A) \[T={{\left( \frac{P}{A\varepsilon \sigma } \right)}^{4}}\]
B) \[T=\left( \frac{P}{A\varepsilon \sigma } \right)\]
C) \[T={{\left( \frac{A\varepsilon \sigma }{P} \right)}^{\frac{1}{4}}}\]
D) \[T={{\left( \frac{P}{A\varepsilon \sigma } \right)}^{\frac{1}{4}}}\]
Correct Answer: D
Solution :
Radiant power P is denned as energy radiated per unit area. ie, \[P=\frac{Q}{t}=A\varepsilon \sigma {{T}^{4}}\] \[\Rightarrow \] \[T={{\left( \frac{P}{A\varepsilon \sigma } \right)}^{1/4}}\]You need to login to perform this action.
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