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. `40 Omega`
B. `30 Omega`
C. `20 Omega`
D. `10 Omega`

Answer

C

Explanation

In first case, `l_(1)=l_(2) therefore R_(2)=R_(1)+10`
In second case, `(R_(1))/(R_(1)+10)=(40)/(60)=(2)/(3)`
`therefore 2R_(1)+20=3R_(1) therefore R_(1)=20 Omega`