The energy needed for breaking a liquid drop | vedclass

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Question

(A) The volume of the large drop is equal to the sum of the volumes of the $n$ small droplets: $\frac{4}{3} \pi R^3 = n \cdot \frac{4}{3} \pi r^3$,whi

Vedclass · Accepted Answer

$4 \pi T R^2 \left[\frac{R}{r} - 1\right]$

Vedclass · Answer

(A) The volume of the large drop is equal to the sum of the volumes of the $n$ small droplets: $\frac{4}{3} \pi R^3 = n \cdot \frac{4}{3} \pi r^3$,which implies $R^3 = n r^3$ or $n = \frac{R^3}{r^3}$.The initial surface area of the large drop is $A_i = 4 \pi R^2$.The final surface area of the $n$ small droplets is $A_f = n \cdot 4 \pi r^2$.The change in surface area is $\Delta A = A_f - A_i = n(4 \pi r^2) - 4 \pi R^2$.Substituting $n = \frac{R^3}{r^3}$,we get $\Delta A = \left(\frac{R^3}{r^3}\right) 4 \pi r^2 - 4 \pi R^2 = 4 \pi R^3 \left(\frac{1}{r}\right) - 4 \pi R^2 = 4 \pi R^2 \left(\frac{R}{r} - 1\right)$.The energy required is $W = T \cdot \Delta A = 4 \pi T R^2 \left(\frac{R}{r} - 1\right)$.

The energy needed for breaking a liquid drop of radius $R$ into $n$ droplets each of radius $r$ is (where $T$ is the surface tension of the liquid).

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