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Question: A large number of liquid drops each of radius r coalesce to form a single drop of radius R. the ener...

A large number of liquid drops each of radius r coalesce to form a single drop of radius R. the energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big is (given surface tension of liquid T, densityρ\rho ):
A. Tρ(1r−1R)\sqrt {\dfrac{T}{\rho }\left( {\dfrac{1}{r} - \dfrac{1}{R}} \right)}
B. 2Tρ(1r−1R)\sqrt {\dfrac{{2T}}{\rho }\left( {\dfrac{1}{r} - \dfrac{1}{R}} \right)}
C. 4Tρ(1r−1R)\sqrt {\dfrac{{4T}}{\rho }\left( {\dfrac{1}{r} - \dfrac{1}{R}} \right)}
D. 6Tρ(1r−1R)\sqrt {\dfrac{{6T}}{\rho }\left( {\dfrac{1}{r} - \dfrac{1}{R}} \right)}

Explanation

Solution

Concept of surface tension and surface energy. Also as a large number of drops coalesce to form a bigger drop so the volume of all smaller drops will be the same as that of the bigger drop.

Formula used:
(1) Increase in surface energy or change in surface energy ==Surface tension ×\times Increase in area.
(2) Kinetic energy =12mv2 = \dfrac{1}{2}m{v^2}
(3) Volume of sphere =43πR3 = \dfrac{4}{3}\pi {R^3}
(4) Surface area of sphere =4πR2 = 4\pi {R^2}
Where R is radius of sphere,
V is velocity
M is mass

Complete step by step answer:
Let us consider n small drops of radius r coalesce to form a single drop of radius R.
In such processes, the total volume before and after conversion remains same so,
n×n \times volume of small drops ==volume of large drop
n×43πr3=43πR3n \times \dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi {R^3}
(Because drops are in form of sphere and volume of sphere is 43πR3\dfrac{4}{3}\pi {R^3})
⇒nr3=R3\Rightarrow n{r^3} = {R^3}
⇒n=R3r3⇒n=(Rr)3\Rightarrow n = \dfrac{{{R^3}}}{{{r^3}}} \Rightarrow n = {\left( {\dfrac{R}{r}} \right)^3}…. (i)
Now, change in surface energy == surface tension ×\times change in area =T×= T \times (initial area – area) … (ii)
Where T is the surface tension.
Initial area == surface area of n small drops
=n×= n \timesSurface area of one drop
=n×4πr2= n \times 4\pi {r^2}
Final area == area of large drop
=4πR2= 4\pi {R^2}
So, equation (ii) becomes,
Change in energy =T×(n4πr2−4πR2) = T \times \left( {n4\pi {r^2} - 4\pi {R^2}} \right)
=4πT(nr2−R2)= 4\pi T\left( {n{r^2} - {R^2}} \right)
Put equation (i) in it, we get,
Change in energy =4πT(R3r3×r2−R2) = 4\pi T\left( {\dfrac{{{R^3}}}{{{r^3}}} \times {r^2} - {R^2}} \right)
Put equation (i) in it, we get, change in energy =4πT(R3r3×r2−R2) = 4\pi T\left( {\dfrac{{{R^3}}}{{{r^3}}} \times {r^2} - {R^2}} \right)
=4πT(R3r−R2)= 4\pi T\left( {\dfrac{{{R^3}}}{r} - {R^2}} \right)
Now, according to question,
Kinetic energy == change in energy
12mv2=4πT(R3r−R2)\dfrac{1}{2}m{v^2} = 4\pi T\left( {\dfrac{{{R^3}}}{r} - {R^2}} \right)… (iii)
Where m is the mass of bigger drop and v is its velocity as density =MassVolume = \dfrac{{Mass}}{{Volume}}
⇒\Rightarrowmass == volume ×\times density
m=43πR3×ρm = \dfrac{4}{3}\pi {R^3} \times \rho … (where ρ\rho is density)
Hence equation (iii) becomes,
12×43πR3×v2=4πT(R3r−R2) 16R3ρv2=TR3(1r−1R) ρ×v26=T(1r−1R) ⇒v2=6Tρ(1r−1R) ⇒v=6Tρ(1r−1R)  \dfrac{1}{2} \times \dfrac{4}{3}\pi {R^3} \times {v^2} = 4\pi T\left( {\dfrac{{{R^3}}}{r} - {R^2}} \right) \\\ \dfrac{1}{6}{R^3}\rho {v^2} = T{R^3}\left( {\dfrac{1}{r} - \dfrac{1}{R}} \right) \\\ \dfrac{{\rho \times {v^2}}}{6} = T\left( {\dfrac{1}{r} - \dfrac{1}{R}} \right) \\\ \Rightarrow {v^2} = \dfrac{{6T}}{\rho }\left( {\dfrac{1}{r} - \dfrac{1}{R}} \right) \\\ \Rightarrow v = \sqrt {\dfrac{{6T}}{\rho }\left( {\dfrac{1}{r} - \dfrac{1}{R}} \right)} \\\

So, the correct answer is “Option D”.

Note:
Here the change is energy == Kinetic energy, this change in energy is not an increase in energy but this change is (initial – final) energy.