a solid weighs 120gf in air and 105gf when it is completely immersed in water calculate the relative density
Answers
Answer:
the relative density of the solid is 8
Explanation:
weight of the solid in air = 120 gf
weight of the solid when completely immersed in water = 105 gf
relative density of solid = weight of solid in air /weight of the solid in air - weight of solid in water
relative density of water
= 120/ 120 - 105 * 1 = 8
mass of the body =70 kg
volume of water displaced= 20,000cm or 0.02m
density in C.G.S = mass /volume = 70* 1000/20,000=3.5g cm
relative density of the solid = density in C.G.S without unit =3.5
Given,
Weight of solid in air = 120 gf
Weight of solid when completely immersed in water = 105 gf
To find,
The relative density.
Solution,
Weight of solid in air = 120 gf
Weight of solid when completely immersed in water = 105 gf
1. According to Archimedes principle, the buoyant force acting on a solid is equal to the weight of the fluid displaced by the solid.
2. Because of the upward buoyant force acting on the solid when it is submerged in water,the solid apparently weighs less in water compared to air.
3. The apparent decrease in the weight of a solid is equal to the weight of the fluid displaced by the solid.
∴ Weight of displaced water = Apparent decrease in weight
= 120 gf− 105 gf
= 15 gf
∴ Mass of displaced water = 15 g
Density of water = 1 g/cm^3
Now,
Volume of solid = Volume of water displaced
= ( Mass of displaced water/Density of Water )
= ( 15 g / 1 g/cm^3 )
= 15 cm^3
Now,
Actual weight of solid = weight of solid in air = 120 gf (since buoyant force due to air is negligible).
We have,
Mass of solid = 120 g
∴ Density of solid = ( Mass of solid / Volume of solid )
= ( 120 g / 15 g/cm^3 )
= 8 g/cm^3
∴ Relative Density = ( Density of solid/ Density of water )
= ( 8 g/cm^3 / 1 g/cm^3)
= 8.
∴ The Relative Density of the solid is 8.