Correct Answer - a
Balance condition in meter bridge is given by
`(P)/(Q)=(l)/(100-l)`
where P is the resistance in left gap and Q is the resistance in right gap and L is length of wire from one end (i.e left end ) where null or balance point is obtained
here `20/30 =(l)/(100-l) rarr l =40 cm`
Now `P=(20)/(2)=10` cm
`therefore 10/230 =(l)/(100-I)rarr I=25 cm`
so balance point shifts by (40-25) cm i.e 15 cm to the left