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

Options

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

Explanation

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

### Step 1: Set up the equations based on the meter bridge principle.
In a meter bridge, the ratio of the resistances is equal to the ratio of the lengths from the ends of the bridge to the balancing point.

Given that the balancing point is at 80 cm from the left, we can write:
\[ \frac{X}{Y} = \frac{L}{100 - L} = \frac{80}{20} = 4 \]
This gives us our first equation:
\[ X = 4Y \quad \text{(Equation 1)} \]

### Step 2: Analyze the second scenario with the 10 Ω resistor.
When a 10 Ω resistor is connected in parallel with resistance X, the equivalent resistance \( R_{eq} \) can be calculated as:
\[ R_{eq} = \frac{X \cdot 10}{X + 10} \]
In this case, the balancing point shifts to 50 cm from the left, so we can write:
\[ \frac{R_{eq}}{Y} = \frac{50}{50} = 1 \]
This implies:
\[ R_{eq} = Y \quad \text{(Equation 2)} \]

### Step 3: Substitute Equation 1 into Equation 2.
From Equation 1, we have \( Y = \frac{X}{4} \). Substituting this into Equation 2 gives:
\[ \frac{X \cdot 10}{X + 10} = \frac{X}{4} \]

### Step 4: Cross-multiply and simplify the equation.
Cross-multiplying gives:
\[ 4 \cdot (X \cdot 10) = X \cdot (X + 10) \]
This simplifies to:
\[ 40X = X^2 + 10X \]
Rearranging gives:
\[ X^2 - 30X = 0 \]

### Step 5: Factor the equation.
Factoring out X gives:
\[ X(X - 30) = 0 \]
This implies \( X = 0 \) or \( X = 30 \). Since resistance cannot be zero, we have:
\[ X = 30 \, \Omega \]

### Step 6: Find the value of Y.
Using Equation 1:
\[ Y = \frac{X}{4} = \frac{30}{4} = 7.5 \, \Omega \]

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