A hemispherical bowl is made of 0.2cm thick steel. The inner diameter of the bowl is 8cm. Find the outer curved surface area of the bowl. Also find the cost of polishing its outer surface at the rate of rs2 per cm
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112
inner diameter =8cm
inner radius =4cm
thickness=0.2cm
outer radius =4cm=0.2cm=4.2cm
outer c.s.a =2πr^2
outer c.s.a =2×22×4.2×4.2/7
=110.88cm2
cost of polishing 1cm =Rs 2
cost of polishing 110.88cm2 =Rs 2×110.88cm2
=Rs221.76
inner radius =4cm
thickness=0.2cm
outer radius =4cm=0.2cm=4.2cm
outer c.s.a =2πr^2
outer c.s.a =2×22×4.2×4.2/7
=110.88cm2
cost of polishing 1cm =Rs 2
cost of polishing 110.88cm2 =Rs 2×110.88cm2
=Rs221.76
Answered by
9
Given:
The inner diameter = 8cm
The thickness of the steel = 0.2cm
To Find:
The outer curved surface area of the bowl
The cost of polishing its outer surface at the rate of Rs. 2 per cm²
Solution:
The outer curved surface area of the bowl is 110.88 cm² and the cost of polishing its outer surface is Rs. 221.76.
Given that the inner diameter of the bowl is 8cm
⇒ The inner radius = 8 / 2 = 4cm
The outer radius of the bowl = The inner radius + The thickness of the steel
= 4cm + 0.2cm
=4.2cm
We know that the outer curved surface area of a hemisphere =2πr²
(Here r is the outer radius of the hemisphere)
Taking π = 22/7,
The outer curved surface area = 2 X 22 X 4.2 X 4.2 / 7
= 110.88 cm²
Given that the cost of polishing 1cm² = Rs. 2
So using the unitary method, the cost of polishing 110.88cm² = Rs. 2 X 110.88cm².
= Rs. 221.76
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