A solid sphere of radius $R$ and a hollow sphere of inner radius $r$ and outer radius $R$ made of copper are heated to the same temperature and are allowed to cool in the same environment. Then, choose the $CORRECT$ statement
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Rate of loss of heat $=\frac{\mathrm{d} \mathrm{Q}}{\mathrm{dt}}=\sigma e \mathrm{A}\left(\theta^{4}-\theta_{\mathrm{s}}^{4}\right)$ is equal
$\left(\frac{\mathrm{d} \theta}{\mathrm{dt}}\right)=\frac{\sigma \mathrm{e} \mathrm{A}}{\mathrm{ms}}\left(\theta^{4}-\theta_{\mathrm{s}}^{4}\right)$
since mass of hollow sphere is less its $\frac{\mathrm{d} \theta}{\mathrm{dt}}$ is more
Hollow sphere. Cools faster
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