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The length of a potentiometer wire is 1200cm and it carries a current of 60mA. For a cell of emf 5V and internal resistance of 20\Omega, the null point on it is found to be at 1000cm. The resistance of whole wire is (in \ \Omega):  
Option: 1 100
Option: 2 80
Option: 3 604
Option: 7 120


Answers (1)







In the potentiometer, a battery of known emf  E is connected in the secondary circuit. A constant current I is flowing through AB from the driver circuit. The jockey is a slide on potentiometer wire AB to obtain null deflection in the galvanometer. Let  l be the length at which galvanometer shows null deflection.

Since the potential of wire AB (V) is proportional to the length AB(L). 

 Similarly E \ \ \alpha \ \ \ l

So we get 




given current through potentiometer wire =60mA


\\60\times10^{-3}R=6\\\Rightarrow R=100\Omega 

So the correct option is 4.


Posted by

vishal kumar

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