Step 1: What the Problem Wants:
We are told the strip must reduce from 23 mm to 20 mm in one pass on rolls of diameter 200 mm, and we are given four possible friction values.
Instead of finding the least friction needed first, this approach directly computes how much reduction each lubricant CAN produce, and compares that with the 3 mm reduction that is actually needed.
Step 2: Formula for Maximum Draft:
The largest thickness drop that a given friction coefficient can support is
\[ \Delta h_{max} = \mu^2 R \]
with $R = 100$ mm (half of the 200 mm roll diameter).
A lubricant works only if its $\Delta h_{max}$ is at least the needed 3 mm.
Step 3: Testing Each Lubricant One at a Time:
For P, $\mu = 0.05$, so $\Delta h_{max} = (0.05)^2 (100) = 0.25$ mm. This is much smaller than 3 mm, so P cannot roll the strip down to 20 mm.
For Q, $\mu = 0.10$, so $\Delta h_{max} = (0.10)^2(100) = 1.0$ mm. Still less than 3 mm, so Q also fails.
For R, $\mu = 0.20$, so $\Delta h_{max} = (0.20)^2(100) = 4.0$ mm. Since 4.0 mm is more than the 3 mm required, R is capable of the pass.
For S, $\mu = 0.25$, so $\Delta h_{max} = (0.25)^2(100) = 6.25$ mm. This comfortably covers the 3 mm requirement, so S also works.
Final Answer:
Comparing the maximum possible draft of each lubricant with the 3 mm actually needed shows that only R (4.0 mm capacity) and S (6.25 mm capacity) are sufficient.
\[ \boxed{\Delta h_{max}(R) = 4 \text{ mm}, \ \Delta h_{max}(S) = 6.25 \text{ mm}, \text{ both} \ge 3 \text{ mm}} \]