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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
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The correct answer is:
$0.25\mathrm{~m}$
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 $l=\frac{3}{12}=0.25 \mathrm{~m}$
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 $l=\frac{3}{12}=0.25 \mathrm{~m}$
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