A) 10%
B) 40%
C) 46%
D) 100%
Correct Answer: C
Solution :
Emitted radiation\[E\propto {{T}^{4}}\] \[\frac{{{E}_{1}}}{{{E}_{2}}}={{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{4}}\] \[\begin{align} & ^{=}{{\left( \frac{T}{T+\frac{T\times 10}{100}} \right)}^{4}} \\ & ={{\left( \frac{10}{11} \right)}^{4}}=\frac{10000}{14641} \\ \end{align}\] Increase in emitted radiation \[\begin{align} & =\frac{{{E}_{2}}-{{E}_{1}}}{{{E}_{1}}}\times 100 \\ & =\frac{14641-10000}{100}=46%\,(App) \\ \end{align}\]
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