Question

A large number of liquid drops each of radius $a$ are merged to form a single spherical drop of radius $b$. The energy released in the process is converted into kinetic energy of the big drop formed. The speed of the big drop is: [$\rho=$ density of liquid, $T=$ surface tension of liquid]
      \[\begin{align}
  & A.{{\left[ \dfrac{6T}{\rho }\left( \dfrac{1}{a}-\dfrac{1}{b} \right) \right]}^{\dfrac{1}{2}}} \\
 & B.{{\left[ \dfrac{6T}{\rho }\left( \dfrac{1}{b}-\dfrac{1}{a} \right) \right]}^{\dfrac{1}{2}}} \\
 & C.{{\left[ \dfrac{\rho }{6T}\left( \dfrac{1}{a}-\dfrac{1}{b} \right) \right]}^{\dfrac{1}{2}}} \\
 & D.{{\left[ \dfrac{\rho }{6T}\left( \dfrac{1}{b}-\dfrac{1}{a} \right) \right]}^{\dfrac{1}{2}}} \\
\end{align}\]

Answer 1

Hint: To find the velocity, we must find the surface energy of the individual drops and then find the change in surface energy first. As it is given that the kinetic energy of the big drop is equal to the change in surface energy. Using this relation, we can calculate the velocity of the big drop.

Formula used:
 $KE=\dfrac{1}{2}mv^{2}$ and $A=V\times T$

Complete step by step answer:
Let $n$ small drops of radius $a$ are merged to form a single big drop of radius $b$. Then the volume of the drops will be equal,
i.e. $n\times \dfrac{4\pi a^{3}}{3}=\dfrac{4\pi b^{3}}{3}$
or, $n=\left(\dfrac{b}{a}\right)^{3}$
To find the energy that is converted to kinetic energy we need to calculate the surface energy of the spheres.
The surface energy given as, $A=V\times T$ ,where $T$ is the surface tension of the drop and $V$ is the volume.
The surface energy of $n$ small drops of radius $a$ be $A_{a}=n(4T\pi a^{2})$ and surface energy big drop of radius $b$be$A_{b}=4T\pi b^{2}$, where $T$ is the surface tension of the drop.
Then the charge in surface energy is given by $\Delta A=A_{a}-A_{b}$
Or $\Delta A=4T\pi( na^{2}-b^{2})$, substituting the value of $n$, we get,$\Delta A=4T\pi\left( \left(\dfrac{b}{a}\right)^{3}a^{2}-b^{2}\right)=4T\pi b^{3}\left(\dfrac{1}{a}-\dfrac{1}{b}\right)$
This change in surface energy is the kinetic energy of a big drop of radius $b$.
Also, $KE=\dfrac{1}{2}mv^{2}$ we can write $m=\rho \times V$ where $m$ is the mass, $V$ is the volume and $v$ is the velocity of the of big drop of radius $b$.
$KE=\dfrac{1}{2}\rho\left(\dfrac{4\pi b^{2}}{3}\right)v^{2}$
Since $KE=\Delta A$
We get, $\Delta A=\dfrac{1}{2}\rho\left(\dfrac{4\pi b^{2}}{3}\right)v^{2}$
Or $4T\pi b^{3}\left(\dfrac{1}{a}-\dfrac{1}{b}\right)=\dfrac{1}{2}\rho\left(\dfrac{4\pi b^{2}}{3}\right)v^{2}$
Then, rearranging we get, $v=\left[ \dfrac{6T}{\rho }\left( \dfrac{1}{a}-\dfrac{1}{b} \right) \right]^{\dfrac{1}{2}}$
Hence the answer is \[A.{{\left[ \dfrac{6T}{\rho }\left( \dfrac{1}{a}-\dfrac{1}{b} \right) \right]}^{\dfrac{1}{2}}}\]

Note:
Since $n$ small drops form the new big drop, we can assume their volumes are equal provided there is no loss of volume during the formation of the new drop. Also note, that though we require mass to calculate the \[KE\], using the simple equation $m=\rho \times V$, we can overcome the problem. Also be careful with the powers and calculations.