In this problem use π = 3.14. Last time this problem was asked in cbse 2015 board exam. Question number 29 [ii] from RS Aggarwal book page number 289, chapter volume and surface area of solid, exercise 17A

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# A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of 5 per 100 sq cm.

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Given data,

Side of cubical block = 10 cm

∴ Largest diameter = 10 cm and r = 5 cm

Total Area = 6a² – πr² + 2πr²

= (6 * [10]²) – ( [3.14] [5]² ) + ( 2 [3.14] [5]² )

= 678.5 cm²

∴ Total surface area of solid = 678.5 cm²

Now,

cost = 5 per 100 sq. cm

∴ 5/100

Per sq. cm = 5/100

∴ 678.5 * [5/100]

= 33.925

∴ Total cost is 33.925 [ approximately 34 ]