A sphere having temperature 600 text{ K} is l | vedclass

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(D) According to Newton's law of cooling, the rate of cooling $\frac{dT}{dt}$ is proportional to $(T^4 - T_0^4)$ for radiation, where $T$ is the tempe

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$35$

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(D) According to Newton's law of cooling, the rate of cooling $\frac{dT}{dt}$ is proportional to $(T^4 - T_0^4)$ for radiation, where $T$ is the temperature of the body and $T_0$ is the temperature of the surroundings.Given $T_1 = 600 	ext{ K}$, $T_0 = 300 	ext{ K}$, and rate of cooling $R = k(T_1^4 - T_0^4)$.$R = k(600^4 - 300^4) = k(300^4)(2^4 - 1^4) = k(300^4)(15)$.Now, at $T_2 = 400 	ext{ K}$, the rate of cooling $R' = k(400^4 - 300^4) = k(300^4)((\frac{4}{3})^4 - 1^4) = k(300^4)(\frac{256}{81} - 1) = k(300^4)(\frac{175}{81})$.Dividing $R'$ by $R$:$\frac{R'}{R} = \frac{k(300^4)(\frac{175}{81})}{k(300^4)(15)} = \frac{175}{81 	imes 15} = \frac{175}{1215} = \frac{35}{243}$.Given $R' = \frac{x}{243}R$, we have $\frac{x}{243} = \frac{35}{243}$, so $x = 35$.

$A$ sphere having temperature $600 \text{ K}$ is losing heat due to radiation. At this temperature its rate of cooling is $R$. The rate of cooling of this sphere at $400 \text{ K}$ is $\frac{x}{243}R$. The value of $x$ is (Temperature of surrounding is $300 \text{ K}$)

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