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
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a
38E
b
8116E
c
916E
d
8164E
answer is D.
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Detailed Solution
Energy radiated by body per second Qt=AσT4 or Qt∝l×b×T4 ( Area =l×b)∴ E2E1=l2l1×b2b1×T2T14=l1/2l1×b1/2b1×6004004=12×12×324⇒E2=8164E