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. `5 Omega`
B. `10 Omega`
C. `15 Omega`
D. `7.5 Omega`

Answer

B

Explanation

Internal resistance, `r= ((l_(1)-l_(2))/(l_(2)))R`
`therefore` For the same cell,
`r= ((l_(1)-2)/(2))xx5 = ((l_(1)-3)/(3))10`
`therefore (l_(1)-2)/(2)=(l_(1)-3)(2)/(3)`
`therefore 3(l_(1)-2)=4(l_(1)-3) therefore 3 l_(1)-6 = 4l_(1)-12 therefore l_(1)=6m`
`therefore r= ((l_(1)-2)/(2))xx5 = ((6-2)/(2))xx5 = 10 Omega`