Question

A body cools in a surrounding which is at a constant temperature of $\theta$. 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 meets 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

The correct option is B $\frac{\tan {\varphi }_2}{\tan {\varphi }_1}=\frac{{\theta }_2-{\theta }_0}{{\theta }_1-{\theta }_0}$

The equation given by Newton's law of cooling is :
$\frac{d\theta }{dt}=-bA(\theta -{\theta }_0)=$ Slope of the graph = $tan\Phi$
$tan{\Phi }_1=-bA({\theta }_1-{\theta }_0)$
$tan{\Phi }_2=-bA({\theta }_2-{\theta }_0)$
i.e.
$\frac{tan{\Phi }_2}{tan{\Phi }_1}=\frac{{\theta }_2-{\theta }_0}{{\theta }_1-{\theta }_0}$