Question

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

• 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 $\frac{81}{64}E$

By $Stefa{n}^{\prime }s$ Law

$Power=\sigma A{T}^4$

$A=length\times breadth$

$\frac{P_2}{E}=\frac{l_2{b}_2{T}_2^4}{l_1{b}_1{T}_1^4}$

Here$\frac{l_2}{l_1}=\frac{1}{2}$ $\frac{b_2}{b_1}=\frac{1}{2}$ $\frac{T_2}{T_1}=\frac{3}{2}$

$⟹P_2=\frac{81}{64}E$