An aluminium alloy used for making cans is cold rolled into a strip of thickness 0.3mm and width 1m. it is coiled round a drum of diameter 15cm, and the outer diameter of the coil is 1m. in the cold rolled condition, the dislocation density is approximately 1015 m-2. estimate: (i) the mass of aluminium on the coil; (ii) the total length of strip on the coil; (iii) the total length of dislocation in the coiled strip.

1. Mass = 2070 kg

2.Total length of strip = 2556 m

3. Total length of dislocation = 7.67 X m

Explanation:

Given:

Aluminium coil thickness, t = 0.3 mm

                                              = 0.3 X m

Width of the coil,w = 1 m

Drum diameter, d = 15 cm

                              = 0.15 m

Coil outer diameter, d = 1 m

Dislocation density =

1). Area of the coil, A = ( -)

                           A =

                           A = 0.767

Volume of the coil,V = A X w

                                  = 0.767 X 1

                                  = 0.767

We know density of aluminum at STP = 2.7 X

Therefore, mass of the aluminum coil is,

Mass,m = Density of aluminium X Volume

             = 2.7 X X 0.767

             = 2070 kg

Mass = 2070 kg

2). Total length of trip of coil is given by

          L = Volume of coil / area of strip

             =

              = 2556 m

Total length of strip = 2556 m

3). Total length of dislocation of the coiled strip = volume X dislocation density

                                                                             = 0.767 X

                                                                              = 7.67 X

Total length of dislocation = 7.67 X m

answer:

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