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 \(\left(\frac{1}{3}\right)^{\mathrm{rd}}\) of initial value and temperature is raised to \(327^{\circ} \mathrm{C}\)
then energy emitted per second becomes
- A \(2 \mathrm{E} / \mathrm{9}\)
- B \(\mathrm{E} / \mathrm{9}\)
- C \(16 \mathrm{E} / \mathrm{9}\)
- D \(4 \mathrm{E} / \mathrm{9}\)
Answer & Solution
Correct Answer
(C) \(16 \mathrm{E} / \mathrm{9}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E}=\sigma \mathrm{AT}^{4}\)
\(\therefore \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}} \cdot\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}\)
\(\mathrm{~A}_{2}=\frac{\mathrm{A}_{1}}{9} \quad \therefore \frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}=\frac{1}{9}\)
\(\mathrm{~T}_{1}=27+273=300 \mathrm{k}\)
\(\mathrm{T}_{2}=327+273=600 \mathrm{k}\)
\(\therefore \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{600}{300}=2\)
\(\therefore \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{1}{9} \times(2)^{4}=\frac{16}{9}\)
\(\therefore \mathrm{E}_{2}=\frac{16}{9} \mathrm{E}_{1}=\frac{16}{9} \mathrm{E}\)
\(\therefore \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}} \cdot\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}\)
\(\mathrm{~A}_{2}=\frac{\mathrm{A}_{1}}{9} \quad \therefore \frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}=\frac{1}{9}\)
\(\mathrm{~T}_{1}=27+273=300 \mathrm{k}\)
\(\mathrm{T}_{2}=327+273=600 \mathrm{k}\)
\(\therefore \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{600}{300}=2\)
\(\therefore \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{1}{9} \times(2)^{4}=\frac{16}{9}\)
\(\therefore \mathrm{E}_{2}=\frac{16}{9} \mathrm{E}_{1}=\frac{16}{9} \mathrm{E}\)
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