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The energy released when $1000$ small drops each of radius $10^{-7}\;m$ coalesce to from a single big drop is (Surface tension of liquid is $14 \times 10^{-2}\;N/m$)

\[\begin {array} {1 1} (a)\;7.9 \times 10^{12}\;J \\ (b)\;1.58 \times 10^{-11}\;J \\ (c)\;7.9 \times 10^{-11}\;J \\ (d)\;1.58 \times 10^{-12}\;J \end {array}\]

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Let $R$ be radius of big drop
Then $\large\frac{4}{3} $$\pi R^3=1000\large\frac{4}{3}$$ \times \pi \times (10^{-7})^3$
$R= 10^{-6}$
Changes in surface area
$\Delta A= 4 \pi R^2- 1000 \times 4 \pi r^2$
$\qquad= 4 \pi \bigg[(10^{-6})^2 -10^3 \times (10^{-7})^2\bigg]$
$\qquad =-36 \pi (10^{-12}) m^2$
Energy released $=T \times |\Delta A|$
$\qquad=14 \times 10^{-2} \times 36 \times \large\frac{22}{7}$$ \times 10^{-12}$
$\qquad=1.58 \times 10^{-11}\;J$
Hence b is the correct answer.

 

answered Oct 8, 2013 by meena.p
edited Feb 18, 2014 by meena.p
 

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