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. Determine the Maximum Shaft Diameter before Coating:
From the image provided, the shaft diameter with tolerances is specified as:
${65}_{-0.05}^{+0.01}\text{mm}$
The maximum limit of the shaft diameter before coating occurs at the upper tolerance limit:
$d_{\text{max, uncoated}}=65.00+0.01=65.01\text{mm}$

2. Calculate the Maximum Coating Thickness:
The nominal TiN coating thickness is 50 μm with an allowed variation of ±5 μm. The maximum possible coating thickness on one side of the shaft surface is:
$t_{\text{max}}=50+5=55\text{μm}$
Converting this thickness into millimeters:
$t_{\text{max}}=\frac{55}{1000}=0.055\text{mm}$

3. Determine the Maximum Coated Shaft Diameter:
Since the shaft is circular and the coating is applied uniformly around the entire circumference, the total increase in the diameter is twice the coating thickness:
$d_{\text{max, coated}}=d_{\text{max, uncoated}}+2\times t_{\text{max}}$
Substituting the calculated values:
$d_{\text{max, coated}}=65.01+2\times 0.055$
$d_{\text{max, coated}}=65.01+0.11=65.12\text{mm}$

4. Determine the Minimum Hole Diameter for Clearance Fit:
To ensure a clearance fit under all manufacturing circumstances, the minimum hole diameter must be at least equal to the maximum diameter of the coated shaft:
$D_{\text{hole, min}}=d_{\text{max, coated}}=65.12\text{mm}$