9. (a) State the Principle of Archimedes - Leaving Cert Applied Maths - Question 9 - 2012

To find the volume of the metal, we can use the formula for buoyant force:

ho g V $$ Where: - $B$ is the buoyant force, - $ ho$ is the density of the fluid (water), - $g$ is the acceleration due to gravity, and - $V$ is the volume. Given that the weight in water is 21 N, the buoyant force is equal to: $$ B = 26 N - 21 N = 5 N $$ We substitute values into the formula: $$ 5 = (1000)(10)V $$ Solving for $V$: $$ V = 0.0005 \,m^3 $$

Relative density (RD) is calculated by the formula:

$RD = \frac{\text{Weight in air}}{\text{Weight in water}}$

For the metal, the weight in air is 26 N and in water is 21 N. Therefore:

$RD = \frac{26 N}{5 N} = 5.2$

To find the tension in the string, we start with the buoyant force acting on the cylinder:

$B = \rho_{liquid} g V_{cylinder}$

Given:

• The density of the liquid is $0.9 \times 1000 \, kg/m^3 = 900 \, kg/m^3$,
• The volume of the cylinder is:

$V_{cylinder} = \pi r^2 h = \pi (0.08)^2 (0.18) = 0.0036 \, m^3$

Now, substituting values into the buoyant force equation:

$B = 900 \times \pi (0.08)^2 (0.18) (10) = 106.368 \, N$

The weight of the cylinder (W) is:

$W = 3000 \times \pi (0.08)^2 (0.18) = 345.67 \, N$

Using the equilibrium condition:

$T + B = W$

Thus, $T = W - B$ Substituting the values we found:

$T = 345.67 - 106.368 \pi = 241.92 \approx 76 \, N$