Question

A drop of liquid of diameter \[2.8mm\] breaks up into \[125\] identical drops. The change in energy is nearly (S.T. of liquid\[ = 75dyne/cm\])
A. zero
B. \[19erg\]
C. \[46erg\]
D. \[74erg\]

Answer 1

Hint: The volume of the big drop is equal to the total volume of all the small drops. From this concept, the radius of each small drop can be calculated. And, the radius will help to measure the area. As for this problem one can calculate the change in energy by the product of surface tension and increased area.
Formula used:
${\text{Volum}}{{\text{e}}_{{\text{big - drop}}}}{\text{ = }}n{\text{Volum}}{{\text{e}}_{{\text{small - drop}}}}$
$n$is the number of the small drop.
Change in energy\[ = \]Surface tension\[ \times \]Increased area.

Complete answer:
A drop of liquid breaks up into several identical drops.
Hence,
${\text{Volum}}{{\text{e}}_{{\text{big - drop}}}}{\text{ = }}n{\text{Volum}}{{\text{e}}_{{\text{small - drop}}}}$
$ \Rightarrow = \dfrac{4}{3}\pi {R^3} = n \times \dfrac{4}{3}\pi {r^3}$
$n$is the number of the small drop.
$R$ the radius of the big drop, \[r\]be the radius of each small drop.
given, the radius of the liquid \[R = \dfrac{{2.8}}{2}mm\] or, \[R = 1.4mm = 0.14cm\], $n = 125$
\[ \Rightarrow \dfrac{4}{3}\pi {\left( {0.14} \right)^3} = 125 \times \dfrac{4}{3}\pi {r^3}\]
\[ \Rightarrow r = \dfrac{R}{5} = \dfrac{{0.14}}{5} = 0.028cm\]
As we know, Change in energy\[ = \]Surface tension\[ \times \]Increased area
\[E = 75 \times \left[ {125 \times 4\pi {r^2} - 4\pi {R^2}} \right]erg\]
\[E = 75 \times 4\pi \left[ {125 \times {{\left( {0.028} \right)}^2} - {{(0.14)}^2}} \right]erg\]
\[E = 73.89erg \simeq 74erg\]

So, the answer is option D.

Note:
All matter is made of molecules that have varying degrees of attraction. This intermolecular attraction is the cause of the molecules of a liquid staying together in a glass or as a puddle on the top of a table rather than moving apart into the air. The attractive force of the molecules on the surface of a liquid between each other is known as the surface tension of that liquid.
Surface tension is measured by the force per unit length of the surface. The SI unit of surface energy is \[N/m\].
The dimension of surface energy is \[\left[ {M{T^{ - 2}}} \right]\].
Surface tension and surface energy are useful since they are measurable numbers. and, they define the strongness of the attraction between molecules. The substances that have High surface energy are of high boiling points. they have to be heated up much to exceed the intermolecular attractions to create vapor or steam.