261. If a particular Fe-C alloy contains less than 0.83% carbon, it is called

1. 1. high speed steel

   
   
   

   2. hypoeutectoid steel

   
   
   

   3. hypereutectoid steel

   
   
   

   4. cast iron

   
   
   

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1. View Hint View Answer Discuss in Forum

   

   Correct Option: B

   

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262. In a electrochemical machining (ECM) operation, a square hole of dimensions 5 mm × mm is drilled in a block of copper. The current used is 5000 A. Atomic weight of copper is 63 and valence of dissolution is 1. Faradays constant is 96500 coulomb. The material removal rate (g/s) is

1. 1. 0.326

   
   
   

   2. 3.26

   
   
   

   3. 3.15 × 10³

   
   
   

   4. 3.15 × 10⁵

   
   
   

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1. View Hint View Answer Discuss in Forum

   MRR =	AI	kg/s.
   	ZF

   
   given:
   A = 63
   I = 5000 A
   F = 96500 couloms.
   Z = 1
   So,
   

   MRR =	63 × 5000	= 3.26 gm/s.
   	96500 × 1

   Correct Option: B

   MRR =	AI	kg/s.
   	ZF

   
   given:
   A = 63
   I = 5000 A
   F = 96500 couloms.
   Z = 1
   So,
   

   MRR =	63 × 5000	= 3.26 gm/s.
   	96500 × 1

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263. In an AJM process, if Q = flow rate of abrasives and d = mean diameter of the abrasives, then MRR is proportional to

1. 1. Q/d²

   
   
   

   2. Qd

   
   
   

   3. Qd²

   
   
   

   4. Qd³

   
   
   

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1. View Hint View Answer Discuss in Forum

   In abrasive jet machining.MRR ∝ QD³ √V³
   

   Correct Option: D

   In abrasive jet machining.MRR ∝ QD³ √V³
   

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264. Which one of the following process conditions leads to higher MRR in ECM?

1. 1. Higher current, large atomic weight

   
   
   

   2. Higher valency, lower current

   
   
   

   3. Lower atomic weight, lower valency

   
   
   

   4. Higher valency, lower atomic weight

   
   
   

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1. View Hint View Answer Discuss in Forum

   In ECM, high current improve material removal rate, also material removal rate increase with atomic weight of material.

   Correct Option: A

   In ECM, high current improve material removal rate, also material removal rate increase with atomic weight of material.

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265. Average power input (in kW) is

1. 1. 0.373

   
   
   

   2. 0.137

   
   
   

   3. 0.218

   
   
   

   4. 0.5

   
   
   

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1. View Hint View Answer Discuss in Forum

   Total Energy consumed
   E= 1/2 CV² =0.121 J/cycle.
   Average Power,
   

   P =	0.121
   	0.55 × 10-3

   
   = 220 W = 0.22 kW.

   Correct Option: C

   Total Energy consumed
   E= 1/2 CV² =0.121 J/cycle.
   Average Power,
   

   P =	0.121
   	0.55 × 10-3

   
   = 220 W = 0.22 kW.

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