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 calculate the inner radius first by subtracting the thickness from the outer radius.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Volume of material in a hollow hemisphere = Outer Volume - Inner Volume.
Step 2: Detailed Explanation:
Outer Radius \( R = 6 \text{ cm} \).
Thickness = 1 cm.
Inner Radius \( r = 6 - 1 = 5 \text{ cm} \).
Volume = \( \frac{2}{3} \pi R^3 - \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (6^3 - 5^3) \).
\[ V = \frac{2}{3} \pi (216 - 125) = \frac{2}{3} \pi (91) = \frac{182}{3} \pi \]
Step 3: Final Answer:
The volume is \( \frac{182}{3} \pi \text{ cm}^3 \).