question_answer

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. \[\left[ \pi =\frac{22}{7} \right]\]

Answer:

Diameter of each cone \[(d)=3.5\,cm\]

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

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

\[\therefore \] Volume of sphere = Volume of 504 cones

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

\[R=\sqrt[3]{\frac{3\times 3\times 7\times 7\times 7\times 3}{2\times 2\times 2}}\]

\[R=\frac{21}{2}cm\]

Hence, diameter of sphere\[=2R=21cm\].

Now, surface area of sphere \[=4\pi {{R}^{2}}\]

\[=4\times \frac{22}{7}\times \frac{21}{2}\times \frac{21}{2}\]

\[=63\times 22\]

\[=1386\,\,c{{m}^{2}}\]

Hence, surface area of sphere is \[1386\text{ }c{{m}^{2}}\].