On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is `1 kOmega`. How much was the resistance on the left slot before interchanging the resistances?

Options

A. `990 Omega`
B. `505 Omega`
C. `550 Omega`
D. `910 Omega`

Explanation

We have `X+Y=1000 Omega`
Initially `X/I=(1000-X)/(100-I)`……..(i)
When X and Y are intercharged then `(1000-x)/(I-10)=X/(100-(I-10))`
or `(1000-X)/(I-10)=X/(110-I)….(ii)`
From Eqs. (i) and (ii), we get
`(100-I)/I=(I-10)/(110-I)`
`(100-I)(110-I)=(I-10)I`
`11000-100I-100+I^2=I^2-10I implies 11000=200I`
`therefore I=55 cm`
Substituting the value of I in Eq (i) we get
`X/55=(1000-55)/(100-55) implies 20X=11000 therefore X=550 Omega`