A piece of wood of volume 0.6 m^(3)floats in water. Find the volume exposed What force is required to meet completely under water? Density of wood =800 kg//m^(3).

A piece of wood of volume 0.6 m^(3)floats in water. Find the volume ex

1.	A piece of wood of volume 0.6 m^(3)floats in water. Find the volume exposed What force is required to meet completely under water? Density of wood =800 kg//m^(3).
Answer» Solution :` VDG = Vrho g , 0 . 6 xx 800 = v xx 1000, v = 0. 48 m ^(3) , V - v = 0.12 m^(3) , ` Additional force= Upthrust - Weight = `V RHO g - VD g = ( 0. 6 xx 1000 - 0.6 xx 800) g = 1176 N`