MHT CET · Physics · Thermal Properties of Matter
A black rectangular surface of area ' \(a\) ' emits energy ' \(E\) ' per second at \(27^{\circ} \mathrm{C}\). If length and breadth is reduced to \(\left(\frac{1}{3}\right)^{\text {rd }}\) of initial value and temperature is raised to \(327^{\circ} \mathrm{C}\) then energy emitted per second becomes
- A \(\frac{16 \mathrm{E}}{9}\)
- B \(\frac{8 \mathrm{E}}{9}\)
- C \(\frac{4 \mathrm{E}}{9}\)
- D \(\frac{E}{9}\)
Answer & Solution
Correct Answer
(A) \(\frac{16 \mathrm{E}}{9}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E}=\mathrm{e} \sigma \mathrm{AT}^4 \text { and } \mathrm{A}=\ell \mathrm{b}\)
\(\frac{E^{\prime}}{E}=\frac{A^{\prime}(327+273)^4}{A(27+273)^4}\)
\(\frac{E^{\prime}}{E}=\frac{1}{9}\left(\frac{600}{300}\right)^4\)
\(
\begin{aligned}
& \therefore \mathrm{E}^{\prime}=\frac{1}{9} \times(2)^4 \times \mathrm{E} \\
& \Rightarrow \mathrm{E}^{\prime}=\frac{16 \mathrm{E}}{9}
\end{aligned}
\)
\(\frac{E^{\prime}}{E}=\frac{A^{\prime}(327+273)^4}{A(27+273)^4}\)
\(\frac{E^{\prime}}{E}=\frac{1}{9}\left(\frac{600}{300}\right)^4\)
\(
\begin{aligned}
& \therefore \mathrm{E}^{\prime}=\frac{1}{9} \times(2)^4 \times \mathrm{E} \\
& \Rightarrow \mathrm{E}^{\prime}=\frac{16 \mathrm{E}}{9}
\end{aligned}
\)
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