Maths

An iron pipe of internal diameter 2.8 cm and uniform thickness 1 mm is melted and a solid cylindrical rod of the same length is formed. Find the diameter of the rod.

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  1. The cross-section of the pipe has area
    pi/4 (2.9^2-2.8^2) = 0.1425pi cm^2

    The rod of radius r has area
    pi r^2 cm^2

    So, pi r^2 = 0.1425 pi
    r=0.3775 cm
    So, the diameter is 0.755 cm

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  2. Its answer is 1.08 cm diameter

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  3. Internal diameter =2.8cm
    Radius=1.4cm
    Thickness =1mm=0.1cm
    External radius =1.4cm+0.1cm=1.5cm
    (V1) volume of hollow pipe =pie×h×(R+r)(R-r)=(22÷7 )×h×2.9×0.1=0.9hcm^2
    (V) volume of rod with same length=pie×r^2×h=(22÷7)×r^2×h
    V1=V
    or,0.9h=(22÷7)×r^2×h
    or,0.29=r^2
    :.r=0.54cm
    So,diameter of rod =2r=2×0.54=1.08cm

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  4. Is the External Radius = Radius * thickness? How

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  5. Thanku very much.And fantastic job!

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