From each end of a solid metal cylinder, metal was scooped out in hemispherical form of 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. [Use π=22/7]
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The radius of the cylinder is given as 4.2cm, and height of the cylinder is given as 10cm.
They said that two hemispherical metal is scooped out at the top as well as bottom.
So two hemisphers form one sphere, now in the problem they are saying indirectly that from a cylinder volume a sphere volume is removed.
Volume of the solid= Volume of cylinder- volume of the sphere, and this volume is dragged into a wire which is also a cylinder. Whose thickness is 1.4cm means that the diameter of the wire is 1.4cm which inturn says that the radius is 0.7cm.
So the volume of wire=volume of cylinder-volume of sphere
πr2h-4/3πr3 = πr2l (where l is the length of wire)⇒πr2(h-4/3r)=πr2l
⇒(4.2)2(10-4/3x4.2)=(0.7)2xl ( the given thickness is 1.4cm radius of wire is 0.7cm)
⇒l=158.4cm
Length of the wire required is 158.4cm.
They said that two hemispherical metal is scooped out at the top as well as bottom.
So two hemisphers form one sphere, now in the problem they are saying indirectly that from a cylinder volume a sphere volume is removed.
Volume of the solid= Volume of cylinder- volume of the sphere, and this volume is dragged into a wire which is also a cylinder. Whose thickness is 1.4cm means that the diameter of the wire is 1.4cm which inturn says that the radius is 0.7cm.
So the volume of wire=volume of cylinder-volume of sphere
πr2h-4/3πr3 = πr2l (where l is the length of wire)⇒πr2(h-4/3r)=πr2l
⇒(4.2)2(10-4/3x4.2)=(0.7)2xl ( the given thickness is 1.4cm radius of wire is 0.7cm)
⇒l=158.4cm
Length of the wire required is 158.4cm.
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