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 n small drops of water, the energy loss is 3E, where E is the energy of the bigger drop. If R is the radius of the bigger drop and r is the radius of the smaller drop, then find the number of smaller drops (n) .
a. r34R
b. r4R
c. r2R2
d. r24R2
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 n smaller drops.
Formulas used:
- Surface area A of a sphere of radius r is A=n×4πr2
- The energy released in the formation of a drop is E=T×ΔA , where E is the energy of the formed drop, T is its surface tension and ΔA 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 E .
The energy released on the formation of the bigger drop =3E .
n represents the number of smaller drops that combined to form the big drop of water.
R and r represent the radius of the big drop and one small drop respectively.
Let T 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 be the total surface area of n smaller drops and A2 be the surface area of the big drop.
Now, the total surface area of n smaller drops =A1=n×4πr2 where 4πr2 is the surface area of one small drop of radius r .
The surface area of the big drop of water with a radius R is given by, A2=4πR2 where R is the radius of the bigger drop.
The change in surface area is given by, ΔA=A1−A2 (since, A1>A2 )
Substituting for A1=n×4πr2 and A2=4πR2 , we get ΔA=(n×4πr2)−(4πR2)
On simplifying, ΔA=4π(nr2−R2)
Step 3: Express the energy released in terms of surface tension T and change in area ΔA .
It is given that the energy of the bigger drop is E and the energy released in the formation of the bigger drop is 3E .
We can express E and 3E in terms of surface tension and surface area
Energy of big drop=surface tension×surface area of big drop
i.e., E=T×A2=T×4πR2
Energy released=surface tension×change in area
i.e., 3E=T×ΔA=T×4π(nr2−R2)
Step 4: Substitute for E in the left-hand-side of the equation of 3E
We have, 3E=T4π(nr2−R2)
On substituting, 3×T4πR2=T4π(nr2−R2)
Cancel out the similar terms to get 3R2=nr2−R2
On simplifying, nr2=4R2
Finally, n=r24R2
Hence, the correct answer is option (D).
Note: When n 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.