Saved Bookmarks
| 1. |
(b) The internal and external diameters of a hollhemispherical vessel are 24 cm and 25 cm respetively. If the cost of painting 1 cm2 of the surfacearea is R 0.05, find the total cost of painting th |
|
Answer» Given,internal diameter=24cminternal radius=24/2=12cmexternal diameter=25 cmexternal radius =25/2=12.5cmsurface area of internal bowl=2πr²=2(22/7)(12²)=2×22×12×12/7=905.14cm²surface area of external bowl=2πR²=2(22/7)(12.5²)=2×22×12.5×12.5/7=982.14cm²surface area of ring=π(R²-r²)=22/7(12.5²-12²)=22/7(12.25)=22×12.25/7=38.5cm²cost of internal bowl=905.14×0.05=Rs.45.25cost of external bowl=982.14×0.05=Rs.49.1cost of ring=38.5×0.05=Rs.1.925cost of bowl=45.25+49.1+1.9=96.25 rupees approxplease like the solution 👍 ✔️👍✔️👍 |
|