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Q.

The rectangular surface of area 8 cm x 4 cm of a black body at a temperature of 127^oC 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 to 327^oC, the rate of emission of energy will become

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a

$\frac{81}{16} E$

b

$\frac{81}{64} E$

c

$\frac{3}{8} E$

d

$\frac{9}{16} E$

answer is D.

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Detailed Solution

Energy radiated by body per second $\frac{Q}{t} = {AσT}^{4}$

$or \frac{Q}{t} \propto l \times b \times T^{4} (Area = l \times b)$

$\begin{matrix} ∴ \frac{E_{2}}{E_{1}} = \frac{l_{2}}{l_{1}} \times \frac{b_{2}}{b_{1}} \times {(\frac{T_{2}}{T_{1}})}^{4} = \frac{(l_{1} / 2)}{l_{1}} \times \frac{(b_{1} / 2)}{b_{1}} \times {(\frac{600}{400})}^{4} \\ = \frac{1}{2} \times \frac{1}{2} \times {(\frac{3}{2})}^{4} \Rightarrow E_{2} = \frac{81}{64} E \end{matrix}$

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The rectangular surface of area 8 cm x 4 cm of a black body at a temperature of 127oC 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 to 327oC, the rate of emission of energy will become