# Civil Engineering - Hydraulics - Discussion

### Discussion :: Hydraulics - Section 1 (Q.No.42)

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]. 2π [C]. 3π [D]. 4π [E]. π/2

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.