Question

A substance evaporates at a rate proportional to the exposed surface. If a spherical mothball of radius $\frac{\mathbf{1}}{\mathbf{3}}\mathit{c}\mathit{m}$has radius $\mathbf{0}\mathbf{.}\mathbf{4}\mathit{c}\mathit{m}$after 6 months, how long will it take:

(a) For the radius to be $\frac{\mathbf{1}}{4}\mathit{c}\mathit{m}$?

(b) For the volume of the mothball to be half of what it was originally?

Short answer

Answer

A substance evaporates at a rate proportional to the exposed surface. If a spherical mothball of radius $\frac{1}{3}cm$has radius $0.4cm$after $6$months, it will take

a) $15$months for the radius to be $\frac{1}{4}cm$.

b) $6.19$months for the mothball to have a volume equal to half the original volume.

Step-by-step solution

Given information

It is given that a spherical mothball of radius $\frac{1}{3}cm$has radius $0.4cm$after $6$months.

Meaning of volume

The mathematical quantity 'volume' represents the amount of three-dimensional space occupied by an object or a closed surface in mathematics. The volume is measured in cubic units such as ${\mathit{m}}^{\mathbf{3}}\mathbf{,}\mathit{c}{\mathit{m}}^{\mathbf{3}}\mathbf{,}\mathbf{}\mathit{i}{\mathit{n}}^{\mathbf{3}}$, and so on. Volume is sometimes referred to as capacity. The volume of a cylindrical jar, for example, is used to determine how much water it can hold.

Find the time when the radius reduce to 14cm

In this case, when the radius has been reduced $0.1cm$, it took 6 months.

So, if the radius reduces$0.25cm$, it will take

$\mathbf{6}\mathit{m}\mathit{o}\mathit{n}\mathit{t}\mathit{h}\mathit{s}\mathbf{+}\mathbf{6}\mathit{m}\mathit{o}\mathit{n}\mathit{t}\mathit{h}\mathit{s}\mathbf{+}\mathbf{3}\mathit{m}\mathit{o}\mathit{n}\mathit{t}\mathit{h}\mathit{s}\mathbf{=}\mathbf{15}\mathit{m}\mathit{o}\mathit{n}\mathit{t}\mathit{h}\mathit{s}\mathbf{.}$.

Find the time when the mothball has a volume equal to half of the original volume.

First, calculate the original volume of the mothball.

$$\begin{gathered} V{\mathrm{\pi r}}_0=\frac{4}{3}{}^3 \\ =\frac{4}{3}\times \mathrm{\pi }\times {\left(0.5\right)}^3 \\ =\frac{\mathrm{\pi }}{6}{\mathrm{cm}}^3 \end{gathered}$$

Now, find the radius of the ball when the mothball has a volume equal to half of the original volume.

$$\begin{gathered} V=\frac{V_2}{2} \\ =\frac{\mathrm{\pi }}{12} \\ =\frac{4}{3}{\mathrm{\pi r}}^3 \\ \mathrm{r}=\sqrt[3]{\frac{1}{16}}\approx 0.39685\mathrm{cm} \end{gathered}$$

Find the time.

$$\begin{gathered} t=\left(0.5-r\right)\left(60months/cm\right) \\ t=\left(0.5-0.39685\right)\left(60months/cm\right) \end{gathered}$$

Therefore, $t=6.19months$.