Question

A body cools in a surrounding maintained at constant temperature of ${\theta }_0$. Assuming that it obeys Newton's law of cooling. Its temperature is $\theta$ is plotted against time. Tangents are drawn to the curve at points P($\theta ={\theta }_2$) and Q($\theta ={\theta }_1$). These tangents meets the time axis at angles ${\phi }_2$ and ${\phi }_1$ as shown then

• 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. $\frac{tan{\phi }_2}{tan{\phi }_1}=\frac{{\theta }_1}{{\theta }_2}$
• D. $\frac{tan{\phi }_2}{tan{\phi }_1}=\frac{{\theta }_2}{{\theta }_1}$

Answer 1

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

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.
$\frac{tan{\Phi }_2}{tan{\Phi }_1}=\frac{{\theta }_2-{\theta }_0}{{\theta }_1-{\theta }_0}$