Discussion
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.
| Allowance and limit parameter (Column I) | Quantitative value (Column II) | ||
|---|---|---|---|
| P | Allowance | 1 | 0.09 mm |
| Q | Maximum clearance | 2 | 24.96 mm |
| R | Maximum material limit for hole | 3 | 0.04 mm |
| 4 | 25.0 mm | ||
Answer: A
The match is P-3, Q-1, R-4.
The limits are: shaft 24.93 mm to 24.96 mm; hole 25.00 mm to 25.02 mm. The shaft is always smaller, so this is a clearance fit.
\[ \begin{aligned} \text{Allowance} &= 25.00 - 24.96 = 0.04\ \text{mm} \\ \text{Max clearance} &= 25.02 - 24.93 = 0.09\ \text{mm} \end{aligned} \]
Allowance is the minimum clearance, the tightest condition: smallest hole minus largest shaft, 0.04 mm (3). Maximum clearance is largest hole minus smallest shaft, 0.09 mm (1). The maximum material limit for the hole is the size at which the hole holds the most metal, its smallest size, 25.0 mm (4).
24.96 mm is the maximum material limit of the shaft, not of the hole, which rules out every option pairing R with 2; the options pairing P with 1 swap allowance and maximum clearance.
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