Question

A large number of liquid drops of radius ‘a’ coalesce to form a single spherical drop of radius b. The energy released in the process is converted to the kinetic energy of the big drop formed. If d and T are the density and the surface tension respectively than the speed of the big drop formed is:
$\begin{align}
  & a)\sqrt{\dfrac{6T\left( b-a \right)}{d\left( ab \right)}} \\
 & b)\sqrt{\dfrac{4T\left( b-a \right)}{d\left( ab \right)}} \\
 & c)\sqrt{\dfrac{2T\left( b-a \right)}{d\left( ab \right)}} \\
 & d)\sqrt{\dfrac{T\left( b-a \right)}{d\left( ab \right)}} \\
\end{align}$

Answer 1

Hint: The spherical drops possess surface energy due to the surface tension of the drops of the liquid. The surface energy is proportional to the radius (r)of the spherical drop and surface tension(T) of the drop of the liquid. It is given in the question that the small drops of radius ‘a’ coalesce to form a single spherical drop of radius b. During the formation of the bigger drop, some of the surface energy gets converted to the kinetic energy of the bigger drop. Hence taking the difference between the total surface energy of the smaller drops and the bigger drop we will get the kinetic energy of the bigger drop. Since the kinetic energy of the drop is directly proportional to the square of the velocity, finally we will determine its velocity.
Formula used:
${{S}_{E}}=4\pi {{r}^{2}}T$

Complete answer:
The surface energy is given by, ${{S}_{E}}=4\pi {{r}^{2}}T$ where r is the radius of the spherical drop and T is the surface tension of the drop of the liquid.
It is given in the question a large number of liquid drops of radius ‘a’ coalesce to form a single spherical drop of radius b. Let the number of the small spherical drops be ‘n’. Since the volume of the liquid remains constant, we can write
$\begin{align}
  & \text{Volume of the bigger drop = Total volume of the small drops} \\
 & \dfrac{4}{3}\pi {{b}^{3}}=n\dfrac{4}{3}\pi {{a}^{3}} \\
 & n={{\left( \dfrac{b}{a} \right)}^{3}} \\
\end{align}$
It is also given that the energy released when the small drops get coalesced to form a bigger spherical drop gets converted to the kinetic energy of the bigger drop. Let the mass of the bigger drop be m and the velocity with which it moves be ‘v’. hence the kinetic energy of the bigger drop is equal to,
$\dfrac{1}{2}m{{v}^{2}}=n4\pi {{a}^{2}}T-4\pi {{b}^{2}}T$
Substituting the value of n from the above equation we get,
$\begin{align}
  & \Rightarrow \dfrac{1}{2}m{{v}^{2}}=n4\pi {{a}^{2}}T-4\pi {{b}^{2}}T \\
 & \Rightarrow \dfrac{1}{2}m{{v}^{2}}=\dfrac{{{b}^{3}}}{{{a}^{3}}}4\pi {{a}^{2}}T-4\pi {{b}^{2}}T \\
 & \Rightarrow \dfrac{1}{2}m{{v}^{2}}=4\pi T\left( \dfrac{{{b}^{3}}}{a}-{{b}^{2}} \right) \\
\end{align}$
The mass of any substance is equal to the product of its density times the volume. Hence the mass of the bigger drop is, $m=d\left( \dfrac{4}{3}\pi {{b}^{3}} \right)$. Substituting this in the above equation we get,
$\begin{align}
  & \Rightarrow \dfrac{1}{2}m{{v}^{2}}=4\pi T\left( \dfrac{{{b}^{3}}}{a}-{{b}^{2}} \right) \\
 & \Rightarrow \dfrac{1}{2}d\left( \dfrac{4}{3}\pi {{b}^{3}} \right){{v}^{2}}=4\pi T\left( \dfrac{{{b}^{3}}}{a}-{{b}^{2}} \right) \\
 & \Rightarrow \dfrac{1}{6}d\left( {{b}^{3}} \right){{v}^{2}}={{b}^{2}}T\left( \dfrac{b}{a}-1 \right) \\
 & \Rightarrow \dfrac{1}{6}db{{v}^{2}}=T\left( \dfrac{b-a}{a} \right) \\
 & \Rightarrow {{v}^{2}}=\dfrac{6T\left( b-a \right)}{dab} \\
 & \Rightarrow v=\sqrt{\dfrac{6T\left( b-a \right)}{dab}} \\
\end{align}$

So, the correct answer is “Option A”.

Note:
It is given in the question that some of the energy gets converted to the kinetic energy of the bigger drop. Hence we were able to equate the difference in the total surface energy to the kinetic energy of the bigger drop. In reality this does not happen, as some energy is always lost to the surrounding.