Question

The rectangular surface of area $8$ cm $\times$ $4$ cm of a black body at a temperature of ${127}^0$C emits 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 $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 $\frac{81}{64}E$

$E\propto A{T}^4$
Let, $E=kA{T}^4$
$E=k(8\times 4){\left(400\right)}^4=8192\times {10}^8$

$E_1=k(\frac{8}{2}\times \frac{4}{2}){\left(600\right)}^4=10368\times {10}^8$

$\frac{E_1}{E}=\frac{10368}{8192}=\frac{81}{64}$