Question

A resistance wire connected in the left gap of a metre bridge balances a $10\, \Omega$ resistance in the right gap at a point which divides the bridge wire in the ratio $3 : 2$. If the length of the resistance wire is $1.5\,m$, then the length of 1 $\Omega$ of the resistance wire is:

• A. $1.0 \times 10^{-2} m$
• B. $1.0 \times 10^{-1} m$
• C. $1.5 \times 10^{-1} m$
• D. $1.5 \times 10^{-2} m$

Answer 1

The Correct Option is B

The given problem involves a meter bridge setup, which operates on the principle of a Wheatstone bridge. The meter bridge consists of a uniform wire of known length, and the problem provides us the resistance ratios in both gaps and the wire's total length.

1. Identify the known values:
   • The resistance in the right gap, \( R_2 = 10\, \Omega \).
   • The ratio of divisions on the meter bridge, \( \frac{L_1}{L_2} = \frac{3}{2} \).
   • Total length of the resistance wire, \( L = 1.5\, \text{m} \).
2. Apply the concept of the meter bridge, where the resistance \( R_1 \) in the left gap (resistance wire) and the resistance in the right gap \( R_2 \) forms a balanced bridge at the given ratios:
3. Using the formula for a balanced wheatstone bridge: \[ \frac{R_1}{R_2} = \frac{L_1}{L_2} \] Given \( \frac{L_1}{L_2} = \frac{3}{2} \), thus, \[ \frac{R_1}{10\, \Omega} = \frac{3}{2} \] Solving for \( R_1 \): \[ R_1 = 10\, \Omega \times \frac{3}{2} = 15\, \Omega \]
4. Then, find the length of the wire producing \( 1\, \Omega \):
   • Total length \( L = 1.5\, \text{m} \) corresponds to \( 15\, \Omega \).
   • Thus, length per \( 1\, \Omega \): \[ \frac{1.5\, \text{m}}{15\, \Omega} = 0.1\, \text{m} \]
   • Therefore, the length of the resistance wire that corresponds to \( 1\, \Omega \) is \( 0.1\, \text{m} \) or \( 1.0 \times 10^{-1} \text{m} \).

Hence, the correct answer is \(1.0 \times 10^{-1} \text{ m}\).