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Rotational Motion MCQs for NEET — Physics Questions with Answers

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The position of a particle is given by: $\overset{⇀}{r} = (\hat{1} + 2 \hat{j} - \hat{k})$ and momentum $\vec{P} = (3 \hat{i} + 4 \hat{j} - 2 \hat{k})$ . The angular momentum is perpendicular to

X-axis

Y-axis

Z-axis

Line at equal angles to all the three axes

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Explanation

$(a) \vec{L} = \vec{r} \times \vec{p} = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & - 1 \\ 3 & 4 & - 2 \end{array}| = - \hat{j} - 2 \hat{k}$

and the X-axis is given by $\hat{i} + 0 \hat{j} + 0 \hat{k}$

The dot product of these two vectors is zero i.e. angular momentum is perpendicular to X-axis.

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A body falling vertically downward under gravity breaks into two parts of unequal messes. The center of mass of the parts taken together shifts horizontally towards

heavier piece

lighter piece

does not shift horizontally

depends on the vertical velocity at the time of breaking

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Explanation

In the absence of air resistance, the only external force acting on the body is the force of gravity. As for the breaking of the body Into two unequal parts occurs, It is purely due to Internal forces that are exerted mutually by parts of the system and cancel. Hence the center of mass continues to move In the same direction and there is no cause of shift of the centre of mass of the falling body (parts of the body after breaking) Hence choice 3 is correct.

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A solid sphere, disc, and solid cylinder all Of the same mass and made up Of same material are allowed to roll down (from rest) on an inclined plane, then

Solid sphere reaches the bottom first

Solid sphere reaches the bottom late

The disc will reach the bottom first

All of them reach the bottom at the same time

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Explanation

$Because its M . I . (or value of \frac{K^{2}}{R^{2}}) is minimum for sphere$

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If the angular momentum of a rotating body about a fixed axis is increased by 10%, its kinetic energy will be increased by

10%

20%

21%

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Explanation

$K E = \frac{1}{2} I ω^{2}; L = I ω K E = \frac{I^{2} ω^{2}}{2 I} = \frac{L^{2}}{2 I} (K E) = \frac{L^{2}}{2 I}; (K E)' = \frac{{(\frac{110 L}{100})}^{2}}{2 I} \frac{K E' - K E}{K E} = \frac{{(\frac{110}{100})}^{2} \frac{L^{2}}{2 I} - \frac{L^{2}}{2 I}}{\frac{L^{2}}{2 I}} = \frac{121}{100} - 1 = \frac{21}{100} \frac{21}{100} \times 100 = 21 % (C h o i c e 3)$

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A cylinder of 500 g and radius 10 cm has moment of inertia (about its natural axis)

$2.5 \times 10^{- 3} k g - m^{2}$

$2 \times 10^{- 3} k g - m^{2}$

$5 \times 10^{- 3} k g - m^{2}$

$3.5 \times 10^{- 3} k g - m^{2}$

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Explanation

$I = \frac{1}{2} M R^{2} = \frac{1}{2} \times 0.5 \times {(0.1)}^{2} = 2.5 \times 10^{- 3} k g - m^{3}$

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Two rings of the same radius and mass are placed such that their centers are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the center and perpendicular to the plane of one of the rings is (mass of the ring and radius = r)

$\frac{1}{2} {mr}^{2}$

${mr}^{2}$

$\frac{3}{2} {mr}^{2}$

$2 {mr}^{2}$

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Explanation

${mr}^{2} + \frac{{mr}^{2}}{2} = \frac{3}{2} {mr}^{2}$

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The angular velocity of a body changes from $\frac{1}{25}$ rev./sec to 1 rev./sec. without applying any torque. The ratio of its radius of gyration in two situations is :

5: 1

1: 5

1: 25

25: 1

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Explanation

$I_{1} ω_{1} = I_{2} ω_{2}, o r M K_{1}^{2} (2 π n_{1}) = M K_{2}^{2} (2 π n_{2}) \Rightarrow \frac{K_{1}^{2}}{K_{2}^{2}} = \frac{n_{2}}{n_{1}} \Rightarrow \frac{K_{1}}{K_{2}} = \sqrt{\frac{n_{2}}{n_{1}}} = \frac{\sqrt{1}}{\sqrt{\frac{1}{25}}} = \frac{5}{1}$

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A solid cylinder and a solid sphere both having the same mass and radius, are released from a rough inclined plane. Both roll without slipping. Then,

the force of friction that acts on the two is the same

the force of friction is greater in case of a sphere than for a cylinder

the force of friction is greater in case of a cylinder than for a sphere

the force of friction will depend on the nature of the surface of the body that is moving and that of the inclined surface and is independent of the shape and size of the moving body

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Explanation

$In case of pure rolling . Force of friction f = \frac{mg \sin θ}{1 + (\frac{{mR}^{2}}{I})}$

Since a moment of inertia of disc is more, the force of friction on cylinder will be more.

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A rigid body rotates about a fixed axis with variable angular velocity equal to $α - β t$ , at the time t, where $α, β$ are constants. The angle through which it rotates before its stops

$\frac{α^{2}}{2 β}$

$\frac{α^{2} - β^{2}}{2 α}$

$\frac{α^{2} - β^{2}}{2 β}$

$\frac{(α - β) α}{2}$

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Explanation

$ω = α - βt . Comparing with ω = ω_{0} - αt initial angular velocity = α Angular retardation = β ∴ Angle rotated before it stops is \frac{α^{2}}{2 β} from 0 α^{2} - 2 βθ$

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An automobile moves on a road with a speed of 54 kmh $^{- 1}$ . The radius of its wheel is 0.45 m and moment of inertia of the wheel about its axis of rotation is 3 kg m $^{2}$ . If the vehicle is brought to rest in 15s, the magnitude of average torque transmitted by its brakes to the wheel is

8.58 $m^{2} s^{- 2}$

10.86 $m^{2} s^{- 2}$

2.86 $m^{2} s^{- 2}$

6.66 $m^{2} s^{- 2}$

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Explanation

$ω_{i} = \frac{15}{0.45} = \frac{100}{3} ω_{i} = 0 ω_{i} = ω_{i} + αt 0 = \frac{100}{3} + (- α) (15) α = \frac{100}{45} τ = (I) (α) = 3 x \frac{100}{45} = 6.66 N . m .$

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