Title: Class 8 physics numerical problem part 1
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: 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
