Question

There are two spherical balls $A$ and $B$ of the same material with 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

• A. rate of cooling is same in both
• B. rate of cooling of A is four times that of B
• C. rate of cooling of A is twice that of B
• D. rate of cooling of A is $ \frac{1}{4} $ times that of B

Answer 1

Answer: (C)

Rate of cooling $R_{c}=\frac{A \varepsilon \sigma\left(T^{4}-T_{0}^{4}\right)}{m c}$
$=\frac{A \varepsilon \sigma\left(T^{4}-T_{0}^{4}\right)}{V \rho c}$
$\Rightarrow R_{c}=\frac{A}{V} \propto \frac{1}{r} \propto \frac{1}{(\text { Diameter })}$
$(\because m=\rho V)$
Since, diameter of $A$ is half that of $B$,
so its rate of cooling will be doubled that of $B$.