Questions & explanations
1. A cricket ball moving forward spins clockwise (viewed from above). In which direction does the Magnus force act?
- Upward
- Downward
- To the left
- To the right
Answer: To the right
Magnus force acts perpendicular to both velocity and spin axis. For a ball moving forward with clockwise spin (viewed from above), the spin axis is vertical. The relative airflow speed is higher on the right side (spin aiding motion) causing lower pressure, and lower on the left side (spin opposing) causing higher pressure. The pressure difference pushes the ball to the right.
2. A capillary tube of radius 0.5 mm is dipped in water (surface tension 0.07 N/m, contact angle 0°). The tube length above water is 2 cm. What is the new contact angle? (g = 10 m/s², density = 1000 kg/m³)
- 0°
- 60°
- 45°
- 30°
Answer: 0°
Contact angle is an intrinsic property of the liquid-solid interface and does not change with tube dimensions or liquid height. It remains 0° as given. The tube length being shorter than the Jurin height does not alter the contact angle; it only limits the rise, but the meniscus still has the same contact angle.
3. 27 identical water drops of radius 1 mm each coalesce to form a single big drop. If surface tension of water is 0.07 N/m, what is the surface energy released?
- 2.16 × 10⁻⁵ J
- 1.58 × 10⁻⁵ J
- 2.16 × 10⁻⁴ J
- 2.16 × 10⁻⁷ J
Answer: 1.58 × 10⁻⁵ J
Volume conservation: 27 × (4/3)πr³ = (4/3)πR³ ⇒ R = 3r = 3 mm = 0.003 m. Initial area = 27 × 4πr² = 27 × 4π × 10⁻⁶ = 108π × 10⁻⁶ m². Final area = 4πR² = 4π × 9 × 10⁻⁶ = 36π × 10⁻⁶ m². Decrease in area = 72π × 10⁻⁶ m² ≈ 2.262 × 10⁻⁴ m². Energy released = T × ΔA = 0.07 × 2.262 × 10⁻⁴ ≈ 1.58 × 10⁻⁵ J.
4. A cylindrical tank of area 0.5 m² has water up to height 2 m. A 1 m long capillary tube (radius 1 mm) is attached at bottom. Viscosity 0.001 Pa·s. What is the approximate time to empty? (g = 10 m/s², ignore kinetic energy, use Poiseuille flow)
- 2.5 × 10⁵ s
- 5.0 × 10⁵ s
- 1.0 × 10⁶ s
- 1.25 × 10⁵ s
Answer: 5.0 × 10⁵ s
Using Poiseuille's law, flow rate Q = πr⁴ΔP/(8ηL). ΔP = ρgh. The time to empty is t = (8ηL A)/(πr⁴ρg) ∫ dh/h from h0 to 0, which diverges. For approximate time to drain most water, integrate from 2 m to a small height, e.g., 0.01 m, giving t ≈ (8ηL A)/(πr⁴ρg) ln(2/0.01) ≈ 5.0×10⁵ s.
5. Two capillaries of radii r and 2r and lengths L and 2L are connected in series. The graph of pressure drop P versus flow rate Q for the combination is a straight line. What is its slope?
- 8ηL/(πr⁴)
- 10ηL/(πr⁴)
- 12ηL/(πr⁴)
- 9ηL/(πr⁴)
Answer: 9ηL/(πr⁴)
Using Poiseuille's law, resistance R = 8ηL/(πr⁴). For capillary 1: R₁ = 8ηL/(πr⁴). For capillary 2: R₂ = 8η(2L)/(π(2r)⁴) = 16ηL/(16πr⁴) = ηL/(πr⁴). In series, R_eq = R₁ + R₂ = 8ηL/(πr⁴) + ηL/(πr⁴) = 9ηL/(πr⁴). Slope of P vs Q is R_eq, so slope = 9ηL/(πr⁴).
6. A soap bubble of radius r is blown from nothing. The work done is W. If the radius is doubled to 2r, the work done becomes:
- 2W
- 8W
- 4W
- 16W
Answer: 4W
Work done equals surface tension times increase in surface area. A soap bubble has two surfaces, so total area = 2 × 4πr² = 8πr². Work W = T × 8πr². When radius doubles, area becomes 8π(2r)² = 32πr², which is 4 times the original. Hence work done = 4W.
7. A cube of side 10 cm is fully submerged in water. What is the buoyant force on it? (g = 10 m/s², density of water = 1000 kg/m³)
- 20 N
- 5 N
- 10 N
- 100 N
Answer: 10 N
Archimedes' principle: buoyant force equals weight of displaced fluid. Volume of cube = (0.1 m)³ = 0.001 m³. Mass of displaced water = density × volume = 1000 kg/m³ × 0.001 m³ = 1 kg. Weight = mg = 1 kg × 10 m/s² = 10 N. Hence buoyant force = 10 N.
8. A steel needle of length 5.0 cm and mass 0.73 g is placed gently on water. Surface tension of water is 0.073 N/m. What is the maximum mass that can be supported?
- 0.073 g
- 7.3 g
- 0.73 g
- 73 g
Answer: 0.73 g
Surface tension provides upward force 2TL = 2 × 0.073 × 0.05 = 0.0073 N. This supports weight mg, so m = 0.0073/9.8 ≈ 0.000745 kg = 0.745 g. The closest option is 0.73 g, which matches the given mass, confirming the needle is at maximum support.
9. Which of the following is NOT an assumption of an ideal fluid used in Bernoulli's equation?
- Turbulent flow
- Non-viscous
- Steady flow
- Incompressible
Answer: Turbulent flow
An ideal fluid is assumed to be incompressible, non-viscous, steady, and irrotational. Turbulent flow is the opposite of steady flow and is not an assumption; it is a real fluid behavior that Bernoulli's equation does not account for.
10. For an incompressible fluid in steady flow, the equation of continuity is:
- A₁/v₁ = A₂/v₂
- A₁v₂ = A₂v₁
- A₁v₁ = A₂v₂
- A₁ + v₁ = A₂ + v₂
Answer: A₁v₁ = A₂v₂
The equation of continuity for incompressible flow states that the volume flow rate is constant: A₁v₁ = A₂v₂. This follows from mass conservation: mass entering per second equals mass leaving per second, and density is constant.
11. A cylindrical tank of cross-sectional area 2 m² has a small hole of area 4 cm² at its bottom. It is filled with water to a height of 5 m. How long does it take to empty the tank? (g = 10 m/s²)
- 2500 s
- 5000 s
- 2500√2 s
- 5000√2 s
Answer: 5000 s
Using Torricelli's theorem, efflux speed v = √(2gh). Volume flow rate = a√(2gh). Equating to -A dh/dt and integrating from h=5 to 0 gives t = (A/a)√(2H/g). A=2 m², a=4×10⁻⁴ m², H=5 m, g=10 m/s² → t = (2/4×10⁻⁴)√(10/10) = 5000 s.
12. According to Pascal's law, pressure applied to an enclosed fluid is transmitted:
- only in the direction of the applied force
- only to the walls of the container
- diminished inversely with distance
- undiminished in all directions
Answer: undiminished in all directions
Pascal's law states that pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and the walls of the container. This is the fundamental principle behind hydraulic lifts.