In potentiometer experiment, if `l_(1)` is the balancing length for e.m.f. of the cell of internal resistance r and `l_(2)` is the balancing length for its terminal potential difference when shunted with resistance R then :A. `l_(1) = l_(2) ((R + R)/(R ))`B. `l_(1) = l_(2) ((R )/(R - r))`C. `l_(1) = l_(2) ((R )/(R + r))`D. `l_(1) = l_(2) ((R - r)/(R ))`

In potentiometer experiment, if `l_(1)` is the balancing length for e.

1.	In potentiometer experiment, if `l_(1)` is the balancing length for e.m.f. of the cell of internal resistance r and `l_(2)` is the balancing length for its terminal potential difference when shunted with resistance R then :A. `l_(1) = l_(2) ((R + R)/(R ))`B. `l_(1) = l_(2) ((R )/(R - r))`C. `l_(1) = l_(2) ((R )/(R + r))`D. `l_(1) = l_(2) ((R - r)/(R ))`
Answer» Correct Answer - a `l_(1) = l_(2)((R + r)/(R ))` The internal resistance of a cell is given a `r = ((l_(1) - l_(2))/(l_(2)))R` `r/R = (l_(1) - l_(2))/(l_(2))` `:. rl_(2) = Rl_(1) - Rl_(2)` `:. l_(2) (R + r) = Rl_(1)` `:. l_(1) = l_(2) ((R + r))/(R)`