Correct Answer - Option 2 : 50%
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
|
The percentage of fine aggregate (X) to be combined with coarse aggregate is determined by,
\({\rm{X}} = \frac{{{{\left( {{\rm{FM}}} \right)}_{{\rm{CA}}}} - {{\left( {{\rm{FM}}} \right)}_{{\rm{MA}}}}}}{{{{\left( {{\rm{FM}}} \right)}_{{\rm{MA}}}} - {{\left( {{\rm{FM}}} \right)}_{{\rm{FA}}}}}} \times 100\)
(FM)CA and (FM)FA = Fineness modulus of coarse aggregate and fine aggregate respectively
(FM)MA = Fineness modulus of mixed aggregate
Calculation:
Given,
(FM)CA = 7.82, (FM)FA = 2.78, (FM)MA = 6.14
\({\rm{X}} = \frac{{{{\left( {{\rm{FM}}} \right)}_{{\rm{CA}}}} - {{\left( {{\rm{FM}}} \right)}_{{\rm{MA}}}}}}{{{{\left( {{\rm{FM}}} \right)}_{{\rm{MA}}}} - {{\left( {{\rm{FM}}} \right)}_{{\rm{FA}}}}}} \times 100\)
\({\rm{X}} = \frac{{7.82 - 6.14}}{{6.14 - 2.78}} \times 100 = 50{\rm{\% }}\)
Hence the percentage of fine aggregate (X) to be combined with coarse aggregate is 50%
