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).

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

Correct Answer: 3.74

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 $\mu m$: 65% + 15% = 80%
- 300 $\mu m$: 80% + 20% = 100%
Thus, the fineness modulus is:
\[ FM = \frac{30 + 65 + 80 + 100}{100} = 3.75. \] Therefore, the fineness modulus of the sample is 3.75, which corresponds to option (A).