Question

calculate the work done in breaking a drop of water of 2mm into million droplets of same size. The surface tension of water is 72*10^-3 N/m

Answer 1

Initial area of drop =4πR2

Final area of drops =n4πr2

Changes in area =ΔA=nπr2−4πR2

=4π(nr2−R2)

Work done =4πT(nr2−R2)

W=4πR3T(1r−1R)

or =4πR2T[n1/3−1]

=4π×(10−3)2×72×10−3[106/3−1]

=8.95×10−5Joules

Hence a is the right answer.

Answer 2

The work done is $3.582\times10^{-4}\ J$.

Explanation:

Given that,

Surface tension $T=72\times10^{-3}\ N/m$

Inner radius R= 2 mm

Since 1 drop breaks in to million droplets

We need to calculate the radius of small radius

initial volume = 1000000× final volume

$\dfrac{4}{3}\pi\times R^3=10^{6}\times\dfrac{4}{3}\pi\times r^3$

$R^3=10^{6}r^3$

$r^3=\dfrac{R^3}{10^{6}}$

$r=\dfrac{R}{100}$

$r=\dfrac{2\times10^{-3}}{100}$

$r=2\times10^{-5}\ m$

We need to calculate the work done

Using formula of work done

$W=T(\Delta S)$

$W=T(10^{6}\times4\pi r^2-4\pi R^2)$

$W=72\times10^{-3}(4\pi\times10^{6}\times(2\times10^{-5})^2-(4\pi\times(2\times10^{-3})^2))$

$W= 3.582\times10^{-4}\ J$

Hence, The work done is $3.582\times10^{-4}\ J$.

Learn more :

Topic : surface tension

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