Question

A metallic right circular cone 20 cm high and angle is 60 is cut into two parts at the middle of its height . Find the volume of the frustum so obtained

Answer 1

Let ABC is the metallic cone and DECB is the required frustum.

Let the radii of frustum are r1 and r2 .

i.e. DP = r1 and BO= r2

Now from ΔADP and ΔABO,

r2 = h1 *tan30

=> r2 = 10* 1/√3

=> r2 =  10/√3

r1 = (h1 + h2 )tan30

=> r1 = 20* 1/√3

=> r1 =  20/√3

Now volume of the frustum DECB = (Π *h2 /3)*(r12 + r1 * r2 + r22 )  

                                                   = (Π *10/3)*{(20/√3)2  + 10/√3 * 20/√3 + (10/√3)2 }

                                                   = (Π *10/3)*{400/3  + 200/3 + 100/3 }

                                                   =  Π *10/3 * 700/3

                                                   = Π *7000/9

Now let l is the length of the wire.

Given diameter of the wire d = 1/16

So radius of the wire R = d/2 = 1/16 * 1/2 = 1/32

Now volume of the frustum = volume of the wire drawn from it

=> (Π*7000)/9 = Π*R2 *l

=> l = (Π*7000)/(Π*R2 *9)

=> l = (7000/{(1/32)2 *9}

=> l = (7000** 32*32)/9

=> l = 7168000/9

=> l= 796444.444 cm

=> l = 796444.444/100 m

=> l = 7964.444 m            (since 100 cm  = 1 m)

So length of wire is  7964.444 m