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The moment of inertia of a ring about an axis passing through the centre and perpendicular to its plane is ‘I’. It is rotating with angular velocity $^{\prime} \omega^{\prime} .$ 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 is kinetic energy isA) $\frac{\mathrm{I} \omega^{2}}{2}$B) $\frac{\mathrm{I} \omega^{2}}{4}$C) $\frac{\mathrm{I} \omega^{2}}{6}$D) $\frac{\mathrm{I} \omega^{2}}{8}$
The moment of inertia of a ring about an axis passing through the centre and perpendicular to its plane is ‘I’. It is rotating with angular velocity $^{\prime} \omega^{\prime} .$ 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 is kinetic energy is
A) $\frac{\mathrm{I} \omega^{2}}{2}$
B) $\frac{\mathrm{I} \omega^{2}}{4}$
C) $\frac{\mathrm{I} \omega^{2}}{6}$
D) $\frac{\mathrm{I} \omega^{2}}{8}$
Physics
The moment of inertia of a ring about an axis passing through the centre and perpendicular to its plane is ‘I’. It is rotating with angular velocity $^{\prime} \omega^{\prime} .$ 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 is kinetic energy is
A) $\frac{\mathrm{I} \omega^{2}}{2}$
B) $\frac{\mathrm{I} \omega^{2}}{4}$
C) $\frac{\mathrm{I} \omega^{2}}{6}$
D) $\frac{\mathrm{I} \omega^{2}}{8}$

Correct option is B $\frac{\mathrm{I} \omega^{2}}{4}$
Initial angular momentum, $\mathrm{L}_{1}=\mathrm{I} \omega$
$\mathrm{K}_{1}=\frac{1}{2} \mathrm{I} \omega^{2}$
When, second ring is put on it and they start rotating with same angular speed, say $\omega_{0}$.
Now, final angular momentum, $L_{2}=2 \mathrm{I} \omega_{0}$
Conservation of angular momentum:-
$\mathrm{L}_{1}=\mathrm{L}_{2}$
$\Longrightarrow \mathrm{I} \omega=2 \mathrm{I} \omega_{0}$
$\Longrightarrow \omega_{0}=\frac{\omega}{2}$
$\mathrm{K}_{2}=2 \times \frac{1}{2} \mathrm{I} \omega_{0}^{2}=\mathrm{I} \omega_{0}^{2}$
$\begin{array}{l}
\mathrm{K}_{2}=\frac{1}{4} \mathrm{I} \omega^{2} \\
\Longrightarrow \mathrm{K}_{1}-\mathrm{K}_{2}=\frac{1}{4} \mathrm{I} \omega^{2}
\end{array}$

Hence, answer is option-(B).

i

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