A rectangular black body of temperature 27 has surface area 6 cm × 3 cm and rate of radiation E. If its temperature is increased by 300 and its linear dimensions are reduced to (1/3)rd of their initial value then the rate of radiation is (a) 8/9E (b) 16/9E (c) 4/9E (d) 32/9E

Answer: (b). P = σ A T⁴, so P ∝ A T⁴. T₁ = 300 K, T₂ = 600 K: factor 2⁴ = 16. Linear dimensions become 1/3, so area becomes 1/9. P' = 16/9E, so the answer is (b).

A rectangular black body of temperature 27 has surface area 6 cm × 3… | MHT CET 2026 (12 May, Shift 2)

Q26MHT CET 2026 · 12 MAY · SHIFT 2Medium

Thermal Radiation: Black Body, Stefan & Wien

A rectangular black body of temperature $27^\circ\text{C}$ has surface area $6$ cm $\times 3$ cm and rate of radiation $E$ . If its temperature is increased by $300^\circ\text{C}$ and its linear dimensions are reduced to $\left(\frac{1}{3}\right)^{rd}$ of their initial value then the rate of radiation is

Show answer and solution

Answer: (b)

$P = \sigma A T^4$ , so $P \propto A T^4$ .
$T_1 = 300\text{ K}$ , $T_2 = 600\text{ K}$ : factor $2^4 = 16$ .
Linear dimensions become $\frac{1}{3}$ , so area becomes $\frac{1}{9}$ .
$P' = \frac{16}{9}E$ , so the answer is (b).

Report a mistake

Thermal Radiation: Black Body, Stefan & Wien question from MHT CET 2026 (12 May, Shift 2)

More questions on Thermal Radiation: Black Body, Stefan & Wien