The ratio of the depth of the liquid to the depth of the cup is 4/5.
Geometry and Ratio
The formula for the volume of a cone is:
[tex]V=\frac{1}{3}\pi r^{2}h[/tex]
where r is the radius of the base of the cone and h is the height of the cone.
When the conical cup is filled with water, there will be a difference in the radius of the water and the cup and a difference in the height of the water and the height of the cup (can be seen in the picture).
From the difference in radius and height of the water with the conical cup, a ratio can be made of similarity rule of triangle (see picture).
If the radius of water is [tex]r_{1}[/tex] and the height of water is [tex]h_{1}[/tex], radius of cup is [tex]r_{2}[/tex] and the height of cup is [tex]h_{2}[/tex], so:
[tex]\frac{r_{1} }{r_{2} } =\frac{h_{1} }{h_{2} }[/tex]
[tex]r_{1}[/tex] with [tex]r_{2}[/tex] and [tex]h_{1}[/tex] with [tex]h_{2}[/tex] have the same certain ratio. Let's assume the ratio is x, then it can be written:
[tex]r_{2}=x.r_{1} \\h_{2}=x.h_{1}[/tex]
The ratio of the volume of water to the volume of the conical cup is 64/125, meaning that 64 of the water in the conical cup with a volume of 125.
From here we can substitute it in the volume ratio:
[tex]\frac{V_{1} }{V_{2}}=\frac{\frac{1}{3}\pi r_{1}^{2}h_{1} }{\frac{1}{3} \pi r_{2}^{2}h_{2}} =\frac{64}{125} \\\frac{V_{1} }{V_{2}}=\frac{\frac{1}{3}\pi r_{1}^{2}h_{1} }{\frac{1}{3} \pi (xr_{1})^{2}(x.h_{1})} =\frac{64}{125} \\\frac{V_{1} }{V_{2}}=\frac{\frac{1}{3}\pi r_{1}^{2}h_{1} }{\frac{1}{3}\pi r_{1}^{2}h_{1}x^{3} } =\frac{64}{125} \\\frac{V_{1} }{V_{2}}=\frac{1}{x^{3}}=\frac{64}{125}\\ \frac{1}{x^{3}}=\frac{64}{125}\\\\x^{3}=\frac{125}{64} \\x=\frac{5}{4}[/tex]
The ratio between the radius and height of the cup and the water is 5/4. This means that the ratio between the radius and height of the water and the cup is 4/5.
Learn more about geometric volume https://brainly.com/question/13338592
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