1.

Two objects of equal volume V=1 m^(3) and different densities d_(1) =500kg//m^(3) and d_(2) =1000kg // m^(3) are gluod to each other so that their contact surface is flat and has an area A=0.1 m^(2) When the objects are submerged in a certain liquid they float in stable equilibrium the contact surface being parallel to the surface of the liquid (see the diagram) How deep (H in meters) can the contact surface be in the liquid so that the objects are not torn apart? The maximum force that the glue can with stand is F=250 N (Neglect atmospheric pressure)

Answer»


SOLUTION :At the "critical" depth H, the equilibrium condition for the COMPOUND object suggests that the density of the liquid is `d=(d_(1)+d_(2))//2` For the top and the bottom object taken separately, the equilibrium CONDITIONS are, respectively
`d_(1) Vg+F-(dVg-dgAH)=0`
`d_(2)Vg-F-(dVg+dgAH)=0`
(For each object, the ter, in parentheses indicates the "effective" buoyancy force) Combining the last TWO equations yields the answer `H=[(d_(2)-d_(1)) Vg-2F]//[(d_(1)+d_(2)) gA]`


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