 Objectives:

• To determine, by use of experiments, the density of regular shaped objects, irregular shaped objects and liquids.

Before reading the following, have a little think for yourself how you would determine the density of the following objects;    Hopefully you are thinking that some are easier that others to determine, but which and would you determine them each differently?

There are different methods for each – we will leave the liquid until last.

Determining the density of the regular shaped objects (the cube and potentially the sphere):

The equipment that would be needed are:

• balance (to determine the mass)
• ruler (to measure each dimension) – a more precise measuring tool such as a vernier calliper would also be acceptable

For the cube; the ruler should be used to measure the height, the width and the depth (if a true cube, then each of these should of course be the same). Using these measurements and multiplying them together, the volume can be calculated.

For the sphere; it is more difficult to find the volume accurately but it can be done by measuring the diameter of it, this can then be halved to find the radius. The equation for the volume of a sphere can be used – $V_{sphere} = \frac{4}{3} \pi r^{3}$.

For both the sphere and the cube, the mass of each can be determined by separately placing them onto a balance (making sure the balance reads zero first).

With the mass and volume of each known, the values can be placed into the density equation such that;  $\rho _{cube} = \frac{m_{cube}}{V_{cube}}$ $\rho _{sphere} = \frac{m_{sphere}}{V_{sphere}}$

Determining the density of the irregular shaped object:

The equipment that would be required here are:

The procedure is quite straight straight forward but the reason for it are important and are what may score you mark in your examinations.

Firstly, because the model hand is an irregular shaped object you can’t just use a ruler to find its height, width and depth. Instead you have to submerge the object into a beaker of water and measure the displaced water. Here is a method that can be followed;

1. Fill the displacement can up to the spout (such that any more would send water trickling down it).
2. Place an empty measuring cylinder to catch anything that drips out of the spout.
3. Carefully submerge the irregular shaped objects into the displacement can. The water level will rise and will flow through the spout and into the measuring cylinder (okay… if you make need to mount the displacement can on a trip first so it is higher than the measuring cylinder).
4. Collect every drop of water and measure the water level carefully in the measuring cylinder and record this value.
5. Remove the object from the water and dry it. Place it onto a balance (making sure the balance reads zero first) and record the mass shown (alternatively do this step first). So now the mass is known and the volume (in ml) of some water is known. Well the trick here is to use the fact that $1 \ ml$ is equivalent to $1 \ cm^{3}$: $1 \ ml = 1 \ cm^{3}$

The object, when submerged into the water displaces the water by a volume equal to the volume of the object itself. In other words: $\text{Volume of object} = \text{volume of displaced water}$

So, the volume of the hand (the irregular shaped object) is known as well as the mass. The density can therefore be calculated by substituting these into the equation to get; $\rho_{hand} = \frac{m_{hand}{V_{hand}}$ This method can also be used for regular shaped objects however, if you are answering a long questions that clearly shows an irregular as well as regular shaped objects, then describe both methods.

Determining the density of the liquid The equipment that would be required here are:

• balance
• measuring cylinder
5. To calculate the mass of the liquid do $latex mass \ 2 – mass \ 1 Now that the mass of the liquid is known as well as the volume, substitute them both in to the equation to find the density of the liquid;$latex \rho_{liquid} = \frac{m_{liquid}{V_{liquid}}\$