In a room where the temperature is 30^{circ}C | vedclass

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(B) According to Newton's law of cooling,the rate of cooling is proportional to the difference between the temperature of the body and the surrounding

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$6$

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(B) According to Newton's law of cooling,the rate of cooling is proportional to the difference between the temperature of the body and the surrounding temperature: $\frac{dT}{dt} = K(T - T_s)$.For the first case: $\frac{61 - 59}{4} = K \left( \frac{61 + 59}{2} - 30 ight)$.$\frac{2}{4} = K(60 - 30) \implies 0.5 = 30K \implies K = \frac{0.5}{30} = \frac{1}{60}$.For the second case: $\frac{51 - 49}{t} = K \left( \frac{51 + 49}{2} - 30 ight)$.$\frac{2}{t} = K(50 - 30) \implies \frac{2}{t} = 20K$.Substituting $K = \frac{1}{60}$: $\frac{2}{t} = 20 	imes \frac{1}{60} = \frac{1}{3}$.Therefore,$t = 6 	ext{ min}$.

In a room where the temperature is $30^{\circ}C$,a body cools from $61^{\circ}C$ to $59^{\circ}C$ in $4$ minutes. The time (in min) taken by the body to cool from $51^{\circ}C$ to $49^{\circ}C$ will be ....... $\text{min}$.

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