Question

The resistance in the left and right gap of a meter bridge are 20 ohm and 30 ohm respectively when the resistance in the left gap is reduced to half its value the balance point shifts by

Answer 1

we know from the formula of metre bridge,
X/Y=L/(100-L) 
Or,20/30=L/(100-L)
solving,L=15 units
Now,
when the resistance in th left gap is reduced to half i.e.10ohm
then,
10/30=L'/(100-L')
solving,L'=25 units
now ,shift =L'-L=25-15=1o units.
(Check if its wrong)

Answer 2

that's the answer 10units