Question

Radius of the Uranus is 4 times the radius of the earth. Then the ratio of volume of the earth to volume of the Uranus is ${{x}^{-3}}$where x = ______ .

Answer 1

The planets are considered to be spherical in shape, so, we will make use of the volume of the sphere to solve this problem. In the formula of volume of a sphere, other than radius all other parameters are constant, so, while computing the ratio of volume the parameters get cancelled out.
Formulae used:
$V=\dfrac{4}{3}\pi {{r}^{3}}$

Complete step by step answer:
From the given information, we have the data as follows.
The radius of the Uranus planet is 4 times the radius of the planet Earth.
The formula to be used to solve this problem is given as follows.
$V=\dfrac{4}{3}\pi {{r}^{3}}$
Where $V$is the volume of the sphere and $r$is the radius of the sphere.
In the case of planet earth, let the radius be ${{r}_{1}}$ and let the volume be, ${{V}_{1}}$. In the case of the planet Uranus, let the radius be ${{r}_{2}}$ and let the volume be, ${{V}_{2}}$.
The volume of the planet earth having the radius ${{r}_{1}}$ and the volume be, ${{V}_{1}}$ is given as follows.
${{V}_{1}}=\dfrac{4}{3}\pi r_{1}^{3}$
The volume of the planet Uranus having the radius ${{r}_{2}}$ and the volume be, ${{V}_{2}}$ is given as follows.
${{V}_{2}}=\dfrac{4}{3}\pi r_{2}^{3}$
Now, we will compute the ratio of the volume of the earth to the volume of the Uranus. So, consider the formula.
$\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{}^{4}/{}_{3}\pi r_{1}^{3}}{{}^{4}/{}_{3}\pi r_{2}^{3}}$
Substitute the given condition in the above equation, that is, the radius of the Uranus planet is 4 times the radius of the planet Earth and cancel out the common terms.
$\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{r_{1}^{3}}{{{\left( 4{{r}_{1}} \right)}^{3}}}$
Cancel out the common terms.

& \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{1}{{{4}^{3}}} \\\ & \therefore \dfrac{{{V}_{1}}}{{{V}_{2}}}={{4}^{-3}} \\\ \end{aligned}$$ Therefore, the value of the ratio of the volume of the earth to the volume of the Uranus is, $${{4}^{-3}}$$. Thus, the value of x is, $$\begin{aligned} & {{x}^{-3}}={{4}^{-3}} \\\ & \Rightarrow x=4 \\\ \end{aligned}$$ **$$\therefore $$The value of x is 4.** **Note:** There is no need to know the value of the radius of the earth to solve this problem, as the ratio cancels out the common terms. In this case, the planet Uranus is considered, even the planets whose radius is smaller than that of the planet earth can be considered.