Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion
Discussion Forum : Hydraulics and Fluid Mechanics - Section 4 (Q.No. 37)
37.
A uniform body 3 m long, 2 m wide and 1 m deep floats in water. If the depth of immersion is 0.6 m, then the weight of the body is
Discussion:
23 comments Page 3 of 3.
Sanjeet Mishra said:
5 years ago
Weight of body = weight of water displaced = buoyancy force.
= Density x volume x g.
= 1000 x 3 x 2 x 0.6 x 9.81.
= 35316 N = 35.316 KN.
= Density x volume x g.
= 1000 x 3 x 2 x 0.6 x 9.81.
= 35316 N = 35.316 KN.
(2)
Blinku poy Blinku said:
5 years ago
Depth of immersion =0.6.
= sp. Grvty x depth of body.
Depth of body = 1m.
Therefore sp. gravty=0.6 hence density of body, ρ.
=600kg/m^3.
Wt of body =rho V g= 600x 3x2x1x9.81 = 35316N.
= sp. Grvty x depth of body.
Depth of body = 1m.
Therefore sp. gravty=0.6 hence density of body, ρ.
=600kg/m^3.
Wt of body =rho V g= 600x 3x2x1x9.81 = 35316N.
(3)
Vignesh said:
5 years ago
Weight of body =weight of liquid displaced by the body so,
W = 1000 * 9.81 * 2 * 3 * 0.6
W = 35.316KN.
W = 1000 * 9.81 * 2 * 3 * 0.6
W = 35.316KN.
(1)
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