Two unknown resistances are connected in two gaps of a meter-bridge. The null point is obtained at 40 cm from left end. A `30Omega` resistance is connected in series with the smaller of the two resistances, the null point shifts by 20 cm to the right end. The value of smaller resistance in `Omega` is

Answer

24

Explanation

`l _(x) = 40 cm, l _(R) = 60 cm`
`(x)/(R) = ( l _(x))/( l _(R)) = ( 40)/( 60) = (2)/(3) ...(i)`
`therefore (x + 30 )/( R) = ( 60)/( 40) = (3)/(2)`
`therefore (x + 30)/( x) = (9)/(4)`
`9x = 4x + 120`
` 5x = 120 implies x = 25 Omega`