Multiple choice

A hemispherical bowl is made from a metal sheet having thickness $0.3cm$. The inner radius of the bowl is $24.7cm$. Find the cost of polishing its outer surface at the rate of Rs.$4$ per $10{cm}^{2}$. (Take $\pi=3.14$)

  1. Rs.$1570$
  2. Rs.$137$
  3. Rs.$127$
  4. Rs.$107$
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A Correct answer
Explanation

The outer radius is inner radius plus thickness, which is 24.7 + 0.3 = 25 cm. The outer surface area of a hemisphere is 2 * pi * r^2 = 2 * 3.14 * 25 * 25 = 3925 cm^2. At a rate of Rs 4 per 10 cm^2, the cost is (3925 / 10) * 4 = 392.5 * 4 = 1570.

AI explanation

The outer radius of the bowl is the sum of the inner radius and the thickness, giving 24.7 plus 0.3, which equals 25 cm. The curved outer surface area of a hemisphere is two times pi times the radius squared, so we calculate 2 times 3.14 times 25 squared, resulting in 3925 square cm. At a polishing cost of Rs. 4 per 10 square cm, the rate is Rs. 0.4 per square cm. Multiplying the area 3925 by 0.4 gives the total cost of Rs. 1570.

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