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. `19 Omega`
C. `20 Omega`
D. `25 Omega`

Answer

b

Explanation

Initially `(5)/(R )= (l_(1))/((100 -l_(1))`…(i)
Finally `(5)/(R//2)= (1.6l_(1))/((100 -16l_(1)))`
or `(5)/(R )= (1.6l_(1))/(2(100 -16l_(1))`….(ii)
`:.(l_(1))/((100 -l_(1))) = (16l_(1))/(2(100 -16l_(1)))`
On solving we get `l_(1) = 25 cm`
From(i) , `(5)/(R )=(25)/((100 - 25)) = (1)/(3) or R = 15 Omega`