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(A) According to Newton's Law of Cooling,the rate of cooling $R$ is directly proportional to the temperature difference between the body and its surro

Vedclass · Accepted Answer

$R_1:R_2 = 3:2$

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(A) According to Newton's Law of Cooling,the rate of cooling $R$ is directly proportional to the temperature difference between the body and its surroundings,i.e.,$R \propto (	heta - 	heta_0)$.Given:Temperature of body $B_1$,$	heta_1 = 100^oC$Temperature of body $B_2$,$	heta_2 = 80^oC$Temperature of surroundings,$	heta_0 = 40^oC$At $t = 0$,the rates of cooling $R_1$ and $R_2$ are:$R_1 \propto (	heta_1 - 	heta_0) = (100 - 40) = 60$$R_2 \propto (	heta_2 - 	heta_0) = (80 - 40) = 40$Therefore,the ratio is:$\frac{R_1}{R_2} = \frac{60}{40} = \frac{3}{2}$Thus,$R_1:R_2 = 3:2$.

Consider two hot bodies $B_1$ and $B_2$ which have temperatures $100^oC$ and $80^oC$ respectively at $t = 0$. The temperature of the surroundings is $40^oC$. The ratio of the respective rates of cooling $R_1$ and $R_2$ of these two bodies at $t = 0$ will be

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