Q 11-10-087JEE MainJEE Main 2021 (27 Jul, Shift 1)Easy
A body takes $4$ min to cool from $61^\circ\text{C}$ to $59^\circ\text{C}$. If the temperature of the surroundings is $30^\circ\text{C}$, the time taken by the body to cool from $51^\circ\text{C}$ to $49^\circ\text{C}$ is:
Answer: (D) $6$ min
Newton's law of cooling (average form): $\dfrac{\Delta T}{t} = k(T_{\text{avg}} - T_0)$.
First case: $\dfrac{2}{4} = k(60 - 30) \Rightarrow k = \dfrac{1}{60}\ \text{min}^{-1}$.
Second case: $\dfrac{2}{t} = \dfrac{1}{60}(50 - 30) = \dfrac13 \Rightarrow t = 6$ min.
Solution by Sreeraj P, M.Sc Physics