(b) A metallic right circular cone 20 cmhighandwhosvertical angle is 6

1.	(b) A metallic right circular cone 20 cmhighandwhosvertical angle is 60° is cut into two parts at tmiddle of its height by a plane parallel to its basIf the frustum so obtained drawn into a wirediameter 1.16 cm. Find the length of the wire
Answer» Let ABC is the metallic cone and DECB is the required frustum. Let the radii of frustum are r1and r2. i.e. DP = r1and BO= r2 Now from ΔADP and ΔABO, r2= h1tan30 => 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 => (Π7000)/9 = ΠR2l => l = (Π7000)/(ΠR29) => l = (7000/{(1/32)29} => l = (7000* 3232)/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 Let ABC is the metallic cone and DECB is the required frustum. Let the radii of frustum are r1and r2. i.e. DP = r1and BO= r2 Now from ΔADP and ΔABO, r2= h1tan30 => 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 => (Π7000)/9 = ΠR2l => l = (Π7000)/(ΠR29) => l = (7000/{(1/32)29} => 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

(b) A metallic right circular cone 20 cmhighandwhosvertical angle is 60° is cut into two parts at tmiddle of its height by a plane parallel to its basIf the frustum so obtained drawn into a wirediameter 1.16 cm. Find the length of the wire

Answer»

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

Let the radii of frustum are r1and r2.

i.e. DP = r1and 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