Question

The work done in splitting a drop of water of 1mm radius into ${10^6}$ droplets is (surface tension of water $72 \times {10^{ - 3}}N/m$).
A. $5.98 \times {10^{ - 5}}J$
B. $10.98 \times {10^{ - 5}}J$
C. $16.95 \times {10^{ - 5}}J$
D. $8.95 \times {10^{ - 5}}J$

Answer 1

Hint: One drop of water splits into many smaller droplets, so the volume remains constant.Assume the radius of the larger drop to be R and small drops to be r.Using the formula for volume of the sphere,i.e. $V = \dfrac{4}{3}\pi {R^3}$ find the volume for both the larger and smaller drops.So, find out the radius of the new small droplet by equating the volume of the drop before splitting and total volume of drips after splitting.Then find out the increase in surface area $\Delta A$, and substitute into the formula $W = T.\Delta A$ (W= work done, T=Surface Tension, $\Delta A$ =Change in surface area)

Complete step by step answer:
One droplet splits into ${10^6}$ smaller droplets, hence volume remains constant. So
$\dfrac{4}{3}\pi {R^3} = {10^6} \times \dfrac{4}{3}\pi {r^3}$, where R=radius of big drop, and r=radius of small drop.
Substituting values (in SI Units) we get,
$\dfrac{4}{3}\pi {\left( {0.001} \right)^3} = {10^6} \times \dfrac{4}{3}\pi {r^3}$
$ \Rightarrow {r^3} = {10^{ - 15}}$
$ \Rightarrow r = {10^{ - 5}}cm$
As one big drop splits into ${10^6}$ smaller droplets, there is an increase in surface area. We know that the surface area of the sphere is $4\pi {r^2}$ .
So, increase in surface area is-
$\left[ {4\pi \times {{\left( r \right)}^2}} \right]n - \left[ {4\pi \times {{\left( R \right)}^2}} \right]$ , {where n=number of smaller droplets}
$ \Rightarrow \left[ {4\pi \times {{\left( {{{10}^{ - 5}}} \right)}^2} \times {{10}^{^6}}} \right] - \left[ {4\pi \times {{\left( {{{10}^{ - 3}}} \right)}^2}} \right]$
$ \Rightarrow \left[ {4\pi \times \left( {{{10}^{ - 4}}} \right)} \right] - \left[ {4\pi \times \left( {{{10}^{ - 6}}} \right)} \right]$
$ \Rightarrow 4\pi \times {10^{ - 6}}\left( {100 - 1} \right)$
$ \Rightarrow 4\pi \times {10^{ - 6}}\left( {99} \right)$……eq1
Now using the formula, $W = T.\Delta A$ and substituting values, we get:
$W = 72 \times 4\pi \times 99 \times {10^{ - 6}} \times {10^{ - 3}}$
$\therefore W = 8.95 \times {10^{ - 5}}J$

Hence, option D is correct.

Note:Sometimes students get confused about how to calculate the work done and consider the work done by a single drop and thus resulting into a wrong answer.So,always remember to multiply with the total number of drops that results into a work done.