A ball 'A' has twice the diameter as another ball 'B' of the same material and with same surface finish. 'A' and 'B' are both heated to the same temperature and allowed to cool by radiation then,

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rate of cooling of 'A' is half that of 'B'

rate of cooling of 'A' is same as that of 'B'

rate of cooling of 'A' is twice that of 'B'

rate of cooling of 'A' is four times that of B

answer is C.

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Detailed Solution

Rate of cooling $R_{f} = \frac{e_{r} σA (T^{4} - T_{0}^{4})}{ms}$

$R_{f} \propto \frac{A}{m} \propto \frac{A}{V} \propto \frac{1}{r}$

$\frac{R_{f_{1}}}{R_{f_{2}}} = \frac{r_{2}}{r_{1}} = \frac{1}{2} \Rightarrow R_{f_{2}} = 2 R_{f_{1}}$

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