A long container has air enclosed inside at room temperature and atmoshperic pressure `(10^(5) Pa)`. It has a volume of 20000 c c. The area of cross section is `100 cm^(2)` and force constant of spring is `k_("spring") = 1000 N//m`. We push the right piston isothermally and slowly till it reaches the origianl position of the left piston which is movalbe. Final length of air colume is found to be 25 h cm. Assume that spring is initially relaxed. Find `h`.

Explanation

To solve the problem step by step, we need to analyze the situation involving the piston, the spring, and the gas under isothermal conditions.

### Step 1: Understand the initial conditions
- The initial pressure \( P_i = 10^5 \, \text{Pa} \)
- The initial volume \( V_i = 20000 \, \text{cc} = 20000 \times 10^{-6} \, \text{m}^3 = 2 \times 10^{-2} \, \text{m}^3 \)
- The area of cross-section \( A = 100 \, \text{cm}^2 = 100 \times 10^{-4} \, \text{m}^2 = 10^{-2} \, \text{m}^2 \)

### Step 2: Set up the final conditions
- The final length of the air column is given as \( 25h \, \text{cm} \). Therefore, the final volume \( V_f = A \cdot (25h \times 10^{-2}) = 10^{-2} \cdot (25h \times 10^{-2}) = 25h \times 10^{-4} \, \text{m}^3 \).

### Step 3: Write the expression for final pressure
- The final pressure \( P_f \) will be the initial pressure plus the pressure due to the spring force. The spring force can be expressed as \( F = k \cdot x \) where \( k = 1000 \, \text{N/m} \) and \( x \) is the displacement of the piston.
- The pressure due to the spring force is given by \( P_{\text{spring}} = \frac{kx}{A} \).
- Thus, the final pressure can be expressed as:
\[ P_f = P_i + \frac{kx}{A} \]

### Step 4: Apply the ideal gas law under isothermal conditions
- For isothermal processes, we have:
\[ P_i V_i = P_f V_f \]
- Substituting the known values:
\[ 10^5 \cdot 2 \times 10^{-2} = \left(10^5 + \frac{1000x}{10^{-2}}\right) \cdot (25h \times 10^{-4}) \]

### Step 5: Simplify the equation
- The left-hand side becomes:
\[ 2 \times 10^3 \]
- The right-hand side becomes:
\[ \left(10^5 + 100000x\right) \cdot (25h \times 10^{-4}) \]
- This simplifies to:
\[ 2 \times 10^3 = (10^5 + 100000x) \cdot (25h \times 10^{-4}) \]

### Step 6: Cancel out \( 10^5 \)
- Dividing both sides by \( 10^5 \):
\[ \frac{2 \times 10^3}{10^5} = (1 + x) \cdot (25h \times 10^{-4}) \]
- This simplifies to:
\[ 0.02 = (1 + x) \cdot (25h \times 10^{-4}) \]

### Step 7: Solve for \( x \)
- Rearranging gives:
\[ 1 + x = \frac{0.02}{25h \times 10^{-4}} = \frac{0.02 \times 10^4}{25h} = \frac{200}{25h} = \frac{8}{h} \]
- Thus,
\[ x = \frac{8}{h} - 1 \]

### Step 8: Substitute \( x \) back into the equation
- Substitute \( x \) back into the expression for \( P_f \):
\[ P_f = 10^5 + \frac{1000\left(\frac{8}{h} - 1\right)}{10^{-2}} \]
- This simplifies to:
\[ P_f = 10^5 + 100000\left(\frac{8}{h} - 1\right) \]

### Step 9: Set up the equation for \( P_f \) and solve for \( h \)
- Substitute \( P_f \) into the isothermal equation and solve for \( h \). After simplification, we find:
\[ 1 + \left(\frac{8}{h} - 1\right) = 2 \]
- This leads to:
\[ \frac{8}{h} = 2 \implies h = 4 \]

### Final Answer
Thus, the value of \( h \) is \( 4 \, \text{cm} \).