Question

The internal and external diameters of a hollow hemispherical vessel are 21 cm and 25.2 cm respectively. The cost of painting 1 $c{m}^2$ of the surface is 10 paise.Find the total cost to paint the vessel all over.

Answer 1

Given,
internal diameter=21cm
internal radius= $\frac{21}{2}$ =10.5cm
external diameter=25.2cm
external radius=$\frac{25.2}{2}$ =12.6cm

surface area of internal bowl=2πr²
=2 × $\frac{22}{7}$×(10.5²)
= $\frac{2\times 22\times 10.5\times 10.5}{7}$
=693cm²
surface area of external bowl=2πR²
=2 ×$\frac{22}{7}$ × (12.6²)
= $\frac{2\times 22\times 12.6\times 12.6}{7}$
=997.9cm²
Surface area of ring=π(R²-r²)
=$\frac{22}{7}$(12.6²-10.5²)
=$\frac{22}{7}$(48.51)
= $\frac{22\times 48.51}{7}$
=152.46cm²
Cost of internal bowl=693×0.1= Rs. 69.3
Cost of external bowl=997.9×0.1= Rs. 99.79
Cost of ring=152.46×0.1= Rs.15.246
Cost of bowl$=69.3+99.7+15.2=Rs.184.2$