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(C) The radius of the sphere at temperature $T$ is given by $R_T = R_0(1 + \alpha \Delta T)$.For $\Delta T = 100^{\circ}C$,$R_{100} = R_0(1 + 100\alph

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$0.996 \omega_0$

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(C) The radius of the sphere at temperature $T$ is given by $R_T = R_0(1 + \alpha \Delta T)$.For $\Delta T = 100^{\circ}C$,$R_{100} = R_0(1 + 100\alpha)$.The moment of inertia of a solid sphere is $I = \frac{2}{5}MR^2$.Since no external torque acts on the sphere,angular momentum is conserved: $I_0 \omega_0 = I_{100} \omega_{100}$.Substituting the values: $\frac{2}{5}MR_0^2 \omega_0 = \frac{2}{5}MR_{100}^2 \omega_{100}$.$R_0^2 \omega_0 = [R_0(1 + 100\alpha)]^2 \omega_{100}$.$R_0^2 \omega_0 = R_0^2(1 + 200\alpha) \omega_{100}$ (using binomial approximation $(1+x)^n \approx 1+nx$ for small $x$).$\omega_{100} = \frac{\omega_0}{1 + 200\alpha} = \frac{\omega_0}{1 + 200(2.0 	imes 10^{-5})} = \frac{\omega_0}{1 + 0.004} = \frac{\omega_0}{1.004}$.$\omega_{100} \approx 0.996 \omega_0$.

$A$ sphere rotates with an angular velocity $\omega_0$ at a temperature of $0^{\circ}C$. What will be its new angular velocity when the temperature is increased to $100^{\circ}C$? (Given: $\alpha = 2.0 \times 10^{-5} \, ^{\circ}C^{-1}$)

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