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NEET NEET Questions — Free MCQ Practice

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Consider a system of two particles having masses $m_{1} and m_{2}$ . If the particle of mass $m_{1}$ is pushed towards the center of mass of particles through a distance, by what distance would be the particle of the mass $m_{2}$ move so as to keep the center of mass of particles at the original position?

A.

$\frac{m_{1}}{m_{1} + m_{2}} d$

B.

$\frac{m_{1}}{m_{2}} d$

C.

d

D.

$\frac{m_{2}}{m_{1}} d$

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A wheel whose moment of inertia is 12 ${Kgm}^{2}$ has an initial angular velocity of 40 rad/sec. A constant torque of 20 Nm acts on the wheel. The time in which the wheel is accelerated to 100 rad/sec is

A.

72 seconds

B.

16 seconds

C.

8 seconds

D.

36 seconds

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Explanation

$\begin{aligned} τ = lα = \frac{l (ω - ω_{0})}{t}; 20 = \frac{12 (100 - 40)}{t} \\ t = 36 sec (Choice D) \end{aligned}$

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A flywheel is in the form of a uniform circular disc of radius 1 m and mass 2 kg. The work which must be done on it to increase its frequency of rotation from 5 rev $s^{- 1}$ to 10 rev $s^{- 1}$ is approximately

A.

1.5 x $10^{2}$ J

B.

3.0 x $10^{2}$ J

C.

1.5 x $10^{3}$ J

D.

3.0 x $10^{3}$ J

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Explanation

3.

$W = change in rotational kinetic energy = \frac{1}{2} I (ω_{f}^{2} 9 - ω_{i}^{2}) \begin{aligned} = \frac{1}{2} (\frac{1}{2} {mR}^{2}) [{(2 {πf}_{f})}^{2} - {(2 {πf}_{i})}^{2}] \\ = \frac{1}{4} \times 2 \times (1)^{2} \times 4 \times π^{2} (100 - 25) \\ = 1 .48 \times 10^{3} J \approx 1 .5 \times 10^{3} J \end{aligned}$

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A particle of mass m has momentum p. it's kinetic energy will be

A.

mp

B.

$p^{2} m$

C.

$\frac{p^{2}}{m}$

D.

$\frac{p^{2}}{2 m}$

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Explanation

$K = \frac{p^{2}}{2 m} and p = \sqrt{2}$ km are two standard results.

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A constant torque acting on a uniform circular wheel changes its angular momentum from $A_{0}$ to $4 A_{0}$ in 4s. The magnitude of this torque is

A.

$\frac{3 A_{0}}{4}$

B.

$4 A_{0}$

C.

$A_{0}$

D.

$12 A_{0}$

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A solid cylinder rolls down an inclined plane that has friction sufficient to prevent sliding. The ratio of rotational energy to total kinetic energy is

A.

$\frac{1}{2}$

B.

$\frac{1}{3}$

C.

$\frac{2}{3}$

D.

$\frac{3}{4}$

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Explanation

$Fraction, \frac{rotational KE}{rotational KE + translational KE} \begin{array}{l} = \frac{\frac{1}{2} {Iω}^{2}}{\frac{1}{2} {Iω}^{2} + \frac{1}{2} {mv}^{2}} = \frac{\frac{1}{2} \times \frac{{mr}^{2}}{2} ω^{2}}{\frac{1}{2} {Iω}^{2} + \frac{1}{2} {mv}^{2}} \\ = \frac{\frac{1}{2} \times \frac{{mr}^{2}}{2} ω^{2}}{\frac{1}{2} \times \frac{1}{2} {mr}^{2} ω^{2} + \frac{1}{2} {mv}^{2}} \\ = \frac{\frac{1}{4} {mv}^{2}}{\frac{1}{4} {mv}^{2} + \frac{1}{2} {mv}^{2}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} (Cholce B) \end{array}$

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A swimmer while jumping into water from a height easily forms a loop in the air, if

A.

He pulls his arms and legs in

B.

He spreads his arms and legs

C.

He keeps himself straight

D.

None of the above

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Explanation

When a swimmer pulls in their arms and legs while jumping into water, their moment of inertia decreases, causing an increase in their angular speed (due to conservation of angular momentum). This increased angular speed allows them to perform a loop in the air more easily.

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Assertion : A ladder is more apt to slip when you are high up on it than when you just begin to climb

Reason : At the high point on a ladder, the torque is large and on climbing up the torque is small

A. If both the assertion and the reason are true and the reason is a correct explanation of the assertion B. If both the assertion and reason are true but the reason is not a correct explanation of the assertion C. If the assertion is true but the reason is false D. If both the assertion and reason are false

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Explanation

The assertion and reason are true, and the reason correctly explains the assertion. When climbing higher on a ladder, the perpendicular distance from the center of mass to the base increases, resulting in a larger torque acting on the ladder, making it more prone to slipping.

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An inclined plane makes an angle of $30 °$ with the horizontal. A solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration equal to

A.

$\frac{g}{3}$

B.

$\frac{5 g}{7}$

C.

$\frac{2 g}{3}$

D.

$\frac{5 g}{14}$

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Explanation

$\begin{aligned} a & = \frac{g \sin θ}{1 + \frac{k^{2}}{r^{2}}} = \frac{g \sin θ}{1 + \frac{2}{5}} = \frac{g \sin 30^{\circ}}{- \frac{7}{5}} \\ = \frac{5}{7} g \times \frac{1}{2} = \frac{5}{14} g \end{aligned}$

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Which of the following statements is correct?

A.

Kinetic energy and momentum both are conserved in all types of collisions

B.

Total kinetic energy is not conserved but momentum is conserved in inelastic COI-
lisions

C.

Momentum is conserved In elastic collisions but .not in inelastic collisions

D.

Total kinetic energy is conserved in inelastic collisions but momentum is not conserved in elastic collisions

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Explanation

In an inelastic collision, the only momentum is conserved. K.E. is not conserved

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