MHT CET · Physics · Thermal Properties of Matter
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
- A \(\frac{T_1}{\mathrm{~T}_2}=\frac{1}{3}\)
- B \(\frac{T_1}{T_2}=\frac{81}{1}\)
- C \(\frac{T_1}{T_2}=\frac{3^{\frac{1}{4}}}{1}\)
- D \(\frac{T_1}{T_2}=\frac{9^{\frac{1}{4}}}{1}\)
Answer & Solution
Correct Answer
(A) \(\frac{T_1}{\mathrm{~T}_2}=\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
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.
- 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.
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