Q. Two solid spheres $A\ $and $B$ made of the same material have radii $r_{A}\ and\ r_{B}$respectively. Both the spheres are cooled from the same temperature under the conditions valid for Newton’s law of cooling. The ratio of the rate of change of temperature of $A$ and $B$ is
A) $\frac{r_{A}}{r_{B}}$
B) $\frac{r_{B}}{r_{A}}$
C) $\frac{r_{A}^{2}}{r_{B}^{2}}$
D) $\frac{r_{B}^{2}}{r_{A}^{2}}$
Explanation
(b) $\frac{4\pi}{3}r^{3}\rho c\left( - \frac{dT}{dt} \right) = \sigma 4\pi r^{2}(T^{4} - T_{0}^{4})$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( - \frac{dT}{dt} \right) = \frac{3\sigma}{\rho rc}\left( T^{4} - T_{0}^{4} \right) = H$ (say)
Ratio of rates of fall of temperature
$\frac{H_{A}}{H_{B}} = \frac{r_{B}}{r_{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.