On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is `1 kOmega`. How much was the resistance on the left slot before interchanging the resistances?

Options

A. `910 Omega`
B. `990 Omega`
C. `505 Omega`
D. `550 Omega`

Answer

To solve the problem, we need to find the resistance in the left slot before interchanging the resistances using the information provided.

### Step-by-step Solution:

1. **Understanding the Problem**:
We have a meter bridge where two resistances \( R_1 \) and \( R_2 \) are connected. When these resistances are interchanged, the balance point shifts by 10 cm. The total resistance of the series combination is given as \( R_1 + R_2 = 1000 \, \Omega \).

2. **Setting Up the Initial Conditions**:
Let the initial balance point be at a distance \( x \) cm from the left end. Therefore, the distance from the right end will be \( 100 - x \) cm.

3. **Applying the Wheatstone Bridge Principle**:
For the initial configuration, we can write:
\[ R_1 \cdot (100 - x) = R_2 \cdot x \quad \text{(1)} \]

4. **After Interchanging the Resistances**:
After interchanging \( R_1 \) and \( R_2 \), the new balance point shifts to \( x - 10 \) cm. Hence, the new distances are \( x - 10 \) cm from the left and \( 100 - (x - 10) = 110 - x \) cm from the right. We can write:
\[ R_2 \cdot (110 - x) = R_1 \cdot (x - 10) \quad \text{(2)} \]

5. **Dividing the Two Equations**:
To eliminate the resistances, we divide equation (1) by equation (2):
\[ \frac{R_1 \cdot (100 - x)}{R_2 \cdot (110 - x)} = \frac{R_2 \cdot x}{R_1 \cdot (x - 10)} \]
This simplifies to:
\[ \frac{(100 - x)}{(110 - x)} = \frac{x}{(x - 10)} \]

6. **Cross-Multiplying**:
Cross-multiplying gives:
\[ (100 - x)(x - 10) = x(110 - x) \]

7. **Expanding Both Sides**:
Expanding both sides:
\[ 100x - 1000 - x^2 + 10x = 110x - x^2 \]
Simplifying this leads to:
\[ 110x - 100x + 1000 - 10x = 0 \]
\[ 0 = 0 \]
Thus, we can rearrange to find:
\[ 210x = 1100 \]

8. **Solving for \( x \)**:
Dividing both sides by 210:
\[ x = \frac{1100}{210} \approx 52.38 \, \text{cm} \]

9. **Finding the Ratio of Resistances**:
Now, using the ratio of resistances:
\[ \frac{R_1}{R_2} = \frac{x}{100 - x} = \frac{52.38}{47.62} \approx \frac{11}{9} \]

10. **Calculating \( R_1 \)**:
Since \( R_1 + R_2 = 1000 \, \Omega \), we can express \( R_1 \) and \( R_2 \):
\[ R_1 = \frac{11}{20} \times 1000 = 550 \, \Omega \]
\[ R_2 = \frac{9}{20} \times 1000 = 450 \, \Omega \]

Thus, the resistance on the left slot before interchanging the resistances was **550 Ω**.