MHT CET · Physics · Current Electricity
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
- A \(1 \mathrm{~m}\)
- B \(0.75 \mathrm{~m}\)
- C \(0.5 \mathrm{~m}\)
- D \(0.25\mathrm{~m}\)
Answer & Solution
Correct Answer
(D) \(0.25\mathrm{~m}\)
Step-by-step Solution
Detailed explanation
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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