21. A cubic casting of 50 mm side undergoes volumetric solidification shrinkage and volumetric solid contraction of 4% and 6% respectively. No riser is used. Assume uniform cooling in all directions. The side of the cube after solidification and contraction is
    

1. 1. 48.32 mm
      

   
   
   

   2. 49.90 mm
      

   
   
   

   3. 49.94 mm
      

   
   
   

   4. 49.96 mm

   
   
   

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   Side after

   Correct Option: A

   Side after

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22. A cube shaped casting solidifies in 5 min. The solidification time in min for a cube of the same material, which is 8 times heavier than the original casting, will be
    

1. 1. 10
      

   
   
   

   2. 20
      

   
   
   

   3. 24
      

   
   
   

   4. 40

   
   
   

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   t ∝ (V/A)² and m₂ = 8m₁,
   ρV₂ = 8ρV₁
   ⇒ V₂ = 8V₁
   a₂ = 2a₁ ⇒ t₂ = 4t₁

   Correct Option: B

   t ∝ (V/A)² and m₂ = 8m₁,
   ρV₂ = 8ρV₁
   ⇒ V₂ = 8V₁
   a₂ = 2a₁ ⇒ t₂ = 4t₁

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23. A cylindrical riser of 6 cm diameter and 6 cm height has to be designed for a sand casting mould for producing a steel rectangular plate casting of 7 cm × 10 cm × 2 cm dimensions having the total solidification time of 1.36 minute. The total solidification time (in minute) of the riser is _____.

1. 1. 4.133 min

   
   
   

   2. 7.133 min

   
   
   

   3. 3 min

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   	V			=		7 × 10 × 2		= 0.673 cm
   	A		casting			2(7 × 10 + 7 × 2 + 2 × 10)

   
   

   	V			=		(π / 4)d3
   	A		Top riser			{ (π / 4)d2 + πd2 }

   
   

   	V			=	d	=	6	= 1.25
   	A		Top riser		S		5

   
   

   	V			=		(π / 4)d3		=	d	= 1 cm
   	A		Bottom riser			{ (π / 2)d2 + πd2 }			6

   
   

   tTop riser		=	(V / A)2Top riser	=		1.2		2
   tcasting			(V / A)2casting			0.673

   
   t_{Top riser} = 4.133 min
   

   tBottom riser		=	(V / A)2Bottom riser	=		1		2
   tcasting			(V / A)2casting			0.673

   
   t_{Bottom riser} = 3 min
   

   Correct Option: B

   	V			=		7 × 10 × 2		= 0.673 cm
   	A		casting			2(7 × 10 + 7 × 2 + 2 × 10)

   
   

   	V			=		(π / 4)d3
   	A		Top riser			{ (π / 4)d2 + πd2 }

   
   

   	V			=	d	=	6	= 1.25
   	A		Top riser		S		5

   
   

   	V			=		(π / 4)d3		=	d	= 1 cm
   	A		Bottom riser			{ (π / 2)d2 + πd2 }			6

   
   

   tTop riser		=	(V / A)2Top riser	=		1.2		2
   tcasting			(V / A)2casting			0.673

   
   t_{Top riser} = 4.133 min
   

   tBottom riser		=	(V / A)2Bottom riser	=		1		2
   tcasting			(V / A)2casting			0.673

   
   t_{Bottom riser} = 3 min
   

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24. Calculate the bite angle when rolling plates 12 mm thick using work rolls 600 mm diameter and reducing the thickness by 3 mm

1. 1. 1.73°

   
   
   

   2. 4.71°

   
   
   

   3. 4.73°

   
   
   

   4. 5.73°

   
   
   

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1. View Hint View Answer Discuss in Forum

   D = 600 mm , R =	D	= 300 mm
   	2

   
   h_i = 12 mm
   ∆h = 3 mm
   

   θ = tan-1		√R∆h
   		{ R - (∆h / 2) }

   
   

   θ = tan-1		√300 × 3
   		{ 300 - (3/ 2) }

   
   θ = 5.73°

   Correct Option: D

   D = 600 mm , R =	D	= 300 mm
   	2

   
   h_i = 12 mm
   ∆h = 3 mm
   

   θ = tan-1		√R∆h
   		{ R - (∆h / 2) }

   
   

   θ = tan-1		√300 × 3
   		{ 300 - (3/ 2) }

   
   θ = 5.73°

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25. A wire of 0.1 mm dia is drawn from a rod of 15 mm diameter dies giving reductions of 20%, 40% and 80% are available. For minimum error in the final size, the number of stages and reduction at each stage respectively would be
    

1. 1. 3 stages and 80% reduction for all three stages
      

   
   
   

   2. 4 stages and 80% reduction for first three stages followed by a finishing stage of 20% reduction
      

   
   
   

   3. 5 stage and reduction of 80%, 80%, 40%, 40%, 20% is sequence
      

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   % reduction =	Diameter reduced in draw
   	Diameter before draw

   
   

   for option (a) =	d0 - d1	= 0.8
   	d0

   
   d₁ = 2d₀
   Given d₀ = 15 mm , we get d₁ = 3 mm
   Similarly, after 2nd pass, we get d₂ = 0.6 mm and after 3rd pass, we get d₃ = 0.12 mm For option (b), 1st 3 stages are same as in option (a), in 4th pass we have
   

   	d3 - d4	= 0.2
   	d3

   
   ⇒ d₄ = 0.8d₃
   ⇒ d₄ = 0.8 × 0.12 = 0.096 mm
   For option (c), first two stages are same as in option (a), in 3rd pass we have
   

   	d3 - d4	= 0.4
   	d3

   
   ⇒ d₃ = 6 , d₂ = 0.6 × 0.6 = 0.36 mm
   For 4th pass we have
   

   	d3 - d4	= 0.4
   	d3

   
   ⇒ d₄ = 0.6 , d₃ = 0.6 × 0.36 = 0.216 mm
   For 5th stage, we have
   

   	d5 - d4	= 0.2
   	d4

   
   ⇒ d₅ = 0.8 , d₄ = 0.8 × 0.216 = 0.1728 mm
   Output in option (b) is closest to the desired diameter of 0.1 mm
   

   Correct Option: B

   % reduction =	Diameter reduced in draw
   	Diameter before draw

   
   

   for option (a) =	d0 - d1	= 0.8
   	d0

   
   d₁ = 2d₀
   Given d₀ = 15 mm , we get d₁ = 3 mm
   Similarly, after 2nd pass, we get d₂ = 0.6 mm and after 3rd pass, we get d₃ = 0.12 mm For option (b), 1st 3 stages are same as in option (a), in 4th pass we have
   

   	d3 - d4	= 0.2
   	d3

   
   ⇒ d₄ = 0.8d₃
   ⇒ d₄ = 0.8 × 0.12 = 0.096 mm
   For option (c), first two stages are same as in option (a), in 3rd pass we have
   

   	d3 - d4	= 0.4
   	d3

   
   ⇒ d₃ = 6 , d₂ = 0.6 × 0.6 = 0.36 mm
   For 4th pass we have
   

   	d3 - d4	= 0.4
   	d3

   
   ⇒ d₄ = 0.6 , d₃ = 0.6 × 0.36 = 0.216 mm
   For 5th stage, we have
   

   	d5 - d4	= 0.2
   	d4

   
   ⇒ d₅ = 0.8 , d₄ = 0.8 × 0.216 = 0.1728 mm
   Output in option (b) is closest to the desired diameter of 0.1 mm
   

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