Multiple choice

A hemispherical dome is open from its base and is made of iron. The thickness of the dome is 3.5 m. The total cost of painting the dome's outer curved surface is Rs. 2464. If the rate of painting is Rs. 8 per m2, then what is the volume (in m3) of iron used in making dome?

  1. 656.42

  2. 614.21

  3. 524.46

  4. 628.83
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D Correct answer
Explanation

Outer surface area = 2464 / 8 = 308 m^2. 2 * pi * R^2 = 308. R^2 = 308 / (2 * 22/7) = 49. R = 7 m. Inner radius r = 7 - 3.5 = 3.5 m. Volume = (2/3) * pi * (R^3 - r^3) = (2/3) * (22/7) * (343 - 42.875) = (44/21) * 300.125 = 628.83 m^3.

AI explanation

Given the total painting cost of 2464 rupees at a rate of 8 rupees per square metre, the outer curved surface area is 2464 / 8 = 308 square metres. Using the outer curved surface area formula for a hemisphere, 2 * pi * R^2 = 308, and substituting pi as 22/7, we find R^2 = 49, so the outer radius R is 7 metres. The inner radius is 7 - 3.5 = 3.5 metres, and the volume of the iron used is found using the formula for the volume of a hollow hemisphere, which is (2/3) * pi * (R^3 - r^3). Substituting the values gives (2/3) * (22/7) * (7^3 - 3.5^3) = (44/21) * (343 - 42.875) = 628.83 cubic metres. The volume of iron used is 628.83 m3.

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