A potentiometer circuit has been setup for finding. The internal resistance of a given cell. The main battery used a negligible internal resistance. The potentiometer wire itsefl is `4 m` long. When the resistance, `R`, connected across the given cell, has value of
(i) Infinity `9.5 Omega`,
(ii) the 'balancing length' , on the potentiometer wire are found to be `3 m` and `2.85 m`, respectively.
The value of internal resistance of the cell is

Options

A. `0.25Omega`
B. `0.95Omega`
C. `0.5Omega`
D. `0.75Omega`

Answer

C

Explanation

The internal resistance of the cell is
`r=((l_(1))/(l_(2))-1)R`
Here, `l_(1)=3m,l_(2)=2.85m,R=9.5Omega`
`therefore r=((3)/(2.85)-1)(9.5Omega)=(0.15)/(2.85)xx9.5Omega=0.5Omega`