491. A hole of 20 mm diameter is to be drilled in a steel block of 40 mm thickness. The drilling is performed at rotational speed of 400 rpm and feed rate of 0.1 mm/rev. The required approach and over run of the drill together, is equal to the radius of drill. The drilling time (in minute) is

1. 1. 1.00

   
   
   

   2. 1.25

   
   
   

   3. 1.50

   
   
   

   4. 1.75

   
   
   

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1. View Hint View Answer Discuss in Forum

   T =	L
   	fN

   
   L = t + Ap₁
   Ap₁ = 0.5 D (holes diameter)
   = 10 mm
   t = 40 mm
   

   T =	50	= 1.25 min.
   	0.1 × 400

   Correct Option: B

   T =	L
   	fN

   
   L = t + Ap₁
   Ap₁ = 0.5 D (holes diameter)
   = 10 mm
   t = 40 mm
   

   T =	50	= 1.25 min.
   	0.1 × 400

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492. If the Taylor's tool life exponent n is 0.2, and the tool changing time is 1.5 min, then the tool life (in min) for maximum production rate is___

1. 1. 4

   
   
   

   2. 6

   
   
   

   3. 10

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   Topt =		1 − n	× Tc			=	1 − 0.2	× 1.5 = 6 min.
   		n					0.2

   Correct Option: B

   Topt =		1 − n	× Tc			=	1 − 0.2	× 1.5 = 6 min.
   		n					0.2

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493. Details pertaining to an orthogonal metal cutting process are given below:
     

     Chip thickness ratio	0.4
     Undeformed thickness	0.6 mm
     Rake angle	+10°
     Cutting speed	2.5 m/s
     Mean thickness of primary shear zone	25 microns

     
     The shear strain rate in s^–1 during the process is

1. 1. 0.1781 × 105

   
   
   

   2. 0.7754 × 105

   
   
   

   3. 1.0104 × 105

   
   
   

   4. 4.397 × 105

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given: r= 0.4
   t₁ = 0.6 mm
   α = 10°
   V_C = 2.5 m/s
   t_m = 25 µm
   

   tan φ =	rcosα	= 0.4233
   	1 − rsinα

   
   ⇒  φ = 22.9°
   

   Shear strain rate =	VCcosα
   	cos(φ − α) × tm

   
   

   =	2.5 × cos10
   	cos12.9 × 25 × 10−6

   
   = 1.010⁴ × 10⁵/s

   Correct Option: C

   Given: r= 0.4
   t₁ = 0.6 mm
   α = 10°
   V_C = 2.5 m/s
   t_m = 25 µm
   

   tan φ =	rcosα	= 0.4233
   	1 − rsinα

   
   ⇒  φ = 22.9°
   

   Shear strain rate =	VCcosα
   	cos(φ − α) × tm

   
   

   =	2.5 × cos10
   	cos12.9 × 25 × 10−6

   
   = 1.010⁴ × 10⁵/s

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494. In a single pass drilling operation, a through hole of 15 mm diameter is to be drilled in a steel plate of 50 mm thickness. Drill spindle speed is 500 rpm, feed is 0.2 mm/rev and drill point angle is 118°, Assuming 2 mm clearance at approach and exit, the total drill time (in seconds) is

1. 1. 35.1

   
   
   

   2. 32.4

   
   
   

   3. 31.2

   
   
   

   4. 30,1

   
   
   

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1. View Hint View Answer Discuss in Forum

   Tc =	(Lh + A + O + H)
   	Nf

   
   Here, L_h = 50 mm; A = O = 2 mm;
   

   C =	D	cot		β		= 7.5 × cot(59) = 4.5 mm
   	2			2

   
   N = 500 rpm; f = 0.2 mm/rev
   T_c = 0.585 min or 35.1 seconds

   Correct Option: A

   Tc =	(Lh + A + O + H)
   	Nf

   
   Here, L_h = 50 mm; A = O = 2 mm;
   

   C =	D	cot		β		= 7.5 × cot(59) = 4.5 mm
   	2			2

   
   N = 500 rpm; f = 0.2 mm/rev
   T_c = 0.585 min or 35.1 seconds

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495. A single-point cutting tool with 12° rake angle is used to machine a steel work-piece. The depth of cut, i.e. uncut thickness is 0.81 mm. The chip thickness under orthogonal machining condition is 1.8 mm. The shear angle is approximately

1. 1. 22°

   
   
   

   2. 26°

   
   
   

   3. 56°

   
   
   

   4. 76°

   
   
   

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1. View Hint View Answer Discuss in Forum

   Relation between shear angle (φ), chip thickness ratio (r) and rake angle (α) is given by
   

   tan φ =	rcosα
   	1 − rsinα

   
   

   where,   r =	0.81	= 0.45
   	1.8

   
   

   tan φ =	0.45cos12
   	1 − 0.45sin12

   
   ⇒  φ = 26°

   Correct Option: B

   Relation between shear angle (φ), chip thickness ratio (r) and rake angle (α) is given by
   

   tan φ =	rcosα
   	1 − rsinα

   
   

   where,   r =	0.81	= 0.45
   	1.8

   
   

   tan φ =	0.45cos12
   	1 − 0.45sin12

   
   ⇒  φ = 26°

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