Determination of density of solid

For solid objects with regular shapes, the mass can be measured using a beam balance, and the volume can be calculated using geometrical formulas:

Rectangular block:

$V = l \times w \times h \quad \Rightarrow \quad \rho = \frac{m}{l\,w\,h}$

Cylinder:

$V = \pi r^2 h \quad \Rightarrow \quad \rho = \frac{m}{\pi r^2 h}$

Sphere:

$V = \frac{4}{3} \pi r^3 \quad \Rightarrow \quad \rho = \frac{3m}{4 \pi r^3}$

Cube:

$V = l^3 \quad \Rightarrow \quad \rho = \frac{m}{l^3}$

Example

A block has a mass of 150 g and dimensions of 3 cm × 4 cm × 5 cm.

Volume:

$V = 3 \times 4 \times 5 = 60 \, \text{cm}^3$

Density:

$\rho = \frac{150 \, \text{g}}{60 \, \text{cm}^3} = 2.5 \, \text{g/cm}^3 = 2500 \, \text{kg/m}^3$

This matches the density of glass, so the block is made of glass.

For solids with irregular shapes, such as stones, the volume is determined using the displacement method with a Eureka can and measuring cylinder:

Procedure:

1. Measure the mass of the object (e.g., a stone) using a digital balance or beam balance.
2. Fill the Eureka can with water up to the spout.
3. Place a beaker under the spout and gently lower the object into the water using string.
4. The displaced water will overflow and can be collected in the beaker.
5. Pour the displaced water into a measuring cylinder and measure its volume. This volume equals the volume of the object.
6. Apply the formula:

$\rho = \frac{m}{V_{\text{displaced water}}}$

Example

Object X has a mass of 50 g. The initial water volume in the measuring cylinder was 60 cm³, and after immersing the object, the final volume was 70 cm³.

• Volume of object:

  $70 - 60 = 10 \, \text{cm}^3$

• Density:

  $\rho = \frac{50}{10} = 5 \, \text{g/cm}^3$

1. Ensure that the object is fully submerged in the water without touching the bottom.
2. All displaced water must be collected and accurately measured.
3. This method is best for objects that cannot be measured using regular volume formulas.