
Which of the following statements is true in case when two water drops coalesce and make a bigger drop?
A. Energy is released
B. Energy is absorbed
C. The surface area of the bigger drop is greater than the sum of the surface areas of both the drops
D. The surface area of the bigger drop is same as that of the sum of the surface areas of both the drops
Answer
164.1k+ views
Hint: To solve the above question, use the property of surface tension in fluids. Energy is released when the surface area of an object decreases while it is absorbed when the surface area increases. Find out whether the surface area increases or decreases when two water drops combine to make a bigger drop.
Complete answer:
Let the radius of $2$ drops be $r$ and the radius of the one large drop be $R$ .
As it is given that these $2$ coalesce to make one large drop, the total initial and final volume will be same.
Considering the drops to be spherical,
Initial volume $ = 2 \times \dfrac{4}{3}\pi {r^3}$ … (1)
Vinal volume $ = \dfrac{4}{3}\pi {R^3}$ … (2)
As both the initial and final volume are same,
$2 \times \dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi {R^3}$
On simplifying further, we get
$R = \sqrt[3]{2}r$ … (3)
Calculating total surface area of the small drops,
Initial surface area $ = 2 \times 4\pi {r^2}$ … (4)
Calculating total surface area of the large drop,
Final surface area $ = 4\pi {R^2}$
Substituting the value of $R$ from equation (3), we get:
Final surface area $ = {(2)^{\dfrac{2}{3}}} \times 4\pi {r^2}$ … (5)
Comparing equation (4) and (5), as $2 \times 4\pi {r^2} > {(2)^{\dfrac{2}{3}}} \times 4\pi {r^2}$ ,
Initial surface area $ > $ Final surface area
From the above comparison, we can conclude that the surface area decreased, that is, the surface area of the final drop is less than the surface area of the initial small drops.
We know that when surface area decreases, energy is liberated.
Thus, the correct option is A.
Note: In the above question, you need to calculate the initial surface area and the final surface area, using the given data, in order to observe the increase or decrease in the surface area. This will let you know whether energy is absorbed or released throughout the process. This is why a student should use the given data correctly to calculate the surface area.
Complete answer:
Let the radius of $2$ drops be $r$ and the radius of the one large drop be $R$ .
As it is given that these $2$ coalesce to make one large drop, the total initial and final volume will be same.
Considering the drops to be spherical,
Initial volume $ = 2 \times \dfrac{4}{3}\pi {r^3}$ … (1)
Vinal volume $ = \dfrac{4}{3}\pi {R^3}$ … (2)
As both the initial and final volume are same,
$2 \times \dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi {R^3}$
On simplifying further, we get
$R = \sqrt[3]{2}r$ … (3)
Calculating total surface area of the small drops,
Initial surface area $ = 2 \times 4\pi {r^2}$ … (4)
Calculating total surface area of the large drop,
Final surface area $ = 4\pi {R^2}$
Substituting the value of $R$ from equation (3), we get:
Final surface area $ = {(2)^{\dfrac{2}{3}}} \times 4\pi {r^2}$ … (5)
Comparing equation (4) and (5), as $2 \times 4\pi {r^2} > {(2)^{\dfrac{2}{3}}} \times 4\pi {r^2}$ ,
Initial surface area $ > $ Final surface area
From the above comparison, we can conclude that the surface area decreased, that is, the surface area of the final drop is less than the surface area of the initial small drops.
We know that when surface area decreases, energy is liberated.
Thus, the correct option is A.
Note: In the above question, you need to calculate the initial surface area and the final surface area, using the given data, in order to observe the increase or decrease in the surface area. This will let you know whether energy is absorbed or released throughout the process. This is why a student should use the given data correctly to calculate the surface area.
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