A wire of length $3 \mathrm{~m}$ connected in the left gap of a meter-bridge balances $8 \Omega$ resistance…
A wire of length $3 \mathrm{~m}$ connected in the left gap of a meter-bridge balances $8 \Omega$ resistance in the right gap at a point, which divides the bridge wire in the ratio $3: 2$. The length of the wire corresponding to resistance of $1 \Omega$ is
$1 \mathrm{~m}$
$0.75 \mathrm{~m}$
$0.5 \mathrm{~m}$
$0.25\mathrm{~m}$
Solution
Let $R_1$ be the resistance of $3 \mathrm{~m}$ long wire connected in the left gap.
For meter-bridge, $\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{l_1}{l_2}$
$\begin{array}{ll}
\therefore & \frac{\mathrm{R}_1}{8}=\frac{3}{2} \\
\therefore & \mathrm{R}_1=\frac{3}{2} \times 8=12 \Omega
\end{array}$
Length of the wire corresponding to the resistance of $1 \Omega$ is $l=\frac{3}{12}=0.25 \mathrm{~m}$