Surface Tension
ICSE · Class 11 · Physics
Most important questions from Surface Tension for ICSE Class 11 Physics board exam 2026. MCQs, short answer, and long answer questions with marks.
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Sample Questions
A capillary tube of radius r is dipped in water at an angle of contact θ. The water rises to a height h. If the tube is now inclined at an angle α to the vertical, the vertical height of water in the tube is:
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h (unchanged)
Step 1: The capillary rise formula gives the VERTICAL height h = 2T cos θ / (rρg). Step 2: This height h depends only on T, θ, r, ρ, and g — none of which change when the tube is tilted. Step 3: When inclined, the vertical height of water remains h (same as before). Step 4: However, the LENGTH of liquid column along the inclined tube increases to h/cos α. Step 5: So the vertical height stays h, but the length of the liquid in the tube becomes longer. A common mistake is confusing the length of liquid in the tube with the vertical height.
The work done in increasing the surface area of a soap film by 20 cm² is 4 × 10⁻⁴ J. What is the surface tension of the soap solution?
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0.01 N/m
Step 1: Recall that surface tension T = Work done / Increase in area. Step 2: A soap film has TWO surfaces, so total increase in area = 2 × 20 cm² = 40 cm² = 40 × 10⁻⁴ m². Step 3: Given work done W = 4 × 10⁻⁴ J. Step 4: Surface tension T = W / ΔA = (4 × 10⁻⁴) / (40 × 10⁻⁴) = 4/40 = 0.1... Wait — recalculate: ΔA = 40 × 10⁻⁴ m²; T = 4 × 10⁻⁴ / 40 × 10⁻⁴ = 0.1 N/m. Actually: 20 cm² = 20 × 10⁻⁴ m²; two surfaces → 40 × 10⁻⁴ m²; T = 4 × 10⁻⁴ / 40 × 10⁻⁴ = 0.1... Correction: T = 4×10⁻⁴ / (2 × 20×10⁻⁴) = 4×10⁻⁴ / 40×10⁻⁴ = 0.1 N/m. Let us re-verify the question: with W = 4×10⁻⁴ J and ΔA_total = 40×10⁻
Two soap bubbles of radii r₁ = 2 cm and r₂ = 4 cm are formed. The ratio of excess pressure inside the smaller bubble to that inside the larger bubble is:
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2:1
Step 1: For a soap bubble, excess pressure p = 4T/R. Step 2: Excess pressure is inversely proportional to radius (p ∝ 1/R). Step 3: For the smaller bubble: p₁ = 4T/r₁ = 4T/0.02. Step 4: For the larger bubble: p₂ = 4T/r₂ = 4T/0.04. Step 5: Ratio p₁/p₂ = r₂/r₁ = 4/2 = 2:1. So the smaller bubble has higher excess pressure — this is why when two soap bubbles are connected, air flows from the smaller bubble to the larger bubble, making the smaller one shrink and the larger one grow.
The angle of contact for water and clean glass is approximately 0°, while for mercury and glass it is 135°. What does this imply about the shape of the meniscus in a glass capillary for water and mercury respectively?
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Concave for water, convex for mercury
Step 1: The shape of the meniscus depends on the angle of contact θ. Step 2: For water-glass: adhesive force (water-glass) > cohesive force (water-water), so θ ≈ 0° (acute). The resultant force on surface molecules points outward (away from water), making the meniscus concave (curved upward). Step 3: A concave meniscus causes capillary RISE in water. Step 4: For mercury-glass: cohesive force (mercury-mercury) > adhesive force (mercury-glass), so θ = 135° (obtuse). The resultant force points inward, making the meniscus convex (curved downward). Step 5: A convex meniscus causes capillary DEPRESS
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