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Question
A solid metallic cone, with radius 6 cm and height 10 cm, is made of some heavy metal A. In order to reduce its weight, a conical hole is made in the cone as shown and it is completely filled with a lighter metal B. The conical hole has a diameter of 6 cm and depth 4 cm. Calculate the ratio of the volume of metal A to the volume of the metal B in the solid.
Solution
Volume of the whole cone of metal A
= `1/3pir^2h`
= `1/3 xx pi xx 6^2 xx 10`
= 120π
Volume of the cone with metal B
= `1/3pir^2h`
= `1/3 xx pi xx 3^2 xx 4`
= 12π
Final volume of cone with metal A
= 120π – 12π
= 108π
`"Volume of cone with metal A"/"Volume of cone with metal B"`
= `(108pi)/(12pi)`
= `9/1`
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