Question

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A hemispherical bowl is made up of 0.2 cm thick steel. The inner diameter of the bowl is 8 cm. Find the outer curved surface area of the bowl. Also, find the cost of painting it's outer surface @ Rs 2 per cm^2.​

Answer 1

ANSWER:-----

Clean steps

Inner diameter=8cm

inner diameter=8/2=4cm

Outer radius=inner radius + thickness of bowl =4cm+0.2cm=4.2cm

Outer curved surface area of

bowl=2πr²

=2×22/7×4.2×4.2

=110.88cm²

Cost of polishing its outer surface at the rate of rs. 2 per cm³ =rs. 110.88×2

=rs. 221.76cm²

hope it helps:--

T!—!ANKS!!!

Answer 2

$\huge\bf\orange{ANSWER:-}$

Inner radius of the hemispherical bowl= ½×8 = 4 cm

Thickness of the steel = 0.2 cm

Therefore,

The outer radius of the bowl= (4+0.2)= 4.2cm

Outer Curved Surface Area of the bowl= 2πr²

$= 2 \times \frac{22}{7} \times (4.2) {}^{2}$

= 110.88 cm²

As the rate of polishing the surface is Rs 2 per cm²

Therefore,

The cost of polishing the outer surface of the bowl = Rs (2×110.88)

= Rs 221.76

Therefore,

The outer Curved Surface Area is 110.88cm² and the cost of polishing the surface area is Rs 221.76