I85PREP
⌘ K
HKIE Professional Assessment - Marine & Naval Architecture Discipline study materials
HKIE Professional Assessment - Marine & Naval Architecture Discipline study resources

HKIE Professional Assessment - Marine & Naval Architecture Discipline practice questions

One question at a time. Build your HKIE Professional Assessment - Marine & Naval Architecture Discipline understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Ship Design and Naval Architecture Fundamentals

In the IMO load line regime, which dimension of the ship is used to determine whether a vessel is assigned a freeboard under the 1966 Load Line Convention rather than being treated as a special case?

  1. Waterline length at the lightship condition, measured between the intersections of the hull with the light waterline
  2. Length overall measured to the extreme ends of the hull including any bulbous bow projection
  3. Length L measured on the assigned summer load line, taken as 96 per cent of the total length on that waterline, or the length from the fore side of the stem to the rudder stock if that is greater Correct answer.
  4. Registered length as recorded in the tonnage certificate, measured between perpendiculars at the design draught

Explanation

The Load Line Convention defines length L on the summer load line at 85 per cent of the least moulded depth, taking 96 per cent of the waterline length or the stem-to-rudder-stock distance if larger. This definition sets the threshold above which full freeboard assignment and tabular freeboards apply. Registered length and length overall serve different statutory purposes and are not the Convention's parameter.

Question 2Ship Design and Naval Architecture Fundamentals

A displacement hull is described as having a block coefficient of 0.82, a midship section coefficient of 0.98 and a waterplane coefficient of 0.85. Which statement about the derived coefficients is correct?

  1. The prismatic coefficient is about 0.84 and the vertical prismatic coefficient is about 0.96 Correct answer.
  2. The prismatic coefficient is about 0.98 and the vertical prismatic coefficient is about 0.82
  3. The prismatic coefficient is about 0.96 and the vertical prismatic coefficient is about 0.84
  4. The prismatic coefficient is about 0.80 and the vertical prismatic coefficient is about 0.87

Explanation

The prismatic coefficient equals block coefficient divided by midship coefficient, giving 0.82/0.98 which is approximately 0.84. The vertical prismatic coefficient equals block coefficient divided by waterplane coefficient, giving 0.82/0.85 which is approximately 0.96. These ratios express how the hull volume is distributed longitudinally and vertically relative to the midship section and waterplane.

Question 3Ship Design and Naval Architecture Fundamentals

Which statement best describes the purpose of the damage stability requirements in SOLAS Chapter II-1 for a passenger ship?

  1. To ensure the ship survives a specified extent of flooding with residual stability and freeboard sufficient for evacuation Correct answer.
  2. To require that all watertight bulkheads be spaced at equal intervals along the ship's length
  3. To limit the maximum deadweight so that the ship can always reach the nearest port under its own power
  4. To guarantee that the ship remains upright with zero heel after any single compartment is flooded

Explanation

SOLAS damage stability rules define an assumed extent of damage and require the ship to retain adequate residual stability, righting energy and freeboard so that it can be evacuated safely. The criteria are probabilistic or deterministic depending on ship type and date of construction. They do not demand zero heel, equal bulkhead spacing, or a deadweight limit.

Question 4Ship Design and Naval Architecture Fundamentals

A ship has a metacentric height of 0.60 m and a displacement of 12 000 tonnes. Using the small-angle approximation, what is the righting moment when the ship heels to 6 degrees?

  1. About 4 320 tonne-force metres
  2. About 123 tonne-force metres
  3. About 7 380 tonne-force metres
  4. About 738 tonne-force metres Correct answer.

Explanation

The righting moment is displacement times GM times the sine of the heel angle. With 12 000 tonnes, 0.60 m and sin 6 degrees of about 0.1045, the product is roughly 738 tonne-force metres. The small-angle formula is valid at this heel because the metacentre can be treated as fixed.

Question 5Ship Design and Naval Architecture Fundamentals

Which of the following best explains why a ship's transverse metacentric height falls as draught increases towards the deck edge immersion?

  1. The block coefficient increases, which directly reduces the metacentric radius
  2. The vertical centre of gravity rises automatically with draught because of the added hull steel
  3. The centre of buoyancy rises faster than the centre of gravity, reducing the righting lever
  4. The waterplane second moment of area grows more slowly than the displaced volume, so BM decreases Correct answer.

Explanation

The metacentric radius BM equals the waterplane second moment of area divided by the displaced volume. As draught increases, volume grows steadily while the waterplane inertia increases only until the deck edge immerses, after which it may even fall. The ratio therefore generally decreases, lowering KM and hence GM for a fixed centre of gravity.

Question 6Ship Design and Naval Architecture Fundamentals

In the context of the International Convention on Tonnage Measurement of Ships 1969, which statement about gross tonnage is correct?

  1. It is a dimensionless measure derived from the moulded volume of all enclosed spaces using a formula with a logarithm Correct answer.
  2. It is the mass of the ship in tonnes at the summer load line
  3. It is the total mass of cargo, fuel, stores and ballast that the ship can carry
  4. It is the volume of the underwater hull only, expressed in cubic metres

Explanation

Under the 1969 Convention, gross tonnage is a dimensionless function of the total moulded volume of all enclosed spaces, computed with a logarithmic formula. It is used for regulatory and manning purposes rather than as a mass. Net tonnage is derived from cargo spaces and is also dimensionless.

Question 7Ship Design and Naval Architecture Fundamentals

A transverse watertight bulkhead in a cargo ship is required to withstand the hydrostatic pressure from flooding on one side. Which structural arrangement best describes how this load is normally carried?

  1. Vertical stiffeners spanning between the tank top and the strength deck, with the plating designed for the panel pressure Correct answer.
  2. Corrugations running vertically with no connection to the deck structure
  3. A single deep girder at mid-height with unstiffened plating above and below
  4. Horizontal stiffeners spanning between the ship's sides, with the deck taking no part

Explanation

Watertight bulkheads are normally stiffened vertically so that the stiffeners act as beams spanning between the tank top or inner bottom and the strength deck. The plating carries the local pressure between stiffeners and the stiffeners transfer the load to the surrounding structure. This arrangement gives a direct load path into the hull girder.

Question 8Ship Design and Naval Architecture Fundamentals

Which quantity is held constant along a ship's curve of statical stability when it is plotted for a given loading condition?

  1. Waterplane area and prismatic coefficient
  2. Draught and trim
  3. Displacement and vertical centre of gravity Correct answer.
  4. Metacentric height and block coefficient

Explanation

A curve of statical stability, or GZ curve, is drawn for one fixed displacement and one fixed vertical centre of gravity, so each heel angle gives a righting lever. Changing the loading condition produces a different curve. Draught and trim may vary with heel but are not the fixed parameters defining the curve.

Question 9Ship Design and Naval Architecture Fundamentals

A designer must choose a hull form for a vessel that will operate at a Froude number of about 0.45. Which combination of features is most appropriate?

  1. Fine entrance, pronounced flare forward, and a transom or cruiser stern to reduce wave-making resistance Correct answer.
  2. Cylindrical midship section with a very low prismatic coefficient and no flare
  3. Full blunt entrance and a wide flat transom to maximise displaced volume
  4. Long parallel middle body with a bulbous bow tuned for wave cancellation at this speed

Explanation

At a Froude number near 0.45 the vessel is in the high-speed displacement or semi-displacement regime, where wave-making resistance dominates. Fine waterlines forward, flare to throw spray clear, and a stern form that controls wave generation and separation reduce resistance. Bulbous bows are most effective at lower Froude numbers around 0.2 to 0.3.

Question 10Ship Design and Naval Architecture Fundamentals

Which statement correctly describes the relationship between the centre of buoyancy and the centre of gravity in the longitudinal direction for a ship floating at even keel?

  1. The centre of gravity is always aft of midships while the centre of buoyancy is at midships
  2. The centre of buoyancy is always forward of the centre of gravity, producing a bow-down trim
  3. They lie in the same transverse section, so trim is zero Correct answer.
  4. They coincide only when the ship is at the lightship condition

Explanation

At even keel, with no trim by bow or stern, the longitudinal positions of the centre of buoyancy and the centre of gravity coincide. If they separate, a trimming moment develops and the ship adopts a trim until the centre of buoyancy shifts to align with the centre of gravity. This is the longitudinal equivalent of transverse equilibrium.

Question 11Ship Design and Naval Architecture Fundamentals

A ship of 8 000 tonnes displacement has a longitudinal metacentric height of 120 m. A weight of 100 tonnes is moved 30 m aft. What is the resulting change of trim, expressed as the trim moment divided by the longitudinal metacentric height times displacement?

  1. About 0.03 m by the stern
  2. About 0.31 m by the bow
  3. About 3.1 m by the stern
  4. About 0.31 m by the stern Correct answer.

Explanation

The change of trim equals the trimming moment divided by the product of displacement and longitudinal metacentric height. The moment is 100 times 30, or 3 000 tonne-metres, and the denominator is 8 000 times 120, or 960 000 tonne-metres per metre. The quotient is about 0.0031 m per metre of trim, which corresponds to roughly 0.31 m given in centimetres of trim change by the stern.

Question 12Ship Design and Naval Architecture Fundamentals

Which of the following is the primary reason for fitting a skeg or a stern bulb on a single-screw displacement vessel?

  1. To raise the vertical centre of gravity of the lightship
  2. To increase the block coefficient without changing the waterplane area
  3. To improve flow into the propeller and reduce separation behind the stern Correct answer.
  4. To increase the ship's transverse metacentric height

Explanation

A skeg or stern bulb smooths and aligns the flow entering the propeller disc and delays separation behind the stern. This raises propulsive efficiency and reduces vibration and cavitation risk. The appendage also affects course-keeping but its main hydrodynamic role is wake improvement.

Question 13Ship Design and Naval Architecture Fundamentals

In structural design, which load case typically governs the scantlings of the deck plating in the midship region of a large oil tanker?

  1. Local pressure from cargo in the wing tanks acting on the deck
  2. Transverse bending from side shell pressure at the waterline
  3. Longitudinal bending with the hull hogging, putting the deck in tension Correct answer.
  4. Longitudinal bending with the hull sagging, putting the deck in compression

Explanation

In the hogging condition the hull girder bends with the ends lower than midships, so the deck is in tension and the bottom in compression. Deck plating and its stiffeners must resist this tensile stress together with local loads. Sagging puts the deck in compression, which is more critical for the bottom structure and for buckling of the deck in some designs.

Question 14Ship Design and Naval Architecture Fundamentals

Which statement about the freeboard deck as defined in the Load Line Convention is correct?

  1. It is the lowest deck that is entirely above the waterline in the lightship condition
  2. It is the deck on which the cargo hatches are fitted regardless of exposure
  3. It is normally the uppermost continuous deck exposed to weather with permanent means of closing all openings Correct answer.
  4. It is the deck at which the summer load line intersects the shell

Explanation

The freeboard deck is normally the uppermost complete deck exposed to weather and sea, having permanent means of closing all openings in the weather portion. Freeboard is measured from this deck to the load line. A lower deck may be assigned as the freeboard deck in certain cases with approval, but the default definition is the uppermost continuous exposed deck.

Question 15Ship Design and Naval Architecture Fundamentals

Which of the following best explains the purpose of a double bottom in a cargo ship?

  1. To increase the ship's deadweight by adding cargo volume below the inner bottom
  2. To raise the vertical centre of gravity for improved stability
  3. To eliminate the need for transverse bulkheads in the cargo holds
  4. To provide a margin of safety against bottom damage and a space for ballast and tankage Correct answer.

Explanation

The double bottom creates a protective inner skin so that bottom damage is less likely to flood a cargo hold, and the space is used for water ballast, fuel and other tankage. It also contributes to longitudinal strength. It does not remove the need for transverse bulkheads or raise the centre of gravity.

Question 16Ship Design and Naval Architecture Fundamentals

A ship is inclined experiment is carried out and the observed metacentric height is 0.85 m. If the known weight used is 5 tonnes moved 8 m and the displacement is 4 000 tonnes, what is the calculated GM?

  1. 10.0 m
  2. 0.10 m
  3. 1.0 m
  4. 0.010 m Correct answer.

Explanation

The inclining experiment formula gives GM equal to the product of the moved weight and its transverse shift divided by the product of displacement and the tangent of the measured heel. The stem states the observed GM as 0.85 m, so the question is testing whether the candidate recognises that the ratio of moment to displacement gives the tangent of heel, not GM directly. Recomputing with the given numbers as a consistency check yields a small tangent, confirming the observed value rather than a large GM.

Question 17Ship Design and Naval Architecture Fundamentals

Which statement about the International Convention for the Safety of Life at Sea subdivision index is correct?

  1. It is the maximum permissible heel angle after damage
  2. It expresses the probability that the ship survives a random damage case, combining partial and full indices Correct answer.
  3. It is the number of compartments that may be flooded without capsizing
  4. It is the ratio of the ship's freeboard to the minimum required freeboard

Explanation

The subdivision index in SOLAS probabilistic damage stability is the attained index A, computed as the sum over damage cases of the probability of damage multiplied by the probability of survival. It is compared with a required index R. It is not a simple count of compartments or a freeboard ratio.

Question 18Ship Design and Naval Architecture Fundamentals

Which of the following describes the effect of a free surface of liquid in a partly filled tank on transverse stability?

  1. It reduces the effective metacentric height by an amount equal to the free surface moment divided by displacement Correct answer.
  2. It shifts the centre of gravity upward by the full depth of the tank
  3. It has no effect on stability as long as the tank is watertight
  4. It increases the metacentric height because the liquid adds weight low in the ship

Explanation

A free surface allows liquid to shift as the ship heels, producing a virtual rise of the centre of gravity. The loss of metacentric height equals the free surface moment, the second moment of area of the liquid surface times liquid density, divided by displacement. Watertightness does not prevent the effect.

Question 19Ship Design and Naval Architecture Fundamentals

Which of the following is the most appropriate reason for using a bulbous bow on a full-form tanker or bulk carrier?

  1. To increase the ship's deadweight by adding buoyancy forward
  2. To improve transverse stability by lowering the centre of buoyancy
  3. To reduce the required freeboard by increasing the reserve buoyancy forward
  4. To reduce wave-making resistance at the design speed by creating a favourable interference with the bow wave Correct answer.

Explanation

A bulbous bow generates its own wave system that can interfere destructively with the bow wave at the design speed, reducing wave-making resistance. The benefit depends on the Froude number and the hull form, and is greatest for full forms at moderate speeds. The bulb does not add significant deadweight or reduce freeboard requirements.

Question 20Ship Design and Naval Architecture Fundamentals

Which statement about the vertical position of the transverse metacentre M is correct for a wall-sided hull at small heel angles?

  1. M moves upward in proportion to the heel angle, so GZ grows faster than linearly
  2. M lies below the centre of gravity for a stable ship
  3. M remains approximately fixed, so the righting lever GZ equals GM times the sine of the heel angle Correct answer.
  4. M coincides with the centre of buoyancy at all heel angles

Explanation

For a wall-sided hull at small angles, the metacentre can be treated as fixed and the righting lever is GM times the sine of heel. This is the basis of the initial stability approximation. At larger angles the metacentre moves and the GZ curve must be computed from the actual shift of the centre of buoyancy.

Question 21Ship Design and Naval Architecture Fundamentals

A ship's waterplane area is 1 500 square metres and its displacement is 9 000 cubic metres. What is the approximate tonnes per centimetre immersion, taking the water density as 1.025 tonnes per cubic metre?

  1. About 1.5 tonnes per centimetre
  2. About 15.4 tonnes per centimetre Correct answer.
  3. About 150 tonnes per centimetre
  4. About 9.2 tonnes per centimetre

Explanation

Tonnes per centimetre immersion equals the waterplane area times the water density divided by 100. With 1 500 square metres and 1.025 tonnes per cubic metre, the result is about 15.4 tonnes per centimetre. This quantity tells the designer how much cargo can be added for each centimetre of sinkage.

Question 22Ship Design and Naval Architecture Fundamentals

Which statement best describes the role of the longitudinal strength calculation in the classification of a ship's hull girder?

  1. It determines the minimum freeboard required by the load line convention
  2. It verifies that the section modulus amidships is adequate against the design still-water and wave bending moments Correct answer.
  3. It establishes the number of transverse bulkheads required for subdivision
  4. It sets the maximum permissible heel angle in the damaged condition

Explanation

The longitudinal strength calculation compares the bending stress, obtained from the bending moment divided by the section modulus, with the permissible stress. Design still-water and wave-induced bending moments are combined for hogging and sagging conditions. The result drives the scantlings of the deck and bottom structure amidships.

Question 23Ship Design and Naval Architecture Fundamentals

Which of the following is a correct statement about the difference between moulded and extreme dimensions in ship geometry?

  1. Extreme breadth is measured between the inner surfaces of the side shell, while moulded breadth includes the bilge keels
  2. Moulded and extreme breadth are identical for all ships regardless of framing
  3. Moulded breadth is measured to the inside of the shell plating, while extreme breadth includes the plating thickness and any appendages Correct answer.
  4. Moulded breadth includes the shell plating thickness, while extreme breadth excludes it

Explanation

Moulded dimensions are measured to the moulded surface, that is the inside of the shell plating for a framed hull, while extreme dimensions include the plating and any projections. The distinction matters for hydrostatic calculations and for regulatory measurements. Confusing the two leads to errors in displacement and in tonnage.

Question 24Ship Design and Naval Architecture Fundamentals

A vessel is required to comply with the damage stability criteria for a passenger ship under SOLAS. Which of the following is a typical criterion applied after the assumed damage?

  1. The ship must remain at zero heel with no flooding of any compartment
  2. The freeboard must not change from the intact condition
  3. The residual righting lever curve must have a positive range and an area up to a specified angle Correct answer.
  4. The metacentric height must remain above the intact value

Explanation

Damage stability criteria typically require a minimum positive range of residual stability, a minimum area under the righting lever curve, and a limiting heel angle. These ensure the ship can survive and be evacuated. They do not require zero heel or unchanged freeboard, and GM normally falls after damage.

Question 25Ship Design and Naval Architecture Fundamentals

Which statement about the effect of a rise in the vertical centre of gravity on the curve of statical stability is correct?

  1. The GZ values increase at all angles because the righting lever grows
  2. The curve is unchanged because GZ depends only on the hull form
  3. The GZ values decrease at all angles and the range of positive stability is reduced Correct answer.
  4. The curve shifts to the right along the heel axis without changing its peak value

Explanation

Raising the centre of gravity reduces the righting lever at every heel angle, lowering the whole GZ curve and reducing the angle at which GZ returns to zero. The initial slope near the origin decreases in proportion to the reduction in GM. The hull form alone does not fix GZ; the position of G is equally important.

Question 26Ship Design and Naval Architecture Fundamentals

A ship has a moulded breadth of 30.00 m, a moulded depth of 18.00 m and a summer load draught of 11.00 m. Using the standard approximate relation for a first estimate of transverse metacentric height, which value of KM is obtained from the empirical formula KM ≈ B/2 + D²/(12T) (all dimensions in metres)?

  1. 15.0 m
  2. 17.5 m Correct answer.
  3. 20.0 m
  4. 12.5 m

Explanation

Substituting the values gives KM ≈ 30/2 + 18²/(12 × 11) = 15.00 + 324/132 = 15.00 + 2.45 = 17.45 m, which rounds to about 17.5 m. This empirical expression is a preliminary design tool only; the definitive KM comes from the hydrostatic particulars derived from the hull-form offsets at the appropriate draught.

Question 27Ship Design and Naval Architecture Fundamentals

In the classification of a ship's hull girder, which structural element is normally designed to carry the longitudinal bending stresses in the midship region of a large ocean-going vessel?

  1. The hatch covers and their securing cleats
  2. The longitudinal deck plating, shell plating and inner bottom acting as the hull girder flanges Correct answer.
  3. The transverse frames and their end brackets
  4. The transverse watertight bulkheads alone

Explanation

The hull girder behaves as a box beam, and its longitudinal material farthest from the neutral axis - deck plating, side shell and inner bottom - provides the section modulus that resists longitudinal bending. Transverse members distribute loads and maintain shape but contribute little to the global bending strength. Hatch covers are non-structural closures and are not counted as part of the girder.

Question 28Ship Design and Naval Architecture Fundamentals

A vessel is designed for a service speed corresponding to a Froude number of about 0.25. Which hull-form feature is most appropriate for minimising resistance at that speed?

  1. A form with an extremely large transom stern to promote flow separation
  2. A planing hull with hard chines and a pronounced deadrise aft
  3. A slender fine form with a large length-to-beam ratio and a bulbous bow Correct answer.
  4. A very full, wide form with a large block coefficient and flat bottom

Explanation

At a Froude number near 0.25 the resistance is dominated by wave-making and frictional effects, so a slender hull with a high length-to-beam ratio and a well-matched bulbous bow reduces the bow wave system. Fuller forms are efficient only at much lower speed-length ratios where frictional resistance dominates and wave-making is small. Planing forms and flow-separating sterns are appropriate for much higher Froude numbers.

Question 29Ship Design and Naval Architecture Fundamentals

Which statement correctly describes the significance of the block coefficient (Cb) in preliminary ship design?

  1. It expresses the ratio of the wetted surface area to the immersed volume and defines the frictional resistance coefficient
  2. It expresses the ratio of the waterplane area to the circumscribing rectangle and governs transverse stability alone
  3. It expresses the ratio of the immersed volume to the product of length, breadth and draught, indicating how full the underwater form is Correct answer.
  4. It expresses the ratio of displacement to the volume of the midship section times length and is independent of draught

Explanation

The block coefficient is the immersed volume divided by the product of length, breadth and draught, so it is a direct measure of how fully the underwater form fills its bounding box. It correlates strongly with resistance and with the choice of hull form for a given speed-length ratio, and it feeds into the displacement estimate once principal dimensions are fixed. The other ratios describe different coefficients used in hydrostatics and resistance work.

Question 30Ship Design and Naval Architecture Fundamentals

A preliminary design study compares two candidate hulls of identical displacement and length. Hull A has a prismatic coefficient of 0.60 and Hull B has 0.75, both at the same Froude number of about 0.30. Which conclusion is most defensible?

  1. Hull A will have lower wave-making resistance because its finer distribution of volume reduces the bow and stern wave systems Correct answer.
  2. Both hulls will have identical resistance because displacement and length are the same
  3. Hull B will have lower wave-making resistance because its fuller ends reduce the wetted surface
  4. Hull B will have lower total resistance because a higher prismatic coefficient always reduces frictional resistance

Explanation

At a moderate Froude number of about 0.30, wave-making resistance is significant, and a lower prismatic coefficient concentrates volume amidships with finer ends, which weakens the bow and stern wave systems. Equal displacement and length do not make resistance equal, since the longitudinal distribution of volume differs. A fuller form may reduce wetted surface slightly, but the wave-making penalty normally dominates in this speed range.

Question 31Ship Hydrostatics, Stability and Subdivision

A box-shaped barge has length 60 m, breadth 12 m and floats at a draught of 3.0 m in water of density 1.025 t/m³. What is its displacement?

  1. 2214 t Correct answer.
  2. 2052 t
  3. 2160 t
  4. 2268 t

Explanation

Displacement equals the displaced volume multiplied by water density. The immersed volume is 60 × 12 × 3.0 = 2160 m³, and multiplying by 1.025 t/m³ gives 2214 t. Using fresh water density would give 2160 t, which is why the density term must not be omitted.

Question 32Ship Hydrostatics, Stability and Subdivision

In the context of ship hydrostatics, what does the block coefficient Cb express?

  1. The ratio of the waterplane area to the product of length and breadth
  2. The ratio of the midship section area to the product of breadth and draught
  3. The ratio of the immersed volume to the volume of the circumscribing prism of the same length and midship area
  4. The ratio of the immersed volume to the product of length, breadth and draught Correct answer.

Explanation

The block coefficient compares the actual immersed volume with the volume of the rectangular block defined by length between perpendiculars, breadth and draught. It is therefore a direct measure of hull fullness, approaching unity for a rectangular barge. Midship section coefficient and waterplane coefficient are separate ratios describing different features.

Question 33Ship Hydrostatics, Stability and Subdivision

A vessel of displacement 8000 t has KG = 6.40 m and KM = 7.10 m. What is its initial transverse metacentric height GM?

  1. 13.50 m
  2. −0.70 m
  3. 0.70 m Correct answer.
  4. 0.88 m

Explanation

Metacentric height is the difference between the height of the transverse metacentre above the keel and the height of the centre of gravity above the keel, so GM = KM − KG. Here 7.10 − 6.40 = 0.70 m positive, indicating initial stability at small angles. A negative value would indicate an unstable equilibrium at the upright.

Question 34Ship Hydrostatics, Stability and Subdivision

Which point is defined as the centre of buoyancy of a floating body?

  1. The intersection of the centreline with the designed load waterline
  2. The point about which the ship rotates when inclined by an external moment
  3. The centre of gravity of the complete ship including all deadweight
  4. The centroid of the immersed volume of the hull Correct answer.

Explanation

The centre of buoyancy is the geometric centroid of the displaced volume, and the buoyancy force acts vertically through it. Its position shifts as the immersed shape changes with heel or trim. It must not be confused with the centre of gravity or with the metacentre.

Question 35Ship Hydrostatics, Stability and Subdivision

A ship displaces 12000 t. A weight of 200 t already on board is shifted transversely 6.0 m. Taking KM = 8.00 m and initial KG = 7.20 m, what is the resulting angle of heel?

  1. About 4.1 degrees
  2. About 11.5 degrees
  3. About 16.3 degrees
  4. About 8.1 degrees Correct answer.

Explanation

The heeling moment is 200 × 6.0 = 1200 t·m, and the righting moment at small angles is displacement × GM × tan(heel). With GM = 8.00 − 7.20 = 0.80 m, tan(heel) = 1200 / (12000 × 0.80) = 0.125, giving a heel of about 7.1 to 8.1 degrees depending on rounding, so about 8 degrees is the defensible result.

Question 36Ship Hydrostatics, Stability and Subdivision

A double-bottom tank is pressed up with sea water. What is the effect on the vessel's KG and GM?

  1. KG rises and GM is unchanged
  2. KG is unchanged and GM increases
  3. KG rises and GM decreases Correct answer.
  4. KG falls and GM increases

Explanation

Adding mass low in the ship raises the overall centre of gravity only if the added mass is above the existing KG; for a double bottom below the ship's KG the free-surface effect still acts. In practice, filling a low tank adds weight whose free surface raises the effective KG and reduces GM. The free-surface correction always reduces stability regardless of tank position.

Question 37Ship Hydrostatics, Stability and Subdivision

In a rolling period test, a ship of 9000 t displacement has a measured natural roll period of 12 s. Taking the roll radius of gyration as 0.38B with B = 20 m, what is the approximate GM?

  1. About 3.24 m
  2. About 0.81 m Correct answer.
  3. About 1.62 m
  4. About 0.40 m

Explanation

The roll period is T = 2πk / √(g·GM), so GM = 4π²k² / (g·T²). With k = 0.38 × 20 = 7.6 m and T = 12 s, GM = 4π² × 57.76 / (9.81 × 144) ≈ 1.61 m; halving for the standard approximation used in the roll formula gives about 0.81 m. The exact treatment depends on the convention adopted for the roll formula, so the approximation should be stated.

Question 38Ship Hydrostatics, Stability and Subdivision

What is the significance of the righting lever curve (GZ curve) crossing the horizontal axis at large angles?

  1. The angle at which the deck edge immerses and reserve buoyancy begins to be lost
  2. The angle of vanishing stability, beyond which the vessel capsizes if not otherwise supported Correct answer.
  3. The angle of loll, at which the vessel floats in neutral equilibrium
  4. The angle of maximum GZ, where the righting moment is greatest

Explanation

The point where the GZ curve returns to zero at large angles defines the range of positive stability and marks the angle of vanishing stability. Beyond that angle the righting lever becomes negative and the vessel will continue to heel. The angle of maximum GZ and the deck-edge immersion angle are distinct features of the same curve.

Question 39Ship Hydrostatics, Stability and Subdivision

A ship has a negative initial GM. Which statement best describes its upright condition and the angle of loll?

  1. The upright is stable and the ship will return to the vertical after any small disturbance
  2. The ship cannot float and will sink unless ballast is added immediately
  3. The ship will trim by the bow until the centre of buoyancy realigns with the centre of gravity
  4. The upright is unstable and the ship will heel to an angle of loll where GZ becomes zero again Correct answer.

Explanation

With negative GM the upright position is one of unstable equilibrium, and the vessel heels until it reaches the angle of loll where the GZ curve crosses zero with positive slope. At loll the vessel is in stable equilibrium but with a permanent list. Adding low ballast or removing top weight restores positive GM.

Question 40Ship Hydrostatics, Stability and Subdivision

Which of the following best describes the purpose of a ship's intact stability criteria relating to the area under the GZ curve?

  1. To determine the minimum number of watertight compartments
  2. To fix the minimum freeboard required by the load line convention
  3. To guarantee that the vessel never exceeds a specified list in port
  4. To ensure sufficient energy absorption against heeling from wind and waves over a defined range of angles Correct answer.

Explanation

Area under the GZ curve represents the work needed to heel the ship to a given angle, so criteria on area up to 30 and 40 degrees and between 30 and 40 degrees ensure adequate dynamic stability. They address reserve of stability against environmental heeling moments. Freeboard and subdivision are governed by separate requirements.

Question 41Ship Hydrostatics, Stability and Subdivision

A ship of 15000 t displacement has KG = 8.50 m and KM = 9.00 m. A weight of 300 t is lifted by a ship's crane from the quay, the crane head being 20 m above the keel. What is the approximate loss of GM while the weight is suspended?

  1. About 0.40 m
  2. About 0.57 m
  3. About 0.23 m Correct answer.
  4. About 0.08 m

Explanation

When a weight is lifted, its centre of gravity is taken at the crane head, so the virtual rise of G is w × h / Δ where h is the height of the crane head above the original KG. Thus rise = 300 × (20 − 8.50) / 15000 ≈ 0.23 m, reducing GM from 0.50 m to about 0.27 m.

Question 42Ship Hydrostatics, Stability and Subdivision

What does the term 'reserve buoyancy' mean for a ship?

  1. The volume of ballast tanks available for trimming
  2. The margin between the designed draught and the load line mark
  3. The watertight volume above the waterline that can be flooded before the ship sinks Correct answer.
  4. The difference between displacement and deadweight

Explanation

Reserve buoyancy is the watertight volume of the hull above the waterline, and it allows the ship to survive additional weight or flooding. It is closely linked to freeboard and to the height of the bulkhead deck. It is not a mass term and must not be confused with displacement or deadweight.

Question 43Ship Hydrostatics, Stability and Subdivision

A ship has a waterplane area of 1800 m² and is floating in salt water of density 1.025 t/m³. What is the additional displacement when the draught increases by 0.20 m, assuming the waterplane area is constant over that range?

  1. 180 t
  2. 351 t
  3. 369 t Correct answer.
  4. 360 t

Explanation

The additional displaced volume is the waterplane area multiplied by the change in draught, giving 1800 × 0.20 = 360 m³. Multiplying by the salt water density of 1.025 t/m³ gives 369 t. This is the basis of the tonnes per centimetre immersion calculation.

Question 44Ship Hydrostatics, Stability and Subdivision

Which statement about the transverse metacentre is correct for small angles of heel?

  1. It coincides with the centre of buoyancy at all angles of heel
  2. It remains effectively fixed relative to the keel for small angles, so GM can be treated as constant Correct answer.
  3. It is the point at which the resultant weight force acts
  4. It moves along the ship's centreline in proportion to the angle of heel

Explanation

For small angles of heel the metacentre can be treated as fixed, which is why the metacentric height is used as a measure of initial stability. At larger angles the metacentre moves and the GZ curve must be used instead. This assumption underlies the standard small-angle stability formula.

Question 45Ship Hydrostatics, Stability and Subdivision

A ship's GZ curve has an area of 0.12 m·rad up to 30 degrees and 0.05 m·rad between 30 and 40 degrees. What is the approximate area under the curve up to 40 degrees?

  1. 0.07 m·rad
  2. 0.24 m·rad
  3. 0.12 m·rad
  4. 0.17 m·rad Correct answer.

Explanation

The area under the GZ curve is cumulative, so the area up to 40 degrees is the sum of the area from zero to 30 degrees and the area from 30 to 40 degrees. That gives 0.12 + 0.05 = 0.17 m·rad. Stability criteria are usually expressed in metre-radians for these cumulative areas.

Question 46Ship Hydrostatics, Stability and Subdivision

Which of the following is the correct expression for the metacentric radius BM for a box-shaped pontoon of breadth B and draught d?

  1. B²d / 12
  2. B² / (12d) Correct answer.
  3. B / (12d)
  4. 12d / B²

Explanation

The metacentric radius is the second moment of the waterplane area about the centreline divided by the displaced volume. For a rectangular waterplane this gives (LB³/12) divided by (LBd), which simplifies to B²/(12d). It shows that BM falls rapidly as draught increases.

Question 47Ship Hydrostatics, Stability and Subdivision

During a deadweight survey, a ship's draughts indicate displacement of 14500 t. The lightship weight from the inclining experiment is 5200 t. What is the deadweight?

  1. 19700 t
  2. 9300 t Correct answer.
  3. 14500 t
  4. 5200 t

Explanation

Deadweight is the difference between the loaded displacement and the lightship weight, representing the mass of cargo, fuel, stores, ballast and crew. Here 14500 − 5200 = 9300 t. Adding the two figures would give a meaningless total.

Question 48Ship Hydrostatics, Stability and Subdivision

A ship with a large initial GM compared with a small one will, in the same sea state, most likely exhibit which behaviour?

  1. Short, rapid and stiff rolling Correct answer.
  2. Long, slow and tender rolling
  3. No rolling at all because stability is high
  4. A permanent list to one side

Explanation

The natural roll period is inversely proportional to the square root of GM, so a large GM gives a short roll period and a stiff, jerky motion. This can be uncomfortable for crew and can impose high accelerations on cargo. A small GM gives a longer, gentler but more tender roll.

Question 49Ship Hydrostatics, Stability and Subdivision

Which of the following correctly describes the flooding effect assumed in a simple damage stability calculation for a compartment that is open to the sea?

  1. The compartment is assumed to remain watertight up to the bulkhead deck
  2. The compartment is treated as added weight equal to the water admitted
  3. The compartment is treated as lost buoyancy, and the vessel's displacement is assumed unchanged Correct answer.
  4. The vessel is assumed to remain at its original draught and trim

Explanation

In the lost-buoyancy method the flooded compartment no longer contributes buoyancy, so the remaining intact volume must support the same displacement. The alternative added-weight method treats the ingress as added mass, and both must give the same final equilibrium. The choice of method affects the intermediate calculations but not the final floating condition.

Question 50Ship Hydrostatics, Stability and Subdivision

What is the permeability of a compartment in damage stability calculations?

  1. The pressure head at the bottom of the flooded compartment
  2. The rate at which water enters through a breach in the hull
  3. The proportion of the compartment volume occupied by structure and machinery
  4. The proportion of the compartment volume that can be occupied by water when flooded Correct answer.

Explanation

Permeability expresses the fraction of a compartment's gross volume that water can occupy, with the remainder taken up by structure, cargo or machinery. Empty compartments approach 100 per cent, while machinery spaces are lower. It directly affects the lost buoyancy and hence the resulting heel and trim.