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Physics NEET Questions — Free MCQ Practice

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The relation between velocity (v) and time (t) is $vα \sqrt{t}$ , then which one of the following quantity is constant:

A.

Force

B.

Power

C.

Momentum

D.

Kinetic Energy

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Explanation

2.

$v = k \sqrt{t}$ {where k is proportionality constant}

$a = \frac{dv}{dt} = \frac{k}{2 \sqrt{t}}$

velocity, acceleration be the time-dependent, so K.E. force and momentum is also time-varying

$Power = force \times velocity = \frac{mk}{2 \sqrt{t}} k \sqrt{t} = \frac{{mk}^{2}}{2}$

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A steel wire can withstand a load up to 2940 N. A load of 150 kg is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position, so that the wire does not break when the load passes through the position of equilibrium, is (2008 E)

A.

30 $°$

B.

60 $°$

C.

80 $°$

D.

85 $°$

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A body is thrown vertically up with certain initial velocity, the potential and kinetic energies of the body are equal at a point P in its path. If the same body is thrown with double the velocity upwards, the ratio of potential and kinetic energies of the body when it crosses the same point, is

A.

1:1

B.

1:4

C.

1:7

D.

1:8

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Explanation

$W h e n t h e b o d y i s t h r o w n w i t h v e l o c i t y = v; T E = K E_{i n i t i a l} = \frac{1}{2} m {(v)}^{2} = K A t t h e p o i n t, w h e n K E a n d P E a r e e q u a l : K E + P E = T E \Rightarrow K E = P E = \frac{K}{2} W h e n t h e b o d y i s t h r o w n w i t h v e l o c i t y = 2 v; T E' = K E'_{i n i t i a l} = \frac{1}{2} m {(2 v)}^{2} = 4 \times \frac{1}{2} m {(v)}^{2} = 4 K P E a t t h e s a m e p o i n t w i l l b e s a m e = \frac{K}{2} K E a t t h a t p o i n t = 4 K - \frac{K}{2} = \frac{7 K}{2} R a t i o o f P E t o K E = 1 : 7$

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A body is displaced from (0,0) to (1m,1m) along the path x=y by a force $F = (x^{2} \hat{j} + y \hat{i}) N$ . The work done by this force will be :

A.

$\frac{4}{3} J$

B.

$\frac{5}{6} J$

C.

$\frac{3}{2} J$

D.

$\frac{7}{5} J$

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Explanation

$dW = \vec{F} . \vec{dr} = (y \hat{i} + x^{2} \hat{j}) . (dx \hat{i} + dy \hat{j}) = ydx + x^{2} dy From y = x, putting y in terms of x and x in terms of y in above relations dW = xdx + y^{2} dy Integrating both sides - W = \int_{0}^{1} xdx + \int_{0}^{1} y^{2} dy = {[\frac{x^{2}}{2}]}_{0}^{1} + {[\frac{y^{3}}{3}]}_{0}^{1} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6} J$

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A force F is applied on a body which moves with a velocity v in the direction of the force, then the power will be

A.

${Fv}^{2}$

B.

Fv

C.

$F / v^{2}$

D.

F/v

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Explanation

$(2) P = Fv = \frac{W}{t} = F (\frac{s}{t})$

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Three different objects of masses $m_{1}, m_{2}$ and m₃ are allowed to fall from rest and from the same point ‘O’ along three different frictionless paths. The speeds of the three objects, on reaching the ground, will be in the ratio of

A.

$m_{1} : m_{2} : m_{3}$

B.

$m_{1} : 2 m_{2} : 3 m_{3}$

C.

1 : 1 : 1

D.

$\frac{1}{m_{1}} : \frac{1}{m_{2}} : \frac{1}{m_{3}}$

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Explanation

Speed of the object at reaching the ground $v = \sqrt{2 g h}$

If heights are equal then velocity will also be equal.

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When a body moves with a constant speed along a circle

A.

No work is done on it

B.

No acceleration is produced in the body

C.

No force acts on the body

D.

Its velocity remains constant

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Explanation

When speed is constant in circular motion, it means work done by centripetal force is zero.

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A sphere of mass m is tied to end of a string of length l and rotated through the other end along a horizontal circular path with speed v. The work done by centripetal force in full horizontal circle is

A.

0

B.

$(\frac{m v^{2}}{l}) . 2 π l$

C.

$mg . 2πl$

D.

$(\frac{m v^{2}}{l}) . (l)$

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Explanation

Work done by centripetal force in uniform circular motion is always equal to zero.

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A ball is suspended by a thread of length l. What minimum horizontal velocity has to be imparted to the ball for it to reach the height of the suspension:

A.

gl

B.

2 gl

C.

$\sqrt{g l}$

D.

$\sqrt{2 g l}$

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Explanation

To reach the height of suspension l, the particle must have a vertical component of velocity âˆš(2gl) at the highest point. This vertical velocity can be obtained by imparting a horizontal velocity âˆš(2gl) at the lowest point, as the total velocity is âˆš(2) times the horizontal component.

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A body of mass m hangs at one end of a string of length l, the other end of which is fixed. It is given a horizontal velocity so that the string would just reach where it makes an angle of 60° with the vertical. The tension in the string at mean position is

A.

2 mg

B.

mg

C.

3 mg

D.

$\sqrt{3} m g$

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Physics MCQs for NEET — Practice Questions with Answers

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