Question

504 cones each of diameter 3.5 cm and height 3 cm are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area.$[\pi =\frac{22}{7}]$

Answer 1

Diameter of each cone (d)=3.5 cm

Radius of each cone (r)=$\frac{3.5}{2}=\frac{7}{4}cm[\because r=\frac{d}{2}]$

Height of each cone(h)=3 cm

Volume of 504 cones=$504\times \text{Volume of one cone}$

=504 $\times \frac{1}{3}\pi r^{2}h$

=504$\times \frac{1}{3}\times \frac{22}{7}\times \frac{7}{4}\times \frac{7}{4}\times 3$

Let radius of sphere be R cm

$\therefore \text{Volume of sphere}=\text{Volume of 504 cones}$

$\frac{4}{3}\times \frac{22}{7}\times R^3=504\times \frac{1}{3}\times \frac{22}{7}\times \frac{7}{4}\times \frac{7}{4}\times 3$

$R^3=\frac{504\times 7\times 7\times 3}{4\times 4\times 4}$

$R=\frac{21}{2}cm$

Hence,diameter of sphere=2R=21 cm

Now,surface area of sphere=$4\pi R^2$

$=4\times \frac{22}{7}\times \frac{21}{2}\times \frac{21}{2}$

$=63\times 22$

=$1386c{m}^2$

Hence,surface area of sphere is $1386c{m}^2$.