The temperature of a radiating surface changes from 400 K to 1200 K. The ratio of total emissive powers at the higher and lower temperatures would be-

This question was previously asked in
NPCIL SA/ST ME GJ Held on 08/11/2019, Shift-1
View all NPCIL Scientific Assistant Papers >
  1. 3
  2. 9
  3. 27
  4. 81

Answer (Detailed Solution Below)

Option 4 : 81
Free
NPCIL Scientific Assistant Quantum Mechanics Test
10 Qs. 10 Marks 13 Mins

Detailed Solution

Download Solution PDF

Concept:

Stefan-Boltzmann Law

The thermal energy radiated by a black body per second per unit area is proportional to the fourth power of the absolute temperature and is given by:

E ∝ T4

E = σT4

σ = The Stefan – Boltzmann constant = 5.67 × 10-8 W m-2K-4

Calculation:

Given:

T1 = 400 K, T2 = 1200 K

Latest NPCIL Scientific Assistant Updates

Last updated on Mar 27, 2025

-> NPCIL Scientific Assistant Recruitment Notification 2025 is out! 

->The Nuclear Power Corporation of India Limited (NPCIL) has released the NPCIL Scientific Assistant Recruitment notification for 45 vacancies.

-> Candidates can apply online start applying from 12 March 2025 till 1 April 2025.

-> NPCIL Exam Date 2025 is yet to be announced, candidates can keep a check on the official website for latest updates.

-> Candidates with diploma in Civil/Mechanical/Electrical/Electronics with a minimum of 60% marks are eligible to apply. 

Hot Links: teen patti master golden india teen patti classic teen patti gold old version teen patti real