A hollow spherical copper ball just floats in water completely immersed - Leaving Cert Applied Maths - Question 9 - 2015

Using the principle of buoyancy again, the weight of the ship when loaded can be set equal to the weight of the liquid displaced:

1. Find the volume of liquid displaced:

   $V_{displaced} = A \cdot h = 1250 \text{ m}^2 \cdot 0.375 \text{ m}$

   where A is the cross-sectional area.

2. The total weight when loaded with M tonnes is:

   $W = (6500 + M) \times 1000 \text{ kg} \cdot g$

3. Setting the weights equal:

   $(6500 + M) \cdot 1000 \text{ kg} \cdot g = 1030 \cdot (1250 \cdot 0.375) \cdot g$

4. Solving for M:

   $M = \frac{1030 \cdot (1250 \cdot 0.375)}{1000} - 6500 \approx 482.8125 \text{ tonnes}.$

1. Calculate the buoyancy in fresh water when the ship is loaded:

   $6982.8125 \text{ kg} \times 1000 \text{ g} = \rho_{fresh-water} \cdot V_{displaced};$

   where (V_{displaced} = 1250 \cdot h_{fresh} )

2. Using the density of fresh water:

   $6982.8125 \times 1000 = 1000 \cdot (1250 \cdot h_{fresh})$

3. Solve for (h_{fresh}):

   $h_{fresh} \approx 5.58625 \text{ m}.$

4. Now find the difference in height when moving from sea-water to fresh-water:

   $h_{sea} - h_{fresh} = 0.375 - h_{fresh} \approx 0.16 ext{ m}.$