MHTCET 10 May 2024 Shift 1 Paper

Concept:According to Wien's displacement law, the wavelength of maximum emission is inversely proportional to the absolute temperature of a black body.Explanation:Let $\lambda_P$ and $\lambda_Q$ be the wavelengths at which bodies P and Q radiate maximum energy.Given that the wavelength difference is $3 \ \mu \mathrm{m}$, and since Q is cooler, its peak wavelength is longer:$$\lambda_Q - \lambda_P = 3 \ \mu \mathrm{m}$$From Wien's law, $\lambda T = \text{constant}$, so:$$\lambda_P T_P = \lambda_Q T_Q$$It is given that $T_P = 4 T_Q$. Substituting this gives:$$\lambda_P (4 T_Q) = \lambda_Q T_Q$$Cancelling $T_Q$ from both sides:$$4 \lambda_P = \lambda_Q$$Thus, $\lambda_Q = 4 \lambda_P$.Substitute this relation into the wavelength difference equation:$$4 \lambda_P - \lambda_P = 3 \ \mu \mathrm{m}$$$$3 \lambda_P = 3 \ \mu \mathrm{m}$$$$\lambda_P = 1 \ \mu \mathrm{m}$$Now, using $\lambda_Q = 4 \lambda_P$:$$\lambda_Q = 4 \times 1 \ \mu \mathrm{m} = 4 \ \mu \mathrm{m}$$Answer:The wavelength at which body Q radiates maximum energy is $4 \ \mu \mathrm{m}$.Therefore, the correct option is B. $4 \ \mu \mathrm{m}$.