Question

A mercury drop of radius $1cm$ is sprayed into ${10^6}$ drops of equal size. The energy expended in joule is (surface tension of mercury is $460 \times {10^{ - 3}}N{m^{ - 1}}$
A.$0.057$
B.$5.7$
C.$5.7 \times {10^{ - 4}}$
D.$5.7 \times {10^{ - 6}}$

Answer 1

Hint: Recall the concept of surface tension. Surface tension is defined as the ability of a liquid surface, by which the liquids can shrink into a minimum surface area that they can. The liquid molecules act as a stretched elastic membrane and resist effect caused by any external force. This can be observed in any liquid drop with spherical shape.

Complete answer:
Step I:
Let the radius of big drop of mercury be $ = R$
The radius of small drop when it is sprayed be $ = r$
The total volume of the drop remains conserved after it is sprayed. Therefore it can be written that
Step II:
$ \Rightarrow \dfrac{4}{3}\pi {R^3} = n \times \dfrac{4}{3}\pi {r^3}$
Where ‘R’ is the radius of the mercury drops before spraying into drops and ‘r’ is the radius after splitting in drops and ‘n’ is the number of drops into which the mercury drop is broken.
$ \Rightarrow R = {n^{\dfrac{1}{3}}}.r$---(i)
Step III:
Given that the radius of drop is $1cm$, substituting the value and solving
$r = {10^2} \times 1$
$r = 0.01cm$
Step IV:
Energy required will be equal to the force of surface tension and the change in surface area of the drop. Therefore,
$ \Rightarrow E = T \times \Delta A$
$ \Rightarrow E = T \times 4\pi ({R^2} - n{r^2})$
Where ‘T’ is the surface tension
‘E’ is the energy required
Step V:
Substitute the value of ${r^2}$ from equation (i),
$ \Rightarrow E = 460 \times {10^{ - 3}} \times 4 \times 3.14[{(R)^2} - n{(\dfrac{R}{{{n^{\dfrac{1}{3}}}}})^2}]$
$ \Rightarrow E = 460 \times {10^{ - 3}} \times 4 \times 3.14 \times [{(R)^2} - {n^{1 - \dfrac{2}{3}}}{(R)^2}]$
$ \Rightarrow E = 460 \times 4 \times 3.14 \times {10^{ - 3}}{(R)^2}[1 - {n^{\dfrac{1}{3}}}]$
$ \Rightarrow E = 460 \times 4 \times 3.14 \times {10^{ - 3}} \times 1 \times {10^{ - 4}}[1 - {10^{\dfrac{6}{3}}}]$
$ \Rightarrow E = - (460 \times 4 \times 3.14 \times {10^{ - 7}} \times 99)$
$ \Rightarrow E = \dfrac{{571982.4}}{{{{10}^7}}}$
$ \Rightarrow E = 0.057J$
Therefore the energy expanded will be $0.057J$

$ \Rightarrow $ Option A is the right answer.

Note:
It is to be noted that the term energy means the ability to do work. When the mercury drop is broken in parts, energy will be required. It is to be remembered that energy is a property and it can not be created or destroyed. It can be transferred from one form to another.