A wire of resistance `10Omega` is elongated by 10% the resistance of the elongated wire

Options

A. `11Omega`
B. `11.1Omega`
C. `12.1Omega`
D. `13.1Omega`

Answer

B

Explanation

`(R_1)/(R_2)=(l_1)/(100-l_1)`
Rightarrow `(2)/(3)= (l_1)/(100-l_1)`
l-1 = 40 cm
Balancing point is shifted by 22.5cm
Rightarrow `l_1^1=l_1+22.5= 40+22.5= 62.5cm`
On connecting shunt resistance, `R_2` changes to `R_2^1`
`(2)/(R_2^1) = (62.5)/(100-62.5)`
`R_2^1 = 1.2Omega`
`(1)/(R_2^1) = (1)/(R_2)+(1)/(S)`
Rightarrow `(1)/(1.2)= (1)/(3)+(1)/(S)`
Rightarrow `(1)/(S)=0.5`
therefore `S= 2Omega`