Cooling rate of a sphere of $600\,K$ at external environment $(200\,K)$ is $R$ . When the temperature of sphere is reduced to $400\,K$ then cooling rate of the sphere becomes
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$\mathrm{R}_{1} \propto(600)^{4}-(200)^{4}$
$\mathrm{R}_{2} \propto(400)^{4}-(200)^{4}$
$\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{(16+4)(16-4)}{(36+4)(36-4)}=\frac{20 \times 12}{40 \times 32}$
$\mathrm{R}_{2}=\frac{3}{16} \mathrm{R}$
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