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Q.

The rectangular surface of area 8cm × 4 cm of a black body at a temperature of 127^oC emits energy at rate of E per second. If the length and breadth are reduced to half of its initial value and the temperature is raised to 327^oC, the rate of emission of energy will be

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a

$(\frac{3 E}{8})$

b

$(\frac{9 E}{16})$

c

$(\frac{81 E}{16})$

d

$(\frac{81 E}{64})$

answer is D.

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Detailed Solution

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l₁ = 8cm, b₁ = 4cm ; l₂ = 4cm, b₂ = 2cm
T₁ = 127^oC = 400k, T₂ = 327^o C = 600k
$\frac{E_{2}}{E_{1}} = {(\frac{T_{2}}{T_{1}})}^{4} (\frac{l_{2} b_{2}}{l_{1} b_{1}}) \frac{E_{2}}{E} = {(\frac{600}{400})}^{4} (\frac{4 x 2}{8 x 4}) = E_{2} = \frac{81}{64} E$

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The rectangular surface of area 8cm × 4 cm of a black body at a temperature of 127oC emits energy at rate of E per second. If the length and breadth are reduced to half of its initial value and the temperature is raised to 327oC, the rate of emission of energy will be