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`.

Options

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

Answer

a

Explanation

`(2)/(S)= (1)/((100 - 1))`….(i)
and `(S)/(2)=((l+20))/((100-l)-20)=((l+20))/((80-l))`
or `(2)/(S)=((80-1))/((l+20))`…..(ii)
From (i) and (ii), `(l)/((100-l))=((80-l))/((l+20))`
On solving, `l=40 cm`
From (i), `(2)/(S)=(40)/((100-40))=(40)/(60)=(2)/(3)` or `S=3 Omega`