Sep 2021 Nav Arc 1

Q.2 A ship of L = 140m, COF = 2m, FWD of M/Ship, KM = 6.1 m, KG = 5.5 m, MCTC = 130 tm, TPC = 16, W = 5700T, at Drafts F/4.15 m, A/4.85 m, grounds on a rock. 10 abaft her stem. The tide then falls by 0.80 m. Calculate her V GM then and her final

Jan 2022 Nav Arc 1

Q.1 A vessel floating at drafts forward 8.0 m aft 9.0 m, grounds at a point 25 m aft of forward perpendicular. Calculate the drafts and virtual GM of the ship when the tide has fallen by 75 cm. Given MCTC 300 tm, TPC 30 t, KG 7.5 m, KM 8.5 m, Length 165 m, LCF 80 m, forward of

May 2022 Nav Arc 1

Q.2 A vessel of L = 140 mt, W = 16000t, MCTC = 190, TPC = 24t, LCF = 72m, KM = 8.20m, KG = 7.20m, FSM = 1370mt, draft = Fwd = 6.68m, Aft = 8.84m grounds on an isolated rock, the draft then are fwd 5.88 and aft 9.28m. Calculate following:i) The up thrust provided by the rock.ii)

July 2022 Nav Arc 1

Q.2 A ship of L = 150m drawing 5.0m forward and 7.0m aft. Her W = 13,000 t, TPC = 23, LCF = 73.50 m runs aground at a point 20m aft of forward perpendicular. After grounding the drafts are found to be 4.6m forward and 7.0 aft. Find:(a) Rise in tide required for the vessel to refloat. (b) Upthrust

Jan 2025 PM Nav Arc 1

Q.2 A vessel of L 148 m, LCF 70 m, draft forward 8.00 m, draft aft 9.80 m, TPC 32, MCTC 264 lightly grounds on gently sloping seabed. Soundings taken in that instant showed forward depth as 8.00 m and aft depth as 10.9 m. Find the draft after tide falls by (a) 30 cm (b) 2.00 m Answer:

Oct 2025 PM Nav Arc 1

Q.2 A ship of L = 140m, W = 16000t, LCF=72m, MCTC=190tm, FSM=137tm, KM = 8.2m, KG = 7.2m TPC=24t, draft Fwd = 6.70m. Aft = 8.85m grounds lightly on an isolated rock. The drafts now are found to be fwd 5.90m and aft 9.30m. Calculate the virtual GM and the rise of tide required to refloat the vessel. Answer:

Jan 2025 AM Nav Arc 1

Q.1 A vessel of displacement 29,000 t, length 200m, Fwd draft 7m and Aft Draft 8.5m runs aground without any damage to her hull. Her MCTC 250 tm/cm and TPC 20 t/cm, LCF 105m, FSM 1000 tm, KM 8.5m, KG 7.8m. Tide then falls by 50 cm. The drafts then observed were to be Fwd 7.8m and Aft 7.5m. Calculate

July 2023 AM Nav Arc 1

Q.1 M.V. ‘Hindship’ displacement 16160 T, KG 7.850 m, FSM 2420 Tm to be dry docked. If on taking blocks overall the Residual GM (F) is to be 0.20m, determine maximum trim with which she will enter dry dock. Also state arrival drafts F & A. Answer:

July 2024 AM Nav Arc 1

Q.3 A vessel displacing 14000 tonnes enters dry-dock with a clearance of 0.50m over the blocks. Drafts while entering dry-dock are 5.35 forward, 6.77m aft, MCTC 120, TPC 22 tonne, LCF 4.00 m aft of mid-ships, length 150m, KG 6.25M, KM 6.40m. Assume the hydrostatic data to remain constant.Determine:a) The drop in water level required before the vessel takes the

Jan 2024 AM Nav Arc 1

Q.1 A ship of length 200m, Beam 32m, draft Fwd: 6.05m, Aft 6.75m is to be dry docked. CF = 3m aft of midship, MCTC 400 tm, KM = 8.45m, KG = 7.8m, TPC = 29t. Calculate her residual GM and drafts Fwd and Aft when the trim has reduced to 10 cms. Answer:

Prepare with Confidence.
Access previous year Chief Mate MMD Numerical exam

questions with answers.

© Copyright 2025 Chief Mate MMD. All right Reserved .