In an experiment to find the internal resistance of a cell by a potentiometer, a balance was obtained for 50 cm length of the potentiometer wire, with a cell of e.m.f. 2V. When the cell was shunted by a resistance of `2 Omega`, the balancing length of the potentiometer wire was 40 cm. What was the internal resistance of the cell ?

Options

A. `0.25 Omega`
B. `0.75 Omega`
C. `0.5 Omega`
D. `1 Omega`

Answer

C

Explanation

`r= R[(l_(1)-l_(2))/(l_(2))]=2 [(50-40)/(40)]= 0.5 Omega`