If the ship grounds on a level bottom:

• A ground reaction force acts vertically upwards from the seabed. This force counteracts part of the ship's weight and alters the distribution of buoyancy forces.
• The ship’s centre of gravity (G) may appear to rise relative to the metacentre (M) because the upward ground reaction reduces the buoyancy force acting on the underwater volume.
• This virtual rise in G reduces GM, potentially leading to a list.
• If the list becomes excessive and the righting moment is insufficient, the ship may capsize.

If the ship grounds on a pinnacle:

• The ship experiences two forces at the ship's bottom:
  • A downward force due to the ship’s weight.
  • An upward reaction force is concentrated on the pinnacle.
• The resulting force between the grounding pressure and the ship’s centre of buoyancy shifts downward towards the pinnacle.
• This situation is similar to when the ship's stern touches the keel block in a dry dock.
• A virtual loss of GM occurs because the ship’s inclining moment may exceed the maximum righting lever.
• If the inclining moment is too great, the ship may develop an excessive list or even capsize.

(b) Given:

$$Displacement,\:\Delta=8000\:tonnes$$

$$TPC=15\:tonnes$$

$$Initial\:Draught=5.2m$$

$$Final\:Draught=3.2m$$

$$Ship\:KG=4.0m$$

$$KM=5.0m$$

To find GM

$$Uptrust,\:P=TPC\times Fall\:in\:water\:level\:\left(cm\right)$$

$$=15\times\left(520-320\right)$$

$$=15\times200$$

$$P=3000\:tonnes$$

To Find Virtual loss of GM:

$$Virtual\:loss\:of\:GM_1=\frac{P\times KM}{\Delta}$$

$$=\frac{3000\times5}{8000}$$

$$=\frac{15000}{8000}$$

$$GM_1=1.88m$$

Actual KM = 5.0m (given)

$$Virutal\:KM=Actual\:KM-Virtual\:loss\:of\:GM_1$$

$$=5.0-1.88$$

$$=3.12$$

Similarly, Actual KG = 4.0m (given)

$$New\:GM=Virtual\:KM-\:Actual\:KG$$

$$=3.12-4.0$$

$$=-0.88$$