Question

The ratio of the volume and surface area of a sphere of unit radius:
$$\begin{gathered} \left(a\right)4:3 \\ \left(b\right)3:4 \\ \left(c\right)1:3 \\ \left(d\right)3:1 \end{gathered}$$

Answer 1

Hint: In this question, we use the formula of volume and surface area of the sphere. First, find the value of volume and surface area of the sphere then take a ratio. We use volume of sphere $={\displaystyle \frac{4\pi r^3}{3}}$ and also surface area of sphere $$=4\pi r^2$$ where r is the radius of sphere.

Complete step-by-step solution:

Given, radius of sphere r=1
Now use formula of volume of sphere, $V={\displaystyle \frac{4\pi r^3}{3}}$
$$\begin{gathered} \Rightarrow V={\displaystyle \frac{4\pi \times {\left(1\right)}^3}{3}} \\ \Rightarrow V={\displaystyle \frac{4\pi }{3}}.............\left(1\right) \end{gathered}$$
 Now use formula of surface area of sphere, $$S=4\pi r^2$$
$$\begin{gathered} \Rightarrow S=4\pi \times {\left(1\right)}^2 \\ \Rightarrow S=4\pi ................\left(2\right) \end{gathered}$$
Now, take a ratio of the volume and surface area of a sphere.
Divide (1) equation by (2) equation,
$$\begin{gathered} \Rightarrow {\displaystyle \frac{V}{S}}={\displaystyle \frac{{\displaystyle \frac{4\pi }{3}}}{4\pi }} \\ \Rightarrow {\displaystyle \frac{V}{S}}={\displaystyle \frac{4\pi }{4\pi \times 3}} \\ \Rightarrow {\displaystyle \frac{V}{S}}={\displaystyle \frac{1}{3}} \end{gathered}$$
The ratio of the volume and surface area of a sphere is 1:3.
So, the correct option is (c).

Note: In such types of problems we can use two different ways to solve questions in easy ways. First way we already mentioned above and in a second way, we take a ratio of the volume of a sphere to the surface area of the sphere so we get the general ratio of volume to the surface area of a sphere and then put the value of the radius of the sphere.
$\Rightarrow {\displaystyle \frac{V}{S}}={\displaystyle \frac{{\displaystyle \frac{4\pi r^3}{3}}}{4\pi r^2}}={\displaystyle \frac{r}{3}}$
So, we can directly use this ratio and get the solution in just seconds.