Question

504 cones, each of diameter 3.5 cm and height 3 cm, sphere melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area

Answer 1

volume of 504 cones
=504*1/3*pi*r^2*h
=504*1/3*22/7*1. 75^2*3
=504*22/21*49/16*3
=24*22*147/16
=528*147/16
=33*147=4851cm^3
volume of sphere=4851cm^3
4/3*pi*r^3=4851
4/3*22/7*r^3=4851
88/21*r^3=4851
r^3=4851*21/88
r^3=441*21/8
r^3=9261/8
r=21/2=10.5cm
so diameter=10.5*2=21cm
now; its surface area=
4*pi*r^2
4*22/7*10.5^2 cm^2
88/7*441/4 cm^2
(22*63)cm^2
1386cm^2

Answer 2

as we know that the volume of any solid object does not change if it is melted in any other object
suppose the symbol of pie is ¶
volume of cone= 1/3׶×rsquare×h
so volume of 504 cone =1/3×22/7×3.5/2×3.5/2×3×504
=4851 cm cube
volume of sphere = 4851cubic cm
4851= 4/3׶×r cube
r cube =4851×3×7/22×4
r cube = 441×3×7/2×4
= 3×3×7×7×7×3/2×4
r = 3√3×7×7×7×3×3/2×4
r = 21/2
diameter= 21/2×2=21cm
t.s.a= 4׶×rsquare
= 4×22/7×21/2×21/2
= 88×3/2×21/2
=22×3×21 cm square= 1386 square cm