Two resistances are connected in two gaps of slide wire bridge. The balance point is at `40cm` from left end. A resistance `X` is connected in series with smaller resistance `R` and balance point shifts to `40cm` from right end. What is the value of `X` if `R` is `4Omega`?

Options

A. `4Omega`
B. `5Omega`
C. `6Omega`
D. `7Omega`

Explanation

To solve the problem step by step, we will analyze the two scenarios presented in the question regarding the slide wire bridge.

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
- We have a slide wire bridge where two resistances \( R \) and \( S \) are connected.
- The balance point is at \( 40 \, \text{cm} \) from the left end.

2. **Using the Balance Condition**:
- The balance condition for the first case can be written as:
\[ \frac{R}{S} = \frac{40}{60} \]
- Simplifying this gives:
\[ \frac{R}{S} = \frac{2}{3} \]
- From this, we can express \( R \) in terms of \( S \):
\[ R = \frac{2}{3} S \quad \text{(Equation 1)} \]

3. **Introducing Resistance \( X \)**:
- In the second scenario, a resistance \( X \) is connected in series with \( R \), and the balance point shifts to \( 40 \, \text{cm} \) from the right end.
- The new balance condition can be expressed as:
\[ \frac{R + X}{S} = \frac{60}{40} = \frac{3}{2} \]
- Rearranging gives:
\[ R + X = \frac{3}{2} S \quad \text{(Equation 2)} \]

4. **Substituting Known Values**:
- We know \( R = 4 \, \Omega \) (given in the problem).
- Substitute \( R \) into Equation 1 to find \( S \):
\[ 4 = \frac{2}{3} S \implies S = 6 \, \Omega \]

5. **Substituting \( R \) and \( S \) into Equation 2**:
- Now substitute \( R \) and \( S \) into Equation 2:
\[ 4 + X = \frac{3}{2} \times 6 \]
- Calculate the right side:
\[ 4 + X = 9 \]

6. **Solving for \( X \)**:
- Rearranging gives:
\[ X = 9 - 4 = 5 \, \Omega \]

### Final Answer:
The value of \( X \) is \( 5 \, \Omega \).

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