In an experiment with a potentiometer to measure the internal resistance of a cell. when the cell in the secondary circuit is by shounted by `5 Omega`, the null point is at `220 cm`. When the cell is shunted by `20 Omega ` the null point is at `300 cm`. Find the internal resistance of the cell.

Options

A. `2Omega`
B. `4Omega`
C. `6Omega`
D. `8Omega`

Answer

B

Explanation

`(V_(1))/(V_(2))=(l_(1))/(l_(2))`
`V_(1)=[(E)/(R_(1)+r)]R_(1),V_(2)[(E)/(R_(2)+r)]R_(2)`
`(R_(1)(R_(2)+r))/(R_(2)(R_(1)+r))=(l_(1))/(l_(2))impliesr=4Omega`