Question

A cubical vessel of height $1m$ is full of water. Work done by gravity in taking water out of the vessel will be :
1)$5000J$
2)$10000J$
3)$5J$
4)$10J$

Answer 1

Hint As we have to find the work done by the gravitational force, To calculate the work done by the gravitational force, we will adopt a simplifying approach and will assume that air resistance is negligible.

Complete step by step solution
Correct answer is $5000J$
As given
The height of the cubical vessel:
$h = 1m$
Here $h$ is the height of the vessel (total height)
Center of mass (COM) of the vessel is located at $\dfrac{h}{2}$
So, we can write
 $h' = \dfrac{h}{2} = \dfrac{1}{2}m$
Also, the volume of the vessel (water) would be
${h^3} = {1^3} = 1{m^3}$
And mass of the water would be
$m = \rho V$
So we get
$m = 1000 \times 1 = 1000kg$
Because density of the water is $1000kg/{m^3}$
Now as we have to find the work done by the gravitational force
So work done for pumping out water is equal to the negative of change in its potential energy.
Hence,
$W = - \Delta P.E$
Where W is the work done which is equal to change in potential energy
Therefore,
$W = - (P.{E_f} - P.{E_i})$
$P.{E_f}$ is the final potential energy and
$P.{E_i}$ is the initial potential energy
After taking water out the final potential energy $P.{E_f}$ would be 0
And initial potential energy $P.{E_i}$ would be the initial potential energy which would be equal to $mgh'$
So, we get
$W = - (0 - mgh')$
$W = mgh'$
On putting values we get

$W = 1000 \times 10 \times \dfrac{1}{2} = 5000J$

Note Remember while measuring the energy the height should always be measured from the center of mass (COM) and moreover always use units carefully as in this case the value of the density of water must be put carefully.