MCQ
The temperature of the body is increased from $-73^o C\, to \,327^o C$, the ratio of energy emitted per second is :
  • A
    $1:3$
  • $1:81$
  • C
    $1:27$
  • D
    $ 1:9$

Answer

Correct option: B.
$1:81$
b
${T_1} =  - 73 + 273 = 200K$

${T_2} = 327 + 273 = 600\;K$

Energy per second $P\left( { = \frac{Q}{t}} \right) \propto {T^4}$

$\Rightarrow \frac{{{P_2}}}{{{P_1}}} = {\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^4} = {\left( {\frac{{600}}{{200}}} \right)^2} = 81$

$\frac{{{P_1}}}{{{P_2}}} = \frac{1}{{81}}$

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