Ship displacement = 10000 tonne, sea water density 1025 kg/m3, draught 6 m.

KG = 6.25 m, KB = 3.325 m, BM = 4.865 m, TPC = 20.5.

A rectangular tank 12 m long, 10 m wide, 6 m deep, with its base 1 m above the keel, is filled to a depth of 5 m with sea water ballast.

Step 1 - Weight of ballast added.

Volume of ballast = 12 x 10 x 5 = 600 m3.

Weight of ballast = 600 x 1.025 = 615 tonne.

New displacement = 10000 + 615 = 10615 tonne.

Step 2 - Change in draught.

Increase in draught = weight added/TPC = 615/20.5 = 30 cm = 0.30 m.

New draught = 6 + 0.30 = 6.30 m.

Step 3 - New KB.

The ship is wall-sided, so the increase in KB = increase in draught/2 = 0.30/2 = 0.15 m.

New KB = 3.325 + 0.15 = 3.475 m.

Step 4 - New BM.

BM = I/V. For a wall-sided ship, I (second moment of area of waterplane) is constant, so BM varies inversely with volume.

Original volume V1 = 10000/1.025 = 9756.1 m3.

Original BM = 4.865 m, so I = BM x V1 = 4.865 x 9756.1 = 47463 m4.

New volume V2 = 10615/1.025 = 10356.1 m3.

New BM = I/V2 = 47463/10356.1 = 4.583 m.

Step 5 - New KG.

The ballast is added at a height above the keel. The centre of gravity of the ballast is at the centre of the 5 m depth, i.e. at 1 + 5/2 = 3.5 m above the keel.

New KG = (10000 x 6.25 + 615 x 3.5)/10615 = (62500 + 2152.5)/10615 = 64652.5/10615 = 6.091 m.

Step 6 - New KM and GM.

New KM = new KB + new BM = 3.475 + 4.583 = 8.058 m.

New GM = new KM - new KG = 8.058 - 6.091 = 1.967 m.

Step 7 - Original GM.

Original KM = KB + BM = 3.325 + 4.865 = 8.190 m.

Original GM = KM - KG = 8.190 - 6.25 = 1.940 m.

Step 8 - Change in effective metacentric height.

Change in GM = new GM - original GM = 1.967 - 1.940 = +0.027 m.

Answer: The effective metacentric height increases by 0.027 m (GM increases from 1.940 m to 1.967 m).