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

To find the volume of steel used in the hemispherical bowl, we need to determine the difference between the volume of the outer hemisphere and the inner hemisphere.

1. \(Outer\ radius\ (R) = 6\ cm\).
2. \(Thickness\ of\ the\ bowl = 1\ cm\). Therefore, the inner radius \((r)\) of the hemisphere is: 
   \(r = R - 1 = 6\ cm - 1\ cm = 5\ cm\).
3. The volume \((V)\) of a hemisphere is given by the formula: \(V = \frac{2}{3}\pi r^3\).
4. Calculate the volume of the outer hemisphere with radius \(6\ cm\): 
   \(V_{outer} = \frac{2}{3}\pi (6)^3 = \frac{2}{3}\pi \times 216 = \frac{432}{3}\pi\ cm^3 = 144\pi\ cm^3\).
5. Calculate the volume of the inner hemisphere with radius \(5\ cm\): 
   \(V_{inner} = \frac{2}{3}\pi (5)^3 = \frac{2}{3}\pi \times 125 = \frac{250}{3}\pi\ cm^3\).
6. The volume of steel used is the difference between the volumes of the outer and inner hemispheres: 
   \(V_{steel} = V_{outer} - V_{inner} = 144\pi\ cm^3 - \frac{250}{3}\pi\ cm^3\). 
   Simplifying: \(V_{steel} = \left( \frac{432}{3} - \frac{250}{3} \right)\pi\ cm^3 = \frac{182}{3}\pi\ cm^3\).

Therefore, the volume of steel used in the bowl is \(\frac{182}{3}\pi\ \text{cm}^3\).