| ⇦ |
| ⇨ |
If the radius of star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy protection is Q?
Options
(a) Q / 4 πR²σ
(b) (Q / 4 πR²σ)⁻¹/²
(c) (4 πR²Q / σ)¹/⁴
(d) (Q / 4 πR²σ)¹/⁴
Correct Answer:
(Q / 4 πR²σ)¹/⁴
Explanation:
Stefan’s law for black body radiation Q = σe AT⁴
T = [Q / σ(4πR²) ]¹/⁴ Here e= 1
A = 4πR²
Related Questions: - After 300 days, the activity of a radioactive sample is 5000 dps
- Two bodies A(of mass 1kg) and B(of mass 3 kg) are dropped from heights of 16m and 25m
- If an electron and a proton have the same de-Broglie wavelength, then the kinetic
- X-rays when incident on a metal
- Photocells convert
Topics: Properties of Bulk Matter
(130)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- After 300 days, the activity of a radioactive sample is 5000 dps
- Two bodies A(of mass 1kg) and B(of mass 3 kg) are dropped from heights of 16m and 25m
- If an electron and a proton have the same de-Broglie wavelength, then the kinetic
- X-rays when incident on a metal
- Photocells convert
Topics: Properties of Bulk Matter (130)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Leave a Reply