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 isA. `500Omega`B. `290Omega`C. `490Omega`D. `20Omega`

The length of a potentiometer wire is 600 cm and it carries a current

1.	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 isA. `500Omega`B. `290Omega`C. `490Omega`D. `20Omega`
Answer» 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`

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