1.

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 is1. 2.002. 2.603. 2.754. 2.90

Answer» 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


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