Seam 6 OT3P

1. What is the weight of the vessel (W) that is considered to act downwards through G?
The centre of gravity
The center of buoyancy
The center of floatation
The centerline
What is the force of buoyancy that is equal to W and acts upwards through B?
The centre of gravity
The center of floatation
The center of buoyancy
The centerline
An empty compartment amidships be holed below the waterline to such an extent that the water may flow freely into and out of the compartment. A vessel holed in this way is said to be ________.
Punctured
Bilged
Drilled
Flooded
What will happen to the buoyancy provided by the bilged compartment?
Gained
Transferred
Conserved
Lost
What will happen to the drafts when the vessel now floats at the waterline W1L1, where it is again displacing its own weight of water?
The fore draft will increase which results trim by the head.
The aft draft will increase which results trim by the stern.
The draft will decrease evenly.
The draft will increase evenly.
Which of the following is true?
The volume of lost buoyancy (v) is made good by the volumes ‘y’ and ‘z’.
The volume of lost buoyancy (x) is made good by the volumes ‘v’ and ‘z’.
The volume of lost buoyancy (y) is made good by the volumes ‘x’ and ‘v’.
The volume of lost buoyancy (z) is made good by the volumes ‘v’ and ‘x’.
Which of the following shows the increased in draft due to bilging?
Increased in draft is equal to the area of intact waterplane over the volume of lost buoyancy.
Increased in draft is equal to the volume of lost buoyancy over the area of intact waterplane.
Increased in draft is equal to the area of the bilged waterplane over the volume of lost buoyancy.
Increased in draft is equal to the volume of lost buoyancy over the area of the bilged waterplane
If this compartment is bilged, what will be the final draught?
5.304 m
5.542 m
4.696 m
5.589 m
What will be the change in GM?
0.512 m reduction
0.423 mreduction
0.602 m reduction
0.741 mreduction
If this compartment is bilged, what will be the sinkage?
0.550 m
0.525 m
0.615 m
0.635 m
What will be the draughts fore and aft if this compartment is bilged?
10.625 m, 4.765 m
10.485 m, 4.345 m
10.655 m, 4.425 m
10.595 m, 4.855 m
What will be the new mean draught?
7.216 m
7.189 m
7.104 m
7.255 m
What is the value of the shift of buoyancy (BB1)?
0.658 m
0.586 m
0.699 m
0.558 m
What is the moment of inertia of the bilged waterplane using parallel axis theorem?
284135 m4
298567 m4
292457 m4
282088 m4
What will be the GM?
3.845 m
4.125 m
3.731 m
4.565 m
What will be the angle of heel?
9 degrees
10 degrees
11 degrees
12 degrees
 ____ is the amount of water that can enter a compartment or tank after it has been bilged.
Permeability
Bilging
Flooding
Grounding
The value of ____ does not change throughout the bilging process since there is no weight added, discharged or transferred.
GM
KM
KG
KB
Which of the following is true?
The lower the value of the permeability for a bilged compartment, the greater will be a ship’s loss of buoyancy when the ship is bilged.
The higher the value of the permeability for a bilged compartment, the lower will be a ship’s loss of buoyancy when the ship is bilged.
The higher the value of the permeability for a bilged compartment, the greater will be a ship’s loss of buoyancy when the ship is bilged.
The higher the value of the permeability for a bilged compartment, the greater will be a ship’s loss of permeability when the ship is bilged.
A box-shaped vessel 100 metres long x 20 metres wide x 12 metres deep is floating in salt water on an even keel at 6 metres draft. A forward compartment is 10 metres long, 12 metres wide and extends from the outer bottom to a watertight flat, 4 metres above the keel. The compartment contains cargo of permeability 25 per cent. What will be the new drafts if this compartment is bilged?
Aft 6.15 m, Fore 6.32m
Aft 6.25 m, Fore 6.42m
Aft 5.90 m, Fore 6.22m
Aft 5.75 m, Fore 6.12m
As the water level falls in the drydock there is no effect on the ship’s stability so long as the ship is completely waterborne, but after the stern lands on the blocks ____________.
The draft aft will decrease and the trim will change by the head.
The draft forward will decrease, and the trim will change by the stern.
Both a and b
Neither a nor b.
It is the interval of time between the stern post landing on the blocks and the ship taking the blocks overall.
Critical period
Upthrust phase
Grounding period
Docking phase
What will happen when positive effective metacentric height cannot be maintained throughout the drydocking period?
The ship will destroy the blocks since it cannot hold the weight of the ship
The ship will heel over and perhaps slip off the blocks with disastrous results.
The propeller will be destroyed since it will be the first to land when there is no positive stability.
All of the above
What is the initial hydrostatic draft?
4.100m
4.267m
4.302m
4.067m
What is the upthrust force?
438.2 T
428.3 T
483.2 T
432.8 T
What is the new displacement?
7369.9 T
7639.9 T
7939.9 T
7396.9 T
What is the value of KM from the new displacement?
9.788m
9.887m
9.798m
9.878m
What is the GM fluid?
0.724m
0.742m
0.774m
0.747m
What is the virtual loss of GM?
0.509m
0.806m
0.608m
0.580m
Which of the following will be course of action possible to increase the GM at Critical Instant?
Increase the GM by lowering weights or ballasting a double bottom.
Reduce the initial trim by transfer or ballasting so that the loss of GM is reduced.
Both a and b.
Neither a nor b.
What is the reduction of pressure energy which causes the vessel to sink further and the UKC reduces?
Squat
Trim
Sinkage
Grounding
What is the law that governs this phenomenon of the vessel’s sinkage due to the reduction of pressure?
Newton’s Law
Galileo’s Law
Bowditch’s Law
Bernoulli’s Law
Along with the reduction in UKC, what other particulars changes?
Vessel’s trim also changes
Vessel’s displacement also changes
Vessel’s center of gravity also changes
Vessel’s metacentric height also changes
The reduction in UKC varies approximately with the square of _____.
Ship's speed
Ship's inertia
Ship's draft
Ship's block coeffiecient
Vessels having higher block coefficient will also have a higher ______.
Trim
Squat
Freeboard
Speed
A container ship has a Cb of 0.575 and is proceeding at a speed of 6 kts. When static, the ship was on even keel.
 
What is the squat of the ship in open waters?
0.207m
0.306m
0.405m
0.504m

A container ship has a Cb of 0.575 and is proceeding at a speed of 6 kts. When static, the ship was on even keel.

What is the squat of the ship in shallow waters?

0.414m
0.612m
0.810m
0.504m

A container ship has a Cb of 0.575 and is proceeding at a speed of 6 kts. When static, the ship was on even keel.

What is the squat if the constant K value = 1.752?
0.343 m
0.383 m
0.363 m
0.323 m

A boxed shaped vessel floating on an even keel in saltwater has the following particulars:

Displacement  = 20,000 T             Length = 146 m                Breadth = 29.4 m             Draught = 6.52 m

KG = 9.90 m                       FSM=2621.43

The vessel has a ballast tank with a full capacity of 800 T and is empty. The vertical center of gravity of the tank is 2.0 m. An allision happened and the tank was flooded with seawater ¼ full.

What is the VCG of the ballast tank at ¼ full?
0.50m
1.01m
0.75m
0.25m

A boxed shaped vessel floating on an even keel in saltwater has the following particulars:

Displacement  = 20,000 T             Length = 146 m                Breadth = 29.4 m             Draught = 6.52 m

KG = 9.90 m                       FSM=2621.43

The vessel has a ballast tank with a full capacity of 800 T and is empty. The vertical center of gravity of the tank is 2.0 m. An allision happened and the tank was flooded with seawater ¼ full.

What is the distance between ship’s COG and the tank’s COG when flooded at ¼ full?
9.4m
9.5m
9.6m
9.7m

A boxed shaped vessel floating on an even keel in saltwater has the following particulars:

Displacement  = 20,000 T             Length = 146 m                Breadth = 29.4 m             Draught = 6.52 m

KG = 9.90 m                       FSM=2621.43

The vessel has a ballast tank with a full capacity of 800 T and is empty. The vertical center of gravity of the tank is 2.0 m. An allision happened and the tank was flooded with seawater ¼ full.

What is the change of position of ship’s center of gravity due to added weight?
0.903m
0.093m
0.309m
0.039m

A boxed shaped vessel floating on an even keel in saltwater has the following particulars:

Displacement  = 20,000 T             Length = 146 m                Breadth = 29.4 m             Draught = 6.52 m

KG = 9.90 m                       FSM=2621.43

The vessel has a ballast tank with a full capacity of 800 T and is empty. The vertical center of gravity of the tank is 2.0 m. An allision happened and the tank was flooded with seawater ¼ full.

What will be the KG after flooding?
9.993 m
9.708 m
9.807 m
9.939 m
The KG after flooding indicates that the GM increased due to __________.
Added weights below the ship’s center of gravity.
Free surface effect of the water inside the tank.
Lost of buoyancy of the tank.
In and out of water inside the tank.
What is the final KG after applying free surface effect?
9.397 m
9.937 m
9.739 m
9.973 m
This process considers the flooded water to be a weight added to a certain point in the ship. It provides more accurate results and is hence used by most stability analysis software?
Added Weight Method
Lost Buoyancy Method
Permeability Method
Free Surface Correction Method
How can the subdivision of a ship preserves it buoyancy?
A ship is divided transversely into a number of watertight compartments to restrict the flooding to one or more compartments in case of damage
A ship is divided longitudinally into a number of watertight compartments to restrict the flooding to one or more compartments in case of damage
This prevents progressive flooding across the entire ship’s breadth in case of a damage at any location.
All of the above
Which of the following indicates that the margin line would submerge if the following compartments were flooded together?
Both CD and DE.
Both DE and EF.
Both BC and CD.
Both AB and EF.
The margin line would remain above the waterline for simultaneous flooding in which following compartments?
Both CD and DE.
Both AB and BC.
Both BC and CD.
Both EF and FG.
It is the maximum portion of the ship’s length with the point ‘P’ at the center that can be completely flooded symmetrically without immersing the margin line
Floodable length
Permissible length
Permeable length
Longitudinal length
This method assumes that the damaged compartment does not contribute to the total buoyancy of the ship. Hence, the ship loses a part of its total waterplane, and its buoyancy, therefore reducing stability.
Added Weight Method
Lost Buoyancy Method
Permeability Method
Free Surface Correction Method
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