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

A

`1.96V`

B

`1.8V`

C

`1.64V`

D

`0.96V`

Text Solution

Verified by Experts

The correct Answer is:
A
`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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Knowledge Check

  • 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`

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    A
    `2 V`
    B
    `2.04 V`
    C
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    D
    `1.0 V`

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    B
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    C
    `3//5 Vm^(-1)`
    D
    `4//5 Vm^(-1)`