# 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

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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.

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.

Last updated date: 03rd Oct 2023

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