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The length of a potentiometer wire is 60...

The length of a potentiometer wire is 600 cm and it carries a current of `40 mA`. For a cell of emf `2V` and internal resistance `10Omega`, the null point is found to be at `500 cm`. On connecting a voltmeter acros the cell, the balancing length is decreased by `10 cm`
The resistance of the voltmeter is

A

`500Omega`

B

`290Omega`

C

`490Omega`

D

`20Omega`

Text Solution

Verified by Experts

The correct Answer is:
C
`r=R(l_1/l_2-1)`
`:. 10=R(500/490-1)`
Solvng this equation we get
`R=490 Omega`
Further `r = R(E/V-1)`
or `10=490(2/V-1)`
Solvinng we get `V=1.96V`
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