⇦ | ⇨ |
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 meter stick of mass 400 g is pivoted at one end and displaced through an angle 60°.
- A 1Ω resistance in series with an ammeter is balanced by 75 cm of potentiometer wire
- A sound absorber attenuates the sound level by 20 dB. The intensity decreases
- A parallel plate air capacitor is charged to a potential difference of V volts.
- A coin of mass m and radius r having moment of inertia I about the axis passes
Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A meter stick of mass 400 g is pivoted at one end and displaced through an angle 60°.
- A 1Ω resistance in series with an ammeter is balanced by 75 cm of potentiometer wire
- A sound absorber attenuates the sound level by 20 dB. The intensity decreases
- A parallel plate air capacitor is charged to a potential difference of V volts.
- A coin of mass m and radius r having moment of inertia I about the axis passes
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