A solid piece of metal has a weight of 15 N - Leaving Cert Applied Maths - Question 9 - 2018

To find the volume of the metal, we first calculate the buoyant force (B):

$B = ext{Weight in air} - ext{Weight in water} = 15 ext{ N} - 11 ext{ N} = 4 ext{ N}$

Using the formula for buoyancy:

ho_{water} imes g imes V$$ where: - $ ho_{water} = 1000 ext{ kg m}^{-3}$ (density of water) - $g$ is the acceleration due to gravity (approximately $9.81 ext{ m s}^{-2}$) Substituting values: $$4 ext{ N} = 1000 ext{ kg m}^{-3} imes 9.81 ext{ m s}^{-2} imes V$$ Solving for V, $$V = rac{4 ext{ N}}{1000 ext{ kg m}^{-3} imes 9.81 ext{ m s}^{-2}} = 4 imes 10^{-4} ext{ m}^3$$

To find the density ($\rho$) of the metal, we use the formula:

$\rho = \frac{\text{Weight}}{\text{Volume} \times g}$

where:

• Weight of the metal = 15 N
• Volume = $4 \times 10^{-4} m^3$

Thus,

$\rho = \frac{15 \text{ N}}{4 \times 10^{-4} \text{ m}^3 \times 9.81 \text{ m s}^{-2}} = 3750 \text{ kg m}^{-3}$

Using the equilibrium condition for the cylinder:

$T + W = B$

Where:

• T is the tension in the string
• W is the weight of the cylinder (Weight = density x volume x g)
• B is the buoyant force (calculated based on its relative density)

First, we calculate the weight of the cylinder:

The volume of the cylinder:

$V = \pi r^2 h = \pi (0.03 ext{ m})^2 (0.12 ext{ m})$

Calculating this gives: $V = \pi (0.0009) \times (0.12) \approx 3.39 \times 10^{-4} \text{ m}^3$

Weight of the cylinder: $W = 0.6 \times 1000 \text{ kg/m}^3 \times g \times V\n= 700 \text{ g} \cdot 9.81 \approx 6.87 ext{ N}$

Next, we can use the buoyancy in the liquid:

$B = \text{Relative Density} \times \text{Volume} \times g = 1.3 \times W = 1.3 imes 6.87 ext{ N} \approx 8.94 ext{ N}$

Putting this into the force balance equation:

$T + 6.87 ext{ N} = 8.94 ext{ N}$

Solving gives: $T = 8.94 ext{ N} - 6.87 ext{ N} = 2.07 ext{ N}$