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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Similar questions

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

(e)

and absorptivity

(a)

of the two bodies.

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)