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 step by step, we will use the principles of a meter bridge and the information given in the question.

### Step 1: Understand 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 to the left by 10 cm. The total resistance of the series combination of \( R_1 \) and \( R_2 \) is given as \( 1 \, k\Omega \) (or \( 1000 \, \Omega \)).

### Step 2: Set Up the Equations
Let the original balance point be at a distance \( L \) cm from the left end. According to the meter bridge principle, the ratio of the resistances is proportional to the lengths on the bridge:

1. Before interchanging:
\[ \frac{R_1}{R_2} = \frac{L}{100 - L} \]

2. After interchanging, the balance point shifts to \( L - 10 \) cm:
\[ \frac{R_2}{R_1} = \frac{L - 10}{100 - (L - 10)} = \frac{L - 10}{110 - L} \]

### Step 3: Equate the Ratios
From the first equation, we can express \( R_2 \) in terms of \( R_1 \):
\[ R_2 = R_1 \cdot \frac{100 - L}{L} \]

Substituting \( R_2 \) into the second equation gives:
\[ \frac{R_1 \cdot \frac{100 - L}{L}}{R_1} = \frac{L - 10}{110 - L} \]

This simplifies to:
\[ \frac{100 - L}{L} = \frac{L - 10}{110 - L} \]

### Step 4: Cross Multiply and Solve for \( L \)
Cross multiplying gives:
\[ (100 - L)(110 - L) = L(L - 10) \]

Expanding both sides:
\[ 11000 - 100L - 110L + L^2 = L^2 - 10L \]

This simplifies to:
\[ 11000 - 210L = -10L \]
\[ 11000 = 200L \]
\[ L = \frac{11000}{200} = 55 \, \text{cm} \]

### Step 5: Find the Resistance Values
Now that we have \( L = 55 \, \text{cm} \), we can use the relationship \( R_1 + R_2 = 1000 \, \Omega \) and the ratio we found earlier.

From the ratio:
\[ \frac{R_1}{R_2} = \frac{55}{45} \]

Let \( R_1 = 55k \) and \( R_2 = 45k \). Then:
\[ R_1 + R_2 = 55k + 45k = 100k = 1000 \, \Omega \]
Thus, \( k = 10 \).

Calculating \( R_1 \) and \( R_2 \):
\[ R_1 = 55 \times 10 = 550 \, \Omega \]
\[ R_2 = 45 \times 10 = 450 \, \Omega \]

### Conclusion
The resistance in the left slot before interchanging the resistances was \( R_1 = 550 \, \Omega \).

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