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.

Explanation

To solve the problem, we will use the principles of a meter bridge and the concept of balancing the bridge to find the relationships between the resistances \( R_1 \), \( R_2 \), and \( X \).

### Step-by-Step Solution

1. **Understanding the Meter Bridge Setup**:
- In a meter bridge, the bridge is balanced when the ratio of the resistances is equal to the ratio of the lengths from the null point to the ends of the bridge.
- Let \( L \) be the total length of the bridge (1 meter or 100 cm).

2. **First Condition with Resistance \( R_1 \)**:
- When \( R_1 \) is in the left gap and \( X \) in the right gap, the null point is at 40 cm from the left.
- This means the length on the right side is \( 100 - 40 = 60 \) cm.
- According to the meter bridge principle:
\[ \frac{R_1}{X} = \frac{40}{60} = \frac{2}{3} \]
- Rearranging gives:
\[ 3R_1 = 2X \quad \text{(1)} \]

3. **Second Condition with Resistance \( R_2 \)**:
- When \( R_2 \) is in the left gap and \( X \) in the right gap, the null point is at 50 cm from the left.
- The length on the right side is \( 100 - 50 = 50 \) cm.
- Again, using the meter bridge principle:
\[ \frac{R_2}{X} = \frac{50}{50} = 1 \]
- This implies:
\[ R_2 = X \quad \text{(2)} \]

4. **Substituting Equation (2) into Equation (1)**:
- From equation (2), substitute \( X \) in equation (1):
\[ 3R_1 = 2R_2 \]
- Rearranging gives:
\[ R_1 = \frac{2}{3} R_2 \quad \text{(3)} \]

5. **Finding the Positions of \( R_1 \) and \( R_2 \)**:
- Now, we will analyze the two cases: (i) when \( R_1 \) and \( R_2 \) are in series, and (ii) when \( R_1 \) and \( R_2 \) are in parallel.

#### (i) When \( R_1 \) and \( R_2 \) are in Series:
- The total resistance in the left gap is:
\[ R_s = R_1 + R_2 \]
- Using equation (3):
\[ R_s = \frac{2}{3}R_2 + R_2 = \frac{5}{3}R_2 \]
- The null point will shift, and we can find the new position using the meter bridge principle:
\[ \frac{R_s}{X} = \frac{L_1}{L_2} \]
where \( L_1 \) is the length on the left and \( L_2 \) is the length on the right.

#### (ii) When \( R_1 \) and \( R_2 \) are in Parallel:
- The equivalent resistance in the left gap is:
\[ R_p = \frac{R_1 R_2}{R_1 + R_2} \]
- Substituting \( R_1 \) from equation (3):
\[ R_p = \frac{\left(\frac{2}{3}R_2\right)R_2}{\frac{2}{3}R_2 + R_2} = \frac{\frac{2}{3}R_2^2}{\frac{5}{3}R_2} = \frac{2}{5}R_2 \]
- Again, use the meter bridge principle to find the new null point position.