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Twenty seven droplets of water, each of radius 0.1 mm coalesce into a single drop. Find the change in surface energy. Surface tension of water is 0.072 N/m.

in Chapter2:Mechanical Properties of Fluids by Expert (10.7k points) | 179 views

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Correct answer = 1.628 x 10-7 J 

Explaination ::

Given :: r=0.1mm =0.1 x 10-3m , T=0.072 N/m 

Let R be the radius of the single drop formed due to the coalescence of 27 droplets of mercury.
Volume of 27 droplets = volume of the single drop as the volume of the liquid remains constant. 

\(\therefore 27 \times \frac{4}{3} \pi r^3 =\frac{4}{3} \pi R^3\) 

\(\therefore 27 r^3= R^3\) 

\(\therefore 3r=R\)                ---{taking cube root}

surface area of 27 droplets =\(27\times 4\pi r^2\) 

surface area of single droplet =\(4\pi R^2\) 

Decrease in surface area =\(27\times r^2 4\pi - 4\pi R^2\)

\( = 4\pi (27r^2-R^2)=4\pi[27r^2-(3r)^2]\) 

=\(4\pi \times 18r^2\) 

∴ The energy released = surface tension × decrease in surface area 

=\(T\times 4\pi \times 18 r^2=\) \(0.072\times 4\times 3.142\times18 \) \(\times (1\times10^{-4})^2\) 

= 1.628 x 10-7 J 

 

by Expert (10.7k points)

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