Two small drops of liquid of same radius coalesce to form a big drop. The ratio of the total surface energies after and before the change is

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Aicharya · Accepted Answer

$2 	imes \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3} \quad 	herefore \quad R=2^{\frac{1}{3}} r$ Ratio of energies $\frac{E_{2}}{E_{1}}=\frac{4 \pi R^{2} T}{2 	imes \pi r^{2} 	imes T}$ $\frac{E_{2}}{E_{1}}=\frac{R^{2}}{2 r^{2}}=\frac{2^{\frac{2}{3}} r^{2}}{2 r^{2}}=2^{\frac{2}{3}-1}=2^{-\frac{1}{3}}$

Two small drops of liquid of same radius coalesce to form a big drop. The ratio of the total surface…

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