Question

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

Answer 1

Use the formula for the work done on the water. This formula gives the relation between the force on the object and displacement of the water. Use the formula for the force acting on the water. Substitute the value of the mass of the water in terms of density of the water and the volume of the water in terms of the area and thickness of the water spread.

Formulae used:
The work done on an object is given by
$W = Fs$ …… (1)
Here, $W$ is the work done, $F$ is the force on the object and $s$ is the displacement of the object.
The force acting on an object is
$F = ma$ …… (2)
Here, $F$ is the force acting on the object, $m$ is the mass of the object and $a$ is the acceleration of the object.
The density of an object is given by
$\rho = \dfrac{M}{V}$ …… (3)
Here, $\rho$ is the density of the object, $M$ is the mass of the object and $V$ is the volume of the object.
The volume of an object is given by
$V = At$ …… (4)
Here, $V$ is the volume of the object, $A$ is the area of the object and $t$ is the thickness of the object.

Complete step by step answer:
We have given that the height of the cloud is $1000\,{\text{m}}$ above the surface of the earth.
$s = 1000\,{\text{m}}$
After the rainfall, the water covers an area of ${10^6}\,{{\text{m}}^2}$ and thickness $2\,{\text{cm}}$.
$A = {10^6}\,{{\text{m}}^2}$
$t = 2\,{\text{cm}}$
Let $M$ be the mass of the water to be raised.
The acceleration of the water will be equal to the acceleration due to gravity.
We can determine the work done in raising the water to the height of the clouds using equations (1), (2), (3) and (4).
Rearrange equation (3) for the mass $M$ of the water.
$M = \rho V$
Substitute $Mg$ for $F$ in equation (1).
$W = Mgs$
Substitute $\rho V$ for $M$ in the above equation.
$W = \rho Vgs$
Substitute $At$ for $V$ in the above equation.
$W = \rho Atgs$
Substitute ${10^3}\,{\text{kg}} \cdot {{\text{m}}^{ - 3}}$ for $\rho$, ${10^6}\,{{\text{m}}^2}$ for $A$, $2\,{\text{cm}}$ for $t$, $9.8\,{\text{m}} \cdot {{\text{s}}^{ - 2}}$ for $g$ and $1000\,{\text{m}}$ for $s$ in the above equation.
$W = \left( {{{10}^3}\,{\text{kg}} \cdot {{\text{m}}^{ - 3}}} \right)\left( {{{10}^6}\,{{\text{m}}^2}} \right)\left( {2\,{\text{cm}}} \right)\left( {9.8\,{\text{m}} \cdot {{\text{s}}^{ - 2}}} \right)\left( {1000\,{\text{m}}} \right)$
$\Rightarrow W = \left( {{{10}^3}\,{\text{kg}} \cdot {{\text{m}}^{ - 3}}} \right)\left( {{{10}^6}\,{{\text{m}}^2}} \right)\left[ {\left( {2\,{\text{cm}}} \right)\left( {\dfrac{{{{10}^{ - 2}}\,{\text{m}}}}{{1\,{\text{cm}}}}} \right)} \right]\left( {9.8\,{\text{m}} \cdot {{\text{s}}^{ - 2}}} \right)\left( {1000\,{\text{m}}} \right)$
$\therefore W = 1.96 \times {10^{11}}\,{\text{J}}$

Therefore, the work done in raising the water to the height of the clouds is $1.96 \times {10^{11}}\,{\text{J}}$.Hence, the correct option is A.

Note: One can also solve the same question by another method. The work done in raising the water to the height of the clouds is equal to the negative of the change in potential energy of the water. The terms mass and volume of water coming in the formula can be replaced by the product of the density and volume of the water and the product of area and thickness of the earth covered by the water respectively.