A solid copper sphere (density p and specific heat c) of radius .r at an initial temperature 200 K is suspended inside a chamber whose walls are at almost O K. The time required for the temperature of the sphere to drop to 100 K is

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$\frac{7 rρc}{72 σ} \times 10^{- 6} \sec$

$\frac{7 {rρc}^{2}}{72 σ} \times 10^{- 6} \sec$

$\frac{r^{2} ρc}{72 σ} \times 10^{- 6} \sec$

$\frac{{rρ}^{2} c^{2}}{σ^{2}} \times 10^{- 6} \sec$

answer is A.

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Detailed Solution

$mc (- \frac{dT}{dt}) = {σT}^{4} A ∴ dt = - \frac{mcdT}{{σAT}^{4}}$
Integrating, we get
$t = - \frac{mc}{σA} \int_{200}^{100} T^{- 4} dT ∴ t = \frac{{(\frac{4}{3} {πr}^{3} ρ)}^{c}}{σA} {[\frac{T^{- 3}}{(- 3)}]}_{200}^{100} \begin{array}{l} = \frac{4 {πr}^{3} ρc}{9 σ 4 {πr}^{2}} [\frac{1}{(100)^{3}} - \frac{1}{(200)^{3}}] \\ = \frac{rρc}{9 σ} [\frac{1}{(100)^{3}} - \frac{1}{(200)^{3}}] \\ = [\frac{7 rρc}{72 σ} \times 10^{- 6}] \sec \end{array}$