The temperature of a body is increased from -73^ C to 327^ C . Then the ratio of emissive power is

Correct answer: D. 81 Radiation from a body =R R=e σ T^4 × time × area, where e= emissive power e T^4= constant or e=( constant T^4 ) or (e_1)/(e_2)=((T_2)/(T_1))^4 or (e_1)/(e_2)=((273+327)/(273-73))^4 =((600)/(200))^4=((3)^4)/(1)=(81)/(1) or (e_1)/(e_2)=(81)/(1)

AIIMS 2016

Exam: AIIMS
Year: 2016
Subject: Physics
Chapter: Thermal Properties of Matter

AIIMS 2016 · Physics · Thermal Properties of Matter

The temperature of a body is increased from -73^ C to 327^ C . Then...

The temperature of a body is increased from

-73^{\circ} \mathrm{C}

327^{\circ} \mathrm{C}

. Then the ratio of emissive power is

Options

A $1 / 9$
B $1 / 27$
C $27$
D $81$

Correct answer

D. $81$

Step-by-step solution

Radiation from a body

=R R=e \sigma T^4 \times

time

\times

area, where

e=

emissive power

\therefore e T^4=

constant or

e=\left(\frac{\text{ constant }}{T^4}\right) or \frac{e_1}{e_2}=\left(\frac{T_2}{T_1}\right)^4 or \frac{e_1}{e_2}=\left(\frac{273+327}{273-73}\right)^4 =\left(\frac{600}{200}\right)^4=\frac{(3)^4}{1}=\frac{81}{1} or \frac{e_1}{e_2}=\frac{81}{1}

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