Answer: SG = 2.67
Specific gravity of the sand is 2.67
Explanation:
Specific gravity = density of material/density of water
Given;
Mass of sand m = 100g
Volume of sand = volume of water displaced
Vs = 537.5cm^3 - 500 cm^3
Vs = 37.5cm^3
Density of sand = m/Vs = 100g/37.5 cm^3
Ds = 2.67g/cm^3
Density of water Dw = 1.00 g/cm^3
Therefore, the specific gravity of sand is
SG = Ds/Dw
SG = (2.67g/cm^3)/(1.00g/cm^3)
SG = 2.67
Specific gravity of the sand is 2.67
A cylinder contains 500 cm³ of water, and after 100 g of sand is poured, the water level rises to 537.5 cm³. The specific gravity of the sand is 2.67.
A cylinder contains 500 cm³ of water, and after 100 grams (m) of sand is poured, the water level rises to 537.5 cm³.
The volume of the sand is equal to the difference in the levels of water.
[tex]V = 537.5 cm^{3} - 500 cm^{3} = 37.5 cm^{3}[/tex]
The density of the sand (ρs) is the ratio of its mass to its volume.
[tex]\rho s = \frac{m}{V} = \frac{100g}{37.5 cm^{3} } = 2.67 g/cm^{3}[/tex]
The specific gravity of the sand (γ) is the ratio of the density of the sand (ρs) and the density of the water (ρw).
[tex]\gamma = \frac{\rho s }{\rho w} = \frac{2.67 g/cm^{3} }{1.00 g/cm^{3}} = 2.67[/tex]
A cylinder contains 500 cm³ of water, and after 100 g of sand is poured, the water level rises to 537.5 cm³. The specific gravity of the sand is 2.67.
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