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Physics MCQs for NEET — Practice Questions with Answers

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A ship A is moving westwards with a speed of 10 km $h^{- 1}$ and a ship B, 100 km south of A is moving northwards with a speed of 10 km $h^{- 1}$ . The time after which the distance between them becomes the shortest, is:

5 hr

$5 \sqrt{2}$ hr

$10 \sqrt{2}$ hr

0 hr

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A particle projected with kinetic energy $k_{0}$ with an angle of projection $θ$ . Then the variation of kinetic K with vertical displacement y is

linear

parabolic

hyperbolic

periodic

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A body is thrown horizontally with a velocity $\sqrt{2 g h}$ from the top of a tower of height h. It strikes the level ground through the foot of the tower at a distance x from the tower. The value of x is:

$\frac{h}{2}$

$\frac{2 h}{3}$

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Explanation

$x = u \sqrt{\frac{2 h}{g}} = \sqrt{2 g h} \sqrt{\frac{2 h}{g}} x = 2 h$

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A particle starts from the origin at t=0 and moves in the x-y plane with constant acceleration 'a' in the y direction. Its equation of motion is $y = b x^{2}$ . The x component of its velocity (at t=0) is:

variable

$\sqrt{\frac{2 a}{b}}$

$\frac{a}{2 b}$

$\sqrt{\frac{a}{2 b}}$

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Explanation

$y = b x^{2} \frac{d y}{d t} = 2 b x \frac{d x}{d t} \Rightarrow \frac{d^{2} y}{d t^{2}} = 2 b [{(\frac{d x}{d t})}^{2}] \sqrt{\frac{a}{2 b}} = v_{x}$

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A body projected with velocity u with an angle of projection $θ$ . Change in velocity after the time (t) from the projection is:

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

u sin $θ$

u cos $θ$

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Explanation

$\vec{u} = u \cos θ \hat{i} + u \sin θ \hat{j} A f t e r t i m e t : \vec{v} = u \cos θ \hat{i} + (u \sin θ - g t) \hat{j} C h a n g e i n v e l o c i t y, ∆ \vec{v} = \vec{v} - \vec{u} = u \cos θ \hat{i} + (u \sin θ - g t) \hat{j} - (u \cos θ \hat{i} + u \sin θ \hat{j}) \Rightarrow ∆ \vec{v} = (- g t) \hat{j} \Rightarrow |∆ \vec{v}| = (g t)$

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A particle has initial velocity $(3 \hat{i} + 4 \hat{j})$ and has acceleration $(0.4 \hat{i} + 0.3 \hat{j})$ . Its speed after 10 s

7 units

$7 \sqrt{2}$ units

8.5 units

10 units

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Explanation

$\vec{v} = \vec{u} + \vec{a} t = (3 \hat{i} + 4 \hat{j}) + (0.4 \hat{i} + 0.3 \hat{j}) \times 10 = 7 \hat{i} + 7 \hat{j} | \vec{v} | = \sqrt{7^{2} + 7^{2}} = 7 \sqrt{2} m / s$

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The gravity in space is given by $\vec{g} = - 10 \hat{j} m s^{- 2}$ . Two particles are simultaneously projected with velocity $\vec{u_{1}} = (10 m s^{- 1}) \hat{i} + (10 m s^{- 1}) \hat{j}$ and $\vec{u_{2}} = (20 m s^{- 1}) \hat{i} + (10 m s^{- 1}) \hat{j}$ . Then, the ratio of their times of flight

1:1

1:2

2:1

none

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Explanation

$T i m e o f f l i g h t = \frac{2 u_{y}}{g} \frac{T_{1}}{T_{2}} = \frac{u_{y, 1}}{u_{y, 2}} = \frac{10}{10} = 1$

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The x and y coordinates of a particle at any time t are given by $x = 7 t + 4 t^{2}$ and y = 5t, where x and y are in metre and t in seconds. The acceleration of particle at t = 5s is

Zero

8 m/s²

20 m/s²

40 m/s²

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Explanation

$a = \sqrt{a_{x}^{2} + a_{y}^{2}} = {[{(\frac{d^{2} x}{d t^{2}})}^{2} + {(\frac{d^{2} y}{d t^{2}})}^{2}]}^{\frac{1}{2}}$

Here $\frac{d^{2} y}{d t^{2}} = 0$ . Hence $a = \frac{d^{2} x}{d t^{2}} = 8 m / s^{2}$

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A boat is sent across a river in perpendicular direction with a velocity of 8 km/hr. If the resultant velocity of boat is 10 km/hr, then velocity of the river is :

10 km/hr

8 km/hr

6 km/hr

4 km/hr

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A boat moves with a speed of 5 km/h relative to water in a river flowing with a speed of 3 km/h and having a width of 1 km. The minimum time taken around a round trip is

5 min

60 min

20 min

30 min

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Explanation

For the round trip he should cross perpendicular to the river

∴ Time for trip to that side $= \frac{1 k m}{4 k m / h r} = 0 .25 h r$

To come back, again he take 0.25 hr to cross the river.

Total time is 30 min, he goes to the other bank and come back at the same bank.

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