A large number of droplets, each of radius, r coalesce to form a...

A large number of droplets, each of radius,

r

coalesce to form a bigger drop of radius,

R

. 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=

surface tension,

\rho=

density)

Options

A $\left[\frac{T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{1 / 2}$
B $\left[\frac{6 T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{1 / 2}$
C $\left[\frac{3 T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{1 / 2}$
D $\left[\frac{2 T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{1 / 2}$

Correct answer

B. $\left[\frac{6 T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{1 / 2}$

Step-by-step solution

When small droplets coelesce to form a bigger drop, energy released in this process is given by, 4

\pi

R^3 T

\left[\frac{1}{r}-\frac{1}{R}\right]

Where,

R=

radius of big drop

r=

radius of small drop

T=

surface tension According to question

\begin{aligned}& \frac{1}{2}m v²=4 \piR^3 T\left[\frac{1}{r}-\frac{1}{R}\right]\\ \Rightarrow & \frac{1}{2} \left[\frac{4}{3} \piR^3 \rho\right]v^2=4 \piR^3 T \left[\frac{1}{r}-\frac{1}{R}\right]\\ \Rightarrow V² & =\frac{6 T}{\rho}\left[\frac{1}{r}-\frac{1}{R}\right] \RightarrowV=\left[\frac{6 T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{\frac{1}{2}} \end{aligned}

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