Q118 / 120 · VITEEE 2018 · Physics
A body cools from $50.0^{\circ} \mathrm{C}$ to $49.9^{\circ} \mathrm{C}$ in $5 \mathrm{~s}$. How long will it take to cool from $40.0^{\circ} \mathrm{C}$ to $39.9^{\circ} \mathrm{C}$ ? Assume the temperature of surroundings to be $30.0^{\circ} \mathrm{C}$ and Newton's law of cooling to be valid
- A$2.5 \mathrm{~s}$
- Correct answerB$10 \mathrm{~s}$
- C$20 \mathrm{~s}$
- D$5 \mathrm{~s}$
Solution
$$
\begin{aligned}
& \frac{50-49.9}{5}=K\left(\frac{50+49.9}{2}-30\right)....(i) \\
& \frac{40-39.9}{t}=K\left[\frac{40+39.9}{2}-30\right]....(ii)
\end{aligned}
$$
From equations (i) and (ii), we get $\mathrm{t} \approx 10 \mathrm{~s}$.
Hence, the correct answer is option: 'B'.