The resistance in the left and right gaps of a balanced meter bridge are `R_(1)` and `R_(1)`. The balanced point is `50 cm`. If a resistance of `24 Omega` is connected in parallel to `R_(2)`, the balance point is `70 cm`. The value of `R_(1)` or `R_(2)` is.

Options

A. `12 Omega`
B. `8 Omega
C. `16 Omega`
D. `32 Omega`

Answer

D

Explanation

` (R_(1))/(R_(2)) = (50)/(50) = 1`
`:. R_(1) = R_(2) = R`…(i)
When `24 Omega` is connected in parallel with `(R_(2)`,then the balance point is ` 70 cm`, so
`(R_(P))/R=(((24R)/(24 + R)))/(R) = (30)/(70)`
`(because R_(P) lt R)` lt brgt `:. (24)/(24 +R)=3/7`
`:. 168 = 72 +3R`
`:. 96 = 3R`
`:. R = 32 Omega`.