A metallic cylinder of radius 3cm and height of 5cm. To reduce its weight a conical hole is drilled in the cylinder.the conical hole has radius of 3/2cm and its depth 8/9 cm calculate the ratio of the volume of metal left in cylinder to the volume of metal taken out in conical shape
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therefore volume of cylinder is πr²h
22/7×3×3×5 is the volume of cylinder
volume of cylinder removed = volume of cylinder - volume of part removed
=π((3×3×5)-(3/2×3/2×8/9))
=π{(45-2)}
=π(43)
=43π
therefore ratio=43π/2π
=43/2
22/7×3×3×5 is the volume of cylinder
volume of cylinder removed = volume of cylinder - volume of part removed
=π((3×3×5)-(3/2×3/2×8/9))
=π{(45-2)}
=π(43)
=43π
therefore ratio=43π/2π
=43/2
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Original volume of the cylinder = pi*3*3*5
volume of conical hole = 1/3 * pi * 3/2 * 3/2 * 8/9
ratio of volume left / volume taken out
= (original volume of cylinder - volume of conical hole) / volume of conical shape
volume of conical hole = 1/3 * pi * 3/2 * 3/2 * 8/9
ratio of volume left / volume taken out
= (original volume of cylinder - volume of conical hole) / volume of conical shape
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