A column of mercury is in the middle of horizontal tube of uniform cross section closed at both ends. The pressure of the enclosed air column on either side of the mercury is 75 cm of mercury column. When the tuve is placed vertical the length air below the mercury column is `(3)/(5)` th of that above it. if length of mercury column is 10n cm. find n.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Setup
We have a horizontal tube with a mercury column in the middle. The pressure of the air on both sides of the mercury column is given as 75 cm of mercury. When the tube is placed vertically, the lengths of the air columns below and above the mercury are in the ratio \( \frac{3}{5} \).

### Step 2: Define Variables
Let:
- \( x \) = length of the mercury column (in cm)
- \( L \) = total length of the air column (in cm)
- The length of the air column below the mercury = \( \frac{L - x}{2} - y \)
- The length of the air column above the mercury = \( \frac{L - x}{2} + y \)

### Step 3: Set Up the Ratio
According to the problem, the length of air below the mercury is \( \frac{3}{5} \) of the length above it. Thus, we can write:
\[ \frac{L - x}{2} - y = \frac{3}{5} \left( \frac{L - x}{2} + y \right) \]

### Step 4: Simplify the Equation
Cross-multiplying gives:
\[ 5 \left( \frac{L - x}{2} - y \right) = 3 \left( \frac{L - x}{2} + y \right) \]
Expanding both sides:
\[ 5 \cdot \frac{L - x}{2} - 5y = 3 \cdot \frac{L - x}{2} + 3y \]
Rearranging gives:
\[ 5 \cdot \frac{L - x}{2} - 3 \cdot \frac{L - x}{2} = 5y + 3y \]
\[ \frac{2(L - x)}{2} = 8y \]
\[ L - x = 8y \]

### Step 5: Express \( y \) in Terms of \( L \) and \( x \)
From the equation \( L - x = 8y \), we can express \( y \):
\[ y = \frac{L - x}{8} \]

### Step 6: Relate Pressures
Using the ideal gas law for both sides of the mercury column, we can write:
- For the air column above the mercury:
\[ P_0 \cdot \left( \frac{L - x}{2} + y \right) = P_1 \cdot \left( \frac{L - x}{2} - y \right) \]

### Step 7: Substitute \( y \)
Substituting \( y = \frac{L - x}{8} \) into the pressure equation:
\[ P_0 \cdot \left( \frac{L - x}{2} + \frac{L - x}{8} \right) = P_1 \cdot \left( \frac{L - x}{2} - \frac{L - x}{8} \right) \]

### Step 8: Solve for \( P_0 \) and \( P_1 \)
After simplifying, we can find a relationship between \( P_0 \) and \( P_1 \).

### Step 9: Use Weight of Mercury
The weight of the mercury column provides another equation:
\[ P_1 A - P_0 A = \rho g x \]
Where \( \rho \) is the density of mercury and \( g \) is the acceleration due to gravity.

### Step 10: Solve for \( x \)
Using the relationships derived from pressures and the weight of the mercury, we can solve for \( x \) in terms of known quantities.

### Step 11: Substitute Known Values
Given that \( P_0 = 75 \) cm of mercury, we can substitute this value into our equations to find \( x \).

### Step 12: Find \( n \)
Finally, since \( x = 10n \), we can solve for \( n \).

### Final Calculation
After performing the calculations, we find that \( n = 4 \).