Question

A cubic vessel of height $2m$ is full of water. The work done in pumping the water out of the vessel is?
A) $72.3kJ$
B) $78.4kJ$
C) $64.5kJ$
D) $57.9kJ$

Answer 1

Recall that the work done is defined as the measure of the amount of transfer of energy. It is said to be done when an external force is applied on an object or particle in order to move it over a distance in a specific direction. It is a vector quantity because it has both magnitude and direction.

Complete step by step solution:
Given that the height of the cubical vessel is, $h = 2m$
Suppose That the length of the cubic vessel is ‘l’
So its volume will be $v = {l^3}$
The centre of mass of the water in the cubical vessel from the ground is $h = \dfrac{l}{2}$
It is known that the density of the substance is the ratio of mass per unit volume. So, mass of the cubical vessel can be written as
$Mass = Density \times Volume$
$m = v\rho$
Substituting the value of volume, the mass becomes
$m = {l^3}\rho$
Work done in pumping out the water will be equal to the potential energy required to pump the water to the required height. So work done is given by
$w = mgh$
$w = mg\dfrac{l}{2}$
$\Rightarrow w = {l^3}\rho g\dfrac{l}{2}$
$\Rightarrow w = \dfrac{{\rho g{l^4}}}{2}$
Substituting the given values in the above equation, and solving
$\Rightarrow w = \dfrac{1}{2} \times 1000 \times 10 \times {(2)^4}$
$\Rightarrow w = 4900 \times 16$
$\Rightarrow w = 78400J = 78.4kJ$
The work done in pumping the water out of the vessel is $= 78.4kJ$`

Option B is the right answer.

Note: It is to be noted that the centre of mass is defined as the point, where the whole mass of the particles of the body or system is said to be concentrated. When there are a lot of particles in motion that are to be dealt with, then this concept is used.