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 $R_c=\frac{A\epsilon \sigma \left(T^4-T_0^4\right)}{mc}$
$=\frac{A\epsilon \sigma \left(T^4-T_0^4\right)}{V\rho c}$
$\Rightarrow R_c=\frac{A}{V}\propto \frac{1}{r}\propto \frac{1}{(\text{ Diameter })}$
$(\because m=\rho V)$
Since, diameter of $A$ is half that of $B$,
so its rate of cooling will be doubled that of $B$.