A resistance of `2 Omega` is connected across one gap of a meter bridge (the length of the wire is `100 cm` and an unknown resistance, greater than ` 2 Omega`, is connected across the other gap. When these resistance are interchanged, the balance point shifts by `20cm`. Neglecting any corrections, the unknown resistance is.
(a)` 3 Omega`
(b) `4 Omega`
( c) `5 Omega`
(d) `6 Omega`.

Answer

` 3 Omega`

Explanation

`(2)/(X) = (1)/((100 - 1))` …..(i)
When `R` and `X` are interchanged then
`(X)/(2) = ((1 + 20))/((100 - 1 - 20)) = ((1 + 20))/((80 - 1))`
`(2)/(X) = ((80 - 1))/((1 + 20))`…..(ii) On solving from (i) and (ii) we shall get
`1 = 40 cm and X = 3 Omega`