86. The maximum interference in mm after assembly between a bush of size 30^+0.06+0.03 and a shaft of size 30^+0.04+0.02 is

1. 1. 0.07

   
   
   

   2. 0.05

   
   
   

   3. 0.02

   
   
   

   4. 0.01

   
   
   

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1. View Hint View Answer Discuss in Forum

   

   Correct Option: D

   

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87. Holes 1, 2, 3, 4 drilled in sequence, with angular positions and tolerances are shown in figure. Evaluate closing angle X with its tolerance band.
    

1. 1. 0.44 mm

   
   
   

   2. 0.22 mm

   
   
   

   3. 0.02 mm

   
   
   

   4. 2.22 mm

   
   
   

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   X_max = 360 - [79.9 + 10.88 + 109.7] = 60.6
   X_min = 360 - [80.1 + 110.2 + 110.3] = 59.4
   Clearance = X_max – X_min = 1.2
   

   Xmean =	Xmax + Xmin	=	60.6 + 59.4
   	2		2

   
   X_mean = 60
   X = 60 ± 60
   when dia is minimum = 59.9mm
   

   (OC)2 =	59.9/2	= 34.58 mm
   	sin 60°

   
   Error in depth = 2[(OC) – (OC)₂] = 0.22 mm.

   Correct Option: B

   X_max = 360 - [79.9 + 10.88 + 109.7] = 60.6
   X_min = 360 - [80.1 + 110.2 + 110.3] = 59.4
   Clearance = X_max – X_min = 1.2
   

   Xmean =	Xmax + Xmin	=	60.6 + 59.4
   	2		2

   
   X_mean = 60
   X = 60 ± 60
   when dia is minimum = 59.9mm
   

   (OC)2 =	59.9/2	= 34.58 mm
   	sin 60°

   
   Error in depth = 2[(OC) – (OC)₂] = 0.22 mm.

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88. A hole is specified as φ 30± 0.04 mm with – φ0.01 mm. The virtual diameter of the hole (i.e., the maximum diameter of the pin that can enter the hole) is _______.

1. 1. 9.95 mm

   
   
   

   2. 29.95 mm

   
   
   

   3. 39.95 mm

   
   
   

   4. 49.95 mm

   
   
   

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   In maximum material condition
   Dia_hole = 30 – 0.04 – 0.01 = 29.95 mm
   This is virtual dia of the hole.

   Correct Option: B

   In maximum material condition
   Dia_hole = 30 – 0.04 – 0.01 = 29.95 mm
   This is virtual dia of the hole.

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89. Abbe's principle of alignment is used to be followed in

1. 1. vernier calipers

   
   
   

   2. depth vernier

   
   
   

   3. internal caliper micrometer

   
   
   

   4. height vernier

   
   
   

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   NA

   Correct Option: C

   NA

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90. The diameter of a hole is given as 50.00^+0.015-0.0 mm. The upper limit on the dimensions in mm, of the shaft for achieving maximum interference of 50 microns is _______.

1. 1. 10 microns

   
   
   

   2. 30 microns

   
   
   

   3. 40 microns

   
   
   

   4. 28 microns

   
   
   

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1. View Hint View Answer Discuss in Forum

   In worst assembly condition, clearance will maximum
   Clearance (max) = Max hole dia – Minimum shaft diameter
   = 20.025 – 19.995 = 0.030 = 30 microns

   Correct Option: B

   In worst assembly condition, clearance will maximum
   Clearance (max) = Max hole dia – Minimum shaft diameter
   = 20.025 – 19.995 = 0.030 = 30 microns

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