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MAP821S-MATERIALS PHYSICS-JAN 2020 |
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NAMIBIA UNIVERSITY
OF SCIENCE AND TECHNOLOGY
FACULTY OF HEALTH AND APPLIED SCIENCES
DEPARTMENT OF NATURAL AND APPLIED SCIENCES
QUALIFICATION : BACHELOR OF SCIENCE HONOURS
QUALIFICATION CODE: O08BOSH
LEVEL: 8
COURSE NAME: MATERIALS PHYSICS
COURSE CODE: MAP821S
SESSION: JANUARY 2020
DURATION: 3 HOURS
PAPER: THEORY
MARKS: 100
SUPPLEMENTARY/SECOND OPPORTUNITY EXAMINATION QUESTION PAPER
EXAMINER(S)
Prof Dipti R. Sahu
MODERATOR:
Dr Zivayi Chiguvare
INSTRUCTIONS
1. Answer any 5 of the 6 questions given.
2. Write clearly and neatly.
3. Number the answers clearly.
PERMISSIBLE MATERIALS
Non-programmable calculator
THIS QUESTION PAPER CONSISTS OF 5 PAGES (Including front page and formula sheet)
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Question 1
[20]
1.1 When making hardness measurements, what will be the effect of making an indentation (5)
very close to pre-existing indentation? Why?
1.2 For a brass alloy, the stress at which plastic deformation begins is 345 MPa, and the
modulus of elasticity is 103 GPa.
(a)
What is the maximum load that may be applied to a specimen with a cross-
sectional area of 130 mm? without plastic deformation?
(5)
(b)
If the original specimen length is 76 mm, what is the maximum length to
which it may be stretched without causing plastic deformation?
(5)
1.3
(a) What is tensile testing?
(3)
(b) Why is tensile testing important?
(2)
Question 2
[20]
2.1 Explain why, on a cold day, the metal door handle of an automobile feels colder to the (5)
touch than a plastic steering wheel, even though both are at the same temperature.
2.2 Railroad tracks made of 1025 steel are to be laid during the time of year when the (5)
temperature averages 4°C. If a joint space of 5.4 mm is allowed between the standard
11.9 m long rails, what is the highest possible temperature that can be tolerated without
the introduction of thermal stresses?
For these railroad tracks, each end can expand one-half of the joint space distance, or
the track may expand a total of this distance (4.6 mm). The value of a for the 1025 steel
is 12.0 x 10° (°C)?.
2.3
(a)
Define thermal stress.
(3)
(b)
Explain how thermal stress resulting from restrained thermal expansion
(7)
and contraction varies.
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Question 3
[20]
3.1 Which of the following oxide materials when added to fused silica (SiO2) will increase its (5)
index of refraction: AlzO3, TiO2z, NiO, MgO? Why?
3.2. The fraction of non-reflected light that is transmitted through a 200-mm thickness of
(5)
glass is 0.98. Calculate the absorption coefficient of this material.
3.3. (a)
Define photoconductivity.
(3)
(b)
State two applications of photoconductivity
(2)
(c)
Explain, giving reasons, whether the semiconductor Zinc Selenide (ZnSe), which (5)
has a band gap of 2.58 eV, is, or is not, photoconductive when exposed to visible
light radiation.
Question 4
[20]
4.1 Explain the following:
(a)
dielectric loss
(2)
(b) dielectric break down
(2)
(c) local electric field
(3)
(d) polarizability
(3)
4.2 10%? m? arsenic atoms are added to high-purity silicon.
(a)
Is the resulting material n-type or p-type? Explain your answer.
(2)
(b)
Given that the charge of electron = -1.6 x 10°C, and that electron mobility = (3)
0.07 m?/V/s; calculate the room-temperature electrical conductivity of this
material.
4.3 Briefly explain why the ferroelectric behaviour of BaTiO3 ceases above its ferroelectric (5)
Curie temperature.
Question 5
[20]
5.1 Given that the saturation magnetization for Fe2O3 is 5.0 x 10° A/m, and that the unit
(5)
cell edge length of ferrite is 0.839 nm, design a cubic mixed-ferrite magnetic material
that has a saturation magnetization of 5.25 x 10°A/m.
5.2 Schematically sketch on a single plot the B-versus-H behaviour for a ferromagnetic
material
(a)
at OK,
(2)
(b)
at a temperature just below its Curie temperature, and
(2)
(c)
at a temperature just above its Curie temperature.
(2)
(d)
Briefly explain why these curves have different shapes.
(4)
5.3. State the differences between soft magnetic materials from hard magnetic materials. (5)
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Question 6
[20]
6.1 In the table below are listed four hypothetical aligned fibre-reinforced composites (5)
(labelled A, B, D, and D), along with their characteristics. On the basis of these data,
rank the four composites from highest to lowest strength in the longitudinal direction,
and then justify your ranking.
Composite
A
B
C
D
Fibre
type
Glass
Glass
Carbon
Carbon
Vol.
Fraction
fibre
0.20
0.35
0.40
0.30
Fibre
Strength
(MPa)
3.5 x 10°
3.5 x 10°
5.5 x 10°
5.5 x 10°
Ave. Fibre
length
(mm)
8
12
8
8
Critical
length
(mm)
0.70
0.75
0.40
0.50
6.2 Differentiate between polymorphism and isomerism.
(5)
6.3 Sketch the repeat structure for each of the following alternating copolymers:
(a) poly (ethylene-propylene)
(3)
(b) poly(butadienestyrene)
(c) poly(isobutylene-isoprene).
(4)
END
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Formula Sheet: Materials Physics
Mechanical properties: stress o=F/A, strain ¢=(I-lo)/lo=Al/lo , stress-strain curve o = f(e) = Ee
shear stress t=F/A, shear strain Ab/h=tan y, t = G tan y, compressibility AV/Vo = -Kp = -p/K
K = E/(3(1-2v)) G=E/(2(1+v)) E/G = 9/(3+(G/K))
Elastic energy Weer= E = J. F(s)ds = i Dsds = +2 DS? orE=222E02 = +2 8¢?
Thermal properties: Heat capacity C = AQ/AT, specific heat capacity c = AQ/(mAT)
Thermal expansion | - lo = (Ti - To), AV = yYAT
Heat conductivity and heat transition: = =O = -as |grad(T)| = ASAT = as (T1—T2)
H. transfer: “2 = @ = @AAT = aA(T1—T2), h. transition: 2 = Q =kAAT =kA(T1-T2)
_ 2897,8 pm k
Stefan-Boltzmann law: = oAT* , Wien‘s displacement law: “"** —
T
Optical properties: Snell‘s law: nisin(c) = nasin(B),
Some of Fresnel‘sj laws: . reflecti.on coeff. % =__ ttaann((aat—pB))
3 _= __siSnn(was-fp)y
transmission coeff. ts=r;+1 , natp=ni(rp+1) , reflectivity p=r’, transmittivity t=(n2cosB)/(nicosa)t?,
Brewster angle: tan Og =n2/Mn:. critical angle: sin &¢ =n2/n1, spectr. reflectivity RA) = Le (A)
Lambert-Beer law: J,(x,A) =1,(A) exp|-@(A)x| I,(x, A) = Ip(A)10~-?
I,(A)
-Ig(I/Iy) = ODBel
Abbe number: v=(n(green)-1)/(n(blue)-n(red))
Electrical properties: resistance R = pL/A, electrical conductivity o = 1/p, p(T) = p(To)[1 + B(T — To)]
Current densityj = I/A = Q/tA = neAL/At = nev, electron mobility v = pteE, Lorentz force:
F = q(@ x B), capacity ofa plate capacitor Cy = & “, E= ErE05 flux density D=e,e9E
Susceptibility ye = €-1, P=E CeE,
Magnetic properties: MF of a straight wire: H(r) = Up “ey , coil: H = ~ magn. flux density:
Bo = HoH, B = uo,, B= uo + uM =po(H+M) = uo(H+%mH) = woh + xm)
Faraday effect: 8 = VdB
Metallic materials: Force on charged particle in field E: Fo = qE = mo drift velocity: vp = =TF
;
2
Conductivity o = = = &= thermo voltage Urn = (Ss — Sa) DT
Magnetic materials: magn. moment:m = IAa ,m = mug, = Ge be;
Etchin. g: Ani° sotropy: A= 1 — —v#27@1 4 = 1 — 44Vv.
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