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

Answer

B

Explanation

Balance condition in meter bridge is given by
`(P)/(Q)=(l)/(100-l)`
Where P is the resistance in right gap and l is length of wire from one end (i.e., left and) where null or balance point is obtained.
Here, `(20)/(30)=(l)/(100-l) rArr l = 40 cm`
Now, `P'=(20)/(2)=10 cm`
`therefore " " (10)/(30)=(l')/(100-l') rArr l' = 25 cm`
So, balance point shifts by (40-25) cm i.e., 15 cm to the left.