The length of a potentiometer wire `AB` is `600 cm`, and it carries a constant current of `40 mA` from `A` to `B`. For a cell of emf `2 V` and internal resistance `10 Omega`, the null point is found at `500 cm` from `A`. When a voltmeter is connected across the cell, the balancing length of the wire is decreased by `10 cm`.

Potential gradient along `AB` is

Options

A. `1//5 Vm^(-1)`
B. `2//5 Vm^(-1)`
C. `3//5 Vm^(-1)`
D. `4//5 Vm^(-1)`

Answer

B

Explanation

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`