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(C) The rate of cooling $R_C$ is given by the formula: $R_C = \frac{dQ}{dt} \cdot \frac{1}{mc} = \frac{A \varepsilon \sigma (T^4 - T_0^4)}{mc}$.Since

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Rate of cooling of $A$ is twice that of $B$

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(C) The rate of cooling $R_C$ is given by the formula: $R_C = \frac{dQ}{dt} \cdot \frac{1}{mc} = \frac{A \varepsilon \sigma (T^4 - T_0^4)}{mc}$.Since $m = ho V$,where $ho$ is density and $V$ is volume,we have $R_C = \frac{A \varepsilon \sigma (T^4 - T_0^4)}{V ho c}$.For a sphere,$A = 4 \pi r^2$ and $V = \frac{4}{3} \pi r^3$,so $\frac{A}{V} = \frac{3}{r}$.Thus,$R_C \propto \frac{1}{r} \propto \frac{1}{	ext{Diameter}}$.Given that the diameter of $A$ is half that of $B$ $(D_A = \frac{1}{2} D_B)$,the rate of cooling of $A$ will be twice that of $B$ $(R_{C,A} = 2 R_{C,B})$.

There are two spherical balls $A$ and $B$ of the same material with the same surface,but the diameter of $A$ is half that of $B$. If $A$ and $B$ are heated to the same temperature and then allowed to cool,then

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