Question

A solid consisting of a right circular cone of height 12 cm and radius 6 cm standing on a hemisphere of radius 6 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of the water displaced out of the cylinder, if the radius of the cylinder is 6 cm and height is 18 cm.


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Answer 1

Answer:

Your answer is given below:-

Step-by-step explanation:

Volume of water displaced out = Volume of the solid immersed in.

Volume of the solid = Volume of the cone + Volume of the hemisphere

∴ The volume of water displaced out = Volume of the solid

           = (1) + (2)

           = 905.14 cm3

Answer 2

$\large\color{yellow}\boxed{\colorbox{blue}{Answer : - }}$

Given :

For cone:

height = 12 cm

radius = 6 cm

For hemisphere:

radius = 6 cm

For cylinder:

radius = 6 cm

height = 18 cm

$\sf{Find:}$

Volume of the water displaced by the cylinder.

We know that,

$\color{red}Volume \: of \: water \: displaced \: = \: [ Volume \: of \: cone + Volume \: of \: Hemisphere ]$

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

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

$= \frac{1}{3} \times \frac{22}{7} \times 6 \times 6 \: (12 + 2 \times 6)$

$= \frac{22}{7} \times 12 \: (12 + 12)$

$= \frac{22}{7} \times 12 \times 24$

$= 905.14 \: {cm}^{3}$