krit.club logo

Gravitation - Archimedes’ Principle and Buoyancy

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Buoyancy is the upward force exerted by a fluid on an object immersed in it. This force is also referred to as upthrust (FBF_B).

•

Archimedes’ Principle states that when a body is immersed fully or partially in a fluid, it experiences an upward force that is equal to the weight of the fluid displaced by it.

•

The magnitude of the buoyant force depends on the density (ρ\rho) of the fluid and the volume (VV) of the immersed part of the body.

•

Density is defined as mass per unit volume. The SI unit of density is kg m−3kg\,m^{-3}.

•

Relative Density (R.D.R.D.) is the ratio of the density of a substance to the density of water. It is a unitless quantity.

•

An object sinks in a liquid if its density is greater than the density of the liquid (ρobject>ρliquid\rho_{object} > \rho_{liquid}).

•

An object floats in a liquid if its density is less than or equal to the density of the liquid (ρobject≤ρliquid\rho_{object} \le \rho_{liquid}).

📐Formulae

Density (ρ)=Mass (m)Volume (V)\text{Density } (\rho) = \frac{\text{Mass } (m)}{\text{Volume } (V)}

Relative Density=Density of substanceDensity of water\text{Relative Density} = \frac{\text{Density of substance}}{\text{Density of water}}

Buoyant Force (FB)=Vdisplaced×ρfluid×g\text{Buoyant Force } (F_B) = V_{displaced} \times \rho_{fluid} \times g

Apparent Weight=Actual Weight−Buoyant Force\text{Apparent Weight} = \text{Actual Weight} - \text{Buoyant Force}

💡Examples

Problem 1:

The volume of a 50 g50\,g sealed packet is 20 cm320\,cm^3. Will the packet float or sink in water if the density of water is 1 g cm−31\,g\,cm^{-3}? What will be the mass of the water displaced by this packet?

Solution:

Density of the packet ρ=mV=50 g20 cm3=2.5 g cm−3\rho = \frac{m}{V} = \frac{50\,g}{20\,cm^3} = 2.5\,g\,cm^{-3}. Since 2.5 g cm−3>1 g cm−32.5\,g\,cm^{-3} > 1\,g\,cm^{-3}, the packet will sink. Mass of water displaced = Volume of packet ×\times Density of water = 20 cm3×1 g cm−3=20 g20\,cm^3 \times 1\,g\,cm^{-3} = 20\,g.

Explanation:

The packet sinks because its density is higher than that of water. According to Archimedes' principle, it displaces a volume of water equal to its own volume.

Problem 2:

Relative density of silver is 10.810.8. The density of water is 103 kg m−310^3\,kg\,m^{-3}. What is the density of silver in SI units?

Solution:

Relative Density=Density of silverDensity of water\text{Relative Density} = \frac{\text{Density of silver}}{\text{Density of water}}. Therefore, Density of silver=Relative Density×Density of water=10.8×103 kg m−3=1.08×104 kg m−3\text{Density of silver} = \text{Relative Density} \times \text{Density of water} = 10.8 \times 10^3\,kg\,m^{-3} = 1.08 \times 10^4\,kg\,m^{-3}.

Explanation:

We use the definition of relative density to find the absolute density of the substance by multiplying the R.D.R.D. with the density of the reference substance (water).