Question

A jar made of 3/16-inch-thick glass has an inside radius of 3.00 inches and a total height of 6.00 inches (including the bottom thickness of glass). The glass has a density of 165 lb/ft3. The jar is placed in water with a density of 62.5 lb/ft3.
Assume the jar sits upright in the water without tipping over. How far will the empty jar sink into the water?

What is the volume of the glass shell of the jar? Precision 0.00

What is the weight of the jar? Precision 0.00

What is the weight of water the empty jar will displace? Precision 0.00

What is the volume of water the empty jar will displace? Precision 0.00

How far will the empty jar sink?

Answer 1

We can calculate the volume of a glass shell of the jar by calculating the total volume of a jar and then subtract the volume of air that fits in the jar (the space bounded by the shell).
The total volume of the jar is:
$V_t=Bh=\pi h(r+a)^2$
Where h is the height of the jar, a is the thickness and r is the inner radius.
The volume of empty space, bounded by the jar, is:
$V_e=B(h-a)=\pi r^2(h-a)$
The volume of a jar shell is the difference between these two:
$V_s=V_t-V_e=\pi h(r+a)^2-\pi r^2(h-a)$
When we plug in all the number we get:
$V_s=27.17 { in}^3$
The weight of the jar is simply its volume times its density. We have to convert density of the glass:
$165 \frac{lb}{ft^3}=\frac{165}{1728}\frac{lb}{in^3}=0.0949\frac{lb}{in^3}$
Now we can find the mass:
$m_j=V_s\rho_j=27.17\cdot 0.0949=2.58 lb$
The volume of water jar would displace depends how far the jar would sink.
There two forces acting on the jar when you put in the water. The first one is gravity and it's pulling the jar down. The second force is the buoyancy the and this force pushes the jar upward. In order for that jar to be stable, these two forces must be the same.
The buoyancy is given with this formula:
$B=\rho gV_{disp}$
The volume of water that would be displaced by the jar can be expressed by this formula:
$V_{disp}=\pi h'(r+a)^2$
Where h' is the height of the jar that is submerged in the water.
So the buoyancy force is:
$B=\rho_w g\pi h'(r+a)^2$
This force must be equal to the force of gravity acting on the jar: 
$F_g=B\\ m_jg=\rho_w g\pi h'(r+a)^2\\ m_j=\rho_w \pi h'(r+a)^2$
We solve for h':
$h'=\frac{m_j}{\rho_w \pi(r+a)^2}$
If we plug in the number we get:
$h'=2.24 in$
Now we can calcuate the amount of water discplaced and its mas(keep in mind that we also converted density of water into lb/in^3):
$V_{disp}=\pi h'(r+a)^2=71.50 {in}^3\\ m_v=\rho_w\cdot V_{disp}=2.58 lb$
We get the same mass as the mass of the jar, which makes perfect sense:
$m_jg=\rho_w V_{disp}g\\ m_j=m_{wdisp}$