Question

If a black body radiation enclosure expands, so that its volume becomes 64 times, the temperature of the enclosure becomes:

• A. One fourth
• B. Four times
• C. 16 times
• D. 8 times

Hint:

For black body radiation, during adiabatic expansion, temperature follows \( T \propto V^{-1/3} \).

Answer 1

The Correct Option is A

For an **adiabatic expansion of a black body**, the relation between temperature \( T \) and volume \( V \) is:
\[ T V^{1/3} = \text{constant} \] If the volume increases 64 times: \[ T_2 = T_1 \times \left( \frac{1}{64} \right)^{1/3} \] \[ T_2 = T_1 \times \frac{1}{4} \] Thus, the correct answer is:
\[ Option 1: \frac{T_1}{4} \text{ (One fourth)}. \]