Question

metallic cylinder having height 17.5 cm and diameter 10 cm is melted to form 11 identical cubes. Find the side length of the cube​

Answer 1

Answer:

a = 5

Step-by-step explanation:

volume of cylinder = volume of 11 cubes

22/7 * 17.5 * 5*5 = 11 * a^3

Answer 2

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

Given that,

Metallic cylinder having height 17.5 cm and diameter 10 cm is melted to form 11 identical cubes.

We know,

When one object is melted and recast in to other object, then volume of first object is equals to volume of other object.

So, it means volume of cylinder is equals to volume of 11 cubes.

We have,

Height of metallic cylinder, h = 17.5 cm

Diameter of metallic cylinder = 10 cm

So, Radius of metallic cylinder, r = 5 cm

Let assume that side of the cube be x cm.

Now,

$\rm \: Volume_{(cylinder)} \: = \:11 \times Volume_{(cube)}$

$\rm \: \pi \: {r}^{2}h \: = \: 11 \times {x}^{3}$

$\rm \: \dfrac{22}{7} \times 5 \times 5 \times 17.5 = {11x}^{3}$

$\rm \: 2 \times 5 \times 5 \times 2.5 = {x}^{3}$

$\rm \: {x}^{3} = 5 \times 5 \times 5$

$\rm \: {x}^{3} = {5}^{3}$

$\rm\implies \: \boxed{ \rm{ \: \:x \: = \: 5 \: cm \: }}$

So, when metallic cylinder having height 17.5 cm and diameter 10 cm is melted to form 11 identical cubes, the edge of the cube is 5 cm

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

Additional Information :-

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