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Here you can find the 2020 GATE Material Science (XE-C) paper with answer key and full explanation in pdf given below:
This set of Engineering Materials Science "Problems & Answers" are based on “GATE 2020 (XE-C)”
GATE 2020 (XE-C)
QUESTIONS, ANSWERS & EXPLANATION
UNIT - II (QUESTION 10 to 22)
10) Read the two statements related to sintering and select the correct option.
Statement-1: Sintering in a vacuum leads to improved densification as compared to sintering under ambient (at atmospheric pressure) conditions.
Statement-2: Closed pores formed during sintering inhibit full densification
Correct Answer is "B"
Sintering is the process of compacting and forming a solid mass of material by heat or pressure without melting it to the point of liquefaction.
(1) If we put a powder sample in a vacuum chamber then all the trapped gases will be evacuated from the chamber under heat and/or pressure. So there will be very small pores in the compacted solid containing very little volume. So density is more in this case i.e. statement – 1 is true.
(2) If closed pores formed during sintering then the volume increase due to the presence of pores but the mass is constant so density decreases i.e. statement – 2 is true.
(1) If we put a powder sample in a vacuum chamber then all the trapped gases will be evacuated from the chamber under heat and/or pressure. So there will be very small pores in the compacted solid containing very little volume. So density is more in this case i.e. statement – 1 is true.
(2) If closed pores formed during sintering then the volume increase due to the presence of pores but the mass is constant so density decreases i.e. statement – 2 is true.
11) Select the correct option that appropriately matches the process of the material/product that can be fabricated using them.
|
Process |
Material/Product |
|
(I) Powder processing |
(P) Organic semiconductor thin films |
|
(II) Spin coating |
(Q) Single-crystal silicon |
|
(III) Czochralski process |
(R) Poly-silicon |
|
(IV) Chemical vapor deposition |
(S) Porous bronze bearings |
Correct Answer is "C"
We know that,
(1) By powder processing or sintering we get porous product
(2) By spin coating we get a thin film
(3) In the Czochralski (CZ) process we use single crystal silicon to produce large cylindrical single-crystal ingots for the electrical industry
(4) In chemical vapor deposition poly-silicon is used.
(1) By powder processing or sintering we get porous product
(2) By spin coating we get a thin film
(3) In the Czochralski (CZ) process we use single crystal silicon to produce large cylindrical single-crystal ingots for the electrical industry
(4) In chemical vapor deposition poly-silicon is used.
12) Consider a FCC structured metal with lattice parameter a = 3.5 Å. If the material is irradiated using X-rays of wavelength X = 1.54056 Å, the Bragg angle (2𝜃) corresponding to the fourth reflection will be:
Correct Answer is "D"
From Bragg’s diffraction law we know that,
n𝝀 = 2dhkl 𝐬𝐢𝐧𝜽
where,
n = order of reflection for x-ray diffraction
λ = X-ray wavelength = 1.54056 Å
dhkl = Interplanar spacing for crystallographic planes having indices h,k and l
dhkl = 𝒂/√(𝒉2 + 𝒌2 + 𝒍2) , 𝒂 = lattice parameter = 3.5 Å
Crystallographic
structure
Value
of (𝒉2 + 𝒌2 + 𝒍2) for different reflection
1st
2nd
3rd
4th
5th
6th
7th
8th
SC
1
2
3
4
5
6
7
8
BCC
2
4
6
8
10
12
14
16
FCC
3
4
8
11
12
16
19
20
So for 4th reflection of FCC crystal metal, 𝒉2 + 𝒌2 + 𝒍2 = 11
Hence, dhkl =𝒂/√(𝒉2 + 𝒌2 + 𝒍2) = √𝟏𝟏/𝟑.𝟓
Now,
1 × 1.54056 = 2 × 𝟑.𝟓 /√ 𝟏𝟏 × 𝐬𝐢𝐧𝜽
⇒ 𝐬𝐢𝐧𝜽 = 0.73
⇒ 𝜽 = 𝐬𝐢𝐧-1(𝟎.𝟕𝟑) = 46.9°
⇒ 2𝜽 = 93.80°
n𝝀 = 2dhkl 𝐬𝐢𝐧𝜽
where,
n = order of reflection for x-ray diffraction
λ = X-ray wavelength = 1.54056 Å
dhkl = Interplanar spacing for crystallographic planes having indices h,k and l
dhkl = 𝒂/√(𝒉2 + 𝒌2 + 𝒍2) , 𝒂 = lattice parameter = 3.5 Å
Crystallographic
structure
Value
of (𝒉2 + 𝒌2 + 𝒍2) for different reflection
1st
2nd
3rd
4th
5th
6th
7th
8th
SC
1
2
3
4
5
6
7
8
BCC
2
4
6
8
10
12
14
16
FCC
3
4
8
11
12
16
19
20
13) The number of Schottky defects per mole of KCI at 300 °C under equilibrium condition will be:
Given:
Activation energy for the formation of Schottky defect = 250 kJ.mol-1
Avogadro number = 6.023 × 1023 mol-1
Universal Gas Constant = 8.314 J.K-1.mol-1
Correct Answer is "D"
We know that,
Number of Schottky defects per unit volume,
Nv = N exp(-Q/2kT)
Number of Schottky defect per mole,
Nm = NA exp(-Q/2RT) Where,
NA = Avogadro number = 6.023 × 1023 mol-1
Q = Activation energy for the formation of Schottky defect = 250 kJ.mol-1 = 250000 J.mol-1
R = Universal gas constant = 8.314 J.K-1.mol-1
So, Nm = 6.023 × 1023 × exp{-250000/(2 × 8.314 × 573)} = 2.42 × 1012
Number of Schottky defects per unit volume,
Nv = N exp(-Q/2kT)
Number of Schottky defect per mole,
Nm = NA exp(-Q/2RT) Where,
NA = Avogadro number = 6.023 × 1023 mol-1
Q = Activation energy for the formation of Schottky defect = 250 kJ.mol-1 = 250000 J.mol-1
R = Universal gas constant = 8.314 J.K-1.mol-1
So, Nm = 6.023 × 1023 × exp{-250000/(2 × 8.314 × 573)} = 2.42 × 1012
14) In industry, the probability of an accident occurring in a given month is 1/100. Let P(n) denote the probability that there will be no accident over a period of 'n' months. Assume that the events of individual months are independent of each other. The smallest integer value of 'n' such that P(n) ≤ 1/2 is ______(round off to the nearest integer).
Correct Answer is "69 - 69"
Let, A: occurring an accident in a given month
So the probability of occurring an accident in a given month, P(A) = 1/100 (given)
Probability of not occurring an accident over a period of ‘n’ month = P(n) ≤ 1/2 (given)
Now, P(n) = Ac for n months...................(1)
Where, Ac: Not occurring an accident in a given month
So, P(Ac) = 1 – P(A) = 1 – (1/100) = 99/100
From (1)
P(n) = P(Ac) × P(Ac) × P(Ac) × P(Ac)...........n times
⇒ P(n) = 𝟗𝟗/𝟏𝟎𝟎 × 𝟗𝟗/𝟏𝟎𝟎 × 𝟗𝟗/𝟏𝟎𝟎 × 𝟗𝟗/𝟏𝟎𝟎 ×........... n times
⇒ P(n) = (𝟗𝟗/𝟏𝟎𝟎)n
now for the smallest integer value of n
(𝟗𝟗/𝟏𝟎𝟎)n = P(n) = 1/2
Taking log on both sides,
n log (𝟗𝟗/𝟏𝟎𝟎) = log(1/2)
⇒ n = 68.967 = 69 (rounded off to the nearest integer)
So the probability of occurring an accident in a given month, P(A) = 1/100 (given)
Probability of not occurring an accident over a period of ‘n’ month = P(n) ≤ 1/2 (given)
Now, P(n) = Ac for n months...................(1)
Where, Ac: Not occurring an accident in a given month
So, P(Ac) = 1 – P(A) = 1 – (1/100) = 99/100
From (1)
P(n) = P(Ac) × P(Ac) × P(Ac) × P(Ac)...........n times
⇒ P(n) = 𝟗𝟗/𝟏𝟎𝟎 × 𝟗𝟗/𝟏𝟎𝟎 × 𝟗𝟗/𝟏𝟎𝟎 × 𝟗𝟗/𝟏𝟎𝟎 ×........... n times
⇒ P(n) = (𝟗𝟗/𝟏𝟎𝟎)n
now for the smallest integer value of n
(𝟗𝟗/𝟏𝟎𝟎)n = P(n) = 1/2
Taking log on both sides,
n log (𝟗𝟗/𝟏𝟎𝟎) = log(1/2)
⇒ n = 68.967 = 69 (rounded off to the nearest integer)
15) For a FCC metal, the ratio of surface energy of (111) surface to (100) surface is ______(round-off to two decimal places). Assume that only the nearest neighbour broken bonds contribute to the surface energy.
Correct Answer is "0.84 – 0.90"
The surface energy of a plane,
Shkl = (𝑼0/𝑵A) × Nb × PDhkl) × 1/2
Where,
U0) = Bond Energy for NA) atom
NA) = Avogadro’s number
Nb) = number of bond being broken for a given plane
PDhkl) = Planar density of the given plane
½ factor is used for eliminating the other surface
According to the given problem,
𝐒111)/𝐒100) = {[(𝑼0)/(𝑵A)] × 𝐍b111 × 𝐏𝐃111 × (𝟏/𝟐)}/ {[(𝑼0)/(𝑵0A)] × 𝐍b100 × 𝐏𝐃100 × (𝟏/𝟐)} = {(𝐍𝐛𝟏𝟏𝟏) × (𝐏𝐃111)}/{(𝐍b100) × (𝐏𝐃100)}
We know that,
FCC CRYSTAL
Plane indices
Number of bond
associated with the plane
(1 1 1)
3
(1 0 0)
4
(1 1 0)
5
So, Nb111 = 3 & Nb100 = 4
PD111 CALCULATION:
PD111 = number of atoms centred on (1 1 1) / planearea of (1 1 1) plane
⇒PD111 = {𝟏/𝟔 ×𝟑 (no. of corner atoms)} + {𝟏/𝟐 × 𝟑(no. of center atoms)}] / {(√𝟑/𝟒) × (√𝟐a)𝟐}
⇒ PD111 = {(𝟏/𝟐)+ (𝟑/𝟐)} / {(√𝟑/𝟒) × (√𝟐a)𝟐} = 𝟐 / {(√𝟑/𝟒) × (√𝟐a)𝟐} = 𝟒 / {√𝟑a𝟐}
PD100 CALCULATION:
PD100 = number of atoms centred on (1 0 0) / planearea of (1 0 0) plane
⇒PD100 = [{(𝟏/𝟒) ×𝟒 (no. of corner atoms)} + 𝟏(no. of center atoms)]/a𝟐
⇒ PD100 = (𝟏+𝟏)/a𝟐 = 𝟐/a𝟐
Hence , S111/S100 = Nb111 × PD111Nb100 × PD100= {𝟑 × (𝟒/ √𝟑a𝟐)}/{𝟒 × (𝟐/a𝟐)} = √𝟑/𝟐 = 0.866 = 0.87 (rounded off to two decimal places)
Shkl = (𝑼0/𝑵A) × Nb × PDhkl) × 1/2
Where,
U0) = Bond Energy for NA) atom
NA) = Avogadro’s number
Nb) = number of bond being broken for a given plane
PDhkl) = Planar density of the given plane
½ factor is used for eliminating the other surface
According to the given problem,
𝐒111)/𝐒100) = {[(𝑼0)/(𝑵A)] × 𝐍b111 × 𝐏𝐃111 × (𝟏/𝟐)}/ {[(𝑼0)/(𝑵0A)] × 𝐍b100 × 𝐏𝐃100 × (𝟏/𝟐)} = {(𝐍𝐛𝟏𝟏𝟏) × (𝐏𝐃111)}/{(𝐍b100) × (𝐏𝐃100)}
We know that,
FCC CRYSTAL
Plane indices
Number of bond
associated with the plane
(1 1 1)
3
(1 0 0)
4
(1 1 0)
5
16) Pure silicon (Si) has a band gap (Eg) of 1.1 eV. This Si is doped with 1 ppm (parts per million) of phosphorus atoms. Si contains 5 x1028 atoms per m3 in pure form. At temperature T = 300 K, the shift in Fermi energy upon doping with respect to intrinsic Fermi level of pure Si will be ____eV (with appropriate sign and round-off to two decimal places).
Intrinsic carrier concentration of Si, ni, is given as:
ni = {2(2𝜋𝑚𝑘B𝑇 / ℎ2)3/2} exp (- 𝐸g/2𝑘B𝑇)
Given: Mass of an electron, m = 9.1 x 10-31kg
Charge of an electron, e =1.6 x 10-19 C
Boltzmann constant, kB= 1.38 × 1023 J.K-1
Planck's constant, h = 6.6×10-34 J.s
Correct Answer is "0.33 - 0.45"
Shift of Fermi energy upon doping with respect to intrinsic Fermi level,
Efn - Efi = kBTln(n/ni)
Where,
Efn = Fermi energy for semiconductor
Efi= Intrinsic Fermi energy
kB = Boltzmann constant = 1.38 × 1023 J.K-1
n = no, of free electrons per m3
ni = Intrinsic charge carrier concentration
Since Si is tetravalent and P is pentavalent so after doping one electron is free per P atom per cubic meter
Total no. of free electron per cubic meter, n = 1ppm × 5 × 1028 = (1/106) × 5 × 1028 = 5 × 1022
Calculation of ni from the given data:
ni = {2(2𝜋mkBT/h2)3/2} exp (- 𝐸g/2𝑘B𝑇)
⇒ ni = 2 × {[(𝟐×𝝅×𝟗.𝟏×𝟏𝟎−𝟑𝟏×𝟏.𝟑𝟖×𝟏𝟎−𝟐𝟑×𝟑𝟎𝟎)/(𝟔.𝟔×𝟏𝟎−𝟑𝟒)𝟐)]𝟑/𝟐} exp {(- 𝟏.𝟏)/(𝟐×𝟖.𝟔𝟐×𝟏𝟎−𝟓×𝟑𝟎𝟎 )}
⇒ ni = 1.4668 × 1016
Now,
Efn - Efi = 𝟖.𝟔𝟐×𝟏𝟎-5 ×𝟑𝟎𝟎 × ln{(𝟓×𝟏𝟎22)/(𝟏.𝟒𝟔𝟔𝟖×𝟏𝟎16)}
⇒ Efn - Efi = 0.3889 eV = 0.39 eV(rounded off to two decimal places)
Efn - Efi = kBTln(n/ni)
Where,
Efn = Fermi energy for semiconductor
Efi= Intrinsic Fermi energy
kB = Boltzmann constant = 1.38 × 1023 J.K-1
n = no, of free electrons per m3
ni = Intrinsic charge carrier concentration
Since Si is tetravalent and P is pentavalent so after doping one electron is free per P atom per cubic meter
Total no. of free electron per cubic meter, n = 1ppm × 5 × 1028 = (1/106) × 5 × 1028 = 5 × 1022
Calculation of ni from the given data:
ni = {2(2𝜋mkBT/h2)3/2} exp (- 𝐸g/2𝑘B𝑇)
⇒ ni = 2 × {[(𝟐×𝝅×𝟗.𝟏×𝟏𝟎−𝟑𝟏×𝟏.𝟑𝟖×𝟏𝟎−𝟐𝟑×𝟑𝟎𝟎)/(𝟔.𝟔×𝟏𝟎−𝟑𝟒)𝟐)]𝟑/𝟐} exp {(- 𝟏.𝟏)/(𝟐×𝟖.𝟔𝟐×𝟏𝟎−𝟓×𝟑𝟎𝟎 )}
⇒ ni = 1.4668 × 1016
Now,
Efn - Efi = 𝟖.𝟔𝟐×𝟏𝟎-5 ×𝟑𝟎𝟎 × ln{(𝟓×𝟏𝟎22)/(𝟏.𝟒𝟔𝟔𝟖×𝟏𝟎16)}
⇒ Efn - Efi = 0.3889 eV = 0.39 eV(rounded off to two decimal places)
17) The schematic diagram shows the light of intensity Io incident on a material (shaded grey) of thickness, x, which has an absorption coefficient, α and reflectance, R. The intensity of transmitted light is I. The reflection of light (of a particular wavelength) occurs at both the surfaces (surfaces indicated in the diagram).
The transmittance is estimated to be ______(round-off to three decimal places).
Given that for the wavelength used, α = 103 m-1 and R = 0.05.
Correct Answer is "0.330 - 0.334"
We know that,
Intensity of the transmitted radiation (when reflection losses taken into account),
IT = Io (1 -R)2 e-ax
Where,
Io = Intensity of the incident radiation
R = Reflectance = 0.05
α = Absorption coefficient = 103 m-1
x = material thickness = 1mm = 1×10-3 m
Hence,
Transmittance = IT/Io = (1 - R)2 e-ax = (1 - 0.05)2 e(-103×1×10-3) = 0.33201 = 0.332 (rounded off to three decimal places)
Intensity of the transmitted radiation (when reflection losses taken into account),
IT = Io (1 -R)2 e-ax
Where,
Io = Intensity of the incident radiation
R = Reflectance = 0.05
α = Absorption coefficient = 103 m-1
x = material thickness = 1mm = 1×10-3 m
Hence,
Transmittance = IT/Io = (1 - R)2 e-ax = (1 - 0.05)2 e(-103×1×10-3) = 0.33201 = 0.332 (rounded off to three decimal places)
18) 𝐹𝑒3𝑂4 (also represented as FeO. 𝐹𝑒2𝑂3) is a FCC structured inverse spinel (𝐴𝐵2𝑂4) material where 1/8 of tetrahedral sites are occupied by half of B cations and 1/2 of the octahedral sites are occupied by remaining B and A cations. The magnetic moments of cations on octahedral sites are anti-parallel with respect to those on tetrahedral sites. Atomic number of Fe is 26 and that of 0 is 8. The saturation magnetic moment of 𝐹𝑒3𝑂4 per formula unit in terms of Bohr magnetons (μB) will be ______ μB. Ignore contribution from orbital magnetic moments.
Correct Answer is " 4 - 4"
Given that, 𝑭𝒆3𝑶4 ≡ FeO. 𝑭𝒆2𝑶3
Here the valency of Fe in FeO is +2 and in 𝑭𝒆2𝑶3 is +3
According to the given problem,
𝑭𝒆3𝑶4 is a FCC structured inverse spinal 𝑨𝑩2𝑶4 type material.
So here, A ≡ 𝑭𝒆𝟐+ and B ≡ 𝑭𝒆3+
No. Of A i.e. (𝒏A) = 1 and no. Of B i.e. (𝒏B) = 2
Further we know that,
FCC has 8 tetrahedral sites and 4 octahedral sites in each unit cell.
According to the given problem,
1/8 of tetrahedral sites are occupied by half of B cations
i.e. (1/8) × 8 = 1 tetrahedral site is occupied by (1/2) × (𝒏B) = (1/2) × 2 = 1 𝑭𝒆3+ cation
And, 1/2 of octahedral sites are occupied by remaining B and A cations
i.e. (1/2) × 4 = 2 octahedral sites are occupied by (2-1) = 1 𝑭𝒆3+ cation and 1 𝑭𝒆𝟐+ cation
Sites
Occupied cations
Tetrahedral
1 𝑭𝒆3+cation
Octahedral
1 𝑭𝒆3+ cation and 1 𝑭𝒆2+cation
Now, the magnetic moments of 1 𝑭𝒆3+ cation in the tetrahedral site and 1 𝑭𝒆3+ cation in the octahedral site are anti-parallel (given) so the magnetic moment contribution (spin only) from 𝑭𝒆3+ cation is zero as they cancel out each other.
The spin only magnetic moment of 𝑭𝒆𝟐+ cation in the tetrahedral site only contribute to the total magnetic moment of 𝑭𝒆3𝑶4.
Now the electronic configuration of 𝑭𝒆𝟐+ cation is,
𝟏𝒔𝟐𝟐𝒔𝟐𝟐𝒑6𝟑𝒔𝟐𝟑𝒑6𝟑𝒅4
So the number of unpaired electron is 4.
Hence the required magnetic moment is 4 Bohr magneton.
Here the valency of Fe in FeO is +2 and in 𝑭𝒆2𝑶3 is +3
According to the given problem,
𝑭𝒆3𝑶4 is a FCC structured inverse spinal 𝑨𝑩2𝑶4 type material.
So here, A ≡ 𝑭𝒆𝟐+ and B ≡ 𝑭𝒆3+
No. Of A i.e. (𝒏A) = 1 and no. Of B i.e. (𝒏B) = 2
Further we know that,
FCC has 8 tetrahedral sites and 4 octahedral sites in each unit cell.
According to the given problem,
1/8 of tetrahedral sites are occupied by half of B cations
i.e. (1/8) × 8 = 1 tetrahedral site is occupied by (1/2) × (𝒏B) = (1/2) × 2 = 1 𝑭𝒆3+ cation
And, 1/2 of octahedral sites are occupied by remaining B and A cations
i.e. (1/2) × 4 = 2 octahedral sites are occupied by (2-1) = 1 𝑭𝒆3+ cation and 1 𝑭𝒆𝟐+ cation
Sites
Occupied cations
Tetrahedral
1 𝑭𝒆3+cation
Octahedral
1 𝑭𝒆3+ cation and 1 𝑭𝒆2+cation
19) A piezoelectric ceramic with piezoelectric coefficient (dzz) value of 100 × 10-12 C.N-1 is subjected to a force, Fz, of 10 N, applied normal to its x-y face, as shown in the figure.
If relative dielectric constant (𝜀r) of the material is 1100, the voltage developed along the z-direction of the sample will be ____ Volts (round-off to two decimal places). Ignore any nonlinear effects.
Given: Permittivity of free space (𝜀0) is 8.85 × 10-12 F.m-1
Correct Answer is "0.97 - 1.09"
We know that,
Voltage across the capacitor, V = 𝑸/𝑪 = 𝑭z×𝒅zz/{(𝜺0𝜺r𝑨)/𝒕}
Where,
Q = charge
C = Capacitance of a capacitor
Fz = force along z-direction = 10N
dzz =piezoelectric coefficient = 100 × 10-12 C.N-1
𝜺0 = Permittivity of free space = 8.85 × 10-12 F.m-1
𝜺r = Relative dielectric constant = 1100
A = Area of the sample = 1 cm × 1 cm = 1 cm𝟐 = 1 × 10-4 m𝟐
t = thickness of the sample = 1 mm = 1×10-3 m
Hence,
V = (𝟏𝟎×𝟏𝟎𝟎×𝟏𝟎-12)/{(𝟖.𝟖𝟓×𝟏𝟎-12×𝟏𝟏𝟎𝟎×𝟏×𝟏𝟎-4)/(𝟏×𝟏𝟎-3)} = 1.0272V = 1.02V (rounded off to two decimal place)
Voltage across the capacitor, V = 𝑸/𝑪 = 𝑭z×𝒅zz/{(𝜺0𝜺r𝑨)/𝒕}
Where,
Q = charge
C = Capacitance of a capacitor
Fz = force along z-direction = 10N
dzz =piezoelectric coefficient = 100 × 10-12 C.N-1
𝜺0 = Permittivity of free space = 8.85 × 10-12 F.m-1
𝜺r = Relative dielectric constant = 1100
A = Area of the sample = 1 cm × 1 cm = 1 cm𝟐 = 1 × 10-4 m𝟐
t = thickness of the sample = 1 mm = 1×10-3 m
Hence,
V = (𝟏𝟎×𝟏𝟎𝟎×𝟏𝟎-12)/{(𝟖.𝟖𝟓×𝟏𝟎-12×𝟏𝟏𝟎𝟎×𝟏×𝟏𝟎-4)/(𝟏×𝟏𝟎-3)} = 1.0272V = 1.02V (rounded off to two decimal place)
20) Silicon carbide (SiC) particles are added to aluminium (Al) matrix to fabricate particle reinforced Al-SiC composite. The resulting composite is required to process specific modulus (E/𝜌; E: elastic modulus, 𝜌: density) three times that of pure Al. Assuming iso-strain condition, the volume fraction of SiC particles in the composite will be _____ (round off to two decimal place)
|
Material |
E(GPa) |
|
|
Al |
69 |
2.70 |
|
SiC |
379 |
2.36 |
Correct Answer is "0.36 - 0.45"
Given that,
𝑬𝑪/𝝆𝑪 = 3× 𝑬M/𝝆M ................(1)
Where,
𝑬𝑪 = Elastic modulus of composite (Al – SiC)
𝑬M = Elastic modulus of Matrix (Al)
𝝆𝑪 = density of composite (Al – SiC)
𝝆M = density of Matrix (Al)
We know that,
𝑬𝑪 = 𝑬M𝑽M + 𝑬F𝑽F ....................(2)
𝝆𝑪 = 𝝆M𝑽M + 𝝆F𝑽F .....................(3)
Where,
𝑬F = Elastic modulus of fabricated reinforced particle (SiC)
𝝆F = Density of fabricated reinforced particle (SiC)
𝑽M & 𝑽F = Volume fraction of Matrix and reinforced particle respectively
Now from (1), (2) & (3) we get,
(𝑬M𝑽M + 𝑬F𝑽F)/(𝝆M𝑽M + 𝝆F𝑽F )= 3× 𝑬M/𝝆M
⇒ {𝑬M(𝟏−𝑽F) + 𝑬F𝑽F}/{𝝆M(𝟏−𝑽F) + 𝝆F𝑽F} = 3× 𝑬M/𝝆M (since 𝑽M + 𝑽F = 1)
⇒ 𝑽F = 𝟐𝝆M𝑬M/{(𝟐𝝆M𝑬M)+(𝝆M𝑬F)−( 𝟑𝝆F𝑬M)} = (𝟐×𝟐.𝟕𝟎×𝟔𝟗)/ {(𝟐×𝟐.𝟕𝟎×𝟔𝟗) + (𝟐.𝟕𝟎×𝟑𝟕𝟗) − (𝟑×𝟐.𝟑𝟔×𝟔𝟗)}
⇒ 𝑽F = 𝟑𝟕𝟐.𝟔𝟗𝟎𝟕.𝟑𝟖 = 0.4106 =0.41(rounded off to two decimal place)
Where,
𝑬𝑪 = Elastic modulus of composite (Al – SiC)
𝑬M = Elastic modulus of Matrix (Al)
𝝆𝑪 = density of composite (Al – SiC)
𝝆M = density of Matrix (Al)
We know that,
𝑬𝑪 = 𝑬M𝑽M + 𝑬F𝑽F ....................(2)
𝝆𝑪 = 𝝆M𝑽M + 𝝆F𝑽F .....................(3)
Where,
𝑬F = Elastic modulus of fabricated reinforced particle (SiC)
𝝆F = Density of fabricated reinforced particle (SiC)
𝑽M & 𝑽F = Volume fraction of Matrix and reinforced particle respectively
Now from (1), (2) & (3) we get,
(𝑬M𝑽M + 𝑬F𝑽F)/(𝝆M𝑽M + 𝝆F𝑽F )= 3× 𝑬M/𝝆M
⇒ {𝑬M(𝟏−𝑽F) + 𝑬F𝑽F}/{𝝆M(𝟏−𝑽F) + 𝝆F𝑽F} = 3× 𝑬M/𝝆M (since 𝑽M + 𝑽F = 1)
⇒ 𝑽F = 𝟐𝝆M𝑬M/{(𝟐𝝆M𝑬M)+(𝝆M𝑬F)−( 𝟑𝝆F𝑬M)} = (𝟐×𝟐.𝟕𝟎×𝟔𝟗)/ {(𝟐×𝟐.𝟕𝟎×𝟔𝟗) + (𝟐.𝟕𝟎×𝟑𝟕𝟗) − (𝟑×𝟐.𝟑𝟔×𝟔𝟗)}
⇒ 𝑽F = 𝟑𝟕𝟐.𝟔𝟗𝟎𝟕.𝟑𝟖 = 0.4106 =0.41(rounded off to two decimal place)
21) Isothermal weight gain per unit area (ΔW/A, where ΔW is the weight gain (in mg) and A is the area (in cm2)) during oxidation of a metal at 600 °C follows parabolic rate law, where, ΔW/A = 1.0 mg.cm-2 after 100 min of oxidation. The ΔW/A after 500 min at 600 °C will be _____ mg.cm-2 (round-off to two decimal places).
Correct Answer is "2.20 - 2.28"
According to the given problem the parabolic rate law during oxidation is given below,
𝑾𝟐 = 𝑲𝟏t + 𝑲𝟐 .......(1)
Where,
W = weight gain per unit area = Δ𝑾/𝑨 = 1.0 mg.𝒄𝒎-2
t = time of oxidation = 100 min
𝑲𝟏 & 𝑲𝟐 = rate constant
Given that, 1.0 = 𝑲𝟏 × 100 + 𝑲2 .....(2)
From initial condition, 0.0 = 𝐾𝟏 × 0 + 𝐾2 ⇒ 𝐾2 = 0
So, from (2) we get,
𝐾𝟏 = 1/100
Now, from (1) 𝑊2 = 1/100 ×500 + 0 = 5
⇒ W = 5 = 2.236 = 2.24 (rounded off to two decimal places)
𝑾𝟐 = 𝑲𝟏t + 𝑲𝟐 .......(1)
Where,
W = weight gain per unit area = Δ𝑾/𝑨 = 1.0 mg.𝒄𝒎-2
t = time of oxidation = 100 min
𝑲𝟏 & 𝑲𝟐 = rate constant
Given that, 1.0 = 𝑲𝟏 × 100 + 𝑲2 .....(2)
From initial condition, 0.0 = 𝐾𝟏 × 0 + 𝐾2 ⇒ 𝐾2 = 0
So, from (2) we get,
𝐾𝟏 = 1/100
Now, from (1) 𝑊2 = 1/100 ×500 + 0 = 5
⇒ W = 5 = 2.236 = 2.24 (rounded off to two decimal places)
22) A plain carbon steel sample containing 0.1 wt% carbon is undergoing carburization at 1100 °C in a carbon rich surroundings with fixed carbon content of 1.0 wt% all the time. The carburization time necessary to achieve a carbon concentration of 0.46 wt% at a depth of 5 mm at 1100 °C is ____hour (round off to the nearest integer).
Given: Diffusivity of carbon in iron at 1100 °C is 6.0 × 10−11 𝑚2.𝑠−1 and
|
erf(z) |
z |
|
0.56 |
0.55 |
|
0.60 |
0.60 |
|
0.64 |
0.65 |
|
0.68 |
0.70 |
Correct Answer is "77 - 83"
This a non-steady-state diffusion problem which can be solved by using the following formula
(𝐶𝑥− 𝐶0)/(𝐶s− 𝐶0)} = 1 – erf(𝑥2 /√𝐷𝑡) .....(1)
Where,
𝐶0 = initial concentration of the diffusing species = 0.1 wt%
𝐶𝑥 = concentration at position x after diffusion time t = 0.46 wt%
𝐶s = surface concentration of diffusing species = 1 wt%
D = diffusion coefficient or diffusivity = 6.0 × 10-11 𝑚2.𝑠-1
erf(𝑥2 /√𝐷𝑡) = Gaussian error function
x = depth = 5mm = 5×10-3 m
t = diffusion time = ?
According to the given problem, 𝐶− 𝐶𝑥𝐶𝑥− 𝐶0 = 0.46−0.11−0.1 = 0.4
From (1) we get,
erf(𝑥2/ 𝐷𝑡) = 1 - 𝐶𝑥− 𝐶𝑥𝐶𝑥− 𝐶𝑥 = 1 – 0.4 = 0.6
From the given data, when erf(𝑥2/√𝐷𝑡) = 0.6
Then, (𝑥2/ 𝐷𝑡) = 0.6
Dt = {𝑥2/(22×0.62)} = {(5×10-3)2}{2-3×0.62} = 1.736 × 10-5
t = {1.736 × 10-5/(6.0 × 10-11} = 289351.85 s = 80.37 hour = 80 hour (rounded off to the nearest integer)
(𝐶𝑥− 𝐶0)/(𝐶s− 𝐶0)} = 1 – erf(𝑥2 /√𝐷𝑡) .....(1)
Where,
𝐶0 = initial concentration of the diffusing species = 0.1 wt%
𝐶𝑥 = concentration at position x after diffusion time t = 0.46 wt%
𝐶s = surface concentration of diffusing species = 1 wt%
D = diffusion coefficient or diffusivity = 6.0 × 10-11 𝑚2.𝑠-1
erf(𝑥2 /√𝐷𝑡) = Gaussian error function
x = depth = 5mm = 5×10-3 m
t = diffusion time = ?
According to the given problem, 𝐶− 𝐶𝑥𝐶𝑥− 𝐶0 = 0.46−0.11−0.1 = 0.4
From (1) we get,
erf(𝑥2/ 𝐷𝑡) = 1 - 𝐶𝑥− 𝐶𝑥𝐶𝑥− 𝐶𝑥 = 1 – 0.4 = 0.6
From the given data, when erf(𝑥2/√𝐷𝑡) = 0.6
Then, (𝑥2/ 𝐷𝑡) = 0.6
Dt = {𝑥2/(22×0.62)} = {(5×10-3)2}{2-3×0.62} = 1.736 × 10-5
t = {1.736 × 10-5/(6.0 × 10-11} = 289351.85 s = 80.37 hour = 80 hour (rounded off to the nearest integer)
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