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

Answer

b

Explanation

For first case `(R_(1) + 10)/(R_(2)) = (50)/(100 - 50) = 1`
or `R_(2) = R_(1) + 10`…..(i)
When `10 Omega ` resistance the balance point will shift toward lower length
`:.` New balancing length `= 50 - 10 = 40 cm` for second case
`(R_(1))/(R_(2)) = (40)/(100 - 40) = (40)/(60) = (2)/(3) :. R_(2) = (3)/(2)R_(1)`
From (i) `(3)/(2)R_(1) = R_(1) = 10`
or `R_(1) = 20 Omega `