Question

Calculate the surface energy released if eight small drops of water each of radius 6cm combines to from a big drop? plzz do fast....frnd..

Answer 1

$\large\underline\mathrm\red{Solution}$

• Let r and R be the radius of small drop and big drop respectively then as the volume remains constant,

$\large\mathrm{4/3 \:\:π\: R³ \:\: = \:\: n 4/3 \:\: πr³. \:\:\:\:\:}$ $\large\mathrm\blue{( R \:\: = \:\:\: n^1/3 r.)}$

$\large\mathrm{Total \: surface \: area \: of \: small \: drops \: = \: 4πr²\: n}$

$\large\mathrm\red{And}$

$\large\mathrm{Surface \: area \: of \: big \: drop \: = \: 4πR²}$

$\large\mathrm\red{Total \: release \: of \: surface \: energy}$

$\implies$ [4πr²n - 4πR²]

$\implies$ 4π[r²n - n^2/3.r2]T

$\implies$ 4πr²[n - n^2/3]

$\implies$ 4πr²[8 - (2³)²/³]T

$\implies$ 4πr²[8 - 4]T

$\implies$ 16πr²T.

$\large\underline\mathrm\red{Note}$

$\large\mathrm\red{(1)}$. Release of surface energy if small drops combine to form a a single big drop. $\implies$ 4πr²[n - n²/³]T

$\large\mathrm\red{(2)}$. For soap bubble; energy released when small bubbles combine to form a big bubble.

$\implies$ 2 × 4πr²[n - n²/³]T

$\large\mathrm{When \: n = \: No. \: of \: small \: drops}$

$\large\mathrm{r \: = \: radius \: of \: small \: drops.}$

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