Question

The rectangular surface of area $8cm\times 4cm$ of a black body at a temperature of ${127}^{0}C$ emits energy at the rate of $E$ per second. If both length and breadth of the surface are reduced to half of its initial value, and the temperature is raised to ${327}^{0}C$, then the rate of emission of energy will become :

• A. $\frac{3}{8}E$
• B. $\frac{81}{16}E$
• C. $\frac{9}{16}E$
• D. $\frac{81}{64}E$

Answer 1

The correct option is D $\frac{81}{64}E$

Let $A_1=32$ as given.

Let $A_2$ be the area when length and breadth are reduced by half. Thus the area will be $\frac{1}{4}$th of $A_1$

$\therefore A_2=\frac{1}{4}\ast 32=8$

Given $T_1={127}^{0}C={400}^{0}K$

Given $T_2={327}^{0}C={600}^{0}K$

From Stefan's law $E=\sigma A{T}^4$

$\therefore \frac{E_1}{E_2}=\frac{A_1{T_1}^4}{A_2{T_2}^4}=\frac{32\ast (400{)}^4}{8\ast (600{)}^4}=\frac{64}{81}$

$\therefore \frac{E_2}{E_1}=\frac{81}{64}$