Question

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`

Answer 1

Correct Answer - 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`