A body cools from a temperature $3 T$ to $2 T$ in 10 minutes. The room temperature is $T$. Assume th

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A body cools from a temperature $3 T$ to $2 T$ in 10 minutes. The room temperature is $T$. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be\

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(b) : According to Newton's law of cooling, 
$ \frac{d T}{d t}=K\left(T-T_{s}ight) $ 
For two cases, $\frac{d T_{1}}{d t}=K\left(T_{1}-T_{s}ight)$ and $\frac{d T_{2}}{d t}=K\left(T_{2}-T_{s}ight)$\ Here, $T_{s}=T, T_{1}=\frac{3 T+2 T}{2}=2.5 T$ and $\frac{d T_{1}}{d t}=\frac{3 T-2 T}{10}=\frac{T}{10}$ 
$T_2=\frac{2 T+T^{\prime}}{2}$ and $\frac{d T_2}{d t}=\frac{2 T-T^{\prime}}{10}$ ...(i)
 So, $\quad \frac{T}{10}=K(2.5 T-T)$\ and $\frac{2 T-T^{\prime}}{10}=K\left(\frac{2 T+T^{\prime}}{2}-Tight)$\ Dividing eqn. (i) by eqn. (ii), we get\ $\frac{T}{2 T-T^{\prime}}=\frac{(2.5 T-T)}{\left(\frac{2 T+T^{\prime}}{2}-Tight)}$ or, $\frac{2 T+T^{\prime}}{2}-T=\left(2 T-T^{\prime}ight) 	imes \frac{3}{2}$\ $T^{\prime}=3\left(2 T-T^{\prime}ight)$ or, $4 T^{\prime}=6 T ; \quad 	herefore \quad T^{\prime}=\frac{3}{2} T$\