Subject

Mathematics

Class

CBSE Class 10

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 Multiple Choice QuestionsLong Answer Type

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31. From each end of a solid metal cylinder, metal was scooped out in the hemispherical form of the same diameter. The height of the cylinder is 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into a cylindrical wire of 1.4 cm thickness. Find the length of the wire. ( π = 22/7)


Height of the solid metal cylinder, h = 10 cm
Radius of the solid metal cylinder, r = 4.2 cm

therefore,

Radius of each hemisphere = radius of the solid metal cylinder, r = 4.2 cm

Now, 

Volume of the rest of the cylinder = Volume of cylinder - 2 x volume of each hemisphere

 space equals space πr squared straight h space minus space 2 space straight x space 2 over 3 space πr cubed
space equals space πr squared space open parentheses straight h space minus 4 over 3 straight r close parentheses
equals straight pi space straight x space left parenthesis 4.2 space right parenthesis squared space open parentheses 10 space minus space 4 over 3 space straight x space 4.2 close parentheses
space equals straight pi space straight x space left parenthesis 4.2 right parenthesis squared space straight x space left parenthesis 4.4 right parenthesis space cm cubed

Thickness of the cylindrical wire = 1.4 cm
Therefore, 
Radius of the cylindrical wire, R = 1.4/2 = 0.7 cm

Let the length of the wire be H cm.

It is given that the rest of the cylinder is melted and converted into a cylindrical wire.

therefore,

Volume of the cylindrical wire = Volume of the rest of the cylinder

⇒ π x 0.7 x 0.7 x H = π x (4.2)2 x (4.4)

rightwards double arrow space straight H space equals space fraction numerator 4.2 space straight x space 4.2 space straight x space 4.4 over denominator 0.7 space straight x space 0.7 end fraction space equals space 158.4 space cm

Hence, the length of the wire is 158.4 cm



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