Question

A water tank filled with water up to a height H hole is made in the tank wall at a depth h from the surface of water. Then the velocity of efflux
\[\begin{align}
  & A.\,\sqrt{gh} \\
 & B.\,\sqrt{2gh} \\
 & C.\,2gh \\
 & D.\,\rho gh \\
\end{align}\]

Answer 1

Hint: This question can be solved using the concept of either Bernoulli’s theorem or Torricelli’s law. The atmosphere at the hole of the tank will be different from the atmosphere at the top of the tank, thus, the pressure will also be different, so, using this condition, we will solve this question.
Formula used:
\[v=\sqrt{2gh}\]

Complete answer:
From the given information, we have the data as follows.
A water tank filled with water up to a height H hole is made in the tank wall at a depth h from the surface of the water.
The atmosphere at the hole of the tank will be different from the atmosphere at the top of the tank, thus, the pressure will also be different.
The liquid at the top surface of the tank will be at rest, whereas, the liquid at the hole of the tank will be in motion. Let v be the velocity of efflux. Let the pressure at the top of the tank be P and let the pressure at the hole of the tank be, \[{{P}_{0}}\]. Thus, using Bernoulli’s equation, we can represent the equations as,
\[{{P}_{0}}+\dfrac{1}{2}\rho {{v}^{2}}+\rho gh=P+\rho gH\]
The velocity of the efflux can be expressed as, \[v=\sqrt{\dfrac{2(P-{{P}_{0}})}{\rho }+2gh}\]
As the pressure equals, when the water starts to flow, so, \[P={{P}_{0}}\].
\[\begin{align}
  & v=\sqrt{\dfrac{2({{P}_{0}}-{{P}_{0}})}{\rho }+2gh} \\
 & \Rightarrow v=\sqrt{\dfrac{2(0)}{\rho }+2gh} \\
 & \therefore v=\sqrt{2gh} \\
\end{align}\]
\[\therefore \] The velocity of efflux is \[\sqrt{2gh}\].

Note:
Torricelli’s law – in the case of an open tank, the speed of the liquid horizontally flowing out through a hole at a distance below the surface equals the speed acquired by a vertically freely falling object from a distance. Bernoulli’s theorem deals with the study of fluids in steady or streamlines flow.