Question

A hemispherical bowl of internal diameter 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.
(A) 50 cm (B) 5.2 cm (C) 5.4 cm (D) 6.0 cm.

Answer 1

Hint: Start with finding the total volume of liquid by finding the volume of the bowl. Then compare it with the total volume of cylindrical bottles ensuring that 10% of liquid is lost while transferring it.

According to the question, the diameter of the bowl is 36 cm. Then its radius is 18 cm.
$ \Rightarrow r = 18cm$
We know that the volume of a hemispherical bowl is $\dfrac{2}{3}\pi {r^3}$. Using this formula, we’ll get:
$
   \Rightarrow V = \dfrac{2}{3} \times \dfrac{{22}}{7} \times {\left( {18} \right)^3}, \\
   \Rightarrow V = \dfrac{{44 \times 5832}}{{21}}, \\
   \Rightarrow V = 12219.43c{m^3} \\
$
Thus, the total volume of liquid is $12219.43c{m^3}$.
Given, the diameter of the cylinder is 6 cm. Thus, its radius is 3 cm.
We know that the volume of the cylinder is $\pi {r^2}h$. And we have 72 cylindrical bottles. So, total volume is:
$
   \Rightarrow V = 72 \times \pi \times {\left( 3 \right)^2}h, \\
   \Rightarrow V = 72 \times \dfrac{{22}}{7} \times 9h \\
$
Now, from the total volume of liquid calculated above (i.e. $12219.43c{m^3}$), 10% is lost while transferring it. The volume of liquid left is 90% of $12219.43c{m^3}$. And this volume must be equal to the total cylindrical bottle’s volume. So, we have:
$
   \Rightarrow 72 \times \dfrac{{22}}{7} \times 9h = 90\% {\text{ of }}12219.43c{m^3}, \\
   \Rightarrow 72 \times \dfrac{{22}}{7} \times 9h = \dfrac{{90}}{{100}} \times 12219.43, \\
   \Rightarrow h = \dfrac{{7 \times 12219.43}}{{72 \times 22 \times 10}}, \\
   \Rightarrow h = \dfrac{{85536.01}}{{15840}}, \\
   \Rightarrow h = 5.4cm \\
$
Thus the height of each cylindrical bottle is 5.4 cm. (C) is the correct option.
Note: Volume of liquid is conserved in the above question.
$ \Rightarrow $Volume of liquid stored in the bowl $ = $volume of liquid transferred in bottles $ + $volume of liquid lost while transferring it.
This is the key principle for solving such types of problems.