The resistance in the two arms of the meter bridge are `5 Omega` and `R Omega`, respectively. When the resistance `R` is shunted with an equal resistance, the new balance point is `1.6 l_(1)`.
The resistance `R` is

Options

A. `10 Omega`
B. `15 Omega`
C. `20 Omega`
D. `25 Omega`

Answer

B

Explanation

(b) Initially, `(5)/(l_(1)) = (R )/(100 - l_(1))`
Finally, `(5)/(1.6 l_(1)) = (R )/(2 (100 - 1.6l_(1))`
`implies (R )/(1.6 (100 - l_(1))) = (R )/(2 (100 = 1.6 l_(2)))`
`implies 160 - 1.6 l_(1) = 200 - 3.2 l_(1)`
`implies 1.6 l_(1) = 40`
`implies l_(1) = 25`
From Equaiton (i), `(5)/(25) = (R )/(75) implies R = 15 Omega`