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Volume of a Right Circular Cone

An open cylin...

Question

An open cylindrical vessel of internal diameter 7 cm and height 8 cm stands on a horizontal table. Inside this is placed a solid metallic right circular cone, the diameter of whose base is $3 \frac{1}{2}$ cm and height 8 cm. Find the volume of water required to fill the vessel. If this cone is replaced by another cone, whose height is $1 \frac{3}{4}$ cm and the radius of whose base is 2 cm, find the drop in the water level.

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Solution

Cylinder: $Height = 8 \ cm, Radius = 3.5 \ cm$

Cone: $Height = 8 \ cm, Radius = 1.75 \ cm$

Volume of water required $= Volume \ of \ Cylinder - Volume \ of \ Cone$

$=$ $\frac{22}{7}$ $(3.5^2 \times 8 -$ $\frac{1}{3}$ $\times 1.75^2 \times 8) = 282.33 \ cm^3$

If the cone was replaced by another cone with $Height = 1.75 \ cm, Radius = 2 \ cm$ , then let the drop in water level $= h$ . Therefore

$\pi \times 3.5^2 \times h =$ $\frac{1}{3}$ $\pi (1.75^2 \times 8 - 2^2 \times 1.75)$

$\Rightarrow h =$ $\frac{1.75^2 \times 8 - 2^2 \times 1.75}{3 \times 3.5^2}$ $= 0.4762 \ cm$

$\\$

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