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 \(cm^3\)) is :

• A. 182
• B. 1823
• C. 6823
• D. 3643

Hint:

Always double-check whether the given radius is the inner radius or the outer radius.
For hollow objects, thickness is subtracted from the outer radius to find the inner radius.
Keep \(\pi\) as a symbol throughout the calculation since the options are expressed in terms of \(\pi\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This problem asks for the volume of material (steel) used to make a hollow hemispherical bowl.
We are given the outer radius and the thickness of the steel.

Step 2: Key Formula or Approach:
Let \(R\) be the outer radius of the hemispherical bowl and \(r\) be the inner radius.
The inner radius is calculated by subtracting the thickness (\(t\)) from the outer radius:
\[ r = R - t \]
The volume of a solid hemisphere of radius \(x\) is given by:
\[ V = \frac{2}{3}\pi x^3 \]
The volume of steel used is the difference between the outer volume and the inner volume:
\[ V_{\text{steel}} = \frac{2}{3}\pi R^3 - \frac{2}{3}\pi r^3 = \frac{2}{3}\pi (R^3 - r^3) \]

Step 3: Detailed Explanation:
1. Given, the outer radius of the bowl, \(R = 6\text{ cm}\).
2. The thickness of the steel, \(t = 1\text{ cm}\).
3. Calculate the inner radius, \(r\):
\[ r = R - t = 6 - 1 = 5\text{ cm} \]
4. Write down the formula for the volume of the hollow hemispherical shell:
\[ V_{\text{steel}} = \frac{2}{3}\pi (R^3 - r^3) \]
5. Substitute the values of \(R\) and \(r\) into the formula:
\[ V_{\text{steel}} = \frac{2}{3}\pi (6^3 - 5^3) \]
6. Calculate the cubes of the radii:
\[ 6^3 = 216 \]
\[ 5^3 = 125 \]
7. Find the difference:
\[ 216 - 125 = 91 \]
8. Multiply by the constant fraction:
\[ V_{\text{steel}} = \frac{2}{3}\pi (91) = \frac{182}{3}\pi\text{ cm}^3 \]

Step 4: Final Answer:
Hence, the correct option is (B).