An unknown resistance `R_(1)` is connected is series with a resistance of `10 Omega`. This combination is connected to one gap of a meter bridge, while other gap is connected to another resistance `R_(2)`. The balance point is at `50 cm` Now , when the `10 Omega` resistance is removed, the balanced point shifts to `40 cm` Then the value of `R_(1)` is.

Options

A. `60 Omega`
B. `40 Omega`
C. `20 Omega`
D. `10 Omega`

Answer

C

Explanation

`(R_(1) + 10)/(R_2) =(50)/(50)= 1`
`:. R_1 + 10 = R_2`
Again, `(R_1)/(R_2)=(40)/(60)= 2/3`
Substituting the value of `R_(2)` from Eq. (i), we get
`3R_(1) = 2(R_(1) + 10)`
`:. R_(1) = 20 Omega`.