HC Verma Concepts Of Physics Objective Solutions Of Chapter 3 (Rest and Motion:Kinematics)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 3 (Rest and Motion: Kinematics):

OBJECTIVE –I

1. A motor car is going due north at a speed of 50 km/h. It makes a 90° left turn without changing the speed. The change in the velocity of the car is about

(a) 50 km/h towards west

(b) 70 km/h towards south-west

(c) 70 km/h towards north-west

(d) Zero.

Sol: (b)

Change in velocity, Δ V

→ Δ V = V₂ – V₁

Or, Δ V = √ [50² + (– 50)²]

Or, Δ V = √ (5000)

Or, Δ V = 70.7 ≈ 70 km/h towards south-west.

2. Figure (3-Q2) shows the displacement-time graph of a particle moving on the X-axis.

(a) The particle is continuously going in positive x direction

(b) The particle is at rest

(c) The velocity increases up to a time t₀, and then becomes constant

(d) The particle moves at a constant velocity up to a time t₀, and then stops.

Sol: (d)

Slope of the tangent of a curve at a point in displacement-time graph is equal to the velocity. Here up to a time t₀, slope of the curve remain constant. After that slope is zero. So, the particle moves at a constant velocity up to a time t₀, and then stops.

3. A particle has a velocity u towards east at t = 0. Its acceleration is towards west and is constant. Let X_A and X_B be the magnitude of displacements in the first 10 seconds and the next 10 seconds

(a) X_A < X_B

(b) X_A = X_B

(c) X_A > X_B

(d) The information is insufficient to decide the relation of x_A with x_B.

Sol: (d)

Data is insufficient to decide the relation of x_A with x_B.

4. A person travelling on a straight line moves with a uniform velocity v₁ for some time and with uniform velocity v₂ for the next equal time. The average velocity v is given by

(a) v = (v₁ + v₂)/2

(b) v = √( v₁ * v₂)

(c) 2/v = (1/v₁) + (1/v₂)

(d) 1/v = (1/v₁) + (1/v₂)

Sol: (a)

Given: t₁ = t₂ = t.

Total distance travelled, S = v₁t + v₂t

Total time taken, T = t + t = 2t

So, the average velocity, v = S/T

Or, v = (v₁ + v₂)/2.

5. A person travelling on a straight line moves with a uniform velocity v₁ for a distance x and with a uniform velocity v₂ for the next equal distance. The average velocity v is given by

(a) v = (v₁ + v₂)/2

(b) v = √( v₁ * v₂)

(c) 2/v = (1/v₁) + (1/v₂)

(d) 1/v = (1/v₁) + (1/v₂)

Sol: (c)

Time taken by person for travelling 1^st x distance, t₁ = x/v₁.

Time taken by person for travelling next x distance, t₂ = x/v₂.

Total time taken, t = t₁ + t₂ = x {(1/v₁) + (1/v₂)}

Total distance travelled, S = x + x = 2x

So, the average velocity, v = S/t

Or, v = 2/ {(1/v₁) + (1/v₂)}

Or, 2/v = (1/v₁) + (1/v₂).

6. A stone is released from an elevator going up with an acceleration a. The acceleration of the stone after the release is

(a) a upward
(b) (g – a) upward

(c) (g – a) downward
(d) g downward.

Sol: (d)

After the release of the stone, only acceleration due to gravity acts downward.

7. A person standing near the edge of the top of a building throws two balls A and B. The ball A is thrown vertically upward and B is thrown vertically downward with the same speed. The ball A hits the ground with a speed v_A and the ball B hits the ground with a speed v_B. We have

(a) v_A > v_B 

(b) v_A < v_B 

(c) v_A = v_B

(d) the relation between v_A and v_B depends on height of the building above the ground.
Sol: (c) 

Let height of the building be h.

Ball A projected upwards with velocity u falls back to building top with velocity u downwards. It completes its journey to ground under gravity.

Therefore, (v_A)² = u² + 2gh ------- (1)

Ball B starts with downwards velocity u and reaches ground after travelling a vertical distance h.

Therefore, (v_B)² = u² + 2gh ------- (2)

From equation 1 and 2

We get, v_A = v_B.

8. In a projectile motion the velocity

(a) is always perpendicular to the acceleration

(b) is never perpendicular to the acceleration

(c) is perpendicular to the acceleration for one instant only

(d) is perpendicular to the acceleration for two instants.

Sol: (c) 

Only one instant velocity is perpendicular to the acceleration.

9. Two bullets are fired simultaneously, horizontally and with different speeds from the same place. Which bullet will hit the ground first?

(a) the faster one

(b) the slower one

(c) both will reach simultaneously

(d) depends on the masses.

Sol: (c) 

Both will reach simultaneously, because both have same vertical motion.

Initial velocity and acceleration in vertical direction are same for both the cases.

10. The range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be

(a) 25 m 
(b) 37 m
(c) 50 m 
(d) 100 m.

Sol: (d) 

Given: for both the cases initial velocity (u) is same.

We know,

Range, R = u² sin 2θ/g

For 1^st case:

→ 50 = u² sin 2*15 /g

Or, u²/g = 100

For 2^nd case:

→ R = u² sin (2*45)/g

Or, R = u²/g = 100 m.

11. Two projectiles A and B are projected with angle of projection 15° for the projectile A and 45° for the projectile B. If R_A and R_B be the horizontal range for the two projectiles, then
(a) R_A < R_B

(b) R_A = R_B

(c) R_A > R_B

(d) the information is insufficient to decide the relation of R_A with R_B.

Sol: (d) 

We have required the value of initial velocity (u) to decide the relation of R_A with R_B.

12. A river is flowing from west to east at a speed of 5 metres per minute. A man on the south bank of the river, capable of swimming at 10 metres per minute in still water, wants to swim across the river in the shortest time. He should swim in a direction

(a) due north

(b) 30° east of north

(c) 30° north of west

(d) 60° east of north.

Sol: (a) 

He should swim in a direction due north to across the river in the shortest time. 

13. In the arrangement shown in figure (3-Q3), the ends P and Q of an inextensible string move downwards with uniform speed u. Pulleys A and B are fixed. The mass M moves upwards with a speed

(a) 2u cos θ

(b) u/cos θ

(c) 2u/cos θ

(d) u cos θ.

Sol: (b) 

OBJECTIVE –II

1. Consider the motion of the tip of the minute hand of a clock. In one hour

(a) the displacement is zero

(b) the distance covered is zero

(c) the average speed is zero

(d) the average velocity is zero.

Sol: (a), (d)

Initial and final position of the tip of the minute hand of a clock is same in one hour. So, the displacement of the tip of the minute hand is zero.

And, velocity = displacement/time = zero.

2. A particle moves along the X-axis as

x = u (t – 2s) + a(t – 2 s)².

(a) the initial velocity of the particle is u

(b) the acceleration of the particle is a

(c) the acceleration of the particle is 2a

(d) at t = 2 s particle is at the origin.

Sol: (c), (d)

Given: x = u (t – 2s) + a (t – 2 s)².

We have, dx/dt = u + 2a (t – 2 s)

And, d²x/dt² = 2a

So, initial velocity = dx/dt|_{t = 0} = u – 4a.

Therefore, option (a) is wrong.

Acceleration, a = d²x/dt² = 2a

Therefore, option (b) is wrong and option (c) is correct.

Position of particle at t = 2 s is given by

→ x = u (t – 2s) + a(t – 2 s)²

Or, x = u (2 s – 2s) + a(2 s – 2 s)²

Or, x = 0

Therefore, option (d) is correct.

3. Pick the correct statements:

(a) Average speed of a particle in a given time is never less than the magnitude of the average velocity.

(b) It is possible to have a situation in which |dv/dt| ≠ 0 but d|v|/dt = 0.

(c) The average velocity of a particle is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval.

(d) The average velocity of a particle moving on a straight line is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval. (Infinite accelerations are not allowed.)

Sol: (a), (b), (c)

4. An object may have

(a) varying speed without having varying velocity

(b) varying velocity without having varying speed

(c) nonzero acceleration without having varying velocity

(d) nonzero acceleration without having varying speed.

Sol: (b), (d)

If speed of an object is changes then velocity also changes. So option (a) is wrong. In case of uniform circular motion speed remain constant but direction changes. So it velocity changes. Therefore, option (b) is correct.

Rate of change of velocity is acceleration. So, option (c) is wrong.

In case of uniform circular motion speed remain constant but it has an acceleration. So, option (d) is correct.

5. Mark the correct statements for a particle going on a straight line:

(a) If the velocity and acceleration have opposite sign, the object is slowing down.

(b) If the position and velocity have opposite sign, the particle is moving towards the origin.

(c) If the velocity is zero at an instant, the acceleration should also be zero at that instant.

(d) If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.

Sol: (a), (b), (d)

6. The velocity of a particle is zero at t = 0.

(a) The acceleration at t = 0 must be zero.

(b) The acceleration at t = 0 may be zero.

(c) If the acceleration is zero from t = 0 to t = 10s speed is also zero in this interval.

(d) If the speed is zero from t = 0 to t = 10s the acceleration is also zero in this interval.

Sol: (b), (c), (d)

7. Mark the correct statements:

(a) The magnitude of the velocity of a particle is equal to its speed.

(b) The magnitude of average velocity in an interval is equal to its average speed in that interval.

(c) It is possible to have a situation in which the speed of a particle is always zero but the average speed is not zero.

(d) It is possible to have a situation in which the speed of the particle is never zero but the average speed in an interval is zero.

Sol: (a)

8. The velocity-time plot for a particle moving on a straight line is shown in the figure (3-Q4).

(a) The particle has a constant acceleration.

(b) The particle has never turned around.

(c) The particle has zero displacement.

(d) The average speed in the interval 0 to 10 s is the same as the average speed in the interval 10 s to 20 s.

Sol: (a), (d)

9. Figure (3-Q5) shows the position of a particle moving on the X-axis as a function of time.

(a) The particle has come to rest 6 times.

(b) The maximum speed is at t = 6 s.

(c) The velocity remains positive for t = 0 to t = 6 s.

(d) The average velocity for the total period shown is negative.

Sol: (a)

10. The accelerations of a particle as seen from two frames S₁ and S₂ have equal magnitude 4 m/s².

(a) The frames must be at rest with respect to each other.

(b) The frames may be moving with respect to each other but neither should be accelerated with respect to the other.

(c) The acceleration of S₂, with respect to S₁ may either be zero or 8 m/s².

(d) The acceleration of S₂ with respect to S₁ may be anything between zero and 8 m/s².

Sol: (d)

Previous year’s chapter-wise questions and solutions of Kinematics: Click Here

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