In a meter bridge the balancing length rom the left end (standard resistance of `1Omega` is in the right gap) is found to be 20 cm . The value of the unkown resistance is

Options

A. `0.3Omega`
B. `0.2Omega`
C. `0.25Omega`
D. `0.5Omega`

Answer

C

Explanation

Let `1_1=l`
`l_2= l+10%` of `l`
`l+(10l)/(100)`
= `(110l)/(100)`
We know that , `R alpha l^2`
`(R_1)/(R_2)= ((l_1)/(l_2))^2`
Rightarrow `(10)/(R_2)= ((100)/(110))^2`
therefor `R_2= 12.1Omega`