The inner and outer radii of a hemispherical wooden bowl are 6 cm and 8 cm, respectively. Its entire surface has to be polished and the cost of polishing π cm2 is ₹50. How much will it cost to polish the bowl?
Show answer & explanation
Correct answer: D
Explanation
The correct answer is Option 4 i.e. ₹11,400.
Given Inner radius = 6 cm, and Outer radius = 8 cm
We know that,
Outer curved surface area (CSA) = 2πR2
here, R = outer radius
Inner curved surface area (CSA) = 2πr2
here, r = inner radius
So, Base area = πR2 - πr2
Now,
Total Surface Area(TSA) = Outer CSA + Inner CSA + Base area
⇒ TSA = 2πR2 + 2πr2 + πR2 - πr2
⇒ TSA = 3πR2 + πr2
⇒ TSA = 3π(8)2 + π(6)2
⇒ TSA = 192π + 36π
⇒ TSA = 228π cm2
The cost of polishing is Rs. 50 per π cm2
Total cost = 228 × 50 = ₹11400