571. In a rolling process, thickness of a strip is reduced from 4 mm to 3 mm using 300 mm diameter rolls rotating at 100 rpm. The velocity of the strip (in m/sec) at the neutral point is

1. 1. 1.57

   
   
   

   2. 3.14

   
   
   

   3. 47.10

   
   
   

   4. 94.20

   
   
   

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1. View Hint View Answer Discuss in Forum

   V =	πDN
   	60

   
   At Neutral Point, Roll velocity and plate velocity are equal.

   Correct Option: A

   V =	πDN
   	60

   
   At Neutral Point, Roll velocity and plate velocity are equal.

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572. The maximum possible percentage reduction in area per pass during wire drawing of an ideal plastic material without friction is of the order of

1. 1. 37

   
   
   

   2. 50

   
   
   

   3. 63

   
   
   

   4. 75

   
   
   

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1. View Hint View Answer Discuss in Forum

   σd = σ0		Ai
   		Af

   
   

   Ai	= R → Extrusion Ratio
   Af

   
   If σ_d = σ₀ {Maximum possible Reduction}
   

   1 = ln		Ai
   		Af

   
   

   e =	Ai
   	Af

   
   

   % Reduction =	Ai − Af	× 100 =		1 −	1		× 100
   	Ai				e

   
   % Reduction = 63%

   Correct Option: C

   σd = σ0		Ai
   		Af

   
   

   Ai	= R → Extrusion Ratio
   Af

   
   If σ_d = σ₀ {Maximum possible Reduction}
   

   1 = ln		Ai
   		Af

   
   

   e =	Ai
   	Af

   
   

   % Reduction =	Ai − Af	× 100 =		1 −	1		× 100
   	Ai				e

   
   % Reduction = 63%

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573. A copper strip of 200 mm width and 300 mm thickness is to be rolled to a thickness of 295 mm. The roll of radius 300 mm rotates at 100 rpm. The average shear strength of the work material is 180 MPa. The roll strip contact length and the roll force are

1. 1. 15.8 mm and 0.569 MN

   
   
   

   2. 18.97 mm 0.683 MN

   
   
   

   3. 38.73 mm and 1.395 MN

   
   
   

   4. 38.73 mm and 2.09

   
   
   

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1. View Hint View Answer Discuss in Forum

   h_i = 300 mm h_f = 295 mm
   Δh = h_i – h_f = 5 mm
   R = 300 mm, N = 100 rpm.
   τ_{avg = 180 MPa
   Roll strip contact length
   L = √ΔhR
   = √5 × 300 = √1500 = 38.73 mm
   Rolling force = τavg.Lf × width = 1.394 MN}

   Correct Option: C

   h_i = 300 mm h_f = 295 mm
   Δh = h_i – h_f = 5 mm
   R = 300 mm, N = 100 rpm.
   τ_{avg = 180 MPa
   Roll strip contact length
   L = √ΔhR
   = √5 × 300 = √1500 = 38.73 mm
   Rolling force = τavg.Lf × width = 1.394 MN}

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574. In order to reduce roll pressure in strip rolling, back tension can be applied to strip. (T/F)

1. 1. T

   
   
   

   2. F

   
   
   

   3. T or F

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   Back Tension is applied. It decreases the amount of force in Rollers laterally.

   Correct Option: A

   Back Tension is applied. It decreases the amount of force in Rollers laterally.

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575. If the elongation factor during rolling of an ingot is 1.22. The minimum number of passes needed to produce a section 250 x 250 mm from an ingot of 750 × 750 mm are

1. 1. 8

   
   
   

   2. 9

   
   
   

   3. 10

   
   
   

   4. 12

   
   
   

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1. View Hint View Answer Discuss in Forum

   Elongation factor =	Ai	= 1.22
   	Ar

   
   Let n → number of passes
   

   750 × 750	= 250 × 250
   (1.11)n

   
   (1 – 22)ⁿ = 9
   n/n 1.22 = 9
   n = 11.04

   Correct Option: C

   Elongation factor =	Ai	= 1.22
   	Ar

   
   Let n → number of passes
   

   750 × 750	= 250 × 250
   (1.11)n

   
   (1 – 22)ⁿ = 9
   n/n 1.22 = 9
   n = 11.04

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