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 `1k Omega`. How much was the resistance on the left slot before interchanging the resistances ?

Options

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

Answer

A

Explanation

Let `R_(1)` be the resistance in the left slot and `R_(2)` be the resistance in the right slot initially. Let in this case balance point is at a distance x cm from the left end. We can write the following equation: When resistance are interchanged then balanced point is shifted to the left hence balance length becomes `(x-10)` cm. We can write the following equation for this case:
`(R_(2))/(R_(1)) = ((x-10))/(100-(x-10))`
`rArr (R_(2))/(R_(1)) = (x-10)/(110-x)`...(ii)
Multiplying equations (i) and (ii) we can write the following: `(x)/(100-x) xx (x-10)/(110-x)=1`
`rArr x^(2) - 10x = 11000 -100x - 110x + x^(2)`
`rArr 200x = 11000 rArr x= 55cm`
Substituting in (i) we get
`(R_(1))/(R_(2)) = (55)/(45) rArr (R_(1))/(R_(2)) = (11)/(9) rArr R_(2) = (9)/(11) R_(1)` ...(iii)
Resultant of the two resistors in series combination is given to be `1000 Omega` hence we can write the following:
`R_(1) + R_(2) = 1000`
Substituting from equation (iii) we get,
`R_(1) + (9)/(11)R_(1) = 1000 rArr (20)/(11) R_(1) = 1000 rArr R_(1) = 550 Omega`
Hence option (a) is correct.