physics : class11 Surface Energy and Contact Angle : Surface Energy : Properties of Bulk Matter MCQ Questions and Answers
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Surface Energy and Contact Angle
MCQ questions and answers.
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MCQ Questions
1. At a liquid–solid–air interface, the surface tensions are γSV = 0.075 N m⁻¹, γSL = 0.025 N m⁻¹, and γLV = 0.050 N m⁻¹. According to Young’s equation, the equilibrium contact angle of the liquid is
A.
0°
B.
30°
C.
60°
D.
90°
Correct Answer:
0°
Explanation:
Young’s equation is γSV = γSL + γLV cos θ. Hence cos θ = (0.075 − 0.025)/0.050 = 1, giving θ = 0°. This corresponds to complete wetting.
2. A liquid has liquid–vapour surface energy 0.060 J m⁻² and makes a contact angle of 60° with a solid surface. The work of adhesion between the liquid and solid is
A.
0.030 J m⁻²
B.
0.060 J m⁻²
C.
0.090 J m⁻²
D.
0.120 J m⁻²
Correct Answer:
0.090 J m⁻²
Explanation:
The work of adhesion is W_A = γLV(1 + cos θ). Therefore W_A = 0.060(1 + 0.5) = 0.090 J m⁻².
3. A liquid has γLV = 0.050 N m⁻¹ and an equilibrium contact angle of 120° on a solid. Its spreading coefficient on the solid is
A.
−0.075 N m⁻¹
B.
−0.025 N m⁻¹
C.
0.025 N m⁻¹
D.
0.075 N m⁻¹
Correct Answer:
−0.075 N m⁻¹
Explanation:
The spreading coefficient is S = γSV − γSL − γLV = γLV(cos θ − 1). With cos 120° = −0.5, S = 0.050(−0.5 − 1) = −0.075 N m⁻¹.
4. A liquid drop on a solid surface has γLV = 0.040 N m⁻¹ and contact angle 60°. If γSV = 0.090 N m⁻¹, the solid–liquid interfacial energy γSL is
A.
0.030 N m⁻¹
B.
0.050 N m⁻¹
C.
0.070 N m⁻¹
D.
0.110 N m⁻¹
Correct Answer:
0.070 N m⁻¹
Explanation:
From Young’s equation, γSL = γSV − γLV cos θ. Thus γSL = 0.090 − 0.040 × 0.5 = 0.070 N m⁻¹.
5. Two identical spherical liquid drops, each of radius r, coalesce into one spherical drop. If the liquid surface energy is γ per unit area, the decrease in surface energy is
A.
4πγr²(2 − 2^(2/3))
B.
4πγr²(2^(2/3) − 2)
C.
4πγr²(2 − 2^(1/3))
D.
4πγr²(2^(1/3) − 1)
Correct Answer:
4πγr²(2 − 2^(2/3))
Explanation:
Volume conservation gives the final radius R = 2^(1/3)r. The initial area is 8πr² and the final area is 4πR² = 4π2^(2/3)r². Therefore the decrease in surface energy is γ(8πr² − 4π2^(2/3)r²) = 4πγr²(2 − 2^(2/3)).
6. For a liquid drop on a solid, the work of adhesion equals twice the liquid–vapour surface energy. The equilibrium contact angle is
A.
0°
B.
60°
C.
90°
D.
180°
Correct Answer:
90°
Explanation:
Using W_A = γLV(1 + cos θ), the condition W_A = 2γLV gives 1 + cos θ = 2, so cos θ = 1 and θ = 0°. Wait: this condition actually corresponds to complete wetting, so the correct option is 0°.
7. A liquid has γSV = 0.100 N m⁻¹, γSL = 0.060 N m⁻¹, and γLV = 0.080 N m⁻¹. The equilibrium contact angle is closest to
8. A liquid completely wets a clean solid surface under equilibrium conditions. Which relation among the surface energies must hold for the limiting case of zero contact angle?
A.
γSV = γSL − γLV
B.
γSV = γSL + γLV
C.
γSL = γSV + γLV
D.
γLV = γSV + γSL
Correct Answer:
γSV = γSL + γLV
Explanation:
For θ = 0°, Young’s equation γSV = γSL + γLV cos θ reduces to γSV = γSL + γLV. This is the limiting condition for complete wetting.
9. For a liquid with γLV = 0.072 N m⁻¹, the work of adhesion to a solid is 0.108 J m⁻². The contact angle of the liquid is
A.
30°
B.
60°
C.
90°
D.
120°
Correct Answer:
60°
Explanation:
The relation W_A = γLV(1 + cos θ) gives 0.108 = 0.072(1 + cos θ). Thus 1 + cos θ = 1.5, so cos θ = 0.5 and θ = 60°.
10. A liquid forms a contact angle of 150° on a solid surface. Which statement about the relative interfacial energies is necessarily correct?
A.
γSV − γSL is positive and equal to 0.866γLV
B.
γSV − γSL is negative and equal to 0.866γLV
C.
γSV − γSL is positive and equal to 0.500γLV
D.
γSV − γSL is zero
Correct Answer:
γSV − γSL is negative and equal to 0.866γLV
Explanation:
Young’s equation gives γSV − γSL = γLV cos θ. Since cos 150° = −√3/2 ≈ −0.866, γSV − γSL = −0.866γLV. The large contact angle indicates poor wetting.
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