Q. A body cools in a surrounding which is at a constant temperature of $\theta_{0}$. Assume that of obeys Newton’s law of cooling. Its temperature $\theta$ is plotted against time $t$. Tangents are drawn to the curve at the points $P(\theta = \theta_{2})$ and $Q(\theta = \theta_{1})$. 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) $\frac{\tan\phi_{1}}{\tan\phi_{2}} = \frac{\theta_{2}}{\theta_{1}}$
Explanation
(b) For $\theta t$ plot, rate of cooling $= \frac{d\theta}{dt} =$ slope of the curve
At $P,\frac{d\theta}{dt} = \tan\phi_{2} = k\left( \theta_{2} - \theta_{0} \right),\ $where$\ k =$ constant
At $Q,\frac{d\theta}{dt} = \tan\phi_{1} = k\left( \theta_{1} - \theta_{0} \right) \Rightarrow \frac{\tan\phi_{2}}{\tan\phi_{1}} = \frac{\theta_{2} - \theta_{0}}{\theta_{1} - \theta_{0}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.