In a meter-bridge experiment with a resistance `R_(1)` in left gap and a resistance `X` in a right gap. null point is obtained at `40 cm` from the left emf. With a resistance `R_(2)` in the left gap, the null point is obtainned at `50 cm` from left hand. Find the position of the left gap is containing `R_(1)` and `R_(2)` (i) in series and (ii) in parallel.

Answer

To solve the problem step by step, we will analyze the two cases: when the resistances \( R_1 \) and \( R_2 \) are in series and when they are in parallel.

### Given:
1. Null point with \( R_1 \) in left gap and \( X \) in right gap is at \( 40 \, \text{cm} \).
2. Null point with \( R_2 \) in left gap and \( X \) in right gap is at \( 50 \, \text{cm} \).

### Step 1: Finding the relationship between \( R_1 \) and \( X \)

Using the principle of the meter bridge, we have:
\[ \frac{R_1}{X} = \frac{40}{60} \]
This simplifies to:
\[ R_1 = \frac{2}{3}X \quad \text{(Equation 1)} \]

### Step 2: Finding the relationship between \( R_2 \) and \( X \)

Similarly, for the second case:
\[ \frac{R_2}{X} = \frac{50}{50} \]
This simplifies to:
\[ R_2 = X \quad \text{(Equation 2)} \]

### Step 3: Case (i) - \( R_1 \) and \( R_2 \) in series

When \( R_1 \) and \( R_2 \) are in series, the equivalent resistance \( R_{eq} \) is:
\[ R_{eq} = R_1 + R_2 = \frac{2}{3}X + X = \frac{5}{3}X \]

Using the balance condition again:
\[ \frac{R_{eq}}{X} = \frac{L_1}{100 - L_1} \]
Substituting \( R_{eq} \):
\[ \frac{\frac{5}{3}X}{X} = \frac{L_1}{100 - L_1} \]
This simplifies to:
\[ \frac{5}{3} = \frac{L_1}{100 - L_1} \]

### Step 4: Cross-multiplying to solve for \( L_1 \)

Cross-multiplying gives:
\[ 5(100 - L_1) = 3L_1 \]
Expanding and rearranging:
\[ 500 - 5L_1 = 3L_1 \implies 500 = 8L_1 \implies L_1 = \frac{500}{8} = 62.5 \, \text{cm} \]

### Step 5: Case (ii) - \( R_1 \) and \( R_2 \) in parallel

For resistances in parallel, the equivalent resistance \( R_{eq} \) is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
Substituting \( R_1 \) and \( R_2 \):
\[ \frac{1}{R_{eq}} = \frac{3}{2X} + \frac{1}{X} = \frac{3 + 2}{2X} = \frac{5}{2X} \]
Thus,
\[ R_{eq} = \frac{2X}{5} \]

Using the balance condition:
\[ \frac{R_{eq}}{X} = \frac{L_2}{100 - L_2} \]
Substituting \( R_{eq} \):
\[ \frac{\frac{2X}{5}}{X} = \frac{L_2}{100 - L_2} \]
This simplifies to:
\[ \frac{2}{5} = \frac{L_2}{100 - L_2} \]

### Step 6: Cross-multiplying to solve for \( L_2 \)

Cross-multiplying gives:
\[ 2(100 - L_2) = 5L_2 \]
Expanding and rearranging:
\[ 200 - 2L_2 = 5L_2 \implies 200 = 7L_2 \implies L_2 = \frac{200}{7} \approx 28.57 \, \text{cm} \]

### Final Answers:
1. When \( R_1 \) and \( R_2 \) are in series, the null point is at \( L_1 = 62.5 \, \text{cm} \).
2. When \( R_1 \) and \( R_2 \) are in parallel, the null point is at \( L_2 \approx 28.57 \, \text{cm} \).