A sphere maintained at temperature 600 K, has cooling rate R in an external environment of 200 K temperature. If its temperature falls to 400 K, the cooling rate will be:

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$\frac{3}{16} R$

$\frac{16}{3} R$

$\frac{9}{27} R$

$\frac{3}{64} R$

answer is A.

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Detailed Solution

Cooling rate $\infty T^{4} - T_{0}^{4}$
So, $\frac{R'}{R} = \frac{400^{4} - 200^{4}}{600^{4} - 200^{4}} = \frac{4^{4} - 2^{4}}{6^{4} - 2^{4}} = \frac{3}{16} .$

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