A `6Omega` resistance is connected in the left gap of a meter bridge. In the second gap `3Omega` and `6Omega` are joined in parallel. The balance point of the bridge is at _______

Options

A. 75 cm
B. 60 cm
C. 30 cm
D. 25 cm

Answer

To solve the problem, we need to find the balance point of a meter bridge with a 6Ω resistance on one side and a parallel combination of 3Ω and 6Ω on the other side.

### Step-by-Step Solution:

1. **Identify the resistances**:
- Left gap resistance, \( R_1 = 6Ω \)
- Right gap consists of \( R_2 = 3Ω \) and \( R_3 = 6Ω \) in parallel.

2. **Calculate the equivalent resistance of the right gap**:
- The formula for resistances in parallel is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_2} + \frac{1}{R_3} \]
- Substituting the values:
\[ \frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{6} \]
- Finding a common denominator (which is 6):
\[ \frac{1}{R_{eq}} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \]
- Therefore, the equivalent resistance \( R_{eq} \) is:
\[ R_{eq} = 2Ω \]

3. **Set up the balance condition of the meter bridge**:
- According to the meter bridge principle, when balanced:
\[ \frac{R_1}{l_1} = \frac{R_{eq}}{l_2} \]
- Where \( l_1 \) is the length of the left gap and \( l_2 \) is the length of the right gap.
- We also know that:
\[ l_1 + l_2 = 100 \text{ cm} \]

4. **Substituting the known values**:
- From the balance condition:
\[ \frac{6}{l_1} = \frac{2}{l_2} \]
- Rearranging gives:
\[ 6 \cdot l_2 = 2 \cdot l_1 \]
- From \( l_2 = 100 - l_1 \):
\[ 6(100 - l_1) = 2l_1 \]

5. **Solving for \( l_1 \)**:
- Expanding the equation:
\[ 600 - 6l_1 = 2l_1 \]
- Bringing all terms involving \( l_1 \) to one side:
\[ 600 = 6l_1 + 2l_1 \]
\[ 600 = 8l_1 \]
- Dividing both sides by 8:
\[ l_1 = \frac{600}{8} = 75 \text{ cm} \]

6. **Finding \( l_2 \)**:
- Using \( l_2 = 100 - l_1 \):
\[ l_2 = 100 - 75 = 25 \text{ cm} \]

### Final Answer:
The balance point of the bridge is at \( 75 \text{ cm} \).