A potentiometer wire of length `L` and resistance `10 Omega` is connected in series with a battery of emf `2.5 V` and a resistance in its primary circuit. The null point corresponding to a cell of a emf `1 V` is obtained at a distance `L//2`. If the resistance in the primary circuit is doubled then the position of new null point will be.

Options

A. 0.4 L
B. 0.5 L
C. 0.6 L
D. 0.8 L

Answer

C

Explanation

Potential gradient `= ((2.5)/(10 + R)) xx (10)/(L) = (25)/((10 + R)R)`
`(25)/((10 + R)L) xx (L)/(2) = 1 rArr 10 + R = 12.5`
`R = 2.5 Omega`
Now potential gradient `=((2.5)/(10 + 2R))xx (10)/(L)= (25)/((10 + 5)L) = (5)/(3 L)`
`(5)/(3 L) x = 1 rArr x = 0.6 L`.