Question

A hemispherical bowl is made of steel of thickness 1 cm. The outer radius of the bowl is 6 cm. The volume of steel used (in \(\text{cm}^3\)) is :

• A. \(182 \pi\)
• B. \(\frac{182}{3} \pi\)
• C. \(\frac{682}{3} \pi\)
• D. \(\frac{364}{3} \pi\)

Hint:

Always read carefully to see if the problem gives you the inner radius or the outer radius, and whether it mentions "radius" or "diameter".
Writing down \(R = 6\) and \(r = 5\) first avoids simple subtraction errors under exam conditions.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a hemispherical bowl with a steel thickness of 1 cm. The outer radius of this bowl is 6 cm. We need to determine the volume of steel used to construct the bowl.

Step 2: Key Formula or Approach:
1. The volume of steel used is the difference between the outer volume and the inner volume of the hemispherical bowl:
\[ V_{\text{steel}} = V_{\text{outer}} - V_{\text{inner}} \]
2. The formula for the volume of a hemisphere of radius \(r\) is:
\[ V = \frac{2}{3}\pi r^3 \]
Thus, the volume of a hollow hemisphere is:
\[ V = \frac{2}{3}\pi (R^3 - r^3) \]
where \(R\) is the outer radius and \(r\) is the inner radius.

Step 3: Detailed Explanation:
1. Identify the given parameters:
- Outer radius, \(R = 6\ \text{cm}\)
- Thickness of the steel, \(t = 1\ \text{cm}\)
2. Calculate the inner radius \(r\):
The inner radius is the outer radius minus the thickness:
\[ r = R - t = 6 - 1 = 5\ \text{cm} \]
3. Set up the volume calculation:
\[ V = \frac{2}{3}\pi (R^3 - r^3) \]
\[ V = \frac{2}{3}\pi (6^3 - 5^3) \]
4. Calculate the cubic values:
\[ 6^3 = 216 \]
\[ 5^3 = 125 \]
5. Subtract the volumes:
\[ 6^3 - 5^3 = 216 - 125 = 91 \]
6. Multiply by the coefficients:
\[ V = \frac{2}{3}\pi \times 91 = \frac{182}{3}\pi\ \text{cm}^3 \]
This matches option (B).

Step 4: Final Answer:
The volume of steel used is \(\frac{182}{3} \pi\ \text{cm}^3\), which corresponds to option (B).