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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

`400 Omega`

B

`500 Omega`

C

`510 Omega`

D

`490 Omega`

Text Solution

Verified by Experts

The correct Answer is:
D
Since `2 V` is balanced across a length of `500 cm = 5 m`, so the potential gradient is `k = 2//5 Vm^(-1)`
Reading of voltmeter = potential difference across voltmeter
`= V_(A) - V_(D) = k(4.9)`

Reading of the voltmeter `= (2)/(5) xx 4.9 = 1.96 V`
Terminal potential difference of the battery will also be `1.96 V`. So
`1.96 = 2 - i xx 10` or `i = 4 xx 10^(-3) A`
So, resistance of the voltmeter is
`R_(V) = 1.96//(4 xx 10^(-3)) = 490 Omega`
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