In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9Ω is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell.

Given:
Initial null point is at length l1 = 350 cm
Final null point is at length l2 = 300 cm
External resistance R = 9Ω
Let r be the internal resistance
We know that in potentiometer
E/V = l1/l2
and V = ER/(R+r)
From both the equations, we get
E/(ER/(R+r)) = l1/l2
(R+r)/R = l1/l2
R+r = (l1/l2)R
r = (l1/l2)R - R
r = (l1 - l2)R/l2
r = (350 - 300)9/300
r = 1.5Ω