Answer
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Hint: The disruption of intermolecular bonds that occurs when a surface is formed is measured by surface energy. Surface free energy or interfacial free energy are other names for it. In layman's terms, surface energy is the work performed per unit area by the force that produces the new surface.
Complete answer:
Surface tension is the phenomenon that occurs when the surface of a liquid comes into contact with another phase, according to the description (it can be a liquid as well). Liquids prefer to have as little surface area as possible. The liquid's surface behaves like an elastic sheet.
Let us come to the problem:
Since the diameter of drop$ = 4mm$Radius of drop$ = 2mm = 2 \times {10^{ - 3}}$
S=Surface tension=$0.07\,N{m^{ - 1}}$
a large drop in volume$ = 1000 \times $ volume of each droplet
Let R be the major drop's radius and r be the small droplet's radius.
$\dfrac{4}{3}\pi {R^3} = 1000 \times \dfrac{4}{3}\pi {r^3}$
$
{R^3} = {(10r)^3} \\
R = 10r \\
$
liquid drop's initial energy$ = {E_1} = T.A$
${E_1} = T \times 4\pi {R^2}$ T= surface tension, A=area
Final energy of 1000 droplets $ = {E_2} = 1000T.a$
${E_2} = 1000T \times 4\pi {r^2}$ T= surface tension, a=area
Now, change in energy$ = {E_2} - {E_1}$
$ = 1000T \times 4\pi {r^2}$-$T \times 4\pi {R^2}$
\[
= T \times 4\pi [1000{r^2} - {R^2}] \\
= T \times 4\pi [1000{\left( {\dfrac{R}{{10}}} \right)^2} - {R^2}]\{ \because R = 10r\} \\
= T \times 4\pi [10{R^2} - {R^2}] \\
= T \times 4\pi \times 9{R^2} \\
= 36\pi T{R^2} \\
\]
Now, let us put the value of T and R
$ = 36 \times 3.14 \times 0.07\,\,J{m^{ - 2}} \times (2 \times {10^{ - 3}})$
$ = 3.165 \times {10^{ - 5}}\,\,J{m^{ - 2}}$
Note:
The surface tension of a liquid refers to the attractive force that the molecules present at its surface exert on each other. The equivalent attractive force present between the molecules at the surface of a solid material is known as surface energy.
Complete answer:
Surface tension is the phenomenon that occurs when the surface of a liquid comes into contact with another phase, according to the description (it can be a liquid as well). Liquids prefer to have as little surface area as possible. The liquid's surface behaves like an elastic sheet.
Let us come to the problem:
Since the diameter of drop$ = 4mm$Radius of drop$ = 2mm = 2 \times {10^{ - 3}}$
S=Surface tension=$0.07\,N{m^{ - 1}}$
a large drop in volume$ = 1000 \times $ volume of each droplet
Let R be the major drop's radius and r be the small droplet's radius.
$\dfrac{4}{3}\pi {R^3} = 1000 \times \dfrac{4}{3}\pi {r^3}$
$
{R^3} = {(10r)^3} \\
R = 10r \\
$
liquid drop's initial energy$ = {E_1} = T.A$
${E_1} = T \times 4\pi {R^2}$ T= surface tension, A=area
Final energy of 1000 droplets $ = {E_2} = 1000T.a$
${E_2} = 1000T \times 4\pi {r^2}$ T= surface tension, a=area
Now, change in energy$ = {E_2} - {E_1}$
$ = 1000T \times 4\pi {r^2}$-$T \times 4\pi {R^2}$
\[
= T \times 4\pi [1000{r^2} - {R^2}] \\
= T \times 4\pi [1000{\left( {\dfrac{R}{{10}}} \right)^2} - {R^2}]\{ \because R = 10r\} \\
= T \times 4\pi [10{R^2} - {R^2}] \\
= T \times 4\pi \times 9{R^2} \\
= 36\pi T{R^2} \\
\]
Now, let us put the value of T and R
$ = 36 \times 3.14 \times 0.07\,\,J{m^{ - 2}} \times (2 \times {10^{ - 3}})$
$ = 3.165 \times {10^{ - 5}}\,\,J{m^{ - 2}}$
Note:
The surface tension of a liquid refers to the attractive force that the molecules present at its surface exert on each other. The equivalent attractive force present between the molecules at the surface of a solid material is known as surface energy.
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