Question

One-third of the solid matter in a sludge containing 90 % water is composed of fixed mineral solids with specific gravity 2.5, and two-third is composed of volatile solids with specific gravity 1.0.
Specific gravity of all solids lies between

• A. 1.2 and 1.3
• B. 1.5 and 1.6
• C. 1.7 and 1.8
• D. 2.0 and 2.1

Hint:

Combine specific gravities of the two solid fractions using the volume-additivity relation 1/Gs = (mass fraction 1)/G1 + (mass fraction 2)/G2, ignoring the water content.

Answer 1

The Correct Option is A

Concept:
Specific gravity of a solid equals its mass divided by the mass of an equal volume of water. For a mix of two kinds of solids, find the actual mass and actual volume of each part separately, add them up, and divide total mass by total volume to get the combined specific gravity. The 90 % water content of the sludge is extra information about the whole sludge and plays no role in finding the specific gravity of the solids alone.


Step 1: Assume a convenient total mass of solids.
Take the total mass of solids as $M = 300$ units, chosen so the thirds come out as whole numbers. Then:
Mass of fixed mineral solids, $M_f = \frac{1}{3} \times 300 = 100$ units, with $G_f = 2.5$.
Mass of volatile solids, $M_v = \frac{2}{3} \times 300 = 200$ units, with $G_v = 1.0$.


Step 2: Find the volume of each fraction.
Volume is mass divided by specific gravity, taking density of water as the reference unit:
\[ V_f = \frac{M_f}{G_f} = \frac{100}{2.5} = 40 \text{ units} \]
\[ V_v = \frac{M_v}{G_v} = \frac{200}{1.0} = 200 \text{ units} \]


Step 3: Add up total mass and total volume.
Total mass, $M = 100 + 200 = 300$ units.
Total volume, $V = V_f + V_v = 40 + 200 = 240$ units.


Step 4: Divide total mass by total volume.
\[ G_s = \frac{M}{V} = \frac{300}{240} = 1.25 \]
This confirms the specific gravity of the combined solids is 1.25, which sits between 1.2 and 1.3, and rules out the other three ranges, since none of them contain 1.25.


Step 5: Final answer.
\[ \boxed{G_s = 1.25} \]