Question

A shaft of diameter \(25^{-0.04}_{-0.07}\) mm is assembled in a hole of diameter \(25^{+0.02}_{+0.00}\) mm. Match the allowance and limit parameter in Column I with its corresponding quantitative value in Column II for this shaft–hole assembly.

• A. P-3, Q-1, R-4
• B. P-1, Q-3, R-2
• C. P-1, Q-3, R-4
• D. P-3, Q-1, R-2

Hint:

Allowance uses the minimum hole and maximum shaft sizes, while clearance uses the maximum hole and minimum shaft sizes.

Answer 1

The Correct Option is A

We are given: Shaft diameter \(25^{-0.04}_{-0.07}\) mm Hole diameter \(25^{+0.02}_{+0.00}\) mm Step 1: Determine maximum and minimum sizes.
Shaft maximum size = \(25 - 0.04 = 24.96\) mm
Shaft minimum size = \(25 - 0.07 = 24.93\) mm
Hole maximum size = \(25 + 0.02 = 25.02\) mm
Hole minimum size = \(25 + 0.00 = 25.00\) mm Step 2: Calculate allowance.
Allowance = minimum hole size − maximum shaft size
= \(25.00 - 24.96 = 0.04\) mm So P corresponds to value 3. Step 3: Calculate maximum clearance.
Maximum clearance = maximum hole size − minimum shaft size
= \(25.02 - 24.93 = 0.09\) mm So Q corresponds to value 1. Step 4: Maximum material limit for hole.
This is the minimum hole size = \(25.00\) mm So R corresponds to value 4. Thus the matching is: P–3, Q–1, R–4. Final Answer: (A)