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 2 of 3.

Harish Thakur said:   8 years ago
Thanks @Pradeep Gk.

Shubam said:   7 years ago
Pressure = density * gravity * height.
9.81*1000*0.6 =5886N/M2,
Then force =pressure *area,
5886 * 3 * 2 = 35.3 KN.

Gokulkumar said:   7 years ago
Let's take the body as the submerged body then the formula is given by;

W*A*X.
9810*6*0.6.
35314N = 35.3 KN.

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

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.

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)

Sai varma said:   6 years ago
Thanks @Vishwas.

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

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

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)


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