Question

The graph shown in the figure, represents the fall of temperature $\left(T\right)$ of two bodies $x$ and $y$ having the same surface area, with time $\left(t\right)$ due to radiation. Choose the correct relationship between emissive power $\left(E\right)$ and absorptive power $\left(a\right)$ of the two bodies.

• A. $E_x>E_y;a_x<a_y$
• B. $E_x<E_y;a_x>a_y$
• C. $E_x>E_y;a_x>a_y$
• D. $E_x<E_y;a_x<a_y$

Answer 1

The correct option is D $E_x<E_y;a_x<a_y$


Rate of cooling$(\frac{-dT}{dt})\propto \text{emissivity}\left(e\right)\propto \text{Emissive power}\left(E\right)$
From the graph,
Temperature of $y$ falls more rapidly with time than that of $x$.
$\Rightarrow {\theta }_2>{\theta }_1$
And slope of graph represents the rate of cooling.
So, $(\frac{-dT}{dt}{)}_y>(\frac{-dT}{dt}{)}_x$
Hence, $E_y>E_x$

As per Kirchoff's law, a good emitter is also a good absorber.
So, $a_y>a_x$
Thus, option (d) is the correct option.