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$$\frac {50} \pi$$​ per $$cm^2$$​ is Rs.50. How much will it cost to polish the bowl?

Correct option is D

Given:
Radius of inner surface of hemisphere $$r_2$$​=6cm.
Radius of outer surface of hemisphere $$r_1$$​  =8cm.
Cost of polishing = Rs. $$\frac{50}{π} cm^2$$​​
Formula Used:

Area of concentric circles= $$=π[r_2^2 -{r_1}^2]$$​​
Surface Area of hemisphere = $$2πr^2$$​​
Total Cost = Price $$\times$$​Area
Solution:
Hemispherical Surface to be painted = $$2 \times π\times 6^2 + 2 \timesπ \times 8^2 = 2 \timesπ \times (6^2+ 8^2) = 200π (cm)^2$$​​
Area of boundary = $$π(8^2 -6^2) =28π cm^2$$​​
Total Cost =  $$\frac{50}{π} \times$$​ (200π + 28π) = Rs 11400