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

Answer

B

Explanation

In case of internal resistance measurement by potentiometer,
`(V_(1))/(V_(2)) = (l_(1))/(l_(2)) = ([ER_(1)//(R_(1) + r)])/([ER_(2)//(R_(2) + r)]) = (R_(1)(R_(2) + r))/(R_(2)(R_(1) + r))`
Here `l_(1) = 2m, l_(2) = 3 m, R_(1) = 5 Omega`, and `R_(2) = 10 Omega.`So
`(2)/(3) = (5(10 + r))/(10(5 + r))` or `r = 10 Omega`