The balancing point in meter bridge experiment is obtained at 30 cm from the left. If the right gap contains `3.5 Omega`, what is the resistance in the left gap?

Explanation

`(x)/(R ) = (l_(1))/(100 - l_(1)) implies (x)/(3.5) = (30)/(70) implies x = (30 xx 3.5)/(70) = 1.5` `:. X = 1.5 Omega`
The resistance in the left gap in meter bridge is `x = 1.5 Omega`