In an orthogonal machining with a single point cutting tool of rake angle 10°, the uncut chip thickness and the chip thickness are 0.125 mm and 0.22 mm, respectively. Using Merchant’s first solution for the condition of minimum cutting force, the coefficient of friction at the chip-tool interface is (round off to two decimal places).

The correct answer is 0.74.

Step-by-Step Explanation:

1. Identify the given parameters:
Rake angle ($\alpha$) = 10°
Uncut chip thickness ($t$) = 0.125 mm
Chip thickness ($t_c$) = 0.22 mm

2. Calculate the chip thickness ratio ($r$):
The chip thickness ratio is defined as:
$r=\frac{t}{t_c}$
Substituting the given values:
$r=\frac{0.125}{0.22}\approx 0.5682$

3. Determine the shear angle ($\phi$):
The relation between the shear angle ($\phi$), rake angle ($\alpha$), and chip thickness ratio ($r$) is given by:
$\tan \phi =\frac{r\cos \alpha }{1-r\sin \alpha }$
Substituting the values:
$\tan \phi =\frac{0.5682\cos \mathrm{10°}}{1-0.5682\sin \mathrm{10°}}$
$\tan \phi =\frac{0.5682\times 0.9848}{1-0.5682\times 0.1736}$
$\tan \phi =\frac{0.5596}{1-0.0986}$
$\tan \phi =\frac{0.5596}{0.9014}\approx 0.6208$
$\phi ={\tan }^{-1}\left(0.6208\right)\approx \mathrm{31.83°}$

4. Use Merchant's first solution to find the friction angle ($\beta$):
Merchant's first solution for the condition of minimum cutting force is:
$\phi =\mathrm{45°}+\frac{\alpha }{2}-\frac{\beta }{2}$
Rearranging the equation to solve for the friction angle ($\beta$):
$\frac{\beta }{2}=\mathrm{45°}+\frac{\alpha }{2}-\phi$
$\beta =\mathrm{90°}+\alpha -2\phi$
Substituting $\alpha =\mathrm{10°}$ and $\phi =\mathrm{31.83°}$:
$\beta =\mathrm{90°}+\mathrm{10°}-2\left(\mathrm{31.83°}\right)$
$\beta =\mathrm{100°}-\mathrm{63.66°}=\mathrm{36.34°}$

5. Calculate the coefficient of friction ($\mu$):
The coefficient of friction at the chip-tool interface is related to the friction angle by:
$\mu =\tan \beta$
$\mu =\tan \left(\mathrm{36.34°}\right)\approx 0.7356$

Rounding off to two decimal places, we get:
$\mu \approx 0.74$