The thickness of a sheet is reduced by rolling (without any change in width) using 600 mm diameter rolls. Neglect elastic deflection of the rolls and assume that the coefficient of friction at the roll-workpiece interface is 0.05. The sheet enters the rotating rolls unaided. If the initial sheet thickness is 2 mm, the minimum possible final thickness that can be produced by this process in a single pass is______mm (round off to two decimal places).

To find the minimum possible final thickness that can be produced in a single pass of rolling without external aid, we use the maximum draft relationship.

The maximum possible reduction in sheet thickness (also known as the maximum draft, $\Delta h_{\max }$) during rolling under unaided entry is governed by the coefficient of friction ($\mu$) and the roll radius ($R$). This relationship is given by:
$\Delta h_{\max }=h_0-h_{f,\min }={\mu }^{2}R$
where:
- $h_0$ is the initial sheet thickness,
- $h_{f,\min }$ is the minimum possible final sheet thickness,
- $\mu$ is the coefficient of friction at the roll-workpiece interface, and
- $R$ is the radius of the rolls.

From the problem description, we are given:
- Roll diameter, $D=600\text{mm}$
- Therefore, roll radius, $R=\frac{D}{2}=\frac{600}{2}=300\text{mm}$
- Coefficient of friction, $\mu =0.05$
- Initial sheet thickness, $h_0=2\text{mm}$

Now, we substitute these values into the maximum draft equation to solve for $h_{f,\min }$:
$2-h_{f,\min }={\left(0.05\right)}^2\times 300$

Calculate the right-hand side of the equation:
${\mu }^{2}R=0.0025\times 300=0.75\text{mm}$

Substitute this back to find the minimum final thickness:
$2-h_{f,\min }=0.75$
$h_{f,\min }=2-0.75=1.25\text{mm}$

Thus, the minimum possible final thickness that can be produced is 1.25 mm.