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9. (a) A buoy in the form of a hollow spherical shell of external radius 0.7 m and internal radius 0.65 m floats in water - Leaving Cert Applied Maths - Question 9 - 2018

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9. (a) A buoy in the form of a hollow spherical shell of external radius 0.7 m and internal radius 0.65 m floats in water. The density of the material of the shell ... show full transcript

Worked Solution & Example Answer:9. (a) A buoy in the form of a hollow spherical shell of external radius 0.7 m and internal radius 0.65 m floats in water - Leaving Cert Applied Maths - Question 9 - 2018

Step 1

What percentage of the volume of the buoy is immersed?

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Answer

To find the percentage of the volume of the buoy immersed, we need to calculate the buoyant force and equate it to the weight of the buoy.

  1. Volume of the Buoy (V):

    The volume of the hollow sphere is given by:

    V=43π(R3r3)V = \frac{4}{3} \pi (R^3 - r^3) Where:

    • R=0.7mR = 0.7 \, \text{m} (external radius)
    • r=0.65mr = 0.65 \, \text{m} (internal radius)

    Hence:

    V=43π(0.730.653)V = \frac{4}{3} \pi (0.7^3 - 0.65^3)

  2. Weight of the Buoy (W):

    The weight can be calculated using the density of the shell:

    W=density×V=3430×VW = \text{density} \times V = 3430 \times V

  3. Buoyant Force (B):

    The buoyant force is given by the formula:

    B=Volume submerged×Density of water×gB = \text{Volume submerged} \times \text{Density of water} \times g

Since the buoy is floating:

B=WB = W

From the above, we can derive:

1000×(13π(0.7)2hs)=3430×(43π(0.7)343π(0.65)3)1000 \times \left(\frac{1}{3} \pi (0.7)^2 h_s \right) = 3430 \times \left(\frac{4}{3} \pi (0.7)^3 - \frac{4}{3} \pi (0.65)^3 \right)

  1. Percentage Immersed:

    Finally, the percentage of the volume immersed:

    Percentage=(VsV)×100\text{Percentage} = \left( \frac{V_s}{V} \right) \times 100 where VsV_s is the submerged volume.

Step 2

Find the relative density of the rod.

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Answer

To find the relative density of the rod:

  1. Identify the Geometry:

    • Using the geometry of the inclined rod and equilibrium conditions under gravitational forces, the balance of moments can be used.
  2. Use Trigonometry:

    • At an angle of 45°, cos(45°) = 12\frac{1}{\sqrt{2}} will be used in calculations.
  3. Relate lengths and weights:

    • The weight WW can be expressed in terms of the lengths:

    B=2x2h2x(W)(1)B = \frac{2x}{2h-2x}(W)(1)

  4. Connect to the Known Values:

    • Set up the equation for equilibrium involving the weights and distances:

    Whsin45=B×Whssin45W h \sin 45 = B \times \frac{W}{hs} \sin 45

  5. Final Calculation:

    • After rearranging and substituting the known values, compute the relative density using:

    ss2=2h(2hs)\text{s}_s^2 = 2h \cdot (2h - s) that leads to the conclusion on relative density as 12\frac{1}{2}.

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