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 Omega` resistance and is at a length of `3 m` when the cell is shunted by a `10 Omega` resistance, the internal resistance of the cell is then

Options

A. `1 Omega`
B. `1.5 Omega`
C. `10 Omega`
D. `15 Omega`

Answer

c

Explanation

`(Er_(1))/((R_(1)+r)) = kl_(1) and (Er_(2))/((R_(2)+r)) = kl_(1)`
`:. (Er_(1)//(R_(1)+r))/(Er_(2)//(R_(2)+r)) = (l_(1))/(l_(2)) or (R_(1)(R_(2)+r))/(R_(2)(R_(1)+r)) = (l_(1))/(l_(2))`
`:. (5(10+r))/(10(5+r)) = (2)/(3)`
On solving `r = 10 Omega`