Question

The rectangular surface of area $8cm\times 4cm$ of a black body at a temperature of ${127}^{o}C$ emitts energy at the rate of $E$ per second. IF the length and breadth of the surface are each reduced to half of the initial value and the temperature is raised to ${327}^{o}C$, 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 $\frac{81}{64}E$

The radiant power of black body is given by

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

Here, $A_2=\frac{A_1}{4},{\ell }_2=\frac{{\ell }_1}{2},b_2=\frac{b_1}{2}{T}_1=127℃=273+127=400K{T}_1=327℃=273+327=600K\frac{P_2}{E}=\frac{1}{4}\times {\left(\frac{600}{400}\right)}^4=\frac{1}{4}\times \frac{81}{16}=\frac{81}{64}\therefore P_2=\frac{81E}{64}$