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Physics MCQs for NEET — Practice Questions with Answers

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A body freely falling from the rest has a velocity ‘v’ after it falls through a height ‘h’. The distance it has to fall down for its velocity to become double, is

2 h

4 h

6 h

8 h

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Velocity of a body on reaching the point from which it was projected upwards, is

$v = 0$

$v = 2 u$

$v = 0 .5 u$

$v = - u$

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Explanation

Body reaches the point of projection with same speed in opposite direction.

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A body projected vertically upwards with a velocity u returns to the starting point in 4 seconds. If g = 10m/sec², the value of u is

5 m/sec

10 m/sec

15 m/sec

20 m/sec

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Explanation

Time of flight $T = \frac{2 u}{g} = 4 \sec \Rightarrow u = 20 m / s$

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Time taken by an object falling from rest to cover the height of h₁ and h₂ is respectively t₁ and t₂ . Then the ratio of t₁ to t₂ is

h₁ : h₂

$\sqrt{h_{1}} : \sqrt{h_{2}}$

h₁ : 2h₂

2h₁ : h₂

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Explanation

$t = \sqrt{\frac{2 h}{g}} \Rightarrow \frac{t_{1}}{t_{2}} = \sqrt{\frac{h_{1}}{h_{2}}}$

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A body is thrown vertically up from the ground. It reaches a maximum height of 100m in 5sec. After what time it will reach the ground from the maximum height position

1.2 sec

5 sec

10 sec

25 sec

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Explanation

Time of ascent = Time of descent = 5 sec

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A body thrown vertically upwards with an initial velocity u reaches maximum height in 6 seconds. The ratio of the distances travelled by the body in the first second and the seventh second is

1 : 1

11 : 1

1 : 2

1 : 11

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Explanation

Time of ascent $= \frac{u}{g} = 6 \sec \Rightarrow u = 60 m / s$

Distance in first second $h_{first} = 60 - \frac{g}{2} (2 \times 1 - 1) = 55 m$

Distance in seventh second will be equal to the distance in first second of vertical downward motion $h_{seventh} = \frac{g}{2} (2 \times 1 - 1) = 5 m$ ⇒ $h_{first} / h_{seventh} = 11 : 1$

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A particle is thrown vertically upwards. If its velocity at half of the maximum height is 10 m/s, then maximum height attained by it is (Take g = 10 m/s²)

8 m

10 m

12 m

16 m

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Explanation

Let particle thrown with velocity u and its maximum height is H then $H = \frac{u^{2}}{2 g}$

When particle is at a height H/2, then its speed is 10 m/s

From equation $v^{2} = u^{2} - 2 g h$

${(10)}^{2} = u^{2} - 2 g (\frac{H}{2}) = u^{2} - 2 g \frac{u^{2}}{4 g} \Rightarrow u^{2} = 200$

Maximum height $\Rightarrow H = \frac{u^{2}}{2 g} = \frac{200}{2 \times 10} = 10 m$

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A body, thrown upwards with some velocity, reaches the maximum height of 20m. Another body with double the mass thrown up, with double initial velocity will reach a maximum height of

200 m

16 m

80 m

40 m

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Explanation

Mass does not affect on maximum height.

$H = \frac{u^{2}}{2 g} \Rightarrow H \propto u^{2}$ , So if velocity is doubled then height will become four times. i.e. $H = 20 \times 4 = 80 m$

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A balloon starts rising from the ground with an acceleration of 1.25 m/s² . After 8s, a stone is released from the balloon. The stone will (g = 10 m/s²)

Reach the ground in 4 second

Begin to move down after being released

Have a displacement of 50 m

Cover a distance of 40 m in reaching the ground

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Explanation

When the stone is released from the balloon. Its height $h = \frac{1}{2} a t^{2} = \frac{1}{2} \times 1 .25 \times {(8)}^{2} = 40 m$ and velocity

$v = a t = 1 .25 \times 8 = 10 m / s$

$T i m e t a k e n b y t h e s t o n e t o r e a c h t h e g r o u n d : h = u t + \frac{1}{2} a t^{2} - 40 = 10 t + \frac{1}{2} (- 10) t^{2} t^{2} - 2 t - 8 = 0 (t - 4) (t + 2) = 0 t = 4 s e c$

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A body is thrown vertically upwards with a velocity u. Find the true statement from the following

Both velocity and acceleration are zero at its highest point

Velocity is maximum and acceleration is zero at the highest point

Velocity is maximum and acceleration is g downwards at its highest point

Velocity is zero at the highest point and maximum height reached is $u^{2} / 2 g$

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Explanation

At highest point $v = 0$ and $H_{\max} = \frac{u^{2}}{2 g}$

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