To calculate the radius of the spherical ball based on the rise in water level, we need to equate the volume of the displaced water to the volume of the spherical ball. Here's how we can calculate it:

Given:

• Radius of the cylinder $R = 10 \, \text{cm}$
• Rise in water level $h = 4 \, \text{cm}$

Step 1: Volume of water displaced

The volume of water displaced is equal to the volume of the cylinder that corresponds to the rise in water level. This is given by the formula for the volume of a cylinder:

$V_{\text{water}} = \pi R^2 h$

Substituting the known values:

$V_{\text{water}} = \pi (10 \, \text{cm})^2 (4 \, \text{cm}) = 400 \pi \, \text{cm}^3$

Step 2: Volume of the spherical ball

The volume of the spherical ball is given by the formula:

$V_{\text{ball}} = \frac{4}{3} \pi r^3$

Where $r$ is the radius of the ball. Since the ball displaces an equal amount of water, we set the volume of the ball equal to the volume of the water displaced:

$\frac{4}{3} \pi r^3 = 400 \pi$

Step 3: Solve for the radius $r$

Cancel out $\pi$ on both sides of the equation:

$\frac{4}{3} r^3 = 400$

Multiply both sides by $\frac{3}{4}$:

$r^3 = \frac{3}{4} \times 400 = 300$

Now, take the cube root of both sides:

$r = \sqrt[3]{300}$

Using the cube root:

$r \approx 6.67 \, \text{cm}$

Final Answer:

The radius of the spherical ball is approximately 6.67 cm.

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Would you like more details or have any further questions?

Here are 5 related questions to expand your understanding:

1. How do you calculate the volume of a cylinder with different dimensions?
2. What happens if the cylinder is not perfectly vertical?
3. How does the formula for a sphere's volume differ from other geometric shapes?
4. If the water level rises by 8 cm instead of 4 cm, how would the radius change?
5. How does the density of the ball affect the water displacement?

Tip: Whenever working with volumes in geometry, remember that displacement principles allow you to calculate complex shapes through simpler measurements, like water levels!