A resistance of `5Omega` is connected across the left gap of a metrebridge and unknown resistance which is greater than `5Omega` is connected across the right gap. When these resistances are interchanged the balancing length is found to change by 20cm. What is the value of the unknown resistance?

Options

A. `5Omega`
B. `15//2Omega`
C. `4 Omega`
D. `10//7 Omega`

Explanation

To solve the problem, we will use the concept of a meter bridge and the principle of Wheatstone bridge. The meter bridge is a device that helps to determine unknown resistances by balancing two arms of a bridge circuit.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- A resistance of \( R_1 = 5 \, \Omega \) is connected across the left gap of the meter bridge.
- An unknown resistance \( R_2 \) (which is greater than \( 5 \, \Omega \)) is connected across the right gap.
- Let the initial balancing length be \( x \) cm.

2. **Applying the Balance Condition**:
- According to the meter bridge principle, the ratio of the resistances is equal to the ratio of the lengths:
\[ \frac{R_1}{R_2} = \frac{x}{100 - x} \]
- Substituting \( R_1 = 5 \, \Omega \):
\[ \frac{5}{R_2} = \frac{x}{100 - x} \]
- Cross-multiplying gives:
\[ 5(100 - x) = R_2 x \]
- Rearranging this, we get:
\[ R_2 = \frac{500 - 5x}{x} \]

3. **Interchanging the Resistances**:
- When the resistances are interchanged, the new balancing length is \( x + 20 \) cm.
- The new balance condition is:
\[ \frac{R_2}{R_1} = \frac{x + 20}{100 - (x + 20)} \]
- Substituting \( R_1 = 5 \, \Omega \):
\[ \frac{R_2}{5} = \frac{x + 20}{80 - x} \]
- Cross-multiplying gives:
\[ R_2 (80 - x) = 5(x + 20) \]
- Rearranging this, we get:
\[ R_2 = \frac{5(x + 20)}{80 - x} \]

4. **Setting the Two Expressions for \( R_2 \) Equal**:
- Now we have two expressions for \( R_2 \):
\[ \frac{500 - 5x}{x} = \frac{5(x + 20)}{80 - x} \]
- Cross-multiplying gives:
\[ (500 - 5x)(80 - x) = 5x(x + 20) \]

5. **Expanding and Simplifying**:
- Expanding both sides:
\[ 40000 - 500x - 400x + 5x^2 = 5x^2 + 100x \]
- Simplifying:
\[ 40000 - 900x = 100x \]
- Rearranging gives:
\[ 40000 = 1000x \]
- Solving for \( x \):
\[ x = 40 \, \text{cm} \]

6. **Finding the Unknown Resistance \( R_2 \)**:
- Substitute \( x = 40 \) cm into either expression for \( R_2 \):
\[ R_2 = \frac{500 - 5(40)}{40} = \frac{500 - 200}{40} = \frac{300}{40} = 7.5 \, \Omega \]

### Final Result:
The value of the unknown resistance \( R_2 \) is \( 7.5 \, \Omega \).