The internal and external diameters of a hollow hemispherical vessel are 16 𝑐𝑚 and 12 𝑐𝑚 respectively. If the cost of painting 1 cm² of the surface area is rupees 5. 00, find the total cost of painting the vessel all over. (Use π = 3. 14)

We are given the internal and external diameters of a hollow hemispherical bowl, and we have to find how much it will cost to paint the bowl's surface area.
While painting, the inner area, the outer area and the upper thickness of the bowl will be painted.
So, we have to find the curved surface area of the internal and external parts of the hemisphere and the upper part.
Here, the diameter of the external part, 𝐷 = 16 𝑐𝑚
So, the radius of the external part, 𝑅 = 𝐷/2 
⇒ 𝑅 = 16/2  
⇒ 𝑅 = 8 𝑐𝑚
And the diameter of the internal part, 𝑑 = 12 𝑐𝑚
So, the radius of the external part, 𝑟 = 𝑑/2
⇒ 𝑟 = 12/2
⇒ 𝑟 = 6 𝑐𝑚
The surface area of the hemispherical bowl,

A = $2{\mathrm{\pi R}}^2 + 2{\mathrm{\pi r}}^2 + \mathrm{\pi }({\mathrm{R}}^2 - {\mathrm{r}}^2)$                                                              
⇒ 𝐴 = 2 × 3. 14 × 8 × 8 + 2 × 3. 14 × 6 × 6 + 3. 14(8 × 8 − 6 × 6)
⇒ 𝐴 = 401. 92 + 226. 08 + 3. 14(64 − 36)
⇒ 𝐴 = 628 + 3. 14 × 28
⇒ 𝐴 = 628 + 87. 92
⇒ 𝐴 = 715. 94 𝑐𝑚²
The cost of painting the surface area per 𝑐𝑚2 is 5. 00 rupees. So, cost of painting 715. 94 𝑐𝑚2 = 715. 94 × 5
= 3579. 6
Hence, the cost of painting the hemispherical bowl is rupees. 3579. 6