11. 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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12. 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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13. Chip speed along the tool rake face will be

1. 1. 0.83 m/s

   
   
   

   2. 0.53 m/s

   
   
   

   3. 1.2 m/s

   
   
   

   4. 1.88 m/s

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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14. The shear angle is

1. 1. 14°

   
   
   

   2. 25°

   
   
   

   3. 29°

   
   
   

   4. 75°

   
   
   

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1. View Hint View Answer Discuss in Forum

   α = 15°
   W = 2 mm
   t₂ = 0.4mm
   f_r = 0.2 mm/rev = t₁
   

   r =	t1	=	0.2	=	1
   	t2		0.4		2

   
   r = 0.5
   

   φ = tan-1		r cos ∝
   		1 - r sin∝

   
   

   φ = tan-1		0.5 cos 15
   		1 - 0 sin 15

   
   φ = 0

   Correct Option: C

   α = 15°
   W = 2 mm
   t₂ = 0.4mm
   f_r = 0.2 mm/rev = t₁
   

   r =	t1	=	0.2	=	1
   	t2		0.4		2

   
   r = 0.5
   

   φ = tan-1		r cos ∝
   		1 - r sin∝

   
   

   φ = tan-1		0.5 cos 15
   		1 - 0 sin 15

   
   φ = 0

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15. If the cutting force and the thrust force are 900 N and 810 N, the mean strength in MPa

1. 1. 137.94

   
   
   

   2. 477.91

   
   
   

   3. 500.58

   
   
   

   4. 635.84

   
   
   

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1. View Hint View Answer Discuss in Forum

   F_c = 900 N,
   F_t = 810 N
   

   tanβ =	FT	+ tanα
   	Fc
   	1 -	FT	tanα
   		Fc

   
   

   tanβ =	810	+ tan 15°
   	900
   	1 -	810	tan 15°
   		900

   
   β = 56.98°
   

   Fs	=	Rcos(φ + β - ∝)
   Fc		Rcos(β - ∝)

   
   F_s = 394.56 N
   

   Cs =	Fs	× sinφ =	394.56	× sin 29
   	Wt1		2 × 0.2

   
   C_s = 478 mpa

   Correct Option: B

   F_c = 900 N,
   F_t = 810 N
   

   tanβ =	FT	+ tanα
   	Fc
   	1 -	FT	tanα
   		Fc

   
   

   tanβ =	810	+ tan 15°
   	900
   	1 -	810	tan 15°
   		900

   
   β = 56.98°
   

   Fs	=	Rcos(φ + β - ∝)
   Fc		Rcos(β - ∝)

   
   F_s = 394.56 N
   

   Cs =	Fs	× sinφ =	394.56	× sin 29
   	Wt1		2 × 0.2

   
   C_s = 478 mpa

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