Question

A solid metal cone of base radius 5 cm and slant height 13 cm is melted and recast into cones of base radius 2.5 cm and height 4 cm . how many such small cones are made ?

Answer 1

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

Given that,

A solid metal cone of base radius 5 cm and slant height 13 cm.

It means,

Radius of cone, r = 5 cm

Slant height of cone, l = 13 cm

Let assume that the height of cone be h cm.

So, using the relationship,

$\: {l}^{2} = \rm \: {r}^{2} + {h}^{2}$

So, on substituting the values, we get

$\: {13}^{2} = \rm \: {5}^{2} + {h}^{2}$

$\: 169 = \rm \: 25 + {h}^{2}$

$\rm \: {h}^{2} = 169 - 25$

$\rm \: {h}^{2} = 144$

$\rm \: {h}^{2} = {12}^{2}$

$\rm\implies \:h \: = \: 12 \: cm$

Now, Dimensions of small cone

Radius of cone, R = 2.5 cm

Height of cone, H = 4 cm

Let assume that number of small cones be n.

Now, According to statement

A solid metal cone of base radius, r = 5 cm and height, h = 12 cm is melted and recast into n small cones of base radius R = 2.5 cm and height, H = 4 cm.

Thus,

$\rm \: Volume_{(Big\:cone)} \: = \: n \times Volume_{(small\:cone)}$

$\rm \: \dfrac{1}{3}\pi \: {r}^{2}h \: = \: n \times \dfrac{1}{3}\pi {R}^{2} H$

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

On substituting the values, we get

$\rm \: {5}^{2} \times 12 \: = \: n \times {(2.5)}^{2} \times 4$

$\rm\implies \:n \: = \: 12$

Thus, number of small cones = 12

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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

Question :-

• A solid metal cone of base Radius 5 cm and slant height 13 cm which is melted and recast into cones of base radius 2.5 cm and height 4 cm. How many such small cones are made ?

Answer :-

• Number of Small Cones = 12 .

Explanation :-

• Here, Radius of cone is given 5 cm and the Height of Cone is 13 cm . When the cone is melted, the New Radius of Cone is 2.5 cm and the New Height of Cone is 4 cm . And, we have to find the number of Cones , which will formed after melted .

◆ In the first case :-

• Radius of Cone is 5 cm .
• Slant Height of Cone is 13 cm .

● Using Concept :-

• L² = R² + H²

∴ By substituting the given values :-

$\Longrightarrow \: \: \sf{13{}^{2} = {5}^{2} + H {}^{2} }$

$\Longrightarrow \: \: \sf{169 = 25 + H}$

$\Longrightarrow \: \: \sf{ {H}^{2} = 169 - 25 }$

$\Longrightarrow \: \: \sf{ {Height}^{2} = 144 }$

$\Longrightarrow \: \: \sf{Height = \sqrt{144} }$

$\Longrightarrow \: \: \sf{Height = 12 \: cm}$

◆ Now, in the second case :-

• Radius of Cone is 2.5 cm .
• Height of Cone is 4 cm .

● Using Concept :-

$\sf{Volume \: _{Big\:Cone} = n \times Volume \: _{Small\:Cone}}$

$\Longrightarrow \: \: \sf{ \frac{1}{3} \: \pi \: {r}^{2} \: h \: \: = \: \: n \times \frac{1}{3} \: \pi \: {r}^{2} \: h }$

$\Longrightarrow \: \: \sf{ {r}^{2} \: h \: \: = \: \: n \times {r}^{2} \: h }$

∴ By substituting the given values :-

$\Longrightarrow \: \: \sf{ {5}^{2} \times 12 \: \: = \: \: n \times (2.5 {}^{2}) \times 4 }$

$\Longrightarrow \: \: \sf{25 \times 12 \: \: = \: \: n \times (6.25) \times 4}$

$\Longrightarrow \: \: \sf{25 \times 12 \: \: = \: \: n \times 25}$

$\Longrightarrow \: \: \sf{n =12}$

Hence :-

• Value of small cones = 12 .

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