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 to parallel to X,the balancing point is 50 cm 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 will use the principles of a meter bridge and the concept of resistances in series and parallel.

### Step 1: Understand the setup
We have two unknown resistances, X and Y, connected to a meter bridge. The balancing point is initially at 80 cm from the left side.

### Step 2: Use the meter bridge formula
From the meter bridge, we know that the ratio of the resistances is equal to the ratio of the lengths from the balancing point:
\[ \frac{X}{Y} = \frac{80}{20} \]
This simplifies to:
\[ \frac{X}{Y} = 4 \quad \text{(Equation 1)} \]
From this, we can express X in terms of Y:
\[ X = 4Y \]

### Step 3: Analyze the second scenario
When a 10 Ω resistance is connected in parallel to X, the new balancing point is at 50 cm from the left. The effective resistance of X in parallel with 10 Ω is given by:
\[ R_{eff} = \frac{10X}{10 + X} \]
Now, using the meter bridge again, we have:
\[ \frac{R_{eff}}{Y} = \frac{50}{50} \]
This simplifies to:
\[ R_{eff} = Y \quad \text{(Equation 2)} \]

### Step 4: Substitute for \( R_{eff} \)
Substituting the expression for \( R_{eff} \) into Equation 2 gives:
\[ \frac{10X}{10 + X} = Y \]

### Step 5: Substitute \( X \) from Equation 1
Now substitute \( X = 4Y \) into the equation:
\[ \frac{10(4Y)}{10 + 4Y} = Y \]
This simplifies to:
\[ \frac{40Y}{10 + 4Y} = Y \]

### Step 6: Cross-multiply to solve for Y
Cross-multiplying gives:
\[ 40Y = Y(10 + 4Y) \]
Expanding the right side:
\[ 40Y = 10Y + 4Y^2 \]
Rearranging the equation:
\[ 4Y^2 - 30Y = 0 \]

### Step 7: Factor the equation
Factoring out Y gives:
\[ Y(4Y - 30) = 0 \]
This gives us two solutions:
1. \( Y = 0 \) (not valid)
2. \( 4Y - 30 = 0 \) leading to \( Y = \frac{30}{4} = 7.5 \, \Omega \)

### Step 8: Find X using Y
Now substitute \( Y = 7.5 \, \Omega \) back into Equation 1 to find X:
\[ X = 4Y = 4 \times 7.5 = 30 \, \Omega \]

### Final Answer
Thus, the values of X and Y are:
\[ X = 30 \, \Omega, \quad Y = 7.5 \, \Omega \]