Civil Engineering - Hydraulics - Discussion
Discussion Forum : 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
Discussion:
8 comments Page 1 of 1.
Basem Mohammed said:
5 years ago
How to solve this problem? Describe in detail.
(2)
Subhasmita Mishra said:
6 years ago
How to solve this problem without a calculator?
Mukul Paul said:
6 years ago
Time period = 2π*{k^2/GM*g}^0.5.
T=3π.
T=3π.
Aman Singla said:
8 years ago
Agreed @Kanji Dodiya.
Kanji dodiya said:
8 years ago
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.
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.
(5)
Dinesh prasad sao said:
9 years ago
Here 2π(K^2/M.t x g)^.5.
Abhay said:
10 years ago
Time period = 2π (K^2/GM. g).
Where, K = Radius of gyration.
GM = Metacentric height.
T = 2π (4^2/0.725x9.81) = 3π answer.
Where, K = Radius of gyration.
GM = Metacentric height.
T = 2π (4^2/0.725x9.81) = 3π answer.
(2)
John wick said:
1 decade ago
Simply put all values in formula.
Time period = 2*π*(4^2/9.80665*0.725)^1/2.
Ans = 3π
Time period = 2*π*(4^2/9.80665*0.725)^1/2.
Ans = 3π
(1)
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