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] 4 hours ago, Commander said: (2) How many times the volume of *S2* is that of *S1* [math]\frac{ \frac{4}{3} \pi \sqrt{ 3 \frac{x}{2}^2}^3 }{\frac{4}{3} \pi \frac{x}{2}^3}=5.2[/math] 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.
TheVat Posted September 10, 2023 Posted September 10, 2023 This may be a shortcut? Spoiler Just take diagonal of a unit cube, which is 1.73, and cube that, so 5.178
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.
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.
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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