Question

If the elongation factor during rolling of an ingot is 1.22. The minimum number of passes needed to produce a section 250 $\times$ 250 mm from an ingot of 750 $\times$ 750 mm are

• A. 8
• B. 9
• C. 10
• D. 12

Answer 1

The correct option is D 12

$\text{Elongation factor}=\frac{A_0}{A_1}=1.22$

Let number of passes be n, in order to get desired section of 250 $\times$ 250 $m{m}^2$ from ingot of 750 $\times$ 750 $m{m}^2$

$\text{Here, we have}\frac{750\times 750}{(1.22{)}^n}=250\times 250$

$\Rightarrow (1.22{)}^n=\frac{750\times 750}{250\times 250}=9$

Taking $lo{g}_e$ on both sides, we get

$nlo{g}_{e}1.22=lo{g}_{e}9$

or n = 11.04 pass means 12 pass