A pop- gun consists of a cylindrical barrel `3cm^(2)` in cross section closed at one end by a cork and having a well fitting piston at the other. If the piston is pushed slowly, in the cork is finally ejected, giving a pop, the frequency of which is found to be `512Hz`. Assuming that the initial distance between the cork and the piston was `25cm` and that there is no leaking of air, calculate the force required to eject the cork. Atmospheric pressure `=1kg.cm^(2)`, `v=340m//s` (in kg. wt).

Explanation

To solve the problem, we need to calculate the force required to eject the cork from the pop-gun. We will follow these steps:

### Step 1: Understand the system
The pop-gun consists of a cylindrical barrel with a cork at one end and a piston at the other. When the piston is pushed, it creates pressure that eventually ejects the cork. The frequency of the sound produced when the cork is ejected is given as 512 Hz.

### Step 2: Determine the wavelength
Since the cork is at one end (open) and the other end (piston) is closed, the system behaves like a closed pipe. The fundamental frequency of a closed pipe is given by:

\[ f = \frac{v}{4L} \]

Where:
- \( f \) = frequency (512 Hz)
- \( v \) = speed of sound (340 m/s)
- \( L \) = length of the air column (initially 25 cm)

### Step 3: Calculate the effective length of the air column
Rearranging the formula for \( L \):

\[ L = \frac{v}{4f} \]

Substituting the values:

\[ L = \frac{340 \, \text{m/s}}{4 \times 512 \, \text{Hz}} = \frac{340}{2048} \approx 0.165 \, \text{m} \approx 16.6 \, \text{cm} \]

### Step 4: Calculate the change in length of the air column
The initial length of the air column was 25 cm, and the effective length when the cork is ejected is approximately 16.6 cm. The change in length (\( \Delta L \)) is:

\[ \Delta L = 25 \, \text{cm} - 16.6 \, \text{cm} = 8.4 \, \text{cm} \]

### Step 5: Calculate the pressure difference
The pressure at the cork when it is ejected can be calculated using the formula for pressure in a closed system:

\[ P_1 = P_{atm} + \Delta P \]

Where:
- \( P_{atm} = 1 \, \text{kg/cm}^2 \) (atmospheric pressure)
- \( \Delta P = \frac{P_{atm} \cdot L}{L'} \)

Substituting the values:

\[ \Delta P = \frac{1 \, \text{kg/cm}^2 \cdot 25 \, \text{cm}}{16.6 \, \text{cm}} \approx 1.5 \, \text{kg/cm}^2 \]

### Step 6: Calculate the force
The force required to eject the cork can be calculated using the formula:

\[ F = P \cdot A \]

Where:
- \( A = 3 \, \text{cm}^2 \) (cross-sectional area)

Substituting the values:

\[ F = 1.5 \, \text{kg/cm}^2 \cdot 3 \, \text{cm}^2 = 4.5 \, \text{kg} \]

### Final Answer
The force required to eject the cork is **4.5 kg**.

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