Question

A hollow cylinder has solid hemisphere inward at one end and on the other end it is closed with a flat circular plate. The height of water is $10cm$ when flat circular surface is downward. Find the level of the water, when it is inverted upside down, common diameter is $7cm$ and height of the cylinder is $20cm$.

Answer 1

Given

A hollow cylinder has a solid hemisphere inward at one end and on the other end, it id closed with a flat circular plate.

Height of the water when flat circular surface is downward $\left(h\right)=10cm$

diameter $\left(d\right)=7cm$

Height of the cylinder $\left(h\right)=20cm$

Also, given to find the level of the water when it is inverted upside down.

$[\pi =\frac{22}{7}]$

Volume of water $\pi r^{2}h$

$=\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\times 10$

Volume of water $=11\times 7\times 5$

Volume of water $=385c{m}^3$ ......... $eq1$

Now, to find the level opf the water when it is inverted upside down.

Now, the hemisphere will be down and water is inverted down

let the height of the water $=h$

[Volume = Volume of cylinder volume of hemisphere]

[$\therefore$ volume of cylinder $=\pi r^{2}h$ volume of hemisphere $=\frac{2}{3}\pi r^3$ ]

$v=\pi .r^2.h-\frac{2}{3}.\pi .r^3$

$\pi =\frac{22}{7}$

$V=\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\times h-\frac{2}{3}\times \frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\times \frac{7}{2}$

$V=\frac{77}{9}h-\frac{11\times 7\times 7}{2\times 2}\iff v=\frac{77}{2}h-\frac{77}{2}\times \frac{7}{3}$

$[\because v=385]$ from $eq\left(1\right)$

$V=\frac{77}{2}[h-\frac{7}{3}]\iff 385=\frac{77}{2}[h-\frac{7}{3}]$

$\Rightarrow 385\times \frac{2}{77}=h-\frac{7}{3}\iff 10=\frac{3h-7}{3}$

$30=3h-7\iff 3h=30+7\iff h=\frac{37}{3}cm$