Q125 / 150 · MHT CET 2026 (15 April Shift 2) · Physics
A black rectangular surface of area A emits energy E per second at $127^\circ$C. If length and breadth is reduced to half of initial value and temperature is raised to $527^\circ$C then energy emitted becomes
- A$E$
- B$2E$
- Correct answerC$4E$
- D$8E$
Solution
Using Stefan-Boltzmann law, the energy emitted per second by a black body is given by $E = \sigma A T^4$.
Initial temperature $T_1 = 127^\circ\text{C} = 400\text{ K}$
Initial area $A_1 = A$
Initial energy emitted $E_1 = \sigma A (400)^4 = E$
Final temperature $T_2 = 527^\circ\text{C} = 800\text{ K}$
Since length and breadth are reduced to half, the new area $A_2 = \dfrac{L}{2} \times \dfrac{B}{2} = \dfrac{A}{4}$
Final energy emitted $E_2 = \sigma A_2 T_2^4$
$E_2 = \sigma \left(\dfrac{A}{4}\right) (800)^4$
$E_2 = \dfrac{1}{4} \times 16 \times \sigma A (400)^4$
$E_2 = 4 \sigma A (400)^4 = 4E$
Answer: $4E$
Initial temperature $T_1 = 127^\circ\text{C} = 400\text{ K}$
Initial area $A_1 = A$
Initial energy emitted $E_1 = \sigma A (400)^4 = E$
Final temperature $T_2 = 527^\circ\text{C} = 800\text{ K}$
Since length and breadth are reduced to half, the new area $A_2 = \dfrac{L}{2} \times \dfrac{B}{2} = \dfrac{A}{4}$
Final energy emitted $E_2 = \sigma A_2 T_2^4$
$E_2 = \sigma \left(\dfrac{A}{4}\right) (800)^4$
$E_2 = \dfrac{1}{4} \times 16 \times \sigma A (400)^4$
$E_2 = 4 \sigma A (400)^4 = 4E$
Answer: $4E$
Hence, the correct answer is option: 'C'.