Question

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

Class X - NCERT Question

Surface Area and Volume chapter

Answer 1

$\huge \bf \green{Hey \: there !! }$

Let OAB be the cone in which $\angle$AOB = 60° .

Clearly $\angle$DOE = 30= , OE = 10 cm , OF = 20 cm .

Let ED = r cm and FB = R cm .

$\mathsf {\therefore \frac{ED}{OE} = \tan 30 \degree. }\\ \\ \ \sf \implies \frac{ED}{10} = \frac{1}{ \sqrt{3} } . \\ \\ \sf \implies ED = (10 \times \frac{1}{ \sqrt{3} } )cm. \: \\ \\ \sf \implies r = \frac{10}{ \sqrt{3} } cm. \\ \\ \sf and, \frac{FB}{OF} = \tan30 \degree \\ \\ \sf \implies \frac{FB}{20} = \frac{1}{ \sqrt{3} } . \\ \\ \sf \implies FB = (20 \times \frac{1}{ \sqrt{3} } )cm. \\ \\ \sf \implies R = \frac{20}{ \sqrt{3} } cm.$

Also, EF = 10 cm .

Thus , ABCD is the frustum of a cone in which

$\sf R = \frac{20}{ \sqrt{3} } cm, \: r = \frac{10}{ \sqrt{3} } cm \: \: and \: h = 10cm.$

$\sf Volume \: of \: this \: frustum = \frac{1}{3} \pi h( {R}^{2} + {r}^{2} + Rr). \\ \\ = \sf \frac{1}{3} \pi10 \: ( \frac{400}{3} + \frac{100}{3} + \frac{200}{3} ) {cm}^{3} . \\ \\ = \sf ( \frac{\pi \times10 }{3} \times \frac{700}{3} ) {cm}^{3} . \\ \\ \sf = ( \frac{7000\pi}{9} ) {cm}^{3} .$

Let the length of the wire be l .

Radius of the wire , $\sf r_1 = \frac{1}{32} cm$ .

$\sf Volume \: of \: the \: wire = \pi {r_1}^{2} l = \pi \times ( { \frac{1}{32} )}^{2} \times l . \\ \\ \sf \therefore \frac{7000 \cancel\pi}{9} = \frac{ \cancel\pi l}{32 \times 32} . \\ \\ \sf \implies l = ( \frac{7000 \times 32 \times 32}{9} )cm. \\ \\ \sf \implies l = ( \frac{7000 \times 32 \times 32}{9 \times 100} )m. \\ \\ \sf \implies l = ( \frac{71680}{9} )m. \\ \\ \huge \boxed{ \boxed{ \pink{ \sf\therefore l = 7964.44m.}}}$

✔✔ Hence, it is solved ✅✅.

THANKS

#BeBrainly.

Answer 2

Hi dear!!!?
:-) ... hope it helps