Multiple choice

A soild trophy, consisting of two parts, has been designed in the following manner: the bottom part is a frustum of a cone with the bottom radius 30 cm, the top radius 20 cm, and height 40 cm, while the top part is a hemisphere with radius 20 cm. Moreover, the flat surface of the hemisphere is the same as the top surface of the frustum. If the entire trophy is to be gold-plated at the cost of Rs. 40 per square cm, what would the cost for gold plating be closest to?

  1. Rs. 1,12,000

  2. Rs. 3,60,000

  3. Rs. 4,73,000

  4. Rs. 5,23,000

  5. Rs. 3,72,000
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The surface area includes the lateral area of the frustum, the base of the frustum, and the curved surface of the hemisphere. Calculating these areas and multiplying by the cost per square cm yields approximately 473,000.

AI explanation

The total surface area to be plated is the sum of the base area of the frustum, the curved surface area of the frustum, and the curved surface area of the hemisphere. Using pi as 3.1416, the base area is pi times 30 squared which is 2827.44, the frustum curved surface area is pi times 40 times 30 plus 20 which is 6283.19, and the hemisphere curved surface area is 2 pi times 20 squared which is 2513.27. The total surface area is 11623.9, and multiplying by the rate of 40 rupees per square cm gives a total cost of 464956 rupees. The closest value is Rs. 4,73,000.

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