A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Given:

A solid consisting of a right circular cone of height \( 120 \mathrm{~cm} \) and radius \( 60 \mathrm{~cm} \) standing on a hemisphere of radius \( 60 \mathrm{~cm} \) is placed upright in a right circular cylinder full of water such that it touches the bottoms. 

The radius of the cylinder is \( 60 \mathrm{~cm} \) and its height is \( 180 \mathrm{~cm} \).

To do:

We have to find the volume of water left in the cylinder.

Solution:

The radius of the conical part $= 60\ cm$

Height of the conical part $h = 120\ cm$

Therefore,

Total volume of the solid $=\frac{1}{3} \pi r^{2} h+\frac{2}{3} \pi r^{3}$

$=\frac{1}{3} \pi r^{2}(h+2 r)$

$=\frac{1}{3} \times \frac{22}{7} \times(60)^{2}(120+2 \times 60)$

$=\frac{22}{21} \times 3600(120+120)$

$=\frac{22}{21} \times 3600 \times 240$

$=\frac{6336000}{7}$

$=905142.857 \mathrm{~cm}^{3}$

Height of the cylinder $\mathrm{H}=180 \mathrm{~cm}$

Radius of the cylinder $r=60 \mathrm{~cm}$

Volume of cylinder $=\pi r^{2} \mathrm{H}$

$=\frac{22}{7} \times(60)^{2} \times 180$

$=\frac{22}{7} \times 3600 \times 180$

$=\frac{14256000}{7}$

$=2036571.429 \mathrm{~cm}^{3}$

The volume of water left $=$ Difference in volumes

$=2036571.429-905142.857$

$=1131428.572 \mathrm{~cm}^{2}$

$=\frac{1131428.572}{100 \times 100 \times 100} \mathrm{~m}^{3}$

$=1.131 \mathrm{~m}^{3}$

The volume of water left in the cylinder is $1.131\ m^3$. 

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Updated on: 10-Oct-2022

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