Multiple choice

A metal recycling plant receives an iron cannonball with a radius of 2.1 decimeters from an old battleship. The company decides to melt the cannonball and reshape it into a cylindrical iron support rod for a bridge. The rod is designed with a height of 28 cm, ensuring that no iron is wasted. Find the ratio of the total surface area of the cylindrical rod to that of the original cannonball.

  1. 6 : 5

  2. 7 : 4

  3. 7 : 5

  4. 7 : 6

  5. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Cannonball r = 2.1 dm = 21 cm. Volume = (4/3) * pi * 21^3. Surface area = 4 * pi * 21^2. Cylinder: Volume = pi * r_c^2 * 28 = (4/3) * pi * 21^3. r_c^2 = (4 * 9261) / (3 * 28) = 441. r_c = 21. Surface area = 2 * pi * r_c * (r_c + h) = 2 * pi * 21 * (21 + 28) = 2 * pi * 21 * 49. Ratio = (2 * pi * 21 * 49) / (4 * pi * 21^2) = (2 * 49) / (4 * 21) = 98 / 84 = 7/6.

AI explanation

The cannonball radius is 21 cm, so its volume is (4/3) * pi * 21^3 = 12348 * pi cubic cm. Equating this to the cylinder volume pi * r^2 * 28 gives a cylinder radius of 21 cm. The cannonball surface area is 4 * pi * 21^2 = 1764 * pi, while the cylinder total surface area including both ends is 2 * pi * 21 * 28 + 2 * pi * 21^2 = 2058 * pi. The ratio of the cylinder surface area to the cannonball surface area is 2058 * pi to 1764 * pi, which simplifies to 7 to 6.

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