NTA Abhyas JEE Main2020PhysicsThermal Properties of MatterPractice
A black rectangular surface of area A emits energy E per second at 27 ° C . If length and breadth are reduced to 1 3 rd of initial value and temperature is raised to 327 ° C , then energy emitted per second becomes
Options
- A4 E 9
- B7 E 9
- C10 E 9
- D16 E 9
Correct answer
D. 16 E 9
Step-by-step solution
E = e σ . A T 4 - T 0 4 a n d A = l b When l and b changes to l 3 and b 3 A → A 9 E ′ E = A ′ A 327 + 273 4 27 + 273 4 ∴ E ′ E = 1 9 600 300 4 ∴ E ′ = 1 9 × 2 4 × E E ′ = 16 E 9