
A sphere and a cube have equal surface areas. The ratio of volume of sphere to that of the cube is:
A.$\sqrt 6 :\sqrt \pi $
B.$\sqrt 2 :\sqrt \pi $
C.$\sqrt \pi :\sqrt 3 $
D.$\sqrt \pi :\sqrt 5 $
Answer
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Hint: In geometry, a cube is a three-dimensional solid object bounded by six square faces, facets or sides, with three meeting at each vertex. The cube is the only regular hexahedron and is one of the five Platonic solids. It has 6 faces, 12 edges, and 8 vertices.
Formulae used
I.Surface area of the cube $ = 6{a^2}$
II.Surface area of the sphere $ = 4\pi {r^2}$
Complete step-by-step answer:
Surface area of the cube $ = 6{a^2}$
Surface area of the sphere $ = 4\pi {r^2}$
Given that both the surface areas are equal –
$4\pi {r^2} = 6{a^2}$
Term multiplicative on one side is moved to the opposite side then it goes to the denominator and vice-versa.
\[\dfrac{{{r^2}}}{{{a^2}}} = \dfrac{6}{{4\pi }}\]
Take square-root on both the sides of the equation –
\[\sqrt {\dfrac{{{r^2}}}{{{a^2}}}} = \sqrt {\dfrac{6}{{4\pi }}} \]
Square and square-root cancel each other.
\[\dfrac{r}{a} = \sqrt {\dfrac{6}{{4\pi }}} \] ….. (A)
Ratio of the volumes can be given by,
$\dfrac{{Volume\,{\text{of sphere}}}}{{Volume\,of\;cube}} = \dfrac{{\dfrac{4}{3}\pi {r^3}}}{{{a^3}}}$
Now, the term in the denominator of the numerator goes to the denominator.
$\dfrac{{Volume\,{\text{of sphere}}}}{{Volume\,of\;cube}} = \dfrac{{4\pi {r^3}}}{{3{a^3}}}$
Place value from the equation (A)
$\dfrac{{Volume\,{\text{of sphere}}}}{{Volume\,of\;cube}} = \dfrac{4}{3}\pi \sqrt {\dfrac{6}{{4\pi }}} \times \dfrac{6}{{4\pi }}$
Common factors from the numerator to the denominator cancel each other.
$\dfrac{{Volume\,{\text{of sphere}}}}{{Volume\,of\;cube}} = \dfrac{1}{{3 \times 2}}\sqrt {\dfrac{6}{\pi }} \times \dfrac{6}{1}$
Simplify the above expression –
$\dfrac{{Volume\,{\text{of sphere}}}}{{Volume\,of\;cube}} = \sqrt {\dfrac{6}{\pi }} $
From the given multiple choices – the option A is the correct answer.
So, the correct answer is “Option A”.
Note: Volume is the quantity of three-dimensional space enclosed by a closed surface, for example, the space that a substance or 3D shape occupies or contains. Volume is often quantified numerically using the SI derived unit, the cubic metre
Formulae used
I.Surface area of the cube $ = 6{a^2}$
II.Surface area of the sphere $ = 4\pi {r^2}$
Complete step-by-step answer:
Surface area of the cube $ = 6{a^2}$
Surface area of the sphere $ = 4\pi {r^2}$
Given that both the surface areas are equal –
$4\pi {r^2} = 6{a^2}$
Term multiplicative on one side is moved to the opposite side then it goes to the denominator and vice-versa.
\[\dfrac{{{r^2}}}{{{a^2}}} = \dfrac{6}{{4\pi }}\]
Take square-root on both the sides of the equation –
\[\sqrt {\dfrac{{{r^2}}}{{{a^2}}}} = \sqrt {\dfrac{6}{{4\pi }}} \]
Square and square-root cancel each other.
\[\dfrac{r}{a} = \sqrt {\dfrac{6}{{4\pi }}} \] ….. (A)
Ratio of the volumes can be given by,
$\dfrac{{Volume\,{\text{of sphere}}}}{{Volume\,of\;cube}} = \dfrac{{\dfrac{4}{3}\pi {r^3}}}{{{a^3}}}$
Now, the term in the denominator of the numerator goes to the denominator.
$\dfrac{{Volume\,{\text{of sphere}}}}{{Volume\,of\;cube}} = \dfrac{{4\pi {r^3}}}{{3{a^3}}}$
Place value from the equation (A)
$\dfrac{{Volume\,{\text{of sphere}}}}{{Volume\,of\;cube}} = \dfrac{4}{3}\pi \sqrt {\dfrac{6}{{4\pi }}} \times \dfrac{6}{{4\pi }}$
Common factors from the numerator to the denominator cancel each other.
$\dfrac{{Volume\,{\text{of sphere}}}}{{Volume\,of\;cube}} = \dfrac{1}{{3 \times 2}}\sqrt {\dfrac{6}{\pi }} \times \dfrac{6}{1}$
Simplify the above expression –
$\dfrac{{Volume\,{\text{of sphere}}}}{{Volume\,of\;cube}} = \sqrt {\dfrac{6}{\pi }} $
From the given multiple choices – the option A is the correct answer.
So, the correct answer is “Option A”.
Note: Volume is the quantity of three-dimensional space enclosed by a closed surface, for example, the space that a substance or 3D shape occupies or contains. Volume is often quantified numerically using the SI derived unit, the cubic metre
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