Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

What is the work done in splitting a drop of water 1 mm radius into $10^6$ droplets (surface tension of water $= 72 \times 10^{-3} \; J/m^2$)

\[\begin {array} {1 1} (a)\;8.95 \times 10^{-5}\;J \\ (b)\;2.45 \times 10^{-5}\;J \\ (c)\;3.25 \times 10^{-4}\;J \\ (d)\;2.10 \times 10^{-4}\;J \end {array}\]

Can you answer this question?

1 Answer

0 votes
$R$- radius of big drop
$r$- radius of smaller drops
$\large\frac{4}{3} $$\pi R^3=n \large\frac{4}{3} $$\pi r^3$
$n= \bigg( \large\frac{R^3}{r^3}\bigg)$
Initial area of drop $=4 \pi R^2$
Final area of drops $=n 4 \pi r^2$
Changes in area $=\Delta A= n \pi r^2-4 \pi R^2$
$\qquad= 4 \pi(nr^2-R^2)$
Work done $= 4 \pi T(nr^2-R^2)$
$W= 4 \pi R^3T \bigg(\large\frac{1}{r} -\frac{1}{R}\bigg)$
 or $\qquad= 4 \pi R^2 T [n^{1/3}-1]$
$\qquad= 4 \pi \times (10^{-3})^2 \times 72 \times 10^{-3} [10^{6/3}-1]$
$\qquad= 8.95 \times 10^{-5}Joules$
Hence a is the right answer.


answered Oct 8, 2013 by meena.p
edited Mar 14, 2014 by thagee.vedartham

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App