Class 8 Physics – Density Numerical Problems (Part 1, 2 & 3)

Title: Density – Numerical Problems (Class 8 Physics) – Step-by-Step
Complete Package

Question: 1

The density of air is 1.28 g litre⁻¹. Express it in : (a) g cm⁻³ (b) kg m⁻³.

Answer :

(a)g cm⁻³

1 liter (L) = 1000 cm³

1.28 g litre⁻¹= 1.28 g / litre

Given :

Density = 1.28 g / Litre.

Substitute the (L) value.

1.28 g / 1000 cm³ = 0.00128 g cm⁻³


Final Answer : 0.00128 g cm⁻³


(b) kg m⁻³

We need to convert g to kg
1g = 1/ 1000 kg
= 0.001 kg

Convert L to m³
1L = 1/ 1000 m³
= 0.001 m³
1.28 g litre⁻¹ = 1.28 g / L

Substitute gram and litre value
= 1.28× 0.001 kg / 0.001m³

, = 1.28 kg / m³
= 1.28 kg m⁻³

Final Answer: 1.28 kg m⁻³


Question: 2

The dimensions of a hall are 10m × 7m × 5m . If the density of air is 1.11 kgm⁻³ . Find
the mass of air in the hall.

Answer:

Given :

Dimension of the hall.

Length (L) = 10 m
Breadth (B)= 7m
Height ( H) = 5m

Density of the air D = 1.11kgm⁻³

Mass of the air in the hall = ?

Step:1 Find Volume of the hall

Formula

Volume= Length x Breadth × Height
= 10m ×7m×5m
= 350m³
V = 350m³

Step : 2 Find Mass of air

Formula for mass

Mass = Density x Volume