The graph, shown in the 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 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 B $E_{x} > E_{y}$ and $a_{x} > a_{y}$
The graph shows that for the same temperature difference $(T_{2} - T_{1})$ , less time is taken for $x$ . This means the emissivity is more for $x$ . According to Kirchhoff's law, a good emitter is a good absorber as well.

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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)