The resistance in the left and right gaps of a balanced meter bridge are `R_(1)` and `R_(1)`. The balanced point is `50 cm`. If a resistance of `24 Omega` is connected in parallel to `R_(2)`, the balance point is `70 cm`. The value of `R_(1)` or `R_(2)` is.

Options

A. `12 Omega`
B. `8 Omega
C. `16 Omega`
D. `32 Omega`

Explanation

To solve the problem, we will use the principles of a meter bridge and the concept of resistances in parallel.

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
In a balanced meter bridge, the ratio of the resistances is equal to the ratio of the lengths from the ends of the bridge to the balance point. Given that the balance point is at 50 cm, we can set up our first equation:
\[ \frac{R_1}{R_2} = \frac{50}{100 - 50} = \frac{50}{50} = 1 \]
This implies:
\[ R_1 = R_2 \]

2. **Introducing the Parallel Resistance**:
When a resistance of 24 Ω is connected in parallel to \( R_2 \), the equivalent resistance \( R_{eq} \) of \( R_2 \) and the 24 Ω resistor can be calculated using the formula for resistors in parallel:
\[ R_{eq} = \frac{R_2 \cdot 24}{R_2 + 24} \]

3. **Setting Up the New Balance Point**:
With the new balance point at 70 cm, we can set up the second equation:
\[ \frac{R_1}{R_{eq}} = \frac{70}{100 - 70} = \frac{70}{30} = \frac{7}{3} \]

4. **Substituting \( R_1 \) with \( R_2 \)**:
Since we established that \( R_1 = R_2 \), we can substitute \( R_1 \) in the second equation:
\[ \frac{R_2}{\frac{R_2 \cdot 24}{R_2 + 24}} = \frac{7}{3} \]

5. **Cross-Multiplying to Solve for \( R_2 \)**:
Cross-multiplying gives us:
\[ R_2 \cdot (R_2 + 24) = \frac{7}{3} \cdot 24 \]
Simplifying the right side:
\[ R_2 \cdot (R_2 + 24) = 56 \]

6. **Expanding and Rearranging**:
Expanding the left side:
\[ R_2^2 + 24R_2 - 56 = 0 \]

7. **Using the Quadratic Formula**:
We can solve this quadratic equation using the quadratic formula:
\[ R_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 24, c = -56 \):
\[ R_2 = \frac{-24 \pm \sqrt{24^2 - 4 \cdot 1 \cdot (-56)}}{2 \cdot 1} \]
\[ R_2 = \frac{-24 \pm \sqrt{576 + 224}}{2} \]
\[ R_2 = \frac{-24 \pm \sqrt{800}}{2} \]
\[ R_2 = \frac{-24 \pm 20\sqrt{2}}{2} \]
\[ R_2 = -12 \pm 10\sqrt{2} \]

8. **Calculating the Positive Root**:
Since resistance cannot be negative, we take the positive root:
\[ R_2 = -12 + 10\sqrt{2} \]

9. **Finding \( R_1 \)**:
Since \( R_1 = R_2 \), we conclude:
\[ R_1 = R_2 = -12 + 10\sqrt{2} \approx 32 \, \Omega \]

### Final Answer:
Both \( R_1 \) and \( R_2 \) are approximately 32 Ω.