Question 53 1AE01 - First Assistant Engineer

The Correct Answer is A **Explanation for Option A (0.0005 inch):** Option A, **0.0005 inch**, represents a very tight, precision tolerance typically required for critical dimensions like the outside diameter (OD) of a bushing. Bushings are often press-fit into housings or must mate precisely with other components (like bearings or shafts). The specific illustration (GS-0017, although not visible here) almost certainly shows a dimension requiring a bilateral tolerance. A tolerance of $\pm 0.0005$ inch (or a total tolerance band of $0.001$ inch) is a common high-precision requirement in machining and aligns with the expected function of an accurate bushing OD dimension. **Why the other options are incorrect:** * **B) 0.002 inch:** This value represents a total tolerance band of $0.002$ inch (if specified as $\pm 0.001$ inch). While $0.002$ inch is a reasonable machining tolerance, it is less precise than $0.0005$ inch. If the part drawing specified a very tight tolerance (as implied by the correct answer A), $0.002$ inch would be too loose for the intended precision fit of the bushing OD. * **C) 0.060 inch:** This value is an extremely large tolerance for a machined feature like the outside diameter of a bushing. A total tolerance band of $0.060$ inch (approximately $1/16$ of an inch) would make the part unusable for any precision application, such as press fits or accurate locating. This magnitude of tolerance is usually reserved for non-critical features, raw stock, or general castings, not precision ODs. * **D) 1.6015 inches:** This is not a tolerance; it is an actual dimension (a specific size) or the upper limit of a dimension. A tolerance must be a permissible variation or range (a difference between limits, such as $0.0005$ inch or $0.002$ inch). It cannot be an absolute measurement value like $1.6015$ inches.