Que. I : A mass of 10kg is suspended from the end of a steel of length 2m and radius 1mm, what is the elongation of the rod beyond its original length?

Que 2 : A pressure of sea water increases by 1.0atm for each 10metres increase in the depth. by what what percentage is the density of water increased in the deepest ocean of about 12km; compressibility = 5.0 × 10^-5 ​

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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