Question

a solid sphere made of nickel alloy , with diameter 21 cm is melted and recast into cylindrical 2 rupee coins of diameter 3 cm and thickness 2 mm. find the total value of coins obtained.

Answer 1

In the given question;

For a solid sphere made of nickel;
Diameter = 21 cm
Radius (r) = $\frac{21}{2}$ cm
= 10.5 cm

Now using the formula $Volumeofsphere=\frac{4}{3}{\mathrm{\pi r}}^3$;
$$\begin{gathered} \mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{o}\mathrm{f}solidmetallic\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}=\frac{4}{3}\mathrm{\pi }{\left(10.5\right)}^3 \\ \mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{i}\mathrm{d}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{c}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}=\frac{4}{3}\mathrm{\pi }\left(10.5\right)\left(10.5\right)\left(10.5\right) \\ \mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{i}\mathrm{d}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{c}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}=\frac{4}{3}\mathrm{\pi }\left(1157.63\right) \\ \mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{i}\mathrm{d}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{c}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}=1543.5\mathrm{\pi } \end{gathered}$$

For a cylindrical Rs 2 coin;
Diameter = 3 cm
Radius (r) = $\frac{3}{2}$ cm
= 1.5 cm
Height (h) = 2 mm
= 0.2 cm

Now, using the formula $Volumeofcylinder={\mathrm{\pi r}}^2\mathrm{h}$
$$\begin{gathered} \mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{cylindrical}\mathrm{coin}=\mathrm{\pi }{\left(1.5\right)}^2\left(0.2\right) \\ \mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{cylindrical}\mathrm{coin}=\mathrm{\pi }\left(2.25\right)\left(0.2\right) \\ \mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{cylindrical}\mathrm{coin}=0.45\mathrm{\pi } \end{gathered}$$

Now, let the number of coins formed be n.

We know that volume of a solid metallic sphere will be equal to the volume of n Rs 2 coins. So,
$$\begin{gathered} 1543.5\mathrm{\pi }=n\left(0.45\mathrm{\pi }\right) \\ n=\frac{1543.5\mathrm{\pi }}{0.45\mathrm{\pi }} \\ n=3430 \end{gathered}$$

Therefore, 3430 Rs 2 coins can be formed from a solid metallic sphere of diameter 21 cm.