Two metallic spheres P and Q are of same surface finish and same density. Weight of P is twice that of Q. Both spheres are heated to same temperature and are left in room to cool by radiation. The ratio of initial rate of cooling of P to that of Q is

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$\sqrt{2} : 1$

${(2)}^{2 / 3} : 1$

$1 : \sqrt{2}$

$1 : {(2)}^{1 / 3}$

answer is C.

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Detailed Solution

Let the mass of Q be m.

$∴$ mass of P = 2m

$\begin{array}{l} Now, - m . s . \frac{dT}{dt} = eσ . (4 {πR}^{2}) (T^{4} - {T_{0}}^{4}) \\ \Rightarrow - \frac{dT}{dt} = \frac{eσ . (4 {πR}^{2})}{(\frac{4}{3} {πR}^{3} ρ) S} (T^{4} - {T_{0}}^{4}) \\ \Rightarrow - \frac{dT}{dt} = (\frac{3 eσ}{ρS}) \frac{1}{R} (T^{4} - {T_{0}}^{4}) \\ ∴ - \frac{dT}{dt} \propto \frac{1}{R} \\ Now \frac{4}{3} {πR}_{p}^{3} ρ = 2 \times \frac{4}{3} {πR}_{Q}^{3} ρ \\ \Rightarrow R_{p} = {(2)}^{1 / 3} . R_{Q} \\ ∴ \frac{Initial rate of cooling of P}{Initial rate of cooling of Q} = (\frac{R_{Q}}{R_{P}}) = \frac{1}{{(2)}^{1 / 3}} \end{array}$