Question

If T is the surface tension of a liquid, then what will be energy needed to break a liquid drop of radius R into 64 drops ?
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Answer 1

Explanation:

Let 'r' be the radius of each small droplets.

Then,

Volume of big drop = 64 x volume of each small droplets

According to Question

$= > \frac{4}{3} \pi {R}^{3} = 64 \times \frac{4}{3} \pi {r}^{3} \\ = > \cancel\frac{4}{3} \cancel\pi {R}^{3} = 64 \times \cancel\frac{4}{3} \cancel\pi {r}^{3} \\ = > {R}^{3} = 64 {r}^{3} \\ = > \boxed{R = 4r} \: ............(i)$

$Surface \: area \: of \: big \: drop = 4\pi {R}^{2} \\ Surface \: area \: of \: 64 \: small \: droplets = 64 \times 4\pi {r}^{2}$

So,

$Increase \: in \: surface \: area = 64 \times 4\pi {r}^{2} - 4\pi {R}^{2} \\ Increase \: in \: surface \: area = 4\pi(64 {r}^{2} - {R}^{2} ) \\ Increase \: in \: surface \: area = 4\pi(4 {R}^{2} - {R}^{2} ) \\ (by \: using \: equation \: i) \\ Increase \: in \: surface \: area = 4\pi(3 {R}^{2} ) \\ Increase \: in \: surface \: area = 12\pi {R}^{2}$

By using formula

Energy needed = Surface tension x Increase in surface area

$= > T \times 12\pi {R}^{2} \\ = > 12\pi {R}^{2} T$