physics : class11 Energy Change of a Liquid Surface : Surface Energy : Properties of Bulk Matter MCQ Questions and Answers
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Energy Change of a Liquid Surface
MCQ questions and answers.
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MCQ Questions
1. A spherical liquid drop of radius R breaks into n identical spherical droplets without any loss of liquid. If the surface tension is T, what is the increase in surface energy?
A.
4πTR²(n - 1)
B.
4πTR²(n^(1/3) - 1)
C.
4πTR²(n^(2/3) - 1)
D.
4πTR²(n^(1/2) - 1)
Correct Answer:
4πTR²(n^(1/3) - 1)
Explanation:
Volume conservation gives the radius of each small droplet as R/n^(1/3). Hence the total final area is n × 4π(R/n^(1/3))² = 4πR²n^(1/3). The increase in surface energy is T times the increase in area, equal to 4πTR²(n^(1/3) - 1).
2. Two identical spherical liquid drops, each of radius r, coalesce into one spherical drop. If the surface tension is T, what is the decrease in surface energy?
A.
4πTr²(2 - 2^(1/3))
B.
4πTr²(2 - 2^(2/3))
C.
4πTr²(2^(2/3) - 1)
D.
4πTr²(2^(1/3) - 1)
Correct Answer:
4πTr²(2 - 2^(2/3))
Explanation:
The initial area is 2 × 4πr² = 8πr². Conservation of volume gives the final radius as 2^(1/3)r, so the final area is 4πr²2^(2/3). The decrease in surface energy is T(A_initial - A_final) = 4πTr²(2 - 2^(2/3)).
3. A liquid surface of area 0.020 m² is increased reversibly to 0.035 m² at constant surface tension 0.050 N m⁻¹. What is the increase in surface energy?
A.
2.5 × 10⁻⁴ J
B.
5.0 × 10⁻⁴ J
C.
7.5 × 10⁻⁴ J
D.
1.1 × 10⁻³ J
Correct Answer:
7.5 × 10⁻⁴ J
Explanation:
For a liquid surface at constant surface tension, the change in surface energy is ΔE = TΔA. Here ΔA = 0.035 - 0.020 = 0.015 m², so ΔE = 0.050 × 0.015 = 7.5 × 10⁻⁴ J.
4. A soap film has two surfaces, each of area A. If its total area of each surface increases by ΔA while the surface tension remains T, what is the increase in surface energy?
A.
TΔA
B.
2TΔA
C.
T(A + ΔA)
D.
2TA + TΔA
Correct Answer:
2TΔA
Explanation:
A soap film has two liquid surfaces. Each surface gains energy TΔA, so the total increase in surface energy is TΔA + TΔA = 2TΔA.
5. A spherical liquid drop of radius R is divided into 8 identical spherical droplets. If the surface tension is T, the increase in surface energy is:
A.
4πTR²
B.
8πTR²
C.
12πTR²
D.
28πTR²
Correct Answer:
4πTR²
Explanation:
For n = 8, the total surface area becomes 8^(1/3) = 2 times the original area. Thus the increase in area is 4πR², and the increase in surface energy is T × 4πR² = 4πTR².
6. A spherical liquid drop of radius 3 mm splits into 27 identical droplets. If its surface tension is 0.040 N m⁻¹, what is the increase in surface energy?
A.
9.05 × 10⁻⁷ J
B.
1.36 × 10⁻⁶ J
C.
2.71 × 10⁻⁶ J
D.
5.43 × 10⁻⁶ J
Correct Answer:
2.71 × 10⁻⁶ J
Explanation:
For 27 droplets, the total area becomes 27^(1/3) = 3 times the original area. Therefore ΔA = 2 × 4πR² = 8π(3 × 10⁻³)² m². Hence ΔE = 0.040 × 8π × 9 × 10⁻⁶ ≈ 2.71 × 10⁻⁶ J.
7. A liquid surface is expanded from area A to 4A while its surface tension remains constant at T. The work required in this reversible expansion is:
A.
TA
B.
2TA
C.
3TA
D.
4TA
Correct Answer:
3TA
Explanation:
The work required equals the increase in surface energy. Since the area increase is 4A - A = 3A, the work is ΔE = TΔA = 3TA.
8. A spherical drop of liquid of radius R coalesces with another drop of radius 2R made of the same liquid. If the surface tension is T, what is the decrease in surface energy?
A.
4πTR²(5 - 5^(2/3))
B.
4πTR²(5 - 3^(2/3))
C.
4πTR²(5 - 5^(1/3))
D.
4πTR²(3^(2/3) - 1)
Correct Answer:
4πTR²(5 - 5^(2/3))
Explanation:
The combined volume corresponds to a final radius 5^(1/3)R because R³ + (2R)³ = 9R³, wait: the correct combined volume is 9R³, so the final radius is 9^(1/3)R. Therefore the initial area is 4πR² + 16πR² = 20πR² and the final area is 4πR²9^(2/3). The decrease is 4πTR²(5 - 9^(2/3)).
9. A liquid surface is increased in area by 0.010 m², requiring 6.0 × 10⁻⁴ J of work. Assuming the surface tension is constant, its value is:
A.
0.006 N m⁻¹
B.
0.030 N m⁻¹
C.
0.060 N m⁻¹
D.
0.600 N m⁻¹
Correct Answer:
0.060 N m⁻¹
Explanation:
The work done reversibly in increasing a liquid surface is ΔE = TΔA. Thus T = ΔE/ΔA = (6.0 × 10⁻⁴)/0.010 = 0.060 N m⁻¹.
10. A spherical liquid drop of radius R is split into n identical droplets. By what factor does its surface energy change, assuming constant surface tension and no loss of liquid?
A.
It becomes n times its original value
B.
It becomes n^(1/2) times its original value
C.
It becomes n^(1/3) times its original value
D.
It becomes n^(2/3) times its original value
Correct Answer:
It becomes n^(1/3) times its original value
Explanation:
The radius of each droplet is R/n^(1/3). The total final area is n × 4π(R/n^(1/3))² = 4πR²n^(1/3). Since surface energy equals T times surface area, the final surface energy is n^(1/3) times the original surface energy.
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