Commander Posted September 10, 2023 Posted September 10, 2023 If *C* is a Cube and *S1* is the Sphere Inscribed and *S2* is the Sphere Superscribed then (1) What is the fraction of the volume of *C* is contained in *S1* (2) How many times the volume of *S2* is that of *S1*
Sensei Posted September 10, 2023 Posted September 10, 2023 (edited) Fixed version: Spoiler Volume of cube is [math]x^3[/math] The volume of a sphere inscribed in a cube is: [math]\frac{4}{3} \pi \frac{x}{2}^3[/math] The volume of a sphere superscribed is: [math]\frac{4}{3} \pi \sqrt{ 3 \frac{x}{2}^2}^3[/math] 4 hours ago, Commander said: (1) What is the fraction of the volume of *C* is contained in *S1* [math]\frac{\frac{4}{3} \pi (\frac{x}{2})^3}{x^3}=\frac{\pi}{6}= 52.4\%[/math] Â Â 4 hours ago, Commander said: (2) How many times the volume of *S2* is that of *S1* Â [math]\frac{ \frac{4}{3} \pi \sqrt{ 3 \frac{x}{2}^2}^3 }{\frac{4}{3} \pi \frac{x}{2}^3}=5.2[/math] Â Edited September 10, 2023 by Sensei
Genady Posted September 10, 2023 Posted September 10, 2023 Spoiler Straightforward application of known formulas for volume of a cube and a sphere. Take a length of the cube edge to be 2a to simplify the algebra. Â
TheVat Posted September 10, 2023 Posted September 10, 2023 This may be a shortcut? Â Spoiler Just take diagonal of a unit cube, which is 1.73, and cube that, so 5.178 Â Â
Genady Posted September 10, 2023 Posted September 10, 2023 Let's make this a bit more interesting. Consider a cylindrical glass and four identical balls packed into it compactly, as shown: How much water is needed to exactly cover the balls compared to the volume occupied by the balls?
Sensei Posted September 10, 2023 Posted September 10, 2023 8 minutes ago, Genady said: How much water is needed to exactly cover the balls compared to the volume occupied by the balls? Spoiler The answer depends on the density of the balls.. 😛 If their density is less than 1g/mL of water, they will float on the surface and will only be partially submerged. Â
Genady Posted September 10, 2023 Posted September 10, 2023 4 minutes ago, Sensei said:  Reveal hidden contents The answer depends on the density of the balls.. 😛 If their density is less than 1g/mL of water, they will float on the surface and will only be partially submerged.  Spoiler Correct. The answer also depends on their chemical composition. Â
TheVat Posted September 10, 2023 Posted September 10, 2023 Spoiler let x=cylinder radius and x/2 is ball radius So cylinder v at given height = hπx2 then v ball is 4/3π(x/2)3 h=x/2 + x/2 + x/2(21/2) (simple Pythagorean geometry) So then plug in a number for x, like 10. Then h = 5+5+5(21/2)=17.07, ergo undisplaced volume is roughly 5363. balls volume is 4 x (4/3π53) = 4x523.6=2094 Then, take 5363 - 2094 = 3269 So....cylinder has 3269 units of submerging liquid with balls volume of 2094. (assuming balls denser than liquid) Roughly 1.56. (some rounding errors may have accrued)      1
Genady Posted September 10, 2023 Posted September 10, 2023 8 minutes ago, TheVat said:  Hide contents let x=cylinder radius and x/2 is ball radius So cylinder v at given height = hπx2 then v ball is 4/3π(x/2)3 h=x/2 + x/2 + x/2(21/2) (simple Pythagorean geometry) So then plug in a number for x, like 10. Then h = 5+5+5(21/2)=17.07, ergo undisplaced volume is roughly 5363. balls volume is 4 x (4/3π53) = 4x523.6=2094 Then, take 5363 - 2094 = 3269 So....cylinder has 3269 units of submerging liquid with balls volume of 2094. (assuming balls denser than liquid) Roughly 1.56. (some rounding errors may have accrued)      Yes! +1
Commander Posted September 11, 2023 Author Posted September 11, 2023 Yes TY all for answers ! Pi/6 is the Fraction and 3xSqr Rt 3 times is the Volume of the Bigger Sphere to that of the Smaller one ! Quite Straight forward !
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