Question

The internal and the external diameter of a hollow hemispherical vessel are 21cm and 25.2cm respectively. The cost of painting of 1cm of the surface is 10paisa. Find the total cost to paint the vessel all over.

Answer 1

given,
internal diameter=21cm
internal radius=21/2=10.5cm
external diameter=25.2cm
external radius=25.2/2 =12.6cm
surface area of internal bowl=2πr²
=2(22/7)(10.5²)
=2×22×10.5×10.5/7
=693cm²
surface area of external bowl=2πR²
=2(22/7)(12.6²)
=2×22×12.6×12.6/7
=997.9cm²
surface area of ring=π(R²-r²)
=22/7(12.6²-10.5²)
=22/7(48.51)
=22×48.51/7
=152.46cm²
cost of internal bowl=693×0.1=69.3
cost of external bowl=997.9×0.1=99.79
cost of ring=152.46×0.1=15.246
cost of bowl=69.3+99.7+15.2=184.2




MAYBE IT IS CORRECT

Answer 2

Answer:

Step-by-step explanation:

given,

internal diameter=21cm

internal radius=21/2=10.5cm

external diameter=25.2cm

external radius=25.2/2 =12.6cm

surface area of internal bowl=2πr²

=2(22/7)(10.5²)

=2×22×10.5×10.5/7

=693cm²

surface area of external bowl=2πR²

=2(22/7)(12.6²)

=2×22×12.6×12.6/7

=997.9cm²

surface area of ring=π(R²-r²)

=22/7(12.6²-10.5²)

=22/7(48.51)

=22×48.51/7

=152.46cm²

cost of internal bowl=693×0.1=69.3

cost of external bowl=997.9×0.1=99.79

cost of ring=152.46×0.1=15.246

cost of bowl=69.3+99.7+15.2=184.2