Question

Q.a cube of wood 20cm on each side,floats in water so that 30% is above the surface of the water,and 70% is below.what is the density of the wood(justify the answer)
i.what mass of lead mass has to be placed on the top of the block of wood,so that it wi;; just be totally submerged.
ii.what volume of lead mass has to be placed on the top of wood.so that it will just be totally submerged.
iii.instead of placing weights on the top of the wood,how much force would have to be exerted on the wood to submerge it.
densities in gm/cm3: water-1.0,Iron-7.8,lead-11.3

Answer 1

Vb = 0.7 * 20^3 = 5600 cm^3. = Vol. below surface.

Vb = (Dc/Dw) * Vc. = 5600
(Dc/1) * 20^3 = 5600
Dc = 5600/20^3 = g/0.7 cm^3 = Density of
the cube of wood.

1. Mass of cube = 20^3cm^3 * 0.7g/cm^3=
5600 Grams. = 5.6 kg.
To totally submerge the block, its' density must be increased to 1:
(5600+g)/20^3 = 1.0
5600 + g = 20^3
g = 20^3-5600 = 2400 Grams added.

2. V = 2400g * 1cm^3/11.34g = 212 cm^3.

3. The force = The wt. of the water
displaced:

8000cm^3 * 1g/cm^3 = 8000g = 8 kg of water displaced.
F = m*g = 8kg * 9.8N/kg = 78.4 N.

Answer 2

Correction: Dc = 0.7g/cm^3.