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Physics NEET Questions — Free MCQ Practice

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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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If a gymnast, sitting on a rotating stool, with his arms outstretched, suddenly lowers his hands

A.

the angular velocity increases

B.

his moment of inertia increases

C.

the angular velocity stays constant

D.

the angular momentum increases

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Explanation

When mass is placed nearer the axis of rotation, M.I. decreases, angular velocity increases.

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Moment of inertia of a uniform circular disc about a diameter is I. Its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim will be

A.

5 I

B.

3 I

C.

6 I

D.

4 I

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Explanation

M.I. of a disc about any diameter = $\frac{{MR}^{2}}{4} = I$

M.I. of the disc about an axis passing through a point on its rim and perpendicular to its plane

$= \frac{3}{2} {MR}^{2} = \frac{3}{2} \times 4 I = 6 I$

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A disk and a sphere of the same radius but different masses roll off on two inclined planes Of the same altitude and length Which one of the two objects gets to the bottom of the plane first

A.

Disk

B.

Sphere

C.

Both reach at the same time

D.

Depends on their masses

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Explanation

(b)

$\begin{aligned} a = \frac{gsin θ}{1 + \frac{K^{2}}{R^{2}}} \\ for disc; \frac{K^{2}}{R^{2}} = \frac{1}{2} = 0 .5 \\ for sphere; \frac{K^{2}}{R^{2}} = \frac{2}{5} = 0 .4 \end{aligned}$

a(sphere)>a(disc)

$∴$ sphere reaches first

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A ball hits the floor and rebounds after an inelastic collision. In this case

A.

the momentum of the ball just after the collision is the same as that just before the collision

B.

the mechanical energy of the ball remains the same in the collision

C.

the total momentum of the ball and the earth is conserved

D.

the total energy of the ball and the earth is conserved

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Explanation

Net force on ball and earth system is zero.

Physics MCQs for NEET — Practice Questions with Answers

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