Question

Surface tension of mercury is \[35\times {{10}^{-3}}N{{m}^{-1}}\]. The energy spent in spraying a drop of mercury of radius 1cm into 1 million drops of equal size is (in microjoule)
A. 4356
B. 4.356
C. 43.56
D. 0.4356

Answer 1

Hint: The mercury drop given in the question is sprayed into a million drops of equal size. In this spraying process, the volume of the total mercury will remain constant and this will give us the radius of the sprayed droplets. Now the work done in spraying the mercury drop into a million pieces would be the difference between the initial and final surface energy of the drops. The formulae for surface area and energy spent is given below.

Formula used:
\[V=\dfrac{4}{3}\pi {{r}^{3}}\]
Energy Spent\[=T\times \Delta A\]
\[A=4\pi {{r}^{2}}\]

Complete step-by-step answer:
The mercury drop given in the question is sprayed into a million drops of equal size. In this spraying process, the volume of the total mercury will remain constant.
The volume of a sphere is given by
\[V=\dfrac{4}{3}\pi {{r}^{3}}\]
The initial volume of mercury drop with 1cm radius will be.
\[{{V}_{i}}=\dfrac{4}{3}\pi {{\left( \dfrac{1}{100} \right)}^{3}}\]
Now the mercury drop is sprayed into a million drops of radius ‘r’. And the total volume or amount of mercury would remain constant. So,
\[\begin{align}
  & \dfrac{4}{3}\pi {{\left( \dfrac{1}{100} \right)}^{3}}=\dfrac{4}{3}\pi {{r}^{3}} \\
 & \Rightarrow r=\dfrac{1}{10000}m \\
\end{align}\]
Now the work done in spraying the mercury drop into a million pieces would be the difference between the initial and final surface energy of the drops.
Now the change in surface energy is given by surface tension times the change in surface area.
The surface area of the sphere is given as \[A=4\pi {{r}^{2}}\]
So, the change in energy or the energy spent would be
Energy Spent\[=T\times \Delta A\]
The Surface tension is given to be \[35\times {{10}^{-3}}\], plugging in these values
Energy Spent\[=\left( 35\times {{10}^{-3}} \right)\left( \left( 4\pi {{\left( \dfrac{1}{10000} \right)}^{2}} \right)\times {{10}^{6}}-\left( 4\pi {{\left( \dfrac{1}{100} \right)}^{2}} \right) \right)\]
\[\Rightarrow \]Energy Spent\[=\left( 35\times {{10}^{-3}} \right)\left( 4\times 3.14 \right)\left( \left( \dfrac{1}{100} \right)-{{\left( \dfrac{1}{100} \right)}^{2}} \right)\]
 \[\Rightarrow \]Energy Spent\[=4.356\times {{10}^{-3}}J\]
\[\Rightarrow \]Energy Spent\[=4356\times {{10}^{-6}}J\]

So, the energy spent will be 4356 micro joules.
Therefore, Option A is correct.

So, the correct answer is “Option A”.

Note: Surface energy of a drop is basically the quantification of the disruption of intermolecular bonds that occurs when a surface is created. It is defined as work done per unit surface area to create the new surface. To damage or break the surface a force has to do work equal to the surface energy of the given surface.