9. (a) State the principle of Archimedes - Leaving Cert Applied Maths - Question 9 - 2014

To find the volume of the metal, we use the difference in weight in air and water.

The buoyant force can be calculated as:

$B = W_{air} - W_{water} = 35 ext{ N} - 27 ext{ N} = 8 ext{ N}$

From Archimedes’ principle, we have:

ho_{fluid} imes g imes V$$ Using the density of water, $ ho_{water} = 1000 ext{ kg m}^{-3}$ and $g \approx 10 ext{ m s}^{-2}$: $$8 = 1000 imes 10 imes V$$ Thus, the volume of the metal is: $$V = \frac{8}{10000} = 0.0008 ext{ m}^3$$

The density of the metal can be found using the formula:

$\rho = \frac{W_{air}}{V}$

Substituting the values:

$\rho = \frac{35 ext{ N}}{0.0008 ext{ m}^3} = 43750 ext{ kg m}^{-3}$

To find the tension in the string, we first determine the buoyant forces on the cone.

1. Calculate the weight of the cone: Given the relative density of the cone is 0.9: $W = 0.9 \times \rho_{water} \times V_{cone}$ Where volume of cone, $V_{cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (0.04)^2 (0.12) = 0.00020106 ext{ m}^3$Therefore:$W = 0.9 \times 1000 \times 0.00020106 = 0.180954 ext{ N}$$

2. Calculate the buoyant force in the liquid of relative density 1.3: $B = 1.3 \times 1000 \times 0.00020106 = 0.261378 ext{ N}$

3. Using the equation of motion: $T + W = B$ $T = B - W$ $T = 0.261378 - 0.180954 = 0.080424 ext{ N}$

Thus, the tension in the string is approximately 0.80 N.