The resistances in left and right gap of a meter brigdge are 20 `omega` and 30 `omega` respecitively when the resistance in the left gap is reduced to half its value then balance point shifts by

Options

A. 15 cm to the right
B. 15 cm to the left
C. 20 cm to the right
D. 20 cm to left

Answer

B

Explanation

`(x)/(R)=(l_(x))/(l_(r))`
Case I `:(20)/(30)=(l_(x))/(100-l_(x))`
`:.l_(x)=40 cm`
Case II : `(10)/(30)=(l_(x))/(100-l_(x))`
`:.l_(x)=25cm`
Now shift in null point `=40-25=15` cm towards left.