Question

16. A right circular cone is 3.6 cm high and the radius of its base is 1.6 cm.It is melted and recast into a right circular cone having base radius 1.2 cm. Find its height.​

Answer 1

AnswEr :

We're given that, a right circular cone is 3.6 cm high and the radius of its base is 1.6 cm.

Diagram :

$\setlength{\unitlength}{0.8cm}\begin{picture}(6, 4)\linethickness{0.26mm} \qbezier(5.8,2.0)(5.8,2.3728)(4.9799,2.6364)\qbezier(4.9799,2.6364)(4.1598,2.9)(3.0,2.9)\qbezier(3.0,2.9)(1.8402,2.9)(1.0201,2.6364)\qbezier(1.0201,2.6364)(0.2,2.3728)(0.2,2.0)\qbezier(0.2,2.0)(0.2,1.6272)(1.0201,1.3636)\qbezier(1.0201,1.3636)(1.8402,1.1)(3.0,1.1)\qbezier(3.0,1.1)(4.1598,1.1)(4.9799,1.3636)\qbezier(4.9799,1.3636)(5.8,1.6272)(5.8,2.0)\put(0.2,2){\line(1,0){2.8}}\put(3.2,4){\sf{3.6 cm}}\put(3,2){\line(0,2){4.5}}\put(1.4,1.6){\sf{1.6 cm}}\qbezier(.185,2.05)(.7,3)(3,6.5)\qbezier(5.8,2.05)(5.3,3)(3,6.5)\put(3,2.02){\circle*{0.15}}\put(2.7,2){\dashbox{0.01}(.3,.3)}\end{picture}$

Also, it is given that, the above cone is melted and recast into a right circular cone having base radius 1.2 cm.

Diagram :

$\setlength{\unitlength}{0.8cm}\begin{picture}(6,4)\linethickness{0.26mm} \qbezier(5.8,2.0)(5.8,2.3728)(4.9799,2.6364)\qbezier(4.9799,2.6364)(4.1598,2.9)(3.0,2.9)\qbezier(3.0,2.9)(1.8402,2.9)(1.0201,2.6364)\qbezier(1.0201,2.6364)(0.2,2.3728)(0.2,2.0)\qbezier(0.2,2.0)(0.2,1.6272)(1.0201,1.3636)\qbezier(1.0201,1.3636)(1.8402,1.1)(3.0,1.1)\qbezier(3.0,1.1)(4.1598,1.1)(4.9799,1.3636)\qbezier(4.9799,1.3636)(5.8,1.6272)(5.8,2.0)\put(0.2,2){\line(1,0){2.8}}\put(3.2,4){\sf{h}}\put(3,2){\line(0,2){4.5}}\put(1.4,1.6){\sf{1.2 cm}}\qbezier(.185,2.05)(.7,3)(3,6.5)\qbezier(5.8,2.05)(5.3,3)(3,6.5)\put(3,2.02){\circle*{0.15}}\put(2.7,2){\dashbox{0.01}(.3,.3)}\end{picture}$

In the above diagram, we're supposed to find out the height of the cone.

We know that,

Volume of cone = $\sf \dfrac{1}{3}$πr²h

Now,

Let us assume that h be the height of the new formed cone. So,

Volume of 1st cone = Volume of 2nd cone [Cone is only melted and recasted]

$\longrightarrow \sf \: V_1 = V_2$

$\longrightarrow \sf \: \dfrac{1}{3} \times \pi \times {(1.6)}^{2} \times 3.6 = \dfrac{1}{3} \times \pi \times {1.2}^{2} \times h$

$\longrightarrow \sf \: {(1.6)}^{2} \times 3.6 = {1.2}^{2} \times h$

$\longrightarrow \sf \: 2.56 \times 3.6 =1.44 \times h$

$\longrightarrow \sf \: h = \dfrac{2.56 \times 3.6}{1.44}$

$\longrightarrow \sf \: h = \dfrac{9.216}{1.44}$

$\longrightarrow \sf \: { \large{ \underline{ \boxed{ \sf{ \purple{h = 6.4 \: cm}}}}}} \: \: \bigstar$

Therefore, height of the new formed cone is 6.4 cm .

⠀⠀⠀Additional information :

• Curved surface area of a cone = πrl

• Total surface area of a cone = πr(l+r)

• Value of l = √h² + r²