Question

A solid consisting of a right circular cone, standing on a hemisphere, is placed upright, in a right circular cylinder, full of water, and touches the bottom. Find the volume of water left in the cylinder, having given that the radius of the cylinder is 3 cm and its height is 6 cm; the radius of the hemisphere is 2 cm and the height of the cone is 4 cm. Give your answer to the nearest $c{m}^3$ (Take $\pi =\frac{22}{7}$)

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Answer 1

Radius of the cylinder = 3 cm and its height = 6 cm.

Volume of water in the cylinder, when full = $[\pi \times (3{)}^2\times 6]c{m}^3=(54\pi )c{m}^3$

.Volume of solid consisting of cone and hemisphere = (Volume of hemi-sphere) + (Volume of cone)
= $[\frac{2}{3}\pi \times (2{)}^3+\frac{1}{3}\pi \times (2{)}^2\times 4]c{m}^3=\left(\frac{32\pi }{3}\right)c{m}^3$

Volume of water displaced from cylinder = Volume of solid consisting of cone and hemisphere

= $\left(\frac{32\pi }{3}\right)c{m}^3$
Volume of water left in the cylinder after placing the solid into it

= $(54\pi -\frac{32\pi }{3})c{m}^3=\left(\frac{130\pi }{3}\right)c{m}^3=(\frac{130}{3}\times \frac{22}{7})c{m}^3=136.19c{m}^3$

Hence, the volume of water left in the cylinder to the nearest $c{m}^3$ is 136 $c{m}^3$.