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Question
A metallic cylinder has a radius of 3 cm and a height of 5 cm. It is made of metal A. To reduce its weight, a conical hole is drilled in the cylinder, as shown and it is completely filled with a lighter metal B. The conical hole has a radius of `3/2` cm and its depth is `8/9` cm. Calculate the ratio of the volume of the metal A to the volume of metal B in the solid.
Solution
Volume of metal A = Volume of the cylinder - Volume of the cone
= `π (3)^2 x 5 - 1/3 π (3/2)^2 8/9`
= `π ( 45 - 2/3)`
= `133/3` π cm3
Volume of metal B = Volume of the conical cavity
= `1/3 xx (3/2)^2 . 8/9 = 2/3 π`
Hence, ratio of the volume of the metal A to the volume of the metal B.
= `(133/3 π)/(2/3 π) = 133/2`
= 66.5: 1.
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