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
B. 40
C. 20
D. 10

Answer

C

Explanation

First balance condition-
`R_(2)/(R_(1) + 10) = 50/50`
Second balance condition-
`R_(1)/R_(2) = 40/60 rArr R_(1)/(10 + R_(1)) = 2/3 rArr R_(1) = 20 Omega`