physics : class11 Relation Between Surface Tension and Surface Energy : Surface Tension : Properties of Bulk Matter MCQ Questions and Answers
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Relation Between Surface Tension and Surface Energy
MCQ questions and answers.
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MCQ Questions
1. A liquid surface has surface tension 0.072 N m⁻¹. What minimum work is required to increase its area by 250 cm², assuming temperature remains constant?
A.
1.8 × 10⁻³ J
B.
1.8 × 10⁻² J
C.
2.88 × 10⁻² J
D.
7.2 × 10⁻⁴ J
Correct Answer:
1.8 × 10⁻³ J
Explanation:
For a single liquid surface, the increase in surface energy equals the work done: W = TΔA. Here, ΔA = 250 × 10⁻⁴ m² = 0.025 m², so W = 0.072 × 0.025 = 1.8 × 10⁻³ J.
2. A soap film has surface tension 0.030 N m⁻¹ and one edge is a movable wire of length 0.12 m. If the wire is displaced by 5.0 cm quasistatically, what is the increase in surface energy of the film?
A.
1.8 × 10⁻⁴ J
B.
3.6 × 10⁻⁴ J
C.
7.2 × 10⁻⁴ J
D.
9.0 × 10⁻⁵ J
Correct Answer:
3.6 × 10⁻⁴ J
Explanation:
A soap film has two surfaces. The total increase in area is 2lx, so the increase in surface energy is ΔE = 2Tlx = 2 × 0.030 × 0.12 × 0.05 = 3.6 × 10⁻⁴ J.
3. The surface energy of a liquid surface increases by 0.012 J when its area increases by 300 cm². Assuming constant temperature, the surface tension of the liquid is
A.
0.04 N m⁻¹
B.
0.4 N m⁻¹
C.
4.0 N m⁻¹
D.
0.004 N m⁻¹
Correct Answer:
0.4 N m⁻¹
Explanation:
For a single surface, surface tension equals surface energy increase per unit area: T = ΔE/ΔA. Since 300 cm² = 0.03 m², T = 0.012/0.03 = 0.4 N m⁻¹.
4. Which statement correctly expresses the relation between surface tension T and surface energy per unit area for a liquid surface at constant temperature?
A.
Surface energy per unit area is T/2 because the surface has two directions of tension
B.
Surface energy per unit area is 2T because work is done along two dimensions
C.
Surface energy per unit area is equal to T
D.
Surface energy per unit area is T²
Correct Answer:
Surface energy per unit area is equal to T
Explanation:
For a single liquid surface, an increase in area dA requires work dW = T dA. Therefore, the surface energy per unit area is dW/dA = T.
5. A rectangular soap film has dimensions 10 cm × 8 cm and surface tension 0.025 N m⁻¹. If its length is increased by 2 cm while its breadth remains unchanged, what is the increase in surface energy?
A.
4.0 × 10⁻⁵ J
B.
8.0 × 10⁻⁵ J
C.
1.0 × 10⁻⁴ J
D.
2.0 × 10⁻⁵ J
Correct Answer:
8.0 × 10⁻⁵ J
Explanation:
The increase in area of one face is ΔA = breadth × increase in length = 0.08 × 0.02 = 1.6 × 10⁻³ m². A soap film has two faces, so ΔE = 2TΔA = 2 × 0.025 × 1.6 × 10⁻³ = 8.0 × 10⁻⁵ J.
6. A movable wire of length l forms one side of a soap film. If the surface energy increases at the rate P while the wire moves outward with speed v, the surface tension of the film is
A.
P/(lv)
B.
P/(2lv)
C.
2P/(lv)
D.
Pl/(2v)
Correct Answer:
P/(2lv)
Explanation:
For a soap film, the increase in surface energy for displacement dx is dE = 2Tl dx. Dividing by dt gives P = dE/dt = 2Tlv, hence T = P/(2lv).
7. A liquid surface is expanded reversibly from area A to 3A, and its surface tension remains constant at T. The increase in surface energy is
A.
TA
B.
2TA
C.
3TA
D.
T/A
Correct Answer:
2TA
Explanation:
The change in area is ΔA = 3A − A = 2A. For one surface, the increase in surface energy is ΔE = TΔA = 2TA.
8. A soap film is stretched so that the area of each of its two surfaces increases by ΔA. If its surface tension is T, the external work required in a quasistatic process is
A.
TΔA
B.
2TΔA
C.
T(ΔA)²
D.
ΔA/(2T)
Correct Answer:
2TΔA
Explanation:
The film has two surfaces, and each surface gains area ΔA. The total area increase is 2ΔA, so the increase in surface energy and the required quasistatic work are 2TΔA.
9. For a liquid interface, the force associated with surface tension can be obtained from surface energy by which relation, when the change occurs over a distance x and the relevant length remains constant?
A.
F = x(dE/dA)
B.
F = dE/dx
C.
F = A(dE/dx)
D.
F = dA/dE
Correct Answer:
F = dE/dx
Explanation:
The mechanical work done during a displacement dx is dW = F dx. Since this work becomes surface energy, dE = F dx, giving F = dE/dx. For a film, this leads to F = 2Tl when dE = 2Tl dx.
10. A liquid drop of fixed volume is divided into 8 identical smaller spherical drops. If the original surface energy is E and surface tension remains unchanged, what is the total surface energy of the smaller drops?
A.
E/2
B.
2E
C.
4E
D.
8E
Correct Answer:
2E
Explanation:
If the original radius is R, each of the 8 smaller drops has radius R/2 because volume is conserved. The total area becomes 8 × 4π(R/2)² = 2 × 4πR², so total surface energy, equal to T times total area, becomes 2E.
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