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CGP EDU Academic Team
Published on: September 12, 2026
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
Text Solution
Verified by ExpertsThe correct answer is:
C
Rate of cooling $R_C = \frac{A \epsilon \sigma (T^4 - T_0^4)}{mc} = \frac{A \epsilon \sigma (T^4 - T_0^4)}{V \rho c}$ $\Rightarrow R_C \propto \frac{A}{V} \propto \frac{1}{r} \propto \frac{1}{\text{Diameter}} \quad (\therefore m = \rho V)$ Since diameter of A is half that of B so it's rate of cooling will be doubled that of B
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