# 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

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by jhapoultani

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by jhapoultani

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=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

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