A cylindrical road roller is 1 m long. Its inner diameter is 54 cm and thickness of the iron sheet rolled into the road roller is 9 cm. Find the weight of the roller, if 1 cm^3 iron weighs 8 gm (pi = 3.14).
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⬇SOLUTION⬇
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The length of the road roller = 1 m = 100 cm.
Therefore, height of the cylindrical roller = 100 cm
Inner radius of the cylinder (roller) = r = 54/2 = 27 cm
Thickness of the iron sheet is 9 cm, so outer radius of the cylinder,
R = inner radius + thickness of sheet = (27 + 9) cm = 36 cm
Therefore, volume of the iron sheet = (pi × R^2 × h - pi × r^2 × h) cm^3 = pi × (R^2 - r^2) × h cm^3
= 3.14 × [(36)^2 - (27)^2] × 100 cm^3
= 3.14 × 100 [(36 + 27) (36 - 27)] cm^3
= 3.14 × 63 × 9 × 100 cm^3 = 178038 cm^3
Therefore, weight of the roller = volume × density = 178038 × 8/1000 kg = 1424.304 kg
__________________________
Hope it helps you out ^_^
⬇SOLUTION⬇
__________________________
The length of the road roller = 1 m = 100 cm.
Therefore, height of the cylindrical roller = 100 cm
Inner radius of the cylinder (roller) = r = 54/2 = 27 cm
Thickness of the iron sheet is 9 cm, so outer radius of the cylinder,
R = inner radius + thickness of sheet = (27 + 9) cm = 36 cm
Therefore, volume of the iron sheet = (pi × R^2 × h - pi × r^2 × h) cm^3 = pi × (R^2 - r^2) × h cm^3
= 3.14 × [(36)^2 - (27)^2] × 100 cm^3
= 3.14 × 100 [(36 + 27) (36 - 27)] cm^3
= 3.14 × 63 × 9 × 100 cm^3 = 178038 cm^3
Therefore, weight of the roller = volume × density = 178038 × 8/1000 kg = 1424.304 kg
__________________________
Hope it helps you out ^_^
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