A cylinder is filled with non viscous liquid of density d to a hegiht `h_(0)` and a hole is made at a height `h_(1)` from the bottom of the cylinder. The velocity of liquid issuing out of the hole is

Options

A. `sqrt((2gh_(0)))`
B. `sqrt(2g(h_(0)-h_(1)))`
C. `sqrt(dgh_(1))`
D. `sqrt(dgh_(0))`

Answer

To find the velocity of the liquid issuing out of the hole in the cylinder, we can use Bernoulli's equation. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have a cylinder filled with a non-viscous liquid of density \( d \) to a height \( h_0 \). A hole is made at a height \( h_1 \) from the bottom of the cylinder. We need to determine the velocity of the liquid as it exits through the hole.

### Step 2: Identify Points for Bernoulli's Equation
We will apply Bernoulli's equation between two points:
- Point 1: At the free surface of the liquid (height \( h_0 \))
- Point 2: At the hole (height \( h_1 \))

### Step 3: Write Bernoulli's Equation
Bernoulli's equation states that:
\[ P_1 + \frac{1}{2} \rho V_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho V_2^2 + \rho g h_2 \]
Where:
- \( P_1 \) and \( P_2 \) are the pressures at points 1 and 2 respectively.
- \( V_1 \) and \( V_2 \) are the velocities at points 1 and 2 respectively.
- \( h_1 \) and \( h_2 \) are the heights at points 1 and 2 respectively.

### Step 4: Assign Values
- At point 1 (free surface), the pressure \( P_1 \) is atmospheric pressure \( P_0 \), and since the liquid is at rest, \( V_1 \approx 0 \).
- At point 2 (the hole), the pressure \( P_2 \) is also atmospheric pressure \( P_0 \).

Thus, we can simplify the equation:
\[ P_0 + 0 + \rho g h_0 = P_0 + \frac{1}{2} \rho V_2^2 + \rho g h_1 \]

### Step 5: Cancel Out Terms
Since \( P_0 \) appears on both sides, we can cancel it out:
\[ \rho g h_0 = \frac{1}{2} \rho V_2^2 + \rho g h_1 \]

### Step 6: Rearrange the Equation
Rearranging gives:
\[ \rho g h_0 - \rho g h_1 = \frac{1}{2} \rho V_2^2 \]
Factoring out \( \rho g \):
\[ \rho g (h_0 - h_1) = \frac{1}{2} \rho V_2^2 \]

### Step 7: Cancel \( \rho \)
We can cancel \( \rho \) from both sides (since the liquid is non-viscous and incompressible):
\[ g (h_0 - h_1) = \frac{1}{2} V_2^2 \]

### Step 8: Solve for \( V_2 \)
Multiplying both sides by 2:
\[ 2g (h_0 - h_1) = V_2^2 \]
Taking the square root:
\[ V_2 = \sqrt{2g (h_0 - h_1)} \]

### Final Answer
The velocity of the liquid issuing out of the hole is:
\[ V = \sqrt{2g (h_0 - h_1)} \]

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