Solveeit Logo
Login

Question

Question: The change in energy when a big drop is split in small n droplets is: (A) \[4\pi {R^2}({n^{\dfrac{...

The change in energy when a big drop is split in small n droplets is:
(A) 4πR2(n23−1)T4\pi {R^2}({n^{\dfrac{2}{3}}} - 1)T
(B) 4πR2(n13−1)T4\pi {R^2}({n^{\dfrac{1}{3}}} - 1)T
(C) 4πR2(n−13−1)T4\pi {R^2}({n^{\dfrac{{ - 1}}{3}}} - 1)T
(D) 4πR2(n−23−1)T4\pi {R^2}({n^{\dfrac{{ - 2}}{3}}} - 1)T

Explanation

Solution

We will first apply the condition that Volume of the droplet will remain the same before and after splitting and obtain a relation between the radius of bigger and smaller droplets. Then we will calculate the change in energy using the following formula: -
ΔU=ΔAT\Delta U = \Delta AT
Where,
ΔU=\Delta U = Required change in energy
ΔA=\Delta A = Change in surface area
T= Surface Tension of the body.

Complete step by step Solution

When a big drop is split in small n droplets, the total volume of all the n small droplets after splitting will be equal to the volume of the bigger drop.
Let the radius of bigger drop = R
And the radius of smaller drop = r
Thus, Volume from bigger drop =4πR33 = \dfrac{{4\pi {R^3}}}{3}
And Total volume of smaller droplets n×4πr33n \times \dfrac{{4\pi {r^3}}}{3}
Equating these two we get: -
4πR33=n×4πr33\dfrac{{4\pi {R^3}}}{3} = n \times \dfrac{{4\pi {r^3}}}{3}
R3=n×r3{R^3} = n \times {r^3}
R=n13rR = {n^{\dfrac{1}{3}}}r
⇒r=n−13R\Rightarrow r = {n^{\dfrac{{ - 1}}{3}}}R . . . (1)
Using formula for Surface energy: -
U=TAU = TA
Where,
U = Surface Energy of the body.
T= Surface Tension of the body.
A= Surface Area of the body.
Case 1:
U1=A1{U_1} = {A_1} T . . . (2)
Case 2:
U2=A2{U_2} = {A_2} T . . . (3)
Subtracting (3) from (2): -
U2−U1=(A2−A1)T{U_2}-{U_1} = \left( {{A_2} - {A_1}} \right)T
ΔU=ΔAT\Delta U = \Delta AT
Where,
ΔU=\Delta U = Required change in energy
ΔA=\Delta A = change in surface area
A1{A_1} = Initial surface area of the drop =4πR2 = 4\pi {R^2}
A2{A_2} = Total final surface area of all smaller drops =n×4πr2 = n \times 4\pi {r^2}
ΔA=A2−A1\Delta A = {A_2} - {A_1}
ΔA=n×4πr2−4πR2\Delta A = n \times 4\pi {r^2} - 4\pi {R^2}
T = Surface Tension
We will now plug in the values in the formula: -
ΔU=(n×4πr2−4πR2)T\Delta U = (n \times 4\pi {r^2} - 4\pi {R^2})T
ΔU=−4π(R2−nr2)T\Delta U = - 4\pi ({R^2} - n{r^2})T
ΔU=−4π(R2−n(n−13R)2)T\Delta U = - 4\pi ({R^2} - n{({n^{\dfrac{{ - 1}}{3}}}R)^2})T (Using equ-(1) we substitute the value for ‘r’ in terms of ‘R’)
ΔU=−4π(R2−n1−(23)R2)\Delta U = - 4\pi ({R^2} - {n^{1 - (\dfrac{2}{3})}}{R^2})
ΔU=−4π(R2−n13R2)T\Delta U = - 4\pi ({R^2} - {n^{\dfrac{1}{3}}}{R^2})T
Here, the negative sign represents that Energy is released as final surface energy is less than initial surface energy.
⇒Change In Energy=4π(R2−n13R2)T\Rightarrow Change\,In\,Energy = 4\pi ({R^2} - {n^{\dfrac{1}{3}}}{R^2})T

Hence, option (b) is correct.

Note When a bigger droplet splits into smaller droplets, Surface energy reduces and as a result energy is released. Similarly, when multiple smaller droplets are joined together to form one single bigger drop, Surface energy increases and as a result energy is absorbed. But, in the question change in energy is asked, so the sign has to be omitted.
Another thing to note is that in the question, we have neglected the viscosity of the drops since no energy is lost in that and we assumed all the energy loss is because of the change in surface tension of the system.