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 ast `500 cm`. On connecting a voltmeter acros 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`

Answer 1

Correct Answer - 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`