Q191 / 200 · MHT CET - Medical 2026 (24 April Shift 1) · Physics
Three spheres P, Q and R, having radii of 2 m, 3 m and 4 m respectively, are coated with lamp-black on their outer surfaces. The wavelengths corresponding to maximum intensity are 200 nm, 300 nm and 400 nm respectively. If the powers radiated by them are $P_P$, $P_Q$ and $P_R$ respectively, then
- Correct answerA$P_P$ is maximum
- B$P_Q$ is maximum
- C$P_R$ is maximum
- D$P_P=P_Q=P_R$
Solution
Using Wien's displacement law, $\lambda_{max} T = b \Rightarrow T = \dfrac{b}{\lambda_{max}}$.
Using Stefan-Boltzmann law, the power radiated by a black body is $P = \sigma A T^4 = \sigma (4\pi r^2) T^4$.
Substituting $T$, we get $P = \sigma (4\pi r^2) \left(\dfrac{b}{\lambda_{max}}\right)^4 \Rightarrow P \propto \dfrac{r^2}{\lambda_{max}^4}$.
For sphere P: $P_P \propto \dfrac{2^2}{(200)^4} \propto \dfrac{4}{16 \times 10^8} \propto \dfrac{1}{4 \times 10^8}$
For sphere Q: $P_Q \propto \dfrac{3^2}{(300)^4} \propto \dfrac{9}{81 \times 10^8} \propto \dfrac{1}{9 \times 10^8}$
For sphere R: $P_R \propto \dfrac{4^2}{(400)^4} \propto \dfrac{16}{256 \times 10^8} \propto \dfrac{1}{16 \times 10^8}$
Comparing the values, $P_P > P_Q > P_R$.
Thus, $P_P$ is maximum.
Answer: $P_P$ is maximum
Using Stefan-Boltzmann law, the power radiated by a black body is $P = \sigma A T^4 = \sigma (4\pi r^2) T^4$.
Substituting $T$, we get $P = \sigma (4\pi r^2) \left(\dfrac{b}{\lambda_{max}}\right)^4 \Rightarrow P \propto \dfrac{r^2}{\lambda_{max}^4}$.
For sphere P: $P_P \propto \dfrac{2^2}{(200)^4} \propto \dfrac{4}{16 \times 10^8} \propto \dfrac{1}{4 \times 10^8}$
For sphere Q: $P_Q \propto \dfrac{3^2}{(300)^4} \propto \dfrac{9}{81 \times 10^8} \propto \dfrac{1}{9 \times 10^8}$
For sphere R: $P_R \propto \dfrac{4^2}{(400)^4} \propto \dfrac{16}{256 \times 10^8} \propto \dfrac{1}{16 \times 10^8}$
Comparing the values, $P_P > P_Q > P_R$.
Thus, $P_P$ is maximum.
Answer: $P_P$ is maximum
Hence, the correct answer is option: 'A'.