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

Answer

B

Explanation

By using r=`((l_(1)-l_(2))/(l_(2)))r` `rArr` r=`((i_(2)-2)/(2))xx5`.......(i)
and r=`((l_(3)-3)/(3))xx10`.....(ii)
On "solving" (i) and (ii) r = `10Omega`.