Practice important
Capillary Forces
MCQ questions and answers.
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MCQ Questions
1. A narrow capillary tube is inserted into water, which wets its wall. If the tube radius is reduced to one-fourth while all other conditions remain unchanged, how will the capillary rise change?
A.
It will become one-fourth of the original rise
B.
It will become four times the original rise
C.
It will become twice the original rise
D.
It will remain unchanged because surface tension is constant
Correct Answer:
It will become four times the original rise
Explanation:
Capillary rise is inversely proportional to the tube radius, as given by h = 2T cosθ/(ρgr). Reducing the radius to one-fourth therefore increases the height fourfold.
2. Which combination most directly produces upward movement of water in a fine capillary pore of a dry hydrophilic cell wall?
A.
Adhesion between water and the wall, together with cohesion among water molecules
B.
Repulsion between water and the wall, together with reduced surface tension
C.
Cohesion between water and the wall, together with evaporation alone
D.
Gravity acting upward and adhesion among water molecules
Correct Answer:
Adhesion between water and the wall, together with cohesion among water molecules
Explanation:
Adhesion causes water to spread along and rise against the capillary wall, while cohesion transmits this upward movement through the continuous water column.
3. Two capillary tubes have the same radius and contain the same liquid at the same temperature. In tube X the liquid makes a contact angle of 0°, whereas in tube Y the contact angle is 60°. Which statement is correct about their capillary rises?
A.
The rise in X is twice that in Y
B.
The rise in X is half that in Y
C.
The rises are equal because the tube radii are equal
D.
The liquid falls in both tubes because contact angles are nonzero
Correct Answer:
The rise in X is twice that in Y
Explanation:
Capillary height is proportional to cosθ. Since cos0° = 1 and cos60° = 1/2, the rise in tube X is twice the rise in tube Y.
4. A liquid rises spontaneously in a narrow pore only when the upward component of surface-tension force exceeds the downward gravitational force on the liquid column. At equilibrium, what happens to these forces?
A.
The upward capillary force equals the weight of the liquid column
B.
The upward capillary force becomes zero
C.
The liquid weight becomes zero because of adhesion
D.
The upward capillary force continues increasing without limit
Correct Answer:
The upward capillary force equals the weight of the liquid column
Explanation:
Capillary rise stops when the vertical component of surface tension balances the gravitational force on the raised liquid column. This force balance determines the equilibrium height.
5. A plant cell wall contains extremely fine hydrophilic pores. Compared with a wider pore, the same pore filled with water will generally support a taller water column because the fine pore has:
A.
A smaller surface-area-to-volume ratio and weaker wall interaction
B.
A greater capillary force per unit weight of liquid
C.
A lower water density caused by narrower geometry
D.
A complete absence of gravitational force on the water
Correct Answer:
A greater capillary force per unit weight of liquid
Explanation:
The capillary rise varies inversely with pore radius. Fine pores provide a larger capillary force relative to the weight of the liquid column, allowing water to rise higher.
6. If a liquid does not wet the wall of a capillary tube, which change in the meniscus and liquid level is expected?
A.
A concave meniscus and capillary rise
B.
A convex meniscus and capillary depression
C.
A flat meniscus and unlimited capillary rise
D.
A concave meniscus and capillary depression
Correct Answer:
A convex meniscus and capillary depression
Explanation:
When cohesion within the liquid exceeds adhesion to the wall, the liquid forms a convex meniscus and is depressed below the external liquid level rather than rising.
7. A porous seed coat imbibes water through microscopic spaces. Which alteration would most strongly reduce capillary entry of water without changing the amount of water surrounding the seed?
A.
Increasing the pore radius substantially
B.
Increasing the number of hydrophilic pore surfaces
C.
Decreasing the contact angle between water and the pore wall
D.
Maintaining narrow pores while increasing wall wettability
Correct Answer:
Increasing the pore radius substantially
Explanation:
Capillary rise and capillary driving effect increase as pore radius decreases. Enlarging the pores weakens the capillary effect, even if the surrounding water supply is unchanged.
8. For a wetting liquid in a cylindrical capillary, the capillary height is described by h = 2T cosθ/(ρgr). Which simultaneous change produces the greatest increase in h?
A.
Doubling r and doubling ρ
B.
Halving T and doubling r
C.
Doubling T and halving r
D.
Increasing θ from 0° to 60° while doubling r
Correct Answer:
Doubling T and halving r
Explanation:
Height is directly proportional to surface tension T and inversely proportional to radius r. Doubling T and halving r increases h by a factor of four, assuming the contact angle and density remain unchanged.
9. Why can capillary forces help distribute water through the closely spaced cell-wall spaces of plant tissues even before a continuous bulk flow is established?
A.
Surface tension and adhesion draw water into narrow spaces and maintain a continuous film
B.
Gravity creates an upward force in every cell wall
C.
Water molecules lose cohesion when entering narrow spaces
D.
Capillary forces operate only in completely dry, non-wettable walls
Correct Answer:
Surface tension and adhesion draw water into narrow spaces and maintain a continuous film
Explanation:
In narrow hydrophilic spaces, adhesion pulls water onto the surfaces and surface tension, arising from cohesive forces, sustains the advancing water film. This promotes distribution through the tissue.
10. A capillary tube is placed in water on Earth and then in the same water under conditions where gravitational acceleration is one-fourth of Earth's value. Assuming temperature, tube radius, and wetting remain unchanged, what happens to the equilibrium capillary rise?
A.
It becomes one-fourth as high
B.
It becomes twice as high
C.
It becomes four times as high
D.
It remains unchanged because capillary force is independent of gravity
Correct Answer:
It becomes four times as high
Explanation:
The capillary height is inversely proportional to gravitational acceleration, h = 2T cosθ/(ρgr). Reducing g to one-fourth therefore increases the equilibrium rise fourfold.
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