Question

If 1000 gms of sand is sieved through 4.75 mm, 2.36 mm, 1.18 mm, 600 micron, 300 micron and 150 micron standard sieves and the weights retained are 0 gms, 100 gms, 150 gms, 200 gms, 350 gms and 200 gms respectively, then the fineness modulus of the sand is

1. 2.00
2. 2.60
3. 2.75
4. 2.90

Answer 1

Correct Answer - Option 2 : 2.60

Concept:

Fineness modulus (F.M)

It is a numerical index of fineness or grading of aggregate. It gives some idea of the average size of the aggregate. The value of fineness modulus is higher for coarse aggregate.

\({\rm{F}}.{\rm{M}} = \frac{{{\rm{Cummulative\;percentage\;of\;material\;retained\;on\;each\;sieve}}}}{{100}}\)

Classification based on Fineness modulus

Sr no	Type of aggregate	Fineness modulus
1	Fine aggregate	2.2 – 2.6
2	Medium aggregate	2.6 – 2.9
3	Coarse aggregate	2.9 – 3.2

Calculation:

Sieve Size	Weight of sand Retained (g)	The cumulative weight of sand retained (g)	Cumulative percentage of sand retained (%)
4.75 mm	0	0	(0/1000) × 100 = 0%
2.36 mm	100	0 +100 = 100	(100/1000) × 100 = 10%
1.18 mm	150	100 + 150 = 250	(250/1000) × 100 = 25%
600 μ	200	250 + 200 = 450	(450/1000) × 100 = 45%
300 μ	350	450 + 350 = 800	(800/1000) × 100 = 80%
150 μ	200	800 + 200 = 1000	(1000/1000) × 100 = 100%
Total	1000		260%

Fineness modulus is given by

\({\rm{F}}.{\rm{M}} = \frac{{{\rm{Cummulative\;percentage\;of\;material\;retained\;on\;each\;sieve}}}}{{100}}\)

\({\rm{F}}.{\rm{M}} = \frac{{{\rm{260}}}}{{100}}=2.60\)

Hence the fineness modulus of the sand is 2.60