A solid copper sphere (density $ρ$ and specific heat capacity c) of radius r at an initial temperature 200 K is suspended inside a chamber whose walls are at almost 0 K. The time required (in $μ$ s) for the temperature of the sphere to drop to 100 K is

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$\frac{27}{7} \frac{rρc}{σ}$

$\frac{72}{7} \frac{rρc}{σ}$

$\frac{7}{80} \frac{rρc}{σ}$

$\frac{7}{72} \frac{rρc}{σ}$

answer is B.

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Detailed Solution

$\begin{array}{l} (\frac{4}{3} π r^{3}) \times ρ C \frac{d T}{d t} = σ (4 π r^{2}) T^{4} \\ \Rightarrow \frac{r}{3} \times ρ C \frac{d T}{d t} = σ T^{4} \Rightarrow \frac{r ρ C}{3 σ T^{4}} d T = d t \\ \begin{array}{l} \Rightarrow \frac{r ρ C}{3 σ} \int_{100}^{200} \frac{d T}{T^{4}} = \int_{0}^{t} d t \\ \Rightarrow \frac{r ρ C}{3 σ} \int_{100}^{200} \frac{d T}{T^{4}} = \int_{0}^{t} d t \\ \Rightarrow \frac{r ρ C}{9 σ \times 10^{6}} (1 - \frac{1}{8}) = t \\ ∴ t = \frac{7 r ρ C}{72 σ \times 10^{6}} \end{array} \end{array}$