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 $1.5m$, 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
$\Rightarrow \frac{15}{1}=\frac{1.5}{l_2}$
$\Rightarrow l_2=0.1m$
$=1.0\times 10^{-1}m$
$\therefore$ Length of $1\Omega$ resistance wire is $1.0\times 10^{-1}m$