A metal sphere of radius r and specific heat | vedclass

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(A) The rotational kinetic energy $(KE)$ of a solid sphere is given by $KE = \frac{1}{2} I \omega^2$. For a solid sphere,the moment of inertia $I = \f

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$\frac{2 \pi^2 n^2 r^2}{5 S}$

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(A) The rotational kinetic energy $(KE)$ of a solid sphere is given by $KE = \frac{1}{2} I \omega^2$. For a solid sphere,the moment of inertia $I = \frac{2}{5} m r^2$. The angular velocity $\omega = 2 \pi n$. Substituting these values: $KE = \frac{1}{2} 	imes (\frac{2}{5} m r^2) 	imes (2 \pi n)^2 = \frac{1}{5} m r^2 	imes 4 \pi^2 n^2 = \frac{4}{5} m r^2 \pi^2 n^2$. Given that $50 \%$ of this energy is used to increase the temperature,the heat produced $\Delta Q$ is: $\Delta Q = \frac{1}{2} 	imes KE = \frac{1}{2} 	imes (\frac{4}{5} m r^2 \pi^2 n^2) = \frac{2}{5} m r^2 \pi^2 n^2$. Using the relation $\Delta Q = m S \Delta t$,where $S$ is the specific heat and $\Delta t$ is the rise in temperature: $m S \Delta t = \frac{2}{5} m r^2 \pi^2 n^2$. Solving for $\Delta t$: $\Delta t = \frac{2 \pi^2 n^2 r^2}{5 S}$.

$A$ metal sphere of radius $r$ and specific heat $S$ is rotated about an axis passing through its centre at a speed of $n$ rotations per second. It is suddenly stopped and $50 \%$ of its energy is used in increasing its temperature. Then,the rise in temperature of the sphere is

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