9. (a) A solid piece of metal has a weight of 23 N - Leaving Cert Applied Maths - Question 9 - 2016

The loss of weight when the metal is immersed in water is equal to the buoyant force, which is the weight of the water displaced.

The loss of weight = Weight in air - Weight in water = 23 N - 17 N = 6 N.

Using the formula for buoyant force:

ho g V$$ where: - $B = 6 \, \text{N}$ (the buoyant force) - $ ho = 1000 \, \text{kg m}^{-3}$ (density of water) - $g = 10 \, \text{m s}^{-2}$ (acceleration due to gravity) We can rearrange this to find the volume: $$V = \frac{B}{\rho g} = \frac{6}{1000 \times 10} = 0.0006 \, \text{m}^3.$

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

$\rho = \frac{m}{V}$

where:

• $m = 23 \, \text{N} / g = \frac{23}{10} = 2.3 \, \text{kg}$ (mass of the metal)
• $V = 0.0006 \, \text{m}^3$ (volume found previously)

Now substituting:

$$\rho = \frac{2.3}{0.0006} = 3833.3 , \text{kg m}^{-3}.$

In this case, we evaluate the forces acting on the sphere.

The weight of the sphere ($W$) can be calculated as:

$W = \rho_{sphere} \times V \times g$

The volume of the sphere ($V$) can be found using the formula for the volume of a sphere:

$$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (0.05)^3 = 0.0005236 , \text{m}^3.$

Thus:

$$W = 800 \times 0.0005236 \times 10 = 4.189 , \text{N}.$

Now, we also need to account for the buoyant force ($B$):

$$B = \rho_{liquid} \times V \times g = 1200 \times 0.0005236 \times 10 = 6.283 , \text{N}.$

The net upward force is the sum of the buoyant force and the tension in the string ($T$):

$T + W = B$

Substituting for $W$ and $B$ we derive:

$T + 4.189 = 6.283$

Solving for $T$:

$$T = 6.283 - 4.189 = 2.094 , \text{N}.$