When two or more mercury drops are brought in contact with each other they form a single drop? Why?
Answer
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Hint:Mercury is an atomic number 80 chemical element with the symbol Hg. It was formerly known as hydrargyrum and is now known as quicksilver. Mercury, also known as elemental or metallic mercury, is a gleaming silver-white metal that is liquid at room temperature and was once known as quicksilver. Older thermometers, neon light fixtures, and even electronic switches contain it.
Complete step by step answer:
Of all geometric shapes, a spherical shape has the smallest surface area to volume ratio. When two liquid drops collide, the cohesive forces in their molecules coalesce the drops into a single larger drop. This is since, with the same amount of material, the surface area of the resulting single drop is less than the sum of the smaller drops' surface areas. Heat is absorbed into the air as a result of the drop in surface energy.
Prove it by seeing n droplets of radius r coalesce into a single drop of radius $R$. Amount of the reduction = volume of n droplets since the volume of the liquid stays unchanged.
\[\dfrac{4}{3}\pi {{\rm{R}}^3} = {\rm{n}} \times \dfrac{4}{3}\pi {{\rm{r}}^3}\]
\[\Rightarrow {{\bf{R}}^3} = {\bf{n}}{{\bf{r}}^3}\]
\[\Rightarrow {\bf{R}} = \sqrt[3]{{{\bf{nr}}}}\]
Surface area of n droplets = \[{\bf{n}} \times 4\pi {{\rm{r}}^2}\]
Surface area of the drop = \[4\pi {{\rm{R}}^2} = {{\bf{n}}^{2/3}} \times 4\pi {{\rm{r}}^2}\]
Surface area change = surface area of drop - surface area of n droplets
\[\text{Surface area change}= 4\pi {{\rm{R}}^2} \\
\therefore \text{Surface area change}= \left( {{{\rm{n}}^{2/3}} - {\rm{n}}} \right)\]
There is a reduction in surface area and a decrease in surface energy since the bracketed word is negative.
Note:Because of the strong force of cohesion that acts within the drops and surface tension, when two or more small mercury drops come into contact with each other, they join to create a larger drop. The larger drop created has a smaller surface area and lower surface tension.
Complete step by step answer:
Of all geometric shapes, a spherical shape has the smallest surface area to volume ratio. When two liquid drops collide, the cohesive forces in their molecules coalesce the drops into a single larger drop. This is since, with the same amount of material, the surface area of the resulting single drop is less than the sum of the smaller drops' surface areas. Heat is absorbed into the air as a result of the drop in surface energy.
Prove it by seeing n droplets of radius r coalesce into a single drop of radius $R$. Amount of the reduction = volume of n droplets since the volume of the liquid stays unchanged.
\[\dfrac{4}{3}\pi {{\rm{R}}^3} = {\rm{n}} \times \dfrac{4}{3}\pi {{\rm{r}}^3}\]
\[\Rightarrow {{\bf{R}}^3} = {\bf{n}}{{\bf{r}}^3}\]
\[\Rightarrow {\bf{R}} = \sqrt[3]{{{\bf{nr}}}}\]
Surface area of n droplets = \[{\bf{n}} \times 4\pi {{\rm{r}}^2}\]
Surface area of the drop = \[4\pi {{\rm{R}}^2} = {{\bf{n}}^{2/3}} \times 4\pi {{\rm{r}}^2}\]
Surface area change = surface area of drop - surface area of n droplets
\[\text{Surface area change}= 4\pi {{\rm{R}}^2} \\
\therefore \text{Surface area change}= \left( {{{\rm{n}}^{2/3}} - {\rm{n}}} \right)\]
There is a reduction in surface area and a decrease in surface energy since the bracketed word is negative.
Note:Because of the strong force of cohesion that acts within the drops and surface tension, when two or more small mercury drops come into contact with each other, they join to create a larger drop. The larger drop created has a smaller surface area and lower surface tension.
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