Thermal Properties of Matter - Result Question 11
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11. The power radiated by a black body is $P$ and it radiates maximum energy at wavelength, $\lambda_0$. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength $\frac{3}{4} \lambda_0$, the power radiated by it becomes $n P$. The value of $n$ is
======= ####11. The power radiated by a black body is $P$ and it radiates maximum energy at wavelength, $\lambda_0$. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength $\frac{3}{4} \lambda_0$, the power radiated by it becomes $n P$. The value of $n$ is
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/thermal-properties-of-matter/thermal-properties-of-matter—result-question-11.md (a) $\frac{3}{4}$
(b) $\frac{4}{3}$
(c) $\frac{81}{256}$
(d) $\frac{256}{81}$
[2018]
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Answer:
Correct Answer: 11. (d)
Solution:
- (d) From Wien’s law
$\lambda _{\text{max }} T=$ constant
i.e., $\lambda _{\max _1} T_1=\lambda _{\max _2} T_2$
$\Rightarrow \lambda_0 T=\frac{3 \lambda_0}{4} T^{\prime}$
$\Rightarrow T^{\prime}=\frac{4}{3} T$
Power radiated $P \propto T^{4}$
So, $\frac{P_2}{P_1}=n=(\frac{T^{\prime}}{T})^{4}=(\frac{4}{3})^{4}=\frac{256}{81}$
