Q 11-10-104JEE MainJEE Main 2020 (3 Sep, Shift 2)Medium
A metallic sphere cools from $50^\circ\text{C}$ to $40^\circ\text{C}$ in $300\ \text{s}$. If atmospheric temperature around is $20^\circ\text{C}$, then the sphere's temperature after the next $5$ minutes will be close to:
Answer: (B) $33^\circ\text{C}$
Newton's law of cooling: $\theta - \theta_0$ decreases by the same factor in equal times.
First $300$ s: $(40 - 20)/(50 - 20) = \dfrac23$.
Next $300$ s: $\theta - 20 = \dfrac23\times20 \approx 13.3$, so $\theta \approx 33^\circ\text{C}$.
(The average-temperature form of the law gives the same result: $40 - \theta = 0.2\,\theta$, $\theta \approx 33.3^\circ\text{C}$.)
Solution by Sreeraj P, M.Sc Physics