Question

A solid brass sphere of radius 2.1 dm is converted into a right circular rod to length 7cm . the ratio of total surface areas of the rod to the sphere is

Answer 1

Solution :-

A sphere is converted (I think the sphere is melted and recast) into a right circular cylindrical rod of length 7 cm. The volume of the sphere and the cylindrical rod will be equal.

Volume of sphere = 4/3πr³

Radius of sphere = 2.1 dm or 21 cm

⇒ 4/3*22/7*21*21*21

Volume of sphere = 38808 cm³

Volume of sphere = Volume of cylindrical rod

⇒ 38808 = πr²h

⇒ 38808 = 22/7*r²*7

⇒ 22r² = 38808

⇒ r² = 38808/22

⇒ r² = 1764

⇒ r = √1764

⇒ r = 42 cm

So, radius of the cylindrical rod is 42 cm.


Total surface area of the sphere = 4πr²

⇒ 4*22/7*21*21

⇒ 5544 cm²

Total surface area of sphere is 5544 cm²

Total surface area of cylindrical rod = 2πr(r + h)

⇒ 2*22/7*42*(42 + 7)

⇒ 12936 cm²

So, total surface area of cylindrical rod is 12936 cm²

Now, Ratio of the total surface areas of cylindrical rod to the sphere

= 12936 : 5544

12936 = 2*2*2*3*7*7*11

5544 = 2*2*2*3*3*7*11

Common Factors = 2*2*2*3*7*11

HCF = 1848

HCF of 12936 and 5544 is 1848

⇒ 12936/1848 : 5544/1848

⇒ 7 : 3

So, the ratio of total surface areas of the cylindrical rod to the sphere is 7 : 3

Answer.