Q. In an experiment to measure the internal resistance of a cell by a potentiometer, it is found that the balance point is at a length of 2 m when the cell is shunted by a 5 Ω resistance and is at a length of 3m when the cell is shunted by a 10Ω resistance, the internal resistance of the cell is then

(a) 1.5 Ω

(b) 10 Ω

(c) 15 Ω

(d) 1 Ω

**Ans: (b)**

**Solution : **

$\large \frac{E_1}{E_2} = \frac{l_1}{l_2}$

$\large \frac{(\frac{E R_1}{R_1 + r})}{(\frac{E R_2}{R_2 + r})} = \frac{2}{3}$

$ \large \frac{R_1(R_2 + r)}{R_2 (R_1 + r)} = \frac{2}{3}$

R_{1} = 5 Ω , R_{2} = 10 Ω

$ \large \frac{5(10 + r)}{10 (5 + r)} = \frac{2}{3}$

$ \large \frac{(10 + r)}{2 (5 + r)} = \frac{2}{3}$

30 + 3 r = 20 + 4 r

r = 10 Ω