In the petentiometer arrangemeter shown, the driving cell `A` has emf `epsilon` and internal resistance `r`. The emf of the cell `B` is `(epsilon)/(2)` and internal resistance `2r`. The petentiometer wire `CD` is `100 cm` long. If balance is obtained with length `CJ = l`, then A. `l = 50 cm`B. `l gt 50 cm`C. `l lt 50 cm`D. Balance cannot be obtained

In the petentiometer arrangemeter shown, the driving cell `A` has emf

1.	In the petentiometer arrangemeter shown, the driving cell `A` has emf `epsilon` and internal resistance `r`. The emf of the cell `B` is `(epsilon)/(2)` and internal resistance `2r`. The petentiometer wire `CD` is `100 cm` long. If balance is obtained with length `CJ = l`, then A. `l = 50 cm`B. `l gt 50 cm`C. `l lt 50 cm`D. Balance cannot be obtained
Answer» Correct Answer - B (b) As potential gradient `k` in the potentiometer wire will be less than `(epsilon//100)` `(v//cm)` because of presence of `r`. So for the battery of emf `(epsilon//2)` the balance point `l` will be greater than `50 cm` as shown. `l = (epsilon//2)/(k(lt epsilon//100)) implies l gt 50 cm`

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