In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the emf of two cells, ε2ε1 is:
(A)34
(B)58
(C)23
(D)35✓ Correct
Answer: (D)
Solution
With the galvanometer at the first point, only the first cell lies between the tapping point and the end A, so that cell alone is balanced against the wire:
ε1=k×250
where k is the potential gradient along the wire. Shifting the galvanometer to the second point brings both cells, connected in series, into the balanced loop:
ε1+ε2=k×400
Subtracting gives ε2=k×150, so
ε2ε1=150250=35
The ratio of the emfs is 35.
JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Single correct option
