SCS Class - IV

SCS Class - IV

95SCS-4

Sr. No. 2

EXAMINATION OF MARINE ENGINEER OFFICER

Function: Controlling Operation of ship and care for persons on board at Operational Level

SHIP CONSTRUCTION & STABILITY

CLASS IV

(Time allowed - 3 hours)

INDIA (2001) Afternoon Paper Total Marks 100

N.B. - (1) All Questions are compulsory

(2) All Questions carry Equal Marks.

(3) Neatness in handwriting and clarity in expression carries weightage

  1. An undivided double bottom tank 24 meters long, with vertical sides of depth 1.5m has ½ ordinates commencing at the after perpendicular as follows:

Station / 0 / 1 / 2 / 3 / 4 / 5 / 6
½ Ordinate (m) / 6.0 / 6.0 / 5.9 / 5.7 / 5.4 / 4.9 / 4.3

Find

(1)the volume of the tank,

(2)the distance of the centroid of the tank from the AP.

  1. A vessel displacing 16000 tonnes has a TPC 20 tonnes/cm and is floating at a draft of 9.4m in water of density 1.010 tonnes/m3 in a summer zone. If the vessel’s summer load draft is 9.5m, calculate amount of cargo she may load.
  1. A vessel displacement 12500 tonnes has KG 9.6m. On completion of loading she is required to have KG of 9.5m.

1000 tonnes is loaded at Kg 5.5m.

850 tonnes is loaded at Kg 13.6m.

find the Kg at which to load a further 1600 tonnes of cargo to produce the required final KG.

  1. A box vessel has length, 95m; breadth, 8m; depth, 6.5m and is floating at draft 4m in salt water with KG 3.0m. Find the drafts fore and aft if a full breadth end compartment of length 6m is bilged.
  1. A ship leaves port upright with a full cargo of timber, and with timber on deck. During the voyage, bunkers, stores and fresh water are consumed evenly from each side. If the ship arrives at her destination with a list, explain the probable cause of the list and how this should be remedied.
  1. Sketch transverse section through a ship showing the positions of centre of gravity, centre of buoyancy, and initial matacenter, when the ship is in (a) stable equilibrium (b) unstable equilibrium, and (c) neutral equilibrium
  1. You can generally improve the vessel's stability in a hazardous situation by ______.

(A)pumping double bottoms to the forepeak

(B)ballasting deep tanks

(C)transferring ballast athwartships

(D)deballasting double bottoms

Briefly Justify your Answer

  1. Which of the following conditions will occur to the ship's center of gravity if 200 tons of steel is transferred to the ships cargo hold from shore side?

(A)The reserve buoyancy will rise.

(B)The center of gravity will remain in the same position.

(C)The center of gravity will be lowered.

(D)The reserve buoyancy will remain the same.

Briefly Justify Your Answer

  1. If a vessel rolls to the starboard side, and there are no movable or moving weights onboard, the center of gravity will ______.

(A)move to port

(B)move to starboard

(C)move directly down

(D)stay in the same position

Briefly Justify Your Answer

  1. In the event of a collision, watertight integrity may be lost if ______.

(A)the sounding tube cap from a damaged tank is missing

(B)the dogs on a manhole cover are secure

(C)you have recently replaced a gasket in a watertight door

(D)you operate the dewatering system from a flooded compartment

Briefly Justify Your Answer

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95SCS-4

Sr. No. 2

EXAMINATION OF MARINE ENGINEER OFFICER

Function: Controlling Operation of ship and care for persons on board at Operational Level

SHIP CONSTRUCTION & STABILITY

CLASS IV

(Time allowed - 3 hours)

INDIA (2001) Afternoon Paper Total Marks 100

N.B. - (1) All Questions are compulsory

(2) All Questions carry Equal Marks.

(3) Neatness in handwriting and clarity in expression carries weightage

Answers
Answer for Question No. 1

(1) Volume = 397.2m3

(2) Distance of Centroid of tank from AP = 11.4m

Answer for Question No. 2

FWA 200m

DWA 120m

Sinkage 22cm

Cargo to load 433.6 tonnes.

Answer for Question No. 3
Weight (tonnes) / KG / Moment
12500 / 9.6 / 120000
1000 / 5.5 / 5500
850 / 13.6 / 11560
1600 / X / 1600X
15950 / 137060 + 1600X

KG = moment of weight

Displacement

9.5 = 137060 + 1600X

15950

151525 = 137060 + 1600X

1600X = 14465

X = 9.041m

Load 1600 tonnes at Kg 9.041m

(Note. A more partical view is to regard 9.041m as the mean Kg of the 1600 tonnes after it has been distributed between availablespaces.)

Answer for Question No. 4

MCTC 50.48 tonne m/cm

Trim 185.2cm

Draft forward 5.254m; aft 3.404m

Answer for Question No. 7 Correct Answer: B
Answer for Question No. 8 Correct Answer: C
Answer for Question No. 9 Correct Answer: D
Answer for Question No. 10 Correct Answer: A