Resistance of 100 cm long potentiometer wire is 10`Omega`. It is connected with external resistance R and a cell with emf 2V and negligible internal resistance. While balancing 10 mV emf in the secondary circuit , null point is obtained at 40 cm . Find this external resistance.

Explanation

If potential gradient existing on potentiometer wire is k then we have,
`epsilon_(1) = kl_(1)`
` therefore epsilon_(1) = ((epsilon rho )/(R + L rho + r) ) l_(1) " " ` .... (1)
Resistance per unit length of a potentiometer wire,
`rho = ("total resistance of potentionmeter wire")/("total length of potentiometer wire" ) `
`therefore rho = (10 Omega)/(100 xx 10^(-2) m) = 10 (Omega )/(m ) `
Placing given values in equation (1),
`10 xx 10^(-3) = ( (2 xx 10)/(R + 10 + 0)) (0.4)`
`therefore 0.01 = (8)/(R + 10)`
`therefore R + 10 = 800`
`therefore R = 790 Omega`