There is a small hole near the botton of an open tank filled with liquid. The speed of the water ejected does not depend on

Options

A. area of the hole
B. density of the liquid
C. height of the liquid from the hole
D. acceleration due to gravity

Answer

To solve the problem regarding the speed of water ejected from a small hole at the bottom of an open tank, we can use Bernoulli's theorem. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have an open tank filled with liquid, and there is a small hole near the bottom. The height of the liquid above the hole is denoted as \( H \).

**Hint:** Identify the two sections in the tank: the free surface (section 1) and the hole (section 2).

### Step 2: Apply Bernoulli’s Equation
According to Bernoulli's theorem, we can express the energy conservation between the two sections. The equation is given by:

\[ \frac{p_1}{\rho g} + \frac{v_1^2}{2g} + z_1 = \frac{p_2}{\rho g} + \frac{v_2^2}{2g} + z_2 \]

Where:
- \( p_1 \) and \( p_2 \) are the pressures at sections 1 and 2 respectively.
- \( v_1 \) and \( v_2 \) are the velocities at sections 1 and 2 respectively.
- \( z_1 \) and \( z_2 \) are the heights above a reference level at sections 1 and 2 respectively.

**Hint:** Identify the variables for each section and their values based on the setup.

### Step 3: Substitute Known Values
Since the tank is open to the atmosphere:
- \( p_1 = p_2 = P_{atm} \) (atmospheric pressure)
- The height \( z_1 = H \) (height of the liquid above the hole)
- The height \( z_2 = 0 \) (the hole is our reference level)

Thus, the equation simplifies to:

\[ \frac{P_{atm}}{\rho g} + \frac{v_1^2}{2g} + H = \frac{P_{atm}}{\rho g} + \frac{v_2^2}{2g} + 0 \]

**Hint:** Cancel out the atmospheric pressure terms and simplify the equation.

### Step 4: Neglect v1
Since the area of the free surface is much larger than the area of the hole, the velocity \( v_1 \) at the free surface can be considered negligible compared to \( v_2 \). Therefore, we can ignore \( v_1 \):

\[ H = \frac{v_2^2}{2g} \]

**Hint:** Recognize that the height \( H \) is the only significant factor affecting the velocity \( v_2 \).

### Step 5: Solve for v2
Rearranging the equation gives:

\[ v_2^2 = 2gH \]
\[ v_2 = \sqrt{2gH} \]

**Hint:** Identify the variables that \( v_2 \) depends on.

### Step 6: Analyze Dependencies
From the equation \( v_2 = \sqrt{2gH} \), we can see that:
- \( v_2 \) depends on the height \( H \) (the height of the liquid above the hole).
- \( v_2 \) depends on the acceleration due to gravity \( g \).
- \( v_2 \) does not depend on the area of the hole or the density of the liquid.

**Hint:** Conclude which options are correct based on the dependencies identified.

### Final Conclusion
The speed of the water ejected does not depend on:
- Option 1: Area of the hole (Correct)
- Option 2: Density of the liquid (Correct)
- Option 3: Height of the liquid from the hole (Incorrect)
- Option 4: Acceleration due to gravity (Incorrect)

Thus, the correct options are **1 and 2**.