Question

A body cools in a surrounding which is at a constant temperature of ${\theta }_o$ . Assume that it obeys Newtons law of cooling. Its temperature $\theta$ is plotted against time '$t$'. Tangents are drawn to the curve at the points $P(\theta -{\theta }_1)$ and $Q(\theta -{\theta }_2)$. These tangents meet the time axis at angles of ${\varphi }_{2}and{\varphi }_1$ as shown. Then,

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

The correct option is B $\frac{\tan {\varphi }_2}{\tan {\varphi }_1}=\frac{{\theta }_2-{\theta }_o}{{\theta }_1-{\theta }_o}$

From Newton's Law of cooling:
Rate of cooling $\alpha$ $(\theta -(\theta {)}_s)$
At P, $\frac{d\theta }{dt}=\tan (\pi -{\varphi }_2)=K({\theta }_2-{\theta }_0)$ .....(1)

At Q, $\frac{d\theta }{dt}=\tan (\pi -{\varphi }_1)=K({\theta }_1-{\theta }_0)$ .....(2)

Dividing (1) by (2) we get,

$\frac{\tan (\pi -{\varphi }_2)}{\tan (\pi -{\varphi }_1)}=\frac{({\theta }_2-{\theta }_0)}{({\theta }_1-{\theta }_0)}$

$\therefore \frac{\tan {\varphi }_2}{\tan {\varphi }_1}=\frac{({\theta }_2-{\theta }_0)}{({\theta }_1-{\theta }_0)}$