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 $P(\theta ={\theta }_1)$ and $Q(\theta ={\theta }_2)$ 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 $P,\frac{d\theta }{dt}=tan{\varphi }_2=k({\theta }_2-{\theta }_0)$, where $k=$ constant.
At $Q\frac{d\theta }{dt}=tan{\varphi }_1=k({\theta }_1-{\theta }_0)$
$\Rightarrow \frac{tan{\varphi }_2}{tan{\varphi }_1}=\frac{{\theta }_2-{\theta }_0}{{\theta }_1-{\theta }_0}$