Q143 / 150 · MHT CET 2022 (08 Aug Shift 1) · Physics

Three black discs $x, y, z$ have radii $1 \mathrm{~m}, 2 \mathrm{~m}$ and $3 \mathrm{~m}$ respectively. The wavelength corresponding to maximum intensity are $200 \mathrm{~nm}$, $300 \mathrm{~nm}$ and $400 \mathrm{~nm}$ respectively. The relation between emissive power $E_x, E_y$ and $E_z$ is

  • Correct answer
    A
    $E_x$ is maximum
  • B
    $E_y$ is maximum
  • C
    $E_z$ is maximum
  • D
    $E_x=E_y=E_z$

Solution

According to the Stefan's Law: $\begin{aligned} & E=\frac{\sigma A T^4}{A} \\ & \Rightarrow E \propto T^4 \end{aligned}$ According to the Wein's Law: $\begin{aligned} & T=\frac{k}{\lambda_m} \\ & \therefore E \propto \frac{A}{\lambda^4} \\ & E_x: E_y: E_z=\frac{1}{\lambda_x^4}: \frac{1}{\lambda^4}: \frac{1}{\lambda^4}=\frac{1}{200^4}: \frac{1}{300^4}: \frac{1}{400^4} \end{aligned}$ The relation between emissive power $E_x, E_y$ and $E_z$ is as follows: $E x>E y>E z$

Hence, the correct answer is option: 'A'.

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