Question

A resistance wire connected in the left gap of a metre bridge balances a $10\, \Omega$ resistance in the right gap at a point which divides the bridge wire in the ratio $3 : 2$. If the length of the resistance wire is $1.5\,m$, then the length of 1 $\Omega$ of the resistance wire is:

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

Answer 1

Answer: (B)

Initially, $\frac{P}{10}=\frac{l_{1}}{l_{2}}=\frac{3}{2}$
$\Rightarrow P=\frac{30}{2}=15\Omega$
Now resistance, $R=\frac{\rho\,l}{A}$
$\frac{R_{1}}{R_{2}}=\frac{l_{1}}{l_{2}}$: Length of $15\,\Omega$ resistance wire is 1.5 m
$\Rightarrow \frac{15}{1}=\frac{1.5}{l_{2}}$
$\Rightarrow l_{2}=0.1\,m$
$=1.0 \times 10^{-1}\,m$
$\therefore $ Length of $1\,\Omega$ resistance wire is $1.0\times 10^{-1}\,m$