How many iron balls, each of radius 1 cm, can be made from a sphere whose radius is 8 cm?
Volume(V) of the sphere with radius R=8 cm is given by V= $$\frac{4}{3} \pi R^{3}$$.
                                                    = $$\frac{4}{3} \pi(8^{3})$$= $$\frac{4}{3} \pi (512) cm^{3}$$.                                                                                               Â
and Volume (v) of each iron ball with radius r= 1cm is given by v= $$\frac{4}{3} \pi r^{3}$$.
                                                                                                             = $$\frac{4}{3} \pi (1^{3})$$= $$\frac{4}{3} \pi cm^{3}$$.
Let say 'n' iron balls each of volume 'v' are required to form a sphere of volume 'V'.
=> Total volume of 'n' iron balls = Volume of the sphere
=> $$n\times v$$ = V
=> n= $$V\div v$$ = $$\frac{4}{3}Â \pi (512)\div \frac{4}{3}Â \pi$$ = 512.Â
Hence 512 iron balls are required in total to form the sphere.
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