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
3.53 kN
33.3 kN
35.3 kN
none of these
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
23 comments Page 1 of 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.
(1)

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.
(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.
(2)

Sagar Nirmal said:   5 years ago
Press= density * sp.gra * depth.

Press = 1 * 9.81 * 0.6,
Press = 5.88 KN/m2.

Wt of body = press* area.
Wt. Of body= 5.88 * 6.

Ans= 35.5 Kn.
(1)

Narayana sahu said:   6 years ago
You are right, thanks @Gokulkumar.

Sivareddy said:   6 years ago
The weight of the body = (3*2*1) * 9.81*0.6.
= 35.3KN.
(1)

Sai varma said:   6 years ago
Thanks @Vishwas.

Sumit singh thakur said:   6 years ago
Given,
Length 3M.
Whidth 2M.

Depth 1M and depth of immersion 0.6.

3x2x0.6 = 3.6M^3.

Weight of water 9.81x3.6 = 35.3KN.
(1)

Mongzoa said:   6 years ago
Weight of water displaced(W)= density * Volume(V).

= (9.81) * (3 * 2 * 0.6),
= 9.81 * 3.6,
= 35.316 KN.

Debashis said:   7 years ago
Well said, thanks @Shubham.


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