A cylindrical road roller made of iron is 1 m long, Its internal diameter is 54 cm and the thickness of the iron sheet used in making the roller is 9 cm. Find the mass of the roller, if 1 cm³ of iron has 7.8 gm mass. (Use π = 3.14)
Answers
Answer:
The mass of the road roller is 1388.7 kg
Step-by-step explanation:
SOLUTION :
Given :
Height of the cylindrical road roller, h = 1 m = 100 cm
[1m = 100 cm]
Internal Diameter of the cylindrical road roller = 54 cm
Internal radius of the cylindrical road roller , r = 54/2 = 27 cm
Thickness of the road roller = 9 cm
Outer radius of the cylindrical road roller, R = inner radius + thickness
R = 27 + 9
R = 36 cm
Now,
Volume of the iron sheet = π(R² - r²)h
V = π(36² - 27²) × 100
V = π (1296 - 729) × 100
V = π× 567 × 100
V = 3.14 × 567 × 100
V = 1780.38 cm³
volume of the iron sheet , V = 1780.38 cm³
Mass of 1 cm³ of the iron sheet = 7.8 gm
Mass of 1780.38 cm³ of the iron sheet = 1780.38 × 7.8 = 1388696.4 gm = 1388696.4/1000 kg
[1 g = 1/1000 kg]
= 1388.69 = 1388.7 kg
Hence, the mass of the road roller is 1388.7 kg
HOPE THIS ANSWER WILL HELP YOU….
Let r and R be the inner and outer radius if the cylindrical road roller respectively.
R = Inner radius if the cylinder + Thickness of the iron sheet = 27 cm + 9 cm = 36 cm
Volume of iron in the road roller
= π (R2 – r2) h
Given, weight of 1 cm3 of iron = 7.8 gm
∴ Weight of 178200 cm3 of iron = 7.8 × 178200 gm = 1389960 gm = 1389.96 kg
Thus, the weight of the road roller is 1389.96 kg.