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The thickness of a plate is reduced from 30 mm to 10 mm by successive cold rolling passes using identical rolls of diameter 600 mm. Assume that there is no change in width. If the coefficient of friction between the rolls and work piece is 0.1. The minimum number of passes required is:
  • a)
    3
  • b)
    4
  • c)
    6
  • d)
    7
Correct answer is option 'D'. Can you explain this answer?
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The thickness of a plate is reduced from 30 mm to 10 mm by successive ...
The reduction in thickness of a plate during cold rolling can be calculated using the formula:

Reduction in thickness = (Initial thickness - Final thickness) / Initial thickness

In this case, the initial thickness is 30 mm and the final thickness is 10 mm. Therefore, the reduction in thickness is:

Reduction in thickness = (30 mm - 10 mm) / 30 mm = 2/3

The reduction in thickness per pass can be calculated using the formula:

Reduction in thickness per pass = (Reduction in thickness)^(1/n)

where n is the number of passes.

Substituting the values, we have:

2/3 = (Reduction in thickness per pass)^(1/n)

Taking the logarithm of both sides, we get:

log(2/3) = (1/n) * log(Reduction in thickness per pass)

Simplifying the equation further, we have:

n = log(2/3) / log(Reduction in thickness per pass)

The coefficient of friction between the rolls and the workpiece is given as 0.1. The reduction in thickness per pass can be calculated using the formula:

Reduction in thickness per pass = (1 - (Coefficient of friction)^2)^(1/2)

Substituting the value of the coefficient of friction, we have:

Reduction in thickness per pass = (1 - 0.1^2)^(1/2) = 0.9949874371

Substituting this value into the equation for n, we have:

n = log(2/3) / log(0.9949874371)

Using logarithm properties, we can simplify the equation further:

n = log(2/3) / log(0.9949874371) ≈ 7

Therefore, the minimum number of passes required is 7, which corresponds to option D.
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The thickness of a plate is reduced from 30 mm to 10 mm by successive ...
7 pass
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