Question

A liquid of density $800kg/{{m}^{3}}$ is filled in a cylindrical vessel up to a height of $3m$. This cylindrical vessel stands on a horizontal plane. There is a circular hole on the side of the vessel touching the bottom. What should be the minimum diameter of the hole to move the vessel on the floor, if the plug is removed? Take the coefficient of friction between the bottom of the vessel and the plane as $0.5$ and total mass of vessel plus liquid as $95kg$.
$A)\text{ }0.112m$
$B)\text{ }0.053m$
$C)\text{ }0.206m$
$D)\text{ }0.535m$

Answer 1

Hint: This problem can be solved by finding out the force being imparted to the water flowing out by finding the rate of change of its momentum. The same magnitude of force will be applied on the vessel and will try to move the vessel against the opposing force of friction.

Formula used:
${{v}_{efflux}}=\sqrt{2gh}$
$\text{mass = density}\times \text{volume}$
$F=\dfrac{\Delta p}{\Delta t}$
${{F}_{f}}=\mu N$
$p=mv$

Complete step-by-step answer:
We will find the force imparted to the water flowing out from the hole by finding its rate of change in momentum. Then we will equate the force with the opposing force of static friction to know the force while will be enough to move the vessel.
Now, let us analyze the question.
The mass of the liquid plus vessel is $M=95kg$.
The density of the liquid is $\rho =800kg/{{m}^{3}}$.
Liquid is filled up to a height of $h=3m$.
The coefficient of friction between the bottom of the vessel and the surface is $\mu =0.5$.
Let the normal force on the vessel from the ground be $N$.
Let the weight of the vessel be $W=Mg$.
Let the velocity of efflux of the water from the hole be ${{v}_{efflux}}$.
Let the diameter of the circular hole be $D$.
Let the area of the circular hole be $A$.
Let the force imparted to the water being ejected from the circular hole be ${{F}_{efflux}}$.
Let the mass of the liquid coming out from the hole per second be $m$.
Let the momentum of the liquid coming out from the hole be $p$.
Now, the velocity of efflux ${{v}_{efflux}}$ of liquid from a hole at the bottom of the vessel is given by
${{v}_{efflux}}=\sqrt{2gh}$ --(1)
Where $h$ is the height of the liquid in the vessel and $g$ is the acceleration due to gravity.
Using (1), we get
${{v}_{efflux}}=\sqrt{2gh}$ --(2)
Now, the momentum $p$ of a body is the product of its mass $m$ and velocity $v$.
$\therefore p=mv$ --(3)
The force $F$ on a body can be written as the rate of change of its momentum $p$.
$\therefore F=\dfrac{\Delta p}{\Delta t}$ --(4)
Where $\Delta t$ is the time period for which the force acts on the body.
Putting (3) in (4), we get
$F=\dfrac{\Delta \left( mv \right)}{\Delta t}=m\dfrac{\Delta v}{\Delta t}+v\dfrac{\Delta m}{\Delta t}$ --(5)
Now, for the liquid coming out from the vessel, the velocity of efflux remains approximately constant since the area of the hole is negligible in comparison to the area of the liquid surface. Therefore, for the vessel $\Delta v=0$
Hence, using (5), we get
${{F}_{efflux}}=0+{{v}_{efflux}}m=m{{v}_{efflux}}$ --(6)
Now, the mass of the liquid coming out from the hole per second can be written in terms of its density and volume.
$m=\dfrac{\rho V}{t}=\rho A{{v}_{efflux}}$ $\left( \because \text{mass = density}\times \text{volume} \right)$ --(7)
where $V$ is the volume of liquid coming out in time $t$.
Putting (7) in (6), we get
${{F}_{efflux}}=\rho A{{v}_{efflux}}{{v}_{efflux}}=\rho A{{v}_{efflux}}^{2}$ --(8)
Now the area $A$ of a circle can be written in terms of its diameter $D$ as
$A=\dfrac{\pi {{D}^{2}}}{4}$ --(9)
Putting (9) in (8), we get
${{F}_{efflux}}=\rho \dfrac{\pi {{D}^{2}}}{4}{{v}_{efflux}}^{2}$ --(10)
Putting (1) in (10), we get
${{F}_{efflux}}=\rho \dfrac{\pi {{D}^{2}}}{4}{{\left( \sqrt{2gh} \right)}^{2}}=\rho \dfrac{\pi {{D}^{2}}}{4}2gh=\rho \dfrac{\pi {{D}^{2}}}{2}gh$ --(11)
Now, the normal force on the vessel by the ground balances its weight, since there is no acceleration of the vessel in the vertical direction.
$\therefore N=W=Mg$
$\therefore N=Mg$ --(12)
Now, the frictional force ${{F}_{friction}}$ on a body by a surface of coefficient of friction $\mu $ is given by
${{F}_{friction}}=\mu N$ --(13)
Let the force of friction on the vessel by the surface be ${{F}_{friction}}$.
Using (13), we get
${{F}_{friction}}=\mu N$ --(14)
Putting (12) in (14), we get
${{F}_{friction}}=\mu Mg$ --(15)
Now, the force that is given to the liquid flowing out from the vessel is the same as the force imparted by the liquid on the vessel, but in the opposite direction.
Let the force on the vessel due to the liquid flowing out be ${{F}_{vessel}}$.
$\therefore {{F}_{vessel}}={{F}_{efflux}}$ --(16)
This force is along the horizontal direction and tries to move the vessel upon the surface.
The force of friction opposes this.
To just disturb the equilibrium such that the vessel moves, the magnitude of the force on the vessel due to the liquid flowing out must be equal to the force of friction on the vessel.
$\therefore {{F}_{vessel}}={{F}_{friction}}$ --(17)
Using (16) in (17), we get
${{F}_{efflux}}={{F}_{friction}}$ --(18)
Using (11) and (15) in (18), we get
$\rho \pi \dfrac{{{D}^{2}}}{2}gh=\mu Mg$
$\therefore {{D}^{2}}=\dfrac{2\mu M}{\rho \pi h}$
Square rooting both sides we get
$\therefore \sqrt{{{D}^{2}}}=\sqrt{\dfrac{2\mu M}{\rho \pi h}}$
$\therefore D=\sqrt{\dfrac{2\mu M}{\pi \rho h}}$ --(19)
Putting the values of the variables on the right hand side of equation (19), we get
$D=\sqrt{\dfrac{2\times 0.5\times 95}{800\times \pi \times 3}}\approx 0.112m$
Hence, the required minimum diameter of the vessel is $0.112m$.
Therefore, the correct option is $A)\text{ }0.112m$.

Note: Students must keep in mind that the formula for the velocity of efflux is an approximation that is valid only when the area of the hole is negligible in comparison to the area of the top surface of the liquid inside a vessel. However, in most cases it is very accurate. This question can also give a feel for why when filled balloons are left, the air oozes out from its hole very fast and the balloon moves forward. A similar principle is also used in rocket propulsion.