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(C) Given,weight of sphere $P$ is $8$ times the weight of sphere $Q$. i.e.,$m_P = 8 m_Q$.From Stefan's law,the rate of heat radiation from an object i

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$2$

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(C) Given,weight of sphere $P$ is $8$ times the weight of sphere $Q$. i.e.,$m_P = 8 m_Q$.From Stefan's law,the rate of heat radiation from an object is given by $\frac{dQ}{dt} = e \sigma A T^4$ ... $(i)$,where $e$ is emissivity,$\sigma$ is Stefan's constant,$A$ is surface area,and $T$ is temperature.Also,the rate of heat loss is $\frac{dQ}{dt} = mc \frac{dT}{dt}$ ... (ii),where $c$ is specific heat capacity.Equating $(i)$ and (ii): $mc \frac{dT}{dt} = e \sigma A T^4 \implies \frac{dT}{dt} = \frac{e \sigma A T^4}{mc}$.Since the spheres are made of the same material,$c$ and $e$ are constant. For a sphere,$m = \rho \cdot \frac{4}{3} \pi r^3 \implies r \propto m^{1/3}$.Surface area $A = 4 \pi r^2 \propto (m^{1/3})^2 = m^{2/3}$.Thus,the rate of cooling $\frac{dT}{dt} \propto \frac{A}{m} \propto \frac{m^{2/3}}{m} = m^{-1/3}$.Therefore,$\frac{(\frac{dT}{dt})_Q}{(\frac{dT}{dt})_P} = (\frac{m_P}{m_Q})^{1/3} = (\frac{8 m_Q}{m_Q})^{1/3} = (8)^{1/3} = 2$.

Two metallic spheres $P$ and $Q$ are made of the same material and have the same smoothness,but the weight of $P$ is $8$ times that of $Q$. If the two are heated to the same temperature and left to cool,then the ratio of the rate of cooling of $Q$ to that of $P$ is:

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