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. Please explain
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Answer:
Step-by-step explanation: Volume of the whole cone of metal A
=1/3\pi r^2h
=1/3*\pi 6^2*10
=120\pi
volume of cone with metal B
=1/3\pi r^2 h
=1/3*\pi *3^2*4
=12\pi
final volume of cone with metal A
=120\pi-12\pi=108\pi
=>108\pi/12\pi
=>9/1
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