Q142 / 150 · MHT CET 2026 (20 April Shift 2) · Physics
Two stars A and B radiate maximum energy at wavelength $3.9 \times 10^{-7}$ m and $5.2 \times 10^{-7}$ m respectively. The ratio of the temperature of star A to that of star B will be
- A$2 : 3$
- B$3 : 2$
- C$3 : 4$
- Correct answerD$4 : 3$
Solution
Using Wien's displacement law, $\lambda_{m} T = b$ (constant).
Therefore, $\lambda_{A} T_{A} = \lambda_{B} T_{B}$
$\dfrac{T_{A}}{T_{B}} = \dfrac{\lambda_{B}}{\lambda_{A}}$
Substituting the given values:
$\dfrac{T_{A}}{T_{B}} = \dfrac{5.2 \times 10^{-7}}{3.9 \times 10^{-7}}$
$\dfrac{T_{A}}{T_{B}} = \dfrac{52}{39} = \dfrac{4}{3}$
The ratio of the temperature of star A to that of star B is $4 : 3$.
Answer: $4 : 3$
Therefore, $\lambda_{A} T_{A} = \lambda_{B} T_{B}$
$\dfrac{T_{A}}{T_{B}} = \dfrac{\lambda_{B}}{\lambda_{A}}$
Substituting the given values:
$\dfrac{T_{A}}{T_{B}} = \dfrac{5.2 \times 10^{-7}}{3.9 \times 10^{-7}}$
$\dfrac{T_{A}}{T_{B}} = \dfrac{52}{39} = \dfrac{4}{3}$
The ratio of the temperature of star A to that of star B is $4 : 3$.
Answer: $4 : 3$
Hence, the correct answer is option: 'D'.