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 $(e)$ and absorptivity $(a)$ of the two bodies.

e_{x} > e_{y}

and

a_{x} < a_{y}

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e_{x} < e_{y}

and

a_{x} > a_{y}

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e_{x} > e_{y}

and

a_{x} > a_{y}

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e_{x} < e_{y}

and

a_{x} < a_{y}

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Solution

The correct option is C $e_{x} > e_{y}$ and $a_{x} > a_{y}$
Rate of cooling $(- \frac{d T}{d t}) \propto emissivity (e)$
From graph, ${(- \frac{d T}{d t})}_{x} > {(- \frac{d T}{d t})}_{y} \Rightarrow e_{x} > e_{y}$
Further emissivity $(e) \propto$ Absorptive power $(a) \Rightarrow a_{x} > a_{y}$
( $∵$ good absorbers are good emitters)

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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 ( $e$ ) and absorptivity ( $a$ ) power of the two bodies. (IIT JEE 2003)