Question

. A cube of metal with length of edge 8 cm is melted to make smaller cubes of edge 2 cm. How many small cubes can be thus made?

I want it fast plz answer.​

Answer 1

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

Given that,

A cube of metal with length of edge 8 cm is melted to make smaller cubes of edge 2 cm.

Let assume that number of cubes of edge 2 cm be n.

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

$\rm \: Volume_{(Cube \: of \:edge \: 8 \: cm )} = n \times Volume_{(Cube \: of \:edge \: 2\: cm )}$

$\rm \: {8}^{3} = n \times {(2)}^{3}$

$\rm \: 8 \times 8 \times 8 = n \times 8$

$\rm\implies \:n \: = \: 64$

$\rm\implies \:\boxed{ \rm{ \: Number_{(Cube \: of \:edge \: 2 \: cm )}\: = \: 64 \: \: }}$

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

Additional Information :-

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

Answer 2

Step-by-step explanation:

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

Solution−

Given that,

A cube of metal with length of edge 8 cm is melted to make smaller cubes of edge 2 cm.

Let assume that number of cubes of edge 2 cm be n.

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

\begin{gathered}\rm \: Volume_{(Cube \: of \:edge \: 8 \: cm )} = n \times Volume_{(Cube \: of \:edge \: 2\: cm )} \\ \end{gathered}

Volume

(Cubeofedge8cm)

=n×Volume

(Cubeofedge2cm)

\begin{gathered}\rm \: {8}^{3} = n \times {(2)}^{3} \\ \end{gathered}

8

3

=n×(2)

3

\begin{gathered}\rm \: 8 \times 8 \times 8 = n \times 8 \\ \end{gathered}

8×8×8=n×8

\begin{gathered}\rm\implies \:n \: = \: 64 \\ \end{gathered}

⟹n=64

\begin{gathered}\rm\implies \:\boxed{ \rm{ \: Number_{(Cube \: of \:edge \: 2 \: cm )}\: = \: 64 \: \: }} \\ \end{gathered}

⟹

Number

(Cubeofedge2cm)

=64

\rule{190pt}{2pt}

Additional Information :-

\begin{gathered}\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}\end{gathered}

MoreFormulae

MoreFormulae

★CSA

(cylinder)

=2πrh

★Volume

(cylinder)

=πr

2

h

★TSA

(cylinder)

=2πr(r+h)

★CSA

(cone)

=πrl

★TSA

(cone)

=πr(l+r)

★Volume

(sphere)

=

3

4

πr

3

★Volume

(cube)

=(side)

3

★CSA

(cube)

=4(side)

2

★TSA

(cube)

=6(side)

2

★Volume

(cuboid)

=lbh

★CSA

(cuboid)

=2(l+b)h

★TSA

(cuboid)

=2(lb+bh+hl)

$answer}$