9. (a) State the Principle of Archimedes - Leaving Cert Applied Maths - Question 9 - 2010

Given that 61% of the volume is immersed, the volume of the buoy that is submerged can be calculated as:

V_{sub} = rac{61}{100} V_{total}

The total volume of the hollow sphere is given by:

pi (1^3 - 0.8^3)$$ Calculating: $$V_{total} = rac{4}{3} pi (1 - 0.512) = rac{4}{3} pi (0.488)$$ Now, we can find the buoyancy using the principle: $$B = ho_{water} g V_{sub}$$ Where: - $ ho_{water} = 1000 ext{ kg/m}^3$ - $g$ is the acceleration due to gravity. Equating the forces gives:

ho_{shell} g V_{shell} = B$$

Now solving the equations leads to the density of the shell as:

ho_{shell} = rac{610}{0.488} = 1250 ext{ kg/m}^3$$

To find the relative density of the rod, consider the length of the immersed part to be $x$. Using the geometry of the scenario and the position of point P, we have:

$(1.5 - x) ext{ cos } 60° = 0.5$

Thus:

$x = 1.0 ext{ m}$

For moments about point P:

The buoyant force $B$ can be calculated as:

ho_{water} imes (1.0) imes g ext{ sin } 60°$$ And as the rod is in equilibrium: $$B = W rac{(1.5 - x)}{3} ext{ sin } 60°$$ Substituting values and solving gives the relative density: $$ ext{Relative Density} = rac{W}{B}$$

By using the equilibrium condition for forces acting on the rod:

Considering the forces and moments, we have:

$B + R = W$

Substituting the value of $B$ from part (i) leads to:

$R = W - B$

Using the simplified relationship, we can conclude:

R = rac{3W}{5}