Q. Two identical metal balls at temperature $200℃$ and $400℃$ kept in air at $27℃$. The ratio of net heat loss by these bodies is

A) 1/4
B) 1/2
C) 1/16
D) $\frac{473^{4} - 300^{4}}{673^{4} - 300^{4}}$
Explanation

(d) If temperature of surrounding is considered then net loss of energy of a body by radiation
$Q = A\varepsilon\sigma\left( T^{4} - T_{0}^{4} \right)t \Rightarrow Q \propto \left( T^{4} - T_{0}^{4} \right) \Rightarrow \frac{Q_{1}}{Q_{2}} = \frac{T_{1}^{4} - T_{0}^{4}}{T_{2}^{4} - T_{0}^{4}}$
$= \frac{(273 + 200)^{4} - (273 + 27)^{4}}{(273 + 400)^{4} - (273 + 27)^{4}} = \frac{(473)^{4} - (300)^{4}}{(673)^{4} - (300)^{4}}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.