Question

A body cools in a surrounding which is at a constant temperature of $\theta_{0}$. Assume that it obeys Newton's law of cooling. Its temperature $\theta$ is plotted against time $t$. Tangents are drawn to the curve at the points $P\left(\theta=\theta_{1}\right)$ and $Q\left(\theta=\theta_{2}\right)$. These tangents meet the time axis at angles of $\phi_{2}$ and $\phi_{1}$, as shown

• A. $\frac{\tan \phi_{2}}{\tan \phi_{1}}=\frac{\theta_{1}-\theta_{0}}{\theta_{2}-\theta_{0}}$
• B. $\frac{\tan \phi_{2}}{\tan \phi_{1}}=\frac{\theta_{2}-\theta_{0}}{\theta_{1}-\theta_{0}}$
• C. $\frac{\tan \phi_{1}}{\tan \phi_{2}}=\frac{\theta_{1}}{\theta_{2}}$
• D. $\frac{\tan \phi_{1}}{\tan \phi_{2}}=\frac{\theta_{2}}{\theta_{1}}$

Answer 1

Answer: (B)

For $\theta-t$ plot, rate of cooling $=\frac{d \theta}{d t}=$ slope of the curve.
At $P, \frac{d \theta}{d t}=\tan \phi_{2}=k\left(\theta_{2}-\theta_{0}\right)$, where $k=$ constant.
At $Q \frac{d \theta}{d t}=\tan \phi_{1}=k\left(\theta_{1}-\theta_{0}\right)$
$\Rightarrow \frac{\tan \phi_{2}}{\tan \phi_{1}}=\frac{\theta_{2}-\theta_{0}}{\theta_{1}-\theta_{0}}$