MHT CET · Physics · Thermal Properties of Matter
Ordinary bodies 'A' and 'B' radiate maximum energy with wavelength difference \(4 \mu \mathrm{m}\). The absolute temperature of body 'A' is 3 times that of ' \(\mathrm{B}\) '. The wavelength at
which body 'B' radiates maximum energy is
- A \(12 \mu \mathrm{m}\)
- B \(6 \mu \mathrm{m}\)
- C \(4 \mu \mathrm{m}\)
- D \(8 \mu \mathrm{m} .\)
Answer & Solution
Correct Answer
(B) \(6 \mu \mathrm{m}\)
Step-by-step Solution
Detailed explanation
\(\lambda_{2}-\lambda_{1}=4 \mu \mathrm{m} \quad \quad \mathrm{T}_{1}=3 \mathrm{~T}_{2}\)
\(\lambda_{1} \mathrm{~T}_{1}=\lambda_{2} \mathrm{~T}_{2} \quad \therefore \quad \lambda_{1} 3 \mathrm{~T}_{2}=\lambda_{2} \mathrm{~T}_{2}\)
\(\lambda_{2}=3 \lambda_{1}\)
\(3 \lambda_{1}-\lambda_{1}=4 \mu \mathrm{m}\)
\(2 \lambda_{1}=4 \mu \mathrm{m} \quad \therefore \quad \lambda_{2}=6 \mu \mathrm{m}\)
\(\lambda_{1}=2 \mu \mathrm{m} \quad\)
\(\lambda_{1} \mathrm{~T}_{1}=\lambda_{2} \mathrm{~T}_{2} \quad \therefore \quad \lambda_{1} 3 \mathrm{~T}_{2}=\lambda_{2} \mathrm{~T}_{2}\)
\(\lambda_{2}=3 \lambda_{1}\)
\(3 \lambda_{1}-\lambda_{1}=4 \mu \mathrm{m}\)
\(2 \lambda_{1}=4 \mu \mathrm{m} \quad \therefore \quad \lambda_{2}=6 \mu \mathrm{m}\)
\(\lambda_{1}=2 \mu \mathrm{m} \quad\)
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