31. For resistance spot welding of 1.5 mm thick steel sheets, the current required is of the order

1. 1. 10 Amp

   
   
   

   2. 100 Amp

   
   
   

   3. 1000 Amp

   
   
   

   4. 10,000 Amp

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Air Resistance = 3300 ohm
   Current required = 3300 × (– 1.5 + 1.5)
   Current required = 9900 Amperes
   = 10000 Amperes

   Correct Option: D

   
   Air Resistance = 3300 ohm
   Current required = 3300 × (– 1.5 + 1.5)
   Current required = 9900 Amperes
   = 10000 Amperes

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32. The DC power source for are welding has the characteristic 3V + I = 240, where V = Voltage and I = Current in amp. For maximum are power at the electrode, voltage should be set at

1. 1. 20 V

   
   
   

   2. 40 V

   
   
   

   3. 60 V

   
   
   

   4. 80 V

   
   
   

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1. View Hint View Answer Discuss in Forum

   Constant current characteristics:
   3 V + I = 240
   I = 240 – 3 V
   Drooping characteristics:
   P = V I = V × (240 – 3 V)
   

   dP	= 0
   dV

   
   V = 40 V

   Correct Option: B

   Constant current characteristics:
   3 V + I = 240
   I = 240 – 3 V
   Drooping characteristics:
   P = V I = V × (240 – 3 V)
   

   dP	= 0
   dV

   
   V = 40 V

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33. Circular blanks of 10 mm diameter are punched from an aluminum sheet of 2 mm thickness. The shear strength of aluminum is 80 MPa. The minimum punching force required in kN is

1. 1. 2.57

   
   
   

   2. 3.29

   
   
   

   3. 5.03

   
   
   

   4. 6.33

   
   
   

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1. View Hint View Answer Discuss in Forum

   Punching force = πDtτ
   = π × 10 × 2 × 80 = 5.023 KN

   Correct Option: C

   Punching force = πDtτ
   = π × 10 × 2 × 80 = 5.023 KN

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34. A blank of 50 mm diameter is to be sheared from a sheet of 2.5 mm thickness. The required radial clearance between the die and the punch is 6% of sheet thickness. The punch and die diameter (mm) for this blanking operation respectively are

1. 1. 50.00 and 50.30

   
   
   

   2. 50.00 and 50.15

   
   
   

   3. 49.70 and 50.00

   
   
   

   4. 49.85 and 50.00

   
   
   

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1. View Hint View Answer Discuss in Forum

   Radial clearance =	6	× t = 0.1 ≤ 1.5 mm
   	100

   
   Total clearance = 2 × 0.15 = 0.3
   Die Dia will be exact = 50 mm
   Clearance will be provided on punch = 50–0.30 = 49.70 mm

   Correct Option: C

   Radial clearance =	6	× t = 0.1 ≤ 1.5 mm
   	100

   
   Total clearance = 2 × 0.15 = 0.3
   Die Dia will be exact = 50 mm
   Clearance will be provided on punch = 50–0.30 = 49.70 mm

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35. It is required to punch a hole of diameter 10 mm on a sheet of thickness 3 mm. The shear strength of the work material is 420 MPa. The required punch force is

1. 1. 19.78 kN

   
   
   

   2. 39.58 kN

   
   
   

   3. 98.9 kN

   
   
   

   4. 359.6 kN

   
   
   

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1. View Hint View Answer Discuss in Forum

   Punching force = πDtτ
   = π × 10 × 3 × 420 = 39.58 KN

   Correct Option: B

   Punching force = πDtτ
   = π × 10 × 3 × 420 = 39.58 KN

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