Q. A black metal foil is warmed by radiation from a small sphere at temperature $T$ and at a distance $d$. It is found that the power received by the foil is $'P'$. If both the temperature and the distance are doubled, the power received by the foil will be
A) $16P$
B) $4P$
C) $2P$
D) $P$
Explanation
(b) Energy received per second $i.e.,$ power $P \propto (T^{4} - T_{0}^{4})$
$\Rightarrow P \propto T^{4}\ \ \ \ (\because T_{0} < \ < T)$
Also energy received per sec $(P) \propto \frac{1}{d^{2}}$
(inverse square law)
$\Rightarrow P \propto \frac{T^{4}}{d^{2}} \Rightarrow \frac{P_{1}}{P_{2}} = \left( \frac{T_{1}}{T_{2}} \right)^{4} \times \left( \frac{d_{2}}{d_{1}} \right)^{2}$
$\Rightarrow \frac{P}{P_{2}} = \left( \frac{T}{2T} \right)^{4} \times \left( \frac{2d}{d} \right)^{2} = \frac{1}{4} \Rightarrow P_{2} = 4P$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.