A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherial iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?

Answer

To find the radius of the spherical iron ball that raises the water level in the cylindrical tub, we can follow these steps:

### Step 1: Understand the problem
We have a cylindrical tub with a radius of 12 cm and an initial water depth of 20 cm. When a spherical ball is dropped into the tub, the water level rises by 6.75 cm. We need to find the radius of the ball.

### Step 2: Calculate the volume of water displaced
The volume of water displaced by the spherical ball is equal to the volume of the cylinder formed by the rise in water level. The formula for the volume of a cylinder is:
\[ V = \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height.

Here, the radius \( r \) of the cylindrical tub is 12 cm, and the height \( h \) of the water raised is 6.75 cm. Therefore, the volume of water displaced is:
\[ V = \pi (12^2) (6.75) \]

### Step 3: Calculate the numerical value
Now, we can calculate the volume:
\[ V = \pi (144) (6.75) \]
\[ V = 972\pi \, \text{cm}^3 \]

### Step 4: Relate the volume to the sphere
The volume of the sphere can be calculated using the formula:
\[ V = \frac{4}{3} \pi R^3 \]
where \( R \) is the radius of the sphere. Since the volume of water displaced is equal to the volume of the sphere, we can set them equal:
\[ 972\pi = \frac{4}{3} \pi R^3 \]

### Step 5: Cancel out \( \pi \) and solve for \( R^3 \)
Cancelling \( \pi \) from both sides gives:
\[ 972 = \frac{4}{3} R^3 \]
To eliminate the fraction, multiply both sides by \( \frac{3}{4} \):
\[ R^3 = 972 \times \frac{3}{4} \]
\[ R^3 = 729 \]

### Step 6: Find \( R \)
Now, take the cube root of both sides to find \( R \):
\[ R = \sqrt[3]{729} \]
\[ R = 9 \, \text{cm} \]

### Conclusion
The radius of the spherical iron ball is **9 cm**.

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