Q. Consider two hot bodies $B_{1}$ and $B_{2}$ which have temperatures $100℃$ and $80℃$ respectively at $t = 0$. The temperature of the surroundings is $40℃$. The ratio of the respective rates of cooling, $R_{1}$ and $R_{2}$ of these two bodies at $t = 0$ will be
A) $R_{1}:R_{2} = 3:2$
B) $R_{1}:R_{2} = 5:4$
C) $R_{1}:R_{2} = 2:3$
D) $R_{1}:R_{2} = 4:5$
Explanation
(a) Initially at $t = 0$
Rate of cooling $(R) \propto$ Fall in temperature of body $(\theta - \theta_{0})$
$\Rightarrow \frac{R_{1}}{R_{2}} = \frac{\theta_{1} - \theta_{0}}{\theta_{2} - \theta_{0}} = \frac{100 - 40}{80 - 40} = \frac{3}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.