Q. 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 $B$
Explanation
(c) Rate of cooling $R_{C} = \frac{A\varepsilon\sigma(T^{4} - T_{0}^{4})}{mc} = \frac{A\varepsilon\sigma(T^{4} - T_{0}^{4})}{V\rho C}$
$\Rightarrow R_{C} \propto \frac{A}{V} \propto \frac{1}{r} \propto \frac{1}{(Diameter)}\ \ \lbrack\because m = \rho V\rbrack$
Since diameter of $A$ is half that of $B$ so it’s rate of cooling will be doubled that of $B$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.