NEET Physics Question - Thermal Physics

A sphere maintained at temperature $600 \text{ K}$, has cooling rate $R$ in an external environment of $200 \text{ K}$ temperature. If its temperature, falls to $400 \text{ K}$ then its cooling rate will be: (1999)

$\frac{3}{16}R$

$\frac{16}{3}R$

$\frac{9}{27}R$

None

Solution:

Cooling rate $\propto (T^4 - T_0^4)$. $\frac{R'}{R} = \frac{400^4 - 200^4}{600^4 - 200^4} = \frac{2^4 - 1^4}{3^4 - 1^4} = \frac{15}{80} = \frac{3}{16}$. Since $\frac{3}{16} = \frac{9}{48}$, the correct cooling rate would be $\frac{3}{16}R$. The closest option is incorrectly printed as 9/27, the answer is 3/16 R

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