| ⇦ | 
| ⇨ |
The thermo e.m.f. E in volts of a certain thermocouple is found to vary with temperature difference θ in ⁰C between the two junctions according to the relation
E = 30θ – θ² / 15
The neutral temperature for the thermocouple will be
Options
(a) 30⁰C
(b) 450⁰C
(c) 400⁰C
(d) 225⁰C
Correct Answer:
225⁰C
Explanation:
E = 30θ – θ² / 15
For neutral temperature, dE/ dθ = 0
0 = 30 – 2/15 . θ
θ = 15 x 15
= 225⁰C
Related Questions: - A body undergoes no change in volume. Poisson’s ratio is
- In which case positive work has to be done in seperating them?
- what should be the velocity of an electron so that its momentum becomes equal
- The number of beta particles emitted by a radioactive substance is twice the number
- Wires A and B are made from the same material. A has twice the diameter and three times
Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A body undergoes no change in volume. Poisson’s ratio is
- In which case positive work has to be done in seperating them?
- what should be the velocity of an electron so that its momentum becomes equal
- The number of beta particles emitted by a radioactive substance is twice the number
- Wires A and B are made from the same material. A has twice the diameter and three times
Topics: Current Electricity (136)
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