Question

The inner radius of a hemispherical steel bowl is $5$ cm If the thickness of the bowl is $0.25$ m find the volume of steel used in making the bowl

• A. $30.319$ cm${}^3$
• B. $30319$ cm${}^3$
• C. $41.29$ cm${}^3$
• D. $3.0319$ $c{m}^3$

Answer 1

Answer: (C)

Inner radius $=5cm$
Volume of the inner hemisphere $=\frac{2}{3}\pi r^3=\frac{2}{3}\times \frac{22}{7}\times 5\times 5\times 5=\frac{5500}{21}=261.90c{m}^3$
Outer radius = Inner radius + Thickness of steel $=(5+0.25)cm=5.25$ cm
$\therefore$ Volume of the outer hemisphere = $\frac{2}{3}\pi r^3=\frac{2}{3}\times \frac{22}{7}\times {\left(5.25\right)}^3=\frac{4851}{16}=303.19c{m}^3$
$\therefore$ Volume of the steel used $=(303.19-261.90)$$c{m}^2=41.29c{m}^3$