Question

A tube is mounted so that its base is at height 'h' above the horizontal ground. The tank is filled with water to a depth h. A hole is punched in the side of the tank at depth y below the water surface. Then the value of y so that the range of the emerging stream would be maximum is?
A) $h$
B) $h/2$
C) $h/4$
D) $3h/4$

Answer 1

Hint: The velocity of the water coming out from the hole will increase with increasing depth. Higher the velocity of the water, the higher the range it will cover.
Formula used: In this solution, we will use the following formula:
Bernoulli’s Equation: ${P_{atm}} + \rho gh + \dfrac{1}{2}\rho {v^2} = {\text{constant}}$ where $P$ is the pressure of the water at a certain point, ${P_{atm}}$ is the atmospheric pressure, $h$ is the height of the system, and $v$ is the velocity of the water

Complete step by step answer:
The velocity of the water from the hole can be calculated from Bernoulli’s theorem. Since the water coming out of the container will experience atmospheric pressure, using Bernoulli’s theorem inside the tube and just outside the hole where the water is flowing out, we get
$\rho gy = \dfrac{1}{2}\rho {v^2}$
Solving for $v$, we get
$v = \sqrt {2gy} $
We can see that the velocity of the projectile will increase with the increasing depth of the tube. So, the maximum velocity will be at a depth of $h$.

Hence the correct choice is option (A).

Note: The reason for the maximum velocity of water having maximum range is because it acts as half of a projectile motion. Also, the bottom of the tube is mounted at a platform that is at a height $h$ above the ground which is also equal to the height of the tube. If the height of the tube and the platform were not equal, we would have to calculate the range of the water projectile and differentiate it with respect to height to find the depth for maximum range.