731 metallic cubes with a side length of 3.6 cm are available to be melted and formed into a machine part consisting of two solid cones with a base radius of 5.4 cm, joined at their bases. The heights of the cones are 49 cm and 35 cm, respectively. How many cubes will be surplus after the formation of the two cones? (Take π = 3.14)​

Correct option is C

Given:

Number of cubes = 731

Side of each cube = 3.6 cm

Two cones joined at bases with common radius = 5.4 cm

Heights = 49 cm and 35 cm respectively;

Formula Used:

Volume of cube = side³

Volume of cone = 13πr2h\frac{1}{3} \pi r^2 h

Total volume used = volume of both cones

Surplus = total number of cubes – number of cubes used

Solution:

Volume of one cube = (3.6)³ = 46.656 cm³

Radius of both cones = 5.4 cm 

Total volume of both cubes = $$\frac{1}{3} \times 3.14 \times (5.4)^2 [ 49 + 35]$$​​

= $$3.14 \times 29.16 \times 28$$​​

= 2563.75 cm³

Number of cubes used = $$\frac{2563.75}{46.656} \approx 54.95$$​ = 55 cubes 

Surplus = 731 - 55 = 676 cubes