Civil Engineering - Concrete Technology - Discussion
Discussion Forum : Concrete Technology - Section 1 (Q.No. 29)
29.
If 20 kg of coarse aggregate is sieved through 80 mm, 40 mm, 20 mm, 10 mm, 4.75 mm, 2.36 mm, 1.18 mm, 600 micron, 300 micron and 150 micron standard sieves and the weights retained are 0 kg, 2 kg, 8 kg, 6 kg, 4 kg respectively, the fineness modulus of the aggregate, is
Discussion:
44 comments Page 2 of 5.
Dangerous said:
9 years ago
Sieve Retained Reained perce Cumaltive size
80 0 0 0
40 2 10 10
20 8 40 50
10 6 30 80
4.75 4 20 100
2.36 0 0 100
1.18 0 0 100
600 0 0 100
300 0 0 100
150 0 0 100
740
Fineness modulus =740/100,
= 7.4.
80 0 0 0
40 2 10 10
20 8 40 50
10 6 30 80
4.75 4 20 100
2.36 0 0 100
1.18 0 0 100
600 0 0 100
300 0 0 100
150 0 0 100
740
Fineness modulus =740/100,
= 7.4.
Rahul said:
7 years ago
The fm of coarse aggregate is C+500/100.
The fm of fine aggregate is f/100.
So, what we try on sieve 4.75 lower all are fine particles as per code.
The fm of fine aggregate is f/100.
So, what we try on sieve 4.75 lower all are fine particles as per code.
Aamir said:
6 years ago
@Shalindra. Thanks, you explained so easily and scientific I near seen such in whole academic.
(2)
Tanu said:
1 decade ago
But here both 7.35 and 7.40 options are given. Finally what will we do?
Turab said:
1 decade ago
Please explain how the last column how it comes to b 10 50 80 and 100?
Bilal said:
5 years ago
Fineness modulus = cumulative retained/sample size.
= 148/20,
= 7.4.
= 148/20,
= 7.4.
(11)
Rhythm said:
1 decade ago
But the answer is in 7.35 and your answer is 7.4 which is correct?
Vinod said:
9 years ago
Fineness modulus of given course aggregate is 740/100 =7.4.
Hardik said:
1 decade ago
Approximately maximum near by value is taken as A answer.
Nitin said:
1 decade ago
Yes the answer should be 7.40 means option C is correct.
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