A raft is constructed of wood having a density ρ=600
kg/ m3 Its surface area is A=5.7m² and its volume is V=0.06 m3
.When the raft is placed in fresh water of density 1 g/c m3 how
much of it is below water level?
a)0.1162m b)0.03 m c) 0.059m d) 1.89m e) None of these is true
Mass of raft=volume*density.
Here,volume=0.06m3 and denisty=600 kg/m3
So,mass of raft=0.06*600 = 36 kg.
So, gravitational force on raft=mg, where m is mass and g is gravitational acceleration
=36*9.8 = 352.8 N.
Upward buoyant force must balance the gravitational force for equilibrium.
Now, buoyant force=weight of liquid displaced = vdg, where v is volume of displaced liquid, d is density and g is gravitational acceleration.
Volume v=A*h, where A is cross sectional area and h is the height of raft which is inside the water.
So, buoyant force=Ahdg=352.8
Here,A=5.7 m2,d=1 g/cm3=1000 kg/m3
So, 5.7*h*1000*9.8=352.8
=>h=352.8/(5.7*1000*9.8) = 0.006316 m.
So, the answer is e)none of these.
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