In a potentiometer experiment it is found that no current passes through the galvanometer when the terminals of the cell are connected across `0.52 m` of the potentiometer wire. If the cell is shunted by a resistance of `5Omega` balance is obtained when the cell connected across` 0.4m` of the wire. Find the internal resistance of the cell.

Options

A. `2 Omega`
B. `2.5 Omega`
C. `1 Omega`
D. `1.5 Omega`

Answer

D

Explanation

Let `l_(1)` be the balancing length without shunt and `l_(2)` be the balancing length with shunt across the wire then internal resistance of the cell can be written as follows:
`r= R ((l_(1))/(l_(2))-1)`
`= 5((52)/(40) -1) = 5xx(12)/(40) = 1.5 Omega`
Hence option (d) is correct