A circular shaft having diameter mm is manufactured by turning process. A 50 μm thick coating of TiN is deposited on the shaft. Allowed variation in TiN film thickness is ±5 μm. The minimum hole diameter (in mm) to just provide clearance fit is

The correct option is 65.12.

Step-by-Step Explanation:

1. Shaft Diameter before Plating:
From the first image, the limits of the circular shaft diameter are given as:
$${65}^{+0.01}{}_{-0.05}\text{mm}$$
The upper limit (UL) or the maximum diameter of the shaft before plating is:
$$D_{\text{max, unplated}}=65.00+0.01=65.01\text{mm}$$
This is also explicitly indicated in the third image as UL before Plating = 65.01 mm.

2. Maximum Coating Thickness:
The nominal thickness of the TiN coating is $50\text{μm}$ with an allowed variation of $\pm 5\text{μm}$.
The maximum thickness of the coating ($t_{\text{max}}$) on one side of the shaft (as shown in the third image) is:
$$t_{\text{max}}=50+5=55\text{μm}=0.055\text{mm}$$

3. Maximum Diameter of the Plated Shaft:
Since the coating is deposited uniformly all around the circular shaft, the total diameter increases by twice the coating thickness ($2\times t_{\text{max}}$):
$$D_{\text{max, plated}}=D_{\text{max, unplated}}+2\times t_{\text{max}}$$
$$D_{\text{max, plated}}=65.01+2\times 0.055=65.01+0.11=65.12\text{mm}$$

4. Minimum Hole Diameter for Clearance Fit:
A clearance fit requires that the hole diameter is always greater than or equal to the shaft diameter (i.e., the minimum clearance $\text{Min. C}\ge 0$, as illustrated in the second image).
To just provide a clearance fit, the minimum hole diameter ($H_{\text{min}}$) must equal the maximum possible diameter of the plated shaft:
$$H_{\text{min}}=D_{\text{max, plated}}=65.12\text{mm}$$