| ⇦ | 
| ⇨ |
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: - A national roadway bridge over a canal is in the form of an arc of a circle of radius
- The work function of sodium is 2.3 eV. The threshold wavelength of sodium will be
- Function of a rectifier is
- The area of cross-section of one limb of an U-tube is twice that of the other.
- In the Young’s double slit experiment, the intensity of light at a point on the screen
Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A national roadway bridge over a canal is in the form of an arc of a circle of radius
- The work function of sodium is 2.3 eV. The threshold wavelength of sodium will be
- Function of a rectifier is
- The area of cross-section of one limb of an U-tube is twice that of the other.
- In the Young’s double slit experiment, the intensity of light at a point on the screen
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