Question

A mettalic right circular cone 20m high and whose vertical angle is 60 is cut into two parts at the middle of its height by a plane parallel to its base.if the frustrum so obtained be drawn into a wire od diameter 1/16 cm,find the length of the wire

Answer 1

Solution: O'O = VO' = 10 cm [Middle point of VO (20 cm)]

We need to take out volume of frustum. For which we need length of radius, A'O' and AO.

For ∆VO'A',

→ tan 30° = A'O'/VO'

→ 1/√3 = A'O'/10

→ 10/√3 = A'O'

For ∆VOA,

→ tan 30° = AO/VO

→ 1/√3 = AO/20

→ 20/√3 = AO

Volume of frustum = ⅓ πh(r² + R² + Rr)

→ ⅓ × π × 10 × [(10/√3)² + (20/√3)² + 20/√3 × 10/√3]

→ 10/3 × π(100/3 + 400/3 + 200/3)

→ 10π/3 (700/3)

→ 7000π/9

Diameter = 2 × radius of wire

1/16 cm = 2 × radius of wire

1/32 = radius of wire

→ Volume of cylinder = πr²h

→ 7000π/9 = π(1/32)² × h

→ 7000/9 × 32² = h

→ 796444 cm = h

→ 7964.44 m = h

Answer is 7964.44 m

Answer 2

SOLUTION:-

Given:

•A metallic right circular cone 20cm high & whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base.

•If the frastum so obtained be drawn into a wire of diameter 1/16cm.

To find:

The length of the wire.

Explanation:

•ABC a cone is cut out by a plane parallel to the base FG.

•DEFG is the frastum of cone.

•Let O be the centre of the base of cone & O' the centre of the base of the frastum.

We have,

angle BAC= 60° [cut two parts]

angle OAC= 30°

In right angled ∆AOC:

$tan \theta = \frac{Perpendicular}{Base}$

So,

$tan30 \degree = \frac{OC}{OA} \\ \\ \frac{1}{ \sqrt{3} } = \frac{OC}{10} \\ \\ \sqrt{3} OC = 10 \\ \\ OC = \frac{10}{ \sqrt{3} }cm.............(1)$

Since, ∆AOC - ∆AO'F [Using AA similar]

&

$c = \frac{AO}{AO'} = \frac{OC}{O'F} \\ \\ \frac{10}{20} = \frac{ \frac{10}{ \sqrt{3} } }{O'F} \\ \\ 10O'F = \frac{200}{ \sqrt{3} } \\ \\ O'F = \frac{20}{ \sqrt{3} }cm .............(2)$

Height of frustum of the cone:

$P'O = \frac{1}{2} AO' \\ \\ P'O = \frac{1}{2} \times 20cm \\ \\ P'O = 10cm$

Formula of the volume of frustum cone:

$\frac{\pi h}{3} [ {R}^{2} + Rr + {r}^{2} ]\\ \\ \frac{\pi \times 10}{3} ( {OF}^{2} + O'F \times PE + {PE}^{2} ) \\ \\ \frac{\pi \times 10}{3} [( \frac{20}{ \sqrt{3} } ) + \frac{20}{ \sqrt{3} } \times \frac{10}{ \sqrt{3} } + ( \frac{10}{ \sqrt{3} } ) {}^{2} ] \\ \\ \frac{\pi \times 10}{ 3 } [ \frac{400 + 200 + 100}{3} ]\\ \\ \frac{\pi \times 10 \times 700}{9} \\ \\ \frac{\pi7000}{9} {cm}^{3} ..............(3)$

&

We have, diameter of wire= 1/16

Radius of the wire= 1/16/2 = 1/32cm.

Let the length of wire be R cm.

Volume of the wire:

$\pi \times ( \frac{1}{32} ) {}^{2} \times R...............(4)$

Now,

Comparing equation (3) & (4), we get;

$\pi \times ( \frac{1}{32} ) {}^{2} \times R= \frac{\pi \times 7000}{9} \\ \\ R = (\frac{\pi \times 7000}{9} \times \frac{32 \times 32}{\pi} ) {cm} \\ \\ R = (\frac{7000 \times 32 \times 32}{9} )cm \\ \\ R = \frac{7168000}{9} cm \\ \\ R = 796444.44cm$

Thus,

The length of the wire is 796444cm.