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

Answer

A

Explanation

Initial condition, `(5)/(l_1) = ( R)/(100 - l_1)` …(i)
Final condition, `(5)/(1.6 l_1) = (R//2)/(100 - 1.6 l_1)` …(ii)
(i)//(ii) `rArr 1.6 = (2(100 - 1.6 l_1))/(100 - l_1)`
`160 - 1.6 l_1 = 200 - 3.2 l_1`
`1.6 l_1 = 40 rArr l_1 = (400)/(16) = 25`
From (i) `(5)/(25) = (R)/(100 - 25) rArr R = 15 Omega`.