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(D) According to Newton's law of cooling, the rate of cooling $R$ is proportional to the difference between the temperature of the body $(T_b)$ and th

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$\frac{3}{16}R_1$

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(D) According to Newton's law of cooling, the rate of cooling $R$ is proportional to the difference between the temperature of the body $(T_b)$ and the surrounding temperature $(T_s)$, provided the temperature difference is small. However, for larger temperature differences, we use the Stefan-Boltzmann law for the net rate of heat loss: $P = \sigma A e (T_b^4 - T_s^4)$.Given the cooling rate $R = \frac{dQ}{dt} \propto (T_b^4 - T_s^4)$.For the first case, $T_b = 3T$ and $T_s = T$:$R_1 = k ((3T)^4 - T^4) = k (81T^4 - T^4) = 80kT^4$.For the second case, $T_b = 2T$ and $T_s = T$:$R_2 = k ((2T)^4 - T^4) = k (16T^4 - T^4) = 15kT^4$.Now, find the ratio:$\frac{R_2}{R_1} = \frac{15kT^4}{80kT^4} = \frac{15}{80} = \frac{3}{16}$.Therefore, $R_2 = \frac{3}{16}R_1$.

In an external environment of temperature $T$ Kelvin, a sphere at temperature $3T$ Kelvin has a cooling rate $R_1$. When the temperature of that sphere falls to $2T$ Kelvin, the cooling rate of the sphere will become:

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