# Homework Help: Puck in Mercury

1. May 2, 2012

### eagles12

1. The problem statement, all variables and given/known data

A piece of lead has the shape of a hockey puck, with a diameter of 7.5 cm and a height of 2.9 cm. If the puck is placed in a mercury bath, it floats. How deep below the surface of the mercury is the bottom of the lead puck?

2. Relevant equations

ρhg=13534
vp/vt=pp/phg
v=∏r^2h

3. The attempt at a solution

I found the volume of the lead puck by plugging into the formula for the volume of a cyllinder, and got V=128.11
i looked up the denisty of lead which is 11340 and mercury, which is 13534 but I don't know the height or radius of the mercury bath so I am not sure how to proceed

2. May 2, 2012

### HallsofIvy

The puck will sink into the mercury until the mass of the mercury displaced (the volume of the mercury displaced is the volume of the submerged part of the puck- pi r^2 times the depth) is the same as the mass of the entire puck.

3. May 2, 2012

### eagles12

but i don't know anything about the mercury other than the density, so how do i find the volume of the displaced mercury

4. May 2, 2012

### turbo

If you have the density of the mercury, and the density of the lead, you have everything you need to solve this. How much of the lead puck has to be under the surface to displace its mass in mercury?

5. May 2, 2012

### collinsmark

One other thing that I thought I'd add: The problem statement is not asking you to find the absolute depth that the puck sinks (as in how far the puck sinks relative to its original position). For that you would need more information about the mercury bath.

Instead it asks, "How deep below the surface of the mercury is the bottom of the lead puck?"

Sure, the level of mercury will rise a little bit, but that's irrelevant for this problem. All you need to do is find the hypothetical puck-shaped piece of mercury that happens to have the same weight as the lead puck. The only difference in shape between the lead puck and the mercury shaped "puck" is the height.

Archimedes principle states "Any floating object displaces its own weight of fluid."