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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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Two spheres of same material and radii 4 m and 1 m at temperature 1000 K and 2000 K . Find the ratio of the energies radiated by them in one second .

$(a)\;1:1 \qquad(b)\;1 : 2\qquad(c)\; 1 : 4\qquad(d)\;4 : 1$

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1 Answer

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Answer : $ \;1 : 1$
Explanation :
$\large\frac{stefan's\;law}{energy\;radiated\;per\;second}\;U_{4}=e AT^{4}$
$\large\frac{U_{1}}{U_{2}}=\large\frac{A_{1}T_{1}^{4}}{A_{2}T_{2}^{4}}=\large\frac{r_{1}^{2}}{r_{2}^2}\;.\large\frac{T_{1}}{T_{2}}=\large\frac{16}{1}\;\times (\large\frac{1000}{2000})^4$
$= 1: 1$
answered Mar 26, 2014 by yamini.v
 

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