Ship weighs 127 MN. On filling the ship's boats on one side with
Ship weighs 127 MN. On filling the ship's boats on one side with water weighing 600 kN with the mean distance of the boats from the centre line of the ship being 10 m, the angle of displacement of the plumb line is 2°16'. The metacentric height will be nearly (Take sin 2°16 = 0.04, cos 2°16' = 0.9992 and tan 2°16' = 004)
A. 1.73 m
B. 1.42 m
C. 1.18 m
D. 1.87 m
Please scroll down to see the correct answer and solution guide.
Right Answer is: C
SOLUTION
Concept:
Meta Centre is defined as the point about which a body starts oscillating when body is tilted by a small angle. The meta-centre may also be defined as the point at which the line of action of the force of buoyancy will meet the normal axis of the body when the body is given a small angular displacement.
The distance MG i.e. the distance between the meta-centre of a floating body and the centre of gravity of the body is called meta-centric height. It is measured along the line BG.
Calculation:
G = centre of gravity
GM = Meta centre height
GX = distance b/w boat and ship = 10 m
M = Meta centre
W1 = Total weight
W2 = Weight of boat
WTotal = Wship + Wboat = 127000 + 600 = 127600 kN
Take moment of all forces about ‘μ’
WTotal × (GD) = Wboat × (GX)
127600 × x = 600 × 10
∴ x = 4.4 × 10-2 m
In ΔGMD
\(\sin \theta = \frac{x}{{GM}}\)
\(GM = \frac{{4.4 \times {{10}^{ - 2}}}}{{\sin \theta {{16}^2}}} = \frac{{4.4 \times {{10}^{ - 2}}}}{{04}}\)
∴ GM = 1.17 m
