# A hemispherical bowl of internal diameter 36 cm contains a liquid. This liquid is to be filled in cylindrical bottles of radius 3 cm and height 6 cm. How many bottles are required to empty the bowl?

Last updated date: 27th Mar 2023

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Answer

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Hint: Volume of the liquid is the conserved quantity. Hence, find the volume of the hemispherical bowl and the cylindrical bottle and divide the volume of the hemispherical bowl by the volume of the cylindrical bottle to obtain the number of bottles required.

Complete step-by-step answer:

When the liquid is filled from the hemispherical bowl to the cylindrical bottles, the total volume of the liquid remains the same.

Let \[n\] be the number of cylindrical bottles required to empty the bowl.

Let \[{V_H}\] be the volume of the hemispherical bowl and \[{V_C}\] be the volume of a cylindrical bottle.

Then, we have:

\[{\text{Volume of hemispherical bowl = }}n \times {\text{Volume of cylindrical bottle}}\]

\[{V_H} = n \times {V_C}{\text{ }}.........{\text{(1)}}\]

Given that:

\[{\text{Internal diameter of hemispherical bowl, }}{{\text{d}}_H} = 36{\text{ }}cm\]

\[{\text{Radius of cylindrical bottle, }}{{\text{r}}_C} = 3{\text{ }}cm\]

\[{\text{Height of cylindrical bottle, }}{{\text{h}}_C} = 6{\text{ }}cm\]

We now find the radius of the hemispherical bowl as follows:

\[{\text{Radius of hemispherical bowl, }}{{\text{r}}_H} = \dfrac{{{d_H}}}{2}\]

\[ \Rightarrow {\text{ }}{{\text{r}}_H} = \dfrac{{36}}{2}{\text{ }}cm\]

\[ \Rightarrow {\text{ }}{{\text{r}}_H} = 18{\text{ }}cm\]

We know that the volume of a hemisphere of radius \[r\] is given by:

\[V = \dfrac{2}{3}\pi {r^3}{\text{ }}..........{\text{(2)}}\]

We also know that the volume of a cylinder of radius \[r\] and height \[h\] is given by:

\[V = \pi {r^2}h{\text{ }}..........{\text{(3)}}\]

Using equation (2), the volume of the hemispherical bowl is as follows:

\[{V_H} = \dfrac{2}{3}\pi {r_H}^3{\text{ }}..........{\text{(4)}}\]

Using equation (3), the volume of the cylindrical bottle is as follows:

\[{V_C} = \pi {r_C}^2{h_C}{\text{ }}..........{\text{(5)}}\]

Substituting equation (4) and (5) in equation (1), we get:

\[\dfrac{2}{3}\pi {r_H}^3{\text{ = }}n \times \pi {r_C}^2{h_C}\]

Solving for \[n\] , we get:

\[n = \dfrac{{2{r_H}^3}}{{3{r_C}^2{h_C}}}\]

Substituting the values for \[{r_H}\] , \[{r_C}\] and \[{h_C}\] , we get:

\[n = \dfrac{{2{{(18)}^3}}}{{3{{(3)}^2}(6)}}\]

Simplifying we obtain:

\[n = {\text{18}} \times 2 \times 2\]

\[n = 72\]

Therefore, 72 bottles are required to empty the bowl.

Note: You might proceed with the total surface area being equal in both cases but it is wrong, the total volume of the liquid is the conserved quantity. Also note that it is hemispherical bowl and volume of the hemisphere is half of that of volume of sphere, that is, \[{V_H} = \dfrac{2}{3}\pi {r^3}\] .

Complete step-by-step answer:

When the liquid is filled from the hemispherical bowl to the cylindrical bottles, the total volume of the liquid remains the same.

Let \[n\] be the number of cylindrical bottles required to empty the bowl.

Let \[{V_H}\] be the volume of the hemispherical bowl and \[{V_C}\] be the volume of a cylindrical bottle.

Then, we have:

\[{\text{Volume of hemispherical bowl = }}n \times {\text{Volume of cylindrical bottle}}\]

\[{V_H} = n \times {V_C}{\text{ }}.........{\text{(1)}}\]

Given that:

\[{\text{Internal diameter of hemispherical bowl, }}{{\text{d}}_H} = 36{\text{ }}cm\]

\[{\text{Radius of cylindrical bottle, }}{{\text{r}}_C} = 3{\text{ }}cm\]

\[{\text{Height of cylindrical bottle, }}{{\text{h}}_C} = 6{\text{ }}cm\]

We now find the radius of the hemispherical bowl as follows:

\[{\text{Radius of hemispherical bowl, }}{{\text{r}}_H} = \dfrac{{{d_H}}}{2}\]

\[ \Rightarrow {\text{ }}{{\text{r}}_H} = \dfrac{{36}}{2}{\text{ }}cm\]

\[ \Rightarrow {\text{ }}{{\text{r}}_H} = 18{\text{ }}cm\]

We know that the volume of a hemisphere of radius \[r\] is given by:

\[V = \dfrac{2}{3}\pi {r^3}{\text{ }}..........{\text{(2)}}\]

We also know that the volume of a cylinder of radius \[r\] and height \[h\] is given by:

\[V = \pi {r^2}h{\text{ }}..........{\text{(3)}}\]

Using equation (2), the volume of the hemispherical bowl is as follows:

\[{V_H} = \dfrac{2}{3}\pi {r_H}^3{\text{ }}..........{\text{(4)}}\]

Using equation (3), the volume of the cylindrical bottle is as follows:

\[{V_C} = \pi {r_C}^2{h_C}{\text{ }}..........{\text{(5)}}\]

Substituting equation (4) and (5) in equation (1), we get:

\[\dfrac{2}{3}\pi {r_H}^3{\text{ = }}n \times \pi {r_C}^2{h_C}\]

Solving for \[n\] , we get:

\[n = \dfrac{{2{r_H}^3}}{{3{r_C}^2{h_C}}}\]

Substituting the values for \[{r_H}\] , \[{r_C}\] and \[{h_C}\] , we get:

\[n = \dfrac{{2{{(18)}^3}}}{{3{{(3)}^2}(6)}}\]

Simplifying we obtain:

\[n = {\text{18}} \times 2 \times 2\]

\[n = 72\]

Therefore, 72 bottles are required to empty the bowl.

Note: You might proceed with the total surface area being equal in both cases but it is wrong, the total volume of the liquid is the conserved quantity. Also note that it is hemispherical bowl and volume of the hemisphere is half of that of volume of sphere, that is, \[{V_H} = \dfrac{2}{3}\pi {r^3}\] .

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