physics : class11 Surface Energy and Surface Tension : Surface Energy : Properties of Bulk Matter MCQ Questions and Answers
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Surface Energy and Surface Tension
MCQ questions and answers.
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MCQ Questions
1. A liquid has surface tension 0.072 N m⁻¹. If its free surface area is increased reversibly by 2.5 × 10⁻³ m² at constant temperature, the work done is
A.
1.80 × 10⁻⁴ J
B.
3.60 × 10⁻⁴ J
C.
2.88 × 10⁻⁵ J
D.
5.00 × 10⁻² J
Correct Answer:
1.80 × 10⁻⁴ J
Explanation:
For a single liquid surface, the increase in surface energy equals the work done: W = TΔA = 0.072 × 2.5 × 10⁻³ = 1.80 × 10⁻⁴ J.
2. A soap film is stretched so that the area of one of its surfaces increases by 4.0 × 10⁻³ m². If the surface tension of the soap solution is 0.025 N m⁻¹, the increase in surface energy of the film is
A.
1.0 × 10⁻⁴ J
B.
2.0 × 10⁻⁴ J
C.
5.0 × 10⁻⁵ J
D.
4.0 × 10⁻³ J
Correct Answer:
2.0 × 10⁻⁴ J
Explanation:
A soap film has two surfaces. Therefore, ΔE = 2TΔA = 2 × 0.025 × 4.0 × 10⁻³ = 2.0 × 10⁻⁴ J.
3. For a liquid surface at constant temperature, which statement correctly relates surface tension T and surface energy per unit area Eₛ?
A.
Eₛ = T only for a solid surface
B.
Eₛ = 2T for every liquid surface
C.
Eₛ = T for a single liquid surface
D.
Eₛ = T² for a liquid surface
Correct Answer:
Eₛ = T for a single liquid surface
Explanation:
For a single liquid surface, the work required to increase area by ΔA is TΔA. Hence, surface energy per unit area is ΔE/ΔA = T.
4. A liquid drop of radius r breaks into 8 identical smaller spherical drops without loss of liquid. If the surface tension is T, the increase in surface energy is
A.
4πr²T
B.
8πr²T
C.
12πr²T
D.
16πr²T
Correct Answer:
4πr²T
Explanation:
Volume conservation gives each small radius as r/2. The total new area is 8 × 4π(r/2)² = 8πr², while the original area is 4πr². Thus, the increase in energy is T(8πr² − 4πr²) = 4πr²T.
5. Two identical liquid drops of radius r and surface tension T coalesce into one spherical drop. The decrease in surface energy is
A.
4πr²T(2 − 2^(2/3))
B.
4πr²T(2^(2/3) − 2)
C.
4πr²T(2^(1/3) − 1)
D.
8πr²T(2^(2/3) − 1)
Correct Answer:
4πr²T(2 − 2^(2/3))
Explanation:
The final radius is R = 2^(1/3)r. Initial area is 8πr² and final area is 4πR² = 4πr²2^(2/3). The decrease in surface energy is T[8πr² − 4πr²2^(2/3)] = 4πr²T(2 − 2^(2/3)).
6. A rectangular wire frame carries a soap film of surface tension 0.030 N m⁻¹. A movable side of length 0.20 m is pulled outward quasistatically through 5.0 cm. The work required is
A.
3.0 × 10⁻⁴ J
B.
6.0 × 10⁻⁴ J
C.
1.5 × 10⁻⁴ J
D.
3.0 × 10⁻³ J
Correct Answer:
6.0 × 10⁻⁴ J
Explanation:
The area increase of one surface is Lx = 0.20 × 0.050 m². Since the film has two surfaces, W = 2TLx = 2 × 0.030 × 0.20 × 0.050 = 6.0 × 10⁻⁴ J.
7. The surface tension of a liquid is 0.050 N m⁻¹. The excess pressure inside a spherical liquid drop of radius 2.0 mm is
A.
25 Pa
B.
50 Pa
C.
75 Pa
D.
100 Pa
Correct Answer:
50 Pa
Explanation:
For a liquid drop having one surface, the excess pressure is ΔP = 2T/r = 2 × 0.050/(2.0 × 10⁻³) = 50 Pa.
8. A soap bubble has radius 1.0 cm and surface tension 0.025 N m⁻¹. Its excess internal pressure is
A.
2.5 Pa
B.
5.0 Pa
C.
7.5 Pa
D.
10 Pa
Correct Answer:
10 Pa
Explanation:
A soap bubble has two surfaces, so its excess pressure is ΔP = 4T/r = 4 × 0.025/(1.0 × 10⁻²) = 10 Pa.
9. A liquid surface is expanded isothermally from area A to 3A. If its surface tension remains constant at T, the increase in surface energy is
A.
TA
B.
2TA
C.
3TA
D.
6TA
Correct Answer:
2TA
Explanation:
The area increase is ΔA = 3A − A = 2A. Therefore, the increase in surface energy is ΔE = TΔA = 2TA.
10. A capillary tube of internal radius 0.50 mm is dipped vertically in water. Taking surface tension as 0.072 N m⁻¹, density of water as 1000 kg m⁻³, g = 9.8 m s⁻², and contact angle as zero, the capillary rise is approximately
A.
1.47 cm
B.
2.94 cm
C.
4.80 cm
D.
7.20 cm
Correct Answer:
2.94 cm
Explanation:
For complete wetting, h = 2T/(ρgr) = 2 × 0.072/(1000 × 9.8 × 0.50 × 10⁻³) ≈ 0.0294 m = 2.94 cm. This rise results from the upward surface-tension force at the liquid-air contact line.
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