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 emissive power $(E)$ and absorptive power $(a)$ 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 $a_{x} > a_{y}$
Power Emitted $P = ϵ σ A (T^{4} - {T_{0}}^{4})$

so the more the power radiated will show a more rapid decrease in temperature which is more for body x than y.

so, $ϵ_{x} > ϵ_{y}$ .

since the body which radiates more absorbs more so similarly, $a_{x} > a_{y}$

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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)