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 across the cell, the balancing length is decreased by `10 cm`
The voltmeter reading will be

`1.96V`

`1.8V`

`1.64V`

`0.96V`

Verified by Experts

The correct Answer is:

`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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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

`500Omega`

`290Omega`

`490Omega`

`20Omega`

`2 V`

`2.04 V`

`1.96 V`

`1.0 V`

`1//5 Vm^(-1)`

`2//5 Vm^(-1)`

`3//5 Vm^(-1)`

`4//5 Vm^(-1)`