Two bodies ' X ' and ' Y ' at temperatures ' $\mathrm{T}_1$ ' K and ' $\mathrm{T}_2$ ' K respectively have…

Two bodies ' X ' and ' Y ' at temperatures ' $\mathrm{T}_1$ ' K and ' $\mathrm{T}_2$ ' K respectively have the same dimensions. If their emissive powers are same, the relation between their temperatures is
  1. $\frac{T_1}{\mathrm{~T}_2}=\frac{1}{3}$
  2. $\frac{T_1}{T_2}=\frac{81}{1}$
  3. $\frac{T_1}{T_2}=\frac{3^{\frac{1}{4}}}{1}$
  4. $\frac{T_1}{T_2}=\frac{9^{\frac{1}{4}}}{1}$

Solution

Given: - Two bodies $X$ and $Y$ at temperatures $T_1$ and $T_2$. - Same dimensions. - Same emissive power.
Stefan-Boltzmann Law: $E=\sigma e A T^4$
Here: - E: Emissive power, - $\sigma$ : Stefan-Boltzmann constant, - e: Emissivity, - A: Surface area, - T: Absolute temperature.
For the same emissive power: $T_1^4=T_2^4$
If $T_1 / T_2=1 / 3$, then: $T_1^4 / T_2^4=1 / 81 \quad \Rightarrow \quad T_1 / T_2=\frac{1}{3} .$
Answer: $T_1 / T_2=\frac{1}{3}$, Option 1.

Asked in: MHT CET 2024 (04 May Shift 2)

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