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Question 13 : A metallic solid is made up of a solid cylindrical base with a solid cone on its top. The radius of the base of the cone is 5 cm. and the ratio of the height of the cylinder and the cone is 3:2. A cylindrical hole is drilled through the solid with height equal to 2/3rd of the height of solid. What should be the radius (in cm) of the hole so that the volume of the hole is 1/3rd of the volume of the metallic solid after drilling?
Let the height of the cylinder be 3h and the height of the cone be 2h
Volume of the total solid = Π x 52 x 3h + (1/3) x Π x 52 x 2h
= 75Πh + (50/3)Πh = (275/3)Πh
Height of drilling = (2/3) x 5h = 10h/3
Volume of hole = Π x r2 x 10h/3
\\frac{\text { Volume of hole }}{\text { Volume of solid-volume of hole }}) = \\frac{1}{3})
Solving we get r = \\sqrt{\frac{55}{8}})
Choice D is the correct answer.
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