Taking the Ratio of the Two Volumes

If two similar solids have a scale factor of $\frac{a}{b}$ , then corresponding volumes have a ratio of $\frac{a^3}{b^3}$ . Since the height of the liquid in the second glass is half of the height of the liquid in the first glass, then the scale factor is $\frac{1}{2}$ , and the volumes have a ratio of $\frac{1^3}{2^3} = \boxed{\frac{1}{8}}$ .

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Another way to look at it is that since the liquid in the second glass has half the height, it would also have $\frac{1}{2}$ the radius for the circle, which translates to $\frac{1}{4}$ the area for the circle, which means that the volume ( $V = \frac{1}{3}Ah$ ) of the liquid in the second glass is $\frac{1}{4}$ of $\frac{1}{2}$ of the volume of the liquid in the first glass, or $\boxed{\frac{1}{8}}$ .

Let $V_1$ be the volume of the small cone and $V_2$ be the volume of the larger cone. Considering my figure and by similar triangles, we have

$\dfrac{R}{2h}=\dfrac{r}{h}$ $\color{#D61F06}\implies$ $R=2r$

The ratio of their volume is

$\dfrac{V_1}{V_2}=\dfrac{\frac{1}{3}\pi r^2h}{\frac{1}{3}\pi (R^2)(2h)}=\dfrac{r^2}{2R^2}$

However, $R=2r$ . So

$\dfrac{V_1}{V_2}=\dfrac{r^2}{2(2r)^2}=\dfrac{r^2}{8r^2}=\dfrac{1}{8}$ $\color{#D61F06}\implies$ $\boxed{V_1=\dfrac{1}{8}V_2}$

For solids of similar shade, the volume $V$ is directly proportional to $x^3$ , where $x$ is a linear dimension of space (one of $x$ , $y$ or $z$ ). In equations:

$\begin{aligned} V & \propto x^3 \\ V(x) & = {\color{#3D99F6}k} x^3 & \small \color{#3D99F6} \text{where }k \text{ is a constant.} \end{aligned}$

$\begin{aligned} \implies V \left(\frac x2\right) & = k \left(\frac x2\right)^3 = \frac k8 x^3 = \boxed{\dfrac 18} V(x) \end{aligned}$

Height is halved, radius is halved

V =πr2/3

So eighth part of the initial