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 ${\varphi }_2$ and ${\varphi }_1$, as shown

• A. $\frac{\tan {\varphi }_2}{\tan {\varphi }_1}=\frac{{\theta }_1-{\theta }_0}{{\theta }_2-{\theta }_0}$
• B. $\frac{\tan {\varphi }_2}{\tan {\varphi }_1}=\frac{{\theta }_2-{\theta }_0}{{\theta }_1-{\theta }_0}$
• C. $\frac{\tan {\varphi }_1}{\tan {\varphi }_2}=\frac{{\theta }_1}{{\theta }_2}$
• D. $\frac{\tan {\varphi }_1}{\tan {\varphi }_2}=\frac{{\theta }_2}{{\theta }_1}$

Answer 1

Answer: (B)

For $\theta -t$ plot, rate of cooling $=\frac{d\theta }{dt}=$ slope of the curve.
At $P,\frac{d\theta }{dt}=\tan {\varphi }_2=k\left({\theta }_2-{\theta }_0\right)$, where $k=$ constant.
At $Q\frac{d\theta }{dt}=\tan {\varphi }_1=k\left({\theta }_1-{\theta }_0\right)$
$\Rightarrow \frac{\tan {\varphi }_2}{\tan {\varphi }_1}=\frac{{\theta }_2-{\theta }_0}{{\theta }_1-{\theta }_0}$