Question

The rectangular surface of area 8 cm × 4 cm of a black body at temperature 127°C emits energy E per second. If the length and breadth are reduced to half of the initial value and the temperature is raised to 327°C, the rate of emission of energy becomes
(a) $\frac{3}{8}E$
(b) $\frac{81}{16}E$
(c) $\frac{9}{16}E$
(d) $\frac{81}{64}E$

Answer 1

Answer:

D) 81/64E

Explanation:

Area of the rectangle = 8 cm × 4 cm (Given)

Temperature of the black body = 127°C (Given)

Energy emission = E/sec (Given)

(Q)Blackbody = AσT4t Þ Q/t∝P = AσT4

Since, length and the breadth are halved so area will become one fourth,

Rise in temperature = 327°C

Þ P1/P2 = A1/A2 × (T1/T2)'4

A1 (A1/4)×(273+327/273+127)

Þ P2 = 81/64E

Thus, the rate of emission of energy becomes 81/64E.