Viscosity is the internal friction of a flowing fluid — the property that makes honey pour slowly while water pours fast. NCERT Class 11 Physics Chapter 9 (Mechanical Properties of Fluids), page 190, defines viscosity as the resistance to relative motion between adjacent fluid layers. The coefficient of viscosity η has SI unit Pa·s (also called poiseuille; CGS unit: poise, where 1 Pa·s = 10 poise).
The mechanism. When fluid flows over a surface, the layer in contact with the surface is stationary (no-slip condition). Each successive layer above moves faster. The velocity gradient dv/dx between layers produces a viscous force. Newton's law of viscosity states: F/A = η(dv/dx), where F/A is the tangential stress (shear stress) between layers.
Where NEET tests this. The direct application of viscosity in NEET centres on Stokes' law: the drag force on a small sphere moving through viscous fluid is F = 6πηrv (NCERT Chapter 9, page 188). This drag force is linear in both radius r and velocity v — not quadratic. The terminal velocity formula v_t = 2r²g(ρ_s − ρ_f)/(9η) shows that v_t is proportional to r², not r. This is a common confusion: aspirants who remember Stokes' drag is linear in r incorrectly assume terminal velocity is also linear in r. It is not — because the gravitational force driving the fall scales as r³ while drag scales as r, the balance gives r².
Watch-out. When a NEET question says "the radius is doubled," terminal velocity becomes four times larger, not two times. If you pick the "doubled" option, you have confused the drag-force dependence (linear in r) with the terminal-velocity dependence (quadratic in r).
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The SI unit of coefficient of viscosity is:
Answer: C. The coefficient of viscosity η has the SI unit pascal-second (Pa·s), equivalent to kg/(m·s). This follows directly from the definition η = (F/A)/(dv/dx), giving units of (N/m²)/(s⁻¹) = Pa·s (NCERT Class 11 Physics Chapter 9, page 191).
Why A is wrong: N·m is the unit of torque (or energy), not viscosity. This confuses force × distance with the shear-stress-to-velocity-gradient ratio.
Why B is wrong: Pa/s has dimensions of pressure per time, not viscosity. Viscosity is stress divided by velocity gradient, yielding Pa × s, not Pa ÷ s.
Why D is wrong: kg/m² is surface density, not viscosity. The correct dimensional formula for η is ML⁻¹T⁻¹, which gives kg/(m·s), not kg/m².
The CGS unit of viscosity is the poise. How many poise equal 1 Pa·s?
Answer: D. 1 Pa·s = 1 kg/(m·s). Converting to CGS: 1 kg = 1000 g, 1 m = 100 cm. So 1 Pa·s = 1000 g/(100 cm · s) = 10 g/(cm·s) = 10 poise (the poise is not in NCERT; NCERT Class 11 Physics Chapter 9, page 191 gives the SI unit, poiseuille (Pl), also N s m⁻² or Pa s).
Why A is wrong: Choosing 1 assumes SI and CGS viscosity units are identical. They differ by a factor of 10 because the mass-to-length conversion (kg/m → g/cm) introduces 1000/100 = 10.
Why B is wrong: 0.1 inverts the conversion: 1 poise = 0.1 Pa·s, so 1 Pa·s = 10 poise, not 0.1.
Why C is wrong: 100 is the conversion factor for pressure (1 Pa = 10 dyn/cm²), not viscosity. Viscosity carries an extra length dimension that changes the factor to 10.
In Newton's law of viscous flow, the viscous force between fluid layers is proportional to:
Answer: B. Newton's law of viscosity states F = ηA(dv/dx): the viscous force is proportional to both the velocity gradient dv/dx and the contact area A between layers (NCERT Class 11 Physics Chapter 9, page 191).
Why A is wrong: The force depends on the first power of the velocity gradient, not the square. F = ηA(dv/dx) is linear in dv/dx.
Why C is wrong: Layer separation appears in the denominator of the velocity gradient (dv/dx), but the force is not proportional to the cube of the separation. The relationship is F ∝ 1/dx through the gradient.
Why D is wrong: The force depends on the velocity gradient (difference in velocity per unit separation), not on the absolute velocity of any single layer.
A small steel sphere falls through glycerine. The viscous drag force on the sphere, according to Stokes' law, is F = 6πηrv. If the radius of the sphere is doubled while the velocity remains the same, the drag force becomes:
Answer: C. Stokes' law F = 6πηrv shows that drag force is directly proportional to radius r (first power). Doubling r doubles F. η and v are unchanged (NCERT Class 11 Physics Chapter 9, page 188).
Why A is wrong: Half would require F ∝ 1/r, but Stokes' law gives F ∝ r (first power). Drag increases, not decreases, with radius.
Why B is wrong: The drag force changes because it depends on r linearly. Keeping velocity the same does not keep force the same when radius changes.
Why D is wrong: Four times would require F ∝ r². This is the scaling for terminal velocity, not drag force. In Stokes' law, drag is linear in r.
Two identical spheres fall through the same viscous liquid. Sphere P has twice the density difference (ρ_s − ρ_f) compared to sphere Q. The ratio of their terminal velocities v_P : v_Q is:
Answer: A. Terminal velocity v_t = 2r²g(ρ_s − ρ_f)/(9η). For identical spheres in the same liquid, r and η are identical. v_t is directly proportional to (ρ_s − ρ_f). If sphere P has twice the density difference, v_P = 2v_Q, giving ratio 2 : 1 (NCERT Class 11 Physics Chapter 9, page 189).
Why B is wrong: 1 : 1 would mean terminal velocity is independent of density difference, but v_t ∝ (ρ_s − ρ_f) directly.
Why C is wrong: 4 : 1 confuses the r² dependence with the density-difference dependence. Terminal velocity scales as r² but only as the first power of (ρ_s − ρ_f).
Why D is wrong: 1 : 2 inverts the ratio. The sphere with the greater density difference falls faster, not slower.
A sphere of radius r reaches terminal velocity v_t in a viscous fluid. If the sphere is replaced by one of radius 2r (same material, same fluid), the new terminal velocity is:
Answer: D. v_t = 2r²g(ρ_s − ρ_f)/(9η). Terminal velocity is proportional to r². Doubling r gives (2r)² = 4r², so the new terminal velocity is 4v_t (NCERT Class 11 Physics Chapter 9, page 189).
Why A is wrong: 2v_t assumes v_t ∝ r (linear). This is the common confusion between Stokes' drag (F ∝ r) and terminal velocity (v_t ∝ r²). Gravity scales as r³ while drag scales as rv, producing r² dependence for v_t.
Why B is wrong: v_t/2 implies an inverse relationship. A larger sphere falls faster in viscous fluid, not slower, because the gravitational driving force grows faster (r³) than the drag resistance (r).
Why C is wrong: 8v_t assumes v_t ∝ r³. This would be the case if drag force were independent of r, but Stokes' drag is proportional to r, giving a net r² dependence.
A steel ball of radius 1.0 mm falls through a viscous oil with terminal velocity 2.0 cm/s. A second steel ball of radius 2.0 mm falls through the same oil. What is the terminal velocity of the second ball?
Answer: A. Step 1: v_t ∝ r² (from terminal velocity formula, same material and fluid). Step 2: r₂/r₁ = 2.0 mm / 1.0 mm = 2. Step 3: v₂ = v₁ × (r₂/r₁)² = 2.0 cm/s × 4 = 8.0 cm/s (NCERT Class 11 Physics Chapter 9, page 189).
Why B is wrong: 4.0 cm/s assumes v_t ∝ r (linear scaling). This confuses Stokes' drag dependence on r with terminal velocity dependence on r². The ratio of radii is 2, so the velocity ratio is 2² = 4, giving 8.0 cm/s, not 4.0 cm/s.
Why C is wrong: 1.0 cm/s assumes v_t ∝ 1/r (inverse). A larger sphere reaches a higher terminal velocity, not a lower one, because gravitational force grows as r³ while drag grows as r.
Why D is wrong: 16.0 cm/s assumes v_t ∝ r³. The r³ dependence describes the gravitational force alone; when balanced against Stokes' drag (∝ rv), the terminal velocity scales as r², not r³.
A small sphere is released from rest in a tall column of viscous liquid. Which statement correctly describes the sphere's motion?
Answer: B. At release, velocity is zero so viscous drag (F = 6πηrv) is zero, and the sphere accelerates under net downward force (weight minus buoyancy). As velocity increases, drag increases until it equals the net gravitational force. At that point acceleration becomes zero and the sphere moves at constant terminal velocity (NCERT Class 11 Physics Chapter 9, pages 188–189).
Why A is wrong: Uniform acceleration throughout would mean drag never becomes significant. In viscous fluid, drag grows with velocity and eventually balances gravity, so acceleration must decrease and reach zero.
Why C is wrong: Immediate constant velocity would require drag to instantly match gravity. Since drag is proportional to velocity and the sphere starts from rest (v = 0), drag starts at zero and builds up gradually.
Why D is wrong: Deceleration from the start contradicts the initial condition: at v = 0, there is no drag, so the net force is downward and the sphere accelerates. Deceleration would require drag exceeding the driving force, which cannot happen from rest.
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
Given
A glass sphere of radius r₁ = 1.0 × 10⁻³ m and density ρ_s = 2.5 × 10³ kg/m³ falls through a liquid of density ρ_f = 1.0 × 10³ kg/m³ and viscosity η = 0.80 Pa·s. A second glass sphere of radius r₂ = 3.0 × 10⁻³ m falls through the same liquid.
Required
Find (a) the terminal velocity of the first sphere, and (b) the terminal velocity of the second sphere.
Concept
At terminal velocity, the net downward force (weight minus buoyancy) equals the upward Stokes' drag. The terminal velocity formula v_t = 2r²g(ρ_s − ρ_f)/(9η) applies since we are told the flow is in the Stokes regime (low Reynolds number, spherical body, Newtonian fluid).
Formula
v_t = 2r²g(ρ_s − ρ_f) / (9η)
Substitution (sphere 1)
v_t₁ = 2 × (1.0 × 10⁻³)² × 9.8 × (2.5 × 10³ − 1.0 × 10³) / (9 × 0.80)
Calculation
Numerator: 2 × 1.0 × 10⁻⁶ × 9.8 × 1.5 × 10³ = 2 × 9.8 × 1.5 × 10⁻³ = 29.4 × 10⁻³ = 2.94 × 10⁻²
Denominator: 9 × 0.80 = 7.2
v_t₁ = 2.94 × 10⁻² / 7.2 = 4.08 × 10⁻³ m/s ≈ 4.1 × 10⁻³ m/s
Note on exact constants: g = 9.8 m/s² is used as given (exact for this problem). The factor 2/9 is an exact mathematical constant from the derivation. Neither limits the significant figures of the answer. The answer is reported to 2 significant figures, matching the least-precise given value (η = 0.80 Pa·s, 2 sig figs).
Final answer
(a) v_t₁ ≈ 4.1 × 10⁻³ m/s (about 4.1 mm/s)
(b) Since v_t ∝ r², the ratio v_t₂/v_t₁ = (r₂/r₁)² = (3.0 × 10⁻³ / 1.0 × 10⁻³)² = 9.
v_t₂ = 9 × 4.1 × 10⁻³ = 3.7 × 10⁻² m/s ≈ 37 mm/s
Common trap
The temptation is to say v_t₂ = 3 × v_t₁ (treating terminal velocity as linear in r). Terminal velocity scales as r², not r, because gravitational force grows as r³ while Stokes' drag grows as rv — the net balance gives r². The correct factor is 9, not 3.
Similar NEET-style question
A lead shot of diameter d falls through glycerine with terminal velocity v. What is the terminal velocity of a lead shot of diameter 2d falling through the same glycerine? (Answer: 4v, since r doubles and v_t ∝ r².)
---
Property of a fluid that opposes relative motion between adjacent layers. For Newtonian fluid: F = η A (dv/dx), where η is the coefficient of viscosity (Pa·s). Higher η → more 'sticky' fluid.
-- NCERT Class 11 Physics, Ch. 9, p. 190More in Properties of Bulk Matter: 5 exam traps and mistakes · 12 formulas · 5 question patterns from its other lessons.
No question in our NEET 2020–2025 set targets this topic directly.
All 17 past-paper questions from Properties of Bulk Matter →
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →