Rolling of a ship: Given metacentric height GM = 0.6 m and radius of gyration k = 4 m, compute the time period of rolling.

Introduction / Context:
The small-angle rolling motion of a ship can be modeled as an angular oscillation where the hydrostatic restoring moment is proportional to the heel angle. The natural period depends on the metacentric height (initial stability) and the mass moment of inertia about the roll axis (captured by the radius of gyration k).

Given Data / Assumptions:

• Metacentric height GM = 0.6 m.
• Radius of gyration k = 4 m.
• g = 9.81 m/s^2.
• Small-angle (linear) theory is valid.

Concept / Approach:
The standard rolling period relation is T = 2 * π * k / sqrt(g * GM). Greater GM provides stronger restoring moment and hence a shorter period; a larger k (more mass distributed away from the axis) increases the period.

Step-by-Step Solution:

Verification / Alternative check:
Unit check: k in meters, g in m/s^2, GM in meters, so sqrt(g * GM) has units of 1/s, making T in seconds. A sensitivity check shows that a 10% increase in GM would reduce T by about 5%, consistent with the square-root dependence.

Why Other Options Are Wrong:

• 4.1 s and 5.2 s: These are far too short for the given inertia; would imply unrealistically large GM for k = 4 m.
• 14.1 s: Too long; would correspond to a much smaller GM than 0.6 m.

Common Pitfalls:
Using degrees instead of radians (2 * π factor is essential); forgetting to take the square root; confusing k (radius of gyration) with metacentric radius BM; mixing units for g or length.

Final Answer:
10.4 s