Question : A cylindrical roller made of iron is 1.2 m long, Its internal radius is 24 cm, and the thickness of the iron sheet used in making the roller is 15 cm. What is the mass (in kg) of the roller, if $1 ~cm^3$ of iron has 8g mass?
Option 1: 846.72 $\pi$
Option 2: 845.75 $\pi$
Option 3: 892.8 $\pi$
Option 4: 907.2 $\pi$
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Correct Answer: 907.2 $\pi$
Solution : Length of the roller $(h)$ = 1.2 m = 120 cm The internal radius of the cylinder $(r) $= 24 cm The thickness of the cylinder is 15 cm External radius of the cylinder $(R)$ = (24 + 15) = 39 cm The volume of a hollow right circular cylinder = $\pi (R^2 – r^2)\times h$ The volume of the hollow cylinder = $\pi (39^2- 24^2) × 120$ = $113400\pi$ cm 3 Mass of 1 cm 3 iron = 8 g Mass of $113400\pi$ cm 3 iron = $113400\pi × 8\ g$ = $907200\pi\ g$ = $907.2\pi\ kg$ [∵ 1 kg = 1000 g] Hence, the correct answer is 907.2$\pi$
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