Question

What work should be done in order to squeeze all the water from a horizontally located cylinder (figure shown above) during the time t by means of a constant force acting on the piston? The volume of a water in the cylinder is equal to V, the cross-sectional area of the orifice to s, with s being considerably less than the piston area. The friction and the viscosity are negligibly small.

A. $A = \dfrac{1}{2}\rho \dfrac{{{V^3}}}{{{{(St)}^2}}}$
B. $A = \dfrac{3}{2}\rho \dfrac{{{V^3}}}{{{{(St)}^2}}}$
C. $A = \dfrac{5}{2}\rho \dfrac{{{V^3}}}{{{{(St)}^2}}}$
D. None of these

Answer 1

Hint: Velocity at which water ejects out of the orifice is called discharge velocity or volume discharge rate. It is given by $v = \dfrac{{Volume}}{{Area \times time}}$

Formula Used:
1. Volume discharge rate velocity, $v = \dfrac{{Volume}}{{Area \times time}}$ …… (a)
2. Kinetic energy of mass moving with velocity $v$ given by, $K.E = \dfrac{1}{2}m{v^2}$ ……. (b)

Complete step by step answer:
Given,
density of liquid in container $\rho$
Total volume of liquid to expel out $V$
Cross-sectional area of orifice $S$

Step 1 of 5:

From equation (a), Discharge rate $v$ of liquid out of orifice equals
$v = \dfrac{{Volume}}{{Area \times time}}$

$\Rightarrow v = \dfrac{V}{{St}}$ …… (1)

Step 2 of 5:
From work energy theorem, we know change in Kinetic energy is total work done by piston-
$\Rightarrow \Delta K.E = \dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2}$

Step 3 of 5:
But, let’s say initial velocity is 0 (no water coming out before time t=0)
$\Rightarrow K.E = \dfrac{1}{2}m{v^2}$ …… (2)

Step 4 of 5:
Total mass of water coming out through orifice $M = \rho V$…… (3)

Step 5 of 5:
Putting values from equation (1) and (3) in equation (2), we get,
Kinetic Energy, $A = \dfrac{1}{2}\rho \dfrac{{{V^3}}}{{{{(St)}^2}}}$

Correct Answer: A. $A = \dfrac{1}{2}\rho \dfrac{{{V^3}}}{{{{(St)}^2}}}$

Additional Information:
Another approach of solving the problem is by using Bernoulli principle of fluid flow. Here, Force is acting perpendicular to the cross section area. So, pressure can be calculated as $P = \dfrac{F}{A}$where F is a constant force on the piston. Using Bernoulli’s equation force given by,
$F = \dfrac{1}{2}\rho {v^2}A$ where, $v$ is velocity of flow out of orifice. Then, by deducting the value of $v$ from equation (1) and putting in the equation of total work done. We will get the same answer.

Note: In order to use Bernoulli equation mark two points A and B. Bernoulli equation ensures continuity of flow.