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

Calculation of Air Voids in Bituminous Mix

Given Data:

• Mass of specimen: \( W = 1260 \) g
• Volume of specimen: \( V = 525 \) cm³
• Density of water: \( \rho_w = 1 \) g/cm³
• Theoretical maximum specific gravity: \( G_t = 2.51 \)

Step 1: Calculate Mass Specific Gravity (\( G_m \))

The mass specific gravity is given by:

\[ G_m = \frac{\gamma_m}{\gamma_w} = \frac{W / V}{\rho_w} \]

Substituting the values:

\[ G_m = \frac{1260}{525} = 2.4 \]

Step 2: Calculate Percentage of Air Voids (\( V_a \% \))

The formula for the percentage of air voids in the bituminous mix is:

\[ V_a\% = \left( \frac{G_t - G_m}{G_t} \right) \times 100 \]

Substituting the values:

\[ V_a\% = \left( \frac{2.51 - 2.4}{2.51} \right) \times 100 = 4.38\% \]

Final Answer:

Correct Answer: \( \mathbf{4.38\%} \) (rounded to two decimal places).