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Stainless steel is an alloy of iron, chromium and nickel! A piece of stainless steel consists of 70% iron, 20% chromium and 10% nickel by volume - Leaving Cert Applied Maths - Question 9 - 2012

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Stainless steel is an alloy of iron, chromium and nickel! A piece of stainless steel consists of 70% iron, 20% chromium and 10% nickel by volume. The relative densit... show full transcript

Worked Solution & Example Answer:Stainless steel is an alloy of iron, chromium and nickel! A piece of stainless steel consists of 70% iron, 20% chromium and 10% nickel by volume - Leaving Cert Applied Maths - Question 9 - 2012

Step 1

Find the relative density of stainless steel.

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Answer

To find the relative density of stainless steel, we can use the formula for the density of a mixture.

Let the volume of the stainless steel be 1 m³. The contributions of iron, chromium, and nickel can be calculated as follows:

  • Volume of iron = 70% of 1 m³ = 0.7 m³,
  • Volume of chromium = 20% of 1 m³ = 0.2 m³,
  • Volume of nickel = 10% of 1 m³ = 0.1 m³.

The respective densities are:

  • Density of iron, ρ₁ = 7.8 g/cm³ = 7800 kg/m³,
  • Density of chromium, ρ₂ = 7.2 g/cm³ = 7200 kg/m³,
  • Density of nickel, ρ₃ = 8.9 g/cm³ = 8900 kg/m³.

Using the formula:

ho_{stainless ext{ steel}} = rac{(V_1 imes ho_1) + (V_2 imes ho_2) + (V_3 imes ho_3)}{V_{total}}$$ Substituting the values:

ho_{stainless ext{ steel}} = rac{(0.7 imes 7800) + (0.2 imes 7200) + (0.1 imes 8900)}{1}$$

Calculating this gives:

ho_{stainless ext{ steel}} = 7790 ext{ kg/m}³$$ Thus, the relative density of stainless steel, $s = rac{ ho_{stainless ext{ steel}}}{ ho_{water}} = rac{7790}{1000} = 7.79$.

Step 2

Find (i) the length of the immersed part of the rod.

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Answer

Let the length of the immersed part of the rod be x. Based on Archimedes' principle, the weight of the liquid displaced (which equals the weight of the rod) must balance the weight of the rod:

The buoyant force, B, can be expressed as:

B=ximes0.9imesgB = x imes 0.9 imes g

Where 0.9 is the relative density of the liquid. However, since it is given:

B=0.756WB = 0.756W

Setting these equal gives:

0.756W=ximes0.9g0.756 W = x imes 0.9 g

For equilibrium:

W imes ext{sin} heta = B = x imes (2 - rac{x}{2}) imes g

Rearranging this gives:

Wimes2Wimesx=0.756Wimes0.9W imes 2 - W imes x = 0.756 W imes 0.9

Using the given dimensions:

x24x+336=0x^2 - 4x + 336 = 0

Solving the quadratic equation for x yields:

x=1.2extmx = 1.2 ext{ m}.

Step 3

Find (ii) the value of θ.

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Answer

Using the relationship:

ext{cos} θ = rac{0.4}{2 - 1} = rac{0.4}{1}

This simplifies to:

extcosθ=0.4 ext{cos} θ = 0.4

Thus, the angle θ is calculated as:

θ=extcos1(0.4)extorθ=60°.θ = ext{cos}^{-1}(0.4) ext{ or } θ = 60°.

Therefore, the final answer is:

θ=60°.θ = 60°.

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