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Viscosity, Stokes’ law, terminal velocity and fluid-flow regimes: NEET MCQs with answers

50 questions on this topic of Properties of Solids and Liquids (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. The dimensional formula for the coefficient of dynamic viscosity (η) is:

  1. [M L⁻¹ T⁻²]
  2. [M L⁻² T⁻¹]
  3. [M L⁻¹ T⁻¹]
  4. [M L T⁻¹]
Show answer and explanation

Answer: (C) [M L⁻¹ T⁻¹]

From Newton's law of viscosity, shear stress τ = η(dv/dx), so η = F / (A · dv/dx). Substituting dimensions gives [M L T⁻²] / ([L²] · [T⁻¹]) = [M L⁻¹ T⁻¹].

medium

2. A tiny metallic sphere attains a terminal velocity v₀ while falling through a viscous fluid. The rate of heat generated in the fluid due to viscous forces at this terminal velocity is proportional to:

  1. v₀
  2. v₀²
  3. v₀³
  4. v₀⁵
Show answer and explanation

Answer: (B) v₀²

The rate of heat generated equals the rate of work done by the viscous force, P = F_v · v₀. Since the viscous drag is F_v = 6πηrv₀, the power dissipated is P = 6πηr v₀², which is proportional to v₀².

hard

3. A small glass sphere of density 2500 kg m⁻³ and radius r falls with terminal speed v₁ in a liquid of density 1000 kg m⁻³. An identical-sized sphere of unknown material falls in the same liquid with terminal speed v₂ = -0.5 v₁ (the negative sign indicates it moves vertically upward). The density of this unknown material is:

  1. 500 kg m⁻³
  2. 250 kg m⁻³
  3. 750 kg m⁻³
  4. 1750 kg m⁻³
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Answer: (B) 250 kg m⁻³

Terminal speed is proportional to (ρ - σ), where σ = 1000 kg m⁻³ is the liquid density. Thus, v₁ ∝ (2500 - 1000) = 1500. For the second sphere, v₂ / v₁ = (ρ_x - 1000) / 1500 = -0.5. This gives ρ_x - 1000 = -750, so ρ_x = 250 kg m⁻³.

medium

4. Water of dynamic viscosity 1.0 × 10⁻³ Pa s flows through a straight cylindrical pipe of diameter 2.0 cm. The density of water is 1000 kg m⁻³. If the critical Reynolds number for the transition from laminar to turbulent flow in this pipe is 2000, the maximum average flow speed for which the flow remains strictly laminar is:

  1. 1.00 m s⁻¹
  2. 0.20 m s⁻¹
  3. 0.50 m s⁻¹
  4. 0.10 m s⁻¹
Show answer and explanation

Answer: (D) 0.10 m s⁻¹

The Reynolds number is defined as R_e = (ρ v D) / η. Setting R_e = 2000, we get the critical velocity v = (R_e × η) / (ρ × D) = (2000 × 1.0 × 10⁻³) / (1000 × 0.02) = 2.0 / 20 = 0.10 m s⁻¹.

easy

5. A large horizontal plate of surface area 0.5 m² rests on a 2 mm thick layer of castor oil (η = 0.8 Pa s) covering a flat floor. What horizontal force is required to pull the plate at a constant speed of 0.4 m s⁻¹?

  1. 40 N
  2. 160 N
  3. 80 N
  4. 20 N
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Answer: (C) 80 N

From Newton's law of viscosity, F = η · A · (dv/dx). Here, η = 0.8 Pa s, A = 0.5 m², and velocity gradient dv/dx = 0.4 / (2 × 10⁻³) = 200 s⁻¹. Thus, F = 0.8 × 0.5 × 200 = 80 N.

hard

6. A metal sphere of radius R is falling with terminal velocity v through a viscous liquid. The liquid exerts a retarding viscous torque τ on the sphere if it is made to rotate about its diameter with a small angular speed ω while moving down. Dimensionally, the viscous drag force F_d and torque τ must depend on the coefficient of viscosity η. The dimensional formula of the ratio (τ / F_d) is equivalent to that of:

  1. Velocity
  2. Time
  3. Length
  4. Force
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Answer: (C) Length

The dimensions of torque τ are [M L² T⁻²] and the dimensions of force F_d are [M L T⁻²]. Therefore, the ratio [τ / F_d] = [M L² T⁻²] / [M L T⁻²] = [L], which has the dimensions of Length.

easy

7. According to Stokes' law, the viscous drag on a small sphere of radius r moving with velocity v in a fluid of viscosity η is:

  1. 6πηr²v
  2. 4πηrv
  3. πηrv²
  4. 6πηrv
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Answer: (D) 6πηrv

Stokes' law gives F = 6πηrv for a small sphere moving slowly (streamline flow) through a viscous fluid.

NCERT Class 11

easy

8. A spherical steel ball reaches a terminal speed of 20 cm s⁻¹ when falling through a column of glycerin. If the experiment is repeated using an identical ball in another liquid of the same density as glycerin but having three times its coefficient of viscosity, the terminal speed attained by the ball will be:

  1. 60 cm s⁻¹
  2. 20 cm s⁻¹
  3. 2.22 cm s⁻¹
  4. 6.67 cm s⁻¹
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Answer: (D) 6.67 cm s⁻¹

Terminal velocity is given by v_t = 2r²(ρ - σ)g / (9η). Since the radius, density difference (ρ - σ), and g remain unchanged, v_t is inversely proportional to the viscosity η. Hence, v_t' = v_t / 3 = 20 / 3 ≈ 6.67 cm s⁻¹.

easy

9. The SI unit of coefficient of viscosity is:

  1. Pa m
  2. Pa s (N s m⁻²)
  3. N m⁻¹
  4. N m
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Answer: (B) Pa s (N s m⁻²)

η = F/[A(dv/dx)], so its SI unit is N s m⁻², also called the pascal second (Pa s) or poiseuille.

NCERT Class 11

easy

10. Which of the following physical quantities represents the ratio of inertial force to viscous force acting on a fluid element in motion?

  1. Poiseuille number
  2. Prandtl number
  3. Reynolds number
  4. Mach number
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Answer: (C) Reynolds number

The Reynolds number (R_e = ρvd/η) is a dimensionless parameter defined physically as the ratio of inertial forces to viscous forces within a flowing fluid. It dictates whether the flow regime is laminar, transient, or turbulent.

medium

11. A spherical body of density ρ falls through a viscous fluid of density σ (ρ > 3σ). If a₀ is the initial downward acceleration of the body at the instant of release and a_half is its acceleration when its speed equals half the terminal speed, then the ratio a_half / a₀ is:

  1. 1/4
  2. 1/2
  3. 3/4
  4. 1/√2
Show answer and explanation

Answer: (B) 1/2

The equation of motion is m a = m g (1 - σ/ρ) - 6πηrv. Initially at v = 0, a₀ = g (1 - σ/ρ). At terminal velocity v_t, the net force is zero, so 6πηrv_t = m g (1 - σ/ρ). When v = v_t / 2, the viscous force is half the effective weight, so m a_half = m g (1 - σ/ρ) - (1/2) m g (1 - σ/ρ) = (1/2) m g (1 - σ/ρ) = (1/2) m a₀. Thus, a_half / a₀ = 1/2.

medium

12. A metal sphere of radius r falls through a viscous fluid. When it reaches its terminal velocity v_t, the net viscous drag force acting on it is F_v. If another sphere made of the same material but of radius 2r falls through the same fluid, what will be the viscous drag force acting on it when it attains its own terminal velocity?

  1. 2 F_v
  2. 8 F_v
  3. 4 F_v
  4. 16 F_v
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Answer: (B) 8 F_v

At terminal velocity, the net downward acceleration is zero, which means viscous force F_v = Weight - Buoyant force = (4/3)π r³ (ρ - σ)g. Thus, F_v is directly proportional to r³. When the radius is doubled, the viscous force at terminal velocity increases by a factor of 2³ = 8.

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