A tank of height H is fully filled with water. If the water rushing from a hole made in the tank below the free surface , strikes the floor at a maximum horizontal distance , then the depth of the hole from the free surface must be

Options

A. `3/4 H`
B. `2/3H`
C. `1/4H`
D. `1/2H`

Answer

To solve the problem of finding the depth of the hole from the free surface of a tank filled with water, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We have a tank of height \( H \) filled with water. A hole is made below the free surface, and we need to find the depth \( h \) of this hole from the free surface such that the water strikes the floor at the maximum horizontal distance.

2. **Use Bernoulli's Equation**: We can apply Bernoulli's equation between two points: one just inside the hole (point 1) and one just outside the hole (point 2). The equation is given by:
\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 \]
Here, \( P_1 \) and \( P_2 \) are the pressures at points 1 and 2, \( v_1 \) and \( v_2 \) are the velocities at those points, \( \rho \) is the density of water, and \( g \) is the acceleration due to gravity.

3. **Set Up the Equation**: For point 1 (just inside the hole):
- Pressure \( P_1 = P_0 \) (atmospheric pressure)
- Height \( h_1 = H - h \) (height from the bottom of the tank)
- Velocity \( v_1 \) is negligible since the tank is large compared to the hole.

For point 2 (just outside the hole):
- Pressure \( P_2 = P_0 \)
- Height \( h_2 = 0 \) (at the level of the hole)
- Velocity \( v_2 \) is what we need to find.

Thus, the equation simplifies to:
\[ P_0 + \rho g (H - h) = P_0 + \frac{1}{2} \rho v_2^2 \]
Canceling \( P_0 \) and rearranging gives:
\[ \rho g (H - h) = \frac{1}{2} \rho v_2^2 \]
Dividing by \( \rho \) gives:
\[ g (H - h) = \frac{1}{2} v_2^2 \]
Therefore, we can express \( v_2 \) as:
\[ v_2 = \sqrt{2g(H - h)} \]

4. **Calculate the Time of Flight**: The time \( t \) it takes for the water to fall from height \( h \) to the ground can be calculated using the equation of motion:
\[ h = \frac{1}{2} g t^2 \implies t = \sqrt{\frac{2h}{g}} \]

5. **Calculate the Horizontal Range**: The horizontal range \( R \) is given by the product of the horizontal velocity and the time of flight:
\[ R = v_2 \cdot t = \sqrt{2g(H - h)} \cdot \sqrt{\frac{2h}{g}} = 2\sqrt{h(H - h)} \]

6. **Maximize the Range**: To find the maximum range, we can differentiate \( R \) with respect to \( h \) and set it to zero:
\[ R = 2\sqrt{h(H - h)} \]
Let \( R^2 = 4h(H - h) \). Differentiate:
\[ \frac{d(R^2)}{dh} = 4(H - 2h) = 0 \implies H - 2h = 0 \implies h = \frac{H}{2} \]

7. **Conclusion**: The depth of the hole from the free surface for maximum horizontal distance is:
\[ h = \frac{H}{2} \]

### Final Answer:
The depth of the hole from the free surface must be \( \frac{H}{2} \).

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