The resistance in the two arms of the meter bridge are `5 Omega` and `R Omega`, respectively. When the resistance `R` is shunted with an equal resistance, the new balance point is `1.6 l_(1)`.
The resistance `R` is

Explanation

`(5)/( R) = (l_1)/(100 - l_1)` …(i)

Equivalent resistance of `R` and `R` in parallel is `R//2`
`(5)/(R//2) = (1.6 l_1)/(100 - 1.6 l_1)` ….(ii)
(ii)/(i) `rArr 2= (1.6(100 - l_1))/(100 - 1.6 l_1)`
`200 - 3.2 l_1 = 160 - 1.6 l_1`
`1.6 l_1 = 40 rArr l_1 = 20 cm`
`(5)/(R) = (l_1)/(100 - l_1) = (25)/(100 - 25) = (1)/(3)`
`R = 15 Omega`.