Question

..... is the quantity of fine aggregate required per 50 kg of cement of M-15 (1 : 2 : 4) grade of concrete.

• A. 0.340 $m^3$
• B. 0.053 $m^3$
• C. 0.035 $m^3$
• D. 0.070 $m^3$

Answer 1

The correct option is D 0.070 $m^3$

Volume of 50 kg cement - 0.035 $m^3$ for 50 kg of 1 : 2 : 4
total material = (1+2+4) = 0.245 $m^3$
Volume of fine aggregate $=\frac{2}{(1+2+4)}\times 0.245=0.070m^3$
Volume of coarse aggregate $=\frac{4\times 0.245}{(1+2+4)}=0.140m^3$
Commission takes quantity of fine aggregate as 0.070 kg or 70 gm. for M15 grade for 50 kg of cement which is wrong. The quantity of fine aggregate will be 70 $m^3$.