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The moment of inertia of a loop of radius R and mass M, about any tangent line in its plane will be

$\frac{3 M R^{2}}{2}$

$\frac{M R^{2}}{2}$

$M R^{2}$

$\frac{M R^{2}}{4}$

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Explanation

Use theorem of paralle axes.

$I = \frac{M R^{2}}{2} + M R^{2} = \frac{3}{2} M R^{2}$

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A hollow sphere of diameter 0.2 m and mass 2 kg is rolling on an inclined plane with velocity v = 0.5 m/ s. The kinetic energy of the sphere is

0.1 J

0.3 J

0.5 J

0.42 J

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Explanation

$K_{N} = \frac{1}{2} {mv}^{2} (\frac{K^{2}}{R^{2}} + 1) = \frac{1}{2} \times 2 \times (0 .5)^{2} \times (\frac{2}{3} + 1) = 0 .42 J$

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A man of the mass M stands at one end of a plank of length L which lies at rest on a frictionless surfacé. The man walks to the other end of the plank. If the mass of the plank is $\frac{M}{3}$ , the distance that the man moves relative to the ground is

$\frac{3 L}{4}$

$\frac{L}{4}$

$\frac{4 L}{5}$

$\frac{L}{3}$

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Explanation

Let plank moves x distance x opposite direction. Then, the displacement of man relative to the ground will be, (L- x).
Applying

$\begin{aligned} m_{R} x_{R} & = m_{L} x_{L} \end{aligned} or \begin{aligned} M (L - x) & = \frac{M}{3} x \end{aligned} Solving this equation we get, x = \frac{3 L}{4} ∴ Displacement of man relative to ground = L - \frac{3 L}{4} = \frac{L}{4}$

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Ratio of total kinetic energy and rotational kinetic energy in the motion of a disc is

1:1

2:7

1:2

3:1

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Explanation

$\frac{T o t a l k i n e t i c e n e r g y}{R o t a t i o n a l k i n e t i c e n e r g y} = \frac{\frac{1}{2} m v^{2} (1 + \frac{K^{2}}{R^{2}})}{\frac{1}{2} m v^{2} \frac{K^{2}}{R^{2}}} = \frac{1 + \frac{K^{2}}{R^{2}}}{\frac{K^{2}}{R^{2}}} = \frac{1 + \frac{1}{2}}{\frac{1}{2}} = \frac{3 / 2}{1 / 2} = 3$

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A bomb of mass 9 kg explodes into two pieces of mass 3 kg and 6 kg. The velocity of 3 kg mass is 16 m/s. The velocity of 6 kg mass is

4 m/s

8 m/s

16 m/s

32 m/s

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Explanation

$p_{3} = p_{6} o r 3 \times 16 = 6 \times v o r v = 8 m / s . A s p_{i} = 0$ or both the pieces should have equal and opposite momentum.

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A cylinder is rolling over a surface. Which points on it move rectilinearly?

All points on the curved surface of the cylinder.

All points on the flat surfaces of the cylinder.

All points on the axis of the cylinder.

None of the above

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Explanation

Points on the axis move rectilinearly.

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The magnitude of the impulse developed by a mass of 0.2 kg which changes its velocity from $5 \hat{i} - 3 \hat{j} + 7 \hat{k} m / s to 2 \hat{i} + 3 \hat{j} + \hat{k} m / s$ is

2.7 N-s

1.8 N-s

0.9 N-s

3.6 N-s

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Explanation

$\begin{aligned} Impulse J = Δp & = m (v_{r} - v_{i}) \end{aligned} 0 .2 [(2 \hat{i} + 3 \hat{j} + \hat{k}) - (5 \hat{i} - 3 \hat{j} + 7 \hat{k})] = (- 0 .6 \hat{i} + 1 .2 \hat{j} - 1 .2 \hat{k}) ∴ J = \sqrt{(0 .6)^{2} + (1 .2)^{2} + (1 .2)^{2}} = 1 .8 N - s$

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A round disc of the moment of inertia $l_{2}$ about its axis perpendicular to its plane and passing through its center is placed over another disc of the moment of inertial $l_{1}$ rotating with an angular velocity $ω$ about the same axis. The final angular velocity of the combination of discs is

$\frac{l_{2} ω}{l_{1} + l_{2}}$

$ω$

$\frac{l_{1} ω}{l_{1} + l_{2}}$

$\frac{(l_{1} + l_{2}) ω}{l_{1}}$

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Explanation

According to the conservation of angular momentum

$∵ l_{1} ω_{1} = l_{2} ω_{2} \Rightarrow l_{1} ω = (l_{1} + l_{2}) ω_{2} \Rightarrow ω_{2} = \frac{l_{1} ω}{l_{1} + l_{2}}$

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A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same which one of the following will not be affected

Moment of inertia

Angular momentum

Angular velocity

Rotational kinetic energy

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Explanation

Since no external torque acts on sphere, hence its total angular momentum remains constant.

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A particle of mass 5 g is moving with a uniform speed of $3 \sqrt{2}$ cm/s in the x-y plane along the line y = $2 \sqrt{5}$ cm. The magnitude of its angular momentum about the origin in $g - {cm}^{2} / s$ is

$30 \sqrt{2}$

$30 \sqrt{10}$

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Explanation

The angular momentum of a particle about an axis is given by mvr sin(Î¸), where m is the mass, v is the speed, r is the perpendicular distance from the axis, and Î¸ is the angle between v and r. For the given trajectory, r = 2âˆš5 cm and sin(Î¸) = 2/âˆš10. Substituting the values, we get L = 30âˆš10 g-cm^2/s.

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