Solveeit Logo

Question

Question: When a big drop of water is formed by combining \(n\) small drops of water, the energy loss is \(3E{...

When a big drop of water is formed by combining nn small drops of water, the energy loss is 3E,3E{\text{,}} where EE is the energy of the bigger drop. If RR is the radius of the bigger drop and rr is the radius of the smaller drop, then find the number of smaller drops (n)(n) .
a. 4Rr3\dfrac{{4R}}{{{r^3}}}
b. 4Rr\dfrac{{4R}}{r}
c. 2R2r\dfrac{{2{R^2}}}{r}
d. 4R2r2\dfrac{{4{R^2}}}{{{r^2}}}

Explanation

Solution

Assume the shape of the water drops to be spherical. The energy released in the formation of the bigger drop can be expressed in terms of its surface tension and the change in the surface area. The change in the surface area will be the difference between the surface area of the big drop and that of nn smaller drops.

Formulas used:
- Surface area AA of a sphere of radius rr is A=n×4πr2A = n \times 4\pi {r^2}
- The energy released in the formation of a drop is E=T×ΔAE = T \times \Delta A , where EE is the energy of the formed drop, TT is its surface tension and ΔA\Delta A is the change in its surface area

Complete step by step answer:
Step 1: List the data given in the question.
The energy of the bigger drop is EE .
The energy released on the formation of the bigger drop =3E = 3E .
nn represents the number of smaller drops that combined to form the big drop of water.
RR and rr represent the radius of the big drop and one small drop respectively.
Let TT be the surface tension of the big drop.

Step 2: Express the relation for the change in the surface area involved in the formation of the bigger drop.
The water drops are assumed to be spherical in shape.
Let A1{A_1} be the total surface area of nn smaller drops and A2{A_2} be the surface area of the big drop.
Now, the total surface area of nn smaller drops =A1=n×4πr2 = {A_1} = n \times 4\pi {r^2} where 4πr24\pi {r^2} is the surface area of one small drop of radius rr .
The surface area of the big drop of water with a radius RR is given by, A2=4πR2{A_2} = 4\pi {R^2} where RR is the radius of the bigger drop.
The change in surface area is given by, ΔA=A1A2\Delta A = {A_1} - {A_2} (since, A1>A2{A_1} > {A_2} )
Substituting for A1=n×4πr2{A_1} = n \times 4\pi {r^2} and A2=4πR2{A_2} = 4\pi {R^2} , we get ΔA=(n×4πr2)(4πR2)\Delta A = (n \times 4\pi {r^2}) - (4\pi {R^2})
On simplifying, ΔA=4π(nr2R2)\Delta A = 4\pi (n{r^2} - {R^2})

Step 3: Express the energy released in terms of surface tension TT and change in area ΔA\Delta A .
It is given that the energy of the bigger drop is EE and the energy released in the formation of the bigger drop is 3E3E .
We can express EE and 3E3E in terms of surface tension and surface area
Energy of big drop=surface tension×surface area of big drop{\text{Energy of big drop}} = {\text{surface tension}} \times {\text{surface area of big drop}}
i.e., E=T×A2=T×4πR2E = T \times {A_2} = T \times 4\pi {R^2}
Energy released=surface tension×change in area{\text{Energy released}} = {\text{surface tension}} \times {\text{change in area}}
i.e., 3E=T×ΔA=T×4π(nr2R2)3E = T \times \Delta A = T \times 4\pi (n{r^2} - {R^2})

Step 4: Substitute for EE in the left-hand-side of the equation of 3E3E
We have, 3E=T4π(nr2R2)3E = T4\pi (n{r^2} - {R^2})
On substituting, 3×T4πR2=T4π(nr2R2)3 \times T4\pi {R^2} = T4\pi (n{r^2} - {R^2})
Cancel out the similar terms to get 3R2=nr2R23{R^2} = n{r^2} - {R^2}
On simplifying, nr2=4R2n{r^2} = 4{R^2}
Finally, n=4R2r2n = \dfrac{{4{R^2}}}{{{r^2}}}

Hence, the correct answer is option (D).

Note: When nn small drops of water combine to form one big drop of water, energy is released or lost in the formation. This is because the process involves a decrease in the surface area of the drop.