A ship of 3300 tonne displacement is undergoing an inclining experiment in seawater of density 1025 kg/m\(^3\). A mass of 6 tonne is displaced transversely by 12 m as shown in the figure. This results in a 0.12 m deflection of a 11 m long pendulum suspended from the centerline. The transverse metacenter of the ship is located at 7.25 m above the keel. The distance of the center of gravity from the keel is ________ m (rounded off to two decimal places).
Show Hint
In inclining experiments, the deflection of the pendulum helps in calculating the distance of the center of gravity by using the transverse metacentric height and the mass displaced.
Step 1: Understanding the formula for the inclining experiment.
The formula that relates the deflection \( \delta \), the mass displaced \( m \), the length of the pendulum \( L \), the transverse metacentric height \( GM \), and the distance of the center of gravity \( KG \) is:
\[
\delta = \frac{w \cdot GM}{m \cdot L} \times KG
\]
Where:
\( w = {weight of displaced mass} = m_{{displaced}} \times g \),
\( m = {total mass of the ship} = m_{{ship}} \),
\( g = 9.81 \, {m/s}^2 \) (gravitational acceleration).
Rearranging to solve for \( KG \):
\[
KG = \frac{\delta \cdot m \cdot L}{w \cdot GM}
\]
Step 2: Substituting the known values.
Substitute all known values into the formula to calculate \( KG \):
\[
KG = \frac{0.12 \cdot (3300 \times 10^3) \cdot 11}{(6 \times 10^3) \cdot 7.25}
\]
Step 3: Simplifying the expression.
Simplify the expression to calculate:
\[
KG = \frac{0.12 \cdot 3300 \cdot 11}{6 \cdot 7.25} = \frac{43560}{43.5} = 5.20 \, {m}
\]
Final Answer:
The distance of the center of gravity from the keel is \( \boxed{5.20} \, {m} \).
