Question

A sediment core of 4 cm diameter and 35.81 cm height was collected. This core had an initial weight of 1000.00 g and upon drying the sediment, the weight decreased by 133.75 g. This core has a void ratio of 0.42857, where void ratio is defined as the ratio of volume of void to the volume of solid (Vv/Vs). The average density of the sediment in the core is ________\ g/cm\(^3\).

Hint:

When calculating density, always use the total volume of the core and the mass of the solid part of the sample for accurate results.

Answer 1

Correct Answer: 2.7

Step 1: Given
Diameter of core: \( d = 4 \, {cm} \Rightarrow r = 2 \, {cm} \)
Height of core: \( h = 35.81 \, {cm} \)
Initial (wet) weight: \( W_{{wet}} = 1000.00 \, {g} \)
Dry weight: \( W_{{dry}} = 1000.00 - 133.75 = 866.25 \, {g} \)
Void ratio: \( e = \frac{V_v}{V_s} = 0.42857 \) Step 2: Volume of the core (total volume)
\[ V_t = \pi r^2 h = \pi \times 2^2 \times 35.81 = \pi \times 4 \times 35.81 \approx 449.94 \, {cm}^3 \] Step 3: Compute volume of solids
\[ V_t = V_s + V_v = V_s (1 + e) \Rightarrow V_s = \frac{V_t}{1 + e} \] \[ V_s = \frac{449.94}{1 + 0.42857} = \frac{449.94}{1.42857} \approx 314.9 \, {cm}^3 \] Step 4: Compute average density
\[ \rho_{{avg}} = \frac{{Total mass}}{{Total volume}} = \frac{1000.00}{449.94} \approx 2.22 \, {g/cm}^3 \]