Surface Areas and Volumes — Question 12

Given,

The inner radius of the bowl (r) = 5 cm

Thickness of steel = 0.25 cm

Outer radius of the bowl (R) = inner radius of the bowl + thickness of steel = 5 cm + 0.25 cm = 5.25 cm

Outer curved surface area of the hemisphere = 2πR²

= 2 × $\dfrac{22}{7}$ × (5.25)²

= $\dfrac{44}{7} \times 27.5625$

= 44 × 3.9375

= 173.25 cm².

Hence, the outer curved surface area of the bowl is 173.25 cm².