When a small sphere falls through a viscous fluid, the drag force on it is not proportional to velocity squared (as in turbulent flow) but linearly proportional to velocity. This is Stokes' law.
The formula: F = 6πηrv, where η is the fluid's dynamic viscosity, r is the sphere's radius, and v is its velocity. This holds strictly for smooth, slow flow — low Reynolds number — past a spherical body in a Newtonian fluid (NCERT Class 11 Physics, Chapter 9 "Mechanical Properties of Fluids," page 188).
Where NEET uses this: Stokes' law is the engine behind terminal velocity problems. A sphere released in a viscous liquid accelerates initially, but the drag force (∝ v) grows until it balances the net downward force (weight minus buoyancy). At that point, acceleration is zero and the sphere moves at constant terminal velocity:
v_t = (2r²g(ρ_s − ρ_f)) / (9η)
The high-frequency mistake: treating terminal velocity as proportional to r instead of r². Since Stokes drag scales as r·v while gravitational pull scales as r³, the balance yields v_t ∝ r². Doubling the radius quadruples — not doubles — the terminal velocity. This is a common distractor anchor in NEET.
The v(t) curve trap: the velocity-time graph for a falling sphere is not a straight line. It curves upward and asymptotically flattens at v_t. A distractor showing constant acceleration (straight line) or overshoot (oscillation) is wrong — viscous drag produces a smooth, monotonic approach to terminal velocity with no overshoot.
Watch-out: Stokes' law applies only at low Reynolds numbers. If a problem describes turbulent conditions or high-speed flow, this formula does not apply.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
According to Stokes' law, the viscous drag force on a sphere moving through a fluid is proportional to which of the following?
Answer: C. Stokes' law states F = 6πηrv, so the drag force is proportional to the first power of both radius and velocity, i.e., r·v (NCERT Class 11 Physics, Chapter 9, page 192).
Why A is wrong: A is wrong because v² dependence applies to turbulent drag (high Reynolds number), not Stokes' regime.
Why B is wrong: B is wrong because r² appears in the terminal velocity formula (v_t ∝ r²), not in the drag force expression itself.
Why D is wrong: D is wrong because no power of 2 appears for either r or v in Stokes' law; this conflates turbulent drag with Stokes' drag.
What is the SI unit of dynamic viscosity (η) that appears in Stokes' law?
Answer: B. Dynamic viscosity η has SI unit pascal-second (Pa·s), equivalent to kg/(m·s). This follows directly from the dimensions of Stokes' law: F = 6πηrv → η = F/(6πrv) → N/(m·m/s) = Pa·s (NCERT Class 11 Physics, Chapter 9, pages 191–192).
Why A is wrong: A is wrong because Pa/s has dimensions of pressure per time, which does not match the dimensional analysis of η = force/(length × velocity).
Why C is wrong: C is wrong because N·m is the unit of torque or energy, not viscosity.
Why D is wrong: D is wrong because kg/m² has dimensions of surface mass density, not viscosity. The correct unit kg/(m·s) requires the denominator to include both length and time.
Which of the following is NOT a necessary condition for Stokes' law to be valid?
Answer: B. Stokes' law requires a spherical body, low Reynolds number (smooth, slow flow), and a Newtonian fluid. Compressibility is not a requirement — in fact, the derivation assumes incompressible flow (NCERT Class 11 Physics, Chapter 9, page 192).
Why A is wrong: A is wrong because a spherical body is indeed a necessary condition; Stokes' law is derived specifically for a sphere.
Why C is wrong: C is wrong because the fluid must be Newtonian (viscosity independent of shear rate) for the linear drag relation to apply.
Why D is wrong: D is wrong because low Reynolds number (laminar flow) is a necessary condition for Stokes' law to hold.
A steel ball is dropped into a tall cylinder of glycerine. The terminal velocity of the ball is v_t. If the radius of the ball is doubled while all other conditions remain unchanged, the new terminal velocity is:
Answer: A. Terminal velocity v_t = (2r²g(ρ_s − ρ_f))/(9η), so v_t ∝ r². Doubling r gives (2r)² = 4r², hence v_t becomes 4v_t (NCERT Class 11 Physics, Chapter 9, page 192).
Why B is wrong: B is wrong because this assumes v_t ∝ r (linear dependence). The common mistake is forgetting that while Stokes drag ∝ r, the gravitational force ∝ r³, making the terminal velocity scale as r².
Why C is wrong: C is wrong because terminal velocity increases with radius, not decreases. Larger spheres settle faster, not slower.
Why D is wrong: D is wrong because this inverts the r² dependence. Terminal velocity is directly (not inversely) proportional to r².
Two spheres of the same material but radii r and 3r are dropped simultaneously into a viscous liquid. The ratio of their terminal velocities v₁ : v₂ is:
Answer: D. Since both spheres have the same material (same ρ_s) and fall in the same fluid, v_t ∝ r². Therefore v₁/v₂ = r²/(3r)² = 1/9, giving 1 : 9 (NCERT Class 11 Physics, Chapter 9, page 192).
Why A is wrong: A is wrong because this assumes v_t ∝ r. The r² dependence means tripling the radius increases terminal velocity by a factor of 9, not 3.
Why B is wrong: B is wrong because this assumes v_t ∝ r³. Terminal velocity scales as r² (from the balance of Stokes drag and net gravitational force), not as r³.
Why C is wrong: C is wrong because the larger sphere has the greater terminal velocity, not the smaller one. The ratio should favour the sphere with radius 3r.
A sphere of radius 1.0 mm moves at 0.10 m/s through a liquid of viscosity 1.0 × 10⁻³ Pa·s. By Stokes' law, the viscous drag on it is:
Answer: C. C is correct. Stokes' law gives F = 6πηrv = 6π × (1.0 × 10⁻³) × (1.0 × 10⁻³ m) × 0.10 = 6π × 10⁻⁷ ≈ 1.9 × 10⁻⁶ N. The radius must be in metres.
Why A is wrong: A is wrong because 3.1 × 10⁻⁷ N is πηrv, with the 6 of Stokes' law left out.
Why B is wrong: B is wrong because 1.9 × 10⁻³ N uses the radius in millimetres, r = 1, instead of 1.0 × 10⁻³ m: a factor of a thousand too large.
Why D is wrong: D is wrong because 6.0 × 10⁻⁷ N is 6ηrv with the π left out.
A sphere falls from rest into a tall column of viscous oil. Which velocity-time graph correctly represents its motion until it reaches terminal velocity?
Answer: A. Initially, drag is small (v ≈ 0) and the sphere accelerates. As v increases, viscous drag (∝ v) grows, reducing net force and acceleration. The velocity asymptotically approaches v_t — a smooth curve flattening to a horizontal asymptote (NCERT Class 11 Physics, Chapter 9, page 192).
Why B is wrong: B is wrong because constant acceleration implies no drag or constant drag. Stokes drag increases with velocity, so acceleration continuously decreases — the v-t plot cannot be a straight line.
Why C is wrong: C is wrong because the velocity does not overshoot or decrease. Viscous drag provides a monotonically increasing retarding force without oscillation or reversal.
Why D is wrong: D is wrong because v = 0 at t = 0 (released from rest). The sphere must accelerate before reaching terminal velocity; it does not start at v_t.
A sphere of density 2.0 × 10³ kg/m³ and radius 1.0 × 10⁻³ m falls through a liquid of density 1.0 × 10³ kg/m³ and viscosity 1.0 Pa·s. Taking g = 10 m/s² (exact), the terminal velocity is closest to:
Answer: D. v_t = 2r²g(ρ_s − ρ_f)/(9η) = 2 × (1.0 × 10⁻³)² × 10 × (2.0 × 10³ − 1.0 × 10³) / (9 × 1.0). Numerator: 2 × 10⁻⁶ × 10 × 10³ = 2 × 10⁻² = 0.02. Denominator: 9. v_t = 0.02/9 ≈ 2.2 × 10⁻³ m/s. Note: g = 10 m/s² is given as exact and does not limit significant figures (NCERT Class 11 Physics, Chapter 9, page 192).
Why A is wrong: A is wrong because 2.0 × 10⁻² m/s leaves out the 9 in the denominator: 2r²g(ρ_s − ρ_f)/η = 0.02 m/s.
Why B is wrong: B is wrong because 4.4 × 10⁻³ is double the correct answer: it uses the sphere's density ρ_s = 2.0 × 10³ kg/m³ in place of the difference ρ_s − ρ_f = 1.0 × 10³ kg/m³, leaving out the buoyancy.
Why C is wrong: C is wrong because 1.1 × 10⁻³ is exactly half the correct answer. This results from omitting the factor of 2 in the numerator: the formula is v_t = 2r²g(ρ_s − ρ_f)/(9η), not r²g(ρ_s − ρ_f)/(9η).
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Pattern: Terminal velocity scaling (from PYQ pattern NEET pattern: terminal velocity scaling, observed in NEET 2021, 2022)
Given
• r = 2.0 × 10⁻³ m (2 sig figs)• ρ_s = 8.0 × 10³ kg/m³ (2 sig figs)• ρ_f = 1.2 × 10³ kg/m³ (2 sig figs)• η = 0.90 Pa·s (2 sig figs)• g = 9.8 m/s² (exact, problem-defined)
Required
Terminal velocity v_t.
Concept
At terminal velocity, the net force on the sphere is zero: weight = buoyant force + Stokes drag. The resulting expression for v_t follows from setting the net downward force equal to 6πηrv_t and solving for v_t.
Formula
v_t = 2r²g(ρ_s − ρ_f) / (9η)
Substitution
v_t = 2 × (2.0 × 10⁻³)² × 9.8 × (8.0 × 10³ − 1.2 × 10³) / (9 × 0.90)
Calculation
• r² = (2.0 × 10⁻³)² = 4.0 × 10⁻⁶ m²• (ρ_s − ρ_f) = 6.8 × 10³ kg/m³• Numerator = 2 × 4.0 × 10⁻⁶ × 9.8 × 6.8 × 10³ = 2 × 4.0 × 9.8 × 6.8 × 10⁻⁶⁺³
= 2 × 4.0 × 9.8 × 6.8 × 10⁻³
= 2 × 266.56 × 10⁻³
= 533.12 × 10⁻³
= 0.53312• Denominator = 9 × 0.90 = 8.1• v_t = 0.53312 / 8.1 = 0.06582... m/s
Note on exact constants: g = 9.8 m/s² is given as an exact problem-defined value and does not constrain significant figures. The numerical constants 2 and 9 in the formula are exact integers.
Final answer
v_t ≈ 6.6 × 10⁻² m/s (reported to 2 significant figures, matching the least precise given quantity).
Common trap
The most common error is treating v_t ∝ r instead of v_t ∝ r². If the problem asked "what happens when the radius is halved?", the trap answer is v_t/2, but the correct answer is v_t/4 (since v_t scales as r²). Always check the power of r before answering scaling questions.
Similar NEET-style question
A lead shot of radius r has terminal velocity v in glycerine. A second lead shot of radius 2r is dropped in the same glycerine. What is its terminal velocity?
Answer: v_t ∝ r², so new v_t = (2r/r)² × v = 4v.
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Drag force on a sphere of radius r moving with velocity v through a viscous fluid (η): F = 6π η r v. A sphere falling through the fluid reaches a terminal velocity vt = 2a²(ρ − σ)g/(9η), which depends on the square of the radius and inversely on the viscosity.
-- NCERT Class 11 Physics, Ch. 9, p. 192Drag force on a sphere of radius r moving with velocity v through viscous fluid (low Reynolds number).
| Symbol | Quantity | SI Unit |
|---|---|---|
| F | drag force | N |
| eta | viscosity | Pa*s |
| r | sphere radius | m |
| v | velocity | m/s |
More in Properties of Bulk Matter: 5 exam traps and mistakes · 11 formulas · 4 question patterns from its other lessons.
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expects linear acceleration
Default to constant acceleration without recognising drag-induced terminal v
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