Question

A large vessel of height H, is filled with a liquid of density $\rho $ , up to the brim. A small hole of radius r is made at the side vertical face, close to the base. What is the horizontal force required to stop the gushing of liquid?
A) $2\left( {\rho gH} \right)\pi {r^2}$
B) \[\rho gH\]
C) $\rho gH\pi r$
D) $\rho gH\pi {r^2}$

Answer 1

Hint: Using Bernoulli’s principle at the two points i.e. top and bottom (closet to base).
In the second term the Pressure term will equal to the sum of atmospheric pressure and the pressure due to applied force. Mathematically, $P = {P_0} + \dfrac{F}{A}$

Complete step by step answer:We have to use Bernoulli’s principle to find the force applied. Bernoulli’s principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid's potential energy. Mathematically,
$P + \rho gh + \dfrac{1}{2}\rho {v^2} = C$
The small hole close to the base can be assumed to be situated at base only. Let the points at top and the hole be A and B respectively.
Let the atmospheric pressure be ${P_0}$ . Let the heights of points A and B be ${h_1}$ and ${h_2}$ respectively. Let the velocities at point A and B be \[{v_1}\] and ${v_2}$ respectively. Applying Bernoulli’s principle at points A and B,
\[
  {P_1} + \rho g{h_1} + \dfrac{1}{2}\rho v_1^2 = {P_2} + \rho g{h_2} + \dfrac{1}{2}\rho v_2^2 \\
   \Rightarrow {P_0} + \rho gH + 0 = {P_2} + 0 + 0 \\
   \Rightarrow {P_0} + \rho gH = {P_2} \\
 \]
(because the liquid is static, the velocities at both points are zero)
Now, we can see that the pressure at point B (hole) would actually be balancing the atmospheric pressure from outside plus the pressure of the force applied by us. Let the force applied be F.
$\therefore {P_2} = {P_0} + \dfrac{F}{A}$ where A is the area of hole $\left( {A = \pi {r^2}} \right)$
Substituting the above value in the previous equation we get,
\[
  {P_0} + \rho gH = {P_2} \\
   \Rightarrow {P_0} + \rho gH = {P_0} + \dfrac{F}{A} \\
   \Rightarrow F = \rho gH \times A \\
   \Rightarrow F = \rho gH\pi {r^2} \\
 \]
Hence, we get the required force. Therefore, the correct option is D) $\rho gH\pi {r^2}$

Note:We can also do the question by using the trick that the force required to stop the liquid would be equal to the pressure at that point multiplied by the area.
$
  F = P \times A \\
   \Rightarrow F = \rho gH \times \pi {r^2} \\
   \Rightarrow F = \rho gH\pi {r^2} \\
 $