| ⇦ |
| ⇨ |
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 circular road of radius 1000m has banking angle 45°.The maximum safe speed of a car
- Light of frequency 8×10¹⁵ Hz is incident on a substance of photoelectric work
- A thin rod of length L and mass M is bent at its midpoint into two halves
- An electric bulb has a rated power of 50W at 100 V. if it used on an a.c. source
- Which of the following figures represent the variation of particle momentum
Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A circular road of radius 1000m has banking angle 45°.The maximum safe speed of a car
- Light of frequency 8×10¹⁵ Hz is incident on a substance of photoelectric work
- A thin rod of length L and mass M is bent at its midpoint into two halves
- An electric bulb has a rated power of 50W at 100 V. if it used on an a.c. source
- Which of the following figures represent the variation of particle momentum
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