The Buoyancy Formula gives the upward force exerted by a fluid on any object submerged or floating in it, expressed as \( F_b = ho imes g imes V \), and it forms the foundation of fluid mechanics covered in NCERT Class 9 and Class 11 Physics. Also known as the Archimedes Principle, this formula is essential for CBSE board exams and appears regularly in JEE Main and NEET questions on fluids. This article covers the complete buoyancy formula, its derivation, a comprehensive formula sheet, three solved examples at progressive difficulty levels, CBSE exam tips for 2025-26, common mistakes to avoid, and JEE/NEET application strategies.
Key Buoyancy Formulas at a Glance
Quick reference for the most important buoyancy and fluid mechanics formulas.
- Buoyant Force: \( F_b = ho_f imes g imes V_{disp} \)
- Weight of displaced fluid: \( W_{fluid} = ho_f imes g imes V_{disp} \)
- Apparent Weight: \( W_{app} = W_{obj} – F_b \)
- Condition for floating: \( ho_{obj} \leq ho_f \)
- Fraction submerged: \( rac{V_{sub}}{V_{obj}} = rac{ ho_{obj}}{ ho_f} \)
- Pressure at depth h: \( P = P_0 + ho_f g h \)
- Archimedes Principle: \( F_b = W_{displaced fluid} \)
What is the Buoyancy Formula?
The Buoyancy Formula defines the upward force that a fluid exerts on an object placed inside it. This force is called the buoyant force. It acts in the direction opposite to gravity. The formula is rooted in Archimedes’ Principle, discovered by the Greek mathematician Archimedes in the third century B.C.
According to Archimedes’ Principle, the buoyant force on an object equals the weight of the fluid displaced by that object. This principle applies to all fluids, including liquids and gases. Water and air are both fluids. A ship floats on water because the buoyant force equals its weight. A balloon rises in air for the same reason.
In the NCERT curriculum, buoyancy is introduced in Class 9 Science, Chapter 10 — Gravitation. It is revisited in greater mathematical depth in Class 11 Physics, Chapter 10 — Mechanical Properties of Fluids. Understanding the Buoyancy Formula is non-negotiable for CBSE board exams and competitive tests like JEE and NEET.
The buoyant force depends on three factors: the density of the fluid, the volume of fluid displaced, and the acceleration due to gravity. It does not depend on the mass or density of the submerged object directly. This is a critical distinction that students often miss.
Buoyancy Formula — Expression and Variables
The standard expression for the Buoyancy Formula is:
\[ F_b = \rho_f \times g \times V_{disp} \]
Where \( F_b \) is the buoyant force in Newtons, \( \rho_f \) is the density of the fluid in kg/m³, \( g \) is the acceleration due to gravity (9.8 m/s²), and \( V_{disp} \) is the volume of fluid displaced by the object in m³.
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( F_b \) | Buoyant Force | Newton (N) |
| \( \rho_f \) | Density of the fluid | kg/m³ |
| \( g \) | Acceleration due to gravity | m/s² |
| \( V_{disp} \) | Volume of fluid displaced | m³ |
| \( W_{app} \) | Apparent weight of object in fluid | Newton (N) |
| \( W_{obj} \) | True weight of object in air | Newton (N) |
| \( \rho_{obj} \) | Density of the object | kg/m³ |
Derivation of the Buoyancy Formula
Consider a rectangular object submerged in a fluid of density \( \rho_f \). The object has a height \( h \) and cross-sectional area \( A \). The pressure at the bottom face is \( P_{bottom} = P_0 + \rho_f g (d + h) \). The pressure at the top face is \( P_{top} = P_0 + \rho_f g d \), where \( d \) is the depth of the top face.
The net upward pressure force is:
\[ F_b = (P_{bottom} – P_{top}) \times A = \rho_f g h A \]
Since \( h \times A = V_{disp} \) (the volume of the displaced fluid), we get:
\[ F_b = \rho_f \times g \times V_{disp} \]
This equals the weight of the displaced fluid, confirming Archimedes’ Principle. The derivation holds for objects of any shape because pressure acts perpendicular to every surface element.
Complete Fluid Mechanics Formula Sheet
Use this reference table to revise all key formulas related to buoyancy and fluid mechanics for CBSE and competitive exams.
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Buoyant Force (Archimedes’ Principle) | \( F_b = \rho_f g V_{disp} \) | \( \rho_f \)=fluid density, g=gravity, \( V_{disp} \)=displaced volume | N | Class 9, Ch 10; Class 11, Ch 10 |
| Apparent Weight in Fluid | \( W_{app} = W_{obj} – F_b \) | \( W_{obj} \)=true weight, \( F_b \)=buoyant force | N | Class 9, Ch 10 |
| Condition for Floating | \( \rho_{obj} \leq \rho_f \) | \( \rho_{obj} \)=object density, \( \rho_f \)=fluid density | kg/m³ | Class 9, Ch 10 |
| Fraction of Object Submerged | \( \frac{V_{sub}}{V_{obj}} = \frac{\rho_{obj}}{\rho_f} \) | \( V_{sub} \)=submerged volume, \( V_{obj} \)=total volume | Dimensionless | Class 11, Ch 10 |
| Hydrostatic Pressure | \( P = P_0 + \rho_f g h \) | \( P_0 \)=atmospheric pressure, h=depth | Pa (N/m²) | Class 11, Ch 10 |
| Density | \( \rho = \frac{m}{V} \) | m=mass, V=volume | kg/m³ | Class 9, Ch 10 |
| Relative Density (Specific Gravity) | \( RD = \frac{\rho_{obj}}{\rho_{water}} \) | \( \rho_{water} = 1000 \) kg/m³ | Dimensionless | Class 9, Ch 10 |
| Weight of Object | \( W = mg = \rho_{obj} \cdot V_{obj} \cdot g \) | m=mass, g=gravity | N | Class 9, Ch 10 |
| Pascal’s Law | \( \frac{F_1}{A_1} = \frac{F_2}{A_2} \) | F=force, A=area | Pa | 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 | \( P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant} \) | P=pressure, v=velocity, h=height | Pa | Class 11, Ch 10 |
Buoyancy Formula — Solved Examples
Work through these three examples carefully. They progress from basic Class 9-10 level to JEE-level difficulty.
Example 1 (Class 9-10 Level): Finding the Buoyant Force on a Submerged Stone
Problem: A stone of volume 500 cm³ is fully submerged in water. Calculate the buoyant force acting on the stone. (Density of water = 1000 kg/m³, g = 9.8 m/s²)
Given:
- Volume of stone = 500 cm³ = \( 500 \times 10^{-6} \) m³ = \( 5 \times 10^{-4} \) m³
- Density of water, \( \rho_f \) = 1000 kg/m³
- g = 9.8 m/s²
- Since the stone is fully submerged, \( V_{disp} = V_{stone} = 5 \times 10^{-4} \) m³
Step 1: Write the Buoyancy Formula:
\( F_b = \rho_f \times g \times V_{disp} \)
Step 2: Substitute the values:
\( F_b = 1000 \times 9.8 \times 5 \times 10^{-4} \)
Step 3: Calculate:
\( F_b = 1000 \times 9.8 \times 0.0005 = 4.9 \) N
Answer
The buoyant force acting on the stone = 4.9 N. This equals the weight of 500 cm³ of water displaced.
Example 2 (Class 11-12 Level): Apparent Weight and Fraction Submerged
Problem: A wooden block of mass 2 kg and volume \( 4 \times 10^{-3} \) m³ is placed in water (density = 1000 kg/m³). (a) Does the block float or sink? (b) What fraction of the block is submerged? (c) What is the buoyant force at equilibrium? (g = 10 m/s²)
Given:
- Mass of block, m = 2 kg
- Volume of block, \( V_{obj} = 4 \times 10^{-3} \) m³
- \( \rho_f \) (water) = 1000 kg/m³
- g = 10 m/s²
Step 1: Find the density of the wooden block:
\( \rho_{obj} = \frac{m}{V_{obj}} = \frac{2}{4 \times 10^{-3}} = 500 \) kg/m³
Step 2: Compare densities. Since \( \rho_{obj} = 500 \) kg/m³ \( < \rho_f = 1000 \) kg/m³, the block floats.
Step 3: Find the fraction submerged using the floating condition:
\( \frac{V_{sub}}{V_{obj}} = \frac{\rho_{obj}}{\rho_f} = \frac{500}{1000} = 0.5 \)
So 50% of the block is submerged.
Step 4: Find the buoyant force at equilibrium. At equilibrium, \( F_b = W_{obj} \):
\( F_b = mg = 2 \times 10 = 20 \) N
Verification: \( F_b = \rho_f \times g \times V_{sub} = 1000 \times 10 \times (0.5 \times 4 \times 10^{-3}) = 1000 \times 10 \times 2 \times 10^{-3} = 20 \) N ✓
Answer
(a) The block floats. (b) 50% of the block is submerged. (c) Buoyant force = 20 N.
Example 3 (JEE/NEET Level): Object in Two Fluids — Finding Unknown Density
Problem: A solid sphere weighs 0.5 N in air. When fully submerged in a liquid of density 800 kg/m³, its apparent weight is 0.1 N. Calculate (a) the volume of the sphere and (b) the density of the sphere. (g = 10 m/s²)
Given:
- Weight in air, \( W_{air} \) = 0.5 N
- Apparent weight in liquid, \( W_{app} \) = 0.1 N
- Density of liquid, \( \rho_f \) = 800 kg/m³
- g = 10 m/s²
Step 1: Find the buoyant force:
\( F_b = W_{air} – W_{app} = 0.5 – 0.1 = 0.4 \) N
Step 2: Use the Buoyancy Formula to find the volume of the sphere. Since the sphere is fully submerged, \( V_{disp} = V_{sphere} \):
\( F_b = \rho_f \times g \times V_{sphere} \)
\( 0.4 = 800 \times 10 \times V_{sphere} \)
\( V_{sphere} = \frac{0.4}{8000} = 5 \times 10^{-5} \) m³
Step 3: Find the mass of the sphere:
\( m = \frac{W_{air}}{g} = \frac{0.5}{10} = 0.05 \) kg
Step 4: Find the density of the sphere:
\( \rho_{sphere} = \frac{m}{V_{sphere}} = \frac{0.05}{5 \times 10^{-5}} = 1000 \) kg/m³
Answer
(a) Volume of sphere = \( 5 \times 10^{-5} \) m³ = 50 cm³. (b) Density of sphere = 1000 kg/m³ (same as water, so the sphere would just float at the surface in water).
CBSE Exam Tips 2025-26
- State Archimedes’ Principle in full: In CBSE board exams 2025-26, definition-based questions often carry 2 marks. Always state: “The buoyant force on an object equals the weight of the fluid displaced by it.”
- Convert units before substituting: We recommend converting cm³ to m³ (multiply by \( 10^{-6} \)) before using the Buoyancy Formula. Unit errors are the most common cause of lost marks.
- Use the floating condition clearly: When asked whether an object floats or sinks, compare \( \rho_{obj} \) and \( \rho_f \) explicitly. Write the inequality in your answer for full marks.
- Draw a free body diagram: For apparent weight problems, always draw the object with weight \( W \) downward and \( F_b \) upward. This prevents sign errors.
- Remember g = 9.8 m/s² for CBSE: Use 9.8 unless the problem specifies 10 m/s². Using the wrong value costs half-marks in numerical questions.
- Link to real-life examples: CBSE often asks application questions. Ships, submarines, icebergs, and hot-air balloons are classic examples linked to the Buoyancy Formula.
Common Mistakes to Avoid
Students frequently lose marks on buoyancy problems due to avoidable errors. Here are the most common ones and how to correct them.
- Mistake 1 — Using object volume instead of displaced volume: If an object is only partially submerged, \( V_{disp} \neq V_{obj} \). Only use \( V_{obj} \) when the object is fully submerged. For a floating object, use the submerged volume fraction.
- Mistake 2 — Confusing density of object with density of fluid: The Buoyancy Formula uses \( \rho_f \), the fluid density, not the object’s density. Many students accidentally substitute the object’s density.
- Mistake 3 — Forgetting unit conversion: Volumes given in cm³ or litres must be converted to m³ before calculation. 1 litre = \( 10^{-3} \) m³ and 1 cm³ = \( 10^{-6} \) m³.
- Mistake 4 — Assuming buoyancy depends on object mass: The buoyant force depends only on the volume of fluid displaced and the fluid’s density. A hollow steel ball and a solid steel ball of the same outer volume experience the same buoyant force.
- Mistake 5 — Ignoring atmospheric pressure in pressure problems: When calculating pressure at a depth, always add atmospheric pressure \( P_0 = 1.013 \times 10^5 \) Pa unless the problem asks for gauge pressure only.
JEE/NEET Application of Buoyancy Formula
In our experience, JEE aspirants encounter buoyancy in at least one question per paper, often combined with other concepts like surface tension, viscosity, or equilibrium. NEET questions tend to be more conceptual and application-based. Here is how the Buoyancy Formula appears in competitive exams.
Pattern 1: Object Suspended by a String in a Fluid
A common JEE problem shows an object fully submerged and held by a string attached to the bottom of the container. The tension in the string is:
\[ T = F_b – W_{obj} = \rho_f g V_{obj} – m_{obj} g \]
This applies when the object is less dense than the fluid (e.g., a balloon anchored underwater). If the object is denser, the string is attached to the top and \( T = W_{obj} – F_b \).
Pattern 2: Two-Liquid Systems
JEE Advanced frequently features objects floating at the interface of two immiscible liquids. If an object of density \( \rho_{obj} \) floats with volume \( V_1 \) in liquid 1 (density \( \rho_1 \)) and volume \( V_2 \) in liquid 2 (density \( \rho_2 \)), the equilibrium condition is:
\[ \rho_1 g V_1 + \rho_2 g V_2 = \rho_{obj} g (V_1 + V_2) \]
Our experts suggest practising at least five such problems before the JEE exam. This pattern tests both the Buoyancy Formula and algebraic manipulation skills.
Pattern 3: Apparent Weight and Weighing Scale Problems
NEET regularly asks about a body weighed on a spring balance while submerged. The reading on the spring balance gives the apparent weight. The relationship is:
\[ W_{app} = W_{air} \left(1 – \frac{\rho_f}{\rho_{obj}}\right) \]
This formula is derived directly from the Buoyancy Formula and is extremely useful for quick calculations in NEET MCQs. Memorise this form alongside the standard formula.
FAQs on Buoyancy Formula
Explore More Physics Formulas
Strengthen your understanding of related concepts with these resources on ncertbooks.net. The Fluid Mechanics Formula page covers Bernoulli’s equation, Pascal’s Law, and the continuity equation in full detail. For a deeper understanding of forces, visit our Normal Force Formula guide, which explains contact forces and their relationship to weight and buoyancy. You can also explore the Angular Velocity Formula for rotational mechanics. For a complete overview of all Physics formulas, visit our Physics Formulas hub. For official NCERT textbooks and syllabi, refer to the NCERT official website.