MHT CET 2025 21st April Evening Shift | Rotational Motion Question 1 | Physics

The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is I. It is rotating with angular velocity $\omega$. Another identical ring is gently placed on it so that their centres coincide. If both the rings are rotating about the same axis then loss in kinetic energy is

$\frac{\mathrm{I} \omega^2}{16}$

$\frac{\mathrm{I} \omega^2}{8}$

$\frac{\mathrm{I} \omega^2}{4}$

$\frac{\mathrm{I} \omega^2}{2}$

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

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A body is rotating about its own axis. Its rotational kinetic energy is ' x ' and its angular momentum is ' $y$ '. Hence its moment of inertia about its own axis is

$\frac{x^2}{2 y}$

$\frac{y^2}{2 x}$

$\frac{x}{2 y}$

$\frac{y}{2 x}$

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

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Moment of inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is ' $I$ '. If the same rod is bent in the form of ring, its moment of inertia about the diameter is ' $\mathrm{I}_1$ '. If $\mathrm{I}_1=\mathrm{xI}$, then the value of ' x ' is

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

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

$\frac{3 \pi^2}{4}$

$\frac{4}{3 \pi^2}$

MHT CET 2025 21st April Morning Shift

MCQ (Single Correct Answer)

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A disc of mass 25 kg and radius 0.2 m is rotating at 240 r.p.m. A retarding torque brings it to rest in 20 second. If the torque is due to a force applied tangentially on the rim of the disc then the magnitude of the force is

$\frac{\pi}{2} \mathrm{~N}$

$2 \pi \mathrm{~N}$

$\pi \mathrm{N}$

$\quad 4 \pi \mathrm{~N}$

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