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# 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.

Last updated date: 17th Jul 2024
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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.