| ⇦ | 
| ⇨ |
A potentiometer wire of length 100cm has a resistance of 10Ω. It is connected in series with a resistance and a cell of emf 2V having negligible internal reistance. A source of emf 10 mV is balanced against a length of 40 cm of the potentiometer wire.The value of external resistance is
Options
(a) 760Ω
(b) 640Ω
(c) 790Ω
(d) 840Ω
Correct Answer:
790Ω
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - Masses of 3 wires of same metal are in the ratio 1:2:3 and their lengths
- A astronomical telescope has objective and eye-piece of focal lengths
- An α-particle of energy 5 MeV is scattered through 180⁰ by a fixed uranium nucleus.
- Resistances n, each r Ω, when connected in parallel give an equivalent resistance
- A body is placed in an insulating container at temperature 227⁰C.
Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Masses of 3 wires of same metal are in the ratio 1:2:3 and their lengths
- A astronomical telescope has objective and eye-piece of focal lengths
- An α-particle of energy 5 MeV is scattered through 180⁰ by a fixed uranium nucleus.
- Resistances n, each r Ω, when connected in parallel give an equivalent resistance
- A body is placed in an insulating container at temperature 227⁰C.
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
Let the resistance be r and current be I .According to loop rule :
E=V+Ir
2=I(10+r)
And also
I*(40/100)=0.01V
Here you get I=2.5*10^-3ampere
Put this Vue in eqn 2=I(10+r) and get solution