46. A milling cutter having 8 teeth is rotating at 150 rpm. If the feed per tooth is 0.1 mm, the speed in mm per minute is

1. 1. 120

   
   
   

   2. 187

   
   
   

   3. 125

   
   
   

   4. 70

   
   
   

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1. View Hint View Answer Discuss in Forum

   F = f_t ZN
   = 0.1 × 8 × 150
   F = 120mm/ min

   Correct Option: A

   F = f_t ZN
   = 0.1 × 8 × 150
   F = 120mm/ min

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47. The maximum depth to width ratio in EBM welding is

1. 1. 1: 1

   
   
   

   2. 2: 1

   
   
   

   3. 10: 1

   
   
   

   4. 25: 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

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48. The ideal cutting fluid for low speed machining of metals should be one which

1. 1. removes the heat faster from the cutting zone

   
   
   

   2. forms a coating on the cutting tools by Chemical reaction

   
   
   

   3. forms a low shear strength film of work material at the tool chip interface

   
   
   

   4. serves as a dielectric, minimizing thereby reactions due to EMF at the interface

   
   
   

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1. View Hint View Answer Discuss in Forum

   Function of cutting fluid
   (1) Cooling
   (2) Lubrication
   (3) Cleaning
   (4) Save from external contamination.

   Correct Option: A

   Function of cutting fluid
   (1) Cooling
   (2) Lubrication
   (3) Cleaning
   (4) Save from external contamination.

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49. In a shaping process, the number of double strokes per minute is 30 and the quick return ratio is 0.6. If the length of the stroke is 250 mm, the average cutting velocity in mm/min.

1. 1. 3.0

   
   
   

   2. 4.5

   
   
   

   3. 7.5

   
   
   

   4. 12.0

   
   
   

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1. View Hint View Answer Discuss in Forum

   Number of Double stroke = 30
   Quick Return Ratio = 0.6 = M
   Length of Strike = 250 mm = L
   Average cutting velocity = L.N (1 + 19)
   = 30 × 250 (1 + 0.6) = 12 m /min

   Correct Option: D

   Number of Double stroke = 30
   Quick Return Ratio = 0.6 = M
   Length of Strike = 250 mm = L
   Average cutting velocity = L.N (1 + 19)
   = 30 × 250 (1 + 0.6) = 12 m /min

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50. The ultimate shear stress (in N/mm²) of the work material is

1. 1. 235

   
   
   

   2. 139

   
   
   

   3. 564

   
   
   

   4. 380

   
   
   

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1. View Hint View Answer Discuss in Forum

   τs =	Fs	sinφ
   	Wt1

   
   

   =	751	sin 18.43°
   	2.5 × 0.25

   
   τ = 380N/mm²

   Correct Option: D

   τs =	Fs	sinφ
   	Wt1

   
   

   =	751	sin 18.43°
   	2.5 × 0.25

   
   τ = 380N/mm²

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