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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
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which body 'B' radiates maximum energy is
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$6 \mu \mathrm{m}$
$\lambda_{2}-\lambda_{1}=4 \mu \mathrm{m} \quad \quad \mathrm{T}_{1}=3 \mathrm{~T}_{2}$
$\lambda_{1} \mathrm{~T}_{1}=\lambda_{2} \mathrm{~T}_{2} \quad \therefore \quad \lambda_{1} 3 \mathrm{~T}_{2}=\lambda_{2} \mathrm{~T}_{2}$
$\lambda_{2}=3 \lambda_{1}$
$3 \lambda_{1}-\lambda_{1}=4 \mu \mathrm{m}$
$2 \lambda_{1}=4 \mu \mathrm{m} \quad \therefore \quad \lambda_{2}=6 \mu \mathrm{m}$
$\lambda_{1}=2 \mu \mathrm{m} \quad$
$\lambda_{1} \mathrm{~T}_{1}=\lambda_{2} \mathrm{~T}_{2} \quad \therefore \quad \lambda_{1} 3 \mathrm{~T}_{2}=\lambda_{2} \mathrm{~T}_{2}$
$\lambda_{2}=3 \lambda_{1}$
$3 \lambda_{1}-\lambda_{1}=4 \mu \mathrm{m}$
$2 \lambda_{1}=4 \mu \mathrm{m} \quad \therefore \quad \lambda_{2}=6 \mu \mathrm{m}$
$\lambda_{1}=2 \mu \mathrm{m} \quad$
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