Question

If the weights retained on the 2.36 mm, 1.18 mm, 600 $\mu m$, and 300 $\mu m$ sieves are 30%, 35%, 15%, and 20%, respectively, of the total weight of an aggregate sample, then the fineness modulus of the sample is ........... (rounded off to 2 decimal places).

• A. 3.75
• B. 3.74
• C. 3.76
• D. 3.70

Hint:

The fineness modulus is a measure of the particle size distribution of an aggregate sample. A higher value indicates a coarser aggregate, while a lower value indicates a finer aggregate.

Answer 1

The Correct Option is A

The fineness modulus (FM) of an aggregate sample is calculated by summing the cumulative percentage weight retained on the sieves and dividing by 100. The formula is:

\[ {Fineness\ Modulus} = \frac{\sum {percentage\ cumulative\ weight\ retained}}{100} \]

We are given the cumulative percentages for each sieve size as follows:

• 2.36 mm: 30%
• 1.18 mm: \(30\% + 35\% = 65\%\)
• 600 µm: \(65\% + 15\% = 80\%\)
• 300 µm: \(80\% + 20\% = 100\%\)

Thus, the fineness modulus is calculated as:

\[ FM = \frac{30 + 65 + 80 + 100}{100} = 3.75 \]

Therefore, the fineness modulus of the sample is 3.75, which corresponds to option (A).