Question

During determination of the bulk specific gravity of compacted bituminous specimen, the mass in air of the specimen is 1260 g and volume is 525 cm³. The density of water is 1.0 g/cm³. The theoretical maximum specific gravity of mix is 2.510. The percentage air voids in the compacted specimen is ......... (rounded off to 2 decimal places).

Hint:

When calculating the percentage of air voids in a specimen, use the difference between the theoretical maximum specific gravity and the bulk specific gravity, divided by the theoretical maximum specific gravity.

Answer 1

Correct Answer: 4.22

Given,
- Mass of specimen, \( W = 1260 \, {g} \)
- Volume, \( V = 525 \, {cm}^3 \)
- Density of water, \( \rho_w = 1 \, {g/cm}^3 \)
- Theoretical maximum specific gravity, \( G_t = 2.51 \)
To calculate the percentage of air voids, we first calculate the mass specific gravity \( G_m \): \[ G_m = \frac{\gamma_m}{\gamma_w} = \frac{W / V \times 1}{1260 / 525} = 2.4 \] Now, the percentage of air voids in the bituminous mix is given by: \[ V_a% = \left( \frac{G_t - G_m}{G_t} \right) \times 100 = \left( \frac{2.51 - 2.4}{2.51} \right) \times 100 = 4.38% \] Thus, the percentage of air voids in the compacted specimen is \( \boxed{4.38%} \) (rounded to 2 decimal places).