A cylindrical vessel with internal diameter $ 10 \mathrm{~cm} $ and height $ 10.5 \mathrm{~cm} $ is full of water A solid cone of base diameter $ 7 \mathrm{~cm} $ and height $ 6 \mathrm{~cm} $ is completely immersed in water Find the value of water displaced out of the cylinder. (Take $ \pi=22 / 7 $ )

Given:

A cylindrical vessel with internal diameter \( 10 \mathrm{~cm} \) and height \( 10.5 \mathrm{~cm} \) is full of water.

A solid cone of base diameter \( 7 \mathrm{~cm} \) and height \( 6 \mathrm{~cm} \) is completely immersed in water.

To do:

We have to find the value of water displaced out of the cylinder.

Solution:

Internal diameter of the cylindrical vessel $= 10\ cm$

This implies,

Radius of the vessel $r= \frac{10}{2}$

$=5\ cm$

Height of the vessel $h = 10.5\ cm$

Therefore,

Volume of water filled in the cylinder $= \pi r^{2} h$

$=\pi \times(5)^{2} \times 10.5$

$=\frac{22}{7} \times 2.5 \times \frac{105}{10}$

$=825 \mathrm{~cm}^{3}$

Diameter of the cone $=7 \mathrm{~cm}$

This implies,

Radius of the cone $r=\frac{7}{2} \mathrm{~cm}$

Height of the cone $h_{1}=6 \mathrm{~cm}$
Therefore,

Volume of the cone $=\frac{1}{3} \pi r_{1}^{2} h_{1}$

$=\frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 6$

$=77 \mathrm{~cm}^{3}$

Volume of water displaced out of the cylinder $=$ Volume of the cone

$=77 \mathrm{~cm}^{3}$

The value of water displaced out of the cylinder is $77\ cm^3$.

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Updated on: 10-Oct-2022

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