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(A) According to Stefan-Boltzmann law,the rate of heat loss is given by $\frac{dQ}{dt} = e \sigma A (T^4 - T_0^4)$.Since $\frac{dQ}{dt} = -ms \frac{dT

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$e_x > e_y$ and $a_x > a_y$

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(A) According to Stefan-Boltzmann law,the rate of heat loss is given by $\frac{dQ}{dt} = e \sigma A (T^4 - T_0^4)$.Since $\frac{dQ}{dt} = -ms \frac{dT}{dt}$,we have $-ms \frac{dT}{dt} = e \sigma A (T^4 - T_0^4)$.Thus,the rate of cooling $\left( -\frac{dT}{dt} \right) = \frac{e \sigma A}{ms} (T^4 - T_0^4)$.For bodies of the same surface area and mass,the rate of cooling is directly proportional to the emissivity $(e)$,i.e.,$\left( -\frac{dT}{dt} \right) \propto e$.From the graph,the slope of the curve for body $x$ is steeper than that for body $y$ at any given temperature,which means $\left( -\frac{dT}{dt} \right)_x > \left( -\frac{dT}{dt} \right)_y$.Therefore,$e_x > e_y$.According to Kirchhoff's law of radiation,for any body,the emissivity $(e)$ is equal to its absorptive power $(a)$ at a given wavelength and temperature,i.e.,$e = a$.Thus,$e_x > e_y$ implies $a_x > a_y$.

The graph shown in the adjacent diagram represents the variation of temperature $(T)$ of two bodies,$x$ and $y$,having the same surface area,with time $(t)$ due to the emission of radiation. Find the correct relation between their emissivities $(e)$ and absorptive powers $(a)$.

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