MHT CET · Physics · Thermal Properties of Matter
Three discs \(x, y\) and \(z\) having radii \(2 \mathrm{~m}, 2 \mathrm{~m}\) and \(6 \mathrm{~m}\) respectively are coated on outer surfaces. The wavelength corresponding to maximum are power radiated by them respectively then
- A \(P_y\) is maximum
- B \(P_{\mathrm{z}}\) is maximum
- C \(P_x=P_y=P_z\)
- D \(P_x\) is maximum
Answer & Solution
Correct Answer
(A) \(P_y\) is maximum
Step-by-step Solution
Detailed explanation
As we know that from Wien's displacement law \(T \propto \frac{1}{\lambda_{\max }}\)
From Stefan's Law, the emitted power is proportional to temperature raised to power four and directly poroportional to the surface area of the blackbody:
\(P \propto A T^4\)
So, \(P \propto \frac{A}{\lambda_{\max }^4}\) where \(A=4 \pi r^2\) is the surface area of the body with spherical shape of radius \(r\).
Therefore,
\(P_x: P_y: P_z=\frac{2^2}{3^4}: \frac{2^2}{4^4}: \frac{6^2}{5^4}=\frac{4}{81}: \frac{4}{16}: \frac{36}{625}=\frac{32}{648}: \frac{156}{624}: \frac{36}{625}\)
\(\Rightarrow P_y>P_x>P_z\)
From Stefan's Law, the emitted power is proportional to temperature raised to power four and directly poroportional to the surface area of the blackbody:
\(P \propto A T^4\)
So, \(P \propto \frac{A}{\lambda_{\max }^4}\) where \(A=4 \pi r^2\) is the surface area of the body with spherical shape of radius \(r\).
Therefore,
\(P_x: P_y: P_z=\frac{2^2}{3^4}: \frac{2^2}{4^4}: \frac{6^2}{5^4}=\frac{4}{81}: \frac{4}{16}: \frac{36}{625}=\frac{32}{648}: \frac{156}{624}: \frac{36}{625}\)
\(\Rightarrow P_y>P_x>P_z\)
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