Physics · Mechanical Properties Of Fluids · NEET
By dimensional analysis you can only show F is proportional to η·r·v (F = k·ηrv). The constant k cannot come from dimensions alone; it is fixed by solving the full fluid-flow equations for slow, smooth (laminar) flow around a sphere. That exact solution gives k = 6π. So the 6π is a proven mathematical constant, not something you derive from units. For NEET, just remember and use F = 6πηrv.
The Stokes drag itself is proportional to r (first power): F = 6πηrv. Students confuse this with terminal velocity, which is proportional to r squared. They are different quantities. Drag force scales with r; terminal velocity scales with r². Keep them separate.
Three conditions: (1) the body is a small, smooth, rigid sphere; (2) the fluid is infinite (walls far away); (3) the flow is slow and laminar (low velocity, no turbulence). If the sphere moves fast and the flow becomes turbulent, F = 6πηrv no longer applies and drag rises faster than v.
It uses the radius r. If a problem gives diameter, halve it first. NEET loves this trap: a 1 mm diameter ball has r = 0.5 mm = 5×10⁻⁴ m. Plugging the diameter directly doubles your answer and gives a wrong option.
A moving sphere drags the fluid layer touching it, setting up relative motion between fluid layers. Faster motion means a steeper velocity gradient between layers, so the internal friction (viscosity) resists more strongly. Since F is proportional to v, doubling the speed doubles the drag.
The viscous drag acting on a metal sphere of diameter 1 mm, falling through a fluid of viscosity 0.8 Pa·s with a velocity of 2 m/s, is equal to:
The velocity of a small ball of mass M and density d, when dropped in a container filled with glycerine, becomes constant after some time. If the density of glycerine is d/2, then the viscous force acting on the ball will be:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
When a small sphere moves slowly through a fluid, the viscous drag force opposing its motion is F = 6πηrv, where η is the coefficient of viscosity, r is the sphere's radius, and v is its speed. The force acts opposite to the direction of motion.
F is in newtons (N), η in pascal-second (Pa·s) or N·s/m², r in metres (m), and v in metres per second (m/s). Check: (Pa·s)(m)(m/s) = (N·s/m²)(m²/s) = N. The units are consistent.
Yes. The drag always opposes the sphere's motion, so it slows the sphere down. Because F is proportional to v, the drag becomes larger as the sphere moves faster, and it is what leads to terminal velocity.
Only partly. Dimensions show F is proportional to η·r·v, giving F = k·ηrv. The value k = 6π comes from solving the full equations of viscous flow, not from dimensions. NEET expects you to know the final form F = 6πηrv.
At terminal velocity the Stokes drag exactly balances the net downward force (weight minus buoyancy). Setting 6πηrv = (4/3)πr³(ρ − σ)g and solving gives the terminal velocity vₜ = (2/9)r²(ρ − σ)g/η. So Stokes' law is the starting point for the terminal velocity formula.
