On interchanging the resistance the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their seris combination is 1 `k Omega`. How much was the resistance on the left slot before the interchange ?

Options

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

Answer

C

Explanation

550 `Omega`
In first situation `(R_(1))/(R_(2)) = (l)/(100 - l) " " `... (1)
`l_(2) = 100 - l`
By moving the place `(R_(2))/(R_(1)) = (l - 10)/(110 - l) " " `.... (2)
`l_(2) = 100 - (l - 10)`
`= 100 - l + 10`
`= 110 - l `
Now ` (R_(1))/(R_(2)) xx (R_(2))/(R_(1)) = 1 `
`(l)/(100 - l) xx (l - 10)/(110 - l) = 1 `
`therefore l^(2) - 10 l = 11000 - 100 l - 110 l + l^(2)`
`therefore = 200 l = 11000`
`therefore l = 55 ` cm
Now from eq. (1)
`(R_(1))/(100 - R_(1)) = (55)/(100 - 55)`
`[ because R_(1) + R_(2) = 1000 , therefore R_(2) = 1000 - R_(2) ] `
`therefore (R_(1))/(1000 - R_(1)) = (55)/(45)`
= `(11)/(9)` ,
` therefore 9 R_(1) = 11000 - 11 R_(1)`
`therefore 20 R_(1) = 11000`
`therefore R_(1) = 550 Omega`