The Specific Gravity Formula is a fundamental concept in physics and chemistry that compares the density of a substance to the density of a reference substance, usually water at 4°C. Expressed as \( SG = \frac{\rho_{substance}}{\rho_{water}} \), this dimensionless ratio is covered in NCERT Class 9 (Chapter 10: Gravitation) and Class 11 (Chapter 10: Mechanical Properties of Fluids). It is also highly relevant for JEE Main, JEE Advanced, and NEET, where fluid mechanics questions frequently test this concept. This article covers the formula, its derivation, a complete formula sheet, three solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Specific Gravity Formulas at a Glance
Quick reference for the most important specific gravity and related density formulas.
- Basic Specific Gravity: \ ( SG = \frac{\rho_{substance}}{\rho_{water}} \)
- Using Weight: \ ( SG = \frac{W_{substance}}{W_{equal\ volume\ of\ water}} \)
- Density formula: \ ( \rho = \frac{m}{V} \)
- Specific Gravity via buoyancy: \ ( SG = \frac{W_{air}}{W_{air} – W_{liquid}} \)
- Relative Density (same as SG): \ ( RD = \frac{\rho_{substance}}{\rho_{water\ at\ 4°C}} \)
- Density of water at 4°C: \ ( \rho_{water} = 1000\ kg/m^3 = 1\ g/cm^3 \)
What is the Specific Gravity Formula?
The Specific Gravity Formula defines the ratio of the density of a given substance to the density of a reference substance. In most practical cases, the reference substance is pure water at 4°C (277 K), where water has its maximum density of 1000 kg/m³ or 1 g/cm³. Because it is a ratio of two densities, specific gravity is a dimensionless quantity — it has no unit.
Specific gravity is also called Relative Density in NCERT textbooks. You will find this concept in NCERT Class 9 Science, Chapter 10 (Gravitation), and in NCERT Class 11 Physics, Chapter 10 (Mechanical Properties of Fluids). Both chapters emphasise that relative density helps compare how heavy a material is relative to water.
If the specific gravity of a substance is greater than 1, the substance is denser than water and will sink. If it is less than 1, the substance is less dense than water and will float. This simple rule has enormous practical importance in engineering, medicine, and daily life. For example, the specific gravity of ice is about 0.92, which is why icebergs float with roughly 92% of their volume submerged.
Specific Gravity Formula — Expression and Variables
The standard Specific Gravity Formula is written as:
\[ SG = \frac{\rho_{substance}}{\rho_{water}} \]
An equivalent form using mass and weight (since volume cancels) is:
\[ SG = \frac{W_{substance}}{W_{equal\ volume\ of\ water}} \]
When measuring specific gravity experimentally using Archimedes’ principle:
\[ SG = \frac{W_{air}}{W_{air} – W_{liquid}} \]
where \( W_{air} \) is the weight of the object in air and \( W_{liquid} \) is the apparent weight when submerged in the reference liquid.
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( SG \) or \( RD \) | Specific Gravity / Relative Density | Dimensionless (no unit) |
| \( \rho_{substance} \) | Density of the given substance | kg/m³ or g/cm³ |
| \( \rho_{water} \) | Density of water at 4°C | 1000 kg/m³ or 1 g/cm³ |
| \( W_{air} \) | Weight of object in air | Newton (N) |
| \( W_{liquid} \) | Apparent weight of object in liquid | Newton (N) |
| \( m \) | Mass of the substance | kilogram (kg) |
| \( V \) | Volume of the substance | m³ or cm³ |
Derivation of the Specific Gravity Formula
Start with the definition of density: \( \rho = \frac{m}{V} \).
For a substance of mass \( m_s \) and volume \( V \): \( \rho_{substance} = \frac{m_s}{V} \).
For water occupying the same volume \( V \): \( \rho_{water} = \frac{m_w}{V} \).
Taking the ratio of the two densities:
\[ SG = \frac{\rho_{substance}}{\rho_{water}} = \frac{m_s / V}{m_w / V} = \frac{m_s}{m_w} \]
Since weight \( W = mg \) and \( g \) is the same for both, the ratio of masses equals the ratio of weights. Therefore, specific gravity also equals the ratio of the weight of the substance to the weight of an equal volume of water. The volume \( V \) cancels, confirming that specific gravity is dimensionless.
Complete Fluid Mechanics Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Specific Gravity (Relative Density) | \( SG = \frac{\rho_{s}}{\rho_{w}} \) | \(\rho_s\)=density of substance, \(\rho_w\)=density of water | Dimensionless | Class 9, Ch 10 / Class 11, Ch 10 |
| Density | \( \rho = \frac{m}{V} \) | m=mass, V=volume | kg/m³ | Class 9, Ch 10 |
| Buoyant Force (Archimedes’ Principle) | \( F_b = \rho_{fluid} \cdot V_{displaced} \cdot g \) | \(\rho_{fluid}\)=fluid density, V=displaced volume, g=9.8 m/s² | Newton (N) | Class 9, Ch 10 / Class 11, Ch 10 |
| Pressure in a Fluid | \( P = \rho g h \) | \(\rho\)=density, g=gravity, h=depth | Pascal (Pa) | Class 11, Ch 10 |
| Bernoulli’s Equation | \( P + \frac{1}{2}\rho v^2 + \rho g h = constant \) | P=pressure, v=velocity, h=height | Pa | Class 11, Ch 10 |
| Equation of Continuity | \( A_1 v_1 = A_2 v_2 \) | A=cross-sectional area, v=flow velocity | m³/s | Class 11, Ch 10 |
| Pascal’s Law | \( \frac{F_1}{A_1} = \frac{F_2}{A_2} \) | F=force, A=area | Pa | Class 11, Ch 10 |
| Stokes’ Law (Terminal Velocity) | \( F = 6\pi \eta r v \) | \(\eta\)=viscosity, r=radius, v=velocity | N | Class 11, Ch 10 |
| Surface Tension | \( T = \frac{F}{l} \) | F=force, l=length of surface | N/m | Class 11, Ch 10 |
| Capillary Rise | \( h = \frac{2T\cos\theta}{\rho g r} \) | T=surface tension, \(\theta\)=contact angle, r=radius | m | Class 11, Ch 10 |
Specific Gravity Formula — Solved Examples
Example 1 (Class 9-10 Level)
Problem: A piece of iron has a density of 7874 kg/m³. Find its specific gravity. Will it sink or float in water?
Given: \( \rho_{iron} = 7874 \) kg/m³, \( \rho_{water} = 1000 \) kg/m³
Step 1: Write the Specific Gravity Formula: \( SG = \frac{\rho_{substance}}{\rho_{water}} \)
Step 2: Substitute the values: \( SG = \frac{7874}{1000} \)
Step 3: Calculate: \( SG = 7.874 \)
Step 4: Since \( SG > 1 \), iron is denser than water. It will sink.
Answer
The specific gravity of iron is 7.874. Since it is greater than 1, iron sinks in water.
Example 2 (Class 11-12 Level)
Problem: A solid object weighs 120 N in air and 75 N when fully submerged in water. Calculate its specific gravity using Archimedes’ principle.
Given: \( W_{air} = 120 \) N, \( W_{water} = 75 \) N
Step 1: Use the buoyancy-based Specific Gravity Formula: \( SG = \frac{W_{air}}{W_{air} – W_{water}} \)
Step 2: Calculate the loss of weight (buoyant force): \( W_{air} – W_{water} = 120 – 75 = 45 \) N
Step 3: Substitute into the formula: \( SG = \frac{120}{45} \)
Step 4: Simplify: \( SG = 2.\overline{6} \approx 2.67 \)
Step 5: Verify: \( SG > 1 \), so the object sinks. Its density is \( 2.67 \times 1000 = 2670 \) kg/m³, which is consistent with materials like aluminium alloys.
Answer
The specific gravity of the solid is approximately 2.67. Its density is 2670 kg/m³.
Example 3 (JEE/NEET Level)
Problem: A wooden block floats in water with 60% of its volume submerged. When the same block is placed in an unknown liquid, it floats with only 40% of its volume submerged. Find the specific gravity of the unknown liquid.
Given: Volume submerged in water \( = 0.60 V \), Volume submerged in unknown liquid \( = 0.40 V \), \( \rho_{water} = 1000 \) kg/m³
Step 1: For a floating object, buoyant force equals the weight of the object.
Step 2: In water: \( \rho_{water} \times 0.60V \times g = m_{block} \times g \)
Therefore: \( m_{block} = 0.60V \times 1000 = 600V \) kg (per unit volume)
Step 3: Density of the wooden block: \( \rho_{block} = \frac{m_{block}}{V} = 600 \) kg/m³
Step 4: In the unknown liquid: \( \rho_{liquid} \times 0.40V \times g = m_{block} \times g \)
Therefore: \( \rho_{liquid} \times 0.40V = 600V \)
Step 5: Solve for \( \rho_{liquid} \): \( \rho_{liquid} = \frac{600}{0.40} = 1500 \) kg/m³
Step 6: Specific gravity of the unknown liquid: \( SG = \frac{1500}{1000} = 1.5 \)
Answer
The specific gravity of the unknown liquid is 1.5. Its density is 1500 kg/m³, which is consistent with a concentrated salt solution or a light oil-based liquid.
CBSE Exam Tips 2025-26
- Remember it is dimensionless: In CBSE 2025-26 exams, a very common 1-mark question asks for the unit of specific gravity. The answer is always “no unit” or “dimensionless.” Never write kg/m³ as the unit.
- Use the correct reference temperature: Always state that water’s density is measured at 4°C (277 K). CBSE mark schemes specifically look for this condition in 3-mark descriptive answers.
- Distinguish SG from density: Specific gravity is numerically equal to density in CGS units (g/cm³) but not in SI units. We recommend writing both the SG value and the density in kg/m³ in long-answer questions for full marks.
- Archimedes’ principle link: CBSE frequently combines specific gravity with Archimedes’ principle. Practise the formula \( SG = \frac{W_{air}}{W_{air} – W_{liquid}} \) thoroughly for 5-mark problems.
- Float or sink rule: Always conclude your answer by stating whether the object floats (SG < 1) or sinks (SG > 1). CBSE mark schemes award a separate mark for this conclusion.
- Numerical accuracy: In 2025-26 board exams, carry at least two decimal places in your specific gravity calculations. Round off only in the final step.
Common Mistakes to Avoid
- Assigning units to specific gravity: Many students write the unit as kg/m³. This is incorrect. Specific gravity is always dimensionless because it is a ratio of two densities with identical units that cancel out.
- Using water density at room temperature: The reference density of water is 1000 kg/m³ at exactly 4°C. Using a different temperature value (such as 25°C, where \( \rho \approx 997 \) kg/m³) gives a slightly different answer and may cost marks.
- Confusing specific gravity with specific weight: Specific weight is \( \gamma = \rho g \) and has units of N/m³. Specific gravity is dimensionless. These are two different quantities, and mixing them is a common error.
- Forgetting that SG = RD: In NCERT books, the term “relative density” is used. In engineering and competitive exams, “specific gravity” is more common. Both mean exactly the same thing. Do not treat them as different quantities.
- Incorrectly applying the buoyancy formula: When using \( SG = \frac{W_{air}}{W_{air} – W_{liquid}} \), students sometimes subtract \( W_{air} \) from \( W_{liquid} \) (reversing the denominator). Always subtract the apparent weight in liquid from the weight in air.
JEE/NEET Application of Specific Gravity Formula
In our experience, JEE aspirants encounter the Specific Gravity Formula in at least one question per paper, typically embedded within a fluid mechanics or Archimedes’ principle problem. NEET also tests this concept under the “Properties of Bulk Matter” unit, often in the context of biological fluids or medical instruments like the hydrometer.
Application Pattern 1: Floating Body Problems
JEE problems often describe a body floating at the interface of two liquids and ask for the specific gravity of the body. The key principle is that the total buoyant force equals the weight of the body. If a body of specific gravity \( SG \) floats with fraction \( f \) submerged in a liquid of specific gravity \( SG_L \), then:
\[ SG = f \times SG_L \]
This elegant result allows rapid calculation without finding actual densities.
Application Pattern 2: Hydrometer Calibration (NEET)
A hydrometer is a device that directly measures specific gravity. NEET questions ask students to explain why the hydrometer floats higher in denser liquids. The answer lies in the Specific Gravity Formula: a denser liquid (higher SG) needs less displaced volume to support the same weight, so the hydrometer rises.
Application Pattern 3: Mixture Density Problems
JEE Advanced sometimes asks for the effective specific gravity of a mixture. If two liquids of specific gravities \( SG_1 \) and \( SG_2 \) are mixed in equal volumes, the effective specific gravity is:
\[ SG_{mix} = \frac{SG_1 + SG_2}{2} \]
If mixed in equal masses, the result is the harmonic mean. Our experts suggest practising both cases, as JEE examiners deliberately switch between the two to test conceptual clarity.
FAQs on Specific Gravity Formula
Explore more related topics on ncertbooks.net to strengthen your understanding of fluid mechanics and properties of matter. Read our detailed article on the Buoyancy Formula to understand how buoyant force connects to specific gravity. For a broader understanding of fluid dynamics, visit our Bernoulli’s Equation Formula guide. You can also browse the complete Physics Formulas hub for all Class 9-12 and JEE/NEET formula sheets. For official NCERT textbook references, visit ncert.nic.in.