Question

the internal and external diameters of a hollow hemispherical vessel are 24cm and 25cm respectively. I the cost of painting 1cm2units of the surface area is Rs0.05. find the total cost of painting the vessel all over

Answer 1

given,
internal diameter=24cm
internal radius=24/2=12cm
external diameter=25 cm
external radius =25/2=12.5cm
surface area of internal bowl=2πr²
=2(22/7)(12²)
=2×22×12×12/7
=905.14cm²
surface area of external bowl=2πR²
=2(22/7)(12.5²)
=2×22×12.5×12.5/7
=982.14cm²
surface area of ring=π(R²-r²)
=22/7(12.5²-12²)
=22/7(12.25)
=22×12.25/7
=38.5cm²
cost of internal bowl=905.14×0.05=Rs.45.25
cost of external bowl=982.14×0.05=Rs.49.1
cost of ring=38.5×0.05=Rs.1.925
cost of bowl=45.25+49.1+1.9=96.25 rupees approx



I THINK IT IS CORRECT ANSWER.

Answer 2

Answer:

Given,

internal diameter=24cm

internal radius=24/2=12cm

external diameter=25 cm

external radius =25/2=12.5cm

surface area of internal bowl=2πr²

=2(22/7)(12²)

=2×22×12×12/7

=905.14cm²

surface area of external bowl=2πR²

=2(22/7)(12.5²)

=2×22×12.5×12.5/7

=982.14cm²

surface area of ring=π(R²-r²)

=22/7(12.5²-12²)

=22/7(12.25)

=22×12.25/7

=38.5cm²

cost of internal bowl=905.14×0.05=Rs.45.25

cost of external bowl=982.14×0.05=Rs.49.1

cost of ring=38.5×0.05=Rs.1.925

cost of bowl=45.25+49.1+1.9=96.25 rupees approx

I THINK IT IS CORRECT ANSWER.