In meter bridge , the balancing length from left is found to be 20 cm when standard connected of `1 Omega` is in right gap . The value of unknown resistance is

Options

A. `0.25 Omega`
B. `0.4 Omega`
C. `0.5 Omega`
D. `4 Omega`

Answer

To solve the problem, we will use the principles of a meter bridge and the concept of balancing lengths.

### Step-by-Step Solution:

1. **Understanding the Meter Bridge Setup**:
- A meter bridge consists of a wire of length 1 meter (100 cm) that is used to compare two resistances.
- When the bridge is balanced, the ratio of the resistances is equal to the ratio of the lengths from the jockey to the ends of the bridge.

2. **Identifying Given Values**:
- The balancing length from the left (let's call it \( L_1 \)) is given as 20 cm.
- The standard resistance (\( R_s \)) connected in the right gap is 1 ohm.
- The remaining length from the jockey to the right end (let's call it \( L_2 \)) can be calculated as:
\[ L_2 = 100 \, \text{cm} - L_1 = 100 \, \text{cm} - 20 \, \text{cm} = 80 \, \text{cm} \]

3. **Setting Up the Balance Equation**:
- According to the meter bridge principle, the ratio of the resistances is equal to the ratio of the lengths:
\[ \frac{R_x}{R_s} = \frac{L_2}{L_1} \]
- Here, \( R_x \) is the unknown resistance we need to find.

4. **Substituting the Known Values**:
- Substitute \( R_s = 1 \, \Omega \), \( L_2 = 80 \, \text{cm} \), and \( L_1 = 20 \, \text{cm} \) into the equation:
\[ \frac{R_x}{1} = \frac{80}{20} \]

5. **Calculating the Unknown Resistance**:
- Simplifying the right side:
\[ R_x = \frac{80}{20} = 4 \, \Omega \]

### Final Answer:
The value of the unknown resistance \( R_x \) is **4 ohms**.

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