The change in energy when a big drop is split | vedclass

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(B) When a drop of radius $R$ is split into $n$ smaller drops (each of radius $r$),the total surface area of the liquid increases. Therefore,work must

Vedclass · Accepted Answer

$4 \pi R^{2}\left(n^{1 / 3}-1\right) T$

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(B) When a drop of radius $R$ is split into $n$ smaller drops (each of radius $r$),the total surface area of the liquid increases. Therefore,work must be done against the surface tension. Since the volume of the liquid remains constant:$\frac{4}{3} \pi R^{3} = n \left( \frac{4}{3} \pi r^{3} ight) \implies R^{3} = n r^{3} \implies r = R n^{-1/3}$.The change in energy (work done) is given by $W = T 	imes \Delta A$,where $\Delta A$ is the change in surface area.$W = T [n(4 \pi r^{2}) - 4 \pi R^{2}]$Substitute $r = R n^{-1/3}$:$W = T [n(4 \pi (R n^{-1/3})^{2}) - 4 \pi R^{2}]$$W = T [4 \pi R^{2} n (n^{-2/3}) - 4 \pi R^{2}]$$W = 4 \pi R^{2} T [n^{1/3} - 1]$.

The change in energy when a big drop is split into $n$ small droplets is:

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