Science > Physics > Surface Tension >Change in Surface Energy
In this article. we shall study numerical problems on the change in surface energy due to breaking of drop or coalescence of number of drops into a one drop.
Example – 01:
Two mercury drops one of radius 2 mm and the other of radius 1 mm coalesce to form a single drop. Find the change in the free surface energy. S.T. of mercury = 0.544 N/m.
Given: radius of first drop = r1 = 2 mm = 2 × 10-3 m, radius of second bubble = r2 = 1 mm = 1 × 10-3 m, Surface tension = T = 0.544 N/m
To Find: Change in free surface energy = dU =?
Solution:
Let ‘r’ be the radius of new drop formed.
The initial volume of two drops = Final volume of single drop
∴ 4/3 π r13 + 4/3 π r23 = 4/3 π r3
∴ r13 + r23 = r3
∴ 23 + 13 = r3
∴ r3 = 9
∴ r = 2.08 mm = 2.08 × 10-3 m
∴ Change in surface energy = dU = U1 – U2
∴ dU = T. A1 – T.A2 = T. (A1 – A2)
∴ dU = T. [(4πr1² +4πr2²) – 4πr²) = T × 4π [(r1² +r2²) – r²)]
∴ dU = 0.544 × 4π [((2 × 10-3)² + (1 × 10-3)²) – (2.08 × 10-3)²)]
∴ dU = 0.544 × 4π [4 × 10-6 + 1 × 10-6 – 4.3264 × 10-6]
∴ dU = 0.544 × 4 × 3.142 × 0.6736 × 10-6
∴ dU = 4.61 × 10-6 J
Ans: The change in free surface energy is 4.61 × 10-6 J
Example – 02:
How much energy would be liberated if 103 water droplets each 10-8 m in diameter coalesce to produce a single drop? S.T. of water = 0.072 N/m. Assume drops are spherical.
Given: Initial number of drops = n1 = 103, initial diameter of each drop = d1 = 10-8 m, initial radius of each drop = r1 = 0.5 × 10-8 m = = 5 × 10-9 m Final number of drop = n2 = 1, Surface tension = T = 0.072 N/m
To Find: Energy liberated = dU =?
Solution:
Let ‘r’ be the radius of new drop formed.
The initial volume of n1 drops = Final volume of n2 drop
∴ n1 × 4/3 π r13 = n2 × 4/3 π r3
∴ n1 × r13 = n2 × r3
∴ 103 × (5 × 10-9)3 = 1× r3
∴ 10 × 5 × 10-9 = r
∴ r = 5 × 10-8 m
∴ Energy released = dU = U1 – U2
∴ dU = T. A1 – T.A2 = T. (A1 – A2)
∴ dU = T. ( n1 × 4πr1² – n2 × 4πr²) = T × 4π (n1 × r1² – n2 × r²)
∴ dU = 0.072 × 4π [103 × (5 × 10-9)² – 1 × (5 × 10-8)²]
∴ dU = 0.072 × 4 × 3.142 × [103 × 25 × 10-18 – 25 ×10-16]
∴ dU = 0.072 × 4 × 3.142 × [250 × 10-16 – 25 ×10-16]
∴ dU = 0.072 × 4 × 3.142 × 225 × 10-14
∴ dU = 2.036 × 10-14 J
Ans: The energy liberated is 2.036× 10-14 J
Example – 03:
A drop of mercury of radius 0.1 cm is broken into 8 droplets of the same size. Find the work done if the surface tension of mercury is 540 dyne/cm.
Given: Initial number of drops = n1 = 1, initial radius of each drop = r1 = 0.1 cm = 0.1 × 10-2 m = 1 × 10-3 m, Final number of drop = n2 = 8, Surface tension = T =540 dyne/cm = 540 × 10-3 N/m
To Find: Work done = W =?
Solution:
Let ‘r’ be the radius of new drop formed.
The initial volume of n1 drops = Final volume of n2 drop
∴ n1 × 4/3 π r13 = n2 × 4/3 π r3
∴ n1 × r13 = n2 × r3
∴ 1 × (1 × 10-3)3 = 8× r3
∴ 1 × 10-3= 2 r
∴ r = 0.5 × 10-3 m
∴ Work done = Change in surface energy = dU = U1 – U2
∴ dU = T. A2 – T.A1 = T. (A2 – A1)
∴ dU = T. (n2 × 4πr² – n1 × 4πr1² ) = T × 4π (n2 × r² – n1 × r1² )
∴ dU = 540 × 10-3 × 4π [8 × (0.5 × 10-3)² – 1 × (1 × 10-3)²]
∴ dU = 540 × 10-3 × 4π [8 × 0.25 × 10-6 – 1 × 10-6]
∴ dU = 540 × 10-3 × 4π [2 × 10-6 – 1 × 10-6]
∴ dU = 540 × 10-3 × 4 × 3.142 × 1 × 10-6
∴ dU = 6.786 × 10-6 J
Ans: The work done is 6.786 × 10-6 J
Example – 04:
A mercury drop of radius 10-3 m breaks up into 125 small droplets. Calculate the change in energy assuming that the drops are spherical and S.T. of the mercury is 0.55 N/m.
Given: Initial number of drops = n1 = 1, initial radius of each drop = r1 = 10-3 m, Final number of drop = n2 = 125, Surface tension = T =0.55 N/m
To Find: Change in area = dU =?
Solution:
Let ‘r’ be the radius of new drop formed.
The initial volume of n1 drops = Final volume of n2 drop
∴ n1 × 4/3 π r13 = n2 × 4/3 π r3
∴ n1 × r13 = n2 × r3
∴ 1 × (10-3)3 = 125× r3
∴ 1 × 10-3= 5 r
∴ r = 0.2 × 10-3 m
∴ Work done = Change in surface energy = dU = U1 – U2
∴ dU = T. A2 – T.A1 = T. (A2 – A1)
∴ dU = T. (n2 × 4πr² – n1 × 4πr1² ) = T × 4π (n2 × r² – n1 × r1² )
∴ dU = 0.55 × 4π [125 × (0.2 × 10-3)² – 1 × ( 10-3)²]
∴ dU = 0.55 × 4π [125 × 0.04 × 10-6 – 1 × 10-6]
∴ dU = 0.55 × 4π [5 × 10-6 – 1 × 10-6]
∴ dU = 0.55 × 4 × 3.142 × 4 × 10-6
∴ dU = 2.76 × 10-5 J
∴ Change in energy is 2.76 × 10-5 J
Example – 05:
A mercury drop of radius 0.1 cm breaks up into 27 small droplets. Calculate the work done assuming that the drops are spherical and S.T. of the mercury is 540 dyne/cm.
Given: Initial number of drops = n1 = 1, initial radius of each drop = r1 = 0.1 cm = 0.1 × 10-2 m = 1 × 10-3 m, Final number of drop = n2 = 27, Surface tension = T =540 dyne/cm = 540 × 10-3 N/m
To Find: Work done = W =?
Solution:
Let ‘r’ be the radius of new drop formed.
The initial volume of n1 drops = Final volume of n2 drop
∴ n1 × 4/3 π r13 = n2 × 4/3 π r3
∴ n1 × r13 = n2 × r3
∴ 1 × (1 × 10-3)3 = 27× r3
∴ 1 × 10-3= 3 r
∴ r = 0.33 × 10-3 m
∴ Work done = Change in surface energy = dU = U1 – U2
∴ dU = T. A2 – T.A1 = T. (A2 – A1)
∴ dU= T. (n2 × 4πr² – n1 × 4πr1² ) = T × 4π (n2 × r² – n1 × r1² )
∴ dU = 540 × 10-3 × 4π [27 × (0.33 × 10-3)² – 1 × (1 × 10-3)²]
∴ dU = 540 × 10-3 × 4π [27 × 0.1089 × 10-6 – 1 × 10-6]
∴ dU= 540 × 10-3 × 4π [3 × 10-6 – 1 × 10-6]
∴ dU = 540 × 10-3 × 4 × 3.142 × 2 × 10-6
∴ dU = 1.357 × 10-5 J
Ans: The work done is 1.357 × 10-5 J
Example – 06:
Eight droplets of water each of radius 0.2 mm coalesce together to form a single drop, Find the change in total surface energy, given: surface tension of water = 0.072 N/m.
Given: Initial number of drops = n1 = 8, initial radius of each drop = r1 = 0.2 cm = 0.2 × 10-3 m = 2 × 10-4 m, Final number of drop = n2 = 1, Surface tension = T =0.072 N/m
To Find: Change in total surface energy = dU =?
Solution:
Let ‘r’ be the radius of new drop formed.
The initial volume of n1 drops = Final volume of n2 drop
∴ n1 × 4/3 π r13 = n2 × 4/3 π r3
∴ n1 × r13 = n2 × r3
∴ 8 × (2 × 10-4)3 = 1× r3
∴ 64× 10-12= 2 r
∴ r = 4 × 10-4 m
∴ Work done = Change in surface energy = dU = U1 – U2
∴ dU = T. A2 – T.A1 = T. (A1 – A2)
∴ dU = T. (n1 × 4πr1² – n2 × 4πr² ) = T × 4π (n1 × r² – n2 × r² )
∴ dU = 0.072 × 4π [8 × (2 × 10-4)² – 1 × (4 × 10-4)²]
∴ dU = 0.072 × 4π [8 × 4 × 10-8 – 16 × 10-8]
∴ dU = 0.072 × 4π [32 × 10-8 – 16 × 10-8]
∴ dU = 0.072 × 4 × 3.142 × 16 × 10-8
∴ dU = 1.448 × 10-7 J
Ans: The change in surface energy is 1.448 × 10-7 J
Example – 07:
A mercury drop of radius 0.5 cm falls from a height on glass plate and breaks up into million droplets, all of the same size, Find the height from which the drop must have fallen. Density of mercury 13600 kg/m3. Surface tension of mercury = 0.465 N/m
Given: Initial number of drops = n1 = 1, initial radius of each drop = r1 = 0.5 cm = 0.5 × 10-2 m = 5 × 10-3 m, Final number of drop = n2 = 106, Surface tension = T = 0.465 N/m
To Find: Height from which the drop falls = h =?
Solution:
Let ‘r’ be the radius of new drop formed.
The initial volume of n1 drops = Final volume of n2 drop
∴ n1 × 4/3 π r13 = n2 × 4/3 π r3
∴ n1 × r13 = n2 × r3
∴ 1 × (5 × 10-3)3 = 106× r3
∴ 5 × 10-3= 102 r
∴ r = 5 × 10-5 m
∴ Work done = Change in surface energy = dU = U1 – U2
∴ dU = T. A2 – T.A1 = T. (A2 – A1)
∴ dU= T. (n2 × 4πr² – n1 × 4πr1² ) = T × 4π (n2 × r² – n1 × r1² )
∴ dU= 0.465 × 4π [106 × (5 × 10-5)² – 1 × (5 × 10-3)²]
∴ dU = 0.465 × 4π [106 × 25 × 10-10 – 25 × 10-6]
∴ dU = 0.465 × 4π [2500 × 10-6 – 25 × 10-6]
∴ dU= 0.465 × 4 × 3.142 × 2475 × 10-6
∴ dU = 0,01446J
∴ Work done is 0.01446J
Now, by work energy principle
Poential energy = work done
∴ mgh = 0.01446
∴ (Volume × density) gh = 0.01446
∴ 4/3 π r3 × ρ × g ×h = 0.01446
∴ 4/3 × 3.142 × (5 × 10-3)3 × 13600 × 9.8 ×h = 0.01446
∴ h = (3 × 0.01446) / (4 × 3.142 × (5 × 10-3)3 × 13600 × 9.8)
∴ h = (3 × 0.01446) / (4 × 3.142 × 125 × 10-9 × 13600 × 9.8)
∴ h = 0.2072 m
Ans: The height from which the drop may have fallen is 0.2072 m.
Example – 08:
n droplets of equal size of radius r coalesce to form a bigger drop of radius R. If T is the surface tension of the liquid, then show that the energy liberated is 4π R²T (n1/3 – 1)
Solution:
The initial volume of n drops = Final volume of one drop
∴ n× 4/3 π r3 = 1 × 4/3 π R3
∴ n × r3 = R3
∴ R = r × n1/3
∴ R × n– 1/3 = r
∴ Energy liberated = dU = U1 – U2
∴ dU = T. A1 – T.A2 = T. (A1 – A2)
∴ dU = T. ( n × 4πr² – 1 × 4πR²)
∴ dU = T. ( n × 4π(R × n– 1/3)² – 1 × 4πR²)
∴ dU = T. ( n × 4πR² × n– 2/3 – 1 × 4πR²)
∴ dU = T. 4πR² ( n × n– 2/3 – 1)
∴ dU = 4π R²T (n1/3 – 1)
Example – 09:
n droplets of equal size of radius r coalesce to form a bigger drop of radius R. If T is the surface tension of the liquid, then show that the energy liberated is 4πr² T. (n – n2/3)
Solution:
The initial volume of n drops = Final volume of one drop
∴ n× 4/3 π r3 = 1 × 4/3 π R3
∴ n × r3 = R3
∴ R = r × n1/3
∴ Energy liberated = dU = U1 – U2
∴ dU = T. A1 – T.A2 = T. (A1 – A2)
∴ dU = T. ( n × 4πr² – 1 × 4πR²)
∴ dU = T. ( n × 4πr² – 1 × 4π(r × n1/3)²)
∴ dU = T. ( n × 4πr² – 1 × 4π r² × n2/3)
∴ dU = T. 4πr²( n – n2/3)
∴ dU = 4πr² T. ( n – n2/3)
Note:
For competitive exams use following formulae directly
Coalescence of n droplets of radius r into a single drop of radius R, then energy liberated
dU = 4πr² T. ( n – n2/3) or dU = 4π R²T (n1/3 – 1)
Breaking of one drop of radius R into n droplets of radius r, then energy liberated
dU = 4πR² T. ( n – n2/3) or dU = 4πr²T (n1/3 – 1)
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