A spherical object's volume decreases by $30\%$. What is the radius decrease percentage?

Question

I know how to find the decrease percentage of the volume with respect to the decreasing radius, but I'm not quite sure the other way around. I am stuck in cancelling the radius.

Attempt at solution:

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

Decrease factor: $1-0.3=0.7$

$0.7\cdot V = \frac{4}{3}\pi r^3$

I'm not quite sure how to approach this. With a rough reverse-operation, I managed to get around $55$%. Any guidance would be appreciated.

Answer 1

You need the cubic root of $\ 0.7\ $, which is about $\ 0.8879$

So, the radius decreases with a percentage of about $11.21$%

Answer 2

$$ \begin{gathered} V_{\,0} = \frac{4} {3}\pi r_{\,0} ^{\,3} \quad V_{\,1} = \frac{4} {3}\pi r_{\,1} ^{\,3} \hfill \\ \frac{{V_{\,1} }} {{V_{\,0} }} = \left( {\frac{{r_{\,1} }} {{r_{\,0} }}} \right)^{\,3} = 0.7\quad \to \quad \frac{{r_{\,1} }} {{r_{\,0} }} = \sqrt[3]{{0.7}} \approx 0.8879\quad \to \quad \frac{{\Delta \,r}} {{r_{\,0} }} \approx - 11.2\% \hfill \\ \end{gathered} $$

Answer 3

Let $r$ be the initial radius and let $k$ be a "factor" such that $kr$ is the final radius. The initial volume is $(4/3)\pi r^3$ and the final volume is $(4/3)\pi (kr)^3= (4/3)\pi k^3 r^3$. Since the final volume is .7 times the original volume, we have $\frac{(4/3)\pi k^3r^3}{(4/3)\pi r^3}= k^3= 0.7$

Answer 4

$$\frac vV=\frac{\frac43\pi r^3}{\frac43\pi R^3}=\left(\frac rR\right)^3=1-0.30,$$

then

$$\frac rR=\sqrt[3]{1-0.30}=1-(1-\sqrt[3]{1-0.30})=1-0.11.$$