Question

A spherical water drop of radius $R$ is split up into $8$ equal droplets. If $T$ is the surface tension of water, then the work done in this process is
A. \[4\pi {R^2}T\]
B. \[8\pi {R^2}T\]
C. \[3\pi {R^2}T\]
D. \[2\pi {R^2}T\]

Answer 1

Hint: Surface tension is work done per unit area. To calculate change in surface area and multiply it with surface tension.
Volume of sphere$ = \dfrac{4}{3}\pi {r^3}$
Surface area of sphere$ = 4\pi {r^2}$
$W = AT$.

Complete step by step answer:
When a spherical drop is split up into and equal droplets. The total volume of the sphere will not change.
That is, the volume of the larger sphere is equal to the sum of the volumes of all the spherical droplets.
i.e. $\dfrac{4}{3}\pi {r^3}$
 $ = 8 \times \dfrac{4}{3}\pi {r^3}$
Where, $A$ is the radius of larger spherical water drop $r$ is the radius of small spherical water droplets.
$\dfrac{4}{3}\pi {r^3}$is the volume of the sphere.
Simplifying the above equation, we get
${R^3} = 8{r^3}$
$ \Rightarrow {R^3} = {(2r)^3}$
By taking cube root to both the sides
$R = 2r$
$ \Rightarrow r = \dfrac{R}{2}$ . . (1)
Now, we know that, surface tension is work done per unit area
i.e.$\dfrac{W}{A} = T$
$ \Rightarrow W = AT$
Where, $W$ is work done,
$A$ is surface area
$T$ is surface tension.
Surface area of the sphere is $4\pi {r^2}$ and work done in the process is.
${A_2}T - {A_1}T$
$ \Rightarrow W = 8 \times 4\pi {r^2}T - 4\pi {R^2}T$
$ = 4\pi (8\pi {r^2}T - {R^2}T)$
$ = 4\pi \left( {8\dfrac{{{R^2}}}{4} \times T - {R^2}T} \right)\left( {\because v = \dfrac{R}{2}} \right)$.
$ = 4\pi (2{R^2}T - {R^2}T)$
$W = 4 = 4\pi {R^2}T$
Thus, the work done in the process is $4\pi {R^2}T$.

So, the correct answer is “Option A”.

Note:
If you do not understand why $W = {A_2}T - {A_1}T$
Then you can do it like this: surface energy is a product of surface energy and change in surface area and surface tension.
$ \Rightarrow $Change in surface energy $ = {A_2}T - {A_1}T$.
Now, we know that the change in surface energy is the energy used for work done.
Therefore, change in surface energy is equal to work done
$ \Rightarrow W = {A_2}T - {A_1}T$.