Question

A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use \(\pi = 3.14\))

• A. \(314 \sqrt{2}\) \(\text{cm}^{2}\)
• B. \(314\) \(\text{cm}^{2}\)
• C. \(\frac{3140}{3}\) \(\text{cm}^{2}\)
• D. \(3140 \sqrt{2}\) \(\text{cm}^{2}\)

Hint:

For any cone inscribed in a hemisphere to have maximum volume, its height must be equal to its base radius, which are both equal to the radius of the hemisphere. The slant height will always be \(r\sqrt{2}\).

Answer 1

The Correct Option is A

The problem involves finding the curved surface area of a conical cavity that is carved out of a wooden solid hemisphere. The cavity has to be of maximum volume.

Given:

• Radius of the hemisphere, \( R = 10 \, \text{cm} \)
• \(\pi = 3.14\)

To carve out a conical cavity of maximum volume, it's known that the height (\( h \)) and radius (\( r \)) of the cone must be equal, i.e., \( h = r \). This happens when the cone fits perfectly within the hemisphere.

Using the Pythagorean theorem in the semicircular cross-section of the cone and hemisphere:

\(R^2 = r^2 + h^2\)

Since \( h = r \), we have:

\(R^2 = 2r^2 \Rightarrow r^2 = \frac{R^2}{2} \Rightarrow r = \frac{R}{\sqrt{2}}\)

Now, calculate the curved surface area of the cone, which is given by the formula:

\(\pi r l\)

where \( l \) is the slant height of the cone. \( l \) can be calculated as:

\(l = \sqrt{r^2 + h^2} = \sqrt{2r^2} = \sqrt{2}r\)

Substituting \( r = \frac{R}{\sqrt{2}} \) into the formula:

\(l = \sqrt{2} \times \frac{R}{\sqrt{2}} = R\)

The curved surface area of the cone is:

\(\pi r l = \pi \left(\frac{R}{\sqrt{2}}\right) R\)

Substitute the values:

\(\pi r l = 3.14 \times \frac{10}{\sqrt{2}} \times 10\) \(= 3.14 \times 50\sqrt{2}\) \(= 314\sqrt{2} \, \text{cm}^2\)

Thus, the curved surface area of the cavity carved out is \( 314\sqrt{2} \, \text{cm}^2 \).