A hemispherical bowl is made of steel whose thickness is 2 cm. The inside radius of the bowl is 3 cm. Find the volume (in cm³) of the steel used in making the bowl. (Rounded off to 2 decimal places) (Take π = 22/7)
205.33
Let R be the outer radius and r be the inner radius of the hemispherical bowl. Given, thickness = 2 cm and inner radius r = 3 cm. Therefore, outer radius R = r + thickness = 3 + 2 = 5 cm. Volume of the steel used = Volume of outer hemisphere - Volume of inner hemisphere = (2/3)πR³ - (2/3)πr³ = (2/3)π(R³ - r³) = (2/3) * (22/7) * (5³ - 3³) = (2/3) * (22/7) * (125 - 27) = (2/3) * (22/7) * 98 = (2/3) * 22 * 14 = (44/3) * 14 = 616/3 = 205.333... Rounded off to 2 decimal places, the volume is 205.33 cm³.
Important Questions from Simple and Compound Both
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