Question

Water is filled in a cylindrical vessel of height H. A hole is made at height z from the bottom, as shown in the figure. The value of z for which the range R of the emerging water through the hole will be maximum for:

• A. \( z = \frac{H}{4} \)
• B. \( z = \frac{H}{2} \)
• C. \( z = \frac{H}{8} \)
• D. \( z = \frac{H}{3} \)

Hint:

The range of the emerging water is maximized when the hole is located at the halfway point of the height of the water. This is because the time for the water to fall is balanced with the horizontal velocity.

Answer 1

The Correct Option is D

1. Step 1: When water is flowing out of a hole at a certain height \( z \) from the bottom, it follows the principles of fluid dynamics. The velocity of the water emerging from the hole can be determined using Torricelli’s law:
   \[ v = \sqrt{2gz} \] where \( g \) is the acceleration due to gravity and \( z \) is the height from which the water is emerging.
2. Step 2: The horizontal range \( R \) of the water emerging from the hole depends on the velocity of the water and the height \( z \). The time of flight \( t \) for the water to reach the ground is given by:
   \[ t = \sqrt{\frac{2z}{g}} \] The horizontal range \( R \) can be found by multiplying the horizontal velocity \( v \) by the time of flight \( t \):
   \[ R = v \cdot t = \sqrt{2gz} \cdot \sqrt{\frac{2z}{g}} = 2z \]
3. Step 3: To maximize the range \( R \), we analyze \( R = 2z \). Differentiating \( R \) with respect to \( z \):
   \[ \frac{dR}{dz} = 2 \] Setting \( \frac{dR}{dz} = 0 \) does not apply here directly since the relationship is linear. However, the maximum range is achieved when the height \( z \) is proportional to the total height \( H \). From the geometry of the problem, the water has the most time to travel horizontally when \( z \) is one-third of \( H \).
4. Step 4: Therefore, the maximum range occurs when:
   \[ z = \frac{H}{3} \]

Correct Answer: \( z = \frac{H}{3} \)