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Question
Twenty-seven droplets of water, each of radius 0.1 mm coalesce into a single drop. Find the change in surface energy. Surface tension of water is 0.072 N/m.
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Solution 1
Given:
r = 0.1 mm = 0.1 × 10−3 m,
T = 0.072 N/m
To find:
The change in surface energy.
Solution:
Let R be the radius of the single drop formed due to the coalescence of 27 droplets of mercury. Volume of 27 droplets = volume of the single drop as the volume of the liquid remains constant.
∴ `27 xx 4/3pi"r"^3 = 4/3pi"R"^3`
∴ `27"r"^3 = "R"^3`
∴ 3r = R
Surface area of 27 droplets = 27 × 4πr2 Surface area of single drop = 4πR2
∴ Decrease in surface area = `27 xx 4pi"r"^2 - 4pi"R"^2`
= `4pi (27"r"^2 - "R"^2)`
= `4pi[27"r"^2 - ("3r")^2]`
= `4pi xx 18"r"^2`
∴ The energy released = surface tension × decrease in surface area
= T × 4π × 18r2
= 0.072 × 4 × 3.142 × 18 × (1 × 10−4)2
= 1.628 × 10−7 J.
Solution 2
Given:
r = 0.1 mm = 10−4 m,
n = 27,
T = 0.072 N/m
To find:
Change in surface energy (W)
Formula:
W = TdA
Calculation:
Volume of a single drop = `4/3piR^3` and
The volume of a single droplet = `4/3pir^3`
∴ We have, `4/3piR^3 = n xx 4/3pir^3` or R3 = nr3
∴ R = `root3 n r = root3 27 xx 10^-4 = 3 xx 10^-4`m2
From formula,
W = T (n × 4πr2 − 4πR2)
= 4πT(nr2 - R2)
= 4 × 3.142 × 0.072 × [27 × (10−4)2 − (3 × 10−4)2]
= 3.142 × 0.288 × 18 × 10−8
= 1.629 × 10−7 J
The change in the surface energy is 1.629 × 10−7 J.
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