Q129 / 150 · MHT CET 2024 (16 May Shift 1) · Physics
A bucket full of hot water is kept in a room. If it cools from $75^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in $\mathrm{t}_1$ minutes, from $70^{\circ} \mathrm{C}$ to $65^{\circ} \mathrm{C}$ in $\mathrm{t}_2$ minutes and $65^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in $\mathrm{t}_3$ minutes, then
- Correct answerA$\mathrm{t}_1 \lt \mathrm{t}_2 \lt \mathrm{t}_3$
- B$t_1 \gt t_2 \gt t_3$
- C$t_1=t_2=t_3$
- D$\mathrm{t}_1 \lt \mathrm{t}_2=\mathrm{t}_3$
Solution
According to Newton's law of cooling, Rate of cooling $\propto$ Mean temperature difference $\Rightarrow \frac{\text { Fall in temperature }}{\text { Time }(\mathrm{t})} \propto\left(\frac{\theta_1+\theta_2}{2}-\theta_0\right)$
Case 1: $\left(\frac{\theta_1+\theta_2}{2}\right)_1=\left(\frac{75+70}{2}\right)_1=72.5$
Case 2: $\left(\frac{\theta_1+\theta_2}{2}\right)_2=\left(\frac{70+65}{2}\right)_2=67.5$
Case 3: $\left(\frac{\theta_1+\theta_2}{2}\right)_3=\left(\frac{65+60}{2}\right)_3=62.5$
$\begin{aligned}
\therefore \quad & \left(\frac{\theta_1+\theta_2}{2}\right)_1\gt\left(\frac{\theta_1+\theta_2}{2}\right)_2\gt\left(\frac{\theta_1+\theta_2}{2}\right)_3 \\
& \Rightarrow \mathrm{t}_1 \lt \mathrm{t}_2 \lt \mathrm{t}_3
\end{aligned}$
Hence, the correct answer is option: 'A'.