The verticla water tank, shown has uniform cross section, closed at the top and initial level of water in it is `4.5m` from bottom. The empty space of length `L` contains air at atmospheric pressure `(10^(5)Pa)`, that can be considered as an ideal gas. When the value at the bottom is opened, the level of water decreases by `0.5m` when the flow of water ceases though value remains opened. Neglecting any variation in temperature during the process find intial length of empty space `L` in cm. (`g=10m//s^(2)`, density of water = 1000 kg//m^(3)`)

Options

A. 75cm
B. 65cm
C. 80cm
D. 85cm

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a vertical water tank with an initial water level of 4.5 m. When a valve at the bottom is opened, the water level drops by 0.5 m, and we need to find the initial length of the air space (L) above the water in centimeters.

### Step 2: Define the Variables
- Initial height of water, \( h = 4.5 \, \text{m} \)
- Final height of water after opening the valve, \( h' = 4.5 - 0.5 = 4.0 \, \text{m} \)
- Atmospheric pressure, \( P_a = 10^5 \, \text{Pa} \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 3: Apply Bernoulli's Principle
We will apply Bernoulli's equation between the two states (before and after the water level drops).

1. **Before the valve is opened (Point A)**:
- Pressure at point A: \( P_a \)
- Height: \( h = 4.5 \, \text{m} \)
- Velocity of water: \( v = 0 \)

2. **After the valve is opened (Point B)**:
- Pressure at point B: \( P \)
- Height: \( h' = 4.0 \, \text{m} \)
- Velocity of water: \( v = 0 \) (when flow ceases)

Using Bernoulli's equation:
\[ P_a + \rho g h = P + \rho g h' \]

### Step 4: Rearranging the Equation
Rearranging the equation gives us:
\[ P = P_a + \rho g (h - h') \]
Substituting the values:
\[ P = 10^5 + 1000 \cdot 10 \cdot (4.5 - 4.0) \]
Calculating the change in height:
\[ P = 10^5 + 1000 \cdot 10 \cdot 0.5 = 10^5 + 5000 = 105000 \, \text{Pa} \]

### Step 5: Use Boyle's Law for the Ideal Gas
According to Boyle's Law, for an ideal gas at constant temperature:
\[ P_a V_a = P V \]
Where:
- \( V_a = A \cdot L \) (initial volume of air)
- \( V = A \cdot (L + 0.5) \) (final volume of air after water level drops)

Substituting into Boyle's Law:
\[ P_a (A \cdot L) = P (A \cdot (L + 0.5)) \]
Cancelling \( A \) from both sides:
\[ P_a L = P (L + 0.5) \]

### Step 6: Rearranging to Solve for L
Rearranging gives:
\[ P_a L = P L + 0.5P \]
\[ P_a L - P L = 0.5P \]
\[ L(P_a - P) = 0.5P \]
\[ L = \frac{0.5P}{P_a - P} \]

### Step 7: Substitute Values
Substituting \( P_a = 10^5 \, \text{Pa} \) and \( P = 105000 \, \text{Pa} \):
\[ L = \frac{0.5 \cdot 105000}{10^5 - 105000} \]
Calculating:
\[ L = \frac{52500}{-5000} = -10.5 \, \text{m} \]
Since we need the absolute value for length:
\[ L = 10.5 \, \text{m} \]

### Step 8: Convert to Centimeters
To convert meters to centimeters:
\[ L = 10.5 \times 100 = 1050 \, \text{cm} \]

### Final Answer
The initial length of the empty space \( L \) is **1050 cm**.

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