The Kinematic Viscosity Formula defines the ratio of a fluid’s dynamic viscosity to its mass density, expressed as \( u = \mu / ho \). This concept is covered in fluid mechanics topics relevant to CBSE Class 11 Physics and forms a key part of JEE Main and NEET problem sets on properties of fluids. Understanding this formula helps students analyse how fluids flow under gravity and pressure. This article covers the definition, expression, variables, a complete formula sheet, three solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Kinematic Viscosity Formulas at a Glance
Quick reference for the most important kinematic viscosity formulas.
- Kinematic viscosity: \( u = \dfrac{\mu}{ ho} \)
- Dynamic viscosity from Newton’s law: \( \mu = au / (du/dy) \)
- SI unit of kinematic viscosity: m²/s (Stokes in CGS: 1 St = 10−&sup4; m²/s)
- Reynolds number: \( Re = \dfrac{ ho v L}{\mu} = \dfrac{v L}{ u} \)
- Stokes’ law drag force: \( F = 6\pi \eta r v \)
- Poiseuille’s flow rate: \( Q = \dfrac{\pi r^4 \Delta P}{8 \mu L} \)
- Terminal velocity: \( v_t = \dfrac{2r^2( ho_s – ho_f)g}{9\mu} \)
What is Kinematic Viscosity Formula?
The Kinematic Viscosity Formula describes how resistant a fluid is to flow when only gravitational forces act on it. It is defined as the dynamic (absolute) viscosity of a fluid divided by its density at the same temperature and pressure. Kinematic viscosity is denoted by the Greek letter \( u \) (nu).
In NCERT Class 11 Physics, Chapter 10 — Mechanical Properties of Fluids — students study viscosity as a measure of internal friction in fluids. While dynamic viscosity \( \mu \) measures the shear stress per unit velocity gradient, kinematic viscosity \( u \) normalises this by density. This makes it particularly useful when comparing how fluids behave under their own weight.
The concept is critical in engineering and applied physics. It governs pipe flow, lubrication, aerodynamics, and blood flow in medical physics. For CBSE board students, kinematic viscosity appears in numerical problems and derivation questions. For JEE and NEET aspirants, it connects directly to Reynolds number, Stokes’ law, and Poiseuille’s equation.
Kinematic viscosity has SI units of m²/s. In the CGS system, the unit is the Stokes (St), where 1 St = 10−&sup4; m²/s. The centistokes (cSt) is commonly used in engineering: 1 cSt = 10−&sup6; m²/s.
Kinematic Viscosity Formula — Expression and Variables
The kinematic viscosity formula is:
\[ \nu = \frac{\mu}{\rho} \]
Where \( u \) is kinematic viscosity, \( \mu \) is dynamic viscosity, and \( ho \) is the fluid density. This can be rearranged to find dynamic viscosity: \( \mu = u ho \), or density: \( ho = \mu / u \).
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( u \) (nu) | Kinematic viscosity | m²/s |
| \( \mu \) (mu) | Dynamic (absolute) viscosity | Pa·s (or N·s/m²) |
| \( ho \) (rho) | Mass density of the fluid | kg/m³ |
Derivation of the Kinematic Viscosity Formula
Newton’s law of viscosity states that the shear stress \( au \) in a fluid is proportional to the velocity gradient \( du/dy \):
\[ \tau = \mu \frac{du}{dy} \]
Here, \( \mu \) is the proportionality constant called dynamic viscosity. Its SI unit is Pa·s.
Now, in many fluid flow problems, the ratio of inertial forces to viscous forces matters more than viscosity alone. Inertial force per unit volume is proportional to \( ho \). So, we define kinematic viscosity as:
\[ \nu = \frac{\mu}{\rho} \]
Dimensionally: \( [\mu] = ext{Pa} \cdot ext{s} = ext{kg m}^{-1} ext{s}^{-1} \) and \( [ ho] = ext{kg m}^{-3} \). Dividing gives \( [ u] = ext{m}^2 ext{s}^{-1} \), which confirms the SI unit of kinematic viscosity.
Complete Fluid Mechanics Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Kinematic Viscosity | \( u = \mu / ho \) | \( u \)=kinematic viscosity, \( \mu \)=dynamic viscosity, \( ho \)=density | m²/s | Class 11, Ch 10 |
| Newton’s Law of Viscosity | \( au = \mu (du/dy) \) | \( au \)=shear stress, \( \mu \)=dynamic viscosity, \( du/dy \)=velocity gradient | Pa | Class 11, Ch 10 |
| Stokes’ Law | \( F = 6\pi \eta r v \) | \( \eta \)=viscosity, r=radius, v=velocity | N | Class 11, Ch 10 |
| Terminal Velocity | \( v_t = \dfrac{2r^2( ho_s – ho_f)g}{9\eta} \) | r=radius, \( ho_s \)=sphere density, \( ho_f \)=fluid density, g=gravity | m/s | Class 11, Ch 10 |
| Poiseuille’s Equation | \( Q = \dfrac{\pi r^4 \Delta P}{8 \mu L} \) | r=pipe radius, \( \Delta P \)=pressure difference, L=length | m³/s | Class 11, Ch 10 |
| Reynolds Number | \( Re = \dfrac{ ho v L}{\mu} = \dfrac{vL}{ u} \) | v=velocity, L=characteristic length, \( u \)=kinematic viscosity | Dimensionless | Class 11, Ch 10 |
| Continuity Equation | \( A_1 v_1 = A_2 v_2 \) | A=cross-sectional area, v=fluid velocity | m³/s | Class 11, Ch 10 |
| Bernoulli’s Equation | \( P + rac{1}{2} ho v^2 + ho g h = ext{const} \) | P=pressure, v=velocity, h=height | Pa | Class 11, Ch 10 |
| Pressure in a Fluid | \( P = P_0 + ho g h \) | \( P_0 \)=surface pressure, h=depth | Pa | Class 11, Ch 10 |
| Surface Tension | \( T = F / L \) | F=force, L=length of surface | N/m | Class 11, Ch 10 |
Kinematic Viscosity Formula — Solved Examples
Example 1 (Class 9-10 Level): Direct Application
Problem: The dynamic viscosity of water at 20°C is \( 1.002 imes 10^{-3} \) Pa·s and its density is 998 kg/m³. Calculate the kinematic viscosity of water.
Given:
- Dynamic viscosity: \( \mu = 1.002 imes 10^{-3} \) Pa·s
- Density: \( ho = 998 \) kg/m³
Step 1: Write the kinematic viscosity formula: \( u = \mu / ho \)
Step 2: Substitute the values:
\[ \nu = \frac{1.002 \times 10^{-3}}{998} \]
Step 3: Perform the division:
\[ \nu = 1.004 \times 10^{-6} \text{ m}^2/\text{s} \]
Answer
Kinematic viscosity of water at 20°C = \( 1.004 imes 10^{-6} \) m²/s = 1.004 cSt
Example 2 (Class 11-12 Level): Multi-Step Problem
Problem: An oil has a kinematic viscosity of \( 40 imes 10^{-6} \) m²/s and a specific gravity of 0.85. Find its dynamic viscosity in Pa·s. Also, find the shear stress if the velocity gradient is 200 s−¹.
Given:
- Kinematic viscosity: \( u = 40 imes 10^{-6} \) m²/s
- Specific gravity: SG = 0.85 (density of water = 1000 kg/m³)
- Velocity gradient: \( du/dy = 200 \) s−¹
Step 1: Find the density of oil.
\[ \rho_{\text{oil}} = \text{SG} \times \rho_{\text{water}} = 0.85 \times 1000 = 850 \text{ kg/m}^3 \]
Step 2: Find dynamic viscosity using \( \mu = u ho \).
\[ \mu = 40 \times 10^{-6} \times 850 = 0.034 \text{ Pa} \cdot \text{s} \]
Step 3: Find shear stress using Newton’s law of viscosity.
\[ \tau = \mu \frac{du}{dy} = 0.034 \times 200 = 6.8 \text{ Pa} \]
Answer
Dynamic viscosity = 0.034 Pa·s; Shear stress = 6.8 Pa
Example 3 (JEE/NEET Level): Reynolds Number and Flow Regime
Problem: Water flows through a pipe of diameter 0.05 m at a velocity of 2 m/s. The kinematic viscosity of water is \( 1.0 imes 10^{-6} \) m²/s. Determine the Reynolds number and state whether the flow is laminar or turbulent. (Use: laminar if Re < 2000, turbulent if Re > 4000.)
Given:
- Pipe diameter: \( D = 0.05 \) m
- Flow velocity: \( v = 2 \) m/s
- Kinematic viscosity: \( u = 1.0 imes 10^{-6} \) m²/s
Step 1: Recall the Reynolds number formula using kinematic viscosity.
\[ Re = \frac{vD}{\nu} \]
Step 2: Substitute values.
\[ Re = \frac{2 \times 0.05}{1.0 \times 10^{-6}} = \frac{0.1}{1.0 \times 10^{-6}} = 1.0 \times 10^5 \]
Step 3: Compare with critical Reynolds numbers.
Since \( Re = 100{,}000 \gg 4000 \), the flow is fully turbulent.
Step 4: Physical interpretation — at this Reynolds number, inertial forces dominate viscous forces. The fluid forms eddies and irregular mixing patterns.
Answer
Reynolds number = \( 1.0 imes 10^5 \); Flow is turbulent.
CBSE Exam Tips 2025-26
- Memorise the formula and its units: Write \( u = \mu / ho \) with units m²/s. CBSE board questions frequently ask for SI units of kinematic viscosity. We recommend writing the unit derivation once to internalise it.
- Know the CGS unit: The Stokes (St) appears in several NCERT exercises. Remember that 1 St = 10−&sup4; m²/s and 1 cSt = 10−&sup6; m²/s.
- Distinguish dynamic from kinematic viscosity: CBSE examiners often test whether students confuse \( \mu \) (Pa·s) with \( u \) (m²/s). Practice writing both symbols clearly.
- Link to Reynolds number: In 2025-26 board exams, application-based questions connect kinematic viscosity to the Reynolds number formula \( Re = vL/ u \). Practise these combined problems.
- Dimensional analysis questions: Verify \( [ u] = M^0 L^2 T^{-1} \) by dividing dimensions of \( \mu \) by dimensions of \( ho \). This is a common 2-mark question.
- Temperature dependence: Remember that kinematic viscosity of liquids decreases with increasing temperature. Gases show the opposite trend. This is a common MCQ point.
Common Mistakes to Avoid
- Mistake 1 — Inverting the formula: Some students write \( u = ho / \mu \) instead of \( u = \mu / ho \). Always remember: kinematic viscosity is viscosity divided by density, not the other way around.
- Mistake 2 — Using wrong units for \( \mu \): Dynamic viscosity must be in Pa·s (not poise) when working in SI. 1 Poise (P) = 0.1 Pa·s. Forgetting this conversion leads to answers off by a factor of 10.
- Mistake 3 — Confusing \( u \) with \( \mu \) in Reynolds number: The Reynolds number formula can be written as \( Re = ho v L / \mu \) or \( Re = vL / u \). Students sometimes substitute \( \mu \) where \( u \) is required, or vice versa, causing errors.
- Mistake 4 — Ignoring temperature context: Viscosity values depend on temperature. Always check if the problem specifies temperature before using a standard value from a data table.
- Mistake 5 — Wrong dimensional formula: Students sometimes write the dimension of kinematic viscosity as \( MLT^{-1} \). The correct dimensional formula is \( M^0 L^2 T^{-1} \) because it is mass-independent.
JEE/NEET Application of Kinematic Viscosity Formula
In our experience, JEE aspirants encounter the Kinematic Viscosity Formula most frequently in problems combining fluid flow, Reynolds number, and Stokes’ law. NEET aspirants see it in the context of blood flow in arteries and terminal velocity of particles.
Pattern 1: Reynolds Number Problems (JEE Main)
JEE Main regularly asks students to calculate the Reynolds number using \( Re = vL/ u \) and classify flow as laminar or turbulent. A typical question gives pipe diameter, flow speed, and kinematic viscosity, then asks for the flow regime. Practise converting between CGS and SI units before substituting.
Pattern 2: Terminal Velocity and Stokes’ Law (NEET)
NEET problems often involve a spherical particle falling through a viscous fluid. The terminal velocity formula is \( v_t = 2r^2( ho_s – ho_f)g / (9\mu) \). Once \( \mu \) is found, kinematic viscosity follows from \( u = \mu/ ho_f \). This two-step approach is common in NEET 2024 and previous years’ papers.
Pattern 3: Poiseuille’s Equation and Viscosity Comparison (JEE Advanced)
JEE Advanced problems may give two fluids with different kinematic viscosities flowing through identical pipes. Students must compare flow rates using Poiseuille’s equation \( Q = \pi r^4 \Delta P / (8\mu L) \). Since \( \mu = u ho \), the comparison requires substituting both \( u \) and \( ho \) for each fluid. Our experts suggest practising at least five such comparison problems before the exam.
Key values to memorise for JEE/NEET:
- Kinematic viscosity of water at 20°C: \( \approx 1.0 imes 10^{-6} \) m²/s
- Kinematic viscosity of air at 20°C: \( \approx 1.5 imes 10^{-5} \) m²/s
- Critical Reynolds number for pipe flow: Re = 2000 (laminar) to 4000 (turbulent)
FAQs on Kinematic Viscosity Formula
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For the official NCERT syllabus and chapter details, refer to the NCERT official website.