Q98 / 180 · AIIMS 2017 · Physics
A body initially at $80^{\circ} \mathrm{C}$ cools to $64^{\circ} \mathrm{C}$ in $5 \mathrm{~min}$ and to $52^{\circ} \mathrm{C}$ in $10 \mathrm{~min}$. The temperature of the surrounding is
- A$26^{\circ} \mathrm{C}$
- Correct answerB$16^{\circ} \mathrm{C}$
- C$36^{\circ} \mathrm{C}$
- D$40^{\circ} \mathrm{C}$
Solution
According to Newton's law of cooling
$\frac{\theta_1-\theta_2}{t}=K\left[\frac{\theta_1+\theta_2}{2}-\theta_0\right]$
In the first case, $\frac{80-64}{5}=K\left[\frac{80+64}{2}-\theta_0\right]$ or $\quad 3.2=K\left[72-\theta_0\right]$
In the second case, $\frac{64-52}{5}=K\left[\frac{64+52}{2}-\theta_0\right]$ or $\quad 2.4=K\left[58-\theta_0\right]$
Dividing (i) by (ii), we get $\frac{3.2}{2.4}=\frac{72-\theta_0}{58-\theta_0}$ or $\quad 185.6-3.2 \theta_0=172.8-2.4 \theta_0$ or $\theta_0=16^{\circ} \mathrm{C}$
Hence, the correct answer is option: 'B'.