Question

if the radius of a sphere is boubled, what is the ratio of the volume of the first sphere to that of the second sphere​

Answer 1

Given:-

• Radius of a sphere is doubled.

To find:-

• Ratio of volume of first sphere to that of the second sphere.

Formula to be used:-

• Volume of sphere = (4/3)πr³

Answer:-

❖ Let the radius of the first sphere be $\sf r.$

❖ Then, the radius of the second sphere will be $\sf 2r.$

❖Let the volume of the first sphere be $\sf V_1$ and the volume of the second sphere be $\sf V_2.$

❖ For first sphere:

$\sf V_1 = \dfrac{4}{3} \pi r^3\quad\quad \dots (1)$

❖ For second sphere:

Here, as the radius becomes $\sf 2r,$

$\sf V_2 = \dfrac{4}{3} \pi (2r)^3$

$\longrightarrow \sf V_2 = \dfrac{4}{3} \pi 8r^3 \quad \quad \dots (2)$

❖ Ratio:

The equations (1) and (2) give,

$\sf \dfrac{V_1}{V_2} = \dfrac{ \quad\bigg( \dfrac{4}{3} \pi r^3 \bigg) \quad }{ \quad\bigg(\dfrac{4}{3} \pi 8 r^3\bigg) \quad }$

$\sf \longrightarrow \dfrac{V_1}{V_2} = \dfrac{ \quad\bigg( \cancel{\dfrac{4}{3}} \: \: \cancel{\pi} \: \: \cancel{r^3} \bigg) \quad }{ \quad\bigg(\cancel{\dfrac{4}{3}} \: \: \cancel{\pi} \: \: 8 \: \: \cancel{r^3}\bigg) \quad}$

$\sf \longrightarrow \dfrac{V_1}{V_2} = \dfrac{1}{8}$

$\longrightarrow \underline{\underline{\sf{\green{V_1 : V_2 = 1 : 8}}}}$