Two metal spheres have radii r and 2r and the | vedclass

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Question

(A) According to Wien's displacement law, $\lambda_m T = 	ext{constant} (b)$.For the first sphere: $\lambda T_1 = b \implies T_1 = \frac{b}{\lambda}$

Vedclass · Accepted Answer

$4: 1$

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(A) According to Wien's displacement law, $\lambda_m T = 	ext{constant} (b)$.For the first sphere: $\lambda T_1 = b \implies T_1 = \frac{b}{\lambda}$.For the second sphere: $(2\lambda) T_2 = b \implies T_2 = \frac{b}{2\lambda} = \frac{T_1}{2}$.The radiant energy emitted per second is the power $P = \sigma A T^4 = \sigma (4\pi R^2) T^4$.Thus, $P \propto R^2 T^4$.Calculating the ratio $\frac{P_1}{P_2} = \left( \frac{r_1}{r_2} ight)^2 \left( \frac{T_1}{T_2} ight)^4$.Substituting the values: $\frac{P_1}{P_2} = \left( \frac{r}{2r} ight)^2 \left( \frac{T_1}{T_1/2} ight)^4 = \left( \frac{1}{2} ight)^2 (2)^4 = \frac{1}{4} 	imes 16 = 4$.Therefore, the ratio is $4: 1$.

Two metal spheres have radii $r$ and $2r$ and they emit thermal radiation with maximum intensities at wavelengths $\lambda$ and $2\lambda$ respectively. The respective ratio of the radiant energy emitted by them per second will be .........

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