Two unkonwn resistances X and Y are connected to left and right gaps of a meter bridge and the balancing point is obtained at 80cm from left. When a `10 Omega` resistance is connected parallel to X, the balancing point is 50cm from left. The values of X and Y respectively are

Options

A. `40 Omega, 9 Omega`
B. `30 Omega, 7.5 Omega`
C. `20 Omega, 6 Omega`
D. `10 Omega, 3 Omega`

Answer

To solve the problem, we need to find the values of the unknown resistances \( X \) and \( Y \) using the information provided about the meter bridge and the balancing points.

### Step-by-Step Solution

1. **Understanding the Meter Bridge Principle**:
The meter bridge operates on the principle of Wheatstone bridge. The relationship between the resistances and the balancing point is given by:
\[ \frac{X}{Y} = \frac{L}{100 - L} \]
where \( L \) is the distance from the left end to the balancing point.

2. **First Balancing Point**:
From the problem, when the balancing point is at 80 cm from the left, we can substitute \( L = 80 \) cm into the equation:
\[ \frac{X}{Y} = \frac{80}{100 - 80} = \frac{80}{20} = 4 \]
This implies:
\[ X = 4Y \quad \text{(Equation 1)} \]

3. **Second Balancing Point with Parallel Resistance**:
When a \( 10 \, \Omega \) resistor is connected in parallel with \( X \), the new equivalent resistance \( R_{eq} \) of \( X \) and \( 10 \, \Omega \) is given by:
\[ R_{eq} = \frac{X \times 10}{X + 10} \]
The new balancing point is at 50 cm from the left, so we can set up the equation:
\[ \frac{R_{eq}}{Y} = \frac{50}{100 - 50} = \frac{50}{50} = 1 \]
This implies:
\[ R_{eq} = Y \quad \text{(Equation 2)} \]

4. **Substituting \( R_{eq} \) into Equation 2**:
Substitute \( R_{eq} \) from the earlier expression into Equation 2:
\[ \frac{X \times 10}{X + 10} = Y \]

5. **Using Equation 1**:
From Equation 1, we know \( Y = \frac{X}{4} \). Substitute this into the equation:
\[ \frac{X \times 10}{X + 10} = \frac{X}{4} \]

6. **Cross Multiplying**:
Cross-multiply to eliminate the fraction:
\[ 10X = \frac{X}{4}(X + 10) \]
Simplifying gives:
\[ 10X = \frac{X^2 + 10X}{4} \]
Multiply through by 4 to eliminate the fraction:
\[ 40X = X^2 + 10X \]
Rearranging gives:
\[ X^2 - 30X = 0 \]
Factoring out \( X \):
\[ X(X - 30) = 0 \]
Thus, \( X = 30 \, \Omega \) (since resistance cannot be zero).

7. **Finding \( Y \)**:
Substitute \( X \) back into Equation 1 to find \( Y \):
\[ Y = \frac{X}{4} = \frac{30}{4} = 7.5 \, \Omega \]

### Final Answer
The values of \( X \) and \( Y \) are:
\[ X = 30 \, \Omega, \quad Y = 7.5 \, \Omega \]