Question

a piece of wood of volume 200cc and density 0.84 gram per centimetre cube floats in a liquid of density 2 gram per centimetre cube What volume of wood will remain above the surface of liquid what force must be exerted on what to keep a totally submerged​

Answer 1

Let the volume of the block of wood be

Vcm3

and its density be dwgcm−3

So the weight of the block =Vdwg

dyne, where g is the acceleration due to gravity

=980cms−2

The block floats in liquid of density 0.8gcm−3

with 14th of its volume submerged.So the upward buoyant force acting on the block is the weight of displaced liquid=14V×0.8×gdyne.

Hence by cindition of floatation V×dw×g=14×V×0.8×g⇒dw=0.2gcm-3,

Now let the density of oil be dogcm-3

The block floats in oil with 60% of its volume submerged.So the buoyant force balancing the weight of the block is the weight of displaced oil = 60%×V×do×gdyne

Now applying the condition of floatation we get

60%×V×do×g=V×dw×g⇒60

100×V×do×g=V×0.2×g⇒do=0.2×106=13=0.33gcm-3