Civil Engineering - Hydraulics - Discussion

42. 

The radius of gyration of the water line of a floating ship is 4 m and its metacentric height is 72.5 cm. The period of oscillation of the ship, is

[A]. π
[B].
[C].
[D].
[E]. π/2

Answer: Option C

Explanation:

No answer description available for this question.

John Wick said: (Jan 27, 2015)  
Simply put all values in formula.

Time period = 2*π*(4^2/9.80665*0.725)^1/2.

Ans = 3π

Abhay said: (Jan 26, 2016)  
Time period = 2π (K^2/GM. g).

Where, K = Radius of gyration.

GM = Metacentric height.

T = 2π (4^2/0.725x9.81) = 3π answer.

Dinesh Prasad Sao said: (Feb 14, 2017)  
Here 2π(K^2/M.t x g)^.5.

Kanji Dodiya said: (Sep 1, 2017)  
Time period = 2Π (K^2/GM* g)^0.5.
T=2Π (4^2/0.725*9.81)^0.5,
T=2Π *(16/7.11)^0.5,
T=2Π *2.25^0.5,
T=2Π *1.5,
T=3Π -> ANS.

Aman Singla said: (Jan 24, 2018)  
Agreed @Kanji Dodiya.

Mukul Paul said: (Aug 27, 2019)  
Time period = 2π*{k^2/GM*g}^0.5.
T=3π.

Subhasmita Mishra said: (Feb 10, 2020)  
How to solve this problem without a calculator?

Basem Mohammed said: (Aug 9, 2020)  
How to solve this problem? Describe in detail.

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