When a floating body is given a small angular displacement, the body will oscillate about its metacentre. If T denotes the time period of oscillation or rolling (i.e., time for one complete oscillation) of the floating body, then, the relation is: (where GM is the metacentric height, k is the radius of gyration and T is in seconds)

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Question

This question was previously asked in

BHEL Engineer Trainee Civil 24 Aug 2023 Official Paper

Answer 1

Explanation:

Time period of oscillation of a floating body is given by

\({\rm{T}} = 2{\rm{\pi }}\sqrt {\frac{{{{\rm{K}}^2}}}{{{\rm{g}}\left( {{\rm{GM}}} \right)}}} \)

Where,

K = radius of gyration, and GM = Meta centre height

For a given ‘K’, \({\rm{T}} \propto \frac{1}{{\sqrt {{\rm{GM}}} }}\)

⇒ Higher GM, lesser will be oscillation of body.

The line oscillation of a floating body with increase in metacentric height will be lower.