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Question

Physics Question on Surface tension

Two small drops of mercury, each of radius RR, coalesce to form a single large drop. The ratio of the total surface energies before and after the change is

A

1:21/31 : 2^{1/3}

B

21/3:12^{1/3}: 1

C

2:12 : 1

D

1:21 : 2

Answer

21/3:12^{1/3}: 1

Explanation

Solution

Radius of one drop of mercury is RR.
\therefore The volume of one drop=43πR3 =\frac{4}{3}\pi R^{3}
\therefore; Total volume of the two drops,
V=2×43πR3=83πR3V=2\times\frac{4}{3}\pi R^{3}=\frac{8}{3}\pi R^{3}
Let the radius of the large drop formed be RR ' The volume of the large drop is also V.
43πR3=83πR3R3=2R3R=21/3R.\therefore \frac{4}{3}\pi R'^{3}=\frac{8}{3}\pi R^{3} \Rightarrow R '^{3}=2R^{3}\Rightarrow R'=2^{1/3}R.
Now the surface area of the two drops is
S1=2×4πR2=8πR2S_{1}=2\times4\pi R^{2}=8\pi R^{2}
and the surface area of the resultant drop is
S1=2×4πR2=4π22/3R2S_{1}=2\times4\pi R'^{2}=4\pi2^{2/3}R^{2}
Let T be the surface tension of mercury. Therefore the surface energy of the two drops before coalescing is
U1=S1T=8πR2TU_{1}=S_{1}T=8\pi R^{2}T
and the surface energy after coalescing,
U1=S2T=22/3×4πR2TU_{1}=S_{2}T=2^{2/3}\times4\pi R^{2}T
U1U2=8πR2T22/3×4πR2T=222/3=21/3\therefore \frac{U_{1}}{U_{2}}=\frac{8\pi R^{2}T}{2^{2 / 3}\times4\pi R^{2}T}=\frac{2}{2^{2 / 3}}=2^{1 / 3}