A hemispherical bowl of internal radius 15cm contains a liquid. The liquid is to be filled into cylindrical-shaped bottles of diameter 5cm and height 6cm. How many bottles are necessary to empty the bowl?
Answer
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Hint: Here we go through by simply finding the volume of firs hemispherical ball by formula $\dfrac{2}{3}\pi {r^3}$ and then divide it by the volume of cylindrical shape bottle that we find using $\pi {r^2}h$. To find out the number of bottles that are used.
Complete step-by-step answer:
According to the problem.
Given that,
Radius of hemispherical bowl, R = 15 cm.
Hence the volume of hemispherical bowl $ = \dfrac{2}{3}\pi {R^3}$ . As we know the volume of the sphere is $\dfrac{4}{3}\pi {R^3}$ so for the hemisphere it is divided by two.
$\therefore $ Volume of hemispherical bowl$ = \dfrac{2}{3}\pi {R^3}$
$ \Rightarrow \dfrac{2}{3} \times \dfrac{{22}}{7} \times 15 \times 15 \times 15$
And now for the bottle it is given that,
Height of the bottle (h) =6 cm.
Diameter of the cylindrical bottles (d) =5 cm.
Then radius, r$ = \dfrac{d}{2} = \dfrac{5}{2} = 2.5$cm
And we know that the volume of the cylinder is $\pi {r^2}h$
$\therefore $Volume of the cylindrical bottle$ = \pi {r^2}h$
$ \Rightarrow \dfrac{{22}}{7} \times 2.5 \times 2.5 \times 6$
Let the number of bottles that empty the bowl be (n).
Then the total volume of bottles is $n \times \dfrac{{22}}{7} \times 2.5 \times 2.5 \times 6$ that is equal to the volume of the bowl.
$ \Rightarrow n \times \dfrac{{22}}{7} \times 2.5 \times 2.5 \times 6 = \dfrac{2}{3} \times \dfrac{{22}}{7} \times 15 \times 15 \times 15$
$ \Rightarrow n = \dfrac{{\dfrac{2}{3} \times \dfrac{{22}}{7} \times 15 \times 15 \times 15}}{{\dfrac{{22}}{7} \times 2.5 \times 2.5 \times 6}} = 60$
Hence, the bottles that are necessary to empty the bowl are 60.
Note: Whenever we face such a type of question the key concept for solving the question is first calculate the total volume of the bowl by the data given in question. Then find out the volume of one bottle with the help of the data given in the question. Then assume there (n) bottles are used to empty the bowl. Now multiply the volume of one bottle to (n) to find out the total volume of the bottle then equate this value with the volume of the bowl to find the value of (n).
Complete step-by-step answer:
According to the problem.
Given that,
Radius of hemispherical bowl, R = 15 cm.
Hence the volume of hemispherical bowl $ = \dfrac{2}{3}\pi {R^3}$ . As we know the volume of the sphere is $\dfrac{4}{3}\pi {R^3}$ so for the hemisphere it is divided by two.
$\therefore $ Volume of hemispherical bowl$ = \dfrac{2}{3}\pi {R^3}$
$ \Rightarrow \dfrac{2}{3} \times \dfrac{{22}}{7} \times 15 \times 15 \times 15$
And now for the bottle it is given that,
Height of the bottle (h) =6 cm.
Diameter of the cylindrical bottles (d) =5 cm.
Then radius, r$ = \dfrac{d}{2} = \dfrac{5}{2} = 2.5$cm
And we know that the volume of the cylinder is $\pi {r^2}h$
$\therefore $Volume of the cylindrical bottle$ = \pi {r^2}h$
$ \Rightarrow \dfrac{{22}}{7} \times 2.5 \times 2.5 \times 6$
Let the number of bottles that empty the bowl be (n).
Then the total volume of bottles is $n \times \dfrac{{22}}{7} \times 2.5 \times 2.5 \times 6$ that is equal to the volume of the bowl.
$ \Rightarrow n \times \dfrac{{22}}{7} \times 2.5 \times 2.5 \times 6 = \dfrac{2}{3} \times \dfrac{{22}}{7} \times 15 \times 15 \times 15$
$ \Rightarrow n = \dfrac{{\dfrac{2}{3} \times \dfrac{{22}}{7} \times 15 \times 15 \times 15}}{{\dfrac{{22}}{7} \times 2.5 \times 2.5 \times 6}} = 60$
Hence, the bottles that are necessary to empty the bowl are 60.
Note: Whenever we face such a type of question the key concept for solving the question is first calculate the total volume of the bowl by the data given in question. Then find out the volume of one bottle with the help of the data given in the question. Then assume there (n) bottles are used to empty the bowl. Now multiply the volume of one bottle to (n) to find out the total volume of the bottle then equate this value with the volume of the bowl to find the value of (n).
Last updated date: 19th Sep 2023
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