Question

A wire of length 3 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 m
• B. 0.75 m
• C. 0.5 m
• D. 0.25 m

Hint:

Physics Tip: A meter bridge is a practical application of the Wheatstone bridge principle. Always remember that the resistance in the gaps is directly proportional to the lengths of the bridge wire segments at the balance point.

Answer 1

The Correct Option is D

Concept: Physics (Current Electricity) - Meter Bridge.

Step 1: Calculate the total resistance of the 3 m wire. Let $R_1$ be the resistance of the 3 m long wire in the left gap and $R_2 = 8\Omega$ be the resistance in the right gap. For a balanced meter bridge: $$\frac{R_1}{R_2} = \frac{l_1}{l_2} \text{ }$$ Given the ratio $l_1:l_2 = 3:2$: $$\frac{R_1}{8} = \frac{3}{2} \text{ }$$ $$R_1 = \frac{3}{2} \times 8 = 12\Omega \text{ }$$

Step 2: Find the length per unit resistance. We know that a $12\Omega$ resistance corresponds to a wire length of 3 m. Assuming uniform resistance: $$\text{Length per Ohm} = \frac{\text{Total Length}}{\text{Total Resistance}} \text{ }$$ $$\text{Length for } 1\Omega = \frac{3 \text{ m}}{12\Omega} = 0.25 \text{ m }$$ $$ \therefore \text{The length of the wire corresponding to } 1\Omega \text{ is 0.25 m.} \text{ } $$