# Problems Based on Surface Energy

 Science > You are Here

#### Example -1:

• A water film is formed between two straight parallel wires of 10 cm each with a separation of 0.1 cm. If the distance between the wires is increased by 0.1 cm, how much work is done? T = 0.072 N/m
• Solution:
• Given: Length of wire= l = 10 cm = 10 × 10-2 m, Increase in width = 0.1 cm = 0.1 × 10-2 m, Surface tension = T = 0.072 N/m
• To Find: Work done = W = ?

In this case, the film has two surfaces =  2 (l × b) = 2 × 10 × 10-2 × 0.1 × 10-2 m = 2  × 10-4  m2

Work done = Change in surface energy = T. dA = 0.072  × 2  × 10-4  = 1.44 × 10-5  J

Work done is 1.44 × 10-5  J

#### Example – 2:

• A liquid film is formed in a rectangular frame one side of which is a light and thin movable wire of length 4 cm. The frame is held in a vertical plane with the movable wire in the downward position. Find what weight required to hold the movable wire in equilibrium. How much work is done if the film is stretched by pulling the movable wire downwards by 2 cm? T = 0.025 N/m.
• Solution:
• Given: Length of wire= l = 4  cm = 4 × 10-2 m, displacement = dx = 2 cm = 2 × 10-2 m, Surface tension = T = 0.025 N/m.
• To Find: Weight required = F = ?, Work done = W = ?

In case of wire, the water wets it from two sides, Hence, the effective length = 2 l

We have, T = F / effective length

∴   F = T × effective length =  T × 2 l = 0.025  ×  2  × 4 × 10-2  = 0.002 N

In this case, the film has two surfaces =  2 (l × dx) = 2 × 4 × 10-2 × 2 × 10-2 m = 16  × 10-4  m2

Work done = Change in surface energy = T. dA = 0.025  × 16  × 10-4  = 4 × 10-5  J

∴  The weight required is 0.002 N and work done is 4 × 10-5  J

#### Example – 3:

• There is a soap film on  arectangular wire frame of area 4cm × 4 cm. If area of the frame is increased to 4cm x 5 cm, find work done in the process. Surface tension of soap film is 3 × 10-2 N/m.
• Solution:
• Given: Initial area= A1 = 4cm × 4 cm = 16 cm2 = 16 × 10-4 m2, Final area= A2 = 4cm × 5 cm = 20 cm2 = 20 × 10-4 m2, Surface tension = T =3 × 10-2 N/m.
• To Find: Work done = W = ?

In case of rectangular frame, the water wets it from two sides,

Hence, the change in area  =  dA = 2 (A2 – A1)  =  2 × (20 × 10-4  – 16 × 10-4)

∴  dA = 2 (A2 – A1)  =  2 × 4 × 10-4  =  8 × 10-4  m2

Work done = Change in surface energy = T. dA = 3 × 10-2  × 8  × 10-4  = 2.4 × 10-5  J

work done is 2.4 × 10-5  J

#### Example – 4:

• Find the work done in blowing a soap bubble of radius 5 cm. S.T. of soap solution = 0.035 N/m.
• Solution:
• Given: Radius of bubble = r = 5 cm = 5 × 10-2 m, Surface tension = T = 0.035 N/m
• To Find: Work done = W = ?

In case of bubble it has two free surfaces =  dA = 2 (4 π r²) = 8 π r²

Work done = Change in surface energy = T. dA = T  × 8 π r²

∴  Work done = 0.035  × 8  × 3.142 × ( 5 × 10-2)2

∴  Work done = 0.035  × 8  × 3.142 × 25 × 10-4  = 2.2 × 10-3  J

∴  Work done is  2.2 × 10-3  J

#### Example – 5:

• Find the work done in blowing a soap bubble of radius 10 cm. S.T. of soap solution = 30 dyne/cm.
• Solution:
• Given: Radius of bubble = r = 10 cm = 10 × 10-2 m, Surface tension = T =30 dyne/cm = 30 × 10-3 N/m.
• To Find: Work done = W = ?

In case of bubble it has two free surfaces =  dA = 2 (4 π r²) = 8 π r²

Work done = Change in surface energy = T. dA = T  × 8 π r²

∴  Work done = 30 × 10-3  × 8  × 3.142 × ( 10 × 10-2)2

∴  Work done = 30 × 10-3  × 8  × 3.142 × 100 × 10-4  = 7.54 × 10-3  J

∴  Work done is  7.54 × 10-3  J

#### Example – 6:

• Calculate the work done in increasing the radius of a soap bubble in air from 1 cm to 2 cm. The surface tension of a soap solution is 30 dyne/cm.
• Solution:
• Given: Initial radius of bubble = r = 1 cm = 1 × 10-2 m, Final radius of bubble = r2 = 2 cm = 2 × 10-2 m, Surface tension = T = 30 dyne/cm  = 30 × 10-3  N/m.
• To Find: Work done = W = ?

In case of bubble it has two free surfaces =  dA = A2 – A1 = 2 (4 π r2²)  – 2 (4 π r1²)

∴  dA = 8 π (  r2²  –   r1² ) = 8 π (  (2 × 10-2)²  –  (1 × 10-2)²)

∴  dA =  8 π (  4 × 10-4  –  1 × 10-4) =  8 π  × 3 × 10-4

Work done = Change in surface energy = T. dA = 30 × 10-3  × 8 π  × 3 × 10-4

Work done = 30 × 10-3  × 8  × 3.142 × 3 × 10-4   = 2.262 × 10-4  J

Work done is 2.262 × 10-4  J

#### Example – 7:

• 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.
• Solution:
• Given: radius of first drop = r = 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 = ?

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  =   2.1 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

Change in free surface energy is 4.61 × 10-6  J

#### Example – 8:

• 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.
• Solution:
• Given: Initial number of drops = n1 = 103, initial diameter of each drop = d = 10-8 m, initial radius of each drop = r = 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 = ?

Let ‘r’ be the radius of new drop formed.

The initial volume of  n1 drops = Final volume of n2 drop

∴   n ×  4/3 π r13   =   n ×  4/3 π r3

∴   n ×   r13   =   n ×  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. ( n × 4πr1²   – n × 4πr²) = T × 4π (n ×  r1²  –  n × 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

Energy liberated is 2.036  ×  10-14 J

#### Example – 9:

• 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.
• Solution:
• Given: Initial number of drops = n1 = 1, initial radius of each drop = r = 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 = ?

Let ‘r’ be the radius of new drop formed.

The initial volume of  n1 drops = Final volume of n2 drop

∴   n ×  4/3 π r13   =   n ×  4/3 π r3

∴   n ×   r13   =   n ×  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. (n × 4πr²   –   n × 4πr1² ) = T × 4π (n × r²  –  n ×  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

∴    Work done is  6.786 × 10-6 J

#### Example – 10:

• 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.
• Solution:
• Given: Initial number of drops = n1 = 1, initial radius of each drop = r = 10-3 m , Final number of drop = n2 = 125, Surface tension = T =0.55 N/m
• To Find: Change in area  = dU = ?

Let ‘r’ be the radius of new drop formed.

The initial volume of  n1 drops = Final volume of n2 drop

∴   n ×  4/3 π r13   =   n ×  4/3 π r3

∴   n ×   r13   =   n ×  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. (n × 4πr²   –   n × 4πr1² ) = T × 4π (n × r²  –  n ×  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 – 11:

• 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.
• Solution:
• Given: Initial number of drops = n1 = 1, initial radius of each drop = r = 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 = ?

Let ‘r’ be the radius of new drop formed.

The initial volume of  n1 drops = Final volume of n2 drop

∴   n ×  4/3 π r13   =   n ×  4/3 π r3

∴   n ×   r13   =   n ×  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. (n × 4πr²   –   n × 4πr1² ) = T × 4π (n × r²  –  n ×  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

∴    Work done is  1.357 × 10-5 J

#### Example – 12:

• Eight droplets of water each of radius 0.2 mm coalesce together to form a single drop, Find the change in total surface energy, Givenn : surface tension of water = 0.072 N/m.
• Solution:
• Given: Initial number of drops = n1 = 8, initial radius of each drop = r = 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 = ?

Let ‘r’ be the radius of new drop formed.

The initial volume of  n1 drops = Final volume of n2 drop

∴   n ×  4/3 π r13   =   n ×  4/3 π r3

∴   n ×   r13   =   n ×  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. (n × 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

∴    Change in surface energy is  1.448 × 10-7 J

#### Example – 13:

• 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
• Solution:
• Given: Initial number of drops = n1 = 1, initial radius of each drop = r = 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 = ?

Let ‘r’ be the radius of new drop formed.

The initial volume of  n1 drops = Final volume of n2 drop

∴   n ×  4/3 π r13   =   n ×  4/3 π r3

∴   n ×   r13   =   n ×  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. (n × 4πr²   –   n × 4πr1² ) = T × 4π (n × r²  –  n ×  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  =  0,01446 J

∴    Work done is  0.01446 J

Now, by work energy principle

Poential energy = work done

∴    mgh = 0.01446

∴    (Volume × density) gh = 0.01446

∴   4/3 π r× ρ  × g ×h = 0.01446

∴   4/3 × 3.142 × (5 × 10-3)× 13600  × 9.8 ×h = 0.01446

∴   h = (3 × 0.01446) / ( 4 × 3.142 × (5 × 10-3)× 13600  × 9.8)

∴   h = (3 × 0.01446) / ( 4 × 3.142 × 125 × 10-9 × 13600  × 9.8)

∴   h = 0.2072 m

The height from which the drop may have fallen is 0.2072 m.

#### Example – 14:

• The total energy of the free surface of a liquid drop is 2π times the surface tension of the liquid. What is the diameter of the drop? Assume all terms in S.I. unit.
• Solution:

In case of liquid drop  it has one free surfaces =  dA = 4 π r²

Surface energy = T. dA = T  × 4 π r²

∴    2π  T = T  × 4 π r²

∴    2 =  4  r²

∴    0.5 =   r²

∴   r =  0.707 m

∴  diameter = d = 2 r =  2  × 0.707 = 1.414 m

#### Problem – 15:

• A soap bubble of radius 12 cm is blown. The surface tension of soap solution is 30 dynes/cm. Calculate the work done in blowing the soap bubble.(MHSB October – 13-2 M)
• Solution:
• Given: Radius of bubble = r = 12 cm = 12 × 10-2 m, Surface tension = T =30 dyne/cm = 30 × 10-3 N/m.
• To Find: Work done = W = ?

In case of bubble it has two free surfaces =  dA = 2 (4 π r²) = 8 π r²

Work done = Change in surface energy = T. dA = T  × 8 π r²

∴  Work done = 30 × 10-3  × 8  × 3.142 × ( 12 × 10-2)2

∴  Work done = 30 × 10-3  × 8  × 3.142 × 144 × 10-4  = 0.01086  J

∴  Work done is 0.01086  J

#### Problem – 16:

• The energy of the free surface of liquid drop is 5π times the surface tension of the liquid. Find the diameter drop in c.g.s. System.  (MHSB March – 16, October – 16-2 M)
• Solution:

In case of liquid drop, it has one free ssurface=  dA = 4 π r²

Surface energy = T. dA = T  × 4 π r²

∴    5π  T = T  × 4 π r²

∴    5 =  4  r²

∴    1.25 =   r²

∴   r =  1.12 cm

∴  diameter = d = 2 r =  2  × 1.12 = 2.24 cm

#### Problem – 17:

• The total free surface energy of a liquid drop is π √2  times the surface tension of the liquid. Calculate the diameter of the drop in S.I. unit. (MHSB July – 16 – 2 M)
• Solution:

In case of liquid drop, it has one free ssurface=  dA = 4 π r²

Surface energy = T. dA = T  × 4 π r²

∴    π √2   T = T  × 4 π r²

∴    5√2 =  4  r²

∴      5√2 / 4 =   r²

∴   r =  1.33 m

∴  diameter = d = 2 r =  2  × 1.12 = 2.24 cm

#### Problem – 18:

• Two soap bubbles have radii in the ratio 4:3. What is the ratio of work done to blow these bubbles?
• Solution:
• Given:  r1 / r2=4/3, Surface tension = T1 =T2 = T
• To Find: W/ W2  =?

In case of bubble it has two free surfaces =  dA = 2 (4 π r²) = 8 π r²

Work done = Change in surface energy = T. dA = T  × 8 π r²

For First Bubble    W1  = T1  × 8 π r1²  …………. (1)

For second Bubble    W2  = T2  × 8 π r2²  …………. (2)

Dividing equation (1) by (2) we get

W/ W2  = (T1  × 8 π r1² ) / (T2  × 8 π r2² )

W/ W2  = (r/ r2 )²  = = (4 /3 )² = 16/9

∴ The ratio of work done to blow these bubbles is 16 : 9.

#### Example – 19:

• 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 – 20:

• 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 a n droplets of radius r, then energy liberated

dU  = 4πR² T. ( n   –  n2/3) or  dU  = 4πr²T (n1/3   – 1)

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