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

By Newton’s law of cooling, the rate of fall of temperature is proportional to $$\frac{S}{mc}.$$ Since both balls are of the same material, $c$ is same, and $m \propto V \propto d^3$ while $S \propto d^2$. Hence $$\frac{S}{mc} \propto \frac{d^2}{d^3} = \frac{1}{d}.$$ So the rate of cooling is inversely proportional to diameter. Given that the diameter of $A$ is half that of $B$, $$d_A = \frac{d_B}{2}.$$ Therefore $$\frac{\text{rate of cooling of }A}{\text{rate of cooling of }B} = \frac{1/d_A}{1/d_B} = \frac{d_B}{d_A} = 2.$$ Thus the rate of cooling of $A$ is twice that of $B$. On checking the options, this matches option $C$.