To determine the volume occupied by 350 $10 \, \text{mm}$ balls in a $4 \, \text{inch}$ tube, follow these steps:

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Step 1: Convert the dimensions

1. Diameter of each ball: $10 \, \text{mm} = 1 \, \text{cm}$
   Radius ($r$) = $\frac{1}{2} \, \text{cm} = 0.5 \, \text{cm}$.

2. Tube diameter: $4 \, \text{inches}$
   Convert to centimeters: $4 \, \text{inches} = 4 \times 2.54 = 10.16 \, \text{cm}$.

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Step 2: Volume of a single ball

The volume of a sphere is given by: $V = \frac{4}{3} \pi r^3$ Substitute $r = 0.5 \, \text{cm}$: $V = \frac{4}{3} \pi (0.5)^3 = \frac{4}{3} \pi (0.125) = \frac{\pi}{6} \, \text{cm}^3$ Numerical value: $V \approx 0.5236 \, \text{cm}^3$

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Step 3: Volume of 350 balls

Multiply the volume of one ball by 350: $V_{\text{total}} = 350 \times 0.5236 \approx 183.26 \, \text{cm}^3$

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Step 4: Check if it fits in the tube

We need to compare the total ball volume with the volume of the tube.

1. Tube dimensions: Diameter = $10.16 \, \text{cm}$, so radius $R = 5.08 \, \text{cm}$.
   Assume the tube is completely filled along its length, and we only need to check if the balls can occupy a given length.

   The cross-sectional area of the tube: $A_{\text{tube}} = \pi R^2 = \pi (5.08)^2 \approx 81.03 \, \text{cm}^2$ The height ($h$) required for the balls to fit: $h = \frac{V_{\text{total}}}{A_{\text{tube}}} = \frac{183.26}{81.03} \approx 2.26 \, \text{cm}$

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Final Answer

The 350 $10 \, \text{mm}$ balls occupy a height of approximately $2.26 \, \text{cm}$ in the $4 \, \text{inch}$ tube.

Let me know if you would like further clarification or a diagram for visualization!

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Related Questions:

1. What is the maximum number of balls that can fit in a $4 \, \text{inch}$ tube?
2. How does packing density affect the volume occupied by the balls in the tube?
3. What would happen if the balls were $12 \, \text{mm}$ instead of $10 \, \text{mm}$?
4. How does the shape of the container affect the arrangement of the spheres?
5. What is the surface area of all 350 balls combined?

Tip:

Always double-check unit conversions when solving problems that involve multiple measurement systems!