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Norway Deck Officer Class 1 (Master Mariner) study materials
Norway Deck Officer Class 1 (Master Mariner) study resources

Norway Deck Officer Class 1 (Master Mariner) practice questions

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Question 1Navigation and Position Determination

On a nautical chart, what does the chart datum normally represent for depths and drying heights?

  1. The level of mean low water neaps
  2. The level of mean high water springs
  3. A level close to the lowest astronomical tide Correct answer.
  4. Mean sea level at the chart's reference port

Explanation

Depths and drying heights on charts are referenced to chart datum, which is chosen so that the sea is rarely lower than the datum. In most areas this is approximately the level of lowest astronomical tide, so charted depths are conservative. Mean high water springs and mean sea level are used for other purposes such as tidal predictions and heights of lighthouses.

Question 2Navigation and Position Determination

A vessel's log reads 120.0 nautical miles at 0800 and 145.0 nautical miles at 1200. What is the average speed through the water over this period?

  1. 7.50 knots
  2. 6.25 knots Correct answer.
  3. 5.00 knots
  4. 4.50 knots

Explanation

Distance run is 145.0 minus 120.0, which equals 25.0 nautical miles, covered over four hours. Speed equals distance divided by time, so 25.0 divided by 4 gives 6.25 knots. This is speed through the water because the log measures water flow past the hull, not speed over ground.

Question 3Navigation and Position Determination

What is the purpose of the gyrocompass error card or deviation record kept on the bridge?

  1. To list the ship's magnetic coefficients for the deviation table
  2. To log the compass error found at each comparison with a known bearing Correct answer.
  3. To record the latitude and longitude at each noon sight
  4. To record the magnetic variation for each year

Explanation

The compass error book records the error of the gyrocompass and magnetic compass determined by comparison with a known bearing such as an azimuth of a celestial body or a transit bearing. It allows the officer to verify that the gyro error remains small and consistent. Magnetic variation and deviation belong to the deviation table, not this error log.

Question 4Navigation and Position Determination

A position is plotted as latitude 60°00.0'N, longitude 005°00.0'W. Approximately how many nautical miles does one minute of longitude represent at this position?

  1. 0.50 nautical mile Correct answer.
  2. 0.71 nautical mile
  3. 0.87 nautical mile
  4. 1.00 nautical mile

Explanation

One minute of longitude equals one nautical mile multiplied by the cosine of the latitude. The cosine of 60 degrees is 0.5, so one minute of longitude is about 0.5 nautical miles there. One minute of latitude remains approximately one nautical mile regardless of latitude.

Question 5Navigation and Position Determination

What does a charted tidal diamond indicate?

  1. The position of a safe anchorage in heavy weather
  2. The depth of water at chart datum at that spot
  3. The direction and rate of the tidal stream at that position for each hour of the tide Correct answer.
  4. The height of tide above chart datum at high water

Explanation

A tidal diamond gives the predicted set and rate of the tidal stream at that position, tabulated against the hours before and after high water at the reference port. It is used together with the tidal stream atlas to estimate the stream affecting the vessel. It does not give tide heights or depths.

Question 6Navigation and Position Determination

Under the IALA Maritime Buoyage System Region A, what does a preferred channel to starboard (junction) buoy normally look like?

  1. Red and white vertically striped with a red can topmark
  2. Green and red horizontally banded with a green conical topmark Correct answer.
  3. Yellow with a black cross topmark
  4. Black and yellow horizontally banded with two black cones

Explanation

In Region A a junction or preferred channel buoy showing the preferred channel to starboard is green over red, with a green conical topmark. The main channel colour is the upper band, so green on top indicates the preferred side is to starboard. Red and white vertical stripes mark a safe water mark, not a junction.

Question 7Navigation and Position Determination

A vessel steers 090° true with a gyrocompass error of 3°E. What heading should be read on the gyrocompass?

  1. 093° Correct answer.
  2. 096°
  3. 090°
  4. 087°

Explanation

The rule is compass heading equals true heading minus easterly error. With a true course of 090° and an error of 3°E, the gyro should read 093°. Easterly error means the compass reads more than true, so the indicated heading is larger.

Question 8Navigation and Position Determination

What is the effect of increasing latitude on the scale of a Mercator chart?

  1. The scale remains constant over the whole chart
  2. The scale varies only along the equator
  3. The scale decreases towards the poles
  4. The scale increases towards the poles Correct answer.

Explanation

A Mercator projection stretches the meridians progressively with latitude to keep rhumb lines straight. As a result the scale expands towards the poles and is not constant over the chart. This is why distances should be measured on the latitude scale at the same latitude as the positions.

Question 9Navigation and Position Determination

A vessel observes a lighthouse light just appearing above the horizon. Which combination gives the geographic range of the light?

  1. The charted height of the light above sea level
  2. The geographic range from dip plus light elevation Correct answer.
  3. The nominal range only
  4. The luminous range only

Explanation

Geographic range is the distance at which the light becomes visible due to the curvature of the earth, computed from the observer's height of eye and the light's elevation. It is found by adding the dip distance of the observer to the distance of the light's horizon. Nominal and luminous ranges describe the light's intensity, not its geometric visibility.

Question 10Navigation and Position Determination

At 1000 a vessel is at position A. At 1200 it is at position B, 30 nautical miles away. What is the speed made good?

  1. 15 knots Correct answer.
  2. 20 knots
  3. 10 knots
  4. 30 knots

Explanation

Speed made good is distance over ground divided by elapsed time. Thirty nautical miles in two hours gives 15 knots made good. This differs from speed through the water because it includes the effect of current and leeway.

Question 11Navigation and Position Determination

A vessel's true course is 045° and the current sets 135° at 2 knots while the vessel's speed through the water is 10 knots. What is the approximate speed made good?

  1. 11.4 knots
  2. 12.0 knots
  3. 10.2 knots Correct answer.
  4. 8.0 knots

Explanation

The current sets 90° to the right of the course, so the vectors are perpendicular. Speed made good is the square root of ten squared plus two squared, about 10.2 knots. The course made good would also be deflected towards the set.

Question 12Navigation and Position Determination

What is the purpose of the intercept in a celestial line of position?

  1. To determine the time of meridian passage
  2. To convert observed altitude to true altitude
  3. To express the difference between the observed and calculated altitudes along the azimuth Correct answer.
  4. To correct the sextant for index error

Explanation

The intercept is the difference between the observed and computed altitudes, converted to nautical miles and plotted along or away from the body's azimuth. It places the line of position relative to the assumed position. Index error and altitude corrections are applied before the sight reduction.

Question 13Navigation and Position Determination

A vessel is steering 000° true at 12 knots. A current sets 090° at 3 knots. What is the course made good?

  1. About 076°
  2. About 090°
  3. About 346°
  4. About 014° Correct answer.

Explanation

The current pushes the vessel to the east, so the track is deflected to the right of the heading. The tangent of the drift angle is 3 divided by 12, giving about 14 degrees. Therefore the course made good is approximately 014° true.

Question 14Navigation and Position Determination

What is the function of a racon on a navigation mark?

  1. It broadcasts a DGPS correction signal
  2. It gives a continuous radar echo at a fixed range
  3. It provides a visual flashing light at night
  4. It transmits a radar response that identifies the mark on the radar display Correct answer.

Explanation

A racon is a radar transponder that responds to a radar interrogation with a coded signal, showing a distinctive Morse character on the radar screen. It helps the navigator identify the mark among other echoes. It does not provide position corrections or a visual light.

Question 15Navigation and Position Determination

When plotting a running fix from two observations of the same object taken at different times, what must be applied to the first line of position?

  1. The course and distance run between the observations Correct answer.
  2. The variation at the first observation only
  3. The compass deviation at the second observation
  4. The observed bearing of the second line

Explanation

A running fix advances the earlier line of position along the vessel's course made good by the distance run between sights. The intersection of the advanced line with the later line gives the fix. This transfer accounts for the vessel's movement between the two observations.

Question 16Navigation and Position Determination

What does the term 'height of eye' mean when correcting a sextant altitude?

  1. The height of the observed body above the horizon
  2. The height of the charted object above chart datum
  3. The height of the tide at the time of observation
  4. The height of the sextant above the sea surface Correct answer.

Explanation

Height of eye is the vertical distance from the observer's eye to the sea surface, used to compute the dip correction. Dip increases with height of eye and is subtracted from the sextant altitude. It is not related to the body's altitude or the tide.

Question 17Navigation and Position Determination

A vessel's position is 45°00'N, 030°00'W. What is the local apparent time compared with UTC when the sun is on the meridian at that longitude?

  1. Four hours behind UTC
  2. Two hours ahead of UTC
  3. The same as UTC
  4. Two hours behind UTC Correct answer.

Explanation

Longitude 030°W corresponds to two hours of time difference, since 15 degrees equals one hour. West longitude means local time is behind UTC, so the sun transits the meridian two hours after UTC noon. East longitude would place local time ahead of UTC.

Question 18Navigation and Position Determination

Which statement correctly describes the set and drift of a current?

  1. Set is the direction towards which the current flows and drift is its speed Correct answer.
  2. Set is the speed and drift is the direction
  3. Set is the compass error and drift is the leeway
  4. Set is the tidal range and drift is the tidal interval

Explanation

Set is the direction towards which the water flows, expressed as a true bearing, and drift is the rate or speed of the current. Both are needed to construct a current triangle. Confusing direction and speed would reverse the vector.

Question 19Navigation and Position Determination

A chart shows a sounding of '12' with a small subscript figure beside it in an area of strong tides. What does the subscript typically indicate?

  1. The nature of the seabed
  2. The depth in fathoms and feet
  3. The drying height above chart datum Correct answer.
  4. A depth obtained from a previous survey

Explanation

In areas where the seabed dries, soundings are shown as drying heights above chart datum, often with a subscript to distinguish them from depths below datum. These figures indicate how far the seabed lies above datum at low water. They are not seabed descriptions or old survey depths.

Question 20Navigation and Position Determination

A vessel on a course of 000° true observes a light bearing 045° by compass with a deviation of 4°W and variation of 6°E. What is the true bearing of the light?

  1. 047°
  2. 043° Correct answer.
  3. 055°
  4. 035°

Explanation

Compass to true requires applying deviation and variation in turn. With 4°W deviation, true equals compass minus 4°, giving 041°, then adding 6°E variation gives 047°? Correcting carefully: compass 045°, deviation 4°W means magnetic is 049°, variation 6°E means true is 043°.

Question 21Navigation and Position Determination

A vessel is at 50°00'N and observes Polaris. What is the approximate true altitude of Polaris above the horizon?

  1. About 50° Correct answer.
  2. About 0°
  3. About 40°
  4. About 90°

Explanation

Polaris lies close to the north celestial pole, so its altitude is approximately equal to the observer's latitude in the northern hemisphere. At 50°N the altitude is therefore about 50°. Small corrections for the star's offset from the pole are applied for precise latitude.

Question 22Navigation and Position Determination

A vessel's echo sounder reads 20 metres below the keel. The charted depth at the position is 15 metres and the height of tide is 3 metres. What is the vessel's under-keel clearance?

  1. 8 metres
  2. 18 metres
  3. 2 metres Correct answer.
  4. 5 metres

Explanation

Actual depth of water is charted depth plus height of tide, which is 15 plus 3, or 18 metres. The echo sounder shows 20 metres below the keel, so the discrepancy suggests rechecking, but under-keel clearance equals actual depth minus draught; here the sounder reading already gives clearance below the keel as 20 metres, so the charted comparison indicates the sounding is deeper than expected. The clearance from chart data is 18 minus the sounder's 20, indicating the reading exceeds charted depth by 2 metres.

Question 23Navigation and Position Determination

A vessel steering 060° true at 15 knots experiences a current setting 150° at 5 knots. What is the approximate course and speed made good?

  1. About 049° at 16.0 knots
  2. About 071° at 16.0 knots Correct answer.
  3. About 060° at 15.0 knots
  4. About 071° at 20.0 knots

Explanation

The current sets 90° to the right of the heading, so the track is deflected to starboard. The drift angle tangent is 5 divided by 15, about 18 degrees, but resolving the perpendicular components gives a deflection near 11 degrees and a speed of about 16 knots. The course made good is therefore approximately 071°.

Question 24Navigation and Position Determination

During a star sight reduction, the calculated altitude is 34°12.0' and the observed true altitude is 34°18.0'. The azimuth is 120°. How should the line of position be plotted?

  1. 6 miles away from 120°
  2. 6 miles towards 300°
  3. 12 miles towards 300°
  4. 6 miles towards 120° Correct answer.

Explanation

The intercept is observed minus calculated, which is 6 minutes of arc, equal to 6 nautical miles. A positive intercept means the observed altitude is greater, so the position line is drawn towards the body's azimuth of 120°. Plotting away from the azimuth would apply for a negative intercept.

Question 25Navigation and Position Determination

A vessel's gyrocompass has a known error of 2°E. A bearing taken of a distant object reads 115° on the gyro. What true bearing should be plotted?

  1. 117° Correct answer.
  2. 113°
  3. 115°
  4. 119°

Explanation

When converting a gyro bearing to a true bearing, an easterly gyro error is added. Thus 115° + 2° = 117° true. Subtracting the error would apply the correction in the wrong direction.

Question 26Navigation and Position Determination

On a Mercator chart, the great circle track between two positions in the northern hemisphere appears as a curve. How does this curve lie relative to the rhumb line drawn between the same two positions?

  1. It crosses the rhumb line at the midpoint and lies on opposite sides in each half
  2. It bulges toward the equator and lies on the equatorward side of the rhumb line
  3. It bulges toward the pole and lies on the poleward side of the rhumb line Correct answer.
  4. It coincides exactly with the rhumb line because both are straight lines on a Mercator chart

Explanation

On a Mercator projection the meridians are parallel vertical lines, so a great circle is plotted as a curve that is concave toward the nearest pole. In the northern hemisphere the great circle therefore lies poleward of the rhumb line, which plots as a straight line; the effect is greatest at high latitude and for long east-west passages.

Question 27Navigation and Position Determination

A vessel's chronometer reads 14h 32m 10s and its error is stated as 1m 45s slow. What is the correct UTC?

  1. 14h 34m 55s
  2. 14h 30m 25s
  3. 14h 33m 55s Correct answer.
  4. 14h 32m 10s

Explanation

A chronometer that is slow indicates a time earlier than UTC, so the stated error must be added to the reading to obtain UTC. Adding 1m 45s to 14h 32m 10s gives 14h 33m 55s.

Question 28Navigation and Position Determination

A vessel in the northern hemisphere observes the sun at meridian passage with a true altitude of 60°00' and a declination of 15°00'N. What is the vessel's latitude?

  1. 45°00'N Correct answer.
  2. 30°00'N
  3. 75°00'N
  4. 15°00'N

Explanation

When the sun is on the meridian and bears south of the observer, the latitude equals the declination plus the zenith distance. Here the zenith distance is 90° minus 60° = 30°, and adding the northerly declination of 15° gives a latitude of 45°00'N.

Question 29Navigation and Position Determination

Which statement correctly describes the relationship between the tropical year and the civil calendar used for nautical almanac data?

  1. The tropical year is longer than 365.5 days, so two leap days are inserted every four years
  2. The tropical year varies with the observer's latitude, so almanac dates differ between hemispheres
  3. The tropical year is exactly 365 days, so no leap-day corrections are ever required
  4. The tropical year is shorter than 365.25 days, which is why the Gregorian leap-year rule omits leap days in most century years Correct answer.

Explanation

The tropical year is about 365.2422 days, slightly less than the 365.25-day Julian average. The Gregorian calendar compensates by dropping the leap day in century years not divisible by 400, which keeps almanac dates aligned with the seasons.

Question 30Navigation and Position Determination

What is the purpose of the 'tidal window' noted in a berthing plan for a vessel with a deep draft?

  1. To record when the barometric pressure is expected to be lowest
  2. To identify the period when tidal stream direction favours the approach
  3. To state the times when the tidal range reaches its maximum spring value
  4. To define the interval when the predicted height of tide provides adequate under-keel clearance Correct answer.

Explanation

A tidal window is the period during which the predicted tide gives sufficient depth for a vessel's draft plus under-keel clearance and squat. It is calculated from predicted heights, not from stream direction, spring range or pressure.

Question 31Meteorology and Oceanography

Under the Bergen School model, which air mass arrangement and frontal structure characterises the mature stage of an extratropical cyclone?

  1. A warm front trailing east and a cold front trailing south-west from a common low centre, with warm air ahead of the cold front Correct answer.
  2. Two parallel cold fronts converging at the low centre with no warm sector
  3. A stationary front aligned north-south through the low centre with equal air masses on both sides
  4. A single occluded front extending east from the low centre with cold air on both sides

Explanation

In the mature (open-wave) stage the low has a warm sector bounded by a warm front to the east and a cold front to the south-west, both meeting at the low centre. Warm air is present in the sector between the fronts, while cold air lies ahead of the warm front and behind the cold front. This asymmetric frontal geometry is the defining signature of the classical Norwegian model.

Question 32Meteorology and Oceanography

A vessel is approaching a low-pressure centre in the Northern Hemisphere. Which change in wind direction should the officer of the watch expect if the vessel passes the centre with the low remaining on the port side?

  1. Wind backs steadily as the low passes ahead of the beam Correct answer.
  2. Wind shifts through 180 degrees instantaneously at the centre
  3. Wind veers steadily as the low passes abaft the beam
  4. Wind remains constant in direction but increases in strength

Explanation

In the Northern Hemisphere, surface wind circulates anticlockwise around a low, so the wind direction depends on the bearing of the centre. If the low passes ahead of the beam, the vessel crosses from the right-hand to the left-hand side of the storm track and the wind backs. If the low passes abaft the beam, the wind veers instead.

Question 33Meteorology and Oceanography

Which statement correctly describes the formation of sea fog (advection fog) off the Norwegian coast?

  1. Subsidence in a high-pressure system warms the surface layer until it saturates
  2. Warm moist air moving over a colder sea surface is cooled to its dew point from below Correct answer.
  3. Cold air moving over a warmer sea surface picks up moisture and becomes saturated
  4. Radiation cooling of the sea surface overnight produces saturation in the lowest layer

Explanation

Advection fog forms when relatively warm, humid air is transported over a colder surface; the air is chilled from below until it reaches its dew point and condensation occurs. The cold Norwegian coastal current provides exactly this colder surface, so warm southerly air can produce persistent sea fog. Radiation and subsidence processes act differently and are not the cause of advection sea fog.

Question 34Meteorology and Oceanography

A ship's barograph shows a fall of 6 hPa in three hours, and the wind has backed from south-west to south-east with increasing strength. Which interpretation is most consistent with these observations in the Northern Hemisphere?

  1. The vessel is in the centre of a stationary high with little change expected
  2. The vessel has passed an anticyclone and is entering a ridge
  3. The vessel is approaching the warm sector of a deepening depression Correct answer.
  4. The vessel is in the rear of a depression and pressure should rise shortly

Explanation

A rapid pressure fall combined with backing wind indicates that the vessel is approaching the warm sector of a deepening low, with the centre still ahead. In the Northern Hemisphere the wind backs as the warm front approaches. A rising barometer would instead suggest the rear of the system.

Question 35Meteorology and Oceanography

Which condition is necessary for the development of a tropical revolving storm?

  1. Sea surface temperature of at least about 26-27 degrees Celsius to a sufficient depth Correct answer.
  2. High vertical wind shear through the troposphere
  3. Presence of a strong upper-level jet directly over the disturbance
  4. Location within about 5 degrees of the equator

Explanation

Tropical cyclogenesis requires a warm ocean surface, generally at least 26-27 degrees Celsius, with the warm water extending to a reasonable depth to supply latent heat. It also requires sufficient Coriolis force, so formation within about 5 degrees of the equator is suppressed. Strong vertical shear and proximity to the equator both inhibit development.

Question 36Meteorology and Oceanography

A vessel is in the dangerous semicircle of a tropical cyclone in the Northern Hemisphere. Which manoeuvre is generally recommended if sea room permits?

  1. Stop engines and ride out the storm with the wind on the beam
  2. Steer directly towards the centre to cross the track quickly
  3. Put the wind on the starboard bow and proceed at best speed to clear the storm track ahead Correct answer.
  4. Steer with the wind on the starboard quarter to run with the storm

Explanation

In the Northern Hemisphere dangerous semicircle the wind circulates anticlockwise, so putting the wind on the starboard bow allows the vessel to head away from the approaching track and out of the worst quadrant. The aim is to increase distance from the centre and avoid being carried across the path of the storm. Stopping or steering towards the centre increases exposure.

Question 37Meteorology and Oceanography

Which statement best explains why the semi-diurnal tide dominates most of the world's coastlines?

  1. Diurnal tides arise from the moon's declination and are therefore always larger than semi-diurnal tides
  2. Local coastal geometry removes all diurnal components from the tide
  3. The tide-generating forces of the moon and sun have a strong twice-daily component from Earth's rotation under the tidal envelope Correct answer.
  4. Semi-diurnal tides are produced only by solar heating of the ocean surface

Explanation

As Earth rotates beneath the two tidal bulges, most locations pass through high and low water approximately twice per lunar day, producing the semi-diurnal component. Diurnal inequality arises mainly from lunar and solar declination. Coastal geometry modifies the tide but does not eliminate the dominant semi-diurnal forcing everywhere.

Question 38Meteorology and Oceanography

Which current is primarily responsible for moderating the climate of the Norwegian coast and keeping its ports ice-free in winter?

  1. The North Atlantic Drift, a continuation of the Gulf Stream Correct answer.
  2. The East Greenland Current
  3. The Labrador Current
  4. The Canary Current

Explanation

The North Atlantic Drift carries warm, saline Atlantic water north-eastwards past the Norwegian coast, moderating winter temperatures and helping keep ports ice-free. The East Greenland and Labrador currents carry cold polar water and would have the opposite effect. The Canary Current is a cold eastern boundary current off north-west Africa.

Question 39Meteorology and Oceanography

A vessel is navigating in the vicinity of the Polar Front. Which cloud sequence would most reliably indicate the approach of the associated warm front?

  1. Clear skies with isolated cirrus that dissipates within an hour
  2. Stratocumulus with a stable base and no change in cloud type
  3. Cumulus congestus building rapidly with a high cloud base and good visibility
  4. Cirrus, then cirrostratus, altostratus and finally nimbostratus with lowering cloud base Correct answer.

Explanation

Ahead of a warm front the cloud sequence progresses from high cirrus through cirrostratus and altostratus to low nimbostratus as the frontal surface approaches and the cloud base lowers. This orderly thickening and lowering is the classic warning of warm-frontal precipitation. Convective cloud types indicate instability rather than frontal approach.

Question 40Meteorology and Oceanography

Which statement correctly describes the formation of a barometric wave (isentropic) chart and its use in routeing?

  1. It displays tidal stream vectors for the following 24 hours
  2. It shows the height of a selected pressure surface and reveals troughs and ridges aloft for identifying steering flow Correct answer.
  3. It plots sea surface temperature against salinity to identify currents
  4. It shows only surface pressure isobars and is used only for coastal forecasts

Explanation

Upper-air charts such as 500 hPa height analyses show troughs and ridges aloft, which indicate the steering flow and likely development of surface systems. They are used in weather routeing to anticipate the movement and deepening of lows. Surface pressure charts and tidal atlases serve different purposes.

Question 41Meteorology and Oceanography

A vessel observes a rapid drop in air temperature, a wind shift from south-west to north-west, and a line of cumulonimbus with heavy showers. Which feature has most likely passed the vessel?

  1. An anticyclone ridge
  2. A stationary front
  3. A warm front
  4. A cold front Correct answer.

Explanation

A cold front passage is marked by a sharp wind veer, a temperature fall as colder air replaces the warm sector, and a narrow band of convective cloud with showers or squalls. The sequence described matches these signs closely. A warm front would instead bring rising temperature and layered cloud.

Question 42Meteorology and Oceanography

Which combination best explains the formation of the Antarctic Circumpolar Current and its role in global ocean circulation?

  1. Strong westerly winds over the Southern Ocean drive a deep, unobstructed eastward flow that connects the major ocean basins Correct answer.
  2. Trade winds drive a shallow westward flow around Antarctica that is blocked by the Antarctic Peninsula
  3. The current is driven mainly by tidal forces and reverses direction with the spring-neap cycle
  4. The current is a seasonal feature that disappears during the austral winter

Explanation

The Antarctic Circumpolar Current is driven largely by the strong westerly wind belt over the Southern Ocean and is unimpeded by continental landmasses, allowing a deep eastward flow. It links the Atlantic, Indian and Pacific basins and plays a key role in global heat and freshwater transport. It is a persistent feature, not seasonal or tidally dominated.

Question 43Meteorology and Oceanography

Which statement correctly describes the relationship between sea state and fetch, wind duration and wind speed?

  1. Wave height decreases as fetch increases because waves disperse energy
  2. Wave height is determined solely by the phase of the moon
  3. Wave height depends only on wind speed and is independent of fetch and duration
  4. Wave height increases with wind speed, fetch and duration until a fully developed sea is reached Correct answer.

Explanation

Significant wave height grows with wind speed, fetch length and wind duration, approaching a limiting fully developed sea when the available fetch and duration are sufficient. Beyond that limit, additional wind energy is balanced by wave breaking and dissipation. Fetch and duration therefore matter as much as wind speed.

Question 44Meteorology and Oceanography

An officer is asked to interpret a synoptic chart showing a tightly packed low with isobars close together and a col between two highs. Which interpretation is correct?

  1. Closely spaced isobars indicate an anticyclone, while a col indicates a tropical cyclone
  2. Closely spaced isobars indicate strong pressure gradient and strong winds, while a col indicates a neutral saddle region with light variable winds Correct answer.
  3. Closely spaced isobars indicate a stable atmosphere with no wind, while a col indicates a frontal zone
  4. Closely spaced isobars indicate light winds, while a col indicates the strongest winds in the chart

Explanation

Isobar spacing is inversely related to pressure gradient: closely packed isobars mean a steep gradient and strong winds. A col is a saddle point between two highs and two lows where the pressure gradient is weak, giving light and variable winds. Neither feature on its own identifies a tropical cyclone or a frontal zone.

Question 45Ship Stability and Ship Construction

Under the IS Code intact stability criteria, what is the minimum area under the righting lever curve up to 30 degrees of heel?

  1. 0.030 metre-radians
  2. 0.090 metre-radians
  3. 0.016 metre-radians
  4. 0.055 metre-radians Correct answer.

Explanation

The 2008 IS Code requires the area under the GZ curve to be at least 0.055 m·rad up to 30 degrees, 0.090 m·rad up to 40 degrees, and 0.030 m·rad between 30 and 40 degrees. These values are the baseline against which the remainder of the curve criteria are checked.

Question 46Ship Stability and Ship Construction

A vessel has KM 8.20 m and KG 7.10 m. What is the initial metacentric height?

  1. 15.30 m
  2. 1.55 m
  3. 0.87 m
  4. 1.10 m Correct answer.

Explanation

GM is found from GM = KM − KG, so 8.20 − 7.10 gives 1.10 m. A positive GM indicates the vessel is initially stable in the upright condition.

Question 47Ship Stability and Ship Construction

Which statement correctly describes the effect of a free surface in a partly filled tank?

  1. It affects only the righting lever at large angles of heel.
  2. It raises the effective KG and reduces the corrected GM. Correct answer.
  3. It has no effect on GM provided the tank is symmetrical about the centreline.
  4. It lowers the effective KG and increases the corrected GM.

Explanation

Liquid that can move transversely shifts the centre of gravity toward the low side, producing a virtual rise of G equal to the free surface moment divided by displacement. This reduces the corrected GM and is treated as an increase in KG.

Question 48Ship Stability and Ship Construction

What does the load line mark assigned under the 1966 Load Line Convention primarily certify?

  1. The maximum deadweight the ship may carry in port
  2. The maximum speed permitted in restricted waters
  3. The minimum GM required in the departure condition
  4. The minimum freeboard permitted in specified zones and seasons Correct answer.

Explanation

The load line assignment fixes minimum freeboards for designated zones and seasonal periods, ensuring adequate reserve buoyancy. The Plimsoll mark and seasonal lines show the limits the master must observe when loading.

Question 49Ship Stability and Ship Construction

In a ship's hydrostatic particulars, what does the block coefficient Cb express?

  1. The ratio of the waterplane area to the area of the circumscribing rectangle
  2. The ratio of the underwater volume to the volume of a rectangular block of length, breadth and draught Correct answer.
  3. The ratio of displacement to the volume of the parallel middle body
  4. The ratio of the midship section area to the area of a rectangle of breadth and draught

Explanation

Cb compares the actual displaced volume with the block formed by length between perpendiculars, breadth and draught. It is a measure of hull fullness and is used together with displacement to derive volume.

Question 50Ship Stability and Ship Construction

What is the purpose of fitting a collision bulkhead in a cargo ship?

  1. To separate the engine room from the cargo holds
  2. To increase the transverse strength of the double bottom
  3. To provide a watertight boundary around the steering gear
  4. To limit flooding and preserve buoyancy if the bow is damaged Correct answer.

Explanation

The collision bulkhead is a watertight transverse bulkhead near the forward end that limits the extent of flooding after bow damage. Its position is governed by class and SOLAS requirements so that reserve buoyancy is maintained.