| ⇦ | 
| ⇨ |
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: - The dimensions of magnetic field intensity B are
- A closely wound solenoid of 2000 turns and area of cross-section 1.5 x 10⁻⁴ m² carries
- A particle free to move along X-axis has potential energy given as U(X) =k(1-e⁻ˣ²)
- The electric potential at a point (x,y,z) is given by V = – x²y – xz³ + 4.
- A body, which emits radiations of all possible wavelengths, is known as
Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The dimensions of magnetic field intensity B are
- A closely wound solenoid of 2000 turns and area of cross-section 1.5 x 10⁻⁴ m² carries
- A particle free to move along X-axis has potential energy given as U(X) =k(1-e⁻ˣ²)
- The electric potential at a point (x,y,z) is given by V = – x²y – xz³ + 4.
- A body, which emits radiations of all possible wavelengths, is known as
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