A cylindrical tub of radius $ 5 \mathrm{~cm} $ and length $ 9.8 \mathrm{~cm} $ is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed in the tub. If the radius of the hemisphere is immersed in the tub. If the radius of the hemisphere is $ 3.5 \mathrm{~cm} $ and height of the cone outside the hemisphere is $ 5 \mathrm{~cm} $, find the volume of the water left in the tub. (Take $ \pi=22 / 7 $ )

Given:

A cylindrical tub of radius \( 5 \mathrm{~cm} \) and length \( 9.8 \mathrm{~cm} \) is full of water.

A solid in the form of a right circular cone mounted on a hemisphere is immersed in the tub.

The radius of the hemisphere is immersed in the tub.

The radius of the hemisphere is \( 3.5 \mathrm{~cm} \) and height of the cone outside the hemisphere is \( 5 \mathrm{~cm} \)

To do:

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

Solution:

Radius of the cylindrical tub $R = 5\ cm$

Height of the cylindrical tub $H = 9.8\ cm$
Radius of the solid $r = 3.5\ cm$

Height of the cone $h = 5\ cm$

Volume of water in the tub $=\pi \mathrm{R}^{2} H$

$=\frac{22}{7} \times 5^2 \times 9.8$

$=770 \mathrm{~cm}^{3}$

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

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

$=\frac{1}{3} \times \frac{22}{7} \times (3.5)^2(5+2 \times 3.5)$

$=\frac{38.5}{3}(5+7)$

$=\frac{38.5}{3} \times 12$

$=38.5 \times 4$

$=154 \mathrm{~cm}^{3}$

Volume of the water flown out of the tub $=$ Volume of the solid

This implies,

Water flown out of the tub $= 154\ cm^3$

Water left in the tub $= 770 - 154$

$= 616\ cm^3$

The volume of the water left in the tub is $616\ cm^3$.

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Updated on: 10-Oct-2022

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