Question 1; The elongation of the steel is approximately 0.3123 mm
Question 2; The percentage the density of water increased in the deepest
ocean is approximately 6.4%
The strategy of obtaining the above solution is presented as follows;
Que. 1; The given parameters are;
The mass of the suspended block, m = 10 kg
The length of the steel, l = 2 m
The radius of the steel, r = 1 mm = 1 × 10⁻³ m
The modulus of elasticity of steel, E = 200 GPa = 200 × 10⁹ Pa
The stress, σ, on the steel due to the mass, m, is given as follows;
[tex]\mathbf{\sigma = \dfrac{F}{A}}[/tex]
Where;
F = The force acting on the steel = The weight of the mass
A = The cross sectional area of the steel = π·r²
∴ F = 10 kg × 9.81 m/s² = 98.1 N
A = π × (1 × 10⁻³)² = 3.14159 × 10⁻⁶ m²
Therefore;
σ = 98.1 N/(3.14159 × 10⁻⁶ m²) ≈ 31,226,226.2 Pa
We have;
[tex]\mathbf{ E = \dfrac{\sigma}{\epsilon}}[/tex]
From which we have;
[tex]\epsilon = \dfrac{\sigma}{E}[/tex]
Where;
∈ = The tensile strain = Δl/l
Δl = The elongation of the steel
Therefore;
∈ = 31,226,226.2/(200 × 10^9) = 0.00015613113
∴ Δl = 0.00015613113 × 2 m = 0.00031226226 m = 0.31226226 mm
The elongation of the steel, Δl = 0.31226226 mm ≈ 0.3123 mm
Question 2
The given parameters are;
The change in pressure per unit depth, Δp = 1.0 atm per 10 meters
The depth of the ocean = 12 km = 12,000 m
The compressibility = 5.0 × 10⁻⁵
The formula for compressibility, C, is presented as follows;
[tex]C = \dfrac{1}{V} \times \dfrac{\partial V}{\partial P}[/tex]
The change in pressure, [tex]\partial P[/tex] = 12,000 m × 1.0 atm/(10 m) = 1,200 atm
For a unit volume, V = 1 m³
We get;
[tex]5 \times 10^{-5} = \dfrac{1}{1} \times \dfrac{\partial V}{1,200}[/tex]
[tex]\partial V[/tex] = 5 × 10⁻⁵ m³/(atm) × 1,200 = 0.06 m³
The volume occupied 1 m³ at 12,000 km depth = V - [tex]\partial V[/tex]
∴ The volume occupied 1 m³ at 12,000 km depth = 1 m³ - 0.06 m³ = 0.94 m³
The percentage density increase, [tex]\partial[/tex]ρ% = (m/0.94 - m/1)/m/1 × 100
∴ (1/0.94 - 1/1)/1/1 × 100 ≈ 6.4%
The percentage increase in density ≈ 6.4%
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