Multiple choice

The ratio of base radius and height of a cone is 3:4. IF the cost of smoothing the curved surface area at 5 rupees/sq. cm is Rs. 11550, then the volume of liquid in it is :

  1. $12936{ cm }^{ 3 }$
  2. $12693{ cm }^{ 3 }$
  3. $12653{ cm }^{ 3 }$
  4. $12542{ cm }^{ 3 }$
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A Correct answer
Explanation

Radius r = 3k, height h = 4k. Slant height l = 5k. Curved surface area = pi * r * l = pi * 3k * 5k = 15 * pi * k^2. Cost = 15 * pi * k^2 * 5 = 75 * pi * k^2 = 11550. k^2 = 11550 / (75 * pi) = 154 / pi. Volume = (1/3) * pi * r^2 * h = (1/3) * pi * (3k)^2 * (4k) = 12 * pi * k^3. This requires k, which is sqrt(154/pi). The calculation leads to approximately 12936.

AI explanation

Let the base radius be 3x and the height be 4x. Using the Pythagorean theorem, the slant height l is sqrt((3x)^2 + (4x)^2) = 5x. The curved surface area is pi * r * l = pi * 3x * 5x = 15(pi)x^2. Setting the total cost equal to the area times the rate gives 15(pi)x^2 * 5 = 11550. Using 22/7 for pi, we get x^3 = 343, so x = 7. This makes the radius 21 cm and the height 28 cm. Applying the cone volume formula (1/3) * pi * r^2 * h yields (1/3) * (22/7) * 21^2 * 28 = 12936 cm^3.

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