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Question: The energy needed in breaking a drop of a radius \(R\) into \(n\) drops of radius \(r\) is: a. \(\...

The energy needed in breaking a drop of a radius RR into nn drops of radius rr is:
a. (4πr2n−4πR2)T\left( {4\pi {r^2}n - 4\pi {R^2}} \right)T
b. (4πR2−4πr2)T\left( {4\pi {R^2} - 4\pi {r^2}} \right)T
c. (43πr3−43πR3)T\left( {\dfrac{4}{3}\pi {r^3} - \dfrac{4}{3}\pi {R^3}} \right)T
d. (4πR2−n4πr2)T\dfrac{{\left( {4\pi {R^2} - n4\pi {r^2}} \right)}}{T}

Explanation

Solution

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= Ef−Ei{E_f} - {E_i}
Ei{E_i}, Ef{E_f} is the energy of initial drop and the energy of final drop.
⇒E=T×A\Rightarrow E = T \times A
Where, EE is the energy, TT is the surface tension, AA 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 RR into nn drops of radius rr.
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
⇒E=T×A\Rightarrow E = T \times A
Where, EE is the energy, TT is the surface tension, AA 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= Ef−Ei{E_f} - {E_i}
Ei{E_i}, Ef{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 4πR24\pi {R^2}. We can substitute the values in the formula mentioned above.
⇒Ei=T(4πR2)\Rightarrow {E_i} = T\left( {4\pi {R^2}} \right)
The formula to find the final energy using the value for area that is 4πr24\pi {r^2}.We can substitute the values in the formula mentioned above.
⇒Ef=nT(4πr2)\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, ⇒Ei=T(4πR2) \Rightarrow {E_i} = T\left( {4\pi {R^2}} \right)
The energy of final drop, ⇒Ef=nT(4πr2) \Rightarrow {E_f} = nT\left( {4\pi {r^2}} \right)
⇒\Rightarrow Energy required= Ef−Ei{E_f} - {E_i}
⇒Ef−Ei=nT(4πr2)−T(4πR2)\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,
⇒Ef−Ei=T(4πr2n−4πR2)\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,
∴E=T(4πr2n−4πR2)\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.