The Terminal Velocity Formula gives the constant maximum speed attained by a falling object when the net force acting on it becomes zero. Expressed as \( v_t = \sqrt{\frac{2mg}{\rho C_d A}} \) in its general form, this formula is a core concept in Class 11 Physics (NCERT Chapter 10 — Mechanical Properties of Fluids). It is equally important for JEE Main, JEE Advanced, and NEET, where fluid mechanics questions regularly test this concept. This article covers the complete derivation, a formula cheat sheet, three progressive solved examples, CBSE exam tips, common mistakes, and JEE/NEET application patterns.
Key Terminal Velocity Formulas at a Glance
Quick reference for the most important terminal velocity formulas.
- Terminal velocity (Stokes’ Law): \( v_t = \frac{2r^2(\rho – \sigma)g}{9\eta} \)
- General terminal velocity: \( v_t = \sqrt{\frac{2mg}{\rho C_d A}} \)
- Stokes’ drag force: \( F_d = 6\pi\eta r v \)
- Buoyant force: \( F_b = \frac{4}{3}\pi r^3 \sigma g \)
- Weight of sphere: \( W = \frac{4}{3}\pi r^3 \rho g \)
- Net downward force at terminal velocity: \( W – F_b – F_d = 0 \)
- Terminal velocity proportionality: \( v_t \propto r^2 \)
What is Terminal Velocity Formula?
The Terminal Velocity Formula describes the maximum constant velocity reached by an object falling through a viscous fluid. When an object falls through a fluid such as air or water, three forces act on it simultaneously. These are gravity (downward), buoyancy (upward), and viscous drag (upward). Initially, gravity dominates and the object accelerates. As speed increases, the drag force also increases. Eventually, the upward forces exactly balance gravity. At this point, acceleration becomes zero and the object moves at a constant speed called terminal velocity.
In NCERT Class 11 Physics, Chapter 10 (Mechanical Properties of Fluids), terminal velocity is derived using Stokes’ Law for a small spherical body falling through a viscous medium. The Terminal Velocity Formula for a sphere is given by Stokes’ Law as:
\[ v_t = \frac{2r^2(\rho – \sigma)g}{9\eta} \]
Here, \( \rho \) is the density of the sphere, \( \sigma \) is the density of the fluid, \( r \) is the radius of the sphere, \( g \) is acceleration due to gravity, and \( \eta \) is the coefficient of viscosity. This formula is essential for CBSE board exams and competitive exams alike.
Terminal Velocity Formula — Expression and Variables
The standard Terminal Velocity Formula derived using Stokes’ Law is:
\[ v_t = \frac{2r^2(\rho – \sigma)g}{9\eta} \]
The general form applicable to any shape is:
\[ v_t = \sqrt{\frac{2mg}{\rho_{f} C_d A}} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( v_t \) | Terminal velocity | m/s |
| \( r \) | Radius of the falling sphere | m |
| \( \rho \) | Density of the falling body | kg/m³ |
| \( \sigma \) | Density of the fluid medium | kg/m³ |
| \( g \) | Acceleration due to gravity | m/s² |
| \( \eta \) | Coefficient of dynamic viscosity | Pa·s (or N·s/m²) |
| \( m \) | Mass of the falling body | kg |
| \( C_d \) | Drag coefficient (dimensionless) | Dimensionless |
| \( A \) | Cross-sectional area of the body | m² |
| \( \rho_f \) | Density of the fluid | kg/m³ |
Derivation of Terminal Velocity Formula
Consider a small sphere of radius \( r \) and density \( \rho \) falling through a viscous fluid of density \( \sigma \) and viscosity \( \eta \).
Step 1: Write the weight of the sphere: \( W = \frac{4}{3}\pi r^3 \rho g \)
Step 2: Write the buoyant force (Archimedes’ principle): \( F_b = \frac{4}{3}\pi r^3 \sigma g \)
Step 3: Write Stokes’ drag force: \( F_d = 6\pi\eta r v \)
Step 4: At terminal velocity, net force = 0:
\[ W – F_b – F_d = 0 \]
Step 5: Substitute and solve for \( v_t \):
\[ \frac{4}{3}\pi r^3 (\rho – \sigma)g = 6\pi\eta r v_t \]
\[ v_t = \frac{2r^2(\rho – \sigma)g}{9\eta} \]
This is the standard Terminal Velocity Formula as given in NCERT Class 11 Physics.
Complete Fluid Mechanics Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Terminal Velocity (Stokes) | \( v_t = \frac{2r^2(\rho – \sigma)g}{9\eta} \) | r=radius, ρ=body density, σ=fluid density, η=viscosity | m/s | Class 11, Ch 10 |
| Stokes’ Law (Drag Force) | \( F_d = 6\pi\eta r v \) | η=viscosity, r=radius, v=velocity | N | Class 11, Ch 10 |
| Buoyancy Force | \( F_b = \rho_f V g \) | ρ_f=fluid density, V=volume, g=gravity | N | Class 11, Ch 10 |
| Continuity Equation | \( A_1 v_1 = A_2 v_2 \) | A=cross-section area, v=flow velocity | m³/s | Class 11, Ch 10 |
| Bernoulli’s Equation | \( P + \frac{1}{2}\rho v^2 + \rho gh = \text{const} \) | P=pressure, ρ=density, v=velocity, h=height | Pa | Class 11, Ch 10 |
| Poiseuille’s Law | \( Q = \frac{\pi r^4 \Delta P}{8\eta L} \) | r=radius, ΔP=pressure diff, η=viscosity, L=length | m³/s | Class 11, Ch 10 |
| Reynolds Number | \( Re = \frac{\rho v D}{\eta} \) | ρ=density, v=velocity, D=diameter, η=viscosity | Dimensionless | Class 11, Ch 10 |
| Coefficient of Viscosity | \( \eta = \frac{F}{A (dv/dx)} \) | F=force, A=area, dv/dx=velocity gradient | Pa·s | Class 11, Ch 10 |
| Pressure in a Fluid | \( P = P_0 + \rho g h \) | P_0=surface pressure, ρ=density, h=depth | Pa | Class 11, Ch 10 |
| Surface Tension | \( T = \frac{F}{L} \) | F=force, L=length of surface | N/m | Class 11, Ch 10 |
Terminal Velocity Formula — Solved Examples
Example 1 (Class 11 Level — Direct Application)
Problem: A small steel ball of radius 1 mm and density 8000 kg/m³ falls through glycerine of density 1260 kg/m³ and viscosity 1.5 Pa·s. Calculate the terminal velocity of the ball. (Take g = 9.8 m/s².)
Given:
- Radius, \( r = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \)
- Density of steel, \( \rho = 8000 \text{ kg/m}^3 \)
- Density of glycerine, \( \sigma = 1260 \text{ kg/m}^3 \)
- Viscosity, \( \eta = 1.5 \text{ Pa·s} \)
- \( g = 9.8 \text{ m/s}^2 \)
Step 1: Write the Terminal Velocity Formula:
\[ v_t = \frac{2r^2(\rho – \sigma)g}{9\eta} \]
Step 2: Substitute the values:
\[ v_t = \frac{2 \times (10^{-3})^2 \times (8000 – 1260) \times 9.8}{9 \times 1.5} \]
Step 3: Simplify the numerator:
\[ v_t = \frac{2 \times 10^{-6} \times 6740 \times 9.8}{13.5} \]
\[ v_t = \frac{2 \times 10^{-6} \times 66052}{13.5} = \frac{0.132104}{13.5} \]
Step 4: Calculate the final answer:
\[ v_t \approx 9.79 \times 10^{-3} \text{ m/s} \approx 9.79 \text{ mm/s} \]
Answer
Terminal velocity \( v_t \approx 9.79 \times 10^{-3} \) m/s (approximately 9.79 mm/s).
Example 2 (Class 11-12 Level — Comparative Problem)
Problem: Two spheres A and B are made of the same material and fall through the same viscous fluid. The radius of sphere B is twice the radius of sphere A. Find the ratio of terminal velocity of B to terminal velocity of A.
Given:
- Both spheres: same material (ρ is the same), same fluid (σ and η are the same)
- \( r_B = 2r_A \)
Step 1: Write the Terminal Velocity Formula for each sphere:
\[ v_A = \frac{2r_A^2(\rho – \sigma)g}{9\eta} \]
\[ v_B = \frac{2r_B^2(\rho – \sigma)g}{9\eta} \]
Step 2: Divide \( v_B \) by \( v_A \):
\[ \frac{v_B}{v_A} = \frac{r_B^2}{r_A^2} \]
Step 3: Substitute \( r_B = 2r_A \):
\[ \frac{v_B}{v_A} = \frac{(2r_A)^2}{r_A^2} = \frac{4r_A^2}{r_A^2} = 4 \]
Step 4: This confirms that terminal velocity is proportional to \( r^2 \). Doubling the radius quadruples the terminal velocity.
Answer
The ratio \( v_B : v_A = 4 : 1 \). Sphere B reaches a terminal velocity four times greater than sphere A.
Example 3 (JEE/NEET Level — Concept + Calculation)
Problem: A spherical raindrop of radius 0.2 mm falls through air at 20°C. The density of water is 1000 kg/m³, the density of air is 1.2 kg/m³, and the viscosity of air is \( 1.8 \times 10^{-5} \) Pa·s. (i) Calculate the terminal velocity of the raindrop. (ii) State whether Stokes’ Law is applicable here by computing the Reynolds number. (Take g = 9.8 m/s².)
Given:
- \( r = 0.2 \text{ mm} = 2 \times 10^{-4} \text{ m} \)
- \( \rho = 1000 \text{ kg/m}^3 \), \( \sigma = 1.2 \text{ kg/m}^3 \)
- \( \eta = 1.8 \times 10^{-5} \text{ Pa·s} \)
- \( g = 9.8 \text{ m/s}^2 \)
Part (i) — Terminal Velocity:
Step 1: Apply the Terminal Velocity Formula:
\[ v_t = \frac{2r^2(\rho – \sigma)g}{9\eta} \]
Step 2: Substitute values:
\[ v_t = \frac{2 \times (2 \times 10^{-4})^2 \times (1000 – 1.2) \times 9.8}{9 \times 1.8 \times 10^{-5}} \]
Step 3: Compute numerator and denominator separately:
\[ \text{Numerator} = 2 \times 4 \times 10^{-8} \times 998.8 \times 9.8 = 2 \times 4 \times 10^{-8} \times 9788.24 \approx 7.83 \times 10^{-4} \]
\[ \text{Denominator} = 1.62 \times 10^{-4} \]
Step 4: Divide:
\[ v_t \approx \frac{7.83 \times 10^{-4}}{1.62 \times 10^{-4}} \approx 4.83 \text{ m/s} \]
Part (ii) — Reynolds Number Check:
Step 5: Apply the Reynolds number formula:
\[ Re = \frac{\rho_f v_t D}{\eta} = \frac{1.2 \times 4.83 \times 4 \times 10^{-4}}{1.8 \times 10^{-5}} \]
\[ Re = \frac{2.319 \times 10^{-3}}{1.8 \times 10^{-5}} \approx 128.8 \]
Step 6: Since Re ≈ 129, which is greater than 1 but less than 1000, Stokes’ Law gives an approximate result. For precise calculations at this Re, a correction factor (Oseen’s correction) should be applied. In NEET and JEE, Stokes’ Law is still used as an approximation for such problems.
Answer
(i) Terminal velocity \( v_t \approx 4.83 \) m/s. (ii) Reynolds number \( Re \approx 129 \). Stokes’ Law is approximately valid but not strictly accurate at this Re.
CBSE Exam Tips 2025-26
- Memorise the formula correctly: The Terminal Velocity Formula has \( (\rho – \sigma) \) in the numerator and \( 9\eta \) in the denominator. Students often write \( 6\eta \) by confusing it with Stokes’ drag formula. We recommend writing the derivation at least five times before the exam.
- Unit conversion is critical: Always convert radius to metres (not millimetres) before substituting into the formula. A common error is using r in mm, which gives an answer off by a factor of \( 10^6 \).
- Know the proportionality: CBSE frequently asks “how does terminal velocity change if radius is doubled?” Since \( v_t \propto r^2 \), doubling r quadruples \( v_t \). Similarly, \( v_t \propto (\rho – \sigma) \) and \( v_t \propto \frac{1}{\eta} \).
- Force balance diagram: In 3-mark and 5-mark questions, always draw the free-body diagram showing weight downward and both buoyancy and drag upward. This earns presentation marks in CBSE 2025-26 exams.
- Condition for terminal velocity: Clearly state in your answer that terminal velocity is reached when acceleration = 0, i.e., net force = 0. This is a frequently tested conceptual point.
- Negative \( (\rho – \sigma) \): If the body is less dense than the fluid (e.g., a bubble rising in water), \( (\rho – \sigma) \) becomes negative. This means the “terminal velocity” is directed upward. Our experts suggest practising at least two such problems before the board exam.
Common Mistakes to Avoid
| Common Mistake | Why It’s Wrong | Correct Approach |
|---|---|---|
| Using diameter instead of radius | The formula uses \( r \) (radius), not \( d \) (diameter). Using diameter inflates the answer by a factor of 4. | Always halve the given diameter to get radius before substituting. |
| Forgetting to subtract fluid density | Writing \( \rho g \) instead of \( (\rho – \sigma)g \) ignores the buoyant force. This gives a higher, incorrect terminal velocity. | Always use the effective density difference \( (\rho – \sigma) \) in the formula. |
| Confusing \( 9\eta \) with \( 6\eta \) | Stokes’ drag force has \( 6\pi\eta r v \), but the terminal velocity formula has \( 9\eta \) in the denominator after the full derivation. | Derive the formula once to understand where the 9 comes from. It arises from \( \frac{4/3 \times 2}{6} = \frac{4}{9} \). |
| Assuming terminal velocity is always downward | For a bubble or object less dense than the fluid, terminal velocity is directed upward. | Check the sign of \( (\rho – \sigma) \) to determine direction of terminal velocity. |
| Not checking units of viscosity | Viscosity is sometimes given in Poise (CGS). 1 Poise = 0.1 Pa·s. Using Poise directly in SI calculations gives wrong results. | Always convert viscosity to Pa·s (SI) before using the formula. |
JEE/NEET Application of Terminal Velocity Formula
In our experience, JEE aspirants encounter the Terminal Velocity Formula in two major contexts: direct numerical problems and conceptual/reasoning questions. NEET primarily tests the formula through straightforward substitution and proportionality-based MCQs. Here are the key application patterns to master.
Application Pattern 1: Proportionality-Based MCQs
JEE Main and NEET frequently ask questions like: “If the radius of a sphere is halved, by what factor does terminal velocity change?” Since \( v_t \propto r^2 \), halving the radius reduces terminal velocity by a factor of 4. Similarly, questions test \( v_t \propto (\rho – \sigma) \) and \( v_t \propto \frac{1}{\eta} \). Mastering these proportionalities allows you to solve such questions in under 30 seconds without full calculation.
Application Pattern 2: Multi-Body Comparison
JEE Advanced problems sometimes present two spheres of different materials falling through different fluids. You must compare their terminal velocities using the ratio method. Write the formula for each body and divide. Cancel common terms. This approach is fast and avoids errors. In our experience, students who practise at least 10 ratio-based problems score full marks on such questions.
Application Pattern 3: Integration with Buoyancy
The Terminal Velocity Formula connects directly to the Buoyancy Formula. JEE Advanced questions sometimes ask you to find the terminal velocity of a hollow sphere or a composite object. In such cases, you must first calculate the effective density of the object using its mass and volume, then apply the terminal velocity formula with this effective density. NEET 2023 and 2024 included questions linking buoyancy and terminal velocity in the same problem.
Application Pattern 4: Viscosity Determination (Experimental)
Both JEE and NEET include experimental-based questions on the falling sphere method for measuring viscosity. In this experiment, you measure the terminal velocity of a sphere in a fluid and rearrange the Terminal Velocity Formula to find \( \eta \):
\[ \eta = \frac{2r^2(\rho – \sigma)g}{9v_t} \]
Questions may ask about sources of error, the effect of container walls, or the effect of temperature on viscosity. We recommend revising NCERT Class 11 Chapter 10 experiments alongside this formula.
FAQs on Terminal Velocity Formula
Explore More Physics Formulas
Now that you have mastered the Terminal Velocity Formula, strengthen your fluid mechanics and physics foundation with these related resources on ncertbooks.net:
- Learn how objects float and sink with the Buoyancy Formula — essential for understanding the force balance in terminal velocity derivation.
- Deepen your understanding of fluid flow with Bernoulli’s Equation Formula — another key topic from NCERT Class 11 Chapter 10.
- Explore optics fundamentals with the Brewster’s Law Formula for a complete CBSE Class 12 Physics revision.
- Return to the complete Physics Formulas hub for a full list of NCERT-aligned formula articles for Class 6 to 12.
For official NCERT textbook content, refer to the NCERT Class 11 Physics Chapter 10 on the official NCERT website.