Calculating the Surface Area Ratio of a Hollowed-Out Ball

This problem asks us to find the ratio between the surface area of a sphere after a cylindrical hole has been drilled through its center and the original surface area of the sphere. We are given the radius of the ball and the radius of the cylindrical hole.

Original Ball Surface Area Calculation

First, let's calculate the surface area of the original, unmodified ball (sphere). The radius of the ball is given as \(R = 10\) cm. The formula for the surface area of a sphere is:

\(A_{orig} = 4 \pi R^2\)

Substituting the given radius:

\(A_{orig} = 4 \pi (10 \text{ cm})^2 = 4 \pi (100 \text{ cm}^2) = 400 \pi \text{ cm}^2\)

So, the original surface area is \(400 \pi\) square centimeters.

Hollowed-Out Ball Surface Area Calculation

When a cylindrical hole is drilled through the center of the ball, the total surface area changes. The new surface area includes:

• The outer surface area of the ball, excluding the two circular areas where the drill enters and exits.
• The inner surface area created by the cylindrical hole.

Let's define:

• Sphere Radius: \(R = 10\) cm
• Cylinder Radius (hole radius): \(r = 1\) cm

Calculating the Height of the Cylindrical Hole

The cylinder passes through the center of the sphere. We can visualize a right-angled triangle formed by:

• The sphere's radius (\(R\)) as the hypotenuse.
• The hole's radius (\(r\)) as one leg.
• Half the height of the cylinder (\(h/2\)) as the other leg.

Using the Pythagorean theorem (\(R^2 = r^2 + (h/2)^2\)):

\(10^2 = 1^2 + (h/2)^2\)

\(100 = 1 + (h/2)^2\)

\((h/2)^2 = 100 - 1 = 99\)

\(h/2 = \sqrt{99} = \sqrt{9 \times 11} = 3\sqrt{11}\) cm

Therefore, the full height of the cylindrical hole is:

\(h = 2 \times 3\sqrt{11} = 6\sqrt{11}\) cm

Calculating the New Surface Area (\(A_{new}\))

The total surface area of the hollowed-out ball (\(A_{new}\)) is the sum of the remaining outer curved surface area and the inner cylindrical surface area.

The area of the two circular openings removed from the sphere's surface is \(2 \times \pi r^2\). The lateral surface area of the cylinder is \(A_{cyl\_lateral} = 2 \pi r h\). So, the new total surface area is:

\(A_{new} = (\text{Original Surface Area} - \text{Area of 2 circular holes}) + (\text{Inner Cylindrical Surface Area})\)

\(A_{new} = (4 \pi R^2 - 2 \pi r^2) + (2 \pi r h)\)

Substituting the values:

\(A_{new} = (4 \pi (10)^2 - 2 \pi (1)^2) + (2 \pi (1) (6\sqrt{11}))\)

\(A_{new} = (400 \pi - 2 \pi) + 12 \pi \sqrt{11}\)

\(A_{new} = 398 \pi + 12 \pi \sqrt{11} \text{ cm}^2\)

Surface Area Ratio Calculation

We need to find the ratio of the new surface area (\(A_{new}\)) to the original surface area (\(A_{orig}\)).

Ratio \(= \frac{A_{new}}{A_{orig}} = \frac{398 \pi + 12 \pi \sqrt{11}}{400 \pi}\)

Factor out \(\pi\) from the numerator and cancel it with the denominator:

Ratio \(= \frac{\pi (398 + 12 \sqrt{11})}{400 \pi} = \frac{398 + 12 \sqrt{11}}{400}\)

Simplify the fraction by dividing the numerator and denominator by 2:

Ratio \(= \frac{199 + 6 \sqrt{11}}{200}\)

Comparing with Options

To find the closest value, let's approximate \(\sqrt{11}\). We know \(\sqrt{9}=3\) and \(\sqrt{16}=4\), so \(\sqrt{11}\) is between 3 and 4. A closer approximation is \(\sqrt{11} \approx 3.317\).

Now substitute this value back into the ratio:

Ratio \(\approx \frac{199 + 6(3.317)}{200}\)

Ratio \(\approx \frac{199 + 19.902}{200}\)

Ratio \(\approx \frac{218.902}{200} \approx 1.09451\)

Let's check the given options:

• Option 1: \(\frac{1}{10} = 0.1\)
• Option 2: \(\frac{11}{5} = 2.2\)
• Option 3: \(\frac{219}{200} = 1.095\)
• Option 4: \(\frac{221}{200} = 1.105\)

Our calculated ratio, approximately \(1.09451\), is closest to \(\frac{219}{200}\) (\(1.095\)).

Conclusion

The calculation shows that the ratio of the total surface area of the hollowed-out ball to the surface area of the original ball is \(\frac{199 + 6 \sqrt{11}}{200}\), which is approximately \(1.09451\). This value is closest to \(\frac{219}{200}\).