Fluid Mechanics

JEE Main level, single correct, medium

144 of 192

Question 144 of 192

A large number of liquid drops each of radius rr coalesce to form a single drop of the radius RR. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given surface tension of the liquid TT, density ρ\rho)

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Answer: (B) 6Tρ(1r−1R)\sqrt{\dfrac{6T}{\rho }\left(\dfrac{1}{r}-\dfrac{1}{R}\right)}

n×43πr3=43πR3n\times \dfrac{4}{3}\pi {r}^{3}=\dfrac{4}{3}\pi {R}^{3}
∴n=R3r3.....(i)∴n=\dfrac{{R}^{3}}{{r}^{3}}.....\left(i\right)
KE=Ui−UfKE={U}_{i}-{U}_{f}
=n×4πr2T−4πR2T=n\times 4\pi {r}^{2}T-4\pi {R}^{2}T
=4πT(nr2−R2)=4\pi T(n{r}^{2}-{R}^{2})
=4πTR2(Rr−1)=4\pi T{R}^{2}\left(\dfrac{R}{r}-1\right) from (i)\left(i\right)
12(43πR3⋅ρ)υ2=4πTR2(Rr−1)\dfrac{1}{2}\left(\dfrac{4}{3}\pi {R}^{3}\cdot \rho \right){υ}^{2}=4\pi T{R}^{2}\left(\dfrac{R}{r}-1\right)
Rρυ26=T(Rr−1)\dfrac{R\rho {υ}^{2}}{6}=T\left(\dfrac{R}{r}-1\right)
ρ6υ2=T(1r−1R)\dfrac{\rho }{6}{υ}^{2}=T\left(\dfrac{1}{r}-\dfrac{1}{R}\right)
υ=6Tρ(1r−1R)υ=\sqrt{\dfrac{6T}{\rho }\left(\dfrac{1}{r}-\dfrac{1}{R}\right)}

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