If $1000 g m s$ of sand is sieved through $4.75 m m, 2.36 m m, 1.18 m m, 600 m i c r o n, 300 m i c r o n a n d 150 m i c r o n$ standard sieves and the weight's retained are $0 g m s, 100 g m s, 150 g m s, 200 g m s, 350 g m s, a n d 500 g m s$ respectively, then the fineness modulus of the sand is

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Solution

Fineness modulus of sand
$F M = \frac{10 + (10 + 15) + (10 + 15 + 20) + (10 + 15 + 20 + 35) + (10 + 15 + 20 + 35 + 20)}{100}$
$F M = \frac{260}{100} = 2.6$

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If 1000 gms of sand is sieved through4.75 mm,2.36 mm,1.18 mm,600 micron,300 micron and 150 micron standard sieves and the weight's retained are 0 gms,100 gms,150 gms,200 gms,350 gms,and 500 gms respectively, then the fineness modulus of the sand is

Fineness modulus of sand FM=10+(10+15)+(10+15+20)+(10+15+20+35)+(10+15+20+35+20)100 FM=260100=2.6

Fineness modulus of sand
$F M = \frac{10 + (10 + 15) + (10 + 15 + 20) + (10 + 15 + 20 + 35) + (10 + 15 + 20 + 35 + 20)}{100}$
$F M = \frac{260}{100} = 2.6$