Fluid Mechanics

JEE Main level, single correct, medium

153 of 192

Question 153 of 192

A large number of droplets, each of radius, rr coalesce to form a bigger drop of radius, RR. An engineer designs a machine so that the energy released in this process is converted into the kinetic energy of the drop. Velocity of the drop is ( T=T= surface tension, ρ=\rho= density)

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Answer: (B) [6Tρ(1r−1R)]1/2\left[\frac{6 T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{1 / 2}

When small droplets coelesce to form a bigger drop, energy released in this process is given by,
4πR3T[1r−1R]4 \pi R^3 T\left[\frac{1}{r}-\frac{1}{R}\right]
Where, R=R= radius of big drop
r=r= radius of small drop
T=T= surface tension
According to question
12mv2=4πR3T[1r−1R]⇒12[43πR3ρ]v2=4πR3T[1r−1R]⇒V2=6Tρ[1r−1R]⇒V=[6Tρ(1r−1R)]12\begin{aligned} & \frac{1}{2} m v^2=4 \pi R^3 T\left[\frac{1}{r}-\frac{1}{R}\right] \\ \Rightarrow & \frac{1}{2}\left[\frac{4}{3} \pi R^3 \rho\right] v^2=4 \pi R^3 T\left[\frac{1}{r}-\frac{1}{R}\right] \\ \Rightarrow V^2 & =\frac{6 T}{\rho}\left[\frac{1}{r}-\frac{1}{R}\right] \Rightarrow V=\left[\frac{6 T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{\frac{1}{2}} \end{aligned}

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