Question

How many silver coins , \[1.75{\text{ cm}}\] in diameter and of the thickness \[2\,{\text{mm}}\], must be melted to form a cuboid of dimension \[5.5{\text{ cm }} \times {\text{ 10 cm }} \times {\text{ 3}}{\text{.5 cm}}\] ?

Answer 1

Hint: Shape of the coin is cylindrical form therefore, we use the formula of volume of the cylinder here, also find the volume of the cuboid of the given dimensions then compare it with the volume of the coin.

Complete Step-by-step Solution
Diameter \[1.75{\text{ cm}}\] and thickness \[{\text{2 mm}}\] of a coin is given and also dimensions \[5.5{\text{ cm }} \times {\text{ 10 cm }} \times {\text{ 3}}{\text{.5 cm}}\] of a cuboid is given.

As the silver coin is in cylindrical form.
So we will use the formula of volume of the cylinder here.
Volume of cylinder \[ = \pi {r^2}h\]
Here \[r = \]radius of cylinder
And \[h = \] height of cylinder
Diameter of coin \[ = 1.75{\text{ cm}}\]
Radius of coin \[ = \dfrac{{1.75{\text{ }}}}{2} = 0.875{\text{ cm}}\]
Where, Thickness or height of coin \[ = 2{\text{ mm}}\]\[ = 0.2{\text{ cm}}\]
Because \[10{\text{ mm = 1 cm}}\]
And \[{\text{1 mm = 0}}{\text{.1 cm}}\]
Now, substitute the value of radius and height or thickness in the formula of volume of cylinder.
\[
   = \dfrac{{22}}{7} \times {\left( {0.875} \right)^2} \times 0.2 \\
   = 0.48{\text{ c}}{{\text{m}}^3} \\
\]
Now, here we have the volume of the coin which will be required later.
And now, we will find out the volume of the cuboid.
Volume of cuboid \[ = l \times b \times h\]
Where, length of cuboid is \[5.5{\text{ cm}}\], breadth of cuboid is \[{\text{10 cm}}\], and height of cuboid is \[3.5{\text{ cm}}\].
Now, substitute the value of length, breadth, and height in the formula of volume of the cylinder.
Volume of cuboid \[ = 5.5 \times 10 \times 3.5{\text{ c}}{{\text{m}}^3}\]
Which will be equal to \[192.5{\text{ c}}{{\text{m}}^3}\].
let n be the total numbers of coins required.
We need 192.5 cubic cm volume to be filled with melted coins.
We already found each coin fills a volume of 0.48 cubic cm.
\[n \times \]volume of coin \[ = \]volume of cuboid
\[n \times 0.48 = 192.5\]
$ \Rightarrow n = \dfrac{{192.5}}{{0.48}}$ .
\[ = 401.04\]
As our \[n = 401.4\] but our number of coins cannot be in points so we will take \[n\] equal to \[401\].

Hence, our required number of coins is \[401\].

Note:
As the number of coins is asked, therefore we take both volumes equal to each other while multiplying \[n\] with the volume of coin.