Question

A hemisphere bowl is made of steel of $0.25$ cm thickness. The inner radius of the bowl is $5$ cm. The volume of steel used is __________.$(\pi =3.141$)

• A. $42.15{\text{cm}}^3$
• B. $41.52{\text{cm}}^3$
• C. $41.24{\text{cm}}^3$
• D. $40{\text{cm}}^3$

Answer 1

Answer: (C)

Inner radius of bowl $=r=5\text{cm}$

Thickness of bowl $=0.2\text{cm}$

$\therefore$ Outer radius of bowl $=R=5+0.25=5.25$

Volume of steel used $=\frac{2}{3}\pi R^3-\frac{2}{3}\pi r^3$

$=\frac{2}{3}\times 3.141\times ({5.25}^3-5^3)$

$=41.24{\text{cm}}^3$