Math, asked by BrainlyHelper, 1 year ago

The internal and external diameters of a hollow hemispherical vessel are 21 cm and 25.2 cm respectively. The cost of painting 1 cm² of the surface is 10 paise. Find the total cost to paint the vessel all over.

Answers

Answered by Anonymous
27

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

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Answered by nikitasingh79
40

Answer:

The total cost to paint the vessel all over is ₹ 184.34.

Step-by-step explanation:

Given :  

Internal diameter of the hollow hemispherical vessel = 21 cm

Internal radius of the hollow hemispherical vessel , r = 21/2 = 10.5 cm

External diameter of the hollow hemispherical vessel = 25.2 cm

External radius of the hollow hemispherical vessel , R = 25.2/2 =12.6 cm

Total area of the hollow hemispherical vessel to paint = Curved surface area of outer hemisphere + Curved surface area of inner  hemisphere + area of base  of hollow hemisphere

= 2πR² + 2πr² + π(R²− r²)

= 2π(R² + r²) + π(R²− r²)

= π [2(R² + r²) + (R²− r²)]

= π [2(12.6)² + (10.5)² + (12.6² −10.5²)

= π[2 × (158.76 + 110.25) + (158.76 - 110.25

= π [2 × 269.01 + 48.51]

= π [ 538.02 + 48.51]  

= π × 586.53

= 22/7 × 586.53

= 83.79 × 22  

Total area of the hollow hemispherical vessel to paint = 1843.38 cm²

Given : The cost of painting of 1 cm ² of the surface is = 10 p = 10/100 = ₹ 0.1       [1 p = 1/100 ₹]

Total cost of painting = 1848.38 ×  0.10 = ₹ 184.34

Hence, The total cost of painting all over the vessel is ₹ 184.34

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