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 2 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.
W = 1000 * 9.81 * 2 * 3 * 0.6
W = 35.316KN.
(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.
= (9.81) * (3 * 2 * 0.6),
= 9.81 * 3.6,
= 35.316 KN.
Pradeep gk said:
8 years ago
Presur (p)=density*gravity*height.
Force/area=1000*9.81*0.6,
Force=1000*9.81*0.6*6,
F=35.316 KN.
Force/area=1000*9.81*0.6,
Force=1000*9.81*0.6*6,
F=35.316 KN.
Hindu raj said:
9 years ago
You are correct and you explained with the simple concept. Thanks @Vishwas.
Sivareddy said:
6 years ago
The weight of the body = (3*2*1) * 9.81*0.6.
= 35.3KN.
= 35.3KN.
(1)
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
Ramdeen said:
1 decade ago
@Fakhre Alam check your calculations.
Narayana sahu said:
6 years ago
You are right, thanks @Gokulkumar.
Jaysukh said:
9 years ago
Your method is right @Rahul.
Debashis said:
7 years ago
Well said, thanks @Shubham.
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