Question

A football tube in the form of a sphere is inflated so that it radius increases in the ratio 4:3 . Find the ratio in which the volume is increased. ​

Answer 1

Answer:

The volume is increased in a ratio of 64/27

Explanation:

The Volume of a Sphere

The volume of a sphere of radius r is given by:

$\displaystyle V=\frac{4}{3}\cdot \pi\cdot r^3$

If the radius was changed to r', the new volume would be:

$\displaystyle V'=\frac{4}{3}\cdot \pi\cdot r'^3$

Dividing the latter equation by the first one:

$\displaystyle \frac{V'}{V}=\frac{\frac{4}{3}\cdot \pi\cdot r'^3}{\frac{4}{3}\cdot \pi\cdot r^3}$

Simplifying:

$\displaystyle \frac{V'}{V}=\frac{r'^3}{r^3}$

Or, equivalently:

$\displaystyle \frac{V'}{V}=\left(\frac{r'}{r}\right)^3$

Since r':r = 4:3, thus:

$\displaystyle \frac{V'}{V}=\left(\frac{4}{3}\right)^3$

$\displaystyle \frac{V'}{V}=\frac{4^3}{3^3}$

$\displaystyle \frac{V'}{V}=\frac{64}{27}$

The volume is increased in a ratio of 64/27