A resistance of `2Omega` is connected across one gap of a metre-bridge(the length of the wire is 100 cm) and an unknown resistance, greater than `2Omega`, is connected across the other gap. When these resistances are interchanged, the balance points shifts by 20 cm. Neglecting any corrections, the unknown resistance is

Options

A. `3 Omega`
B. `4 Omega`
C. `5 Omega`
D. `6 Omega`

Answer

A

Explanation

Let R be the unknown resistance and x be the balancing length in the first case.
`(2)/(R ) = (x)/(100-x)` …(i)
In shown case, x is less than 50 cm, because R is greater than `2Omega`. Hence, in the second case when resistances are interchanged then balancing length must have become `x+20`. Thus we can write the following equation:
`(R )/(2) = (x+20)/(100-(x-20))`
`rArr (2)/(R ) =(80-x)/(x+20)` ....(ii)
From equation (i) and (ii) we get the following:
`(80-x)/(x+20) =(x)/(100-x)`
`rArr 8,000 -100x +x^(2) -80x=x^(2)+20x`
`rArr 200x=8,000`
`rArr x=40cm`
Substituting value of x in equation (i) we get the following:
`(2)/(R ) = (40)/(100-40) rArr (R )/(2) =(60)/(40) rArr R= 3Omega`
Hence, option (a) is correct