Question

Calculate the work done in raising a stone of mass 6kg of relative density 2 immersed in water...

Answer 1

ANSWER:

The net work done will be 58.8 Joules

EXPLANATION :

: Specific gravity or relative density is defined as ratio of density of the object to ratio of density of water

: Specific gravity or relative density is defined as ratio of density of the object to ratio of density of waterHence , ρ/ρ₀ = 4

: Specific gravity or relative density is defined as ratio of density of the object to ratio of density of waterHence , ρ/ρ₀ = 4if V be the volume of the object, then according to Archimedes principle;

: Specific gravity or relative density is defined as ratio of density of the object to ratio of density of waterHence , ρ/ρ₀ = 4if V be the volume of the object, then according to Archimedes principle;Net weight of the object = weight - buoyant force

: Specific gravity or relative density is defined as ratio of density of the object to ratio of density of waterHence , ρ/ρ₀ = 4if V be the volume of the object, then according to Archimedes principle;Net weight of the object = weight - buoyant force= ρVg - ρ₀Vg

: Specific gravity or relative density is defined as ratio of density of the object to ratio of density of waterHence , ρ/ρ₀ = 4if V be the volume of the object, then according to Archimedes principle;Net weight of the object = weight - buoyant force= ρVg - ρ₀Vg= ρVg(1- ρ₀/ρ)

: Specific gravity or relative density is defined as ratio of density of the object to ratio of density of waterHence , ρ/ρ₀ = 4if V be the volume of the object, then according to Archimedes principle;Net weight of the object = weight - buoyant force= ρVg - ρ₀Vg= ρVg(1- ρ₀/ρ)= 4 x 9.8 x (1 - 1/4) [as ρV = mass of the object = 4 kg

: Specific gravity or relative density is defined as ratio of density of the object to ratio of density of waterHence , ρ/ρ₀ = 4if V be the volume of the object, then according to Archimedes principle;Net weight of the object = weight - buoyant force= ρVg - ρ₀Vg= ρVg(1- ρ₀/ρ)= 4 x 9.8 x (1 - 1/4) [as ρV = mass of the object = 4 kg= 4 x 9.8 x 3/4

: Specific gravity or relative density is defined as ratio of density of the object to ratio of density of waterHence , ρ/ρ₀ = 4if V be the volume of the object, then according to Archimedes principle;Net weight of the object = weight - buoyant force= ρVg - ρ₀Vg= ρVg(1- ρ₀/ρ)= 4 x 9.8 x (1 - 1/4) [as ρV = mass of the object = 4 kg= 4 x 9.8 x 3/4= 29.4 N

: Specific gravity or relative density is defined as ratio of density of the object to ratio of density of waterHence , ρ/ρ₀ = 4if V be the volume of the object, then according to Archimedes principle;Net weight of the object = weight - buoyant force= ρVg - ρ₀Vg= ρVg(1- ρ₀/ρ)= 4 x 9.8 x (1 - 1/4) [as ρV = mass of the object = 4 kg= 4 x 9.8 x 3/4= 29.4 NHence work done = weight x distance

: Specific gravity or relative density is defined as ratio of density of the object to ratio of density of waterHence , ρ/ρ₀ = 4if V be the volume of the object, then according to Archimedes principle;Net weight of the object = weight - buoyant force= ρVg - ρ₀Vg= ρVg(1- ρ₀/ρ)= 4 x 9.8 x (1 - 1/4) [as ρV = mass of the object = 4 kg= 4 x 9.8 x 3/4= 29.4 NHence work done = weight x distance = 29.4 x 2

: Specific gravity or relative density is defined as ratio of density of the object to ratio of density of waterHence , ρ/ρ₀ = 4if V be the volume of the object, then according to Archimedes principle;Net weight of the object = weight - buoyant force= ρVg - ρ₀Vg= ρVg(1- ρ₀/ρ)= 4 x 9.8 x (1 - 1/4) [as ρV = mass of the object = 4 kg= 4 x 9.8 x 3/4= 29.4 NHence work done = weight x distance = 29.4 x 2 = 58.8 Joules

: Specific gravity or relative density is defined as ratio of density of the object to ratio of density of waterHence , ρ/ρ₀ = 4if V be the volume of the object, then according to Archimedes principle;Net weight of the object = weight - buoyant force= ρVg - ρ₀Vg= ρVg(1- ρ₀/ρ)= 4 x 9.8 x (1 - 1/4) [as ρV = mass of the object = 4 kg= 4 x 9.8 x 3/4= 29.4 NHence work done = weight x distance = 29.4 x 2 = 58.8 JoulesHence the net work done will be 58.8 Joules

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