Question

The energy needed in breaking a drop of a radius $R$ into $n$ drops of radius $r$ is:
a. $\left( {4\pi {r^2}n - 4\pi {R^2}} \right)T$
b. $\left( {4\pi {R^2} - 4\pi {r^2}} \right)T$
c. $\left( {\dfrac{4}{3}\pi {r^3} - \dfrac{4}{3}\pi {R^3}} \right)T$
d. $\dfrac{{\left( {4\pi {R^2} - n4\pi {r^2}} \right)}}{T}$

Answer 1

Hint: To solve the given question, we need to apply the concept of the physical phenomenon. Consider the phenomena that tend to shrink the liquid surface into a minimum surface area.

Formula used:
Energy required= ${E_f} - {E_i}$
${E_i}$, ${E_f}$ is the energy of initial drop and the energy of final drop.
$ \Rightarrow E = T \times A$
Where, $E$ is the energy, $T$ is the surface tension, $A$ is the area.

Complete step by step answer:
In the question they have asked us to find the energy that is required to break a drop of radius $R$ into $n$ drops of radius $r$.
As discussed before the hint the given problem can be approached in the view of surface tension. The tension of the surface film of a liquid that is caused due to the attraction of the particles in layer of the liquid by the bulk of liquid that tends to minimize the area of the surface
The required energy of the single drop due to the surface tension
$ \Rightarrow E = T \times A$
Where, $E$ is the energy, $T$ is the surface tension, $A$ is the area.
To solve the given problem. We can calculate the energy difference between the initial and the final drops. This difference will be the energy to break the radius. The energy difference can be calculated as,
Energy required= ${E_f} - {E_i}$
${E_i}$, ${E_f}$ is the energy of initial drop and the energy of final drop.
We have the formula to find the initial energy and we have the value for the area. Area is $4\pi {R^2}$. We can substitute the values in the formula mentioned above.
$ \Rightarrow {E_i} = T\left( {4\pi {R^2}} \right)$
The formula to find the final energy using the value for area that is $4\pi {r^2}$.We can substitute the values in the formula mentioned above.
$ \Rightarrow {E_f} = nT\left( {4\pi {r^2}} \right)$
The value of energy of initial drop and the energy of final drop.
The energy of initial drop, $ \Rightarrow {E_i} = T\left( {4\pi {R^2}} \right)$
The energy of final drop, $ \Rightarrow {E_f} = nT\left( {4\pi {r^2}} \right)$
$ \Rightarrow $ Energy required= ${E_f} - {E_i}$
$ \Rightarrow {E_f} - {E_i} = nT\left( {4\pi {r^2}} \right) - T\left( {4\pi {R^2}} \right)$
On simplifying the given question. We get,
$ \Rightarrow {E_f} - {E_i} = T\left( {4\pi {r^2}n - 4\pi {R^2}} \right)$
The energy break required to break down the radius of the drop,
$\therefore E = T\left( {4\pi {r^2}n - 4\pi {R^2}} \right)$

Hence, the correct answer is option (A).

Note: Remember that the surface tension always depends upon the forces of attraction between the particles within the surface of the liquid and also upon the gas, solid that is in contact with it.