A sniper fires a rifle bullet into a gasoline tank making a hole `53.0 m` below the surface of gasoline. The tank was sealed at 3.10 atm . The stored gasoline has a density of `660 kgm`. The velocity with which gasoline begins to shoot out of the hole is

Options

A. `27.8ms^(-1)`
B. `41.0ms^(-1)`
C. `9.6ms^(-1)`
D. `19.7ms^(-1)`

Answer

To find the velocity with which gasoline begins to shoot out of the hole in the tank, we can use Bernoulli's theorem. Here’s a step-by-step solution:

### Step 1: Understand the scenario
We have a gasoline tank sealed at a pressure of 3.1 atm, and there is a hole 53.0 m below the surface of the gasoline. The density of gasoline is given as 660 kg/m³.

### Step 2: Identify the points in the system
Let’s denote:
- Point A: the surface of the gasoline in the tank.
- Point B: the hole where the gasoline is shooting out.

### Step 3: Apply Bernoulli’s equation
Bernoulli's equation states that for incompressible, non-viscous fluids, the following holds true between two points in a flow:
\[ P_A + \frac{1}{2} \rho v_A^2 + \rho g h_A = P_B + \frac{1}{2} \rho v_B^2 + \rho g h_B \]

Where:
- \( P \) = pressure
- \( \rho \) = density of the fluid
- \( v \) = velocity of the fluid
- \( g \) = acceleration due to gravity (approximately 9.81 m/s²)
- \( h \) = height above a reference point

### Step 4: Set up the equation
At point A (the surface):
- The pressure \( P_A = 3.1 \, \text{atm} = 3.1 \times 10^5 \, \text{Pa} \)
- The height \( h_A = 0 \) (taking the surface as the reference level)
- The velocity \( v_A \approx 0 \) (the fluid at rest)

At point B (the hole):
- The pressure \( P_B = P_{atm} = 1 \, \text{atm} = 1 \times 10^5 \, \text{Pa} \)
- The height \( h_B = -53.0 \, \text{m} \) (53 m below the surface)
- The velocity \( v_B = v_0 \) (the velocity we want to find)

### Step 5: Substitute into Bernoulli’s equation
Substituting the known values into Bernoulli's equation gives:
\[ 3.1 \times 10^5 + 0 + 0 = 1 \times 10^5 + \frac{1}{2} \cdot 660 \cdot v_0^2 - 660 \cdot 9.81 \cdot 53 \]

### Step 6: Simplify the equation
Rearranging the equation:
\[ 3.1 \times 10^5 - 1 \times 10^5 = \frac{1}{2} \cdot 660 \cdot v_0^2 - 660 \cdot 9.81 \cdot 53 \]

This simplifies to:
\[ 2.1 \times 10^5 = \frac{1}{2} \cdot 660 \cdot v_0^2 - 660 \cdot 9.81 \cdot 53 \]

### Step 7: Calculate the gravitational term
Calculating \( 660 \cdot 9.81 \cdot 53 \):
\[ 660 \cdot 9.81 \cdot 53 \approx 344,658 \]

### Step 8: Substitute back and solve for \( v_0^2 \)
Now substituting back:
\[ 2.1 \times 10^5 + 344,658 = \frac{1}{2} \cdot 660 \cdot v_0^2 \]
\[ 534658 = 330 \cdot v_0^2 \]

### Step 9: Solve for \( v_0^2 \)
\[ v_0^2 = \frac{534658}{330} \approx 1624.6 \]

### Step 10: Take the square root to find \( v_0 \)
\[ v_0 = \sqrt{1624.6} \approx 40.31 \, \text{m/s} \]

### Final Answer
The velocity with which gasoline begins to shoot out of the hole is approximately **41 m/s**.

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