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Physics MCQs for NEET — Practice Questions with Answers

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The distance between two particles is decreasing at the rate of 6 m/sec. If these particles travel with same initial speeds and in the same direction, then the separation increases at the rate of 4 m/sec. The particles have speeds as

5 m/sec ; 1 m/sec

4 m/sec ; 1 m/sec

4 m/sec ; 2 m/sec

5 m/sec ; 2 m/sec

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Explanation

When two particles moves towards each other then $v_{1} + v_{2} = 6$ ...(i)

When these particles moves in the same direction then $v_{1} - v_{2} = 4$ ...(ii)

By solving $v_{1} = 5 m / s$ and $v_{2} = 1 m / s$

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An express train is moving with a velocity v₁. Its driver finds another train is moving on the same track in the same direction with velocity v₂. To escape collision, driver applies a retardation a on the train. The minimum time of escaping collision will be

$t = \frac{v_{1} - v_{2}}{a}$

$t_{1} = \frac{v_{1}^{2} - v_{2}^{2}}{2}$

None

Both

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Explanation

As the trains are moving in the same direction. So the initial relative speed $(v_{1} - v_{2})$ and by applying retardation final relative speed becomes zero.

From $v = u - a t$ ⇒ $0 = (v_{1} - v_{2}) - a t$ ⇒ $t = \frac{v_{1} - v_{2}}{a}$

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A stone falls from a balloon that is descending at a uniform rate of 12 m/s. The displacement of the stone from the point of release after 10 sec is

490 m

510 m

610 m

725 m

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Explanation

$u = 12 m / s$ , $g = 9 .8 m / \sec^{2}$ , $t = 10 \sec$

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

$= 12 \times 10 + \frac{1}{2} \times 9 .8 \times 100 = 610 m$

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A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, the average acceleration during contact is

2100 m/sec² downwards

2100 m/sec² upwards

1400 m/sec²

700 m/sec²

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Explanation

Velocity at the time of striking the floor,

$u = \sqrt{2 g h_{1}} = \sqrt{2 \times 9 .8 \times 10} = 14 m / s$

Velocity with which it rebounds.

$v = \sqrt{2 g h_{2}} = \sqrt{2 \times 9 .8 \times 2 .5} = 7 m / s$

∴ Change in velocity $Δ v = 7 - (- 14) = 21 m / s$

∴ Acceleration $= \frac{Δ v}{Δ t} = \frac{21}{0 .01} = 2100 m / s^{2}$ (upwards)

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A body A is projected upwards with a velocity of 98 m/s. The second body B is projected upwards with the same initial velocity but after 4 sec. Both the bodies will meet after

6 sec

8 sec

10 sec

12 sec

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Explanation

Let t be the time of flight of the first body after meeting, then $(t - 4)$ sec will be the time of flight of the second body. Since $h_{1} = h_{2}$

∴ $98 t - \frac{1}{2} g t^{2} = 98 (t - 4) - \frac{1}{2} g {(t - 4)}^{2}$

On solving, we get $t = 12$ seconds

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Two bodies of different masses m_a and m_b are dropped from two different heights a and b. The ratio of the time taken by the two to cover these distances are

a : b

b : a

$\sqrt{a} : \sqrt{b}$

$a^{2} : b^{2}$

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Explanation

$h = \frac{1}{2} g t^{2} \Rightarrow t = \sqrt{2 h / g}$

$t_{a} = \sqrt{\frac{2 a}{g}} and t_{b} = \sqrt{\frac{2 b}{g}} \Rightarrow \frac{t_{a}}{t_{b}} = \sqrt{\frac{a}{b}}$

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A body falls freely from rest. It covers as much distance in the last second of its motion as covered in the first three seconds. The body has fallen for a time of

3 s

5 s

7 s

9 s

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Explanation

$\frac{1}{2} g {(3)}^{2} = \frac{g}{2} (2 n - 1) \Rightarrow n = 5 s$

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A stone is dropped into water from a bridge 44.1 m above the water. Another stone is thrown vertically downward 1 sec later. Both strike the water simultaneously. What was the initial speed of the second stone ?

12.25 m/s

14.75 m/s

16.23 m/s

17.15 m/s

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Explanation

Time taken by first stone to reach the water surface from the bridge be t, then

$h = u t + \frac{1}{2} g t^{2} \Rightarrow 44 .1 = 0 \times t + \frac{1}{2} \times 9 .8 t^{2}$

$t = \sqrt{\frac{2 \times 44 .1}{9 .8}} = 3 s e c$

Second stone is thrown 1 sec later and both strikes simultaneously. This means that the time left for second stone $= 3 - 1 = 2 s e c$

Hence $44 .1 = u \times 2 + \frac{1}{2} 9 .8 {(2)}^{2}$

$\Rightarrow 44 .1 - 19 .6 = 2 u \Rightarrow u = 12 .25 m / s$

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A body is thrown vertically upwards. If air resistance is to be taken into account, then the time during which the body rises is

Equal to the time of fall

Less than the time of fall

Greater than the time of fall

Twice the time of fall

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Explanation

Let the initial velocity of ball be u

Time of rise $t_{1} = \frac{u}{g + a}$ and height reached $= \frac{u^{2}}{2 (g + a)}$

Time of fall $t_{2}$ is given by

$\frac{1}{2} (g - a) t_{2}^{2} = \frac{u^{2}}{2 (g + a)}$

$\Rightarrow t_{2} = \frac{u}{\sqrt{(g + a) (g - a)}} = \frac{u}{(g + a)} \sqrt{\frac{g + a}{g - a}}$

∴ $t_{2} > t_{1}$ because $\frac{1}{g + a} < \frac{1}{g - a}$

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A ball P is dropped vertically and another ball Q is thrown horizontally from the same height and at the same time. If air resistance is neglected, then

Ball P reaches the ground first

Ball Q reaches the ground first

Both reach the ground at the same time

The respective masses of the two balls will decide the time

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Explanation

Vertical component of velocities of both the balls are same and equal to zero. So $t = \sqrt{\frac{2 h}{g}}$

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Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.