If we use `100 Omega` and `200 Omega` in place of `R` and `X` we get null pont deflection, `l = 33cm`. If we intercharge the resistors, the null point length is found to be `67 cm` Find end corrections `prop and beta`.

Explanation

To solve the problem, we need to find the end corrections \( \alpha \) and \( \beta \) using the given formulas and data. Let's break down the steps:

### Step 1: Identify the Given Data
- When \( R = 100 \, \Omega \) and \( X = 200 \, \Omega \):
- Null point length \( L_1 = 33 \, \text{cm} \)
- When \( R = 200 \, \Omega \) and \( X = 100 \, \Omega \):
- Null point length \( L_2 = 67 \, \text{cm} \)

### Step 2: Use the Formula for \( \alpha \)
The formula for \( \alpha \) is given by:
\[ \alpha = \frac{X \cdot L_1 - R \cdot L_2}{R - X} \]
Substituting the values:
- \( X = 200 \)
- \( R = 100 \)
- \( L_1 = 33 \)
- \( L_2 = 67 \)

\[ \alpha = \frac{200 \cdot 33 - 100 \cdot 67}{100 - 200} \]

### Step 3: Calculate \( \alpha \)
First, calculate the numerator:
\[ 200 \cdot 33 = 6600 \]
\[ 100 \cdot 67 = 6700 \]
So,
\[ \alpha = \frac{6600 - 6700}{100 - 200} = \frac{-100}{-100} = 1 \, \text{cm} \]

### Step 4: Use the Formula for \( \beta \)
The formula for \( \beta \) is given by:
\[ \beta = \frac{R \cdot L_1 - X \cdot L_2}{R - X} - 100 \]
Substituting the values:
\[ \beta = \frac{100 \cdot 33 - 200 \cdot 67}{100 - 200} - 100 \]

### Step 5: Calculate \( \beta \)
First, calculate the numerator:
\[ 100 \cdot 33 = 3300 \]
\[ 200 \cdot 67 = 13400 \]
So,
\[ \beta = \frac{3300 - 13400}{100 - 200} - 100 = \frac{-10000}{-100} - 100 = 100 - 100 = 0 \, \text{cm} \]

### Step 6: Finalize the Results
Thus, the end corrections are:
- \( \alpha = 1 \, \text{cm} \)
- \( \beta = 0 \, \text{cm} \)

### Summary
The final answers are:
- \( \alpha = 1 \, \text{cm} \)
- \( \beta = 0 \, \text{cm} \)