
If a liquid from a completely filled spherical container of inner radius $r$ is poured into a cube shaped container, what would be the dimensions of the cube in terms of the radius of the sphere?
A. $\dfrac{\sqrt[3]{36\pi }}{3}r$
B. ${{r}^{3}}$
C. $\dfrac{4\pi r}{3}$
D. $\sqrt{\dfrac{4\pi r}{3}}$
Answer
489.9k+ views
Hint: Here we have to find the dimension of the cube in terms of the radius of the sphere by the information that the liquid is poured from the spherical container to the cube shaped container. So as we know that Volume of the liquid will remain the same , we will put the volume of the sphere formula equal to the volume of the cube formula. Then we will simplify the equation obtained to get the value of the cube side in terms of the sphere radius.
Complete step by step answer:
We have been given that a liquid fills the spherical container with radius $r$ and the same liquid is poured in a cube shaped container. We can draw the sphere as follows:
So we have the volume of both shaped containers as the same.
We know Volume of Sphere is calculated as follows:
$V=\dfrac{4}{3}\pi {{r}^{3}}$…..$\left( 1 \right)$
Let the side of the cube is of the length $a$ so,
Volume of cube will be as follows:
$V={{a}^{3}}$…..$\left( 2 \right)$
As the volume of liquid is same in both the container therefore put equation (1) equal to equation (2) as follows:
${{a}^{3}}=\dfrac{4}{3}\pi {{r}^{3}}$
Cube rooting both sides we get,
$\Rightarrow {{a}^{3\times \dfrac{1}{3}}}=\sqrt[3]{\dfrac{4}{3}\pi {{r}^{3}}}$
$\Rightarrow a=r\times \sqrt[3]{\dfrac{4}{3}\pi }$
Now as we have options with denominator as $3$ so we will multiply and divide by $9$ inside the cube root on the right hand side as follows:
$\Rightarrow a=r\times \sqrt[3]{\dfrac{4}{3}\pi \times \dfrac{9}{9}}$
$\Rightarrow a=r\times \sqrt[3]{\dfrac{36}{27}\pi }$
As $\sqrt[3]{27}=3$ using it above we get,
$\therefore a=r\dfrac{\sqrt[3]{36\pi }}{3}$
So we get the side of the cube in terms of the radius of the sphere as above.
Hence the correct option is A.
Note: In such types of questions we need to find the relation between the two shapes given in order to get the relation among the dimensions of them. Here we have given that the same liquid is poured from a spherical container to a cubic container so we have taken their volume equal. As the options we have have a denominator free from cube root we have done some extra calculation and got our answer.
Complete step by step answer:
We have been given that a liquid fills the spherical container with radius $r$ and the same liquid is poured in a cube shaped container. We can draw the sphere as follows:
So we have the volume of both shaped containers as the same.
We know Volume of Sphere is calculated as follows:
$V=\dfrac{4}{3}\pi {{r}^{3}}$…..$\left( 1 \right)$
Let the side of the cube is of the length $a$ so,
Volume of cube will be as follows:
$V={{a}^{3}}$…..$\left( 2 \right)$
As the volume of liquid is same in both the container therefore put equation (1) equal to equation (2) as follows:
${{a}^{3}}=\dfrac{4}{3}\pi {{r}^{3}}$
Cube rooting both sides we get,
$\Rightarrow {{a}^{3\times \dfrac{1}{3}}}=\sqrt[3]{\dfrac{4}{3}\pi {{r}^{3}}}$
$\Rightarrow a=r\times \sqrt[3]{\dfrac{4}{3}\pi }$
Now as we have options with denominator as $3$ so we will multiply and divide by $9$ inside the cube root on the right hand side as follows:
$\Rightarrow a=r\times \sqrt[3]{\dfrac{4}{3}\pi \times \dfrac{9}{9}}$
$\Rightarrow a=r\times \sqrt[3]{\dfrac{36}{27}\pi }$
As $\sqrt[3]{27}=3$ using it above we get,
$\therefore a=r\dfrac{\sqrt[3]{36\pi }}{3}$
So we get the side of the cube in terms of the radius of the sphere as above.
Hence the correct option is A.
Note: In such types of questions we need to find the relation between the two shapes given in order to get the relation among the dimensions of them. Here we have given that the same liquid is poured from a spherical container to a cubic container so we have taken their volume equal. As the options we have have a denominator free from cube root we have done some extra calculation and got our answer.
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