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NEET NEET Questions — Free MCQ Practice

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A body is thrown vertically upwards with velocity u. The distance travelled by it in the fifth and the sixth seconds are equal. The velocity u is given by (g = 9.8 m/s²)

A.

24.5 m/s

B.

49.0 m/s

C.

73.5 m/s

D.

98.0 m/s

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Explanation

The given condition is possible only when body is at its highest position after 5 seconds

It means time of ascent = 5 sec

and time of flight $T = \frac{2 u}{g} = 10$ ⇒ $u = 50 m / s$

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A body, thrown upwards with some velocity reaches the maximum height of 50 m. Another body with double the mass thrown up with double the initial velocity will reach a maximum height of

A.

100 m

B.

200 m

C.

300 m

D.

400 m

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Explanation

$H_{\max} \propto u^{2}$ , It body projected with double velocity then maximum height will become four times i.e. 200 m.

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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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