Question

A hemispherical bowl of internal diameter is $36$ cm contains liquid. This liquid is filled into $72$ cylindrical bottles of diameter $6$ cm. Find the height of each bottle if $10\% $ liquid is wasted in this transfer.

Answer 1

Hint: First use the formula of volume of hemisphere which is given as-
Volume of hemisphere=$\dfrac{2}{3}\pi {r^3}$where r is radius. Then put the values according to the question. Then find $10\% $ of the volume of the hemispherical bowl. This volume will be equal to the volume of $72$ cylindrical bottles. Find the volume of cylindrical bottles using formula which is given as-
Volume of cylinder=$\pi {r^2}h$ where r is radius and h is height. Put this equal to the volume of remaining liquid transferred and solve to get the height of bottles.

Complete step-by-step answer:
Given the diameter of hemispherical bowl containing liquid =$36$ cm
This liquid is filled into cylindrical bottles having diameter=$6$ cm
The number of bottles = $72$
Liquid wasted in the transfer =$10\% $
We have to find the height of the bottles.
We know that radius is half of diameter. So we can write,
The radius of hemispherical bowl = $\dfrac{{36}}{2} = 18$ cm
The radius of cylindrical bottle = $\dfrac{6}{2} = 3$ cm

Now we know that the Volume of hemisphere is given as-
Volume of hemisphere=$\dfrac{2}{3}\pi {r^3}$where r is radius
Then on putting the given values we get,
Volume of hemispherical bowl= $\dfrac{2}{3} \times \dfrac{{22}}{7} \times {\left( {18} \right)^3}$ -- (i)
On solving we get,
Volume of hemispherical bowl= $\dfrac{{44}}{7} \times 289 \times 6 = 12219.43{\text{ c}}{{\text{m}}^3}$
Now according to the question $10\% $ liquid is wasted of this volume.
So $10\% $ of $12219.43$ =$\dfrac{{10}}{{100}} \times 12219.43$
On solving we get,
$10\% $ of $12219.43$=$1221.943$
Then the remaining volume of liquid which is transferred = Total volume of hemispherical bowl - $10\% $ of the volume of hemispherical bowl
On putting values we get,
The remaining volume of liquid = $12219.43 - 1221.943$
On subtraction we get,
The remaining volume of liquid= $10997.487{\text{ c}}{{\text{m}}^3}$--- (ii)
Now we know that the volume of cylinder is given as-
Volume of cylinder=$\pi {r^2}h$ where r is radius and h is height
On putting the given values we get,
Volume of one cylindrical bottle = $\dfrac{{22}}{7} \times {\left( 3 \right)^2} \times {\text{h}}$
On solving we get,
Volume of one cylindrical bottle =$\dfrac{{594}}{7}{\text{h}}$
Now the volume of $72$ cylindrical bottles= $72 \times \dfrac{{594}}{7}{\text{h}}$ -- (iii)
Now according to question eq. (ii) and (iii) should be equal.
So on putting them equal to each other we get,
$ \Rightarrow 72 \times \dfrac{{594}}{7}{\text{h = 10997}}{\text{.487}}$
On adjusting we get,
$ \Rightarrow {\text{h = }}\dfrac{{{\text{10997}}{\text{.487}} \times 7}}{{72 \times 594}}$
On simplifying we get,
$ \Rightarrow {\text{h = }}\dfrac{{76982.409}}{{42768}}$
On division we get,
$ \Rightarrow {\text{h = 5}}{\text{.4}}$ cm
The height of bottles is $5.4$ cm.

Note: You can also solve this question by using this method-
Since $10\% $ liquid is wasted so only $90\% $ is transferred. Then we can write,
The volume of $72$ cylindrical bottles = $90\% $ of the volume of hemispherical bowl
Then on putting values from eq. (i) and (ii) we get,
$ \Rightarrow 72 \times \dfrac{{594}}{7}h = \dfrac{{90}}{{100}} \times \dfrac{2}{3} \times \dfrac{{22}}{7} \times 18 \times 18 \times 18$
On solving we get,
$ \Rightarrow h = \dfrac{{90}}{{100}} \times \dfrac{2}{3} \times \dfrac{{22}}{7} \times \dfrac{{18 \times 18 \times 18 \times 7}}{{72 \times 594}}$
On simplifying we get,
$ \Rightarrow h = \dfrac{{27}}{5} = 5.4$ cm