66. In a two-stage wire drawing operation, the fractional reduction (ratio of change in crosssectional area to initial cross-sectional area) in the first stage is 0.4. The fractional reduction in the second stage is 0.3. The overall fractional reduction is

1. 1. 0.24

   
   
   

   2. 0.58

   
   
   

   3. 0.60

   
   
   

   4. 1.00

   
   
   

---

1. View Hint View Answer Discuss in Forum

   40% Reduction = 1 × 0.4 = 0.4
   30% Reduction of 0.6 = 0.6 × 0.3 = 0.18
   Total Reduction = 0.58

   Correct Option: B

   40% Reduction = 1 × 0.4 = 0.4
   30% Reduction of 0.6 = 0.6 × 0.3 = 0.18
   Total Reduction = 0.58

---

67. A metal rod of initial length L0 is subjected to a drawing process. The length of the rod at any instant is given by the expression, L(t) = L₀ (1 + t²), where t is the time in minutes. The true strain rate (in min^–1) at the end of one minute is _______.

1. 1. 1 min^–1

   
   
   

   2. 2 min^–1

   
   
   

   3. 3 min^–1

   
   
   

   4. 4 min^–1

   
   
   

---

1. View Hint View Answer Discuss in Forum

   L(t) = L₀ (1 + t²)
   ⇒ dL = L₀ × 2t dt
   

   As d ∈T =	dL
   	L

   
   

   d ∈T =	L02t dt
   	L0 (1 + t²)

   
   

   		d ∈T	=	2t
   		dt		1 + t²

   
   at t = 1 min
   

   		∈T	=	2 × 1	= 1 min- 1
   		T		1 + 1²

   
   

   Correct Option: A

   L(t) = L₀ (1 + t²)
   ⇒ dL = L₀ × 2t dt
   

   As d ∈T =	dL
   	L

   
   

   d ∈T =	L02t dt
   	L0 (1 + t²)

   
   

   		d ∈T	=	2t
   		dt		1 + t²

   
   at t = 1 min
   

   		∈T	=	2 × 1	= 1 min- 1
   		T		1 + 1²

   
   

---

---

68. In a wire drawing operation, diameter of a Steel wire is reduced from 10 mm to 8 mm. The mean flow stress of the material is 400 MPa. The ideal force required for drawing (ignoring friction and redundant work) is

1. 1. 4.48 kN

   
   
   

   2. 8.97 kN

   
   
   

   3. 20.11 kN

   
   
   

   4. 31.41 kN

   
   
   

---

1. View Hint View Answer Discuss in Forum

   D_i = 10 mm, D_i = 8mm
   σ₀
   = 400 MPa
   

   σd = σ0ln			Di			²
   			Df

   
   

   σd = 400ln			10			²
   			8

   
   σd = 178.51 MPa
   

   Ideal force = 178.51 × 106 ×	π	(0.008)²
   	4

   
   I deal force 8.97 KN

   Correct Option: B

   D_i = 10 mm, D_i = 8mm
   σ₀
   = 400 MPa
   

   σd = σ0ln			Di			²
   			Df

   
   

   σd = 400ln			10			²
   			8

   
   σd = 178.51 MPa
   

   Ideal force = 178.51 × 106 ×	π	(0.008)²
   	4

   
   I deal force 8.97 KN

---

69. Cold shut is a defect in casting due to

1. 1. sand sliding from the cope surface

   
   
   

   2. internal voids or surface depression due to excessive gas trapped

   
   
   

   3. discontinuity resulting from hundred contraction

   
   
   

   4. two streams of material that are too cold to fuse properly

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

---

---

70. Manufacturing process
    A. Hot tears
    B. Porosity
    C. Sand inclusions
    D. Shrinkage cavity Property
    1. Mould restraint
    2. Inadequate risering
    3. High pouring temp
    4. Loosely rammed sand mould
    5. Gas entrapment

1. 1. A – 1, B – 5, C – 4, D – 2

   
   
   

   2. A – 2, B – 5, C – 4, D – 1

   
   
   

   3. A – 4, B – 5, C – 1, D – 2

   
   
   

   4. A – 1, B – 4, C – 5, D – 2

   
   
   

---

1. View Hint View Answer Discuss in Forum

   A – 1, B – 5, C – 4, D – 2

   Correct Option: A

   A – 1, B – 5, C – 4, D – 2

---