The length of a potentiometer wires is l.A cell of emf E is balanced at a length `((l)/(3))` from positive end of the wire . If the length of the wire is increased by `((l)/(2))` the distance at which the same cell gives the balancing point is
( Cell in the primary is ideal and no series resistance is present in the primary circuit . )

Options

A. `(2l )/(3)`
B. `(l) /(2)`
C. `(l )/(6)`
D. `(4l )/(3)`

Answer

B

Explanation

According to question the figure is as
` (##ARH_19Y_SP_22_04_19_01_E02_025_S01.png" width="80%">
If K be the potential gradient of the potentiometer wire then emf of the cell which given balancing length `(l)/(3)` is given by
`E= K.(l)/(3)`
Where `E= (V )/(l ) , (l)/(3) = (V )/(3)`
When length of potentiometer wire is increased by `(l )/(2) ` then new length
`l_(1) = 1 + (l )/(2) = (3l )/(2)`
New potential gradient.
`k= (V )/((3l )/(2 )) rArr k . = (2v )/( 3l )`
If l be the new balancing length then
`E = K.l .`
`E = (2V )/( 3l ) l ,`
`(2V )/(3l ) l. = (V )/(3) rArr l. = (1)/(2) `