Question

A metallic right circular cone of 20 cm height 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 frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.

{ALIGARH MUSLIM UNIVERSITY BOARD OF SECONDARY AND SENIOR SECONDARY EDUCATION, SAMPLE PAPER}

Answer 1

HELLO DEAR,,


after pic solution

Let the length of wire =l.

Volume of wire = Area of cross-section × Length

= (πr2) (l)

$\pi( { \frac{1}{32})l }^{2}$


Now,

Volume of frustum = Volume of wire




$\frac{22000}{9} = \pi( { \frac{1}{32} )}^{2} l \\ = > \frac{22000}{9} = \frac{22}{7} \times ( { \frac{1}{32}) }^{2} l \\ = > l = \frac{22000 \times 7 \times {(32)}^{2} }{7 \times 9} \\ = > l = \frac{7000 \times 1024}{9} \\ = > l = 796444.44c {m}^{2}$



I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Hey there,

in ΔAEG,
tan30° = $\frac{EG}{AG}$
EG = AGtan30°
EG = 10 x $\frac{1}{ \sqrt{3} }$ ⇒ = $\frac{10}{ \sqrt{3} }$ cm

In ΔABD, 
tan30° = $\frac{BD}{AD}$
BD = AGtan30°
BD = 20 x $\frac{1}{ \sqrt{3} }$ ⇒ RD =$\frac{20}{ \sqrt{3} }$ cm
Radius of upper end of frustum
r = $\frac{10}{ \sqrt{3} }$ cm
Radius of lower end of container 
R = $\frac{20}{ \sqrt{3} }$ cm
Height of container
H = 10 cm
Volume of frustum
= $\frac{1}{3}$AH (R² + r² + Rr)
= $\frac{1}{3}$ A x 10 [tex]( \frac{20}{ \sqrt{3} }^2) + ( \frac{10}{ \sqrt{3} }^2) + ( \frac{20}{ \sqrt{3} })( \frac{10}{ \sqrt{3} }) [/tex]
$\frac{10}{3} x \frac{22}{7} ( \frac{400}{3} + \frac{100}{3} + \frac{200}{3})$
$= \frac{220}{21} ( \frac{700}{3})$
$= \frac{22000}{9} cm^2$
Radius of wire r₂ = $\frac{1}{2} x \frac{1}{16} = \frac{1}{32} cm$
Let the length of wire = 1 cm
Volume of wire = area of cross-section x length
= (Rr²) x 1
=  $\frac{22}{7} x ( \frac{1}{32})^2 x1$
So,
Volume of frustum = volume of wire
$\frac{22000}{9} = \frac{22}{7} x ( \frac{1}{32})^2 x1$
⇒1 = 796444.44 cm
⇒1 = 7964.44 cm
Hence, length of wire will be 7964.44 cm

Hope this helps!