hi there
please help
Answers
Mass = Density × Volume
m = ρV
Multiplying Both sides with g,
mg = ρVg
Now, Here mg is the weight(Force Applied) of the System, Now, We can Write it as
W = ρVg
Where W is the weight of the System
Case I:- When the System is in Air,
W = 6000N
ρ₁(Iron) = 7.87 g/cm³ = 7.87 × 10³ kg/cm³
In this case we will calculate only weight of iron because air has no weight.
W = ρ₁V₁g
6000 = ρ₁V₁g
V₁ = 6000/ρ₁g
Case I:- When system is in Water
Given that weight of the sytem Cavities in this case is 4000N
We know that By Principal of Archimedes Weight = Volume of the water displaced. (Supposing that iron system has been completely submerged in water.)
If the volume of iron block+cavities = V,Then,
Weight reduced = ρ₂Vg
4000 = ρ₂Vg
V = 4000/ρ₂g
Hence,
Volume of cavities = V - V₁
= 4000/ρ₂g - 6000/ρ₁g
= 2000/g(2/ρ₂ - 3/ρ₁)
= 2000/g(2/(10³) - 3/7.87 × 10³)
= 2/g(2- (3/7.78))
=2/g( 2 - 0.38)
= 2/g × 1.62
= 3.24/9.8
= 0.33 m³
Explanation:
We know that,
Mass = Density × Volume
m = ρV
Multiplying Both sides with g,
mg = ρVg
Now, Here mg is the weight(Force Applied) of the System, Now, We can Write it as
W = ρVg
Where W is the weight of the System
Case I:- When the System is in Air,
W = 6000N
ρ₁(Iron) = 7.87 g/cm³ = 7.87 × 10³ kg/cm³
In this case we will calculate only weight of iron because air has no weight.
W = ρ₁V₁g
6000 = ρ₁V₁g
V₁ = 6000/ρ₁g
Case I:- When system is in Water
Given that weight of the sytem Cavities in this case is 4000N
We know that By Principal of Archimedes Weight = Volume of the water displaced. (Supposing that iron system has been completely submerged in water.)
If the volume of iron block+cavities = V,Then,
Weight reduced = ρ₂Vg
4000 = ρ₂Vg
V = 4000/ρ₂g
Hence,
Volume of cavities = V - V₁
= 4000/ρ₂g - 6000/ρ₁g
= 2000/g(2/ρ₂ - 3/ρ₁)
= 2000/g(2/(10³) - 3/7.87 × 10³)
= 2/g(2- (3/7.78))
=2/g( 2 - 0.38)
= 2/g × 1.62
= 3.24/9.8
= 0.33 m³