient of bivariate random variables.

1st PUC Statistics List of Practicals

  1. Formation of Univariate and Bivariate Frequency Distributions
  2. Preparation of Blank Tables and Tables with Information
  3. Diagrammatic Representation of Data
  4. Graphical Representation of Frequency Distribution
  5. Measures of Central Tendency and Positions
  6. Measures of Dispersion
  7. Measures of Skewness
  8. Correlation
  9. Regression
  10. Association of Attributes, Interpolation and Extrapolation
  11. Probability Applications
  12. Mathematical Expectation
  13. Covariance and Correlation of Random Variables

We hope the given Karnataka 1st PUC Class 11 Statistics Question Bank with Answers Solutions, Notes, Guide Pdf Free Download of 1st PUC Statistics Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2020-2021 in English Medium and Kannada Medium will help you.

If you have any queries regarding Karnataka State Board NCERT Syllabus 1st Year PUC Class 11 Statistics Question Bank with Answers Pdf, drop a comment below and we will get back to you at the earliest.

1st PUC Question Banks with Answers Karnataka

Expert Teachers at KSEEBSolutions.com has created Karnataka 1st PUC Question Banks with Answers Solutions, Notes, Guide Pdf Free Download of 1st PUC Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2020-21 in English Medium and Kannada Medium are part of KSEEB Solutions. Here We have given the Department of Pre University Education (PUE) Karnataka State Board NCERT Syllabus 1st Year PUC Question Banks with Answers Pdf.

Students can also read 1st PUC Model Question Papers with Answers hope will definitely help for your board exams.

Karnataka 1st PUC Question Banks with Answers

Karnataka 1st PUC Question Banks with Answers

We hope the given Karnataka 1st PUC Class 11 Question Banks with Answers Solutions, Notes, Guide Pdf Free Download of 1st PUC Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2020-2021 in English Medium and Kannada Medium will help you.

If you have any queries regarding Karnataka State Board NCERT Syllabus 1st Year PUC Class 11 Question Banks with Answers Pdf, drop a comment below and we will get back to you at the earliest.

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