Two resistances X and Y in the left and right gaps of a metre bridge give a null point dividing the wire I the ratio of `2 : 3`. When each resistance is increased by `30 Omega`, the new null point divides the wire in the ratio of 5 : 6. What are the values of X and Y ?

Options

A. `X=20 Omega, Y=30 Omega`
B. `X=30 Omega, Y=20 Omega`
C. `X=10 Omega, Y=30 Omega`
D. `X=30 Omega, Y=10 Omega`

Explanation

To solve the problem, we will use the principles of the Wheatstone bridge and set up equations based on the given ratios.

### Step-by-Step Solution:

1. **Understanding the Initial Condition:**
The meter bridge gives a null point dividing the wire in the ratio of 2:3. This means:
\[ \frac{X}{Y} = \frac{2}{3} \]
From this, we can express \(Y\) in terms of \(X\):
\[ Y = \frac{3}{2}X \quad \text{(Equation 1)} \]

2. **Understanding the Condition After Increasing Resistances:**
When each resistance is increased by 30 Ω, the new null point divides the wire in the ratio of 5:6. This means:
\[ \frac{X + 30}{Y + 30} = \frac{5}{6} \]
Cross-multiplying gives:
\[ 6(X + 30) = 5(Y + 30) \]
Expanding this, we get:
\[ 6X + 180 = 5Y + 150 \]
Rearranging gives us:
\[ 6X - 5Y = -30 \quad \text{(Equation 2)} \]

3. **Substituting Equation 1 into Equation 2:**
We can substitute \(Y\) from Equation 1 into Equation 2:
\[ 6X - 5\left(\frac{3}{2}X\right) = -30 \]
Simplifying this:
\[ 6X - \frac{15}{2}X = -30 \]
To eliminate the fraction, multiply through by 2:
\[ 12X - 15X = -60 \]
This simplifies to:
\[ -3X = -60 \]
Thus:
\[ X = 20 \, \Omega \]

4. **Finding Y using Equation 1:**
Now, substitute \(X\) back into Equation 1 to find \(Y\):
\[ Y = \frac{3}{2} \times 20 = 30 \, \Omega \]

5. **Final Values:**
Therefore, the values of the resistances are:
\[ X = 20 \, \Omega \quad \text{and} \quad Y = 30 \, \Omega \]

### Summary:
The values of the resistances are:
- \(X = 20 \, \Omega\)
- \(Y = 30 \, \Omega\)