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Physics MCQs for NEET — Practice Questions with Answers

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If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is

$\frac{1}{2} g t^{2}$

$u t - \frac{1}{2} g t^{2}$

$(u - g t) t$

$u t$

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Explanation

The distance covered by the ball during the last t seconds of its upward motion = Distance covered by it in first t seconds of its downward motion

From $h = u t + \frac{1}{2} g t^{2}$

$h = \frac{1}{2} g t^{2}$ [As u = 0 for it downward motion]

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A car travels a distance S on a straight road in two hours and then returns to the starting point in the next three hours. Its average velocity is

S/5

2S/5

S/2 + S/3

None of the above

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Explanation

Average velocity $= \frac{Total displacement}{Time} = \frac{0}{2 + 3} = 0$

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A body has speed V, 2V and 3V in first 1/3 of distance S, second 1/3 of S and third 1/3 of S respectively. Its average speed will be

$\frac{18}{11} V$

$\frac{11}{18} V$

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Explanation

$v_{a v} = \frac{Total distance}{Time taken} = \frac{x}{\frac{x / 3}{v} + \frac{x / 3}{2 v} + \frac{x / 3}{3 v}} = \frac{18}{11} v$

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If the body covers one-third distance at speed v₁, next one third at speed v₂ and last one third at speed v₃, then average speed will be

$\frac{v_{1} v_{2} + v_{2} v_{3} + v_{3} v_{1}}{v_{1} + v_{2} + v_{3}}$

$\frac{v_{1} + v_{2} + v_{3}}{3}$

$\frac{v_{1} v_{2} v_{3}}{v_{1} v_{2} + v_{2} v_{3} + v_{3} v_{1}}$

$\frac{3 v_{1} v_{2} v_{3}}{v_{1} v_{2} + v_{2} v_{3} + v_{3} v_{1}}$

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Explanation

$v_{a v} = \frac{x}{\frac{x / 3}{v_{1}} + \frac{x / 3}{v_{2}} + \frac{x / 3}{v_{3}}} = \frac{3 v_{1} v_{2} v_{3}}{v_{1} v_{2} + v_{2} v_{3} + v_{1} v_{3}}$

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The displacement of the particle varies with time according to the relation $x = \frac{k}{b} [1 - e^{- b t}]$ . Then the velocity of the particle is

$k (e^{- b t})$

$\frac{k}{b^{2} e^{- b t}}$

$k b e^{- b t}$

None of these

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Explanation

$v = \frac{d x}{d t} = \frac{d}{d t} [\frac{k}{b} (1 - e^{- b t})] = \frac{k}{b} [0 - (- b) e^{- b t}] = k e^{- b t}$ .

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The acceleration of a particle starting from rest, varies with time according to the relation A = – aω² sinω t. The displacement of this particle at a time t will be

$- \frac{1}{2} (a ω^{2} \sin ω t) t^{2}$

$a ω \sin ω t$

$a ω \cos ω t$

$a \sin ω t$

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Explanation

Velocity $v = \int A d t = \int (- a ω^{2} \sin ω t) d t = a ω \cos ω t$

Displacement x $= \int v d t = \int a ω \cos ω t d t = a \sin ω t$

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If the velocity of a particle is (10 + 2t²) m/s, then the average acceleration of the particle between 2s and 5s is

2 m/s²

4 m/s²

12 m/s²

14 m/s²

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Explanation

Average acceleration $= \frac{Change in velocity}{Time taken} = \frac{v_{2} - v_{1}}{t_{2} - t_{1}}$

$= \frac{[10 + 2 {(5)}^{2}] - [10 + 2 {(2)}^{2}]}{3} = \frac{60 - 18}{3} = 14 m / s^{2}$

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A bullet moving with a velocity of 200 cm/s penetrates a wooden block and comes to rest after traversing 4 cm inside it. What velocity is needed for travelling distance of 9 cm in same block?

100 cm/s

136.2 cm/s

300 cm/s

250 cm/s

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Explanation

As $v^{2} = u^{2} - 2 a s \Rightarrow u^{2} = 2 a s$ $(∵ v = 0)$

$\Rightarrow u^{2} \propto s$ ⇒ $\frac{u_{2}}{u_{1}} = {(\frac{s_{2}}{s_{1}})}^{1 / 2}$

⇒ $u_{2} = {(\frac{9}{4})}^{1 / 2} u_{1} = \frac{3}{2} u_{1} = 300 c m / s$

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A thief is running away on a straight road in jeep moving with a speed of 9 ms^–1. A police man chases him on a motor cycle moving at a speed of 10 ms^–1. If the instantaneous separation of the jeep from the motorcycle is 100 m, how long will it take for the police to catch the thief

1 s

19 s

90 s

100 s

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Explanation

The relative velocity of policeman w.r.t. thief = 10 – 9 = 1 m/s.

∴ Time taken by police to catch the thief $= \frac{100}{1}$ =100 sec

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A car A is travelling on a straight level road with a uniform speed of 60 km/h. It is followed by another car B which is moving with a speed of 70 km/h. When the distance between them is 2.5 km, the car B is given a deceleration of 20 km/h². After how much time will B catch up with A

1 hr

1/2 hr

1/4 hr

1/8 hr

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Explanation

Let car B catches, car A after ‘t’ sec, then

$60 t + 2 .5 = 70 t - \frac{1}{2} \times 20 \times t^{2}$

⇒ $10 t^{2} - 10 t + 2 .5 = 0$

⇒ $t^{2} - t + 0 .25 = 0$

∴ $t = \frac{1 \pm \sqrt{1 - 4 \times (0 .25)}}{2} = \frac{1}{2} h r$

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