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

`r = ((l)/(l') - 1) xx R = ((l)/(2) -1) xx 5` ...(i)
`r = ((l)/(3) -1) xx 10` ...(ii)
`((l)/(2) -1) xx 5 = ((l)/(3) - 1)xx 10`
`(l)/(2) - 1 = (2 l)/(3) - 2 rArr (2 l)/(3) - (l)/(2) = 1 rArr 1 = 6 cm`
`r = ((l)/(2) -1) xx 5 = ((6)/(2) -1) xx 5 = 10 Omega`.