In a potenliomcter experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cells is shunted by a resistance of 5 `Omega`, a balance is found when the cells is connected across 40 cm of the wire. Find the internal resistance of the cell.

Options

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

Answer

B

Explanation

`1.5 Omega`
Equation of internal resistance r,
r = `((E)/(V) - 1)` R
`= ((sigma l_(1))/(sigma l_(2)) - 1) ` R
` = ((l_(1))/(l_(2)) - 1)`R
`((l_(1) - l_(2))/(2) )`R
` = ((52 - 40)/(40) ) xx 5 `
`= (12)/(40) xx 5 = 1.5 Omega`