Fluid Mechanics Test 2

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Fluid Mechanics Test 2

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Weekly Quiz Competition

Question 1
1 / -0

A ship with metacentric height 0.36 m is cruising in sea. The Captain of the ship asked an engineer to calculate period of oscillation. What is the period of oscillation _____
(Radius of gyration = 0.9 m)

A
2 sec

B
3 sec

C
4 sec

D
5 sec

Solution

Concept:
Period of oscillation,
\(T = 2\pi \sqrt {\frac{{k_G^2}}{{g\;\;\overline {GM} }}} \)
\(\overline {GM} \) = Metacentric height, k_G = Radius of gyration and g = acceleration due to gravity
Calculation:
\(T = 2\pi \sqrt {\frac{{{{\left( {0.9} \right)}^2}}}{{9.81\; \times \;0.36}}} \)
∴ T = 3.01 sec
Question 2
1 / -0

A body is weighing 100 N was found to weigh 50 N in oil (ρ = 800 kg/m³), what is relative density of object with respect to water?

A
0.625

B
1.2

C
1.6

D
2.1

Solution

Concept:
\({\rm{Relative\;density}} = \frac{{{\rho _b}}}{{{\rho _w}}}\)
where, ρ_b = density of body, ρ_w = density of water
Weight by weighing machine = Actual weight – buoyancy force
Calculation:
50 N = 100 N – (ρVg)_oil
∴ (ρVg)_oil = 50 N
\({V_{body}} = \frac{{50}}{{\left( {9.81\; \times \;800} \right)}}\)
\({V_{body}} = 6.37 \times {10^{ - 3}}\;{m^3}\)
Now,
Weight of body = ρ_b V_b g
∴ 100 N = ρ_b × 6.37 × 10^-3 × 9.81
∴ ρ_b = 1600 kg/m³
\({\rm{Relative\;density\;}} = \frac{{{\rho _b}}}{{{\rho _w}}} = \frac{{1600}}{{1000}}\)
∴ Relative density = 1.6
Question 3
1 / -0

The time period of rolling of a ship of weight 28.4 MN in sea water is ten seconds. The centre of buoyancy of ship is 1.6 m below the centre of gravity. The moment of inertia of ship about the water line is 8500 m⁴, and density of sea water is 1025 kg/m³, find the radius of gyration of ship.

A
6.75 m

B
5.91 m

C
6.31 m

D
5.25 m

Solution

Explanation:
Given:
BG = 1.6 m, T = 10 sec, I = 8500 m⁴, ρ_w = 1025 kg/m³, Weight of ship = 28.4 × 10⁶ N
Now,
Volume of water displaced by ship \( = {\rm{\;}}\frac{{Weight\;of\;ship}}{{{\rho _w}g}}\)
\(V = \frac{{28.4 \times {{10}^6}}}{{1025 \times 9.81}}\)
∴ V = 2824.39 m³
Now,
Metacentric height \({\rm{\;}}\left( {GM} \right) = \frac{I}{V} - BG\)
\(GM = \frac{{8500}}{{2824.39}} - 1.6\)
∴ GM = 1.409 m
Now,
\({\rm{Time\;period\;}}\left( T \right) = 2\pi \sqrt {\frac{{{k^2}}}{{g \times GM}}} \)
\(10 = 2\pi \sqrt {\frac{{{k^2}}}{{9.81 \times 1.409}}} \)
∴ Radius of gyration (k) = 5.91 m