State the Principle of Archimedes - Leaving Cert Applied Maths - Question 9 - 2010

To find the volume of the metal, we can use the weight of the water displaced. The weight of the water displaced (B) is given by:

$B = \text{weight of metal} - \text{weight in water} = 14 \, \text{N} - 9 \, \text{N} = 5 \, \text{N}$

Using the density of water ( ho = 1000 , \text{kg m}^{-3}), the volume (V) can be calculated as:

$V = \frac{B}{\rho g} = \frac{5}{1000 \cdot 10} = 5 \times 10^{-4} \, \text{m}^3$

The relative density ( ho_{rel}) of the metal is calculated by comparing its density to that of water. First, we find the density of the metal:

$\text{Weight} = \rho V g \Rightarrow 14 = \rho \times (5 \times 10^{-4}) \times 10 \Rightarrow \rho = \frac{14}{5 \times 10^{-3}} = 2800 \, \text{kg m}^{-3}$

Thus, the relative density is:

$\rho_{rel} = \frac{2800}{1000} = 2.8$

For the cylindrical object:

1. The buoyant force (B) is given by:

$B = \rho_{liquid} \cdot V \cdot g = 0.9 \cdot 1000 \cdot \left(\pi (0.06)^2 (0.15)\right) \approx 15.27 \, \text{N}$

2. The weight (W) of the cylinder is:

$W = 0.7 \cdot 1000 \cdot \left(\pi (0.06)^2 (0.15)\right) \approx 11.88 \, \text{N}$

3. Using equilibrium conditions, the formula becomes:

$T + W = B\Rightarrow T = B - W = 15.27 - 11.88 = 3.39 \, \text{N}$