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 botioms. Find the volume of water left in the cylinder, if the radius of the cylinder is $ 60 \mathrm{~cm} $ and its height is $ 180 \mathrm{~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:

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 $=\pi r^{2} \mathrm{H}$

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

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

$=\frac{14256000}{7}$

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

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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