A strip of thickness 40 mm is to be rolled to a thickness of 20 mm using a two-high mill having rolls of diameter 200 mm. Coefficient of friction and arc length in mm, respectively are

The correct option is 0.45 and 44.72.

1. Understanding the Given Parameters:
Initial thickness of the strip, _h1 = 40 mm
Final thickness of the strip, _h2 = 20 mm
Diameter of the rolls, D = 200 mm
The radius of the rolls (R) is half of the diameter:

$$R=\frac{D}{2}=\frac{200}{2}=100\text{ mm}$$

2. Calculation of Arc Length of Contact (L):
The arc length of contact (L) between the roll and the strip in a rolling process is given by the relation:

$$L=\sqrt{R\times \Delta h}$$
where $\Delta h$ is the draft (reduction in thickness):

$$\Delta h=h_1-h_2=40-20=20\text{ mm}$$
Substituting the values of R and $\Delta h$ into the equation for arc length:

$$L=\sqrt{100\times 20}=\sqrt{2000}\approx 44.72\text{ mm}$$

3. Calculation of Coefficient of Friction (μ):
The maximum possible draft ($\Delta h$) for a given coefficient of friction (μ) in a rolling operation without slipping is given by:

$$\Delta h_{\text{max}}={\mu }^2\times R$$
Assuming that the mill is operating under the condition of maximum draft to achieve this reduction, we can solve for the coefficient of friction (μ):

$$\mu =\sqrt{\frac{\Delta h}{R}}$$
Substituting the values of $\Delta h$ and R:

$$\mu =\sqrt{\frac{20}{100}}=\sqrt{0.2}\approx 0.447$$
Rounding to two decimal places, we get:

$$\mu \approx 0.45$$

Thus, the coefficient of friction is 0.45 and the arc length is 44.72 mm.