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Question: A mercury drop of radius 1 cm in broken into \({10^6}\) droplets of equal size The work done is \(\r...

A mercury drop of radius 1 cm in broken into 106{10^6} droplets of equal size The work done is ρ=35×10−2N/m\rho = 35 \times {10^{ - 2}}N/m
A). 4.35×10−2J4.35 \times {10^{ - 2}}J
B). 4.35×10−3J4.35 \times {10^{ - 3}}J
C). 4.35×10−6J4.35 \times {10^{ - 6}}J
D). 4.35×10−8J4.35 \times {10^{ - 8}}J

Explanation

Solution

Hint –In order to solve this problem we will use the concept of surface tension and surface energy. As we know in order to increase the surface area of a liquid, we need to do some work. The work we will do will be against the force of surface tension. The amount of work done in increasing the surface area of the liquid will be stored as potential energy in the liquid surface. This additional potential energy per unit area of free surface of liquid is called surface energy.
Formula Used:- Surface energy (E)=S×ΔA\left( E \right) = S \times \Delta A
Where. S = surface tension and ΔA\Delta A = increase in surface area.

Complete step-by-step solution -
Surface tension is the property of any liquid by which its free surface area is reduced.
Surface tension S=ForceLength=Work doneChange in areaSurface{\text{ }}tension{\text{ }}S = \dfrac{{Force}}{{Length}} = \dfrac{{Work{\text{ }}done}}{{Change{\text{ }}in{\text{ }}area}}
Its SI unit is Nm−1N{m^{ - 1}} or Jm−2J{m^{ - 2}} and its dimensional formula is [MT−2]\left[ {M{T^{ - 2}}} \right]
Given
Radius of the droplet
R=1cm=1×10−2mR = 1cm = 1 \times {10^{ - 2}}m
As, we know the volume remains constant
43πR3=106×43πr3 (10−2)3=106×r3 r=10−2102=10−4m  \dfrac{4}{3}\pi {R^3} = {10^6} \times \dfrac{4}{3}\pi {r^3} \\\ {\left( {{{10}^{ - 2}}} \right)^3} = {10^6} \times {r^3} \\\ r = \dfrac{{{{10}^{ - 2}}}}{{{{10}^2}}} = {10^{ - 4}}m \\\
Now, we will calculate the increase in the surface area of the mercury drop

ΔS=106×(4πr2)−4πR2 =106×4×π×10−8−4×π×10−4 =4×9.9×π×10−3  \Delta S = {10^6} \times{(4\pi {r^2)} - 4\pi {R^2}} \\\ = {10^6} \times 4 \times \pi \times {10^{ - 8}} - 4 \times \pi \times {10^{ - 4}} \\\ = 4 \times 9.9 \times \pi \times {10^{ - 3}} \\\

Now the work done is calculated as a product of surface area and surface tension.

=4×9.9×π×10−3×35×10−2 =4.35×10−2J  = 4 \times 9.9 \times \pi \times {10^{ - 3}} \times 35 \times {10^{ - 2}} \\\ = 4.35 \times {10^{ - 2}}J \\\

Hence, the correct option is “A”.

Note- In the above numerical we first calculated the change in the surface area and then multiplied the surface tension to find the amount of work done. The surface tension of a material relies entirely on the material itself and is independent of the film surface area or line length. Liquids drops become circular due to the surface tension property.