Question

two Mercury drops each of radius r merge to form a bigger drop. calculate the surface energy released ?​

Answer 1

$\huge\boxed{\pink\star\mathfrak\orange{\large{\underline{\underline{Solution:-}}}} }$

Surface area of one drop before merging = $\large\bold{4\pi {r}^{2} }$

total surface area of both the drops= $\large\bold{8\pi {r}^{2} }$

hence, the surface energy before merging = $\large\bold{8\pi {r}^{2} s.}$

when the drop merge, the volume of the bigger drop,

$\large\bold{ = > 2 \times \frac{4}{3} \pi {r}^{3} = \frac{8}{3} \pi {r}^{3} }$

if the radius of this new drop is R.

$\large\bold{ \frac{4}{3} \pi {r}^{3} = \frac{8}{3} \pi {r}^{3} }$

$\large\bold{ = > R = {2}^{ \frac{1}{3} } r}$

$\large\bold{ = > 4\pi {R}^{2} = 4 \times {2}^{ \frac{2}{3} } \times \pi {r}^{2} }$

hence, the surface energy=

$\large\bold{4 \times {2}^{ \frac{2}{3} } \times \pi \times {r}^{2} }$

the released surface energy =

$\large\bold{8\pi \times {r}^{2} S - 4 \times {2}^{ \frac{2}{3} } \pi \times {r}^{2} S}$

$\large\bold{ = > 1.65\pi \times {r}^{2} S.}$

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