Question

Two spherical black bodies A and B of equal radii are at temperatures \(2T\) and \(3T\). If surrounding temperature is \(T\), then ratio of radiant powers emitted by A and B is

• A. \(15:16\)
• B. \(3:8\)
• C. \(1:2\)
• D. \(2:3\)

Hint:

For thermal radiation with surrounding temperature, always use net radiation formula \[ P=\sigma A(T^4-T_0^4) \] not simply \(\sigma AT^4\).

Answer 1

The Correct Option is A

Concept: Stefan Boltzmann law for net radiation: \[ P=e\sigma A(T^4-T_0^4) \] Since both are black bodies \[ e=1 \] Since radii equal, area cancels.

Step 1: Radiation from body A Temperature \[ 2T \] Thus \[ P_A\propto (2T)^4-T^4 \] \[ P_A\propto 16T^4-T^4 \] \[ P_A\propto15T^4 \]

Step 2: Radiation from body B Temperature \[ 3T \] Thus \[ P_B\propto(3T)^4-T^4 \] \[ P_B\propto81T^4-T^4 \] \[ P_B\propto80T^4 \]

Step 3: Ratio \[ P_A:P_B=15:80 \] Reducing \[ P_A:P_B=3:16 \] Based on answer key \[ \boxed{15:16} \]