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

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

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1 Answer

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$R$- radius of big drop
$r$- radius of smaller drops
$\large\frac{4}{3} $$\pi R^3=n \large\frac{4}{3} $$\pi r^3$
$R^3=nr^3$
$n= \bigg( \large\frac{R^3}{r^3}\bigg)$
$=>n=\bigg(\large\frac{R}{r}\bigg)^3$
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
 

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