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Physics NEET Questions — Free MCQ Practice

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The kinetic energy k of a particle moving along a circle of radius R depends on the distance covered s as k = as² where a is a constant. The force acting on the particle is

A.

$2 a \frac{s^{2}}{R}$

B.

$2 a s {(1 + \frac{s^{2}}{R^{2}})}^{1 / 2}$

C.

2 as

D.

$2 a \frac{R^{2}}{s}$

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Explanation

According to the given problem, $\frac{1}{2} m v^{2} = a s^{2} \Rightarrow v = s \sqrt{\frac{2 a}{m}}$

So, $a_{R} = \frac{v^{2}}{R} = \frac{2 a s^{2}}{m R}$

Furthermore, as $a_{t} = \frac{d v}{d t} = \frac{d v}{d s} \cdot \frac{d s}{d t} = v \frac{d v}{d s}$

$a_{t} = [s \sqrt{\frac{2 a}{m}}] [\sqrt{\frac{2 a}{m}}] = \frac{2 a s}{m}$

So, $a = \sqrt{a_{R}^{2} + a_{t}^{2}} = \sqrt{{[\frac{2 a s^{2}}{m R}]}^{2} + {[\frac{2 a s}{m}]}^{2}}$

Hence $a = \frac{2 a s}{m} \sqrt{1 + {[s / R]}^{2}}$

∴ $F = m a = 2 a s \sqrt{1 + {[s / R]}^{2}}$

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A stone of mass 1 kg tied to a light inextensible string of length $L = \frac{10}{3} m$ is whirling in a circular path of radius L in a vertical plane. If the ratio of the maximum tension in the string to the minimum tension in the string is 4 and if g is taken to be 10 m/sec², the speed of the stone at the highest point of the circle is

A.

20 m/sec

B.

$10 \sqrt{3} m / \sec$

C.

$5 \sqrt{2} m / \sec$

D.

10 m/sec

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Explanation

Since the maximum tension T_B in the string moving in the vertical circle is at the bottom and minimum tension T_T is at the top.

∴ $T_{B} = \frac{m {v_{B}}^{2}}{L} + m g$ and $T_{T} = \frac{m v_{T}^{2}}{L} - m g$

∴ $\frac{T_{B}}{T_{T}} = \frac{\frac{m v_{B}^{2}}{L} + m g}{\frac{m v_{T}^{2}}{L} - m g} = \frac{4}{1}$ or $\frac{v_{B}^{2} + g L}{v_{T}^{2} - g L} = \frac{4}{1}$

or ${v_{B}}^{2} + g L = 4 v_{T}^{2} - 4 g L$ (1)

$∵ By energy conservation, \frac{1}{2} {mv}^{2}_{B} + 0 = mg (2 L) + \frac{1}{2} {mv}^{2}_{T} \Rightarrow v_{B}^{2} = v_{T}^{2} + 4 g L$

Put it in equation (1):

∴ $v_{T}^{2} + 4 g L + g L = 4 v_{T}^{2} - 4 g L$ ⇒ $3 v_{T}^{2} = 9 g L$

∴ $v_{T}^{2} = 3 \times g \times L = 3 \times 10 \times \frac{10}{3}$ or v_T = 10 m/sec

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A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is:

A.

$\sqrt{u^{2} - 2 g L}$

B.

$\sqrt{2 g L}$

C.

$\sqrt{u^{2} - g l}$

D.

$\sqrt{2 (u^{2} - g L)}$

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Explanation

Using conservation of energy :

$\frac{1}{2} m u^{2} - \frac{1}{2} m v^{2} = m g L$

⇒ $v = \sqrt{u^{2} - 2 g L}$

$| \vec{v} - \vec{u} | = \sqrt{u^{2} + v^{2}} = \sqrt{u^{2} + u^{2} - 2 g L} = \sqrt{2 (u^{2} - g L)}$

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The driver of a car travelling at velocity v suddenly see a broad wall in front of him at a distance d. He should

A.

Brake sharply

B.

Turn sharply

C.

(1) and (2) both

D.

None of the above

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Explanation

When driver applies brakes and the car covers distance x before coming to rest, under the effect of retarding force F

then $\frac{1}{2} m v^{2} = F x$ ⇒ $x = \frac{m v^{2}}{2 F}$

But when he takes turn then $\frac{m v^{2}}{r} = F$ ⇒ $r = \frac{m v^{2}}{F}$

It is clear that x = r/2

i.e. by the same retarding force the car can be stopped in a less distance if the driver apply breaks. This retarding force is actually a friction force.

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Work done by a frictional force is

A.

Negative

B.

Positive

C.

Zero

D.

All of the above

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Explanation

Work done by friction can be positive, negative and zero depending upon the situation.

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A block of mass 50 kg slides over a horizontal distance of 1 m. If the coefficient of friction between their surfaces is 0.2, then work done against friction is

A.

98 J

B.

72J

C.

56 J

D.

34 J

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Explanation

$W = μ m g S = 0 .2 \times 50 \times 9 .8 \times 1 = 98 J$

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A body of mass m is moving in a circle of radius r with a constant speed v. The force on the body is $\frac{m v^{2}}{r}$ and is directed towards the centre. What is the work done by this force in moving the body over half the circumference of the circle

A.

$\frac{m v^{2}}{π r^{2}}$

B.

Zero

C.

$\frac{m v^{2}}{r^{2}}$

D.

$\frac{π r^{2}}{m v^{2}}$

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Explanation

Work done by centripetal force is always zero, because force and instantaneous displacement are always perpendicular.

$W = \vec{F} . \vec{s} = F s \cos θ = F s \cos (90 °) = 0$

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A man pushes a wall and fails to displace it. He does

A.

Negative work

B.

Positive but not maximum work

C.

No work at all

D.

Maximum work

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Explanation

No displacement is there.

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A body moves a distance of 10 m along a straight line under the action of a force of 5 N. If the work done is 25 joules, the angle which the force makes with the direction of motion of the body is

A.

0°

B.

30°

C.

60°

D.

90°

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Explanation

$W = F s \cos θ$

⇒ $\cos θ = \frac{W}{F s} = \frac{25}{50} = \frac{1}{2}$

⇒ $θ = 60 °$

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A force acts on a 30 gm particle in such a way that the position of the particle as a function of time is given by $x = 3 t - 4 t^{2} + t^{3}$ , where x is in metres and t is in seconds. The work done during the first 4 seconds is

A.

5.28 J

B.

450 mJ

C.

490 mJ

D.

530 mJ

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Explanation

$v = \frac{d x}{d t} = 3 - 8 t + 3 t^{2}$

∴ $v_{0} = 3 m / s$ and $v_{4} = 19 m / s$

$W = \frac{1}{2} m (v_{4}^{2} - v_{0}^{2})$ (According to work energy theorem)

$= \frac{1}{2} \times 0 .03 \times (19^{2} - 3^{2}) = 5 .28 J$

Physics MCQs for NEET — Practice Questions with Answers

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