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

Answer

C

Explanation

When R is infinity then balancing length `l_(1) = 3m`, is obtained for the emf (E ) of cell and when resistance is `9.5 Omega` then balancing length `l_(2) = 2.85m`, is obtained for terminal potential difference (V) for the cell. Internal resistance of the cell can be written as:
`r= R [(E )/(V) -1] = R [(l_(1))/(l_(2)) -1]`
`rArr r= (9.5) [(3)/(2.85)-1] = (9.5 xx 0.15)/(2.85) = 0.5 Omega`
Hence, option (c) is correct.