How many 9

How many percent volume remain of a sphere if diameter decrease 3×?

Correct answer:

p =  96.2963 %

Step-by-step explanation:

$$\begin{gathered} d=D\mathrm{/}3 \\ r=R\mathrm{/}3 \\ {V}_1={\displaystyle \frac{4}{3}}\pi r^3 \\ {V}_2={\displaystyle \frac{4}{3}}\pi R^3 \\ p=100\cdot {\displaystyle \frac{V_1-V_2}{V_1}} \\ p=100\cdot {\displaystyle \frac{R^3-r^3}{R^3}} \\ p=100\cdot {\displaystyle \frac{R^3-(R\mathrm{/}3{)}^3}{R^3}} \\ p=100\cdot {\displaystyle \frac{1-(1\mathrm{/}3{)}^3}{1}} \\ p=100\cdot (1-(1\mathrm{/}3{)}^3)={\displaystyle \frac{2600}{27}}=96.2963\mathrm{\%} \end{gathered}$$

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