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10-2.Transmission of Heat
medium
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 $B$
Solution
(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}}$
==> ${R_C} \propto \frac{A}{V} \propto \frac{1}{r} \propto \,\frac{1}{{({\rm{Diameter)}}}}$ $(\because \,m = \rho V)$
Since diameter of $A$ is half that of $B $ so it's rate of cooling will be doubled that of B
Standard 11
Physics
