Question

A cluster of clouds at a height of 1000m above the earth burst and enough rain fell to cover an area of \[{10^6}{m^2}\] with a depth of \[2cm\]. How much work have been done in raising water of the height of clouds (Take \[g = 9.8m{s^{ - 1}}\] and density of water \[ = 10kg{m^{ - 3}}\]
(A)\[1.96 \times {10^{11}}J\]
(B) \[1.96 \times {10^9}J\]
(C) \[1.5 \times {10^{11}}J\]
(D) \[2.5 \times {10^{13}}J\]

Answer 1

Hint: In this question, Height, area and depth is given. By using the given values, firstly find the value of volume. Then we can find the mass by using volume and density of water. After finding all the values, substitute all the values in the work done formula. The angle is zero because force and displacement of water is in the same direction.

Complete step by step solution:
Height of cluster clouds are given,\[h = 1000m\], Area and depth is also given as -\[{10^6}{m^2}\],\[2cm\] respectively.
d=\[2 \times {10^{ - 2}}m\]
Volume of water is\[ = A \times d\]
=\[{10^6} \times 2 \times {10^{ - 2}}{m^3}\]
As we know, density of water is ρ=\[1000kg/{m^3}\]
So, mass of water is\[ = v \times \rho \]
=\[{10^6} \times 2 \times {10^{ - 2}} \times {10^3}\] =\[2 \times {10^7}kg\]
It means we want to lift \[2 \times {10^7}kg\] of mass of water.
So, work done=\[FS\operatorname{Cos} \theta \]
Where F is force and S is distance.
Force\[ = mg\] and θ is \[0^\circ \]. Here force is acting in the upward direction as water is moving in an upward direction.
W= \[mg \times 1000 \times \operatorname{Cos} 0^\circ \]
We know,\[\operatorname{Cos} 0^\circ = 1\]
So, work done=\[2 \times {10^7} \times 9.8\]×\[1000 \times 1\]
\[ = 1.96 \times {10^{11}}J\]
So, the value of work done is\[ = 1.96 \times {10^{11}}J\].

So, the correct answer is “Option A”.

Note:
Work is the measurement of the energy that transfers when we apply force on any object and that object moves through a distance.
W=\[FS\]
Also we know that \[F = ma\]
So, work done= \[m \times a \times S\]
Here m is the mass, a is the acceleration and S is the displacement.
\[Density = \dfrac{{mass}}{{volume}}\]
\[Mass = density \times volume\]
As we know that \[volume = area \times depth\]
By taking all these terms into account we can directly find the work done.