Question

Calculate the energy spent in spraying a drop of mercury of radius R into N droplets all of the same size. If the surface tension of mercury is T.

Answer 1

In order to calculate the energy expended, we consider the radius of small droplets to be ‘r’. We establish a relationship between the initial and final radii of droplets by using the law of conservation of mass. The volume of the system before and after conversion is equal. Then we apply the formula of Work done.

Formula Used:
A water droplet is approximately the size of a sphere.
Surface area of a sphere of radius r is$4\pi {{\text{r}}^2}$.
The volume of a sphere is$\dfrac{4}{3}\pi {{\text{r}}^3}$

Complete step by step answer:
Given a drop of mercury of radius A is sprayed into N small droplets. Let us consider the radius of each of the small droplets to be B. The surface tension of mercury is taken as T.
We consider all the small droplets formed are of equal size.
A drop of liquid has the geometrical figure of a sphere.
Before and after conversion, the volume of the entire system should always be equal according to the law of conservation of mass.
We know the volume of a sphere is $\dfrac{4}{3}\pi {{\text{r}}^3}$, where r is the radius of the sphere.
Therefore initial volume of mercury = final volume of mercury.
Initial radius refers to the original water droplet, i.e. large water droplet. Radius = A
Radius of each of the N water droplets, radius = B
$\Rightarrow \dfrac{4}{3}\pi {{\text{A}}^3} = {\text{N}} \times \dfrac{4}{3}\pi {{\text{B}}^3} \\\ \Rightarrow {\text{B = }}\dfrac{{\text{A}}}{{\left( {{{\text{N}}^{\dfrac{1}{3}}}} \right)}}{\text{ - - - - - - (1)}} \\\$
The work done for converting a droplet into different small droplets is given by the formula,
W = T × ∆A, where T is the surface tension of the liquid and ∆A is the change in area it undergoes.
We know the surface area of a sphere is given by $4\pi {{\text{r}}^2}$, where r is the radius of the sphere.
Radius of initial sphere (original water droplet) = A
Radius of each of the N final water droplets (small droplets) = B
The initial surface area of sphere is $4\pi {{\text{A}}^2}$ and the final surface area after spraying becomes N ×$4\pi {{\text{B}}^2}$.
Now ∆A = Final area – initial area = N ×$4\pi {{\text{B}}^2}$ - $4\pi {{\text{A}}^2}$
Work done = T × 4π (N ×${{\text{B}}^2}$ -${{\text{A}}^2}$)
Substituting the value of ‘B’ from equation (1) in the above equation we get,
$\Rightarrow {\text{Work done = T }} \times {\text{ 4}}\pi \left( {{\text{N}} \times {{\left( {\dfrac{{\text{A}}}{{{{\text{N}}^{\dfrac{1}{3}}}}}} \right)}^2} - {{\text{A}}^2}} \right) \\\ \Rightarrow {\text{W = T }} \times {\text{ 4}}\pi \left( {{{\text{N}}^{1 - \dfrac{2}{3}}} \times {{\text{A}}^2} - {{\text{A}}^2}} \right) \\\ \Rightarrow {\text{W = 4}}\pi {\text{T}}{{\text{A}}^2}\left( {{{\text{N}}^{\dfrac{1}{3}}} - {\text{ }}1} \right) \\\$
Where A is the radius of the original water droplet.

Note:
In order to answer this type of question the key is to know the definitions of the given and their formula.
The numerical value of gyromagnetic ratio of an electron is $1.76085963023{\text{ }} \times {\text{ 1}}{{\text{0}}^{11}}{\text{ }}{{\text{s}}^{ - 1}}{{\text{T}}^{ - 1}}$.
The numerical value of specific charge of an electron is $- 1.75882001076{\text{ }} \times {\text{ 1}}{{\text{0}}^{11}}{\text{ C}}{\text{.K}}{{\text{g}}^{ - 1}}$.