Q169 / 200 · MHT CET - Medical 2026 (22 April Shift 1) · Physics
Two spherical black bodies have radii $R_1$ and $R_2$. Their surface temperatures are $T_1$ K and $T_2$ K respectively. If they radiate the same power, the ratio $\dfrac{R_1}{R_2}$ is
- A$\left(\dfrac{T_1}{T_2}\right)^4$
- B$\left(\dfrac{T_1}{T_2}\right)^2$
- C$\left(\dfrac{T_2}{T_1}\right)^4$
- Correct answerD$\left(\dfrac{T_2}{T_1}\right)^2$
Solution
According to the Stefan-Boltzmann law, the power radiated by a black body is given by $P = \sigma A T^4$.
For a spherical body of radius $R$, the surface area is $A = 4 \pi R^2$.
The radiated power is $P = \sigma (4 \pi R^2) T^4$.
Given that both bodies radiate the same power, $P_1 = P_2$.
$\sigma (4 \pi R_1^2) T_1^4 = \sigma (4 \pi R_2^2) T_2^4$
$R_1^2 T_1^4 = R_2^2 T_2^4$
$\dfrac{R_1^2}{R_2^2} = \dfrac{T_2^4}{T_1^4}$
$\dfrac{R_1}{R_2} = \left(\dfrac{T_2}{T_1}\right)^2$
Answer: $\left(\dfrac{T_2}{T_1}\right)^2$
For a spherical body of radius $R$, the surface area is $A = 4 \pi R^2$.
The radiated power is $P = \sigma (4 \pi R^2) T^4$.
Given that both bodies radiate the same power, $P_1 = P_2$.
$\sigma (4 \pi R_1^2) T_1^4 = \sigma (4 \pi R_2^2) T_2^4$
$R_1^2 T_1^4 = R_2^2 T_2^4$
$\dfrac{R_1^2}{R_2^2} = \dfrac{T_2^4}{T_1^4}$
$\dfrac{R_1}{R_2} = \left(\dfrac{T_2}{T_1}\right)^2$
Answer: $\left(\dfrac{T_2}{T_1}\right)^2$
Hence, the correct answer is option: 'D'.