physics : class11 Two-Surface Contribution in Soap Films : Surface Energy : Properties of Bulk Matter MCQ Questions and Answers
Practice important
Two-Surface Contribution in Soap Films
MCQ questions and answers.
Page 1
of
1
|
10
MCQ Questions
1. A soap film is stretched so that its geometrical area increases by 2.0 × 10^-2 m^2. If the surface tension of the soap solution is 3.0 × 10^-2 N m^-1, how much work is required, assuming the process is quasistatic?
A.
6.0 × 10^-4 J
B.
1.2 × 10^-3 J
C.
1.5 × 10^-3 J
D.
3.0 × 10^-3 J
Correct Answer:
1.2 × 10^-3 J
Explanation:
A soap film has two surfaces. Therefore, the increase in surface energy is W = 2γΔA = 2 × 3.0 × 10^-2 × 2.0 × 10^-2 = 1.2 × 10^-3 J.
2. A movable wire of length 0.12 m forms one side of a soap film. If the surface tension is 2.5 × 10^-2 N m^-1, what force must be applied to hold the wire stationary?
A.
1.5 × 10^-3 N
B.
3.0 × 10^-3 N
C.
6.0 × 10^-3 N
D.
1.2 × 10^-2 N
Correct Answer:
6.0 × 10^-3 N
Explanation:
Both surfaces of the soap film pull the movable wire. Hence, the total force is F = 2γl = 2 × 2.5 × 10^-2 × 0.12 = 6.0 × 10^-3 N.
3. The area of one face of a soap film increases from 4.0 × 10^-2 m^2 to 7.0 × 10^-2 m^2. If γ = 5.0 × 10^-2 N m^-1, the work done in increasing the area is
A.
1.5 × 10^-3 J
B.
3.0 × 10^-3 J
C.
4.5 × 10^-3 J
D.
7.0 × 10^-3 J
Correct Answer:
3.0 × 10^-3 J
Explanation:
The increase refers to the geometrical area of one face, but a soap film has two surfaces. Thus W = 2γ(A2 − A1) = 2 × 5.0 × 10^-2 × 3.0 × 10^-2 = 3.0 × 10^-3 J.
4. A rectangular soap film has a movable side of length l. If this side is displaced through a distance x while the surface tension remains constant at γ, which expression gives the increase in surface energy?
A.
γlx
B.
2γlx
C.
γl^2x
D.
2γl^2x
Correct Answer:
2γlx
Explanation:
The area of each surface increases by lx. Since the film has two surfaces, the total surface-energy increase is 2γ(lx) = 2γlx.
5. A soap film initially has an area of 8.0 × 10^-2 m^2 on each face and finally has an area of 5.0 × 10^-2 m^2 on each face. For γ = 4.0 × 10^-2 N m^-1, what is the change in the film's surface energy?
A.
-2.4 × 10^-3 J
B.
-1.2 × 10^-3 J
C.
1.2 × 10^-3 J
D.
2.4 × 10^-3 J
Correct Answer:
-2.4 × 10^-3 J
Explanation:
The change in surface energy is ΔE = 2γ(Af − Ai) = 2 × 4.0 × 10^-2 × (5.0 × 10^-2 − 8.0 × 10^-2) = -2.4 × 10^-3 J. The negative sign indicates that energy is released.
6. A soap film is expanded quasistatically by increasing the area of each face by 6.0 × 10^-2 m^2. If the work required is 4.8 × 10^-3 J, what is the surface tension?
A.
2.0 × 10^-2 N m^-1
B.
4.0 × 10^-2 N m^-1
C.
8.0 × 10^-2 N m^-1
D.
1.6 × 10^-1 N m^-1
Correct Answer:
4.0 × 10^-2 N m^-1
Explanation:
Because both surfaces contribute, W = 2γΔA. Therefore γ = W/(2ΔA) = 4.8 × 10^-3 /(2 × 6.0 × 10^-2) = 4.0 × 10^-2 N m^-1.
7. A movable wire of length 0.20 m is attached to a soap film. The surface tension is 3.5 × 10^-2 N m^-1. If the wire moves outward by 5.0 × 10^-2 m, how much work is done against the film's surface tension?
A.
3.5 × 10^-4 J
B.
7.0 × 10^-4 J
C.
1.4 × 10^-3 J
D.
3.5 × 10^-3 J
Correct Answer:
7.0 × 10^-4 J
Explanation:
The film exerts force F = 2γl. The work required is W = Fx = 2γlx = 2 × 3.5 × 10^-2 × 0.20 × 5.0 × 10^-2 = 7.0 × 10^-4 J.
8. A soap film of geometrical area A is formed between a frame and contains two air–liquid interfaces. If the surface tension is γ, its surface energy is
A.
γA
B.
2γA
C.
γA/2
D.
4γA
Correct Answer:
2γA
Explanation:
Each of the two surfaces has area A and surface energy γA. Therefore, the total surface energy of the soap film is γA + γA = 2γA.
9. A soap film has surface tension 0.030 N m^-1 and is bounded by a movable wire of length 0.15 m. If an external force of 9.0 × 10^-3 N is applied opposite to the film's pull, the wire is in equilibrium. Which statement is correct?
A.
The force from one surface is 9.0 × 10^-3 N
B.
The force from each surface is 4.5 × 10^-3 N
C.
The total force from both surfaces is 4.5 × 10^-3 N
D.
The surface tension must be 0.060 N m^-1
Correct Answer:
The force from each surface is 4.5 × 10^-3 N
Explanation:
The force due to each surface is γl = 0.030 × 0.15 = 4.5 × 10^-3 N. The two surfaces together exert 9.0 × 10^-3 N, balancing the external force.
10. A soap film is formed with an area of 3.0 × 10^-2 m^2 on each face. If its surface tension is 0.025 N m^-1, what is the total surface energy associated with the film?
A.
3.75 × 10^-4 J
B.
7.50 × 10^-4 J
C.
1.50 × 10^-3 J
D.
2.50 × 10^-3 J
Correct Answer:
1.50 × 10^-3 J
Explanation:
The total area of the two surfaces is 2A. Hence, the surface energy is E = γ(2A) = 2 × 0.025 × 3.0 × 10^-2 = 1.50 × 10^-3 J.
Short Questions and Answers
Practice short questions and answers related to
Two-Surface Contribution in Soap Films
.
Short questions are currently unavailable for this page.