Question:
In a rolling operation, a 200 mm wide strip of 23 mm thickness needs to be reduced to 20 mm in a single pass. The roll diameter is 200 mm. Four lubricants P, Q, R, and S with coefficient of friction 0.05, 0.1, 0.2, and 0.25, respectively, are available for use at the roll-strip interface. Assume that the strip width remains constant throughout the process.
Which ONE or MORE among the following lubricants will enable the rolling process to achieve the desired final thickness of the strip in a single pass?
In a rolling operation, a 200 mm wide strip of 23 mm thickness needs to be reduced to 20 mm in a single pass. The roll diameter is 200 mm. Four lubricants P, Q, R, and S with coefficient of friction 0.05, 0.1, 0.2, and 0.25, respectively, are available for use at the roll-strip interface. Assume that the strip width remains constant throughout the process.
Which ONE or MORE among the following lubricants will enable the rolling process to achieve the desired final thickness of the strip in a single pass?
Which ONE or MORE among the following lubricants will enable the rolling process to achieve the desired final thickness of the strip in a single pass?
Show Hint
Use the rolling condition \( \Delta h \le \mu^2 R \) to find the minimum friction needed, then check which lubricants meet it.
Updated On: Aug 5, 2026
- P
- Q
- R
- S
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Verified By Collegedunia
The Correct Option is C, D
Solution and Explanation
Step 1: Understanding the Question:
In rolling, the rolls grip the strip only because of friction between the roll surface and the strip.
If the friction is too low for the amount of thickness reduction being attempted, the rolls simply slip over the strip instead of pulling it through, and the pass cannot be completed.
So the question is really asking which of the four given friction coefficients is high enough to grip the strip for the required 3 mm reduction in one pass.
Step 2: Key Formula or Approach:
For a rolling pass to be geometrically possible, the actual draft (drop in thickness) must not exceed the maximum draft that friction can support.
The maximum possible draft in flat rolling is given by
\[ \Delta h_{max} = \mu^2 R \]
where \( \mu \) is the coefficient of friction and \( R \) is the roll radius.
The rolling condition is \( \Delta h \le \Delta h_{max} \).
Step 3: Calculating the Minimum Required Friction:
Initial thickness \( h_0 = 23 \) mm and final thickness \( h_f = 20 \) mm, so the required draft is
\[ \Delta h = h_0 - h_f = 23 - 20 = 3 \text{ mm} \]
Roll diameter is 200 mm, so roll radius \( R = 100 \) mm.
Substituting into the rolling condition to find the smallest usable \( \mu \):
\[ 3 \le \mu^2 (100) \implies \mu^2 \ge 0.03 \implies \mu \ge \sqrt{0.03} \approx 0.173 \]
So any lubricant giving a coefficient of friction of at least about 0.173 will let the pass go through.
Step 4: Checking Each Lubricant:
Lubricant P gives \( \mu = 0.05 \), which is far below 0.173, so the rolls will slip; P does not work.
Lubricant Q gives \( \mu = 0.10 \), still below 0.173, so Q also fails to grip the strip.
Lubricant R gives \( \mu = 0.20 \), which is above 0.173, so R can complete the pass.
Lubricant S gives \( \mu = 0.25 \), also above 0.173, so S can complete the pass too.
Final Answer:
Only lubricants R and S provide enough friction to achieve the 3 mm reduction in a single pass, so options C and D are correct. \[ \boxed{\mu \ge 0.173, \text{ satisfied only by R} (\mu=0.20) \text{ and S} (\mu=0.25)} \]
In rolling, the rolls grip the strip only because of friction between the roll surface and the strip.
If the friction is too low for the amount of thickness reduction being attempted, the rolls simply slip over the strip instead of pulling it through, and the pass cannot be completed.
So the question is really asking which of the four given friction coefficients is high enough to grip the strip for the required 3 mm reduction in one pass.
Step 2: Key Formula or Approach:
For a rolling pass to be geometrically possible, the actual draft (drop in thickness) must not exceed the maximum draft that friction can support.
The maximum possible draft in flat rolling is given by
\[ \Delta h_{max} = \mu^2 R \]
where \( \mu \) is the coefficient of friction and \( R \) is the roll radius.
The rolling condition is \( \Delta h \le \Delta h_{max} \).
Step 3: Calculating the Minimum Required Friction:
Initial thickness \( h_0 = 23 \) mm and final thickness \( h_f = 20 \) mm, so the required draft is
\[ \Delta h = h_0 - h_f = 23 - 20 = 3 \text{ mm} \]
Roll diameter is 200 mm, so roll radius \( R = 100 \) mm.
Substituting into the rolling condition to find the smallest usable \( \mu \):
\[ 3 \le \mu^2 (100) \implies \mu^2 \ge 0.03 \implies \mu \ge \sqrt{0.03} \approx 0.173 \]
So any lubricant giving a coefficient of friction of at least about 0.173 will let the pass go through.
Step 4: Checking Each Lubricant:
Lubricant P gives \( \mu = 0.05 \), which is far below 0.173, so the rolls will slip; P does not work.
Lubricant Q gives \( \mu = 0.10 \), still below 0.173, so Q also fails to grip the strip.
Lubricant R gives \( \mu = 0.20 \), which is above 0.173, so R can complete the pass.
Lubricant S gives \( \mu = 0.25 \), also above 0.173, so S can complete the pass too.
Final Answer:
Only lubricants R and S provide enough friction to achieve the 3 mm reduction in a single pass, so options C and D are correct. \[ \boxed{\mu \ge 0.173, \text{ satisfied only by R} (\mu=0.20) \text{ and S} (\mu=0.25)} \]
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