Question

How many coins, 1.4 cm in diameter and 0.4 cm thick, are melted to form a right circular cylinder of height 16 cm and diameter 3.5 cm ?​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that n number of coins, 1.4 cm in diameter and 0.4 cm thick, are melted to form a right circular cylinder of height 16 cm and diameter 3.5 cm.

Now, as base of coin is circular and having thickness. So, coin is also in the shape of cylinder.

Dimensions of coin

Thickness or height of coin, h = 0.4 cm

Diameter of coin = 1.4 cm

So, Radius of coin, r = 0.7 cm

Dimensions of right circular cylinder

Height of cylinder, H = 16 cm

Diameter of cylinder = 3.5 cm

So, radius of cylinder, R = 1.75 cm

As n coins are melted to form a right circular cylinder, so volume of n number of coins is equal to volume of cylinder.

$\rm \: n \times Volume_{(coins)} = Volume_{(cylinder)}$

$\rm \: n \times \pi {(r)}^{2}h \: = \: \pi \: {R}^{2}H$

$\rm \: n \times \pi \: \times (0.7)^{2} \times 0.4 = \pi \times {(1.75)}^{2} \times 16$

$\rm \: n \times \cancel\pi \: \times (0.7)^{2} \times 0.4 = \cancel \pi \times {(1.75)}^{2} \times 16$

$\rm \: n \times \dfrac{7}{10} \times \dfrac{7}{10} \times \dfrac{4}{10} = \dfrac{175}{100} \times \dfrac{175}{100} \times 16$

$\rm \: n \times \dfrac{ \cancel7}{ \cancel{10}} \times \dfrac{ \cancel7}{ \cancel{10}} \times \dfrac{ \cancel4}{ \cancel{10}} = \dfrac{ \cancel{175} \: \: \:^{25} }{ \cancel{100}} \times \dfrac{ \cancel{175}\: \: \:^{25}}{ \cancel{100}\: \: \:^{10}} \times \cancel{16}\: \: \:^{4}$

$\rm \: n \: = \: \dfrac{25 \times 25 \times 4}{10}$

$\rm\implies \:n \: = \: 250$

Hence,

250 number of coins, 1.4 cm in diameter and 0.4 cm thick, are melted to form a right circular cylinder of height 16 cm and diameter 3.5 cm.

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$