In a meter bridge, null point is `20 cm`, when the known resistance `R` is shunted by `10 Omega` resistance, null point is found to be shifted by 10 cm. Find the unknown resistance `X`.

Explanation

`R/X=1 l/(100-l)`
`X=((100-l)/1)R`
or `X=((100-20)/20)R=4R`………..i
when known resistance `R` is shunted its net resistance will decreae. Therefore, resistance parallel to this (i.e. `P`) should also decrease or its new null point length should also decrease.
`:. (R')/(X)= (l')/(100-l')`
`=(20-10)/(100-(20-10))=1/9`
or `X=9R`...............ii
From eqn i and ii we have
`4R=9R'=9[(10R)/(10+R)]`
Solving this equation we get
`R=50/4Omega`
Now from eqn i the unknown resistance
`X=4R=4(50/4)`
or `X=50 Omega`