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Assam CEE Surface Tension — practice questions

Explore 66 free Surface Tension questions in Assam CEE Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Surface Tension questions and exercises with answers

Showing questions 1–30 of 66.

The **excess pressure** inside a spherical liquid drop of radius $r$ and surface tension $T$ is:Read the answer and worked solution →The excess pressure inside a **soap bubble** of radius $r$ and surface tension $T$ is:Read the answer and worked solution →A drop of water of radius 1 mm is broken into 1000 droplets of equal size. The ratio of total surface area after : before equals:Read the answer and worked solution →Newton's FIRST law is also called the law of:Read the answer and worked solution →Newton's SECOND law states:Read the answer and worked solution →The SI unit of force is the:Read the answer and worked solution →Frictional force between surfaces ALWAYS acts:Read the answer and worked solution →Static friction f_s satisfies:Read the answer and worked solution →Kinetic friction f_k is given by:Read the answer and worked solution →For a body on a horizontal surface, the NORMAL FORCE N equals:Read the answer and worked solution →In circular motion at radius r and speed v, the centripetal acceleration is:Read the answer and worked solution →The ANGLE OF REPOSE φ is related to coefficient of friction by:Read the answer and worked solution →A 5 kg object experiences a net force of 20 N. Find its acceleration:Read the answer and worked solution →A 10 kg block on rough horizontal surface with μ_s = 0.3 just starts moving. Force applied:Read the answer and worked solution →A block of mass 2 kg on a smooth incline at 30°. Acceleration down the incline (g = 10):Read the answer and worked solution →A 1500 kg car accelerates from 0 to 30 m/s in 10 s. Average force on car:Read the answer and worked solution →In a tug-of-war, team A pulls with 200 N, team B pulls oppositely with 250 N. The rope (light) experiences net force:Read the answer and worked solution →A block of mass 5 kg on a rough incline at 37° with μ_k = 0.4. Acceleration down (g = 10, sin 37° = 0.6, cos 37° = 0.8):Read the answer and worked solution →For a body on a 60° incline with μ = 0.5, the body:Read the answer and worked solution →An object's WEIGHT on Earth (g = 10) is 50 N. Its MASS is:Read the answer and worked solution →A 5 kg block is pushed with 30 N along a frictionless surface. Acceleration:Read the answer and worked solution →A massless pulley system has two blocks: m₁ = 6 kg and m₂ = 4 kg hanging on either side. Find acceleration of system (g = 10):Read the answer and worked solution →The angle of repose between a block and an incline is 30°. The coefficient of static friction is:Read the answer and worked solution →A 100 g object is moving in a vertical circle. At the TOP of the loop, for minimum speed (string just slack), v_min equals:Read the answer and worked solution →Limiting (max static) friction is INDEPENDENT of:Read the answer and worked solution →A 4 kg mass hangs from a spring scale in an elevator. Spring reads 32 N. The elevator is:Read the answer and worked solution →A 3 kg block is pulled with F = 20 N at 30° above horizontal on rough surface (μ_k = 0.4). Find acceleration (g = 10, sin 30° = 0.5, cos 30° = 0.866):Read the answer and worked solution →In a uniform circular motion, the SPEED is constant but velocity changes direction. The acceleration:Read the answer and worked solution →A car turns a corner of radius 50 m at 20 m/s. Minimum coefficient of friction needed (g = 10):Read the answer and worked solution →A 50 kg person stands on a scale in an elevator accelerating UP at 2 m/s². Scale reading:Read the answer and worked solution →

Worked practice questions

The **excess pressure** inside a spherical liquid drop of radius $r$ and surface tension $T$ is:

  1. $T/r$
  2. $2T/r$
  3. $4T/r$
  4. $8T/r$
Show answer and solution

Correct answer: B. $2T/r$

For a single-surface liquid drop, excess pressure = $2T/r$. For a soap bubble (two surfaces — inside and outside), it's $4T/r$. Smaller drops have larger internal pressure — that's why tiny droplets evaporate faster.

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The excess pressure inside a **soap bubble** of radius $r$ and surface tension $T$ is:

  1. $T/r$
  2. $2T/r$
  3. $4T/r$
  4. $8T/r$
Show answer and solution

Correct answer: C. $4T/r$

Soap bubble has TWO air-soap interfaces (inside and outside surface of the film). **So excess pressure doubles:** $\Delta P = 4T/r$. (Liquid drop has one interface ⇒ $2T/r$.)

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A drop of water of radius 1 mm is broken into 1000 droplets of equal size. The ratio of total surface area after : before equals:

  1. 1 : 10
  2. 10 : 1
  3. 1 : 1000
  4. 1000 : 1
Show answer and solution

Correct answer: B. 10 : 1

**Volume conservation:** $\frac{4}{3}\pi R^3 = 1000 \cdot \frac{4}{3}\pi r^3$ ⇒ $r = R/10$. **Surface area before:** $4\pi R^2$. **After:** $1000 \times 4\pi r^2 = 1000 \times 4\pi R^2/100 = 10 \times 4\pi R^2$. Ratio = 10:1 — surface area grows 10×, requiring 9× work against surface tension.

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