Question

Two spherical black bodies of radius \(\mathrm{R}_1\) and \(\mathrm{R}_2\) with surface temperatures \(\mathrm{T}_1\) and \(\mathrm{T}_2\) respectively. If \(T_1=2 T_2\) and they radiate the same power then relation between \(R_1\) and \(R_2\) is

• A. \(R_{2}=8R_{1}\)
• B. \(\mathrm{R}_{2}=\mathrm{R}_{1}\)
• C. \(R_{2}=2R_{1}\)
• D. \(R_{2}=4R_{1}\)

Hint:

Radiated power is proportional to the surface area times T to the fourth power.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Stefan's law gives \(P=\sigma\,4\pi R^{2}T^{4}\) for a black body.

Step 2: Equal power
\(R_{1}^{2}T_{1}^{4}=R_{2}^{2}T_{2}^{4}\).

Step 3: Substitute
With \(T_{1}=2T_{2}\): \(R_{1}^{2}\times16T_{2}^{4}=R_{2}^{2}T_{2}^{4}\Rightarrow R_{2}^{2}=16R_{1}^{2}\).

Step 4: Result
\(R_{2}=4R_{1}\).

Final Answer:
The radius of the cooler body is four times the other. \[ \boxed{\text{(D) }R_{2}=4R_{1}} \]