A wire connected in the left gap of a meter bridge balance a `10Omega` resistance in the right gap to a point, which divides the bridge wire in the ratio 3:2. If the length of the wire is 1m. The length of one ohm wire is

Options

A. 0.057m
B. 0.067m
C. 0.37m
D. 0.134m

Explanation

To solve the problem step by step, we will follow the logical reasoning based on the principles of a meter bridge and the given information.

### Step 1: Understand the Problem
We have a meter bridge where a wire connected in the left gap balances a 10Ω resistance in the right gap. The balance point divides the bridge wire in the ratio 3:2.

### Step 2: Set Up the Ratios
According to the balance condition of a meter bridge, we have:
\[ \frac{P}{Q} = \frac{R}{S} \]
where \( P \) and \( Q \) are the lengths of the bridge wire on the left and right sides, respectively, and \( R \) and \( S \) are the resistances in the left and right gaps, respectively.

Given that the bridge wire is divided in the ratio 3:2, we can denote:
- \( P = 3x \)
- \( Q = 2x \)

Since the total length of the bridge wire is 1 meter, we have:
\[ P + Q = 1 \quad \Rightarrow \quad 3x + 2x = 1 \quad \Rightarrow \quad 5x = 1 \quad \Rightarrow \quad x = \frac{1}{5} \text{ m} \]
Thus:
- \( P = 3x = \frac{3}{5} \text{ m} \)
- \( Q = 2x = \frac{2}{5} \text{ m} \)

### Step 3: Find the Resistance in the Left Gap
We know that the resistance in the right gap \( S = 10Ω \). Using the ratio of lengths and resistances:
\[ \frac{P}{Q} = \frac{R}{S} \quad \Rightarrow \quad \frac{3/5}{2/5} = \frac{R}{10} \]
This simplifies to:
\[ \frac{3}{2} = \frac{R}{10} \]
Cross-multiplying gives:
\[ 3 \times 10 = 2R \quad \Rightarrow \quad 30 = 2R \quad \Rightarrow \quad R = 15Ω \]

### Step 4: Calculate the Length of 1Ω Wire
The total length of the wire is 1 meter, and we found that the resistance \( R \) of the wire is 15Ω. The length of the wire per ohm can be calculated as:
\[ \text{Length of 1Ω wire} = \frac{\text{Total Length}}{\text{Resistance}} = \frac{1 \text{ m}}{15Ω} = \frac{1}{15} \text{ m} \]

### Step 5: Convert to Meters
Calculating \( \frac{1}{15} \):
\[ \frac{1}{15} \approx 0.0667 \text{ m} \quad \text{or} \quad 6.67 \text{ cm} \]

### Final Answer
The length of the 1Ω wire is approximately \( 0.0667 \text{ m} \) or \( 6.67 \text{ cm} \).

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