A balanced bridge is shown in the circuit diagram. The metre bridge wire has resistance $1\ \Omega\text{m}^{-1}$. The current drawn from the battery is (Internal resistance of battery is negligible)
Choose the correct answer from the options given below
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Always convert the wire units to meters first to match the given resistance parameter ($1\ \Omega\text{m}^{-1}$). The total meter bridge wire has a fixed structural length of exactly $1\ \text{m}$, which means its entire end-to-end baseline resistance is simple to track as a constant $1\ \Omega$ branch in parallel with the top resistors.
Step 1: Read the bridge data. The metre bridge is balanced with segments of $40\,\text{cm}$ and $60\,\text{cm}$, wire resistance $1\,\Omega\,\text{m}^{-1}$, a $4\,\Omega$ arm, an unknown arm $Y$, and a $6\,\text{V}$ supply with negligible internal resistance. Step 2: Find the wire-segment resistances. $R_{AD} = 0.40 \times 1 = 0.4\,\Omega$ and $R_{DC} = 0.60 \times 1 = 0.6\,\Omega$. Step 3: Apply the balance condition. For a balanced bridge $\dfrac{4}{0.4} = \dfrac{Y}{0.6}$, so $10 = \dfrac{Y}{0.6}$ and $Y = 6\,\Omega$. Step 4: Reduce each branch. Top branch: $4 + 6 = 10\,\Omega$. Bottom branch (the wire): $0.4 + 0.6 = 1\,\Omega$. Because the bridge is balanced, no current flows in the galvanometer arm, so these two branches are simply in parallel across the battery. Step 5: Combine in parallel. $R_{eq} = \dfrac{10 \times 1}{10 + 1} = \dfrac{10}{11}\,\Omega$. Step 6: Apply Ohm's law for total current. $I = \dfrac{V}{R_{eq}} = \dfrac{6}{10/11} = \dfrac{66}{10} = 0.66\,\text{A}$, which is option (2). \[ \boxed{I = 0.66\ \text{A}} \]
