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In an electro - chemical machining (E C M ) operation, a square hole of diamensions 5 mm x 5 mm is drilled in a block of copper. The current used is 5000 A. Atomic weight of copper is 63 and valency of dissolution is 1. Faraday’s constant is 96,500 Coulomb. The material removal rate (in g/s) is
  • a)
    0.326
  • b)
    3.26
  • c)
    3.15 x 103
  • d)
    3.15 x 105
Correct answer is option 'B'. Can you explain this answer?
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Constant is 96500 C/mol.

First, we need to calculate the volume of copper to be removed in order to drill the square hole. The volume of the square hole can be calculated as follows:

Volume = (5 mm) x (5 mm) x (depth of hole)

Assuming the depth of the hole is 1 mm, the volume of the hole would be:

Volume = (5 mm) x (5 mm) x (1 mm) = 25 mm^3 = 25 x 10^-9 m^3

Next, we need to calculate the mass of copper to be removed. The density of copper is 8.96 g/cm^3, so the mass of copper to be removed is:

Mass = Volume x Density = 25 x 10^-9 m^3 x 8.96 g/cm^3 = 2.24 x 10^-7 g

Now, we can calculate the number of moles of copper to be removed:

Number of moles = Mass / Atomic weight = 2.24 x 10^-7 g / 63 g/mol = 3.56 x 10^-9 mol

Since the valency of dissolution is 1, the number of electrons required for the dissolution of 1 mole of copper is equal to the number of moles of copper to be removed. Therefore, the total charge required can be calculated as:

Q = Number of moles x Faraday constant = 3.56 x 10^-9 mol x 96500 C/mol = 0.343 C

Therefore, a current of 5000 A would need to flow for a time of:

Time = Charge / Current = 0.343 C / 5000 A = 6.86 x 10^-5 seconds

Therefore, it would take approximately 6.86 x 10^-5 seconds to drill a 5 mm x 5 mm square hole in a block of copper using a current of 5000 A in an ECM operation.
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In an electro - chemical machining (E C M ) operation, a square hole of diamensions 5 mm x 5 mm is drilled in a block of copper. The current used is 5000 A. Atomic weight of copper is 63 and valency of dissolution is 1. Faraday’s constant is 96,500 Coulomb. The material removal rate (in g/s) isa)0.326b)3.26c)3.15 x 103d)3.15 x 105Correct answer is option 'B'. Can you explain this answer?
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