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).
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).
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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.
Updated On: Feb 12, 2026
- 3.75
- 3.74
- 3.76
- 3.70
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The Correct Option is A
Solution and Explanation
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).
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