Practice Fluid Dynamics MCQs for JSMU Karachi Entry Test Physics — topic-wise sets with solved answers.
Q1. Water flows through a horizontal pipe of cross-sectional area 4.0 cm² at a speed of 2.0 m/s. What is the speed in a constricted section of 2.0 cm²?
Answer: 4.0 m/s
Explanation: Equation of continuity states speed is inversely proportional to area. Halving the area must double the speed to maintain constant flow rate.
Q2. An open water tank develops a small hole in its side at a depth of 5.0 m below the water surface. Calculate the speed of efflux.
Answer: 9.9 m/s
Explanation: According to Torricelli's theorem, speed equals sqrt(2gh). With g=9.8 and h=5, the calculation yields approximately 9.9 m/s. Option B is just g.
Q3. A spherical fog droplet of radius 'r' falls with a terminal velocity of 0.2 m/s. What is the terminal velocity of a droplet with radius '2r'?
Answer: 0.8 m/s
Explanation: Terminal velocity is directly proportional to the square of the radius. Doubling the radius increases the velocity by a factor of four.
Q4. A tiny sphere of radius 0.02 m moves through a liquid of viscosity 0.01 Ns/m² at 5 m/s. Calculate the viscous drag force acting on it.
Answer: 0.019 N
Explanation: Stokes' law states F = 6πηrv. Substituting the given values (6 * 3.14 * 0.01 * 0.02 * 5) results in approximately 0.019 N.
Q5. Air of density 1.2 kg/m³ flows horizontally. If the speed increases from 10 m/s to 20 m/s, what is the resulting decrease in pressure?
Answer: 180 Pa
Explanation: Using Bernoulli's equation for horizontal flow, ΔP = 0.5 * ρ * (v2² - v1²). Here, 0.5 * 1.2 * (400 - 100) equals 180 Pa.
Q6. Water (1000 kg/m³) enters a pipe of radius 2 cm at 2 m/s. It then enters a region where the radius is 1 cm. Calculate the pressure drop.
Answer: 30,000 Pa
Explanation: By continuity, v2 = v1 * (r1/r2)². Since r2 = 0.5r1, v2 = 2 * 4 = 8 m/s. Bernoulli's then gives ΔP = 0.5 * 1000 * (64 - 4) = 30,000 Pa.
Q7. A pipe has a cross-sectional area of 2.0 cm². If water moves through it at a speed of 5.0 m/s, what is the volume flow rate?
Answer: 0.001 m³/s
Explanation: Flow rate Q = Area * velocity. Converting 2.0 cm² to 0.0002 m², Q = 0.0002 * 5 = 0.001 m³/s. Option B uses cm² directly.
Q8. An aorta of area 2.0 cm² has blood flowing at 30 cm/s. It branches into 20 capillaries, each of area 0.1 cm². Find the capillary blood speed.
Answer: 30.0 cm/s
Explanation: Blood flow rate is conserved (A1v1 = nA2v2). Thus, 2.0 * 30 = 20 * 0.1 * v, which simplifies to 60 = 2v, giving 30 cm/s.
Q9. In a Venturi meter, the pressure difference between the main pipe and the throat is 1000 Pa. If fluid density is 1000 kg/m³, find the throat speed.
Answer: 1.4 m/s
Explanation: Venturi meter uses P1 - P2 = 0.5 * ρ * v2² (if v1 is negligible). Here, 1000 = 0.5 * 1000 * v2², so v2 = sqrt(2) ≈ 1.4 m/s.
Q10. If a force of 10 N is required to move a 2 m² plate at 5 m/s over a 1 m thick liquid layer, what is the viscosity?
Answer: 1.0 Ns/m²
Explanation: Viscosity is defined as η = (Force * distance) / (Area * velocity). Using units: (N * m) / (m² * m/s) = Ns/m². Option C is power.
Q11. A pipe carrying an ideal fluid narrows from a radius of 3 cm to 1 cm. What is the ratio of the final speed to the initial speed?
Answer: 9:1
Explanation: The ratio of radii is 1:3, so the ratio of areas is 1:9. By continuity, the speed ratio is the inverse of the area ratio, 9:1.
Q12. A sphere of radius r falls at 0.1 m/s in a fluid. If it is replaced by a sphere of radius 2r in a fluid with double the viscosity, find vt.
Answer: 0.20 m/s
Explanation: Terminal velocity vt = (2gr²ρ)/(9η). Since vt is inversely proportional to η, doubling η while doubling r (r² factor of 4) results in 2vt.
Q13. Calculate the hydrostatic pressure exerted by a water column of height 10 m at its base, ignoring atmospheric pressure.
Answer: 98000 Pa
Explanation: Pressure P = ρgh. For water (1000 kg/m³), g=9.8, h=10, P = 1000 * 9.8 * 10 = 98,000 Pa. This is approximately 1 atm.
Q14. An airplane wing of area 10 m² experiences air speeds of 50 m/s and 40 m/s on its surfaces. If air density is 1.2 kg/m³, find the lift.
Answer: 5400 N
Explanation: Lift = ΔP * Area. ΔP = 0.5 * ρ * (v_top² - v_bottom²) = 0.5 * 1.2 * (2500 - 1600) = 0.6 * 900 = 540 Pa. Lift = 540 * 10 = 5400 N.
Q15. A water jet of area 0.01 m² strikes a wall at 2 m/s and stops. Calculate the power delivered by the water to the wall.
Answer: 80 W
Explanation: Power = Force * velocity. For a jet, F = (mass/time) * v = (ρAv) * v = ρAv². Power = ρAv³. Substituting values: 1000 * 0.01 * 8 = 80 W.
Q16. Water speed in a pipe increases from 2 m/s to 4 m/s. What is the change in kinetic energy per unit volume of the water?
Answer: 6000 J/m³
Explanation: The change in kinetic energy per unit volume is 0.5 * ρ * (v2² - v1²). Here, 0.5 * 1000 * (16 - 4) = 500 * 12 = 6000 J/m³.
Q17. To maintain a flow rate of 0.001 m³/s at a speed of 10 m/s, what must be the radius of the pipe?
Answer: 5.6 mm
Explanation: Flow rate Q = Area * v = πr²v. Thus 0.001 = 3.14 * r² * 10. r² ≈ 0.0000318, so r ≈ 0.0056 m or 5.6 mm.
Q18. The viscous drag on a sphere moving at 2 m/s is 0.05 N. What is the drag force when the speed is increased to 4 m/s?
Answer: 0.10 N
Explanation: Force is directly proportional to speed (F = 6πηrv). Doubling the speed from 2 m/s to 4 m/s doubles the force from 0.05 N to 0.10 N.
Q19. A closed tank has water at 5 m depth and an air pressure of 50 kPa above water. Find the efflux speed through a small hole.
Answer: 14.1 m/s
Explanation: For a pressurized tank, v = sqrt(2gh + 2P/ρ). Here, v = sqrt(2*10*5 + 2*50000/1000) = sqrt(100 + 100) = sqrt(200) ≈ 14.1 m/s.
Q20. A pipe with diameter 3 cm has water moving at 0.5 m/s. If the pipe diameter reduces to 1 cm, what is the new speed?
Answer: 4.5 m/s
Explanation: Area is proportional to the square of the diameter. If diameter is 1/3, area is 1/9, so speed must be 9 times higher (9 * 0.5 = 4.5).
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