Question

A soil sample in its undisturbed state was found to have a volume of 100 cm³ and a mass of 300 g. After oven drying, the mass got reduced to 200 g. What will be the value of the void ratio if the specific gravity of soil is 2.7? Assume the unit weight of water to be 1 g/cm³

• A. 0.35
• B. 4.4
• C. 1.35
• D. 0.66

Hint:

Always remember the water content formula w = Ww/Ws and check units for consistency to avoid calculation errors.

Answer 1

The Correct Option is A

Step 1: Calculate the volume of solids ($V_s$).
\[V_s = \frac{\text{Mass of solids}}{\text{Specific gravity of soil} \times \text{Unit weight of water}}\]
\[V_s = \frac{200}{2.7 \times 1} = 74.07 \, \text{cm}^3\]
Step 2: Calculate the volume of voids ($V_v$).
\[V_v = \text{Total volume} - \text{Volume of solids}\]
\[V_v = 100 - 74.07 = 25.93 \, \text{cm}^3\]
Step 3: Calculate the void ratio ($e$).
\[e = \frac{V_v}{V_s}\]
\[e = \frac{25.93}{74.07} = 0.35\]