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CBSEENMA10008835
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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 over 16 cm, find the length of the wire.

Solution
In fig. cone ABC is cut out by a plane parallel to the base FG. DEFG is the frustum so obtained. Let O be the centre of the base of the cone and O’ the centre of the base of the frustum.

It is given that ∠BAC = 60° ∠OAC = 30°
In right triangle AOC, tan 30 degree space equals space OC over OA
rightwards double arrow space space Oc space space equals space OA space straight x space tan space 30 degree space space space space space space space space space space space space equals space 10 space straight x space fraction numerator 1 over denominator square root of 3 end fraction equals fraction numerator 10 over denominator square root of 3 end fraction space cm
And, C =  equals fraction numerator 10 over denominator square root of 3 end fraction space cm. space space space space... left parenthesis straight i right parenthesis Since space space space increment AOC space minus space increment AO apostrophe straight F space space space space space space space space space space space space space space left parenthesis Using space AA space similar space condition right parenthesis therefore space space space fraction numerator AO over denominator AO apostrophe end fraction space equals space fraction numerator OC over denominator straight O apostrophe straight F end fraction rightwards double arrow space space space 10 over 20 equals fraction numerator begin display style fraction numerator 10 over denominator square root of 3 end fraction end style over denominator straight O apostrophe straight F end fraction rightwards double arrow space space space space straight O apostrophe straight F space equals space fraction numerator 20 over denominator square root of 3 end fraction space space space space space space space space space... left parenthesis ii right parenthesis
Height of the frustum = P'O = 1 half space AO apostrophe space equals space 10 space cm
therefore volume of the frustum = πh over 3 left square bracket straight R squared plus Rr plus straight r squared right square bracket
                                         equals fraction numerator straight pi space straight x space 10 over denominator 3 end fraction left square bracket OF squared plus straight O apostrophe straight F space straight x space PE space plus space PE squared right square bracket equals space fraction numerator straight pi space straight x space 10 over denominator 3 end fraction open square brackets open parentheses fraction numerator 20 over denominator square root of 3 end fraction close parentheses plus fraction numerator 20 over denominator square root of 3 end fraction straight x fraction numerator 10 over denominator square root of 3 end fraction plus open parentheses fraction numerator 10 over denominator square root of 3 end fraction close parentheses squared close square brackets space space space space space space space space space space space space space space space space space space left square bracket space Using space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis right square bracket equals space fraction numerator πx 10 over denominator 3 end fraction open square brackets fraction numerator 400 plus 200 plus 100 over denominator 3 end fraction close square brackets equals space fraction numerator straight pi space straight x space 10 space straight x space 700 over denominator 9 end fraction space cm cubed equals space fraction numerator straight pi space 7000 over denominator 9 end fraction space cm cubed space space space space space space space space space space space space space space space space space space space space space... left parenthesis iii right parenthesis space space space space space space space space space space space space space space space space space space space space
Radius of the wire = 1 over 32 space cm
Let h be length of the wire
therefore         Volume of wire = straight p space straight x space open parentheses 1 over 32 close parentheses squared space straight x space straight h space space space space space space space space space space space space space space space space.... left parenthesis iv right parenthesis
From (iii) and (iv), we get
straight pi space straight x space open parentheses 1 over 32 close parentheses squared space straight x space straight h space equals space fraction numerator straight pi space straight x space 7000 over denominator 9 end fraction therefore space space space space space space space straight h space equals space fraction numerator straight pi space straight x space 7000 over denominator 9 end fraction space straight x space fraction numerator 32 space straight x space 32 space over denominator straight pi end fraction cm space space space space space space space space space space space space space space space space space space space space equals space 7168000 over 9 equals 796444.4 cm space left parenthesis approx right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space equals space 7964.44 space straight m space space space space space space space space space space space space space space space space space space space space space space space space equals space 7964 space straight m. space left parenthesis approx right parenthesis space

Some More Questions From Statistics Chapter

A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

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