91. A cylindrical pin of diameter 1.996^+0.015-0.015 mm is assembled into a hole of diameter 2.000^+0.015-0.015 mm. The allowance (in mm) provides for this assembly is

1. 1. 0.001

   
   
   

   2. 0.015

   
   
   

   3. 0.025

   
   
   

   4. 0.035

   
   
   

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1. View Hint View Answer Discuss in Forum

   Allowance = minimum hole dia – maximum shaft dia
   = 1.9985 – 1.9975 =- 0.001 mm

   Correct Option: A

   Allowance = minimum hole dia – maximum shaft dia
   = 1.9985 – 1.9975 =- 0.001 mm

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92. A keyway of 5 mm depth is to be milled in a shaft of diameter 60 ± 0.1 by positioning the shaft in a V-block of angle 120 deg and setting the tool with reference to the intersection of the face of the V-block. The error due to shaft tolerance which would be transferred on to the component is

1. 1. 0.09 mm

   
   
   

   2. 0.119 mm

   
   
   

   3. 0.22 mm

   
   
   

   4. 0.3 mm

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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93. Diameter of a hole after plating needs to be controlled between 30^+0.050-0.010 mm. If the plating thickness varies between 10-15 microns, diameter of the hole before plating should be

1. 1. 30^+0.07-0.030 mm

   
   
   

   2. 30^+0.65-0.020

   
   
   

   3. 30^+0.08-0.030 mm

   
   
   

   4. 30^+0.70-0.040

   
   
   

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1. View Hint View Answer Discuss in Forum

   Remember flating comes from both side
   So D_max – 2 × 0.01 = 30.05
   D_max = 30.07mm
   D_min – 2 × 0.015 = 30.01
   D_min = 30.04

   Correct Option: D

   Remember flating comes from both side
   So D_max – 2 × 0.01 = 30.05
   D_max = 30.07mm
   D_min – 2 × 0.015 = 30.01
   D_min = 30.04

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