Question

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

Answer 1

Answer :

Height of cone = 3 cm
And
Diameter of cone = 3.5 cm , So radius of cone = $\frac{3.5}{2}$ cm
And we know Volume of cone = $\frac{\mathrm{\pi }{\mathrm{r}}^2\mathrm{h}}{3}$
So,
Volume of a cone = $\frac{{\displaystyle \frac{22}{7}\times \frac{3.5}{2}\times \frac{3.5}{2}\times 3}}{3}=\frac{11\times 0.5\times 3.5}{2}=\frac{19.25}{2}=\mathbf{}\mathbf{9}\mathbf{.}\mathbf{625}\mathbf{}\mathbf{}\mathit{c}{\mathit{m}}^{{\mathbf{3}}^{\mathbf{}}}\mathbf{}$
And
We know Volume of sphere = $\frac{4\mathrm{\pi }{\mathrm{r}}^3}{3}$ , Here r radius of sphere
From given condition we get

Volume of metallic sphere = 504 ( Volume of a cone ) , So

$$\begin{gathered} \Rightarrow \frac{4\times \frac{22}{7}\times r^3}{3}=504\times 9.625 \\ \Rightarrow \frac{88\times r^3}{21}=4851 \\ \Rightarrow \frac{r^3}{21}=55.125 \\ \Rightarrow r^3=1157.625 \\ \Rightarrow r=\sqrt[3]{1157.625} \\ \Rightarrow r=10.50cm \\ So, \\ \mathrm{Diameter}(\mathrm{d})=2\mathrm{r}=2\times 10.50=\mathbf{21}\mathbf{}\mathbf{cm}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{}\mathit{A}\mathit{n}\mathit{s}\mathbf{}\mathbf{)} \end{gathered}$$
And
we know Surface area of sphere = $4\mathrm{\pi }{\mathrm{r}}^2$
So,
Surface area of our sphere = $4\times \frac{22}{7}\times 10.5\times 10.5=4\times 22\times 1.5\times 10.5=\mathbf{1386}\mathbf{}{\mathbf{cm}}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{}\mathbf{Ans}\mathbf{}\mathbf{)}$