Question

A resistance wire having length $1\text{m}$, when connected in the left gap of a metre bridge balances a $20\Omega$ resistance in the right gap at a point which divides the bridge wire in the ratio $1:2$. The length of the resistance wire which will have resistance of $2\Omega$ is

• A. $2.0\times {10}^{-1}\text{m}$
• B. $1.5\times {10}^{-1}\text{m}$
• C. $1.5\times {10}^{-2}\text{m}$
• D. $1.0\times {10}^{-2}\text{m}$

Answer 1

The correct option is A $2.0\times {10}^{-1}\text{m}$

For balanced Wheatstone bridge.

$\frac{R}{S}=\frac{1}{2}\Rightarrow R=\frac{S}{2}$

Given : $S=20\Omega \Rightarrow R=10\Omega$

Now, length per unit resistance $=\frac{1}{10}=0.1\text{m/}\Omega$

Length of $2\Omega$ resistance $=2\times 0.1=0.2\text{m}$