Question

A pycnometer of volume \( 500 \, \text{cm}^3 \) was used in specific gravity test and following data is observed. Weight of dry empty pycnometer = \( 125 \, \text{g} \), weight of dry soil and pycnometer = \( 500 \, \text{g} \), weight of dry soil and distilled water filled in pycnometer up to top = \( 850 \, \text{g} \). Then the specific gravity of soil solids is

• A. \( 2.00 \)
• B. \( 2.25 \)
• C. \( 2.50 \)
• D. \( 2.75 \)

Hint:

The specific gravity of soil solids ($G_s$) is a fundamental property in soil mechanics. It is determined using the pycnometer method, which relies on the principle of water displacement. The key is to accurately measure the weight of soil solids and the weight of water that occupies the same volume as the soil solids. The formula $G_s = \frac{W_2 - W_1{(W_4 - W_1) - (W_3 - W_2)$ is commonly used, where $W_1$ is weight of empty pycnometer, $W_2$ is weight of pycnometer + dry soil, $W_3$ is weight of pycnometer + dry soil + water, and $W_4$ is weight of pycnometer + water.

Answer 1

The Correct Option is C

Step 1: Identify the given data.
Let \( W_1 \) be the weight of the dry empty pycnometer. Given \( W_1 = 125 \, \text{g} \). Let \( W_2 \) be the weight of the dry soil and pycnometer. Given \( W_2 = 500 \, \text{g} \). Let \( W_3 \) be the weight of the dry soil, pycnometer, and distilled water filled up to the top. Given \( W_3 = 850 \, \text{g} \). The volume of the pycnometer is \( V_p = 500 \, \text{cm}^3 \).
Since the density of water is approximately \( \rho_w = 1 \, \text{g/cm}^3 \), the weight of water that completely fills the pycnometer is \( W_{w,full} = V_p \times \rho_w = 500 \, \text{cm}^3 \times 1 \, \text{g/cm}^3 = 500 \, \text{g} \).
Let \( W_4 \) be the weight of the pycnometer filled with water only. \( W_4 = W_1 + W_{w,full} = 125 \, \text{g} + 500 \, \text{g} = 625 \, \text{g} \).
Step 2: Calculate the weight of dry soil (\( W_s \)).
The weight of dry soil is the difference between the weight of the pycnometer with dry soil and the weight of the empty pycnometer.
$$W_s = W_2 - W_1$$
$$W_s = 500 \, \text{g} - 125 \, \text{g}$$
$$W_s = 375 \, \text{g}$$
Step 3: Calculate the weight of water having the same volume as the soil solids (\( W_w \)).
The specific gravity of soil solids \( G_s \) is defined as the ratio of the weight of a given volume of soil solids to the weight of an equal volume of water.
In the pycnometer method, the weight of water that would occupy the same volume as the soil solids is given by the formula:
$$W_w = (W_4 - W_1) - (W_3 - W_2)$$
Where \( (W_4 - W_1) \) represents the weight of water that the pycnometer can hold.
And \( (W_3 - W_2) \) represents the weight of water that was added to the pycnometer containing the soil. The difference gives the weight of water displaced by the soil solids. $$W_w = (625 \, \text{g} - 125 \, \text{g}) - (850 \, \text{g} - 500 \, \text{g})$$
$$W_w = 500 \, \text{g} - 350 \, \text{g}$$
$$W_w = 150 \, \text{g}$$
Step 4: Calculate the specific gravity of soil solids (\( G_s \)).
The specific gravity of soil solids is calculated as:
$$G_s = \frac{W_s}{W_w}$$
$$G_s = \frac{375 \, \text{g}}{150 \, \text{g}}$$
To simplify the fraction, we can divide both numerator and denominator by common factors. Both are divisible by 75:
$$G_s = \frac{375 \div 75}{150 \div 75} = \frac{5}{2}$$
$$G_s = 2.50$$
Step 5: Select the correct option.
Based on the calculation, the correct value is \( 2.50 \). $$\boxed{2.50}$$