Density, relative density and buoyancy: Class 9 practice
Density = mass ÷ volume. Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Units of mass and volume
The idea behind density, relative density and buoyancy
Density = mass ÷ volume. Relative density = density of substance ÷ density of water. An object floats if its density is less than the liquid's; the buoyant force equals the weight of liquid displaced.
Step-by-step method
- Find density = m/V.
- Compare with the liquid (water = 1000 kg/m³ or 1 g/cm³).
- Buoyant force = ρ_liquid × V_displaced × g.
Worked example 1 (Basic)
An object of mass 204 g has volume 219 cm³. Find its density.
- ρ = 204/219 = 0.93 g/cm³
- Answer: 0.93 g/cm³
Worked example 2 (Standard)
An object of mass 131 g has volume 151 cm³. Find its density and say whether it floats in water.
- ρ = 131/151 = 0.87 g/cm³
- Less than 1 g/cm³ → floats
- Answer: 0.87 g/cm³, floats
Worked example 3 (Challenge)
A 900 cm³ block is fully submerged in water. Find the buoyant force (g = 10 m/s²).
- V = 0.0009 m³
- F = 1000 × 0.0009 × 10 = 9 N
- Answer: 9 N
Common mistakes and misconceptions
- Mixing g/cm³ and kg/m³
- Using the object's density for buoyant force
Practice questions
- (Basic) An object of mass 412 g has volume 276 cm³. Find its density.
- (Basic) An object of mass 313 g has volume 144 cm³. Find its density.
- (Standard) An object of mass 198 g has volume 377 cm³. Find its density and say whether it floats in water.
- (Standard) An object of mass 157 g has volume 40 cm³. Find its density and say whether it floats in water.
- (Challenge) A 100 cm³ block is fully submerged in water. Find the buoyant force (g = 10 m/s²).
- (Challenge) A 500 cm³ block is fully submerged in water. Find the buoyant force (g = 10 m/s²).
Answer key with full working
- Q1: ρ = 412/276 = 1.49 g/cm³ → Answer: 1.49 g/cm³
- Q2: ρ = 313/144 = 2.17 g/cm³ → Answer: 2.17 g/cm³
- Q3: ρ = 198/377 = 0.53 g/cm³ → Less than 1 g/cm³ → floats → Answer: 0.53 g/cm³, floats
- Q4: ρ = 157/40 = 3.92 g/cm³ → More than 1 g/cm³ → sinks → Answer: 3.92 g/cm³, sinks
- Q5: V = 0.0001 m³ → F = 1000 × 0.0001 × 10 = 1 N → Answer: 1 N
- Q6: V = 0.0005 m³ → F = 1000 × 0.0005 × 10 = 5 N → Answer: 5 N
Syllabus note
This skill is listed under "Gravitation" for Class 9 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is density, relative density and buoyancy in?
"Gravitation", Class 9 Physics (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.