e moist unit weight of the soil taken from tt id is $$ 1900 \mathrm{~kg} / \mathrm{m}^{3} $$ and the moisture content $$ .5 \% $$. The unit weight of solid soil is 26 c $$ / \mathrm{m}^{3} $$. $$ \begin{array}{ll} ext { Compute the dry unit weight. } 1804 \mathrm{~kg} / \mathrm{m}^{3} & ext { b. } 1904 \mathrm{~kg} / \mathrm{m}^{3} \ ext { c. } 1704 \mathrm{~kg} / \mathrm{m}^{3} & ext { d. } 1604 \mathrm{~kg} / \mathrm{m}^{3}\end{array} $$

Question

e moist unit weight of the soil taken from tt id is $$ 1900 \mathrm{~kg} / \mathrm{m}^{3} $$ and the moisture content $$ .5 \% $$. The unit weight of solid soil is 26 c $$ / \mathrm{m}^{3} $$. $$ \begin{array}{ll}	ext { Compute the dry unit weight. } 1804 \mathrm{~kg} / \mathrm{m}^{3} & 	ext { b. } 1904 \mathrm{~kg} / \mathrm{m}^{3} \ 	ext { c. } 1704 \mathrm{~kg} / \mathrm{m}^{3} & 	ext { d. } 1604 \mathrm{~kg} / \mathrm{m}^{3}\end{array} $$

UpStudy AI · Accepted Answer

To compute the dry unit weight of the soil, we can use the following formula:

$$
\gamma_d = \frac{\gamma_m}{1 + w}
$$

where:
- $$\gamma_d$$ = dry unit weight (kg/m³)
- $$\gamma_m$$ = moist unit weight (kg/m³)
- $$w$$ = moisture content (as a decimal)

Given:
- $$\gamma_m = 1900 \, \text{kg/m}^3$$
- Moisture content $$w = 0.5\% = 0.005$$ (as a decimal)

Now, substituting the values into the formula:

$$
\gamma_d = \frac{1900}{1 + 0.005} = \frac{1900}{1.005}
$$

Calculating this:

$$
\gamma_d \approx \frac{1900}{1.005} \approx 1881.19 \, \text{kg/m}^3
$$

However, this value does not match any of the options provided. Let's check the calculation again using the relationship between moist unit weight, dry unit weight, and moisture content.

The relationship can also be expressed as:

$$
\gamma_m = \gamma_d (1 + w)
$$

Rearranging gives:

$$
\gamma_d = \frac{\gamma_m}{1 + w}
$$

Substituting the values again:

$$
\gamma_d = \frac{1900}{1 + 0.005} = \frac{1900}{1.005} \approx 1881.19 \, \text{kg/m}^3
$$

This value is still not matching the options. Let's check the calculation using the specific weight of solids.

The specific weight of solids ($$\gamma_s$$) is given as $$26 \, \text{kN/m}^3$$, which is equivalent to $$2600 \, \text{kg/m}^3$$.

Using the formula for dry unit weight:

$$
\gamma_d = \frac{\gamma_s}{1 + w}
$$

Substituting the values:

$$
\gamma_d = \frac{2600}{1 + 0.005} = \frac{2600}{1.005} \approx 2587.14 \, \text{kg/m}^3
$$

This is also not matching the options.

Let's try calculating the dry unit weight using the moist unit weight and moisture content directly:

$$
\gamma_d = \frac{\gamma_m}{1 + \frac{w}{100}} = \frac{1900}{1 + 0.005} = \frac{1900}{1.005} \approx 1881.19 \, \text{kg/m}^3
$$

This confirms our previous calculations.

Since none of the options provided match the calculated dry unit weight, it seems there may be an error in the options or the values provided.

However, based on the calculations, the dry unit weight is approximately $$1881.19 \, \text{kg/m}^3$$.

If we round it, it would be approximately $$1881 \, \text{kg/m}^3$$, which is not listed in the options.

Please verify the values or options provided.
The dry unit weight is approximately $$1881 \, \text{kg/m}^3$$.

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