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.
Ramdeen said:
1 decade ago
@Fakhre Alam check your calculations.
Fakhre Alam said:
1 decade ago
Volume = 3x2x1 = 6m^3.
Volume of water displaced = Weight of body in kN/specific weight of water in kN/m^3.
V = W/9.81 ----------- (1).
We also know that,
Depth of immersion = Volume/Cross sectional area.
0.6 = V/(3x2) -------- (2).
Comparing (1) & (2) we find.
W = 35.3 kN.
Volume of water displaced = Weight of body in kN/specific weight of water in kN/m^3.
V = W/9.81 ----------- (1).
We also know that,
Depth of immersion = Volume/Cross sectional area.
0.6 = V/(3x2) -------- (2).
Comparing (1) & (2) we find.
W = 35.3 kN.
Jaydip Boda said:
1 decade ago
P = 1000*10*.6
= 6000 N/m^2
F = 6000*6
= 36000 N
= 36 KN
= 6000 N/m^2
F = 6000*6
= 36000 N
= 36 KN
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