physics : class11 Energy of a Soap Film : Surface Energy : Properties of Bulk Matter MCQ Questions and Answers
Practice important
Energy of a Soap Film
MCQ questions and answers.
Page 1
of
1
|
10
MCQ Questions
1. A rectangular soap film has surface tension 30 mN m^-1. Its length is increased from 5.0 cm to 8.0 cm while its breadth remains 4.0 cm. Assuming the process is quasistatic, the work done in increasing the area of the film is:
A.
0.036 J
B.
0.072 J
C.
0.144 J
D.
0.288 J
Correct Answer:
0.072 J
Explanation:
A soap film has two surfaces, so the increase in surface energy is W = 2TΔA. Here, ΔA = (0.08 - 0.05)(0.04) = 1.2 × 10^-3 m^2. Thus W = 2(0.030)(1.2 × 10^-3) = 0.072 J.
2. The surface energy of a soap film of area 200 cm^2 is 1.0 × 10^-3 J. The surface tension of the soap solution is:
A.
1.25 × 10^-2 N m^-1
B.
2.50 × 10^-2 N m^-1
C.
5.00 × 10^-2 N m^-1
D.
1.00 × 10^-1 N m^-1
Correct Answer:
2.50 × 10^-2 N m^-1
Explanation:
For a soap film, surface energy E = 2TA because it has two surfaces. With A = 200 cm^2 = 0.020 m^2, T = E/(2A) = (1.0 × 10^-3)/(2 × 0.020) = 2.50 × 10^-2 N m^-1.
3. A soap film is stretched reversibly so that its area increases by 3.0 × 10^-3 m^2. If its surface tension is 40 mN m^-1, the increase in surface energy is:
A.
1.2 × 10^-4 J
B.
2.4 × 10^-4 J
C.
6.0 × 10^-4 J
D.
1.2 × 10^-3 J
Correct Answer:
2.4 × 10^-4 J
Explanation:
The increase in surface energy of a soap film is ΔE = 2TΔA. Therefore, ΔE = 2(40 × 10^-3)(3.0 × 10^-3) = 2.4 × 10^-4 J.
4. A soap film of surface tension 0.030 N m^-1 is stretched so that its final area is three times its initial area. If the increase in surface energy is 0.12 J, the initial area of the film is:
A.
0.50 m^2
B.
1.00 m^2
C.
2.00 m^2
D.
4.00 m^2
Correct Answer:
2.00 m^2
Explanation:
Since the final area is three times the initial area, ΔA = 2A_initial. Using ΔE = 2TΔA, 0.12 = 2(0.030)(2A_initial), giving A_initial = 2.00 m^2.
5. A movable wire in a rectangular soap-film frame is displaced inward, reducing the total area of the film by 5.0 × 10^-4 m^2. If the surface tension is 25 mN m^-1, the energy released is:
A.
1.25 × 10^-5 J
B.
2.50 × 10^-5 J
C.
5.00 × 10^-5 J
D.
1.00 × 10^-4 J
Correct Answer:
2.50 × 10^-5 J
Explanation:
The decrease in surface energy is 2T|ΔA| because the film has two surfaces. Hence, released energy = 2(25 × 10^-3)(5.0 × 10^-4) = 2.50 × 10^-5 J.
6. A soap film is formed on a rectangular frame with a movable side. The movable side is pulled outward through 2.0 cm while the length of the side perpendicular to its motion is 10 cm. If the surface tension is 0.050 N m^-1, the mechanical work required is:
A.
1.0 × 10^-4 J
B.
2.0 × 10^-4 J
C.
1.0 × 10^-3 J
D.
2.0 × 10^-3 J
Correct Answer:
2.0 × 10^-4 J
Explanation:
The increase in area of one surface is ΔA = (0.10)(0.020) = 2.0 × 10^-3 m^2. Since a soap film has two surfaces, W = 2TΔA = 2(0.050)(2.0 × 10^-3) = 2.0 × 10^-4 J.
7. For a soap film of constant surface tension, a graph of its surface energy E against its total area A is plotted. The slope of this graph represents:
A.
T
B.
2T
C.
T/2
D.
4T
Correct Answer:
2T
Explanation:
The surface energy of a soap film is E = 2TA because the film has two exposed surfaces. Therefore, the slope dE/dA of the E-A graph is 2T.
8. A spherical soap bubble of radius 2.0 cm has surface tension 0.025 N m^-1. Its surface energy is approximately:
A.
1.26 × 10^-4 J
B.
2.51 × 10^-4 J
C.
5.03 × 10^-4 J
D.
1.01 × 10^-3 J
Correct Answer:
2.51 × 10^-4 J
Explanation:
A soap bubble has two spherical surfaces, and the area of one spherical surface is 4πr^2. Hence E = 2T(4πr^2) = 8πTr^2. Substituting T = 0.025 N m^-1 and r = 0.020 m gives E ≈ 2.51 × 10^-4 J.
9. A soap bubble expands quasistatically from radius r to radius 2r while its surface tension remains constant at T. The increase in its surface energy is:
A.
8πTr^2
B.
16πTr^2
C.
24πTr^2
D.
32πTr^2
Correct Answer:
24πTr^2
Explanation:
The surface energy of a soap bubble of radius R is E = 8πTR^2. Therefore, the increase from r to 2r is 8πT[(2r)^2 - r^2] = 8πT(3r^2) = 24πTr^2.
10. A circular hole of radius 1.0 mm is made in a large soap film of surface tension 0.030 N m^-1. Neglecting any deformation around the hole, the decrease in surface energy is:
A.
3.0 × 10^-8 J
B.
6.0 × 10^-8 J
C.
9.4 × 10^-8 J
D.
1.9 × 10^-7 J
Correct Answer:
1.9 × 10^-7 J
Explanation:
The removed area is πr^2, and both surfaces of the film lose this area. Thus the decrease in energy is ΔE = 2Tπr^2 = 2(0.030)π(1.0 × 10^-3)^2 ≈ 1.9 × 10^-7 J.
Short Questions and Answers
Practice short questions and answers related to
Energy of a Soap Film
.
Short questions are currently unavailable for this page.