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Physics NEET Questions — Free MCQ Practice

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The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on planet B. A man jumps to a height of 2m on the surface of A. What is the height of jump by the same person on the planet B

A.

18m

B.

6m

C.

$\frac{2}{3} m$

D.

$\frac{2}{9} m$

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Explanation

$H_{\max} = \frac{u^{2}}{2 g} \Rightarrow H_{\max} \propto \frac{1}{g}$

On planet B value of g is $1 / 9$ times to that of A. So value of $H_{\max}$ will become 9 times i.e. $2 \times 9 = 18 m e t r e$

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A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s². He reaches the ground with a speed of 3 m/s. At what height, did he bail out ?

A.

293 m

B.

111 m

C.

91 m

D.

182 m

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When a ball is thrown up vertically with velocity V₀, it reaches a maximum height of 'h'. If one wishes to triple the maximum height then the ball should be thrown with velocity

A.

$\sqrt{3} V_{o}$

B.

3 V₀

C.

9 V₀

D.

3/2 V₀

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Explanation

$H_{\max} \propto u^{2}$ ∴ $u \propto \sqrt{H_{\max}}$

i.e. to triple the maximum height, ball should be thrown with velocity $\sqrt{3} u$ .

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A particle moving in a straight line covers half the distance with speed of 3 m/s. The other half of the distance is covered in two equal time intervals with speed of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during this motion is

A.

4.0 m/s

B.

5.0 m/s

C.

5.5 m/s

D.

4.8 m/s

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Explanation

If $t_{1}$ and $2 t_{2}$ are the time taken by particle to cover first and second half distance respectively.

$t_{1} = \frac{x / 2}{3} = \frac{x}{6}$ …(i)

$x_{1} = 4 .5 t_{2}$ and $x_{2} = 7 .5 t_{2}$

So, $x_{1} + x_{2} = \frac{x}{2} \Rightarrow 4 .5 t_{2} + 7 .5 t_{2} = \frac{x}{2}$

$t_{2} = \frac{x}{24}$ …(ii)

Total time $t = t_{1} + 2 t_{2} = \frac{x}{6} + \frac{x}{12} = \frac{x}{4}$

So, average speed $= 4 m / s e c$ .

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The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity v₀ The distance travelled by the particle in time t will be

A.

$v_{0} t + \frac{1}{3} b t^{2}$

B.

$v_{0} t + \frac{1}{3} b t^{3}$

C.

$v_{0} t + \frac{1}{6} b t^{3}$

D.

$v_{0} t + \frac{1}{2} b t^{2}$

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Explanation

$\frac{d v}{d t} = b t \Rightarrow d v = b t d t \Rightarrow v = \frac{b t^{2}}{2} + K_{1}$

At $t = 0, v = v_{0} \Rightarrow K_{1} = v_{0}$

We get $v = \frac{1}{2} b t^{2} + v_{0}$

Again $\frac{d x}{d t} = \frac{1}{2} b t^{2} + v_{0} \Rightarrow x = \frac{1}{2} \frac{b t^{2}}{3} + v_{0} t + K_{2}$

At $t = 0, x = 0 \Rightarrow K_{2} = 0$

$∴ x = \frac{1}{6} b t^{3} + v_{0} t$

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A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is

A.

$(\frac{α^{2} + β^{2}}{α β}) t$

B.

$(\frac{α^{2} - β^{2}}{α β}) t$

C.

$\frac{(α + β) t}{α β}$

D.

$\frac{α β t}{α + β}$

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Explanation

Let the car accelerate at rate $α$ for time $t_{1}$ then maximum velocity attained, $v = 0 + α t_{1} = α t_{1}$

Now, the car decelerates at a rate $β$ for time $(t - t_{1})$ and finally comes to rest. Then,

$0 = v - β (t - t_{1})$ ⇒ $0 = α t_{1} - β t + β t_{1}$

⇒ $t_{1} = \frac{β}{α + β} t$

∴ $v = \frac{α β}{α + β} t$

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A stone dropped from a building of height h and it reaches after t seconds on earth. From the same building if two stones are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after t₁ and t₂ seconds respectively, then

A.

$t = t_{1} - t_{2}$

B.

$t = \frac{t_{1} + t_{2}}{2}$

C.

$t = \sqrt{t_{1} t_{2}}$

D.

$t = t_{1}^{2} t_{2}^{2}$

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Explanation

If a stone is dropped from height h

then $h = \frac{1}{2} g t^{2}$ …(i)

If a stone is thrown upward with velocity u then

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

If a stone is thrown downward with velocity u then

$h = u t_{2} + \frac{1}{2} g t_{2}^{2}$ …(iii)

From (i), (ii) and (iii) we get

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

$u t_{2} + \frac{1}{2} g t_{2}^{2} = \frac{1}{2} g t^{2}$ …(v)

Dividing (iv) and (v) we get

∴ $\frac{- u t_{1}}{u t_{2}} = \frac{\frac{1}{2} g (t^{2} - t_{1}^{2})}{\frac{1}{2} g (t^{2} - t_{2}^{2})}$

or $- \frac{t_{1}}{t_{2}} = \frac{t^{2} - t_{1}^{2}}{t^{2} - t_{2}^{2}}$

By solving $t = \sqrt{t_{1} t_{2}}$

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A ball is projected upwards from a height h above the surface of the earth with velocity v. The time at which the ball strikes the ground is

A.

$\frac{v}{g} + \frac{2 h g}{\sqrt{2}}$

B.

$\frac{v}{g} [1 - \sqrt{1 + \frac{2 h}{g}}]$

C.

$\frac{v}{g} [1 + \sqrt{1 + \frac{2 g h}{v^{2}}}]$

D.

$\frac{v}{g} [1 + \sqrt{v^{2} + \frac{2 g}{h}}]$

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Explanation

Since direction of v is opposite to the direction of g and h so from equation of motion

$h = - v t + \frac{1}{2} g t^{2}$

$\Rightarrow g t^{2} - 2 v t - 2 h = 0$

$\Rightarrow t = \frac{2 v \pm \sqrt{4 v^{2} + 8 g h}}{2 g}$

$\Rightarrow t = \frac{v}{g} [1 + \sqrt{1 + \frac{2 g h}{v^{2}}}]$

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A particle is dropped vertically from rest from a height. The time taken by it to fall through successive distances of 1 m each will then be

A.

All equal, being equal to $\sqrt{2 / g}$ second

B.

In the ratio of the square roots of the integers 1, 2, 3.....

C.

In the ratio of the difference in the square roots of the integers i.e. $\sqrt{1}, (\sqrt{2} - \sqrt{1}), (\sqrt{3} - \sqrt{2}), (\sqrt{4} - \sqrt{3})$ ....

D.

In the ratio of the reciprocal of the square roots of the integers i.e.,. $\frac{1}{\sqrt{1}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{4}}$

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Explanation

$h = u t + \frac{1}{2} g t^{2} \Rightarrow 1 = 0 \times t_{1} + \frac{1}{2} g t_{1}^{2} \Rightarrow t_{1} = \sqrt{2 / g}$

Velocity after travelling 1m distance

$v^{2} = u^{2} + 2 g h \Rightarrow v^{2} = {(0)}^{2} + 2 g \times 1 \Rightarrow v = \sqrt{2 g}$

For second 1 meter distance

$1 = \sqrt{2 g} \times t_{2} + \frac{1}{2} g t_{2}^{2} \Rightarrow g t_{2}^{2} + 2 \sqrt{2 g} t_{2} - 2 = 0$

$t_{2} = \frac{- 2 \sqrt{2 g} \pm \sqrt{8 g + 8 g}}{2 g} = \frac{- \sqrt{2} \pm 2}{\sqrt{g}}$

Taking +ve sign $t_{2} = (2 - \sqrt{2}) / \sqrt{g}$

∴ $\frac{t_{1}}{t_{2}} = \frac{\sqrt{2 / g}}{(2 - \sqrt{2}) / \sqrt{g}} = \frac{1}{\sqrt{2} - 1}$ and so on.

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A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time (Given $g = 9 .8 m / s^{2})$

A.

At least 0.8 m/s

B.

Any speed less than 19.6 m/s

C.

Only with speed 19.6 m/s

D.

More than 19.6 m/s

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Explanation

Interval of ball throw = 2 sec.

If we want that minimum three (more than two) ball remain in air then time of flight of first ball must be greater than 4 sec.

$T > 4$ sec

$\frac{2 u}{g} > 4 s e c \Rightarrow u > 19 .6 m / s$

for u =19.6. First ball will just strike the ground(in sky)

Second ball will be at highest point (in sky)

Third ball will be at point of projection or at ground (not in sky)

Physics MCQs for NEET — Practice Questions with Answers

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