Question

How many silver coins, $1.75\mathrm{cm}$ in diameter and of thickness $2\mathrm{mm}$, must be melted to form a cuboid of dimensions $5.5\mathrm{cm}\times 10\mathrm{cm}\times 3.5\mathrm{cm}$?

Answer 1

Step 1: Find the volume of a coin

Given: Diameter of a coin $=1.75\mathrm{cm}$

So, radius of a coin $\mathrm{r}=0.875\mathrm{cm}$ [$\because$Radius is half of diameter]

And thickness of coin $h$$=2\mathrm{mm}$

$\Rightarrow$ $\mathrm{h}=0.2\mathrm{cm}$ $\left[\because 1\mathrm{mm}=0.1\mathrm{cm}\right]$

Now volume of a silver coin $={\mathrm{\pi r}}^2\mathrm{h}$ [As a coin is cylindrical in shape and volume of cylinder is ${\mathrm{\pi r}}^2\mathrm{h},\mathrm{r}$ is radius and $\mathrm{h}$ is height of cylinder]

$=\left(\frac{22}{7}\right){\left(0.875\right)}^2\left(0.2\right)$ $\left[\because \mathrm{\pi }=\frac{22}{7}\right]$

$=0.48125{\mathrm{cm}}^3$

Step 2: Find the volume of a cuboid.

As the dimension of cuboid are $5.5\mathrm{cm}\times 10\mathrm{cm}\times 3.5\mathrm{cm}$

So, its length $\mathrm{l}=5.5\mathrm{cm},$ breadth $\mathrm{b}=10\mathrm{cm}$ and height $\mathrm{h}=3.5\mathrm{cm}$

Volume of cuboid $=\mathrm{lbh}$

$=5.5\times 10\times 3.5$

$=192.5{\mathrm{cm}}^3$

Step 3: Find the number of given coins melted to form a cuboid of given dimensions

Let $\mathrm{n}$ number of silver coins melted to form a cuboid of given dimensions.

Then volume of $\mathrm{n}$ coins must be equal to the volume of given cuboid.

$\Rightarrow$$\mathrm{n}\left(0.48125\right)=192.5$

$\Rightarrow$ $\mathrm{n}=\frac{192.5}{0.48125}$

$\Rightarrow$ $\mathrm{n}=400$

Hence, $400$ number of silver coins must be melted to form a cuboid of given dimensions.