A 2 volt battery, a `15 Omega` resistor and a potentiometer of `100 cm` length, all are connected in series. If the resistance resistance of potentiometer wire is `5 Omega`, then the potential gradient of the potentiometer wire is

Answer

(i)`0.02 Vm^(-1)`,(ii)` 0.08 V`

Explanation

Here `l = 10 m , r =10 Omega `
`V = 4 V, R = 290 Omega`
Current through the potentimerter wire is
`l =- (V)/(r + R) = (4)/(300) xx 10 = (2)/(15) V`
(i) potential gradent
`R = (V')/(l) = (2//15)/(10) = (l)/(75) Vm^(-1) = 0.02 Vm^(-1) `
(ii) Unknown emf `epsilon = KT = (1)/(75) xx 6 = 0.08 V`