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 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.
                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.
                = 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.
                = 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.
                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.
                = 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.
                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.
                = (9.81) * (3 * 2 * 0.6),
= 9.81 * 3.6,
= 35.316 KN.
                        Debashis said: 
                         
                        7 years ago
                
                Well said, thanks @Shubham.
                Post your comments here:
 
            
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