Question

A wooden disk with density 700 kg/m

3

floats face down in pure water. It has radius 50.0

cm and thickness 20.0 cm.

(a) Calculate the height of the top of the disk above the water level.

(Answer: 6.0 cm)

(b) A lead weight is loaded onto the disk, so that the disk sits deeper in the water, and

its top face is now only 2.00 cm above the water level. Calculate the mass of the

lead weight.

(Answer: 31.4 kg)

PLEASE SHOW WORK AND DERIVATIONS FROM EQUATIONS

Answer 1

(a) Density of the wooden disk, p = 700 kg/m^3

Density of water, p(w) = 1000 kg/m^3

Suppose h cm is the height of the top of disk from the water level.

So, height of disk submerged into water = (20 - h) cm = (20-h) x 10^-2 m

So, volume of the submerged portion, V = pi*r^2*(20 - h) x 10^-2

So, weight of the displaced volume of water = V*p(w) = pi*r^2*(20 - h) x 10^-2 x 1000 x g N-------------------------(i)

Weight of the disk = pi*r^2*20 x 10^-2 x 700 x g N---------------------------------------(ii)

Equalize (i) and (ii) -

pi*r^2*(20 - h) x 10^-2 x 1000 x g N = pi*r^2*20 x 10^-2 x 700 x g N

=> (20 - h) x 1000 = 20 x 700

=> 20 - h = (20 x 700) / 1000 = 14

=> h = 20 - 14 = 6.00 cm (Answer)

(b) Suppose m is the mass of the lead.

So, in this case -

pi*r^2*(0.20 - 0.02) x 10^-2 x 1000 x g N = pi*r^2*0.20 x 10^-2 x 700 x g N + m x g

=> pi*0.50^2*0.18*10 = pi*0.50^2*0.20*0.70 + m

=> 1.413 = 0.1099 + m

=> m = 1.413 - 0.1099 = 1.3031 kg (Answer)

This does not matches with the answer mentioned below the problem. Please check for that.