Question

A cylindrical woodden float whose base area is 4 m^2 and height 1 m drifts on water surface in vertical position . Density of wood is 500kg/m^3 and that of water is 1000 kg/m^3 .What minimum work must be performed to take the float out of water

Answer 1

Let the volume of the block of wood be Vcm3and 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−3with 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×g dyne

Now applying the condition of floatation we get

60%×V×do×g=V×dw×g

⇒60100×V×do×g=V×0.2×g

⇒do=0.2×106=13=0.33gcm−3

hope this is helpful...