The amount of work done to break a big water | vedclass

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(A) The work done in breaking a drop is equal to the change in surface energy: $W = S \Delta A$.Let $R$ be the radius of the big drop and $r$ be the r

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) The work done in breaking a drop is equal to the change in surface energy: $W = S \Delta A$.Let $R$ be the radius of the big drop and $r$ be the radius of the small drops.Volume conservation: $\frac{4}{3} \pi R^3 = n \cdot \frac{4}{3} \pi r^3$,which gives $r = \frac{R}{n^{1/3}}$.The work done is $W = S(n \cdot 4 \pi r^2 - 4 \pi R^2) = 4 \pi R^2 S (n^{1/3} - 1)$.For $n = 27$: $W_1 = 4 \pi R^2 S (27^{1/3} - 1) = 4 \pi R^2 S (3 - 1) = 8 \pi R^2 S = 10 \ J$.For $n = 64$: $W_2 = 4 \pi R^2 S (64^{1/3} - 1) = 4 \pi R^2 S (4 - 1) = 12 \pi R^2 S$.Taking the ratio: $\frac{W_2}{W_1} = \frac{12 \pi R^2 S}{8 \pi R^2 S} = \frac{12}{8} = 1.5$.Therefore,$W_2 = 1.5 	imes 10 \ J = 15 \ J$.

The amount of work done to break a big water drop of radius $R$ into $27$ small drops of equal radius is $10 \ J$. The work done required to break the same big drop into $64$ small drops of equal radius will be (in $J$)

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