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16. Asolid metallic right circular cone. 20crm high and whose vertical angle 60, is cut into twoparts at the middle of its height by a plane parallel to its base. If the frustum so obtained bedrawn into a wire of diameter 1/12 cm, find the length of the wire. |
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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) Bhai figure |
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