Q.
The graph, shown in the adjacent diagram, represents the variation of temperature $(T)$ of two bodies, $x$ and $y$ having same surface area, with time ($t$) due to the emission of radiation. Find the correct relation between the emissivity and absorptivity power of the two bodies.
A) $E_{x} > E_{y}\ and\ a_{x} < a_{y}$
B) $E_{x} < E_{x}\ and\ a_{x} > a_{y}$
C) $E_{x} > E_{x}\ and\ a_{x} > a_{y}$
D) $E_{x} < E_{x}\ and\ a_{x} < a_{y}$
Explanation
(c) Rate of cooling $( - \frac{dT}{dt}) \propto$ emissivity(e)
From the graph,
$\left( - \frac{dT}{dt} \right)_{x} > \left( - \frac{dT}{dt} \right)_{y}$
$\therefore$ $e_{x} > e_{y}$
Further emissivity ($e$)$\propto$ absorptive power ($a$) (good absorbers are good emitters also)
$\therefore$ $a_{x} > a_{y}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.