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Physics MCQs for NEET — Practice Questions with Answers

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The position of a particle moving along the x-axis at certain times is given below :

t (s)	0	1	2	3
x (m)	-2	0	6	16

Which of the following describes the motion correctly

Uniform, accelerated

Uniform, decelerated

Non-uniform, accelerated

There is not enough data for generalization

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Explanation

Average velocity $v = \frac{Δ x}{Δ t}$

By using the data from the table

$v_{1} = \frac{0 - (- 2)}{1} = 2 m / s . v_{2} = \frac{6 - 0}{1} = 6 m / s$

$v_{3} = \frac{16 - 6}{1} = 10 m / s$

So, motion is non-uniform but accelerated.

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Consider the acceleration, velocity and displacement of a tennis ball as it falls to the ground and bounces back. Directions of which of these changes in the process ?

Velocity only

Displacement and velocity

Acceleration, velocity and displacement

Displacement and acceleration

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Explanation

Only direction of displacement and velocity gets changed, acceleration is always directed vertically downward.

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The displacement of a particle, moving in a straight line, is given by $s = 2 t^{2} + 2 t + 4$ where s is in metres and t in seconds. The acceleration of the particle is

2 m/s²

4 m/s²

6 m/s²

8 m/s²

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Explanation

$s = 2 t^{2} + 2 t + 4, v = \frac{d s}{d t} = 4 t + 2 a = \frac{d v}{d t} = 4 m / s^{2}$

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A body A starts from rest with an acceleration a₁. After 2 seconds, another body B starts from rest with an acceleration a₂. If they travel equal distances in the 5th second, after the start of A, then the ratio a₁ : a₂ is equal to

5 : 9

5 : 7

9 : 5

9 : 7

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Explanation

According to problem

Distance travelled by body A in 5^th sec and distance travelled by body B in 3^rdsec. of its motion are equal.

$0 + \frac{a_{1}}{2} (2 \times 5 - 1) = 0 + \frac{a_{2}}{2} [2 \times 3 - 1]$

$9 a_{1} = 5 a_{2} \Rightarrow \frac{a_{1}}{a_{2}} = \frac{5}{9}$

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The velocity of a bullet is reduced from 200m/s to 100m/s while travelling through a wooden block of thickness 10cm. The retardation, assuming it to be uniform, will be

$10 \times 10^{4}$ m/s²

$12 \times 10^{4}$ m/s²

$13 .5 \times 10^{4}$ m/s²

$15 \times 10^{4}$ m/s²

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Explanation

$u = 200 m / s, v = 100 m / s, s = 0 .1 m$

$a = \frac{u^{2} - v^{2}}{2 s} = \frac{{(200)}^{2} - {(100)}^{2}}{2 \times 0 .1} = 15 \times 10^{4} m / s^{2}$

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A body of 5 kg is moving with a velocity of 20 m/s. If a force of 100N is applied on it for 10s in the same direction as its velocity, what will now be the velocity of the body ?

200 m/s

220 m/s

240 m/s

260 m/s

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Explanation

$v = u + a t = u + (\frac{F}{m}) t = 20 + (\frac{100}{5}) \times 10 = 220 m / s$

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A particle starts from rest, accelerates at 2 m/s² for 10s and then goes for constant speed for 30s and then decelerates at 4 m/s² till it stops. What is the distance travelled by it ?

750 m

800 m

700 m

850 m

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Explanation

Velocity acquired by body in 10sec

$v = 0 + 2 \times 10 = 20 m / s$

and distance travelled by it in 10 sec

$S_{1} = \frac{1}{2} \times 2 \times {(10)}^{2} = 100 m$

then it moves with constant velocity (20 m/s) for 30 sec

$S_{2} = 20 \times 30 = 600 m$

After that due to retardation $(4 m / s^{2})$ it stops

$S_{3} = \frac{v^{2}}{2 a} = \frac{{(20)}^{2}}{2 \times 4} = 50 m$

Total distance travelled $S_{1} + S_{2} + S_{3} = 750 m$

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The engine of a motorcycle can produce a maximum acceleration 5 m/s². Its brakes can produce a maximum retardation 10 m/s². What is the minimum time in which it can cover a distance of 1.5 km

30 sec

15 sec

10 sec

5 sec

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A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is

12m

18m

24m

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Explanation

$S \propto u^{2}$ . Now speed is two times so distance will be four times $S = 4 \times 6 = 24 m$

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A student is standing at a distance of 50 metres from the bus. As soon as the bus begins its motion with an acceleration of 1ms^–2, the student starts running towards the bus with a uniform velocity u. Assuming the motion to be along a straight road, the minimum value of u, so that the student is able to catch the bus is

5 ms^–1

8 ms^–1

10 ms^–1

12 ms^–1

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Explanation

Let student will catch the bus after t sec. So it will cover distance ut.

Similarly distance travelled by the bus will be $\frac{1}{2} a t^{2}$ .

For the given condition;

$u t = 50 + \frac{1}{2} a t^{2} = 50 + \frac{t^{2}}{2}$ [ $a = 1 m / s^{2}$ ]

⇒ u $= \frac{50}{t} + \frac{t}{2}$

To find the minimum value of u

$\frac{d u}{d t} = 0$ , so we get $t = 10 \sec$ , then $u = 10 m / s$

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