ph8151 - engineering physics dept. of physics multiple

PH8151 -Engineering Physics

Dept. of Physics

Multiple ChoiceQuestions (MCQ)

Regulations 2017(Common to All)

UNIT I PROPERTIES OFMATTER

TOPIC 1.1 ELASTICITY

1. The property by which a body returns to itsoriginal shape after removal of the force iscalled __________

a) Plasticity b) Elasticity c) Ductility

d) Malleability

Answer: b Explanation: When an external force acts on

a body, the body tends to undergo somedeformation. If the external force is removedand the body comes back to its original shapeand size, the body is known as elastic bodyand this property is called elasticity.

2. The property of a material by which it canbe beaten or rolled into thin plates is called__________a) Malleabilityb) Plasticityc) Ductilityd) Elasticity

Answer: aExplanation: A material can be beaten intothin plates by its property of malleability.

3. Which law is also called as the elasticitylaw?a) Bernoulli’s lawb) Stress lawc) Hooke’s lawd) Poisson’s law

Answer: cExplanation: The hooke”s law is valid underthe elastic limit of a body. It itself states thatstress is proportional to the strain within theelastic limit.

4. The materials which have the same elasticproperties in all directions are called__________a) Isotropicb) Brittlec) Homogeneousd) Hard

Answer: aExplanation: Same elastic properties in alldirection is called the homogenity of amaterial.

5. A member which does not regain itsoriginal shape after removal of the loadproducing deformation is said __________a) Plasticb) Elasticc) Rigidd) None of the mentioned

Answer: aExplanation: A plastic material does not

Department of Science and Humanities MCQ for Regulations 2017

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regain its original shape after removal of load.An elastic material regain its original shapeafter removal of load.

6. The body will regain it is previous shapeand size only when the deformation causedby the external forces, is within a certainlimit. What is that limit?a) Plastic limitb) Elastic limitc) Deformation limitd) None of the mentioned

Answer: bExplanation: The body only regain itsprevious shape and size only upto its elasticlimit.

7. The materials which have the same elasticproperties in all directions are called__________a) Isotropicb) Brittlec) Homogenousd) Hard

Answer: aExplanation: Isotropic materials have thesame elastic properties in all directions.

8. As the elastic limit reaches, tensile strain__________a) Increases more rapidlyb) Decreases more rapidlyc) Increases in proportion to the stressd) Decreases in proportion to the stress

Answer: aExplanation: On reaching the tensile stress tothe elastic limit after the proportionality limit,the stress is no longer proportional to thestrain. Then the value of strain rapidlyincreases.

9. What kind of elastic materials are derivedfrom a strain energy density function?a) Cauchy elastic materialsb) Hypo elastic materials

c) Hyper elastic materialsd) None of the mentioned

Answer: cExplanation: The hyper elastic materials arederived from a strain energy density function.A model is hyper elastic if and only if it ispossible to express the cauchy stress tensor asa function of the deformation gradient.

10. What the number that measures anobject’s resistance to being deformedelastically when stress is applied to it?a) Elastic modulusb) Plastic modulusc) Poisson’s ratiod) Stress modulus

Answer: aExplanation: The elastic modulus is the ratioof stress to strain.

TOPIC 1.2 STRESS-STRAINDIAGRAM AND ITS USES

1. The slope of the stress-strain curve in theelastic deformation region is ____________

a) Elastic modulus b) Plastic modulus c) Poisson’s ratio

d) None of the mentioned

Answer: a Explanation: The elastic modulus is the ratio

of stress and strain. So on the stress straincurve, it is the slope.

2. What is the stress-strain curve? a) It is the percentage of stress and stain

b) It is the relationship between stress andstrain

c) It is the difference between stress andstrain


Answer: b Explanation: The relationship between stress


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and strain on a graph is the stress strain curve.It represents the change in stress with changein strain.

3. Which point on the stress strain curveoccurs after the proportionality limit?a) Upper yield pointb) Lower yield pointc) Elastic limitd) Ultimate point

Answer: cExplanation: The curve will be stress strainproportional upto the proportionality limit.After these, the elastic limit will occur.

4. Which point on the stress strain curveoccurs after the lower yield point?a) Yield plateaub) Upper yield pointc) Ultimate pointd) None of the mentioned

Answer: aExplanation: The points on the curve comesin the given order,A. proportionality limitB. elastic limitC. upper yield pointD. lower yield pointE. yield plateauF. ultimate pointG. breaking point.

5. Which point on the stress strain curveoccurs after yield plateau?a) lower yield pointb) Upper yield pointc) Ultimate pointd) Breaking point

Answer: cExplanation: After the yield plateau thecurve will go up to its maximum limit ofstress which is its ultimate point.

6. Which point on the stress strain curveoccurs after the ultimate point?

a) Last pointb) Breaking pointc) Elastic limitd) Material limit

Answer: bExplanation: After the ultimate point thevalue of stress will reduce on increasing ofstrain and ultimately the material will break.

7. Elastic limit is the point ____________a) up to which stress is proportional to strainb) At which elongation takes place withoutapplication of additional loadc) Up to which if the load is removed,original volume and shapes are regainedd) None of the mentioned

Answer: cExplanation: The elastic limit is that limit upto which any material behaves like an elasticmaterial.

8. What is the point P shown on the stressstrain curve?

a) Upper yield point b) Yield plateau

c) Elastic limit d) Ultimate point

Answer: d

Explanation: It is the point showing themaximum stress to which the material can besubjected in a simple tensile stress.


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9. What is the point P shown in the stress-strain curve?

a) Lower yield point b) Elastic limit

c) Proportionality limit d) Breaking point

Answer: d

Explanation: The breaking point is the pointwhere the material breaks. The breaking pointwill be the last point on the stress straincurve.

10. What is the point shown in the stressstrain curve?

a) Elastic limit b) Lower yield point

c) Yield plateau d) Lower strain point

Answer: b

Explanation: It is the lower yield point atwhich the curve levels off and plasticdeformation begins.

11. Where is the necking region?a) The area between lower yield point andupper yield pointb) The area between the plastic limit andelastic limitc) The area between the ultimate point andinitial pointd) The area between the ultimate point andrupture

Answer: dExplanation: Necking is a tensile straindeformation which is cased in after theultimate amount of stress occurs in thematerial.

TOPIC 1.3 FACTORSAFFECTING ELASTICMODULUS AND TENSILESTRENGTH

1. What is the elastic modulus of steel? a) 69-79 GPa

b) 41-45 GPa c) 190-217 GPa

d) 330-360 GPa

Answer: c Explanation: Steel has an elastic modulus of

190-217 GPa. It has E higher than aluminumand magnesium alloys. But lower thantungsten and molybdenum alloys.

2. What is the elastic modulus of titaniumalloys?

a) 150-170 GPa b) 180-214 GPa c) 80-130 GPa

d) 41-45 GPa

Answer: c Explanation: Titanium alloys have the elastic

modulus in the range of 80 to 130 GPa. It isgreater than aluminum and magnesium alloysbut lesser than steel.


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3. What is the order of elastic modulus forNickel alloys, Lead alloys, Molybdenumalloys, alumina?a) Pb < Mo < Ni < Al2O3b) Ni < Pb < Mo < Al2O3c) Pb < Ni < Mo < Al2O3d) Pb < Ni < Al2O3 < Mo

Answer: cExplanation: Lowest elastic modulus is ofLead and its alloys of 14-18 GPa. Ni and Moalloys have elastic modulus of 180-214 and330-360 GPa respectively. Al2O3 has highestE among the four of 415 GPa.

4. What property enhances with a decrease inE?a) Flexibilityb) Stiffnessc) Hardnessd) UTS

Answer: aExplanation: The lower the elastic modulusof material, the more flexible it is. Bending ofmaterial becomes easier. On the other hand,stiffness decreases.

5. What is the unit of elastic modulus?a) Mohsb) GPac) Kgd) N

Answer: bExplanation: Elastic modulus is expressed inGPa. Mohs is the unit used for hardness.Stress is measured in MPa.

6. When applied stress is of shear type, themodulus of elasticity is known as ___a) Bulk modulusb) Modulus of resiliencec) Shear modulusd) Stiffness

Answer: cExplanation: When the type of stress appliedis shear, E is known as shear modulus. It isalso known as modulus of rigidity. Stiffness ismeasured in terms of E.

7. Which of the following tensile property isdimensionless?a) Tensile stressb) Elastic modulusc) True straind) Toughness

Answer: cExplanation: True strain is dimensionlessproperty. Tensile strength and elastic modulusare measured in MPa and GPa respectively.

8. What is a factor which controls the elasticmodulus?a) Alloyingb) Heat treatmentc) Interatomic forcesd) Cold working

Answer: cExplanation: On a constant temperature andpressure, E only is function of two factors.First is type of interatomic forces. Second isarrangement of atoms or crystal structure.

9. Stress should not exceed ___ when inservice.a) Yield strengthb) Tensile strengthc) Fracture strengthd) Toughness

Answer: aExplanation: Above yield strength, materialstarts to deform plastically. It causes a changein dimensions and properties of a material. Somaterial should be used below it.

10. Stress should not exceed ___ when inmechanical workinga) Yield strengthb) Tensile strength


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c) Fracture strengthd) Toughness

Answer: bExplanation: Mechanical working processconsists of plastic deformation. So it isperformed above yield strength. But it islimited below tensile strength so as to avoidfracture.

11. Ductility of material is its ability to flowplastically under compressive load.a) Trueb) False

Answer: bExplanation: Ductility is a tensile property. Itis studied under tensile loading. It is theability to plastic flow without rupture.

12. Work per unit volume of the material is aknown modulus of toughness.a) Trueb) False

Answer: aExplanation: Toughness is measured by theamount of work per unit volume of thematerial under static loading. Work per unitvolume of material is called modulus oftoughness.

TOPIC 1.4 TORSIONAL STRESSAND DEFORMATIONS

1. Which of the following cannot bedetermined using a torsion test?

a) Modulus of elasticity in shear b) Torsion yield strength

c) Modulus of rupture d) Young’s modulus

Answer: d

Explanation: Modulus of elasticity in shear,torsion yield strength and modulus of rupturecan all be determined by performing torsiontest on material.

2. What is the use of weight head in a torsiontesting equipment?a) Holding the job onlyb) Holding the job and applying twistingmomentc) Holding the job and measuring the twistingmomentd) It is not a part of torsion testing equipment

Answer: cExplanation: The main job of weight head isto hold the job and measure the twistingmoment.while twisting head holds the otherend of job and applies twisting moment.

3. Which of the following is used to measurehow much the specimen is twisted?a) Micrometerb) Clinometerc) Troptometerd) Tropometer

Answer: cExplanation: Troptometer is an instrumentwhich is used for measuring the angulardistortion of the material. Mocrometer andvernier callipers are used to measure length.Tropometer measures amount of torsion for abone.

4. Torsional stress multiplied with originalcross sectional are is:a) Maximum twisting loadb) Minimum twisting loadc) Minimum shear loadd) Yield shear load

Answer: aExplanation: Torsional stress is given by theratio of maximum twisting load and originalare of cross section of the material. Therefore,torsional stress multiplied with original crosssectional gives us maximum twisting load.

5. Plastic deformation can only occur in caseof torsional force.a) Trueb) False


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Answer: bExplanation: The above given statement isfalse as plastic deformation can occur in caseof tensile, compressive and torsional loadafter a point. After this point, the body cannotrecover its original shape.

6. What is the unit of polar moment ofinertia?a) m2

b) m5

c) m3

d) m4

Answer: dExplanation: Polar moment of inertiadenoted by J, is given by integration of radiussquare with respect to small area of cross-section. Hence the unit is (m)(m)(m)2 whichis equal to m4.

7. Shear stress on a solid bar and hollow baris same for given dimension.a) Trueb) False

Answer: bExplanation: Shear stress for a hollow barand a solid bar are different dimensions as thehollow bar has two dimensions, outer andinner radius because of which calculation isdifferent than the solid bar which has onlyone diameter.

8. In which of the following the angle of twistincreases fast for a small amount of torque?a) Cold working conditionb) Hot working conditionc) Warm working conditiond) The increase is the same for cold working,hot working and warm working

Answer: bExplanation: When the torsion test isconducted in hot working, it is observed thatfor a slight change in torque on the givenspecimen the angle of twist increase fast as

the material becomes soft at hot workingtemperature.

TOPIC 1.5 TWISTING COUPLE

1. Torque is __________ moment. a) Twisting

b) Shear c) Bending

d) Couple

Answer: a Explanation: A cylindrical shaft is subjected

to twisting moment or torque when a force isacting on the member tangentially at someradius in a plane of its cross section.

2. Twisting moment is a product of__________ and the radius.

a) Direction b) Velocity

c) Force d) Acceleration

Answer: c

Explanation: Twisting moment will be equalto the product of force and radius. When ashaft is subjected to a twisting moment, everycross section of the shaft will surelyexperience shear stress.

3. Torsion is denoted by __________ a) R

b) Q c) T d) N

Answer: c Explanation: If the moment is applied in a

plane perpendicular to the longitudinal axis ofthe beam (or) shaft it will be subjected totorsion. Torsion is represented or denoted byT.

4. The SI units for torsion is __________ a) N m

b) N


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c) N/md) m

Answer: aExplanation: As torsion is a product ofperpendicular force and radius, the units willbe N m.Torque is also known as torsion or twistingmoment or turning moment.

5. _____________ torsion is produced whentwisting couple coincides with the axis of theshaft.a) Exactb) Purec) Nominald) Mild

Answer: bExplanation: When a member is subjected tothe equal and opposite twisting moment at itsends, then the member is said to be subjectedunder pure torsion. Pure Torsion is oftenproduced when the axis of the twisting couplecoincides with the axis of the shaft.

6. Which of the following is known as Re-entrant mouthpiece?a) External Mouthpieceb) Convergent Mouthpiecec) Internal Mouthpieced) Cylindrical Mouthpiece

Answer: cExplanation: According to the position,mouthpieces are classified as an externalmouthpiece and internal mouthpiece. If thetube projects inside the tank, it is called aninternal mouthpiece or re-entrant or borda’smouthpiece.

7. Micrometre contraction gauge is used todetermine ___________a) Cvb) Ccc) Cad) Cd

Answer: bExplanation: The coefficient of contractionmay be determined experimentally bymeasuring the radius of jet as vena contactwith the help of micro meter contractiongauge. This method is not accurate because itis very difficult to measure the correct radiusof jet.

8. What is the general value for coefficient ofcontraction?a) 0.64b) 0.67c) 0.66d) 0.7

Answer: aExplanation: The ratio of the area of a jet atvena contracta to the area of orifice is knownas the coefficient of contraction. The value ofCc varies from 0.61 to 0.69 for differentorifices. Generally, for sharp edged orifice thevalue of Cc may be taken as 0. 64.

9. The Cd value for internal mouthpiecerunning free is __________a) 0.6b) 0.5c) 0.7d) 0.8

Answer: bExplanation: The Cd value for internalmouthpiece running free is 0.5.

Type Of Mouthpiece Value ofCd

External cylindricalmouthpiece 0.855

Internal mouthpiece runningfree 0.5

Internal mouthpiece runningfull 0.707

10. _______ is the velocity with which waterreaches the notch or before it flows over it.a) Velocity of contactb) Velocity of moment


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c) Velocity of approachd) Velocity of head

Answer: cExplanation: The velocity of approach isdefined as the velocity with which waterreaches the notch or weir before it flows overit. This velocity of approach creates anadditional head “ha” equal to Va2 / 2g andeffect head over the notch is increased toH+ha.

11. Which of the following formula wasproposed by Bazin?a) m(2g)1/2×LH3/2

b) m(2g) 1/2×H3/2

c) n(2g) 1/2×LH4/3

d) n(2g)1/2×LH3/2

Answer: aExplanation: Bazin proposed the followingformula for the discharge over rectangularweir:Q = m(2g) 1/2× L H3/2.Where m = 0.405 + 0.003/H.

12. For measuring low discharges_____________ notch is preferred.a) Rectangularb) Steppedc) Trapezoidald) Triangular

Answer: dExplanation: A triangular notch is preferredto a rectangular notch due toi. The nappe emerging from a triangular notchhas the same shape for all heads. As such thevalue for the triangular notch is constant forall heads.ii. The expression for discharge for rightangle triangle law not is very simple.

13. Which of the following is also known asV notch?a) Trapezoidalb) Stepped

c) Triangulard) Sharp edged

Answer: cExplanation: A triangular notch also called av notch is of triangle shape with apex down.The expression of the discharge overtriangular notch or weir is Q = 8/15 Cd (2g)1/2 × H 5/2.

14. Calculate the discharge over rectangularWeir of 3 metres length under the head of400mm.Use Francis formula.a) 1.268 m3/sb) 1.396 m3/sc) 1.475 m3/sd) 1.528 m3/s

Answer: bExplanation: Francis formula for dischargeQ = 1.84 LH3/2.Given L = 3m & H = 0.4mQ = 1.84 × 3 × (0.4)3/2.Q = 1.396 m3/s.

15. _____ converts mechanical energy intohydraulic energy.a) Dynamob) Pumpc) Turbined) Generator

Answer: bExplanation: A pump is a mechanical devicewhich converts the mechanical energy intohydraulic energy. The hydraulic energy is inthe form of pressure energy. The pumps aregenerally used for lifting liquid from a lowerlevel to a higher level.

TOPIC 1.6 TORSIONPENDULUM: THEORY ANDEXPERIMENT


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1. Torsional sectional modulus is also knownas _________a) Polar modulusb) Sectional modulusc) Torsion modulusd) Torsional rigidity

Answer: aExplanation: The ratio of polar moment ofinertia to radius of section is called Polarmodulus or Torsional section modulus. Itsunits are mm3 or m3 (in SI).

2. ________ is a measure of the strength ofshaft in rotation.a) Torsional modulusb) Sectional modulusc) Polar modulusd) Torsional rigidity

Answer: cExplanation: The polar modulus is a measureof the strength of shaft in rotation. As thevalue of Polar modulus increases torsionalstrength increases.

3. What are the units of torsional rigidity?a) Nmm2

b) N/mmc) N-mmd) N

Answer: aExplanation: The product of modulus ofrigidity (C) and polar moment of inertia (J) iscalled torsional rigidity. Torsional rigidity is atorque that produces a twist of one radian in ashaft of unit length.

4. The angle of twist can be written as________a) TL/Jb) CJ/TLc) TL/CJd) T/J

Answer: cExplanation: The angle of Twist = TL/CJ

Where T = Torque in NmL = Length of shaftCJ = Torsional rigidity.

5. The power transmitted by shaft SI systemis given by __________a) 2πNT/60b) 3πNT/60c) 2πNT/45d) NT/60 W

Answer: aExplanation: In SI system, Power (P) ismeasured in watts (W) ; P = 2πNT/60Where T = Average Torque in N.mN = rpm= 2πNT/ 45 1 watt = 1 Joule/sec = 1N.m/s.

6. Area of catchment is measured in___________a) mm3

b) Km2

c) Kmd) mm

Answer: bExplanation: Catchment area can be definedas the area which contributes the surpluswater present over it to the stream or river. Itis an area which is responsible formaintaining flow in natural water bodies. It isexpressed in square kilometres.

7. ______ catchment area is a sum of freecatchment area and intercepted catchmentarea.a) Totalb) Additionalc) Combinedd) Overall

Answer: cExplanation: Combined catchment area isdefined as the total catchment area whichcontributes the water in to stream or a tank.Combined Catchment area = Free catchmentarea + intercepted catchment area.


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8. ___________ has steep slopes and givesmore run off.a) Intercepted Catchment Areab) Good Catchment Areac) Combined Catchment Aread) Average Catchment Area

Answer: bExplanation: Good catchment area consistsof hills or rocky lands with steep slopes andlittle vegetation. It gives more run off.

9. How many number of rain gauge stationsshould be installed an area between 250 to500 km2.a) 2b) 4c) 3d) 5

Answer: cExplanation: 3 number of rain gauge stationsshould be installed an area between 250 to500 km2.Area ofBasin(Km2)

Number of Rain gaugestations

< 125 1125 – 250 2250 – 500 3

10. Trend of rainfall can be studied from_______a) Rainfall graphsb) Rainfall recordsc) Rainfall curvesd) Rainfall cumulatives

Answer: bExplanation: Rainfall records are useful forcalculating run off over a basin. By usingrainfall records estimate of design parametersof irrigation structures can be made. Themaximum flow due to any storm can becalculated and predicted.

11. Estimation of run off “R” is 0.85P-30.48.The above formula was coined by _____

a) Laceyb) Darcyc) Khoslad) Ingli

Answer: dExplanation: Run off can be estimated byR= 0.85P-30.48Where R = annual runoff in mmP = annual rainfall in mm.

12. Monsoon duration factor is denoted by________a) Pb) Sc) Fd) T

Answer: cExplanation: Monsoon duration factor isdenoted by F.Class ofMonsoon

Monsoon DurationFactor (F)

Very Short 0.5Standardlength 1.0

Very long 1.5

13. Runoff coefficient is denoted by _______a) Pb) Nc) Kd) H

Answer: cExplanation: The runoff coefficient can bedefined as the ratio of runoff to rainfall.Rainfall and runoff can be interrelated byrunoff coefficient.R = KPK = R/P [K = is a runoff Coefficientdepending on the surface of the catchmentarea].

14. _________ is a graph showing variationsof discharge with time.a) Rising limb graph


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b) Crest graphc) Hydraulic graphd) Gauge graph

Answer: cExplanation: Hydrograph is a graph showingvariations of discharge with time at aparticular point of the stream. Thehydrograph shows the time distribution oftotal run off at a point of measurement.Maximum flood discharge can also becalculated by using hydrograph.

15. Calculate the torque which a shaft of 300mm diameter can safely transmit, if the shearstress is 48 N / mm2.a) 356 kNmb) 254 kNmc) 332 kNmd) 564 kNm

Answer: bExplanation: Given, the diameter of shaft D= 300 mmMaximum shear stress fs = 48 N/mm2.Torque = T = π/16 fs D3

= 254469004.9 Nmm= 254 kNm.

TOPIC 1.7 BENDING OF BEAMS.BENDING MOMENT

1. What is the bending moment at endsupports of a simply supported beam?

a) Maximum b) Minimum c) Zero

d) Uniform

Answer: c Explanation: At the end supports, the

moment (couple) developed is zero, becausethere is no distance to take the perpendicularacting load. As the distance is zero, themoment is obviously zero.

2. What is the maximum shear force, when acantilever beam is loaded with udlthroughout?a) w×lb) wc) w/ld) w+l

Answer: aExplanation: In cantilever beams, themaximum shear force occurs at the fixed end.In the free end, there is zero shear force. Aswe need to convert the udl in to load, wemultiply the length of the cantilever beamwith udl acting upon. For maximum shearforce to obtain we ought to multiply load anddistance and it surely occurs at the fixed end(w×l).

3. Sagging, the bending moment occurs at the_____ of the beam.a) At supportsb) Mid spanc) Point of contraflexured) Point of emergence

Answer: bExplanation: The positive bending momentis considered when it causes convexitydownward or concavity at top. This issagging. In simply supported beams, it occursat mid span because the bending moment atthe supports obviously will be zero hence thepositive bending moment occurs in the midspan.

4. What will be the variation in BMD for thediagram? [Assume l = 2m].

a) Rectangularb) Trapezoidal


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c) Triangulard) Square

Answer: cExplanation: At support B, the BM is zero.The beam undergoes maximum BM at fixedend.By joining the base line, free end andmaximum BM point. We obtain a right angledtriangle.

5. What is the maximum bending moment forsimply supported beam carrying a point load“W” kN at its centre?a) W kNmb) W/m kNmc) W×l kNmd) W×l/4 kNm

Answer: dExplanation: We know that in simplysupported beams the maximum BM occurs atthe central span.Moment at A = Moment at B = 0Moment at C = W/2 × l/2 = Wl/ 4 kNm(Sagging).

6. How do point loads and udl be representedin SFD?a) Simple lines and curved linesb) Curved lines and inclined linesc) Simple lines and inclined linesd) Cant represent any more

Answer: cExplanation: According to BIS, the standardsymbols used for sketching SFD arePoint load = ———–Udl load = \

7. ________ curve is formed due to bendingof over hanging beams.a) Elasticb) Plasticc) Flexurald) Axial

Answer: aExplanation: The line to which thelongitudinal axis of a beam bends or deflectsor deviates under given load is known aselastic curve on deflection curve. Elasticcurve can also be known as elastic line orelastic axis.

8. The relation between slope and maximumbending moment is _________a) Directly proportionb) Inversely proportionc) Relative proportiond) Mutual incidence

Answer: bExplanation: The relationship between slopeand maximum bending moment is inverselyproportional because, For example in simplysupported beams slope is maximum atsupports and zero at midspan of a


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symmetrically loaded beam where as bendingmoment is zero at supports and maximum atmid span. Hence we conclude that slope andmaximum bending moment are inverselyproportional to each other in a case of thesimply supported beam.

9. What is the SF at support B?

a) 5 kNb) 3 kNc) 2 kNd) 0 kN

Answer: dExplanation: Total load = 2×2 = 4kNShear force at A = 4 kN ( same between Aand C )Shear force at C = 4 kNShear force at B = 0 kNMaximum SF at A = 4 kN.

10. Where do the maximum BM occurs forthe below diagram.

a) -54 kNmb) -92 kNmc) -105 kNmd) – 65 kNm

Answer: cExplanation: Moment at B = 0Moment at C = – (10 × 3) × (3/2)= – 45 kNmMoment at A = – (10 × 3) × (1.5 + 2 )

Maximum BM at A = – 105 kNm= 105 Nm (hogging).

TOPIC 1.8 CANTILEVER:THEORY AND EXPERIMENT.

1. The ratio of maximum deflection of a beamto its ___________ is called stiffness of thebeam.

a) Load b) Slope c) Span

d) Reaction at the support

Answer: c Explanation: The stiffness of a beam is a

measure of it’s resistance against deflection.The ratio of the maximum deflection of abeam to its span can be termed as stiffness ofthe beam.

2. Stiffness of the beam is inverselyproportional to the _____ of the beam.

a) Slope b) Support reaction

c) Deflection d) Load

Answer: c

Explanation: Stiffness of a beam is inverselyproportional to the deflection. Smaller thedeflection in a beam due to given externalload, greater is its stiffness.

3. The maximum ____ should not exceed thepermissible limit to the span of the beam.

a) Slope b) Deflection

c) Load dl Bending moment

Answer: b

Explanation: The maximum deflection of aloaded beam should not exceed thepermissible limit in relation to the span of abeam. While designing the beam the designer


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should be keep in mind that both strength andstiffness criteria.

4. In cantilever beam the deflection occurs at______a) Free endb) Point of loadingc) Through outd) Fixed end

Answer: aExplanation: Deflection can be defined asthe perpendicular displacement of a point onstraight access to the curved axis. Incantilever beams, the maximum deflectionoccurs at free end.

5. The maximum deflection in cantileverbeam of span “l”m and loading at free end is“W” kN.

a) Wl3/2EI b) Wl3/3EI c) Wl3/4EI d) Wl2/2EI

Answer: b Explanation: Maximum deflection occurs at

free end distance between centre of gravity ofbending moment diagram and free end is x =2l/3.

As deflection is equal to the slope × “x”. Theslope = Wl2/2EI radians

Maximum deflection (y) = Ax/EI = Wl3/3EI.

6. In an ideal fluid, the ____________stresses are pretend to be absent.

a) Bending b) Shearing

c) Tensiled) Compressive

Answer: bExplanation: An ideal fluid is a fluid wherethere is no resistance to the deformation. IdealFluids are those Fluids which have noviscosity surface tension. The shear stress isalso absent. This fluid is also called as perfectfluid.

7. Air and water are the examples of___________a) Non Newtonian fluidsb) Vortex fluidsc) Real fluidsd) Ideal fluids

Answer: dExplanation: The ideal Fluids are imaginaryfluids in nature, they are incompressible.These fluids possess low viscosity. Air andwater are considered as ideal fluids.

8. _______ fluids are practical fluidsa) Idealb) Realc) Vortexd) Newtonian

Answer: bExplanation: These fluids possess propertiessuch as viscosity, surface tension. They arecompressible in nature. The certain amount ofresistance is always offered by the fluids, theyalso possess shear stress. They are alsoknown as practical fluids.

9. Specific weight of water at 4°C is____________ N/m3.a) 9810b) 9760c) 9950d) 9865

Answer: aExplanation: The specific weight (weightdensity) of a fluid is weight per unit volume.


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It is represented by symbol w & it isexpressed in Newton per metre cube (N/m3).The specific weight of water at 4 degreecentigrade is 9810 N/m3or 9.81 kN/m3.

10. The inverse of specific weight of a fluid is__________a) Specific gravityb) Specific Volumec) Compressibilityd) Viscosity

Answer: bExplanation: Specific volume is the volumeof the fluid by Unit Weight it is the reciprocalof specific weight is denoted by “v”. SI unitsare m3/N.v= 1/specific weight.

11. Calculate the specific gravity of mercury.a) 12.5b) 14.7c) 13.6d) 11.8

Answer: cExplanation: The specific gravity of anyfluid is the ratio of the specific weight of fluidby specific weight of water. For mercury, thespecific weight is 133416 N/m3. For water, w= 9810 N/m3.S = 133416/9810S= 13.6.

12. Specific gravity of water is __________a) 0.8b) 1c) 1.2d) 1.5

Answer: bExplanation: The specific gravity is alsocalled as relative density. It is dimensionlessquantity and it has no units. The specificgravity of water is the ratio of specific weightof fluid to specific weight of water, as both

the numerator and denominator are same. Thevalue is 1.

13. Compute the maximum deflection at freeend of a cantilever beam subjected to udl forentire span of l metres.a) wl4/8EIb) wl4/4EIc) wl3/8EId) wl2/6EI

Answer: aExplanation: The slope at free end = A/EI =wl3/6EIMaximum deflection at free end is Ax/EI; [x=¾ l] y= wl3/6EI × ¾ l = wl4/8EI.

14. Calculate the maximum deflection of acantilever beam with udl on entire span of 3mthe intensity of you udl be 25 kN/m. Take EIas 4000 kN/m2.a) 0.052mb) 0.063mc) 0.076md) 0.09m

Answer: bExplanation: For cantilever beams with udlon entire span, the maximum deflection =wl4/8EIy = wl4/8EI = 25 × 34/ 8 × 4000 = 0.063m.

15. Which of the following is not an exampleof Malleability?a) Wrought Ironb) Ornamental silverc) Torsteeld) Ornamental gold

Answer: cExplanation: Torsteel is an example ofmechanical property ductility. The ductility isa property of a material by which material canbe fractured into thin wires after undergoing aconsiderable deformation without anyrupture.


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TOPIC 1.9 UNIFORM AND NON-UNIFORM BENDING: THEORYAND EXPERIMENT

TOPIC 1.10 I-SHAPED GIRDERS

1. A beam section is provided on the basis of (i) section modulus, (ii) deflection, (iii) shear a) i, ii

b) ii, iii c) i, iii

d) i, ii and iii

Answer: d Explanation: A beam section is provided on

the basis of (i) section modulus, (ii)deflection, (iii) shear. The beam should beeconomical with furnishing required modulusof section.

2. Which of the following is not correct? a) Angles and T section are strong in bending

b) Channels can be used only for light loads c) I sections are most efficient and

economical shapes d) I section with cover plates are provided

when large section modulus is required

Answer: a Explanation: Angles and T section are weak

in bending. Channels can be used only forlight loads. I sections (rolled and built-up) aremost efficient and economical shapes. Isection with cover plates are provided whenlarge section modulus is required. Generally,ISLB or ISMB are provided in such cases.

3. Local buckling can be prevented by a) limiting width-thickness ratio

b) increasing width-thickness ratio c) changing material

d) changing load on member

Answer: c Explanation: Local buckling of compression

members of beam causes loss of integrity of

beam cross section. It is a function of width-thickness ratio and can be prevented bylimiting width-thickness ratio.

4. Which of the following is true?a) in case of rolled section, less thickness ofplate is adopted to prevent local bucklingb) for built-up section and cold formedsection, longitudinal stiffeners are notprovided to reduce width to smaller sizesc) local buckling cannot be prevented bylimiting width-thickness ratiod) in case of rolled section, high thickness ofplate is adopted to prevent local buckling

Answer: dExplanation: In case of rolled section, higherthickness of plate is adopted to prevent localbuckling. Local buckling cannot be preventedby limiting width-thickness ratio. For built-upsection and cold formed section, longitudinalstiffeners are provided to reduce width tosmaller sizes.

5. Which of the following is not true?a) only plastic section can be used inintermediate framesb) slender sections are preferred in hot rolledstructural steelworkc) compact sections can be used in simplysupported beamsd) semi-compact sections can be used forelastic designs

Answer: bExplanation: Only plastic section can beused in intermediate frames which formcollapse mechanism. Compact sections can beused in simply supported beams which failafter reaching Mp at one section. Semi-compact sections can be used for elasticdesigns where section fails after reaching Myat extreme fibres. Slender sections are notpreferred in hot rolled structural steelwork,but they are extensively used in cold formedmembers.


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6. As per IS specification, the beam sectionsshould bea) not symmetrical about any principal axesb) at least symmetrical about one of theprincipal axesc) symmetrical about all principal axesd) unsymmetrical about all principal axes

Answer: bExplanation: The beam sections should be atleast symmetrical about one of the principalaxes as per IS specification. Angle and T-sections are inherently weak in bending whilechannels can only be used for light loads.Rolled I0section is generally preferred asbeam.

7. Which of the following is the designcriteria for beams?(i) Strength in bending (ii) stiffness(iii)economya) ii onlyb) i and iiic) ii and iiid) i, ii and iii

Answer: dExplanation: Beams should be proportionedfor strength in bending keeping in view thelateral and local stability of compressionflange. beam should have adequate strengthto resist applied bending moments andaccompanying shear forces. Beams should beproportioned for stiffness, keeping in mindthe deflections and deformations underservice condition. Beams should beproportioned for economy. Member should besafe against buckling.

8. Which of the following is not true?a) for optimum bending resistance, beammaterial should be near neutral axisb) for optimum bending resistance, beammaterial should be far away from neutral axisc) for optimum bending resistance, web areaof beam has to be adequate for resisting sheard) maximum bending and maximum shearusually occur at different cross section

Answer: aExplanation: For optimum bendingresistance, beam material should be far awayfrom neutral axis and web area of beam has tobe adequate for resisting shear. Maximumbending and maximum shear usually occur atdifferent cross section. in continuous beams,they may occur at same cross section nearinterior supports, but interaction effects arenormally neglected.

TOPIC 1.11 STRESS DUE TOBENDING IN BEAMS.

1. A beam is said to be of uniform strength, if____________a) B.M. is same throughout the beam

b) Shear stress is the same through the beam c) Deflection is the same throughout the beam

d) Bending stress is the same at every sectionalong its longitudinal axis

Answer: d

Explanation: Beam is said to be uniformstrength if at every section along itslongitudinal axis, the bending stress is same.

2. Stress in a beam due to simple bending is____________a) Directly proportional

b) Inversely proportional c) Curvilinearly related


Answer: a Explanation: The stress is directly

proportional to the load and here the load is interms of bending. So the stress is directlyproportional to bending.

3. Which stress comes when there is aneccentric load applied?

a) Shear stress b) Bending stress

c) Tensile stress d) Thermal stress


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Answer: bExplanation: When there is an eccentric loadit means that the load is at some distancefrom the axis. This causes compression in oneside and tension on the other. This causesbending stress.

4. What is the expression of the bendingequation?a) M/I = σ/y = E/Rb) M/R = σ/y = E/Ic) M/y = σ/R = E/Id) M/I = σ/R = E/y

Answer: aExplanation: The bending equation is givenby M/I = σ/y = E/RwhereM is the bending momentI is the moment of inertiay is the distance from neutral axisE is the modulus of elasticityR is the radius.

5. On bending of a beam, which is the layerwhich is neither elongated nor shortened?a) Axis of loadb) Neutral axisc) Center of gravityd) None of the mentioned

Answer: bExplanation: When a beam is in bending thelayer in the direction of bending will be Incompression and the other will be in tension.One side of the neutral axis will be shortenedand the other will be elongated.

6. The bending stress is ____________a) Directly proportional to the distance oflayer from the neutral layerb) Inversely proportional to the distance oflayer from the neutral layerc) Directly proportional to the neutral layerd) Does not depend on the distance of layerfrom the neutral layer

Answer: aExplanation: From the bending equation M/I= σ/y = E/RHere stress is directly proportional to thedistance of layer from the neutral layer.

7. Consider a 250mmx15mmx10mm steel barwhich is free to expand is heated from 15C to40C. what will be developed?a) Compressive stressb) Tensile stressc) Shear stressd) No stress

Answer: dExplanation: If we resist to expand then onlystress will develop. Here the bar is free toexpand so there will be no stress.

8. The safe stress for a hollow steel columnwhich carries an axial load of 2100 kN is 125MN/m2. if the external diameter of thecolumn is 30cm, what will be the internaldiameter?a) 25 cmb) 26.19cmc) 30.14 cmd) 27.9 cm

Answer: bExplanation: Area of the cross section ofcolumn = π/4 (0.302 – d2) m2

Area = load / stress.So, π/4 ( 0.302 – d2) m2 = 21 / 125d = 26.19cm.

Sanfoundry Global Education & LearningSeries – Strength of Materials.

UNIT II WAVES ANDFIBER OPTICS


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TOPIC 2.1 OSCILLATORYMOTION

1. The motion of the earth about its axis isperiodic and simple harmonic.

a) True b) False

Answer: b Explanation: The earth takes 24 hours to

complete its rotation about its axis, but theconcept of to and fro motion is absent, andhence the rotation of the earth is periodic andnot simple harmonic.

2. An object of mass 0.2kg executes simpleharmonic motion along the x-axis with afrequency of (25/π)Hz. At the position x =0.04, the object has kinetic energy of 0.5J andpotential energy 0.4J. The amplitude ofoscillation is?

a) 6cm b) 4cm c) 8cm d) 2cm

Answer: a Explanation: Total energy,

E=2π2 mv2 A2 0.5+0.4=2π2×0.2×(25/π)2 A2

A2=0.9/(0.4×252)A=3/(2×25)=3/50 m=6cm.

3. A spring of force constant 800N/m has anextension of 5cm. The work done inextending it from 5cm to 15cm is?

a) 8J b) 16J

c) 24J d) 32J

Answer: a Explanation: At x1 = 5 cm,

U1=1/2×k(x1)2=1/2×800×0.052=1J

At x2=15cm,

U2=1/2×k(x2)2=1/2×800×0.152=9JW=U2-U1=9-1=8J.

4. A simple pendulum is attached to the roofof a lift. If the time period of oscillation,when the lift is stationary is T, then thefrequency of oscillation when the lift fallsfreely, will be ___________a) Zerob) Tc) 1/Td) ∞

Answer: aExplanation: In a freely falling lift,g=0v=1/2π×√(g/l)=1/2π×√(0/l)=0.

5. There is a simple pendulum hanging fromthe ceiling of a lift. When the lift is standstill,the time period of the pendulum is T. If theresultant acceleration becomes g/4, then thenew time period of the pendulum is?a) 0.8Tb) 0.25Tc) 2Td) 4T

Answer: cExplanation: T=2×√(l/g)T‘=2π×√((l/g)/4)T‘=2T.

6. A lightly damped oscillator with afrequency v is set in motion by a harmonicdriving force of frequency v’. When v’ islesser than v, then the response of theoscillator is controlled by ___________a) Spring constantb) Inertia of the massc) Oscillator frequencyd) Damping coefficient

Answer: aExplanation: Frequency of driving force islesser than frequency v of a dampedoscillator. The vibrations are nearly in phase


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with the driving force and response of theoscillator is controlled by spring constant.

7. Statement: In simple harmonic motion, thevelocity is maximum, when the accelerationis minimum.Reason: Displacement and velocity in simpleharmonic motion is differ in phase by π/2.a) Both statement and reason are true and thereason is the correct explanation of thestatementb) Both statement and reason are true but thereason is not the correct explanation of thestatementc) Statement is true, but the reason is falsed) Statement and reason are false

Answer: bExplanation: Both statement and reason aretrue but the reason is not the correctexplanation of the statement, In fact, thephase difference between velocity andacceleration is π/2.

8. What is the time period of a pendulumhanged in a satellite? (T is the time period onearth)a) Zerob) Tc) Infinited) T/√6

Answer: cExplanation: In a satellite, g= 0T=2π√(l/g)=2π√(l/0)=∞.

9. Which of the following functionsrepresents a simple harmonic oscillation?a) sinωt-cosωtb) sinωt+sin2ωtc) sinωt-sin2ωtd) sin2 ωt

Answer: aExplanation: y=sinωt-cosωtdy/dt=ωcosωt+ωsinωt(d2 y)/dt2 =-ω2 sinωt+ω2 cosωt=-ω2 (sinωt-cosωt)

a=-ω2 y that is a∝yThis satisfies the condition of simpleharmonic motion.

10. The displacement of a simple harmonicmotion doing oscillation when kinetic energy= potential energy (amplitude = 4cm) is?a) 2√2cmb) 2cmc) 1/√2 cmd) √2 cm

Answer: aExplanation: When kinetic energy =potential energy,y=a/√2=4/√2=2√2 cm.

TOPIC 2.2 FORCED ANDDAMPED OSCILLATIONS:DIFFERENTIAL EQUATIONAND ITS SOLUTION

TOPIC 2.3 PLANEPROGRESSIVE WAVES

1. The ratio of velocity of sound in hydrogenand oxygen at STP is __________

a) 16:1 b) 8:1

c) 4:1 d) 2:1

Answer: c Explanation: vH/vO = √(MO/MH)

=√(32/2)=4:1.

2. It takes 2 seconds for a sound wave totravel between two fixed points when the daytemperature is 10°C. If the temperature risesto 30°C, the sound wave travels between thesame fixed parts in __________

a) 1.9s b) 2s

c) 2.1s d) 2.2s


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Answer: aExplanation: v∝√Tt∝1/vt∝1/√Tt2/t1 =√(T1/T2)=√((273+10)/(273+30))=√(283/303)t2=√(283/303)×2s=1.9s.

3. The disc of a siren containing 60holesrotates at a constant speed of 360rpm. Theemitted sound is in unison with a tuning forkof frequency.a) 10Hzb) 360Hzc) 216Hzd) 60Hz

Answer: bExplanation: Frequency of revolution ofdisc=360rpm=360/60rps=60rpsFrequency of emitted sound=6×No.of holes=6×60=360Hz.

4. The quantity which does not change, whensound enters from one medium to another__________a) Wavelengthb) Speedc) Frequencyd) Velocity

Answer: cExplanation: Frequency remains unchangedwhen sound travels from one medium toanother.

5. The equation of a simple harmonic wave isgiven byy=5sin (π/2) (100t-x)Where x and y are in metre and time is insecond. The period of the wave in second willbe __________a) 0.04b) 0.01c) 1d) 5

Answer: aExplanation: y=5sin (π/2) (100t-x)y=Asin(ωt-kx)ω=2π/T=π/2×100T=2/50=0.04s.

6. If wave y=Acos(ωt+kx) Is moving along x-axis, the shapes of a pulse at t=0 and t=2s.a) Are differentb) Are samec) May not be the samed) Unpredictable

Answer: bExplanation: The shapes of y-x graphremains the same at t=0 and t=2s.

7. y1=4sin (ωt+kx), y2=-4cos (ωt+kx), thephase difference is __________a) π/2b) 3π/2c) πd) Zero

Answer: bExplanation: y1=4sin (ωt+kx)y2=-4cos (ωt+kx)=-4sin (ωt+kx+3π/2)∆φ=3π/2.

8. A wave equation is y=0.1sin[100πt-kx] andwave velocity is 100m/s, its number is equalto __________a) 1/mb) 2/mc) π/md) 2π/m

Answer: cExplanation: y=0.1sin [100πt-kx] y=Asin(ωt-kx)ω=100πWave number=ω/v=100π/100=π/m.

9. A particle on the trough of a wave at anyinstant will come to the mean position after atime (T=time period).a) T/2


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b) T/4c) Td) 2T

Answer: bExplanation: Time taken by a particle tomove from trough to the mean position=T/4.

10. A string is tied on a sonometer, secondend is hanging downward through a pulleywith tension T. The velocity of the transversewave produced is proportional to __________a) 1/√Tb) √Tc) Td) 1/T

Answer: bExplanation: v=√(T/m)v∝√T.

11. The fundamental frequency of asonometer wire is n. If the tension is made 3times and length and diameter are alsoincreased 3 times, the new frequency will be__________a) 3nb) n/3√3c) n/3d) √3 n

Answer: bExplanation: n=1/LD×√(T/πρ)n‘=1/(3L×3D)×√(3T/πρ)=√3/9×n=1/(3√3) n.

TOPIC 2.4 WAVE EQUATION.LASERS : POPULATION OFENERGY LEVELS, EINSTEIN'SA AND B COEFFICIENTSDERIVATION

1. Which of the following is the correctexpression for the relation between Einstein’scoefficients A and B?

a)

b) c) d)

Answer: a Explanation: The expression is the

correct expression for the relation betweenthe two Einstein’s coefficients. Thisexpression is known as the Einstein’s relation.

2. What is the relationship between B21 andB12?

a) B12 > B21 b) B12 < B21 c) B12 = B21 d) No specific relation

Answer: c Explanation: B21 is the coefficient for the

stimulated emission while B12 is thecoefficient for stimulated absorption. Both theprocesses are mutually reverse processes andtheir probabilities are equal. Therefore, B12 =B21.

3. Which of the following Einstein’scoefficient represents spontaneous emission?

a) A12 b) A21 c) B12 d) B21 Answer: b

Explanation: A21 represents the spontaneousemission of photons. A12 signifiesspontaneous absorption.B12 is for stimulatedabsorption while B21 is for stimulatedemission.

4. The correct expression for the rate ofstimulated emission is _______________

a) Rse = A21N2 b) Rse = A21uN2

8πv3hc3

8πv2hc3

8πv2hc2

8πhvc3

8πv3hc3


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c) Rse = B21N2d) Rse = B21uN2

Answer: dExplanation: The stimulates emission isdirectly proportional to the energy density u,of the external radiation field. Also,stimulated emission rate increases with theincrease in number N2 of exited atoms.

5. Which law is used for achieving therelation between the Einstein’s coefficients?a) Heisenberg’s Uncertainty Principleb) Planck’s radiation lawc) Einstein’s equationd) Quantum law

Answer: bExplanation: Planck’s radiation law, whichgives the energy density u = , isused as the formula resembles the one for theenergy density of the external radiation fieldin stimulated emission, u = .

6. The probability of spontaneous emissionincreases rapidly with the energy differencebetween the two states.

a) True b) False

Answer: a Explanation: From Einstein’s relation we

know that the ratio of Einstein’s coefficientsis . Thus, the ration of Einstein’scoefficients is proportional to the cube of thefrequency. Hence, the probability ofspontaneous emission increases rapidly withthe energy difference between the two states.

7. If the frequency of emitted photon is 10Hz, the ratio of Einstein’s coefficient is_____________a) 2.177 X 10-51

b) 3.177 X 10-51

c) 5.177 X 10-51

d) 6.177 X 10-51

Answer: dExplanation: We know Einstein’s relation =

v = 10 Hz, h = 6.63 X 10-34 Js, c = 3 X108m/s

Therefore, the ratio of Einstein’s coefficientsis: 6.177 X 10-51.

8. What is the unit of the coefficient ofspontaneous emission?

a) s-1 b) s

c) J-1 d) J

Answer: a

Explanation: For spontaneous emission, theexpression for the rate is = A21N2, where N2is the number of particles in exited state. Asthe unit of rate is Number of particles persecond, the unit of A21 is s-1.

9. What is the unit for the coefficient ofstimulated emission?

a) s-2 b) m3 s-2

c) J−1 m3 d) J−1 m3 s-2

Answer: d

Explanation: For stimulated emission, theexpression for the rate is B21uN2 where ustands for the energy density and N is thenumber of exited atoms. Therefore, the unitof B turns out to be J−1 m3 s-2.

10. Which Einstein’s coefficient should beused in this case?

8πhv3

c31

eα−1

A21

B21(B12B21

eα−1)

8πv3hc3

8πv3hc3


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a) A12 b) A21 c) B12 d) B21 Answer: d

Explanation: The given figure showsstimulated emission. Hence, the Einsteincoefficient for stimulated emission is B21. Ifit had been spontaneous emission, then A21would have been used.

TOPIC 2.5 RESONANT CAVITY,OPTICAL AMPLIFICATION(QUALITATIVE)

1. A cylindrical cavity resonator can beconstructed using a circular waveguide.

a) shorted at both the ends b) open at both the ends

c) matched at both the ends d) none of the mentioned

Answer: a

Explanation: A cylindrical cavity resonatoris formed by shorting both the ends of thecylindrical cavity because open ends mayresult in radiation losses in the cavity.

2. The dominant mode in the cylindricalcavity resonator is TE101 mode.

a) true b) false

Answer: b Explanation: The dominant mode of

propagation in a circular waveguide is TE111mode. Hence, the dominant mode ofresonance in a cylindrical cavity made of acircular waveguide is TE111 mode. In a

cylindrical resonator, the mode ofpropagation depends on the length of thecavity.

3. Circular cavities are used for microwavefrequency meters.a) trueb) false

Answer: aExplanation: Circular cavities are used formicrowave frequency meters. The cavity isconstructed with a movable top wall to allowthe mechanical tuning of the resonantfrequency.

4. The mode of the circular cavity resonatorused in frequency meters is:a) TE011 modeb) TE101 modec) TE111 moded) TM111 mode

Answer: aExplanation: Frequency resolution of afrequency meter is determined from itsquality factor. Q factor of TE011 mode ismuch greater than the quality factor of thedominant mode of propagation.

5. The propagation constant of TEmn modeof propagation for a cylindrical cavityresonator is:a) √ (k2-(pnm/a)2)b) √ pnm/ac) √ (k2+(pnm/a)2)d) none of the mentioned

Answer: aExplanation: The propagation constant for acircular cavity depends on the radius of thecavity, and the wave number. If the mode ofpropagation is known and the dimension ofthe cavity is known then the propagationconstant can be found out.


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6. A circular cavity resonator is filled with adielectric of 2.08 and is operating at 5GHz offrequency. Then the wave number is:a) 181b) 151c) 161d) 216

Answer: bExplanation: Wave number for a circularcavity resonator is given by the expression2πf011√∈r/C. substituting the given values inthe above expression; the wave number of thecavity resonator is 151.

7. Given that the wave number of a circularcavity resonator is 151 (TE011 mode), and thelength of the cavity is twice the radius of thecavity, the radius of the circular cavityoperating at 5GHz frequency is:a) 2.1 cmb) 1.7 cmc) 2.84 cmd) insufficient data

Answer: dExplanation: For a circular cavity resonator,wave number is given by √( (p01/a)2 +(π/d)2).P01 for the given mode of resonance is 3.832.Substituting the given values the radius of thecavity is 2.74 cm.

8. The loss tangent for a circular cavityresonator is 0.0004.Then the unloaded Q dueto dielectric loss is:a) 1350b) 1560c) 560d) 2500

Answer: dAnswer: Unloaded Q due to the dielectricloss in a circular cavity resonator is thereciprocal of the loss tangent. Hence, takingthe reciprocal of the loss tangent, unloaded Qdue to dielectric loss is 2500.

9. A circular cavity resonator has a wavenumber of 151, radius of 2.74 cm, and surfaceresistance of 0.0184Ω. If the cavity is filledwith a dielectric of 2.01, then unloaded Q dueto conductor loss is:a) 25490b) 21460c) 29390d) none of the mentioned

Answer: cExplanation: Unloaded Q of a circularresonator due to conductor loss is given byka/2Rs. is the intrinsic impedance of themedium. Substituting the given values in theequation for loaded Q, value is 29390.

10. If unloaded Q due to conductor loss andunloaded Q due to dielectric loss is 29390 and2500 respectively, then the total unloaded Qof the circular cavity is:a) 2500b) 29390c) 2300d) 31890

Answer: cExplanation: The total unloaded Q of acircular cavity resonator is given by theexpression (Qc

-1+ Qd-1)-1. Substituting the

given values in the above expression, the totalunloaded Q for the resonator is 2300.

TOPIC 2.6 SEMICONDUCTORLASERS: HOMOJUNCTIONAND HETEROJUNCTION

1. A perfect semiconductor crystal containingno impurities or lattice defects is called as__________a) Intrinsic semiconductor

b) Extrinsic semiconductor c) Excitation

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Answer: aExplanation: An intrinsic semiconductor isusually un-doped. It is a pure semiconductor.The number of charge carriers is determinedby the semiconductor material properties andnot by the impurities.

2. The energy-level occupation for asemiconductor in thermal equilibrium isdescribed by the __________a) Boltzmann distribution functionb) Probability distribution functionc) Fermi-Dirac distribution functiond) Cumulative distribution function

Answer: cExplanation: For a semiconductor in thermalequilibrium, the probability P(E) that anelectron gains sufficient thermal energy at anabsolute temperature so as to occupy aparticular energy level E, is given by theFermi-Dirac distribution. It is given by-P(E) = 1/(1+exp(E-EF/KT))Where K = Boltzmann constant, T = absolutetemperature, EF = Fermi energy level.

3. What is done to create an extrinsicsemiconductor?a) Refractive index is decreasedb) Doping the material with impuritiesc) Increase the band-gap of the materiald) Stimulated emission

Answer: bExplanation: An intrinsic semiconductor is apure semiconductor. An extrinsicsemiconductor is obtained by doping thematerial with impurity atoms. These impurityatoms create either free electrons or holes.Thus, extrinsic semiconductor is a dopedsemiconductor.

4. The majority of the carriers in a p-typesemiconductor are __________a) Holesb) Electronsc) Photonsd) Neutrons

Answer: aExplanation: The impurities can be eitherdonor impurities or acceptor impurities.When acceptor impurities are added, theexcited electrons are raised from the valenceband to the acceptor impurity levels leavingpositive charge carriers in the valence band.Thus, p-type semiconductor is formed inwhich majority of the carriers are positive i.e.holes.

5. _________________ is used when theoptical emission results from the applicationof electric field.a) Radiationb) Efficiencyc) Electro-luminescenced) Magnetron oscillator

Answer: cExplanation: Electro-luminescence isencouraged by selecting an appropriatesemiconductor material. Direct band-gapsemiconductors are used for this purpose. Inband-to-band recombination, the energy isreleased with the creation of photon. Thisemission of light is known aselectroluminescence.

6. In the given equation, what does p standsfor?

p = 2πhk

a) Permittivityb) Probabilityc) Holesd) Crystal momentum

Answer: dExplanation: The given equation is a relationof crystal momentum and wave vector. In thegiven equation, h is the Planck’s constant, k isthe wave vector and p is the crystalmomentum.

7. The recombination in indirect band-gapsemiconductors is slow.


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a) Trueb) False

Answer: aExplanation: In an indirect band-gapsemiconductor, the maximum and minimumenergies occur at different values of crystalmomentum. However, three-particlerecombination process is far less probablethan the two-particle process exhibited bydirect band-gap semiconductors. Hence, therecombination in an indirect band-gapsemiconductor is relatively slow.

8. Calculate the radioactive minority carrierlifetime in gallium arsenide when theminority carriers are electrons injected into ap-type semiconductor region which has a holeconcentration of 1018cm-3. Therecombination coefficient for galliumarsenide is 7.21*10-10cm3s-1.a) 2nsb) 1.39nsc) 1.56nsd) 2.12ms

Answer: bExplanation: The radioactive minority carrierlifetime ςrconsidering the p-type region isgiven by-ςr = [BrN]-1 where Br = Recombinationcoefficient in cm3s-1 and N = carrierconcentration in n-region.

9. Which impurity is added to galliumphosphide to make it an efficient lightemitter?a) Siliconb) Hydrogenc) Nitrogend) Phosphorus

Answer: cExplanation: An indirect band-gapsemiconductor may be made into an electro-luminescent material by the addition ofimpurity centers which will convert it into a

direct band-gap material. The introduction ofnitrogen as an impurity into galliumphosphide makes it an effective emitter oflight. Such conversion is only achieved inmaterials where the direct and indirect band-gaps have a small energy difference.

10. Population inversion is obtained at a p-njunction by __________a) Heavy doping of p-type materialb) Heavy doping of n-type materialc) Light doping of p-type materiald) Heavy doping of both p-type and n-typematerial

Answer: dExplanation: Population inversion at p-njunction is obtained by heavy doping of bothp-type and n-type material. Heavy p-typedoping with acceptor impurities causes alowering of the Fermi-level between the filledand empty states into the valence band.Similarly n-type doping causes Fermi-level toenter the conduction band of the material.

11. A GaAs injection laser has a thresholdcurrent density of 2.5*103Acm-2 and lengthand width of the cavity is 240μm and 110μmrespectively. Find the threshold current forthe device.a) 663 mAb) 660 mAc) 664 mAd) 712 mA

Answer: bExplanation: The threshold current isdenoted by Ith. It is given by-Ith = Jth * area of the optical cavityWhere Jth = threshold current densityArea of the cavity = length and width.

12. A GaAs injection laser with an opticalcavity has refractive index of 3.6. Calculatethe reflectivity for normal incidence of theplane wave on the GaAs-air interface.a) 0.61


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b) 0.12c) 0.32d) 0.48

Answer: cExplanation: The reflectivity for normalincidence of the plane wave on the GaAs-airinterface is given by-r = ((n-1)/(n+1))2 where r=reflectivity andn=refractive index.

13. A homo-junction is an interface betweentwo adjoining single-crystal semiconductorswith different band-gap energies.a) Trueb) False

Answer: bExplanation: The photo-emissive propertiesof a single p-n junction fabricated from asingle-crystal semiconductor material arecalled as homo-junction. A hetero-junction isan interface between two single-crystalsemiconductors with different band-gapenergies. The devices which are fabricatedwith hetero-junctions are said to have hetero-structure.

14. How many types of hetero-junctions areavailable?a) Twob) Onec) Threed) Four

Answer: aExplanation: Hetero-junctions are classifiedinto an isotype and an-isotype. The isotypehetero-junctions are also called as n-n or p-pjunction. The an-isotype hetero-junctions arecalled as p-n junction with large band-gapenergies.

15. The ______________ system is bestdeveloped and is used for fabricating bothlasers and LEDs for the shorter wavelengthregion.a) InP

b) GaSbc) GaAs/GaSbd) GaAs/Alga AS DH

Answer: dExplanation: For DH device fabrication,materials such as GaAs, Alga AS are used.The band-gap in this material may be tailoredto span the entire wavelength band bychanging the AlGa composition. Thus, GaAs/Alga As DH system is used for fabrication oflasers and LEDs for shorter wavelengthregion (0.8μm-0.9μm).

TOPIC 2.7 FIBER OPTICS:PRINCIPLE, NUMERICALAPERTURE AND ACCEPTANCEANGLE

1. What is the principle of fibre opticalcommunication?

a) Frequency modulation b) Population inversion

c) Total internal reflection d) Doppler Effect

Answer: c

Explanation: In optical fibres, the lightentering the fibre does not encounter any newsurfaces, but repeatedly they hit the samesurface. The reason for confining the lightbeam inside the fibres is the total internalreflection.

2. What is the other name for a maximumexternal incident angle?

a) Optical angle b) Total internal reflection angle

c) Refraction angle d) Wave guide acceptance angle

Answer: d

Explanation: Only this rays which passwithin the acceptance angle will be totallyreflected. Therefore, light incident on the corewithin the maximum external incident angle


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can be coupled into the fibre to propagate.This angle is called a wave guide acceptanceangle.

3. A single mode fibre has low intermodaldispersion than multimode.a) Trueb) False

Answer: aExplanation: In both single and multimodefibres the refractive indices will be in step bystep. Since a single mode has less dispersionthan multimode, the single mode step indexfibre also has low intermodal dispersioncompared to multimode step index fibre.

4. How does the refractive index vary inGraded Index fibre?a) Tangentiallyb) Radiallyc) Longitudinallyd) Transversely

Answer: bExplanation: The refractive index of the coreis maximum along the fibre axis and itgradually decreases. Here the refractive indexvaries radially from the axis of the fibre.Hence it is called graded index fibre.

5. Which of the following has moredistortion?a) Single step-index fibreb) Graded index fibrec) Multimode step-index fibred) Glass fibre

Answer: cExplanation: When rays travel throughlonger distances there will be some differencein reflected angles. Hence high angle raysarrive later than low angle rays. Therefore thesignal pulses are broadened thereby results ina distorted output.

6. In which of the following there is nodistortion?

a) Graded index fibreb) Multimode step-index fibrec) Single step-index fibred) Glass fibre

Answer: aExplanation: The light travels with differentspeeds in different paths because of thevariation in their refractive indices. At theouter edge it travels faster than near the centreBut almost all the rays reach the exit end atthe same time due to the helical path. Thus,there is no dispersion in the pulses and hencethe output is not a distorted output.

7. Which of the following loss occurs insidethe fibre?a) Radiative lossb) Scatteringc) Absorptiond) Attenuation

Answer: bExplanation: Scattering is a wavelengthdependent loss. Since the glass used in thefabrication of fibres, the disordered structureof glass will make some vibrations in therefractive index inside the fibre. This causesRayleigh scattering.

8. What causes microscopic bend?a) Uniform pressureb) Non-uniform volumec) Uniform volumed) Non-uniform pressure

Answer: dExplanation: Micro-bends losses are causeddue to non-uniformities inside the fibre. Thismicro-bends in fibre appears due to non-uniform pressures created during the cablingof fibre.

9. When more than one mode is propagating,how is it dispersed?a) Dispersionb) Inter-modal dispersion


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c) Material dispersiond) Waveguide dispersion

Answer: bExplanation: When more than one mode ispropagating through a fibre, then inter modaldispersion will occur. Since many modes arepropagating, they will have differentwavelengths and will take different time topropagate through the fibre.

10. A fibre optic telephone transmission canhandle more than thousands of voicechannels.a) Trueb) False

Answer: aExplanation: Optical fibre has largerbandwidth hence it can handle a large numberof channels for communication.

11. Which of the following is known as fibreoptic back bone?a) Telecommunicationb) Cable televisionc) Delay linesd) Bus topology

Answer: dExplanation: Each computer on the networkis connected to the rest of the computers bythe optical wiring scheme called bustopology, which is an application known asfibre optic back bone.

12. Calculate the numerical aperture of anoptical fibre whose core and cladding aremade of materials of refractive index 1.6 and1.5 respectively.a) 0.55677b) 55.77c) 0.2458d) 0.647852

Answer: aExplanation: Numerical aperture =

Numerical aperture = 0.55677.

13. A step-index fibre has a numericalaperture of 0.26, a core refractive index of 1.5and a core diameter of 100micrometer.Calculate the acceptance angle.

a) 1.47° b) 15.07°

c) 2.18° d) 24.15°

Answer: b

Explanation: sin i = (Numerical aperture)/n sin i = 15.07°.

TOPIC 2.8 TYPES OF OPTICALFIBRES (MATERIAL,REFRACTIVE INDEX, MODE)

TOPIC 2.9 LOSSESASSOCIATED WITH OPTICALFIBERS

1. Multimode step index fiber has___________a) Large core diameter & large numericalaperture

b) Large core diameter and small numericalaperture

c) Small core diameter and large numericalaperture

d) Small core diameter & small numericalaperture

Answer: a

Explanation: Multimode step-index fiber haslarge core diameter and large numericalaperture. These parameters provides efficientcoupling to inherent light sources such asLED’s.

2. A typically structured glass multimode stepindex fiber shows as variation of attenuationin range of ___________

√n12 − n22


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a) 1.2 to 90 dB km-1 at wavelength 0.69μmb) 3.2 to 30 dB km-1 at wavelength 0.59μmc) 2.6 to 50 dB km-1 at wavelength 0.85μmd) 1.6 to 60 dB km-1 at wavelength 0.90μm

Answer: cExplanation: A multimode step index fibersshow an attenuation variation in range of 2.6to 50dBkm-1. The wide variation inattenuation is due to the large differences bothwithin and between the two overallpreparation methods i.e. melting anddeposition.

3. Multimode step index fiber has a large corediameter of range is ___________a) 100 to 300 μmb) 100 to 300 nmc) 200 to 500 μmd) 200 to 500 nm

Answer: aExplanation: A multimode step index fiberhas a core diameter range of 100 to 300μm.This is to facilitate efficient coupling toinherent light sources.

4. Multimode step index fibers have abandwidth of ___________a) 2 to 30 MHz kmb) 6 to 50 MHz kmc) 10 to 40 MHz kmd) 8 to 40 MHz km

Answer: bExplanation: Multimode step index fibershave a bandwidth of 6 to 50 MHz km. Thesefibers with this bandwidth are best suited forshort -haul, limited bandwidth and relativelylow-cost application.

5. Multimode graded index fibers aremanufactured from materials with___________a) Lower purityb) Higher purity than multimode step indexfibers.

c) No impurityd) Impurity as same as multimode step indexfibers.

Answer: bExplanation: Multimode graded index fibershave higher purity than multimode step indexfiber. To reduce fiber losses, these fibers havemore impurity.

6. The performance characteristics ofmultimode graded index fibers are___________a) Better than multimode step index fibersb) Same as multimode step index fibersc) Lesser than multimode step index fibersd) Negligible

Answer: aExplanation: Multimode graded index fibersuse a constant grading factor. Performancecharacteristics of multimode graded indexfibers are better than those of multimode stepindex fibers due to index graded and lowerattenuation.

7. Multimode graded index fibers haveoverall buffer jackets same as multimode stepindex fibers but have core diameters___________a) Larger than multimode step index fibersb) Smaller than multimode step index fibersc) Same as that of multimode step indexfibersd) Smaller than single mode step index fibers

Answer: bExplanation: Multimode graded index fibershave smaller core diameter than multimodestep index fibers. A small core diameter helpsthe fiber gain greater rigidity to resistbending.

8. Multimode graded index fibers withwavelength of 0.85μm have numericalaperture of 0.29 have core/cladding diameterof ___________a) 62.5 μm/125 μm


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b) 100 μm/140 μmc) 85 μm/125 μmd) 50 μm/125μm

Answer: bExplanation: Multimode graded index fiberswith numerical aperture 0.29 having acore/cladding diameter of 100μm/140μm.They provide high coupling frequency LED’sat a wavelength of 0.85 μm and have lowcost. They are also used for short distanceapplication.

9. Multimode graded index fibers useincoherent source only.a) Trueb) False

Answer: bExplanation: Multimode graded index fibersare used for short haul and medium to highbandwidth applications. Small haulapplications require LEDs and low accuracylasers. Thus either incoherent or incoherentsources like LED’s or injection laser diodeare used.

10. In single mode fibers, which is the mostbeneficial index profile?a) Step indexb) Graded indexc) Step and graded indexd) Coaxial cable

Answer: bExplanation: In single mode fibers, gradedindex profile is more beneficial as comparedto step index. This is because graded indexprofile provides dispersion-modified-singlemode fibers.

11. The fibers mostly not used nowadays foroptical fiber communication system are___________a) Single mode fibersb) Multimode step fibersc) Coaxial cablesd) Multimode graded index fibers

Answer: aExplanation: Single mode fibers are used toproduce polarization maintaining fibers whichmake them expensive. Also the alternative tothem are multimode fibers which are complexbut accurate. So, single-mode fibers are notgenerally utilized in optical fibercommunication.

12. Single mode fibers allow single modepropagation; the cladding diameter must be atleast ___________a) Twice the core diameterb) Thrice the core diameterc) Five times the core diameterd) Ten times the core diameter

Answer: dExplanation: The cladding diameter in singlemode fiber must be ten times the corediameter. Larger ratios contribute to accuratepropagation of light. These dimension ratiosmust be there so as to avoid losses from thevanishing fields.

13. A fiber which is referred as non-dispersive shifted fiber is?a) Coaxial cablesb) Standard single mode fibersc) Standard multimode fibersd) Non zero dispersion shifted fibers

Answer: bExplanation: A standard single mode fiberhaving step index profile is known as non-dispersion shifted fiber. As these fibers have azero dispersion wavelength of 1.31μm and soare preferred for single-wavelengthtransmission in O-band.

14. Standard single mode fibers (SSMF) areutilized mainly for operation in ___________a) C-bandb) L-bandc) O-bandd) C-band and L-band


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Answer: cExplanation: SSMFs are utilized foroperation in O-band only. It shows highdispersion in the range of 16 to 20ps/nm/kmin C-band and L-band. So SSMFs are used inO-band.

15. Fiber mostly suited in single-wavelengthtransmission in O-band is?a) Low-water-peak non dispersion-shiftedfibersb) Standard single mode fibersc) Low minimized fibersd) Non-zero-dispersion-shifted fibers

Answer: bExplanation: Standard single mode fiberswith a step index profile are called nondispersion shifted fiber and it is particularlyused for single wavelength transmission in O-band and as if has a zero-dispersionwavelength at 1.31μm.

TOPIC 2.10 FIBRE OPTICSENSORS: PRESSURE ANDDISPLACEMENT.

1. OTDR stands for _______________ a) Optical time domain reflectometer

b) Optical transfer data rate c) Optical time data registers


Answer: a Explanation: OTDR is the short form of

optical time domain reflectometer.

2. Which of the following is not correct forfibre optic sensors?

a) Immune to electro magnetic interference b) Immune to radiation hazard

c) Can be used in harsh environments d) None of the mentioned

Answer: d

Explanation: All of the mentioned arequalities of fibre optic sensors.

3. Fluoride glass is used with __________a) IR wavesb) UV raysc) Normal lightd) All of the mentioned

Answer: aExplanation: Flouride glass is suitable for IRrays of wavelength upto 3200 nm.

4. Silica glass of hydroxyl concentration canbe used for ___________ of wavelength.a) 100 nm to 250 nmb) 250 nm to 800 nmc) 800 nm to 1500 nmd) 100 nm to 3400 nm

Answer: bExplanation: Silica glass with hydroxylconcentration is used for wavelength 250 nmto 800 nm.

5. General spectral range for silica glass is_______________a) Less than 200 nmb) Between 200 nm to 2200 nmc) Between 2000 nm to 5000 nmd) Greater than 3000 nm

Answer: bExplanation: General spectral range of silicaglass is 200 nm to 2200 nm.

6. Epoxy material in fibre optics is intendedfor ________________a) Better optical propertiesb) Better reflectionc) Better sealingd) Reducing noise

Answer: cExplanation: Polished epoxy seal providesliquid and air tight seal.

7. Plastics optical cables can be used for________________a) Short rangeb) Medium range of distance


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c) Long range of distanced) Very high range of distance

Answer: aExplanation: Plastic optical cables aremanufacturing for short range purposes.

8. Which of the following represents lossassociated with glass fibres?a) 3 dB/Kmb) 10 dB/Kmc) 0 dB/Kmd) 50 dB/Km

Answer: aExplanation: Glass fibres have a net loss of 3dB on every single kilometre.

9. Loss associated with plastic fibre is lessthan glass fibres.a) Trueb) False

Answer:bExplanation: Loss associated with plasticfibre is about 100-1250 dB/Km and it isseveral times larger than glass fibres.

10. Cladding in glass fibre have highrefractive index than the core.a) Trueb) False

Answer: bExplanation: Cladding in glass fibre isalways kept at a low refractive index than thecore.

UNIT III THERMALPHYSICS

TOPIC 3.1 TRANSFER OF HEATENERGY

1. The transfer of heat between two bodies indirect contact is calleda) radiationb) convectionc) conductiond) none of the mentioned

Answer: cExplanation: This is the definition ofconduction.

2. Heat flow into a system is taken to be____, and heat flow out of the system is takenas ____a) positive, positiveb) negative, negativec) negative, positived) positive, negative

Answer: dExplanation: The direction of heat transfer istaken from the high temperature system to thelow temperature system.

3. In the equation, dQ=TdXa) dQ is an inexact differentialb) dX is an exact differentialc) X is an extensive propertyd) all of the mentioned

Answer: dExplanation: This is because heat transfer isa path function.

4. The transfer of heat between a wall and afluid system in motion is calleda) radiationb) convectionc) conductiond) none of the mentioned

Answer: bExplanation: This is the definition ofconvection.

5. For solids and liquids, specific heata) depends on the processb) is independent of the process


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c) may or may not depend on the processd) none of the mentioned

Answer: bExplanation: It is the property of specificheat.

6. The specific heat of the substance isdefined as the amount of heat required toraise a unit mass of the substance through aunit rise in temperature.a) trueb) false

Answer: aExplanation: c=Q/(m*Δt).

7. Heat and work area) path functionsb) inexact differentialsc) depend upon the path followedd) all of the mentioned

Answer: dExplanation: It is an important point toremember regarding heat and work transfer.

8. Latent heat is taken ata) constant temperatureb) constant pressurec) both of the mentionedd) none of the mentioned

Answer: cExplanation: The latent heat is heat transferrequired to cause a phase change in a unitmass of substance at a constant pressure andtemperature.

9. Which of the following is true?a) latent heat of fusion is not much affectedby pressureb) latent heat of vaporization is highlysensitive to pressurec) both of the mentionedd) none of the mentioned

Answer: cExplanation: It is a general fact about latent

heat.

10. Heat transfer and work transfer area) boundary phenomenab) energy interactionsc) energy in the transitd) all of the mentioned

Answer: dExplanation: It is an important point toremember regarding heat and work transfer.

TOPIC 3.2 THERMALEXPANSION OF SOLIDS ANDLIQUIDS

1. A faulty thermometer has its fixed pointsmarked as 5° and 95°. The temperature of abody as measured by the faulty thermometeris 59°. Find the correct temperature of thebody on a Celsius scale.

a) 60°C b) 40°C c) 20°C d) 0°C

Answer: a

Explanation: (TC-0)/(100-0)=(Temperatureon faulty scale-Lower fixed point)/(Upperfixed point-Lower fixed point)

(TC-0)/100=(59-5)/(95-5)=54/90 TC=60°C.

2. Temperature is a microscopic concept. a) True

b) False

Answer: b Explanation: Temperature is a macroscopic

concept. It is related to the average kineticenergy of a large number of moleculesforming a system. It is not possible to definethe temperature for a single molecule.

3. The thermometer bulb should have___________


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a) High heat capacityb) No heat capacityc) Small heat capacityd) Varying heat capacity

Answer: cExplanation: The thermometer bulb havingsmall heat capacity will absorb less heat fromthe body whose temperature is to bemeasured. Hence the temperature of that bodywill practically remain unchanged.

4. Calorie is defined as the amount of heatrequired to raise the temperature of 1g ofwater by 1°C and it is defined under which ofthe following conditions?a) From 14.5°C to 15.5°C at 760mm of Hgb) From 98.5°C to 99.5°C at 760mm of Hgc) From 13.5°C to 14.5°C at 76mm of Hgd) From 3.5°C to 4.5°C at 76mm of Hg

Answer: aExplanation: One calorie is defined as theheat required to raise the temperature of 1g ofwater from 14.5°C to 15.5° at 760mm of Hg.

5. Compared to burn due to air at 100°C, aburn due to steam at 100°C is ___________a) More dangerousb) Less dangerousc) Equally dangerousd) Not dangerous

Answer: aExplanation: Compared to burn due to air at100°C, a burn due to steam at 100°C is moredangerous due to the additional heatpossessed by steam.

6. 540g of ice at 0°C is mixed with 540g ofwater at 80°C. What is the final temperatureof the mixture?a) 0°Cb) 40°Cc) 80°Cd) Less than 0°C

Answer: aExplanation: Heat gained by ice = Heat lostby water at 80°C540×80+540×1×θ=540×1×(80-θ)θ=0°C.

7. An ideal black body is thrown into afurnace. The black body is room temperature.It is observed that ___________a) Initially, it is darkest body and at latertimes the brightestb) At all times it is the darkest bodyc) It cannot be distinguished at all timesd) Initially, it is the darkest body and at laterit cannot be distinguished

Answer: aExplanation: Initially at lower temperature, itabsorbs the entire radiations incident upon it.So, it is the darkest body. At later times, whenit attains the temperature of the furnace, theblack body radiates maximum energy. Itappears brightest of all bodies.

8. If the sun were to increase in temperaturefrom T to 2T and its radius from R to 2R,then the ratio of the radiant energy receivedon earth to what it was previously, will be___________a) 4b) 16c) 32d) 64

Answer: dExplanation: Energy radiated by the sun persecond,E = σAT4= σ×4πR2×T4

When its radius and temperature change to2R and 2T respectively,E‘=σ×4π(2R)2×(2T)4

E‘/E=64.

9. On a hilly region, water boils at 95°C.What is the temperature expressed inFahrenheit?a) 100°F


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b) 203°Fc) 150°Fd) 20.3°F

Answer: bExplanation: (F-32)/9=C/5=95/5F = 171+32 = 203°F.

10. A composite rod made of copper(1.8×10(-5) K(-1)) and steel (α=1.2×10(-5)

K(-1)) is heated. Then ___________a) It bends with steel on concave sideb) It bends with copper on concave sidec) It does not expandd) Data is insufficient

Answer: aExplanation: αcopper>αsteelCopper expands more than steel. So rodbends with copper on convex side and steelon concave side.

11. Temperatures of two stars are in ratio 3:2.If wavelength of maximum intensity of firstbody is 4000 Å, what is correspondingwavelength of second body?a) 9000 Åb) 6000 Åc) 2000 Åd) 8000 Å

Answer: bExplanation: ((ʎm)‘)/ʎm = T/T‘ = 3/2(ʎm)‘=3/2 ʎm(ʎm)‘=3/2×4000=6000Å.

12. A piece of blue glass heated to a hightemperature and a piece of red glass at roomtemperature, are taken inside a room that isdimly lit, then ___________a) The blue piece will look blue and red willlook as usualb) Red looks brighter and blue looks ordinarybluec) Blue shines like brighter red compared to

the red piecesd) Both the pieces will look equally red

Answer: cExplanation: According to Stefan’s law, E isproportional to T4

As the temperature of blue glass is more thanthat of red glass, so it will appear brighterthan red glass.

TOPIC 3.3 EXPANSION JOINTS

1. Identify the given joint in ConcreteStructures.

a) Horizontal construction joint b) Vertical construction joint

c) Expansion construction joint d) Water tank joint

Answer: a Explanation: The construction joints are

provided at locations were the construction isstopped either at the end of the day or for anyother reason. The provisions of theconstruction joint become necessary to ensureproper bond between the old work and thenew work.



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a) Horizontal construction jointb) Expansion construction jointc) Vertical construction jointd) Water tank joint

Answer: cExplanation: The construction joint maybehorizontal or vertical. For an inclined orcurved member of the joint should be at rightangle to the axis of the member. It isnecessary to determine the location ofconstruction joints well in advance for theviewpoint of structural stability.

3. Identify the given type of joint in ConcreteStructures.

a) L beam construction jointb) T beam construction jointc) Expansion jointd) Contraction Joint

Answer: bExplanation: In case of T-beams, the ribsshould be filled with concrete first and in theslabs forming the flanges can be filled up tothe centre of the ribs. If a construction jointbetween slab and beam becomes unavoidable

especially as in the case of long and deepbeams, that T beams are used.


a) Expansion jointb) Contraction Jointc) Water tank jointd) Vertical construction joint

Answer: cExplanation: For water tanks and otherstructured which store water, the strips ofcopper, aluminium, galvanized iron or othercorrosion resistant material known as waterstops or waterbars, are placed in constructionjoint as shown in given figure above.


a) Partial contraction jointb) Complete Contraction Jointc) Horizontal construction jointd) Dummy joint

Answer: dExplanation: Above figure shows anotherform of contraction joint. It is also known as adummy joint and in this case, a groove of 3mm width is created in the concrete memberto act as a joint. The groove is filled with thejoint filler and its depth is about 1/3 to 1/5 ofthe total thickness of the member.


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6. For water tanks and other structure whichstore water, the strips of copper, aluminium,galvanized iron or other collision resistancematerial, known as the ________a) Jointsb) Waterstopsc) Dowelsd) Fillers

Answer: bExplanation: The function of waterstops is toseal the joint against the passage of water.The waterstop may also be the natural andsynthetic rubber or polyvinyl chloride(PVC).

7. The expansion and contraction jointsgenerally consist of some elastic material,known as ________ which should becompressible, rigid, cellular and Resilient.a) Keysb) Joint fillerc) Keysd) Fillers

Answer: bExplanation: The usual joint filler are built-in strips of metal, bitumen treated felt, canefibre board, cork bound with rubber or resin,dehydrated cork, natural cork, softwood freefrom knots, etc.

8. The _________ are provided in expansionand contraction joints to transfer the load.a) Dowelsb) Fillersc) Joint fillerd) Waterbars

Answer: aExplanation: The contraction joints areinstalled to allow for shrinkage movement inthe structure. It may either be a completecontraction joint or a partial constructionjoint. In the former case, there is completediscontinuity of both concrete and steel.

9. The ___________ is the most effectiveprocess of repairing concrete work which has

been damaged due to inferior work or otherreasons.a) Groutingb) Scrapingc) Dewateringd) Guniting

Answer: dExplanation: In Guniting, the surface to betreated is cleaned and washed. The nozzle isgenerally kept at a distance of about 750 mmto 850 mm from the surface to be treated andvelocity of nozzle values from 120 m/sec to160 m/sec.

10. The _______ is a mixture of cement andsand, the usual proportion being 1:3.a) Mortarb) Slurryc) Gunited) Concrete

Answer: cExplanation: A cement gun is used to depositthe Gunite mixture on the concrete surfaceunder pressure of about 20 N/cm2 to 30N/cm2. The cement is mixed with slightlymoist sand and the necessary water is addedas the mixture comes out from the cementgun.

TOPIC 3.4 BIMETALLIC STRIPS

1. Measurement of elevated temperatures isdefined as ___________

a) Thermometry b) Pyrometry

c) Metallography d) Radiography

Answer: b

Explanation: Pyrometry deals with elevatedtemperatures, generally around 950 F. Theapparatus that is used in this process is knownas a pyrometer. Thermometry generally dealswith the measurement of temperatures below950 F.


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2. What temperature does the dark red colorgenerally deal with?a) 950 Fb) 1150 Fc) 1175 Fd) 1300 F

Answer: bExplanation: Temperature of metals can beestimated by simply looking at the color ofthe hot body. Dark red is assigned atemperature of 1150 F, whereas for faint red,dark cherry, and cherry red it is 950 F, 1175 F,and 1300 F in that order.

3. What temperature is the dark orange colorassociated with?a) 1475 Fb) 1650 Fc) 1750 Fd) 1800 F

Answer: bExplanation: Temperature of metals can beestimated by simply looking at the color ofthe hot body. Dark orange is associated with atemperature of about 1150 F, while for brightcherry, orange, and yellow it is 1475 F, 1750F, and 1800 F correspondingly.

4. Bimetallic strips are employed in________ thermometers.a) Vapor-pressureb) Liquid-expansionc) Metal-expansiond) Resistance

Answer: cExplanation: Bimetallic strips made bybonding of high-expansion and low-expansion metals are used in the commonthermostat. When used as an industrialtemperature indicator, these can be bent into acoil.

5. Bimetallic strips contain _______ as ametal.a) Muntz metal

b) Yellow brassc) Bronzed) Aluminum

Answer: bExplanation: Bimetallic strips include invaras one metal and yellow brass as another. Forhigher temperatures, nickel alloy can be used.These can be used in temperatures rangingfrom -100 F to 1000 F.

6. Why is invar used in bimetallic strips?a) Low densityb) Low coefficient of expansionc) High-temperature resistanced) High abrasion resistance

Answer: bExplanation: Most bimetallic strips arecomposed of invar and yellow brass as metal.Invar has the advantage of low coefficient ofexpansion, whereas yellow brass has theability to be used at low temperatures.

7. _______ is commonly used in liquid-expansion thermometers.a) Bourdon tubeb) Spinning rotor gaugec) McLeod gauged) Manometer

Answer: aExplanation: Liquid-expansion thermometersconsist of a bulb and an expansible device.The bulb is exposed to the temperature thatneeds to be measured and usually, a Bourdontube is used as an expansion device. Theseare connected by capillary tubing and arefilled with a medium.

8. Resistance thermometer generally makesuse of ________ for the measurement ofresistance.a) Potentiometerb) Adruinoc) Diode bridged) Wheatstone bridge


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Answer: dExplanation: Resistance thermometers arebased on the principle of increase in electricalresistance with increasing temperature. Itconsists of a resistance coil mounted in aprotecting tube which is connected to aresistance measuring instrument. Generally,Wheatstone bridge is used in this process.

9. Which of these materials is not used forresistance coils?a) Nickelb) Copperc) Titaniumd) Platinum

Answer: cExplanation: Resistance coils are generallymade of nickel, copper, or platinum. Nickeland copper can be used in the temperaturerange of 150-500 F, whereas platinum can beused between -350 to 1100 F.

10. Liquid expansion thermometers are filledwith ________a) Mercuryb) Amalgamc) Galliumd) Cesium

Answer: aExplanation: The liquid-expansionthermometer has the entire system filled withan organic liquid or mercury. Mercury is usedat a temperature range of -35 to 950 F.Alcohol and creosote are used at -110 to 160F, and 20 to 400 F respectively.

TOPIC 3.5 THERMALCONDUCTION, CONVECTIONAND RADIATION

1. The sun shines on a 150 m2 road surface soit is at 45°C. Below the 5cm thickasphalt(average conductivity of 0.06 W/m K),is a layer of rubbles at 15°C. Find the rate of

heat transfer to the rubbles.a) 5300 Wb) 5400 Wc) 5500 Wd) 5600 W

Answer: bExplanation: There is conduction through theasphalt layer.heat transfer rate = k A ∆T/∆x = 0.06 × 150×(45-15)/0.05= 5400 W.

2. A pot of steel(conductivity 50 W/m K),with a 5 mm thick bottom is filled with liquidwater at 15°C. The pot has a radius of 10 cmand is now placed on a stove that delivers 250W as heat transfer. Find the temperature onthe outer pot bottom surface assuming theinner surface to be at 15°C.a) 15.8°Cb) 16.8°Cc) 18.8°Cd) 19.8°C

Answer: aExplanation: Steady conduction, Q = k A∆T/∆x ⇒ ∆Τ = Q ∆x / kΑ∆T = 250 × 0.005/(50 × π/4 × 0.22) = 0.796T = 15 + 0.796 = 15.8°C.

3. A water-heater is covered with insulationboards over a total surface area of 3 m2. Theinside board surface is at 75°C and theoutside being at 20°C and the conductivity ofmaterial being 0.08 W/m K. Find thethickness of board to limit the heat transferloss to 200 W ?a) 0.036 mb) 0.046 mc) 0.056 md) 0.066 m

Answer: dExplanation: Steady state conductionthrough board.


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Q = k A ∆T/∆x ⇒ ∆Τ = Q ∆x / kΑ∆x = 0.08 × 3 ×(75 − 20)/200 = 0.066 m.

4. On a winter day with atmospheric air at−15°C, the outside front wind-shield of a carhas surface temperature of +2°C, maintainedby blowing hot air on the inside surface. If thewind-shield is 0.5 m2 and the outsideconvection coefficient is 250 W/Km2, findthe rate of energy loss through front wind-shield.a) 125 Wb) 1125 Wc) 2125 Wd) 3125 W

Answer: cExplanation: Q (conv) = h A ∆Τ = 250 × 0.5× [2 − ( −15)] = 250 × 0.5 × 17 = 2125 W.

5. A large heat exchanger transfers a total of100 MW. Assume the wall separating steamand seawater is 4 mm of steel, conductivity15 W/m K and that a maximum of 5°Cdifference between the two fluids is allowed.Find the required minimum area for the heattransfer.a) 180 m2

b) 280 m2

c) 380 m2

d) 480 m2

Answer: dExplanation: Steady conductionQ = k A ∆T/∆x ⇒ Α = Q ∆x / k∆ΤA = 100 × 10^6 × 0.004 / (15 × 5) = 480 m2.

6. The black grille on the back of arefrigerator has a surface temperature of 35°Cwith a surface area of 1 m2. Heat transfer tothe room air at 20°C takes place withconvective heat transfer coefficient of 15W/Km^2. How much energy is removedduring 15 minutes of operation?a) 202.5 kJb) 212.5 kJ

c) 222.5 kJd) 232.5 kJ

Answer: aExplanation: Q = hA ∆T ∆t, Q = 15 × 1 ×(35-20)×15×60 = 202500 J = 202.5 kJ.

7. A small light bulb (25 W) inside arefrigerator is kept on and 50 W of energyfrom the outside seeps into the refrigeratedspace. How much of temperature differenceto the ambient(at 20°C) must the refrigeratorhave in its heat exchanger having an area of 1m2 and heat transfer coefficient of 15 W/Km2

to reject the leak of energy.a) 0°Cb) 5°Cc) 10°Cd) 15°C

Answer: bExplanation: Total energy that goes out =50+25 = 75 W75 = hA∆T = 15 × 1 × ∆T hence ∆T = 5°C.

8. As the car slows down, the brake shoe andsteel drum continuously absorbs 25 W.Assume a total outside surface area of 0.1 m2

with a convective heat transfer coefficient of10 W/Km2 to the air at 20°C. How hot doesthe outside brake and drum surface becomewhen steady conditions are reached?a) 25°Cb) 35°Cc) 45°Cd) 55°C

Answer: cExplanation: ∆Τ = heat / hA hence ∆T = [Τ(BRAKE) − 20 ] = 25/(10 × 0.1) = 25°CΤ(BRAKE) = 20 + 25 = 45°C.

9. A burning wood in the fireplace has asurface temperature of 450°C. Assume theemissivity to be 1 and find the radiantemission of energy per unit area.a) 15.5 kW/m2


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b) 16.5 kW/m2

c) 17.5 kW/m2

d) 18.5 kW/m2

Answer: aExplanation: Q /A = 1 × σ T^4= 5.67 × 10–8×( 273.15 + 450)4

= 15505 W/m2 = 15.5 kW/m2.

10. A radiant heat lamp is a rod, 0.5 m long,0.5 cm in diameter, through which 400 W ofelectric energy is deposited. Assume thesurface emissivity to be 0.9 and neglectingincoming radiation, find the rod surfacetemperature?a) 700Kb) 800Kc) 900Kd) 1000K

Answer: dExplanation: Outgoing power equals electricpowerT4= electric energy / εσA= 400 / (0.9 × 5.67 ×10-8× 0.5 × π × 0.005)= 9.9803 ×10^11 K4 ⇒ T = 1000K.

11. A water-heater is covered up withinsulation boards over a total surface area of 3m2. The inside board surface is at 75°C andthe outside surface is at 20°C and the boardmaterial has a conductivity of 0.08 W/m K.How thick a board should it be to limit theheat transfer loss to 200 W ?a) 0.066 mb) 0.166 mc) 0.266 md) 0.366 m

Answer: aExplanation: Steady state conductionthrough a single layer board.Δx = kA(ΔT)/QΔx = (0.08*3)*(75-20)/200 = 0.066 m.

12. Find the rate of conduction heat transferthrough a 1.5 cm thick hardwood board, k =0.16 W/m K, with a temperature differencebetween the two sides of 20°C.a) 113 W/m2

b) 213 W/m2

c) 230 W/m2

d) 312 W/m2

Answer: bExplanation: . q = .Q/A = k ΔT/Δx = 0.16Wm /K × 20K/0.015 m = 213 W/m2.

13. A 2 m2 window has a surface temperatureof 15°C and the outside wind is blowing air at2°C across it with a convection heat transfercoefficient of h = 125 W/m2K. What is thetotal heat transfer loss?a) 2350 Wb) 1250 Wc) 2250 Wd) 3250 W

Answer: dExplanation: .Q = h A ΔT = 125 W/m2K × 2m2 × (15 – 2) K = 3250 W.

14. A radiant heating lamp has a surfacetemperature of 1000 K with ε = 0.8. Howlarge a surface area is needed to provide 250W of radiation heat transfer?a) 0.0035 m2b) 0.0045 m2c) 0.0055 m2d) 0.0065 m2

Answer: cExplanation: .Q = εσAT^4A = .Q/(εσT4) = 250/(0.8 × 5.67 × 10-8 ×10004)= 0.0055 m2.

TOPIC 3.6 HEATCONDUCTIONS IN SOLIDS


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1. A flat wall of fire clay, 50 cm thick andinitially at 25 degree Celsius, has one of itsfaces suddenly exposed to a hot gas at 950degree Celsius. If the heat transfer coefficienton the hot side is 7.5 W/m2 K and the otherface of the wall is insulated so that no heatpasses out of that face, determine the timenecessary to raise the center of the wall to350 degree Celsius. For fire clay brick

Thermal conductivity = 1.12 W/m K Thermal diffusivity = 5.16 * 10 -7 m2/s

a) 43.07 hours b) 53.07 hours c) 63.07 hours d) 73.07 hours

Answer: a Explanation: t – t a/t 0 – t a = 0.86. Also, α

T/l 2 = 0.32.

2. Glass spheres of 2 mm radius and at 500degree Celsius are to be cooled by exposingthem to an air stream at 25 degree Celsius.Find maximum value of convectivecoefficient that is permissible. Assume thefollowing property values

Density = 2250 kg/m3 Specific heat = 850 J/kg K

Conductivity = 1.5 W/m K a) 245 W/m2K

b) 235 W/m2K c) 225 W/m2K d) 215 W/m2K

Answer: cExplanation: l = volume/surface area = r/3.So, h = (0.1) (k) (3)/r.

3. The transient response of a solid can bedetermined by the equation. (Where, P isdensity, V is volume, c is specific heat and Ais area)a) – 4 p V c = h A (t – t0)b) – 3 p V c = h A (t – t0)c) – 2 p V c = h A (t – t0)d) – p V c = h A (t – t0)

Answer: dExplanation: It can be determined by relatingrate of change of internal energy withconductive heat exchange at the surface.

4. A 2 cm thick steel slab heated to 525degree Celsius is held in air stream having amean temperature of 25 degree Celsius.Estimate the time interval when the slabtemperature would not depart from the meanvalue of 25 degree Celsius by more than 0.5degree Celsius at any point in the slab. Thesteel plate has the following thermal physicalpropertiesDensity = 7950 kg/m3

C P = 455 J/kg KK = 46 W/m Ka) 6548 sb) 6941 sc) 4876 sd) 8760 s

Answer: bExplanation: t – t a/t I – t a = exponential (- hA T/p V c). Now A/V = 100 per meter.

5. An average convective heat transfercoefficient for flow of air over a sphere hasbeen measured by observing the temperature-time history of a 12 mm diameter coppersphere (density = 9000 kg/m3 and c = 0.4 kJ/kg K) exposed to air at 30 degree Celsius.The temperature of the sphere was measured


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by two thermocouples one located at thecenter and the other near the surface. Theinitial temperature of the ball was 75 degreeCelsius and it decreased by 10 degree Celsiusin 1.2 minutes. Find the heat transfercoefficienta) 27.46 W/m2 Kb) 21.76 W/m2 Kc) 29.37 W/m2 Kd) 25.13 W/m2 K

Answer: dExplanation: t – t a/t I – t a = exponential (- hA T/p V c). So, h = 25.13 W/m2 K.

6. Transient condition meansa) Conduction when temperature at a pointvaries with timeb) Very little heat transferc) Heat transfer with a very little temperaturedifferenced) Heat transfer for a short time

Answer: aExplanation: The term transient or unsteadystate designates a phenomenon which is timedependent.

7. Which of the following is not correct in atransient flow process?a) The state of matter inside the controlvolume varies with timeb) There can be work and heat interactionsacross the control volumec) There is no accumulation of energy insidethe control volumed) The rate of inflow and outflow of mass aredifferent

Answer: cExplanation: In transient heat conductionthere is accumulation of energy inside thecontrol volume.

8. A cylindrical stainless steel (k = 25 W/mK) ingot, 10 cm in diameter and 25 cm long,passes through a heat treatment furnace which

is 5 meter in length. The initial ingottemperature is 90 degree Celsius, the furnacegas is at 1260 degree Celsius and thecombined radiant and convective surfacecoefficient is 100 W/m2 K. Determine themaximum speed with which the ingot movesthrough the furnace if it must attain 830degree Celsius temperature. Take thermaldiffusivity as 0.45 * 10 -5 m2/sa) . 000116 m/sb) .000216 m/sc) . 000316 m/sd) . 000416 m/s

Answer: bExplanation: t – t a/t I – t a = exponential (- hA T/p V c). Now, A/V = 2(r + L)/r L = 0.48per cm. Also, T = 1158.53 second so requiredvelocity is 0.25/1158.53.

9. The curve for unsteady state cooling orheating of bodies isa) Hyperbolic curve asymptotic both to timeand temperature axisb) Exponential curve asymptotic both to timeand temperature axisc) Parabolic curve asymptotic to time axisd) Exponential curve asymptotic to time axis

Answer: dExplanation: α/α 0 = exponential [- h A T/p cV], which represents an exponential curve.

10. What is the wavelength band for TV rays?a) 1 * 10 3 to 34 * 10 10 micron meterb) 1 * 10 3 to 2 * 10 10 micron meterc) 1 * 10 3 to 3 * 10 10 micron meterd) 1 * 10 3 to 56 * 10 10 micron meter

Answer: bExplanation: This is the maximum andminimum wavelength for TV rays.

TOPIC 3.7 THERMALCONDUCTIVITY


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TOPIC 3.8 FORBE'S AND LEE'SDISC METHOD: THEORY ANDEXPERIMENT

1. Thermal conductivity is defined as the heatflow per unit time

a) When the temperature gradient is unity b) Across the wall with no temperature

c) Through a unit thickness of the wall d) Across unit area where the temperature

gradient is unity

Answer: d Explanation: Thermal conductivity of a

material is because of migration of freeelectrons and lattice vibrational waves.

2. Mark the matter with least value of thermalconductivity

a) Air b) Water

c) Ash d) Window glass

Answer: a

Explanation: For air, it is .024 W/ m degreei.e. lowest.

3. Which one of the following forms of waterhave the highest value of thermalconductivity?

a) Boiling water b) Steam

c) Solid ice d) Melting ice

Answer: c

Explanation: For ice, it is 2.25 W/m degreei.e. maximum.

4. The average thermal conductivities ofwater and air conform to the ratio

a) 50:1 b) 25:1 c) 5:1

d) 15:1

Answer: bExplanation: For water, it is 0.55-0.7 W/mdegree and for air it is .024 W/m degree.

5. Identify the very good insulatora) Saw dustb) Corkc) Asbestos sheetd) Glass wool

Answer: dExplanation: Glass wool has a lowestthermal conductivity of 0.03 W/m degreeamongst given option.

6. Most metals are good conductor of heatbecause ofa) Transport of energyb) Free electrons and frequent collision ofatomsc) Lattice defectsd) Capacity to absorb energy

Answer: bExplanation: For good conductors, theremust be electrons that are free to move.

7. Heat conduction in gases is due toa) Elastic impact of moleculesb) Movement of electronsc) EM Wavesd) Mixing of gases

Answer: aExplanation: If there is elastic collision thenafter sometime molecules regain its naturalposition.

8. The heat energy propagation due toconduction heat transfer will be minimum fora) Leadb) Waterc) Aird) Copper

Answer: cExplanation: It is because air has lowest


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value of thermal conductivity amongst givenoptions.

9. Cork is a good insulator becausea) It is flexibleb) It can be powderedc) Low densityd) It is porous

Answer: dExplanation: Cork has thermal conductivityin the range of 0.05-0.10 which is very low soit can be porous.

10. Choose the false statementa) For pure metal thermal conductivity ismoreb) Thermal conductivity decreases withincrease in the density of the substancec) Thermal conductivity of dry material islower than that of damp materiald) Heat treatment causes variation in thermalconductivity

Answer: bExplanation: Thermal conductivity increasewith increase in the density of a substance.

TOPIC 3.9 CONDUCTIONTHROUGH COMPOUND MEDIA(SERIES AND PARALLEL)

1. A composite wall generally consists of a) One homogenous layer

b) Multiple heterogeneous layers c) One heterogeneous layer

d) Multiple homogenous layers

Answer: b Explanation: Walls of houses where bricks

are given a layer of plaster on either side.

2. Three metal walls of the same thicknessand cross sectional area have thermalconductivities k, 2k and 3k respectively. Thetemperature drop across the walls (for sameheat transfer) will be in the ratio

a) 3:2:1b) 1:1:1c) 1:2:3d) Given data is insufficient

Answer: aExplanation: As, δ1 = δ2 = δ3 and crosssectional areas are same i.e. temperature dropvaries inversely with thermal conductivity.

3. A composite wall is made of two layers ofthickness δ1 and δ2 having thermalconductivities k and 2k and equal surface areanormal to the direction of heat flow. Theouter surface of composite wall are at 100degree Celsius and 200 degree Celsius. Theminimum surface temperature at the junctionis 150 degree Celsius. What will be the ratioof wall thickness?a) 1:1b) 2:1c) 1:2d) 2:3

Answer: cExplanation: Q = k 1 A 1 d t 1 / δ1 = k 2 A 2 dt 2 / δ2 Also areas are same.

4. Let us say thermal conductivity of a wall isgoverned by the relation k = k0 (1 + α t). In that case the temperature at the mid-plane of the heat conducting wall would bea) Av. of the temperature at the wall facesb) More than average of the temperature atthe wall facesc) Less than average of the temperature at thewall facesd) Depends upon the temperature differencebetween the wall faces

Answer: bExplanation: k0 is thermal conductivity at 0degree Celsius. Here β is positive so it ismore than average of the temperature at thewall faces.


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5. Heat is transferred from a hot fluid to acold one through a plane wall of thickness(δ), surface area (A) and thermal conductivity(k). The thermal resistance isa) 1/A (1/h1 + δ/k + 1/h2)b) A (1/h1 + δ/k + 1/h2)c) 1/A (h1 + δ/k + h2)d) A (h1 + δ/k + 1/h2)

Answer: aExplanation: Net thermal resistance will besummation of resistance through plane walland from left side and right side of the wall.

6. Find the heat flow rate through thecomposite wall as shown in figure. Assumeone dimensional flow and take

k 1 = 150 W/m degree k 2 = 30 W/m degree

k 3 = 65 W/m degree k 4 = 50 W/m degree AB = 3 cm, BC = 8 cm and CD = 5 cm. The

distance between middle horizontal line fromthe top is 3 cm and from the bottom is 7 cm

a) 1173.88 W b) 1273.88 W c) 1373.88 W d) 1473.88 W

Answer: b Explanation: Q = d t/ R T. R T = R 1 + R e q +

R 2 = 0.02 + 0.01469 + 0.1 = 0.2669degree/W.

7. A pipe carrying steam at 215.75 degreeCelsius enters a room and some heat is gainedby surrounding at 27.95 degree Celsius. Themajor effect of heat loss to surroundings willbe due to

a) Conduction b) Convection

c) Radiationd) Both conduction and convection

Answer: cExplanation: As there is temperaturedifference so radiation suits well.

8. “Radiation cannot be affected throughvacuum or space devoid of any matter”. Trueor falsea) Trueb) False

Answer: bExplanation: It can be affected only by airbetween molecules and vacuum of anymatter.

9. A composite slab has two layers havingthermal conductivities in the ratio of 1:2. Ifthe thickness is the same for each layer thenthe equivalent thermal conductivity of theslab would bea) 1/3b) 2/3c) 2d) 4/3

Answer: dExplanation: 2(1) (2)/1+2 = 4/3.

10. A composite wall of a furnace has twolayers of equal thickness having thermalconductivities in the ratio 2:3. What is theratio of the temperature drop across the twolayers?a) 2:3b) 3:2c) 1:2d) log e 2 : log e 3

Answer: bExplanation: We know that temperature isinversely proportional to thermalconductivity, so the ratio is 2:3.


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TOPIC 3.10 THERMALINSULATION

1. According to Indian Standards, it isrecommended that the overall thermaltransmittance of a roof should not be morethan _______ kcal/m2 h deg C.

a) 2.00 b) 5.00 c) 7.00 d) 9.00

Answer: a Explanation: According to Indian Standards,

it is recommended that the overall thermaltransmittance of a roof should not be morethan 2.00 kcal/m2 h deg C. It is alsorecommended that the thermal dampness of aroof should be less than 75%.

2. The process of direct transmission of heatthrough a material is known as __________

a) Conduction b) Radiation

c) Thermal insulation d) Thermal energy

Answer: a

Explanation: The process of directtransmission of heat through a material isknown as conduction. The amount of heattransfer by this process depends on variousfactors like temperature difference, theconductivity of the medium, the time forwhich the flow takes place etc.

3. Thermal insulation keeps the room cool inwinters and hot in summers.

a) True b) False

Answer: b Explanation: Thermal insulation keeps the

room hot in winters and cool in summerswhich results in comfortable living. Itminimises heat transfer and helps in saving

fuel to maintain the desired temperature in theroom.

4. The amount of heat flow through a unitarea of material of unit thickness in one hour,when the temperature difference ismaintained at 1°C is known as ___________of the material.a) Thermal conductivityb) Thermal resistivityc) Thermal conductanced) Thermal resistance

Answer: aExplanation: The amount of heat flowthrough a unit area of material of unitthickness in one hour, when the temperaturedifference is maintained at 1°C is known asthe thermal conductivity. It is denoted by k.Its units are W/(mK).

5. The reciprocal of thermal conductivity isknown as __________a) Thermal conductanceb) Surface resistancec) Specific conductanced) Thermal resistivity

Answer: dExplanation: The reciprocal of thermalconductivity is known as thermal resistivity.It is given by 1/k. Its units are (mK)/W.

6. Which of the following is the correctrelationship between thermal resistance andthermal conductivity?a) R = k/Lb) R = k.Lc) R = L/kd) R = L2/k

Answer: cExplanation: Thermal resistance and thermalconductivity are related by the equation R =L/k. Here, R is the thermal resistance, k is thethermal conductivity and L is the thickness.Thermal resistance is the reciprocal ofthermal conductance.


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7. Surface resistance is the reciprocal of__________a) Surface coefficientb) Surface resistivityc) Surface conductanced) Surface conductivity

Answer: aExplanation: Surface resistance is thereciprocal of the surface coefficient. It isgiven by 1/f where f denotes the surfacecoefficient. Its units are (m2K)/W.

8. Thermal damping is given by the equationD = (T-t/T) x 100. Here, T denotes_________a) Outside temperature rangeb) Inside temperature rangec) Total outside and inside temperature ranged) Thickness

Answer: aExplanation: Thermal damping is given bythe equation D = (T-t/T) x 100. Here, Tdenotes outside temperature range and tdenotes inside temperature range. Also,thermal damping is denoted by D.

9. Which of the following is the correctrelation between thermal transmittance andthermal time constant?a) T = Q x Ub) T = k/Uc) T = Q/Ud) T = U/Q

Answer: cExplanation: Thermal transmittance andthermal time constant are related by theequation T = Q/U. Here, T denotes thethermal time constant, U is the thermaltransmittance and Q is the quantity of heatstored.

10. Which of the following is not a quality ofa good thermal insulating material?a) It should be durableb) It should have a low thermal resistance

c) It should be readily availabled) It should be fireproof

Answer: bExplanation: A good thermal insulatingmaterial should be durable. It should havehigh thermal resistance. It should be fireproofand readily available.

11. Reflective sheet materials used as thermalinsulating material have ___________reflectivity and __________ emissivity.a) High, highb) High, lowc) Low, lowd) Low, high

Answer: bExplanation: Reflective sheet materials usedas thermal insulating material have highreflectivity and low emissivity. Because ofthis property, it offers high heat resistance.

12. Thermal insulation of roofs can beobtained by covering the top exposed surfaceof the roof with ________a) 2.5 cm thick layer of mud mortarb) 2.5 cm thick layer of coconut pitch cementconcretec) 7.5 cm thick layer of mud mortard) 7.5 m thick layer of coconut pitch cementconcrete

Answer: bExplanation: Thermal insulation of roofs canbe obtained by covering the top exposedsurface of the roof with 2.5 cm thick layer ofcoconut pitch cement concrete. Coconut pitchcement concrete is prepared by mixingcoconut pitch with water and cement.

13. Presence of moisture in the thermalinsulating material increases thermalinsulation while the presence of air spacesdecreases thermal insulation.a) Trueb) False


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Answer: bExplanation: Presence of moisture in thethermal insulating material decreases thermalinsulation while the presence of air spacesincreases thermal insulation. The choice ofthe thermal insulating material depends onvarious factors like cost of material, area tobe covered, coat of heating or cooling, etc.

14. According to the Indian Standards, it isrecommended that thermal damping of a wallshould not be less than _______a) 40%b) 50%c) 60%d) 80%

Answer: cExplanation: According to the IndianStandards, it is recommended that the thermaldamping of a wall should not be less than60%. Hence, it is also recommended that thethermal time constant should not be less than16 h.

15. According to Indian Standards, overallthermal transmittance of a wall should not bemore than __________ kcal/m2 h deg C.a) 1.3b) 2.2c) 3.7d) 4.6

Answer: bExplanation: According to Indian Standards,overall thermal transmittance of a wall shouldnot be more than 2.2 kcal/m2 h deg C. Heatinsulation of exposed walls can be achievedby increasing the thickness of the wall.

TOPIC 3.11 APPLICATIONS:HEAT EXCHANGERS,REFRIGERATORS, OVENS ANDSOLAR WATER HEATERS.

1. What is solar water heater?a) Use solar energy to heat waterb) Use solar energy to generate current whichis then used to heat waterc) Use water to generate heatd) Use solar energy to generate steam

Answer: aExplanation: Solar water heater is a systemthat converts sunlight into heat. This heat isthen used to heat water. As the water getsheated, steam may be produced but thepurpose of solar water is to heat water and notproduce steam. It does not generate current.

2. Which of the following determinescomplexity and size of solar water heatingsystem?a) Foodb) Changes in ambient temperaturec) Chemicalsd) Solar radiation constant

Answer: bExplanation: Changes in ambienttemperature during day-night cycle is one ofthe factors that determines the complexity andsize of solar water heating system. Food,chemicals and solar radiation constant doesnot influence the complexity and size of thesystem.

3. What is freeze protection in a solar waterheating system?a) Ensures that the system is frozenb) Prevents the operation of drainback systemc) Prevents damage to system due to freezingof transfer fluidd) Ensures that the transfer fluid is frozen

Answer: cExplanation: Freeze protection in a solarwater system prevents the system beingdamaged due to freezing of transfer fluid. Itdoes not prevent the operation of drainbacksystem.


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4. What are drainback systems in solar waterheating system?a) The system that reverses the direction offlow of transfer fluidb) The system that tracks the sunc) The system that pumps excess transferfluidd) The system that drains the transfer fluid

Answer: dExplanation: Drainback systems are systemsthat drain the transfer fluid particularly toensure freeze protection. This prevents thefreezing of transfer fluid and any unwanteddamage to the system.

5. How does freeze-tolerance work?a) By expansion of pipes carrying transferfluidb) By compression of pipes carrying transferfluidc) By increasing the temperature of pipescarrying transfer fluidd) By increasing the pressure inside pipescarrying transfer fluid

Answer: aExplanation: Freeze-tolerance works byexpansion of pipes carrying the transfer fluid.The low pressure pipes are made of siliconerubber that expands on freezing.

6. Which of the following metals are used tomake pipes of low cost solar water heatingsystem?a) Goldb) Copperc) Polymerd) Silver

Answer: bExplanation: Copper is used to make pipesof low cost solar water heating systems.Though silver and gold are good thermalconductors they are expensive. Polymer is nota metal.

7. Direct solar water heating systems ______a) offer great overheating protectionb) are called pumped systemsc) offer no overheating protectiond) offer great freeze protection

Answer: cExplanation: Direct solar water heatingsystems are also called compact systems.They offer little or no overheating protectionunless they have a heat export pump.

8. How is the heat transferred from transferfluid to potable water in indirect solar waterheating systems?a) By directly exposing the substance tosunlightb) By using an electrical heaterc) By circulating potable water through thecollectord) By using heat exchanger

Answer: dExplanation: An indirect solar water heatingsystem uses a heat exchanger to transfer heatfrom the transfer fluid to the potable water. Itdoes not expose the transfer fluid directly tothe sunlight and does not use an electricalheater.

9. How is water heated in a direct solar waterheating system?a) By circulating potable water through thecollectorb) By directly exposing water to sunlightc) By using convection from a differenttransfer fluidd) By using heat exchanger

Answer: aExplanation: In a direct solar water heatingsystem, the potable water is the transfer fluid.Hence, it is heated by circulating through thecollector. Indirect solar water heating systemsuse a heat exchanger.

10. Passive systems rely on heat-drivenconvection.


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a) Falseb) True

Answer: bExplanation: Passive systems rely on heat-driven convection. If not, they also use heatpipes to circulate the working fluid throughthe collector and heat it. Hence, they arecheap and are easily maintained.

11. Which of the following is an example ofdirect solar water heating system?a) Pressurised antifreeze systemb) Pumped systems to circulate transfer fluidc) Convection heat storage systemd) Drainback system

Answer: cExplanation: Convection heat storage systemis similar to an integrated collector storagesystem. Both these systems are examples ofdirect solar water heating systems.

12. How is the heat transfer fluid (HTF)heated in bubble pump systems?a) By subjecting the closed HTF circuit tohigh pressureb) By subjecting the closed HTF circuit tohigh pressure and by increasing the volumec) By subjecting the closed HTF circuit tolow pressure and by decreasing the volumed) By subjecting the closed HTF circuit tolow pressure

Answer: dExplanation: In a bubble pump system, theheat transfer fluid circuit is subjected to a lowpressure. This causes the liquid to boil at lowtemperatures as the sun heats it. The volumeis not changed.

13. Batch collectors reduce heat loss bythermally insulating the storage tank.a) Trueb) False

Answer: aExplanation: Batch collectors reduce heat

loss by thermally insulating the storage tank.This is done by covering the tank in a glass-topped box that allows heat from sun to reachthe water tank and traps it – greenhouseeffect.

14. Overheat protection is done by passinghot water through collector during night.a) Falseb) True

Answer: bExplanation: Overheat protection is done bypassing hot water through collector duringnight or when there is less sunlight. This isextremely effective in direct or thermal storeplumbing and ineffective in evacuated-tubecollectors.

UNIT IV QUANTUMPHYSICS

TOPIC 4.1 BLACK BODYRADIATION

1. As the wavelength of the radiationdecreases, the intensity of the black bodyradiations ____________

a) Increases b) Decreases c) First increases then decrease

d) First decreases then increase

Answer: c Explanation: In the case of Black Body

radiations, as the body gets hotter thewavelength of the emitted radiationdecreases. However, the intensity firstincreases up to a specific wavelength thanstarts decreasing, as the wavelength continuesto decrease.


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2. The radiations emitted by hot bodies arecalled as ________________a) X-raysb) Black-body radiationc) Gamma radiationsd) Visible light

Answer: bExplanation: The phenomenon of black—body radiations was given by Max Planck. Hestated that hot bodies emit radiation over awide range of wavelengths. An ideal body isthe one that emits and absorbs radiation of allfrequencies. Such a body called a Black Bodyand the radiations are called Black bodyradiations.

3. An iron rod is heated. The colors atdifferent temperatures are noted. Which of thefollowing colors shows that the iron rod is atthe lowest temperature?a) Redb) Orangec) Whited) Blue

Answer: aExplanation: As the body gets hotter, thefrequency of the emitted radiation keeps onincreasing. Blue color has the highestfrequency out of red, orange and white. Thus,as the iron rod gets heated first it wouldbecome red, then orange, then white and thenfinally blue.

4. A black body is defined as a perfectabsorber of radiations. It may or may not be aperfect emitter of radiations.a) Trueb) False

Answer: bExplanation: A black body is defined as theone which is a perfect absorber as well as aperfect emitter of radiations. Such a bodywould absorb all the radiations falling on itand would emit all of them when heated.

5. From the figure, what’s the relationbetween T1, T2, and T3?

a) T1 > T2 > T3 b) T3 > T2 > T2 c) T3 > T1 > T2 d) T2 > T1 > T3 Answer: b

Explanation: We already know, as thetemperature of the body is higher, theintensity of the black body radiations wouldbe higher. Thus, from the graph, theradiations with temperature T3 has the highestintensity followed by the one withtemperature T2 and then T1. Thus, T3> T2 >T1.

6. Electromagnetic wave theory of light couldnot explain Black Body radiations.

a) True b) False

Answer: a Explanation: According to electromagnetic

theory, the absorption and the emissionshould be continuous. As the wavelengthkeeps decreasing, the intensity of the emittedradiations should keep increasing to infinity.Such is not the case with Black BodyRadiations.

7. The unit of absorptive power is_______________


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a) Tb) Ts-1

c) Tsd) No unit

Answer: dExplanation: Absorptive power can bedefined as the ratio of energy absorbed perunit area upon energy incident per unit timeper unit area. For a black body, it’s absorptivepower is equal to one.

8. For an object other than a black body, it’semissivity, e is _______________a) 1b) 0 < e < 1c) e > 1d) e = 0

Answer: bExplanation: Emissivity is the ratio ofemissive power of any object and theemissive power of the black body having thesame temperature and surface area as theobject. Thus, for a black body, it is equal to 1.For any other object, it is less than 1.

9. What relation between emissivity, e, andAbsorptive Power, a, is given by Kirchhoff’slaw?a) e < ab) e > ac) e = ad) no specific relation

Answer: cExplanation: Kirchhoff’s law states that forany object the emissivity is always equal toabsorptive power. For a black body, both ofthem are equal to one.

10. What is the relation between the Energiesas shown in the figure?

a) Er = 0 b) Ea = 0 c) Et = Ei d) Ei = Er

Answer: a Explanation: As a black body is a perfect

absorber, the reflected energy and thetransmitted energy should be zero. Also, theenergy of the incident radiation should beequal to the energy absorbed.

TOPIC 4.2 PLANCK'S THEORY(DERIVATION)

1. The energy emitted by a black surfaceshould not vary in accordance with

a) Wavelength b) Temperature c) Surface characteristics

d) Time

Answer: d Explanation: It is time independent. For a

prescribed wavelength, the body radiatesmuch more energy at elevated temperatures.

2. In the given diagram let r be the length ofthe line of propagation between the radiatingand the incident surfaces. What is the value ofsolid angle W?


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a) A sin α b) A cos α c) 2A cos α d) 2A cos α

Answer: b Explanation: The solid angle is defined by a

region by the rays of a sphere, and ismeasured as A n/r2.

3. Likewise the amount of emitted radiation isstrongly influenced by the wavelength even iftemperature of the body is

a) Constant b) Increasing

c) Decreasing d) It is not related with temperature

Answer: a

Explanation: Temperature must remainconstant in order to emit radiation.

4. A small body has a total emissive power of4.5 kW/m2. Determine the wavelength ofemission maximum

a) 8.46 micron m b) 7.46 micron m c) 6.46 micron m d) 5.46 micron m

Answer: d Explanation: (Wavelength) max t = 2.8908 *

10 -3.

5. The sun emits maximum radiation of 0.52micron meter. Assuming the sun to be a blackbody, Calculate the emissive ability of thesun’s surface at that temperature

a) 3.47 * 10 7 W/m2 b) 4.47 * 10 7 W/m2

c) 5.47 * 10 7 W/m2

d) 6.47 * 10 7 W/m2

Answer: cExplanation: E = σ b t 4 = 5.47 * 10 7 W/m2.

6. The law governing the distribution ofradiant energy over wavelength for a blackbody at fixed temperature is referred to asa) Kirchhoff’s lawb) Planck’s lawc) Wein’s formulad) Lambert’s law

Answer: bExplanation: This law gives a relationbetween energy over wavelength.

7. The Planck’s constant h has the dimensionsequal toa) M L 2 T -1

b) M L T -1

c) M L T -2d) M L T

Answer: aExplanation: It has unit equal to J s and itsvalue is 6.626 * 10 -34.

8. Planck’s law is given bya) (E) b = 2 π c 2 h (Wavelength) -5/[c h/k(Wavelength) T] – 2b) (E) b = π c 2 h [exponential [c h/k(Wavelength) T] – 3].c) (E) b = 2 π c 2 h (Wavelength)-5/exponential [c h/k (Wavelength) T] – 1d) (E) b = 2 c 2 h (Wavelength) -5/exponential[c h/k (Wavelength) T] – 6

Answer: cExplanation: Planck suggested the followinglaw for the spectral distribution of emissivepower.


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9. A furnace emits radiation at 2000 K.Treating it as a black body radiation, calculatethe monochromatic radiant flux density at 1micron m wavelengtha) 5.81 * 10 7 W/m2

b) 4.81 * 10 7 W/m2

c) 3.81 * 10 7 W/m2

d) 2.81 * 10 7 W/m2

Answer: dExplanation: (E) b = C 1 (Wavelength)-5/exponential [C 2/ (Wavelength) T] – 1.

10. A metal sphere of surface area 0.0225 m2

is in an evacuated enclosure whose walls areheld at a very low temperature. Electriccurrent is passed through resistors embeddedin the sphere causing electrical energy to bedissipated at the rate of 75 W. If the spheresurfaces temperature is measured to be 560 K,while in steady state, calculate emissivity ofthe sphere surfacea) 0.498b) 0.598c) 0.698d) 0.798

Answer: bExplanation: E = e A σ b T.

TOPIC 4.3 COMPTON EFFECT:THEORY AND EXPERIMENTALVERIFICATION

1. Which of the following is the characteristicof a black body?

a) A perfect absorber but an imperfectradiator

b) A perfect radiator but an imperfectabsorber

c) A perfect radiator and a perfect absorber d) A perfect conductor

Answer: c

Explanation: When the radiations are made

to pass through a black body, it undergoesmultiple reflections and is completelyabsorbed. When it is placed in a temperaturebath of fixed temperature, the heat radiationswill come out. Thus a black body is a perfectabsorber and a perfect reflector.

2. The energy distribution is not uniform forany given temperature in a perfect blackbody.a) Trueb) False

Answer: aExplanation: At different temperatures, whena perfect black body is allowed to emitradiations, then the distribution of energy fordifferent wavelengths at various temperaturesis not uniform.

3. Rayleigh-Jean’s law holds good for whichof the following?a) Shorter wavelengthb) Longer wavelengthc) High temperatured) High energy

Answer: bExplanation: According to this law, theenergy distribution is directly proportional tothe absolute temperature and is inverselyproportional to the fourth power of thewavelength. Therefore longer the wavelength,greater is the energy distribution.

4. Wien’s displacement law holds good onlyfor shorter wavelength.a) Falseb) True

Answer: bExplanation: This law states that, the productof the wavelength, corresponding tomaximum energy and the absolutetemperature, is constant. If ʎ is less, then 1/ʎwill be great. Therefore e(hc/ʎKT) will begreat.


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5. Which of the following does not affect thephoton?a) Magnetic or electric fieldb) Light wavesc) Gravityd) Current

Answer: aExplanation: Photons have no charge. Theycan interact with charged particles but notwith themselves. This is why photons areneutral and not affected by magnetic orelectric fields.

6. What is Compton shift?a) Shift in frequencyb) Shift in chargesc) Shift in radiationd) Shift in wavelength

Answer: dExplanation: When a photon collides with anelectron at rest, the photon gives its energy tothe electron. Therefore the scattered photonwill have higher wavelength compared to thewavelength of the incident photon. This shiftin wavelength is called Compton shift.

7. Compton shift depends on which of thefollowing?a) Incident radiationb) Nature of scattering substancec) Angle of scatteringd) Amplitude of frequency

Answer: cExplanation: From the theory of Comptoneffect it is deducted that change inwavelengthΔʎ = h/mc (1-cosɵ). This equation shows thatthe change in wavelength is independent ofthe incident radiation as well as the nature ofscattering substance. The shift depends onlyon the angle of scattering.

8. Which of the following is called as non-mechanical waves?a) Magnetic waves

b) Electromagnetic wavesc) Electrical wavesd) Matter waves

Answer: bExplanation: The waves which travel in theform of oscillating electric and magneticwaves are called electromagnetic waves. Suchwaves do not require any material for theirpropagation and are called non-mechanicalwaves.

9. Which of the following is associated withan electron microscope?a) Matter wavesb) Electrical wavesc) Magnetic wavesd) Electromagnetic waves

Answer: aExplanation: The waves associated withmicroscopic particles when they are in motionare called matter waves. Electron microscopemakes use of the matter waves associatedwith fast moving electrons.

10. A radio station broadcasts its programmeat 219.3 metre wavelength. Determine thefrequency of radio waves if velocity of radiowaves is 3×108 m/s.a) 7.31×10-7 Hzb) 1.954×10-6 Hzc) 1.368×106 Hzd) 6.579×1010 Hz

Answer: cExplanation: ʎ = velocity/frequencyFrequency = velocity/ʎTherefore, frequency = 1.368×106 Hz.

11. Calculate the de-Broglie wavelength of anelectron which has been accelerated from reston application of potential of 400volts.a) 0.1653 Åb) 0.5125 Åc) 0.6135 Åd) 0.2514 Å


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Answer: cExplanation: de-Broglie wavelength = h/√(2×m×e×V)De-Broglie wavelength = (6.625×10-14)/√(2×9.11×10-31×1.6×10-19×400)Wavelength = 0.6135 Å.

TOPIC 4.4 WAVE PARTICLEDUALITY

1. When a pebble is dropped into a pond ofstill water, what happens?

a) Particles move b) Waves move

c) The pebble moves d) Water moves

Answer: b

Explanation: When a pebble is thrown in stillwater, a circular pattern of alternate crestsspread out. The kinetic energy makes theparticles to oscillate which comes in contactwith it. The energy gets transferred to theparticles of the next layer which also beginsto oscillate. Thus it is the disturbance orwaves that move forward and not the particlesof the medium.

2. Mechanical waves are called elastic waves. a) True

b) False

Answer: a Explanation: Waves which require a medium

for their propagation are called mechanicalwaves. They are also called elastic wavesbecause they depend on the elastic propertiesof a medium.

3. What are the essential properties a mediummust possess for the propagation ofmechanical waves?

a) Stable pressure b) Maximum friction

c) Constant temperature d) Minimum friction

Answer: dExplanation: The friction force amongst theparticles of the medium should be negligiblysmall so that they continue oscillating for asufficiently long time and the wave travels asufficiently long distance through the medium

4. Transverse waves can be formed in fluids.a) Trueb) False

Answer: bExplanation: Transverse waves travel in theform of crests and troughs. They involvechanges in the shape of the medium. So theycan be transmitted through media which haverigidity. As fluids do not sustain shearingstress, transverse waves cannot be formed inthem.

5. Which of the following waves can betransmitted through solids, liquids and gases?a) Transverse wavesb) Electromagnetic wavesc) Mechanical wavesd) Longitudinal waves

Answer: dExplanation: Longitudinal waves involvechanges in the volume and density of themedium. Since all media can sustaincompressive stress, longitudinal waves can betransmitted through all the three types ofmedia.

6. For an aluminium the modulus of rigidityis 2.1×1010 N/m2 and density is 2.7×103

kg/m3. Find the speed of transverse waves inthe medium.a) 27.9×103 m/sb) 2.79×103 m/sc) 25.14×103 m/sd) 24.1×103 m/s

Answer: bExplanation: Speed = √(Ƞ/ƿ)Speed = 2.79×103 m/s.


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7. Sound travels through a gas under which ofthe following condition?a) Isothermal conditionb) Non-isothermal conditionc) Adiabatic conditiond) Transverse condition

Answer: cExplanation: The compressions andrarefactions are formed so rapidly that theheat generated in the regions of compressionsdoes not get time to pass into the regions ofrarefactions so as to equalize the temperature.So when sound travels through gas, thetemperature remains constant. Therefore, it isadiabatic.

8. What kind of wave is formed in organpipes?a) Transverse stationary wavesb) Electromagnetic wavesc) Mechanical wavesd) Longitudinal stationary waves

Answer: dExplanation: When two identicallongitudinal waves travelling in oppositedirections overlap, a longitudinal stationarywave is formed. Thus, the waves produced inorgan pipes are longitudinal stationary waves.

9. A wave transmits momentum. Can’t ittransfer angular momentum?a) Yesb) No

Answer: bExplanation: A wave transmittingmomentum cannot transmit angularmomentum because a transfer of angularmomentum means the action of a torquewhich causes rotator motion.

10. What is the most fundamental property ofwave?a) Temperatureb) Pressure

c) Frequencyd) Wavelength

Answer: cExplanation: When a wave travels from onemedium to other, its wavelength as well asvelocity may change. This is the reason thatfrequency is the fundamental property of awave.

11. Which of the following is also known aspressure waves?a) Transverse wavesb) Longitudinal wavesc) Mechanical wavesd) Stationary waves

Answer: bExplanation: Longitudinal waves travel in amedium as series of alternate compressionsand rarefactions and hence are called pressurewaves.

12. In which medium sound travels faster?a) Solidb) Liquidc) Gasd) Water vapour

Answer: aExplanation: Sound travels in solid with thehighest speed because the coefficient ofelasticity of solids is much greater than thecoefficient of elasticity of liquids and gases.

TOPIC 4.5 ELECTRONDIFFRACTION

1. TEM and SEM are the same microscopytechniques.

a) True b) False

Answer: a Explanation: Transmission electron

microscopy (TEM) and scanning electronmicroscopy (SEM) are the different types of


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electron microscopy. These use electrons forcasting the images of an object.

2. The resolving power of TEM is derivedfrom _______________a) electronsb) specimensc) powerd) ocular system

Answer: aExplanation: The resolving power of atransmission electron microscope is derivedfrom the wave-like property of electrons thatpass through the specimen. In SEM, theelectrons reflect back from the specimen.

3. The cathode of transmission electronmicroscope consists of a____________________a) tungsten wireb) bulbc) iron filamentd) gold wire

Answer: aExplanation: The cathode of a transmissionelectron microscope (TEM) is located on topof the column, it contains a tungsten wirefilament that is heated to provide the sourceof electrons.

4. The resolution attainable with standardTEM is less than the theoretical value.a) Trueb) False

Answer: aExplanation: The resolution that can beattained with a standard transmission electronmicroscope is about two orders of magnitudeless than the theoretical value. This is due tospherical aberration of electron-focusinglenses.

5. During TEM, a vacuum is created insidethe _________________________a) room of operation

b) specimenc) columnd) ocular system

Answer: cExplanation: To prevent the prematurescattering of electrons by collision with thegas molecules, a vacuum is generated throughwhich the electrons travel, in the column priorto operation.

6. Which of the following component of TEMfocuses the beam of electrons on the sample?a) ocular lensb) condenser lensc) staged) column

Answer: bExplanation: The condenser lens focuses theelectron beam on to the specimen, in case oftransmission electron microscope. Thespecimen is supported on the grid holder andplaced inside the column.

7. Image formation in electron microscope isbased on ___________________________a) column lengthb) electron numberc) differential scatteringd) specimen size

Answer: cExplanation: In case of the electronmicroscope, the image formation is based onthe differential scattering of the electrons byparts of the specimen. The scattering ofelectrons is proportional to the size nuclei ofthe atoms that make up the sample.

8. The biological materials have little intrinsiccapability to ____________________a) scatter electronsb) stainc) remain viabled) be captured


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Answer: aExplanation: The insoluble materials of cellscontain atoms of low atomic number such ascarbon, hydrogen, oxygen and nitrogen. Thebiological materials therefore have very littleintrinsic capability of scattering the electrons.

9. Glutaraldehyde is a ________________a) metalb) fixativec) non-metald) atomic species

Answer: bExplanation: Glutaraldehyde and osmiumtetroxide are common fixatives used in thetransmission electron microscopy for thefixation of biological specimens. They stainas well as keep the sectioned specimens in astate of similarity with the living counterpart.

10. Osmium is a ___________________a) non metalb) heavy metalc) alloyd) light metal

Answer: bExplanation: Osmium is a heavy metal thatreacts with fatty acids leading to thepreservation of membranes. Osmiumtetroxide is used as a fixative in transmissionelectron microscopy.

11. In TEM, the tissue is stained by floatingon drops of ______________________a) hydrocarbonsb) slow-molecular weight stainsc) heavy metal soutionsd) oil immersion

Answer: cExplanation: The tissue is stained by floatingon drops of uranyl acetate and lead citrate(heavy metal solutions). These solutionswhen bound to the specimen, provide thedensity required to scatter the electron beam.

12. Shadow casting is a technique ofvisualizing ___________________a) isolated particlesb) mountsc) shoot tipsd) root tips

Answer: aExplanation: Shadow casting is a techniqueof viewing isolated particles. The particles aremade to cast shadows after their placement insealed chambers. The chamber contains afilament of carbon and a heavy metal.

TOPIC 4.6 CONCEPT OF WAVEFUNCTION AND ITS PHYSICALSIGNIFICANCE

TOPIC 4.7 SCHRODINGER'SWAVE EQUATION

TOPIC 4.8 TIME INDEPENDENTAND TIME DEPENDENTEQUATIONS

1. Which of the following is the correctexpression for the Schrödinger wavefunction?

a) b) c)

d)

Answer: d Explanation: The correct expression for the

Schrödinger wave equation is . Schrodinger

equation is a basic principle in itself.

2. For a quantum wave particle, E =_____________a) &hbar; k

b) &hbar; ω

iℏ dΨdt = −i ℏ

2m∂Ψ∂x + UΨ

iℏ dΨdt

= −i ℏ2m

∂2Ψ∂x2 + UΨ

iℏ dΨdt

= −i ℏz

2m∂Ψ∂x + UΨ

iℏ dΨdt = −i ℏz

2m∂2Ψ∂x2 + UΨ

iℏ dΨdt = −i ℏz

2m∂2Ψ∂x2 + UΨ


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c) &hbar; ω/2d) &hbar; k/2

Answer: bExplanation: The Energy of a wave particleis given as &hbar; ω while the momentum ofthe particle is given as &hbar; k. These arethe desired relation.

3. Schrodinger Wave equation can be derivedfrom Principles of Quantum Mechanics.a) Trueb) False

Answer: bExplanation: Schrodinger equation is a basicprinciple in itself. It cannot be derived fromother principles of physics. Only, it can beverified with other principles.

4. Which of the following can be a wavefunction?a) tan xb) sin xc) cot xd) sec x

Answer: bExplanation: Out of all the given options, sinx is the only function, that is continuous andsingle-valued. All the rest of the functions areeither discontinuous or double-valued.

5. Which of the following is not acharacteristic of wave function?a) Continuousb) Single valuedc) Differentiabled) Physically Significant

Answer: dExplanation: The wave function has nophysical significance. It merely helps indetermining the state of a particle. It is thesquare of the wave function that has aphysical significance.

6. Find the function, f(x), for which X f(x) = where a is the real quantity.

a) ke-x2

b) ke-x2/2a

c) ke-x2/2a2

d) ke-x2/2a

Answer: c Explanation: Now, given that, X f(x) =

X f(x) =

df/f = -xdx/a2 ln f = -x2/2a2 + C

f = ke-x2/2a2

.

7. dΨ/dx must be zero.a) True

b) False

Answer: b Explanation: For a wave function, dΨ/dx,

must be continuous and single-valuedeverywhere, just like Ψ. Also, Ψ must benormalizable.

8. Any wave function can be written as alinear combination of _________________

a) Eigen Vectors b) Eigen Values

c) Eigen Functions d) Operators

Answer: c

Explanation: A wave function describes thestate of a particle. It does not have a physicalsignificance. Moreover, it can be written as alinear combination of Eigen functions, i.e.,Ψ(x) = AF(x) + BG(x).

9. The Schrödinger is a differential equation. a) True

b) False

Answer: b Explanation: The Schrodinger wave equation

− iℏ a

2pxf(x),

− iℏ a

2pxf(x).− i

ℏ a2pxf(x)/dx


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generated is a partial differential equation. Itis a basic principle in itself and cannot bederived from other principles of physics.There are two types of partial differentialequation time dependent form and steady-state form.

10. Which of the following can be a solutionof Schrodinger equation?a)

b)

c)

d)

Answer: c Explanation: Out of the following, only the

below diagram can be the solution of theSchrodinger Wave equation. because otherdiagram does not have a continuous dΨ/dx.Some diagrams are double valued anddiscontinuous also.


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TOPIC 4.9 PARTICLE IN A ONE-DIMENSIONAL RIGID BOX

1. The walls of a particle in a box aresupposed to be ____________

a) Small but infinitely hard b) Infinitely large but soft

c) Soft and Small d) Infinitely hard and infinitely large

Answer: d

Explanation: The simplest quantum-mechanical problem is that of a particle in abox with infinitely hard walls and areinfinitely large.

2. The wave function of the particle lies inwhich region?

a) x > 0 b) x < 0 c) 0 < X < L

d) x > L

Answer: c Explanation: The particle cannot exist

outside the box, as it cannot have infiniteamount of energy. Thus, it’s wave function isbetween 0 and L, where L is the length of theside of the box.

3. The particle loses energy when it collideswith the wall.

a) True b) False

Answer: b Explanation: The total energy of the particle

inside the box remains constant. It does notloses energy when it collides with the wall.

4. The Energy of the particle is proportionalto __________

a) n b) n-1

c) n2 d) n-2

Answer: cExplanation: In a particle inside a box, theenergy of the particle is directly proportionalto the square of the quantum state in whichthe particle currently is.

5. For a particle inside a box, the potential ismaximum at x = ___________a) Lb) 2Lc) L/2d) 3L

Answer: aExplanation: In a box with infinitely highbarriers with infinitely hard walls, thepotential is infinite when x = 0 and when x =L.

6. The Eigen value of a particle in a box is___________a) L/2b) 2/Lc)

d)

Answer: d Explanation: The wave function for the

particle in a box is normalizable, when thevalue of the coefficient of sin is equal to

. It is the Eigen value of the wave function.

7. Particle in a box can never be at rest. a) True

b) False

Answer: a Explanation: If the particle in a box has zero

energy, it will be at rest inside the well and itviolates the Heisenberg’s UncertaintyPrinciple. Thus, the minimum energypossessed by a particle is not equal to zero.

8. What is the minimum Energy possessed bythe particle in a box?

a) Zero

√L/2√2/L

√2/L


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b) c)

d)

Answer: b Explanation: The minimum energy

possessed by a particle inside a box withinfinitely hard walls is equal to . Theparticle can never be at rest, as it will violateHeisenberg’s Uncertainty Principle.

9. The wave function of a particle in a box isgiven by ____________

a)

b)

c)

d)

Answer: bExplanation: The wave function for theparticle in a box is given by: .The Energy possessed by the particle is givenby: .

10. The wave function for which quantumstate is shown in the figure?

a) 1 b) 2 c) 3 d) 4

Answer: b Explanation: The shown wave function is for

the 2nd principal quantum number, i.e., it isthe wave function for the state when n = 2.

11. Calculate the Zero-point energy for aparticle in an infinite potential well for anelectron confined to a 1 nm atom.a) 3.9 X 10-29 Jb) 4.9 X 10-29 Jc) 5.9 X 10-29 Jd) 6.9 X 10-29 J

Answer: cExplanation: Here, m = 9.1 X 10-31 kg, L =10-9m.Therefore, E =

= 3.14 X 3.14 X 1.05 X 1.05 X 10-68/ 2 X 9.1X 10-31 X 10-9

= 5.9 X 10-29 J.

TOPIC 4.10 TUNNELLING(QUALITATIVE)

1. Tunneling is required in case of____________a) Laying pavement

b) Laying road c) On ground passage

d) Underground passage

Answer: d Explanation: Tunnel can be defined as

artificial underground passage, which iscreated for different purposes. It is required incase of highways, railways, sewerage andwater supply.

2. The line at which the tunnel wall breaksfrom sloping outward can be given as_________a) Spring line

b) Oval line c) Centre line

d) Middle line

Answer: a Explanation: Spring line is determined as the

line at which the wall breaks from slopingoutward to sloping inward toward the crown.

π2ℏ2

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√ 2L sin

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L

√ 2Lsin πx

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This acts as a barrier between the outwardand inward regions.

3. Which of the following should beconsidered while aligning a tunnel?a) Atmospheric conditionsb) Hydrological conditionsc) Climatic conditionsd) Surface limits

Answer: bExplanation: The determination of thealignment for tunnel can be done based ongeological and hydrological conditions, cross-section and length of the continuous tunnel,time of consideration and limit of the surface.

4. Among the following, which doesn’tbelong to tunnel classification?a) Firm groundb) Running groundc) Rocky groundd) Soft ground

Answer: cExplanation: Tunneling has been classifiedbased on the type of strata present. It includesfirm ground, soft ground and running ground.These will determine the bearing capacity ofthe soil.

5. Which method can be adopted if full faceexcavation is not possible?a) Back bearing methodb) Plottingc) Trenchingd) Benching

Answer: dExplanation: In case of no possibility of fullface excavation, top heading method isadopted for having a better output. Benchingprocess is also adopted for digging smalltunnels.

6. Among the following, which can beadopted for providing support for soft strata?a) Bents of aluminum

b) Bents of iridiumc) Bents of steeld) Bents of plastic

Answer: cExplanation: The provision of soft stratamust be done for withstanding the excavation.It can be provided by bents of wood, bents ofsteel, liner plates and poling, which areplaced to retain material between the adjacentbents.

7. A steel cylinder which is pushed in the softsoil is determined as ________a) Jar bornb) Shieldc) Rodd) Pole

Answer: bExplanation: In the case of soft grounds,tunneling can be dangerous and cave-ins arecommon. To prevent this, an iron or steelcylindrical element called shield is placed inthe soft soil, which can crave the holeperfectly.

8. Which method can be adopted in case ofrock tunneling?a) Full face methodb) Benchingc) Tracingd) Back bearing method

Answer: aExplanation: Rock tunneling involves thesame principle as of the tunneling in theground. Those include full face method, topheading method and drift method. Based onthe conditions of the area, these can beadopted.

9. Which of the following can act as analternative for blasting?a) Tunnelingb) Continuous blastingc) Sequential blastingd) Fire-setting


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Answer: dExplanation: Blasting is the conventionalmethod which is adopted in case of rocktunneling. In the case of fire-setting, tunnel isheated with fire and then cooled with water.Rapid expansion and contraction weakensand rock and tends to break.

10. Which of the following involves in thesequence of rocky strata?a) Marking profileb) Placing rocksc) Improving foul gasesd) Recording values

Answer: aExplanation: Rocky strata involve usage ofblasting method for developing tunnels. Thesequence contains marking profile, loadingexplosive, removing foul gases, checking,scaling, mucking and bolting.

TOPIC 4.11 SCANNINGTUNNELLING MICROSCOPE.

1. Which of the following is used in electronmicroscope?

a) electron beams b) magnetic fields c) light waves

d) electron beams and magnetic fields

Answer: d Explanation: Electron Microscope uses

electron beams and magnetic fields toproduce the image, whereas the lightmicroscope uses light waves and glass lenses.In electron microscopy, a much higherresolution is obtained with extremely shortwavelength of the electron beam.

2. Electron Microscope can give amagnification up to ___________

a) 400,000X b) 100,000X c) 15000X

d) 100X

Answer: aExplanation: The resolving power of theelectron microscope is more than 100 timesthat of the light microscope, and it producesuseful magnification up to 400,000X. It ispossible to resolve objects as small as 10Angstrom.

3. Which of the following are true forelectron microscopy?a) specimen should be thin and dryb) image is obtained on a phosphorescentscreenc) electron beam must pass through evacuatedchamberd) specimen should be thin and dry, image isobtained on a phosphorescent screen andelectron beam must pass through evacuatedchamber

Answer: dExplanation: Since electrons can travel onlyin high vacuum, the entire electron paththrough the instrument must be evacuated;specimens must be completely dehydratedprior to examination. Only very thinspecimens can be observed in theconventional electron microscope since thepenetrating power of electrons through matteris weak. The magnified image may be viewedon a phosphorescent or fluorescent screen.

4. Degree of scattering in transmissionelectron microscope is a function of__________a) wavelength of electron beam usedb) number of atoms that lie in the electronpathc) number and mass of atoms that lie in theelectron pathd) mass of atoms that lie in the electron path

Answer: cExplanation: In a transmission electronmicroscope, contrast results from thedifferential scattering of electrons by thespecimen, the degree of scattering being a


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function of the number and mass of atomsthat lie in the electron path.

5. Negative Staining is used for examining_____________a) virus particlesb) protein moleculesc) bacterial flagellad) virus particles, protein molecules andbacterial flagella

Answer: dExplanation: In negative-staining theelectron opacity of the surrounding field isincreased by using an electron-dense materialsuch as phosphotungstic acid as a stain.Negative staining is particularly valuable forthe examination of very small structures suchas virus particles, protein molecules andbacterial flagella.

6. Which among the following helps us ingetting a three-dimensional picture of thespecimen?a) Transmission Electron Microscopeb) Scanning Electron Microscopec) Compound Microscoped) Simple Microscope

Answer: bExplanation: The scanning electronmicroscope lacks the resolving powerobtainable with the transmission electronmicroscope but has the advantage ofrevealing a striking three-dimensional picture.The surface topography of a specimen can berevealed with clarity and depth of field notpossible by any other method.

7. The secondary electrons radiated back inscanning microscope is collected by?a) specimenb) anodec) vacuum chamberd) cathode

Answer: bExplanation: In scanning electron

microscope (SEM), the surface of thespecimen is irradiated with a very narrowbeam of electrons. Such irradiations causeslow energy (secondary) electrons to beejected from the specimen which can then becollected on a positively-charged plate oranode thereby generating an electric signal.

8. On what factors do the intensity ofsecondary electrons depend upon?a) shape of the irradiated objectb) chemical composition of the irradiatedobjectc) number of electrons ejectedd) size and chemical composition of theirradiated object, number of electrons ejectedand on the number of electrons reabsorbed bysurrounding

Answer: dExplanation: The irradiations in SEM causessecondary electrons to be ejected from thespecimen thereby generating a signal that isproportional to the number of electronsstriking the anode. The intensity or thenumber of secondary electrons depends onthe shape and the chemical composition ofthe irradiated object and also on the numberof electrons ejected and the number ofelectrons reabsorbed by surrounding.

9. Where do we obtain the magnified imageof the specimen in SEM?a) cathode ray tubeb) phosphorescent screenc) anoded) scanning generator

Answer: aExplanation: In TEM, the image is obtainedon a phosphorescent screen but in SEM themagnified image of the surface topography ofthe specimen is obtained on the cathode raytube. The electronic signals generated scanthe specimen in a raster pattern in the mannerof a television system to produce an image ona cathode ray tube.


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10. Which of the following techniques areused in Transmission Electron Microscopy(TEM) for examining cellular structure?a) Negative-Stainingb) Shadow Castingc) Ultrathin Sectioningd) Negative-Staining, Shadow Casting,Ultrathin Sectioning, Freeze-Etching

Answer: dExplanation: Numerous techniques areavailable for use with electron microscopywhich extends its usefulness in characterizingcellular structure. Some of them areNegative-Staining (which increases theelectron opacity of surrounding), ShadowCasting (helps in producing three-dimensional structure of the object), UltrathinSectioning and Freeze-Etching.

UNIT V CRYSTALPHYSICS

TOPIC 5.1 SINGLECRYSTALLINE,POLYCRYSTALLINE ANDAMORPHOUS MATERIALS

1. Which of the following properties isgenerally exhibited by amorphous solids?

a) Anisotropy b) Glass-transition

c) Equal strength of all bonds d) All of the mentioned

Answer: b

Explanation: Due to random organization ofparticles, amorphous solids have the samephysical properties along all directions, or areisotropic. Random organization of particlesalso results in unequal bond strengths. Uponcooling, amorphous solids turn into a brittle

glass-like state from a flexible rubber-likestate. This is called glass-transition.

2. Metal glass was first prepared at:a) California Institute of Technologyb) Massachusetts Institute of Technologyc) Techniond) University of Michigan

Answer: aExplanation: Metal glass or amorphousmetal was reportedly first produced by W.Klement (Jr.), Willens and Duwez in 1960 atCaltech. It was prepared by rapid cooling (~ 1MK/s) of molten metal alloys.

3. Polycrystalline solids are isotropic.a) Trueb) False

Answer: aExplanation: Anisotropy is a characteristicbehavior shown by ideal crystals. However,the presence of flaws like grain boundariescauses the solid to deviate from crystallineproperties.

4. Metal glasses differ from their crystallinecounterparts in many ways. Chiefapplication(s) of metal glasses include(s):a) Bullet-proof glassesb) Power transformersc) Conducting wiresd) All of the mentioned

Answer: bExplanation: Amorphous metals are nottransparent and have relatively lowerelectrical conductivity. However, most metalglasses possess high magnetic susceptibilityand low coercivity.

5. Consider the following cooling diagram foran amorphous solid.


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Glass-transition temperature is represented as: a) A

b) Bc) C


Answer: b Explanation: An exact melting point does not

exist for amorphous solids. An approximateglass-transition temperature is defined byextrapolating the cooling curve as shown.

6. Soda-lime glass is the most common typeof glass. The component present in largestw/w percentage is:

a) SiO2 b) Al2O3 c) Na2O d) CaO

Answer: a

Explanation: Glass inherits its transparencyfrom crystalline SiO2, also called quartz.However, quartz has very high and narrowglass-transition. To overcome this, smallamount of Na2 is added. CaO is also added toprevent water-solubility imparted by soda.

7. Lead-oxide glass is called “crystal glass”because:

a) It contains crystalline Pb b) It contains SiO2 crystals

c) It contains PbO crystalsd) None of the mentioned

Answer: dExplanation: Well, crystal glass isamorphous, not crystalline. This glass earnsits name from its excellent decorativeproperties and high refractive index.

8. Crystallinity increases with increasing rateof cooling of a liquid.a) Trueb) False

Answer: bExplanation: When a liquid is cooled rapidly,the particles get less time to move andarrange themselves in an orderly fashion andhence the crystallinity of the resulting soliddecreases.

TOPIC 5.2 SINGLE CRYSTALS:UNIT CELL, CRYSTALSYSTEMS, BRAVAIS LATTICES,DIRECTIONS AND PLANES IN ACRYSTAL, MILLER INDICES

1. Which of the following is not true forcrystallographic axes?

a) They must be parallel to the edges of theunit cell

b) They must be perpendicular to each otherc) They must originate at one of the verticesof the cell

d) They form a right-handed co-ordinatesystem

View answer

Answer: b Explanation: The axes must be parallel to the

edges of the unit cell, which in case of somecrystal systems like monoclinic, hexagonaletc. are not mutually perpendicular.

2. The point coordinate indices q, r, and s aremultiples of:


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a) Unit cell edge lengthsb) Distance between nearest neighboursc) Cosine of angles between unit cell edgesd) None of the mentionedView answer

Answer: aExplanation: Point coordinate indices are thefractions which when multiplied by thecorresponding unit cell edge lengths, providethe location of a given point in thecrystallographic coordinate system.

3. The point coordinates of the vertex justopposite to the origin area) 0 0 0b) 0 0 1c) 0 1 1d) 1 1 1View answer

Answer: dExplanation: Since the opposite vertex islocated at distances equal to the edge lengthsalong the coordinate axes.

4. If x, y, and z are three positive axes of thecrystallographic coordinate system withorigin at point A, then which line points in thedirection [1 0 1] ?

a) AD

b) CHc) FBd) GEView answer

Answer: cExplanation: Moving 1 unit along positive x-axis, 0 units along positive y-axis, 1 unitalong negative z-axis points in a directionparallel to line FB.

5. In the following diagram, what is thedirection cosine of the line EB?

a) [1 11]b) [111]c) [1 1 1]d) [1 0 0]View answer

Answer: aExplanation: One can reach from point E toB by traversing 1 unit along each of positivex, negative y, and negative z-axis, where oneunit along any axis equals the correspondingedge length.

6. In cubic crystals, crystallographicdirections are arranged in families. Which ofthe following directions does not belong tothe family <110>?a) [1 0 1]


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b) [11 0]c) [10 1̅]d) None of the mentionedView answer

Answer: dExplanation: Since the cubic lattice issymmetrical about all the three axes, theabove directions are equivalent irrespective oforder & sign and are part of same directionfamily.

7. Convert [2 111] from four-index system tothree-index system.a) [2 1 0]b) [3 0 1]c) [111]d) [12 1]View answer

Answer: bExplanation: [u v t w] can be converted to [UV W] using the formula: i) U = 2u+v ii) V =2v+u iii) W = w. Four-index systems aregenerally used for hexagonal lattices.

8. Miller indices of the hatched plane in thefollowing figure are:

a) (2 3 1) b) (3 2 1) c) (3 2 0)

d) (1 1 1)View answer

Answer: aExplanation: If a plane intercepts thecoordinate axes at distances A, B, and C fromthe origin, then Miller indices are given bymultiplying (a/A b/B c/C) by a suitable factorso as to obtain integers.

9. Which of the following is a property ofMiller indices?a) They uniquely identify a planeb) They are always positivec) They are not fractionsd) None of the mentionedView answer

Answer: cExplanation: Two or more planes can havesame Miller indices which can be negative,zero or positive depending on the intercept onthe axes. If the ratios of intercepts to latticeconstants come out be fractional, then theyare scaled to lowest integers to be representedas Miller indices.

10. Miller indices for perpendicular planesare always the same.a) Trueb) FalseView answer

Answer: bExplanation: It is true only for cubic lattices.For other systems, there is no simplerelationship between planes with the sameMiller indices.

TOPIC 5.3 INTER-PLANARDISTANCES

1. X-rays have larger wavelengths than whichof the following?

a) Gamma rays b) Beta rays

c) Microwave


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d) Visible lightView answer

Answer: aExplanation: Larger wavelengths mean lessenergy. Only gamma rays have higher energy(or shorter wavelengths) than x-rays. Betarays are actually streaming of particles andhave much less energy than x-rays.

2. X-ray diffraction patterns are used forstudying crystal structure of solids becausea) They have very high energy, hence theycan penetrate through solidsb) They are electromagnetic radiation, andhence do not interact with matter (crystals)c) Their wavelengths are comparable to inter-atomic distancesd) Their high frequency enables rapidanalysisView answer

Answer: cExplanation: For diffraction to occur, theobstacle size should be comparable to thewavelength of the incident radiation.

3. For destructive interference to take place,the path difference between the two wavesshould be:a) nλb) 2nλc) (n + 1/2)λd) (2n + 1)λView answer

Answer: aExplanation: Constructive interferenceoccurs when the phase difference betweentwo interfering waves is an integral multipleof 2π. Also, the ratio of path difference towavelength equals that of phase difference to2π.

4. Bragg’s law is not a sufficient condition fordiffraction by crystalline solids.a) True

b) FalseView answer

Answer: aExplanation: Atoms present at non-cornerpositions may result in out-of-phasescattering at Bragg angles.

5. The Miller indices h, k, and l of parallelplanes in a BCC lattice should satisfy whichof the following X-ray diffraction reflectionrules?a) h + k + l should be evenb) h, k, and l should all be either even or oddc) h, k, and l should form Pythagoras tripletd) all planes allow reflectionsView answer

Answer: aExplanation: If the sum of Miller indicesbecomes odd for a BCC lattice, destructiveinterference occurs.

6. Minimum interplanar spacing required forBragg’s diffraction is:a) λ/4b) λ/2c) λd) 2λView answer

Answer: bExplanation: Maximum value of incidentangle can be 90° for which sine is 1. Hence d= λ/2 ( nλ = 2.d.sinθ)

7. Laue’s model pictures XRD as reflectionfrom parallel crystalline planes. Reflection isdifferent from refraction as:a) diffraction occurs throughout the bulkb) intensity of diffracted beams is lessc) diffraction in crystals occurs only atBragg’s anglesd) all of the mentionedView answer

Answer: dExplanation: Reflection is a surface


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phenomenon, and large portions of theincident waves can be reflected. Moreover,reflection can occur at any angle of incidencewhereas diffraction patterns (alternative darkand bright bands) occur only at Bragg’sangles.

8. In Bragg’s equation [nλ = 2.d.sinθ], θ is theangle between:a) specimen surface and incident raysb) normal to specimen surface and incidentraysc) parallel lattice surfaces d distance apart andincident raysd) normal to parallel lattice surfaces ddistance apart and incident raysView answer

Answer: cExplanation: In the following figure, one caneasily deduce that the path difference (PQ +QR) between the two incident waves is2.d.sinθ using simple trigonometry.

9. In the powder method of XRD, theintensities of various bright lines arecompared to determine the crystal structure.For simple cubic lattice the ratio of intensitiesat first two maxima are:

a) 1⁄2 b) 3⁄4 c) 1⁄2 d) None of the mentioned

View answer

Answer: aExplanation: For simple cubic lattice, theintensities at subsequent maxima are in theratio 1:2:3:4:5:6:8

10. K-alpha x-rays have shorter wavelengthsthan K-beta x-rays?a) Trueb) FalseView answer

Answer: bExplanation: K-alpha is formed from atransfer of electrons from L shell to K whileK-beta result from M-to-K transition. HenceK-alpha lines have lower energy (or longerwavelength).

TOPIC 5.4 COORDINATIONNUMBER AND PACKINGFACTOR FOR SC, BCC, FCC,HCP AND DIAMONDSTRUCTURES

1. Lead is a metallic crystal having a _______structure.

a) FCC b) BCC c) HCP d) TCP

Answer: a Explanation: Crystalline solids are classified

as either metallic or non-metallic. Pb, alongwith Cu, Ag, Al, and Ni, has a face-centeredcubic structure.

2. Which of the following has a HCP crystalstructure?

a) W b) Mo

c) Cr d) Zr

Answer: d Explanation: Crystalline solids are classified


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as either metallic or non-metallic. W, Mo, andCr are examples of the body-centered cubicstructure of crystals. The HCP structure isfound in Mg, Zn, Ti, Cd, Zr, and others.

3. Amorphous solids have _______ structure.a) Regularb) Linearc) Irregulard) Dendritic

Answer: cExplanation: Materials in which themolecule is the basic structural solid and hasan irregular structure is known as amorphoussolid. Crystalline solids, on the other hand,usually are arranged in a regular manner.

4. At ________ iron changes its BCCstructure to FCC.a) 308oCb) 568oCc) 771oCd) 906oC

Answer: dExplanation: Similar to metallic crystals, afew non-metallic crystals also change formdue to temperature and pressure differences.This process is termed as polymorphism. Ironchanges from BCC at room temperature toFCC form at 906oC.

5. At room temperature, tin is formed into_________a) Gray tinb) White tinc) Red tind) Yellow tin

Answer: bExplanation: Similar to metallic crystals, afew non-metallic crystals also change formdue to temperature and pressure differences.Tin crystallizes in a non-metallic diamondstructure (gray tin) at low temperatures. At

room temperature, it forms a metallicstructure (white tin).

6. Which of the following is a property ofnon-metallic crystals?a) Highly ductileb) Less brittlec) Low electrical conductivityd) FCC structure

Answer: cExplanation: Non-metallic crystals are lessductile and have low electrical conductivity.On the other hand, metallic crystals arediffering since they are more ductile and havehigh electrical conductivity.

7. Which of the following is not anamorphous material?a) Glassb) Plasticsc) Leadd) Rubbers

Answer: cExplanation: Materials in which themolecule is the basic structural solid and hasan irregular structure are known asamorphous solid. Most amorphous materialsare polymers such as plastics and rubbers.The most common amorphous material isglass.

8. The crystal lattice has a _________arrangement.a) One-dimensionalb) Two-dimensionalc) Three-dimensionald) Four-dimensional

Answer: cExplanation: Lattice is defined as the regulargeometrical arrangement of points in crystalspace. Space or crystal lattice is a three-dimensional network of imaginary linesconnecting the atoms.


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9. The smallest portion of the lattice is knownas __________a) Lattice structureb) Lattice pointc) Bravais crystald) Unit cell

Answer: dExplanation: Lattice is defined as the regulargeometrical arrangement of points in crystalspace. The unit cell is the smallest portion ofthe lattice, which when repeated in alldirections gives rise to a lattice structure.

10. Bravais lattice consists of __________space lattices.a) Elevenb) Twelvec) Thirteend) Fourteen

Answer: dExplanation: There are fourteen ways inwhich points can be arranged in a space sothat each has identical surroundings. Thesefourteen space lattices constitute the Bravaisspace lattices.

11. A unit cell that contains lattice points onlyat the corners is known as _________a) Primitive unit cellb) Secondary unit cellc) Layered unit celld) Derived unit cell

Answer: aExplanation: If a unit cell chosen containslattice points only at its corners, it is called aprimitive or simple unit cell. It contains onlyone lattice point since each point at the eightcorners is shared equally with adjacent unitcells.

12. The axial relationship of a monocliniccrystal system is given as ___________a) a = b = cb) a = b ≠ c

c) a ≠ b = cd) a ≠ b ≠ c

Answer: dExplanation: The crystal system is a formatby which crystal structures are classified.Each crystal system is defined by therelationship between edge lengths a, b, and c.For monoclinic, orthorhombic, and triclinicsystems, the axial relationship is given by a ≠b ≠ c.

13. The axial relationship of a rhombohedralcrystal system is given as ___________a) a = b = cb) a = b ≠ cc) a ≠ b = cd) a ≠ b ≠ c

Answer: aExplanation: The crystal system is a formatby which crystal structures are classified.Each crystal system is defined by therelationship between edge lengths a, b, and c.For cubic and rhombohedral systems, theaxial relationship is given by a = b = c.

14. The interracial angles of a hexagonalcrystal system are given by __________a) α = β = ϒ = 90ob) α = β = 90o ϒ = 120oc) α = β = ϒ ≠ 90od) α ≠ β ≠ ϒ ≠ 90o

Answer: bExplanation: The crystal system is a systemby which crystal structures are classified.Each crystal system is defined by therelationship between edge lengths a, b, and cand interaxial angles α, β, and ϒ. Forhexagonal system, the interaxial angles aregiven by α = β = 90o ϒ = 120o.

15. The interracial angles of a triclinic crystalsystem are given by __________a) α = β = ϒ = 90ob) α = β = 90o ϒ = 120o


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c) α = β = ϒ ≠ 90od) α ≠ β ≠ ϒ ≠ 90o

Answer: dExplanation: The crystal system is a systemby which crystal structures are classified.Each crystal system is defined by therelationship between edge lengths a, b, and cand interaxial angles α, β, and ϒ. For thetriclinic system, the interaxial angles aregiven by α ≠ β ≠ ϒ ≠ 90o.

16. What is the atomic radius of a BCCcrystal structure?a) a/2b) a/4c) a√2/4d) a√3/4

Answer: dExplanation: Atomic radius is defined as halfthe distance between the centers of twoneighboring atoms. The atomic radius of asimple cube and HCP is a/2 respectively,whereas it is a√2/4 and a√3/4 for FCC andBCC respectively.

17. What is the coordination number of asimple cubic structure?a) 6b) 8c) 10d) 12

Answer: aExplanation: Coordination number is definedas the number of nearest neighboring atomsin crystals. The coordination number for thesimple cubic structure is 6, whereas it is 8 and12 for BCC and FCC respectively.

18. What is the atomic packing factor of BCCstructure?a) 0.54b) 0.68c) 0.74d) 0.96

Answer: bExplanation: The density of packing in acrystal is determined using the atomicpacking factor (APF). The APF of FCC andHCP structures is 0.74, and 0.54 for simplecubic structure, whereas it is 0.68 for BCCstructure.

Sanfoundry Global Education & LearningSeries – Engineering Materials &Metallurgy.

TOPIC 5.5 CRYSTALIMPERFECTIONS: POINTDEFECTS, LINE DEFECTS

1. Which of the following is a point defect incrystals?

a) Edge dislocation b) Interstitialcies

c) Grain boundaries d) Cracks

Answer: b

Explanation: Crystal defects are classified aspoint defects, line defects, and boundarydefects. Point defects include vacancies,impurities, interstitialcies, and electronicdefects.

2. How can the number of defects bedetermined?

a)

b)

c)

d)

Answer: a Explanation: The number of defects at

equilibrium at a definite temperature can be


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determined using the equation Ne(-Ed⁄kT). It isdenoted by nd. Here, N stands for the totalnumber of atomic spots and Ed is theactivation energy.

3. The defect that occurs due to adisplacement of an ion is known as__________a) Vacancy defectb) Schottky defectc) Frankel defectd) Interstitial defect

Answer: cExplanation: Frankel defect occurs due to adisplacement of an ion from the crystallattice. It is related to the interstitial defect,where an ion simply occupies a positionbetween regular atoms.

4. Which defect does the following figuredepict?

a) Vacancy defectb) Schottky defectc) Frankel defectd) Interstitial defect

Answer: bExplanation: When a pair of positive andnegative ions both disappear from a crystallattice, the effect is called a Schottky defect.It is closely related to vacancy defects wheresimply an ion is missing.

5. Which of these is a Frankel defect?a)

b)

c)


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d)

Answer: aExplanation: Frankel defect occurs due to adisplacement of an ion from the crystal latticewhile retaining all of the ions, unlike otherdefects. It is related to the interstitial defect,where an ion simply occupies a positionbetween regular atoms.

6. Which defect does the following diagramrepresent?

a) Vacancy defectb) Schottky defectc) Frankel defectd) Interstitial defect

Answer: dExplanation: Interstitial defects occur whenan atom occupies an empty position in acrystal lattice. Self-interstitial effects occurdue to their own atoms, while others occurdue to a foreign substance.

7. _______ occurs when a foreign substancereplaces an atom in a crystal.a) Vacancy defect

b) Substitutional impurityc) Frankel defectd) Interstitial impurity

Answer: bExplanation: A substitutional impurityoccurs due to the occupation of a foreignatom in place of an atom in a crystal. On theother hand, interstitial impurities occur whena regular atom occupies a random space in thecrystal lattice.

8. A disturbance in a region between twoideal parts of a crystal is known as ________a) Boundary defectb) Point defectc) Line defectd) Volume defect

Answer: cExplanation: Line defect is regarded as adisturbed region between two perfect parts ofa crystal. They may be of either edgedislocation type or screw dislocation.

9. In screw dislocation, the Burger’s vectorlies _________ to the dislocation line.a) Perpendicularb) Parallelc) At an angled) Sideways

Answer: bExplanation: The Burger’s vector in screwdislocation lies parallel to the dislocation linealong the axis of a line of atoms in the sameplane. On the other hand, it lies at an anglefor edge dislocation.

10. Generation of dislocations can beidentified using _______a) Schottky mechanismb) Burger’s vectorc) Twistd) Frank-Read mechanism

Answer: dExplanation: The Frank-Read mechanism


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uses the Frank-Read source and its operationas a dislocation multiplier. The Frank-Readsource contains a fixed lined at the X and Ynodes. Under the application of stress, thedislocation line expands and is furtheroperated until it becomes readable.

11. What are one-dimensional defects?a) Boundary defectb) Point defectc) Line defectd) Volume defect

Answer: cExplanation: When compared geometrically,line defects are seen as one-dimensionaldefects. Line defects are also known asdislocations, with common types as edge andscrew dislocations.

12. What are two-dimensional defects?a) Boundary defectb) Point defectc) Line defectd) Volume defect

Answer: aExplanation: The defects that occur on thesurface of a material are known as surface orboundary defects. Geometrically, they areregarded as two-dimensional defects.

13. How is the dislocation energy defined?a) J m-1

b) J m-2

c) m-2

d) N m-1

Answer: aExplanation: Dislocation energy is definedad joule per meter and is denoted by E.Dislocation density is defined as meter percubic meter or simply as per meter square.

TOPIC 5.6 BURGER VECTORS,STACKING FAULTS - ROLE OF

IMPERFECTIONS IN PLASTICDEFORMATION

1. Which one of the following, is a lineimperfection?

a) Grain boundary b) Tilt boundary

c) Dislocation d) Stacking fault

Answer: c

Explanation: Dislocations look like lines inthe lattice, which are one dimensional. So itfalls under the category of line imperfections.While grain boundary, tilt boundary andstacking fault are surface imperfections.

2. What is the range of dislocation density inpure and unhardened metals?

a) 1010 – 1012 m-2 b) 1014 – 1016 m-2 c) 104 – 106 m-2

d) 108 – 1010 m-2

Answer: a Explanation: Pure unhardened metals (real

crystal) contain large number of dislocationsof the order of1010 – 1012 m-2. So these caneasily be deformed.

3. What is unit of dislocation density? a) m

b) m-2 c) kg/m3

d) m-3

Answer: b Explanation: Dislocation density is defined

as number of lines in a unit volume. So theunit of dislocation density is m/m3 or m-2.

4. Dislocations are responsible for________________a) Increase in strength

b) Decrease in strength


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c) Increase or decrease in strength dependingon dislocation densityd) Don’t affect strength of metal

Answer: cExplanation: If dislocation density is inrange 1010 – 1012 m-2, it decreases thestrength by easy plastic deformation. On theother hand, if it exceeds to 1016 m-2, thesecauses strengthening of metal by interactionto imperfections.

5. The number of dislocations increasesdrastically during ______a) Solidificationb) Plastic deformationc) Elastic deformationd) Heat treatment

Answer: bExplanation: Dislocations increase innumber drastically during plasticdeformation. It is due to the formation of aFrank-Read source.

6. Positive edge dislocation is denoted by______a) ↶b) ↷c) ꓕd) ꓔ

Answer: cExplanation: Positive edge dislocation isrepresented as ‘ꓕ’ (inverted ꓔ) because theextra half plane lies above the shear plane. ‘ꓔ’is used for negative edge dislocation. ‘↶’ and‘↷’ are used to represent screw dislocations.

7. Edge dislocation introduces shear strainonly.a) Trueb) False

Answer: bExplanation: Edge dislocations causecompressive, tensile and shear lattice strains.

While screw dislocations cause shear strainonly.

8. What term is used for the defect, producedby an array of dislocations that produces asmall difference in orientation between theadjoining lattices?a) Tilt boundaryb) Twist boundaryc) Free surfaced) Low angle grain boundary

Answer: dExplanation: The array of dislocationsproduces an angular mismatch between thelattices, which is referred to as low anglegrain boundary. These have an angle less than100.

9. Which statement is false?a) Plastic deformation decreases dislocationdensity.b) Strain hardening is the increase ofdislocation density with plastic deformation.c) Slip plane is the crystallographic plane ofdislocation motion.d) Dislocation can change its plane of motionby climb on high temperatures.

Answer: aExplanation: Plastic deformation increasesthe dislocation density. Plastic deformationcauses generation of dislocations due to theFrank-Read source. All other statements arecorrect.

10. Most crystalline materials havedislocations in their as formed state.a) Trueb) False

Answer: aExplanation: Crystalline materials havedislocations due to stresses (mechanical,thermal …) associated with the formingprocesses. Thus forming processes are thesource of dislocations.


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11. Dislocation density can vary from _____to _____ in metals.a) 105 – 1012 cm-2

b) 105 – 1012 m-2

c) 108 – 1010 cm-2

d) 106 – 1010 cm-2

Answer: aExplanation: Dislocation density varies inmetals from 105 cm-2 (in carefully solidifiedmetal) to 1012 cm-2 (in heavily deformedmetal). It depends on various parameters asdegree of deformation, temperature etc.

12. Large plastic deformation corresponds to_____ of grains along the direction of appliedstress.a) Growthb) Rupturec) Recrystallizationd) Elongation

Answer: dExplanation: Larger plastic deformationleads to elongation of grains. This becomespossible due to the presence of dislocations.

TOPIC 5.7 GROWTH OFSINGLE CRYSTALS: SOLUTIONAND MELT GROWTHTECHNIQUES.

1. The diagram given below represents whichof the following method?

a) Bridgman methodb) Stockbarger methodc) Czochralski methodd) Zone melting method

Answer: cExplanation: Czochralski method is basicallya method for the growth of a single crystalfrom a melt of the same composition. It isalso widely used for the growth of the crystalsof semiconducting materials, Si, Ge, GaAs,etc. It has also been used to produce lasergenerator materials such as Ca(NbO3)2 dopedwith neodymium.

2. In the process of Czochralski methodwhich of the following relation is appropriatebetween the melt and the growing crystals?a) Melt and the growing crystals are usuallynot related to each otherb) Melt and the growing crystals are usuallyrotated counterclockwisec) Melt and the growing crystals are usuallyrotated clockwised) Melt and the growing crystals are usuallykept at a constant position

Answer: bExplanation: The melt and the growingcrystals are usually rotated counterclockwiseduring pulling in the process of Czochralski,in order to maintain a constant temperature,melt uniformity, etc.


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3. Which of the following statements isappropriate for Stockbarger method?a) Solidification is achieved by passing themelt through a concentration gradientb) Solidification is achieved by passing themelt through a temperature gradientc) Liquefaction is achieved by passing themelt through a concentration gradientd) Liquefaction is achieved by passing themelt through a temperature gradient

Answer: bExplanation: Stockbarger method is based onsolidification of stoichiometric melt but inthese, oriented solidification of the melt isachieved by effectively passing the meltthrough a temperature gradient such thatcrystallization occurs at the cooler end. Thusthis method is achieved by arranging for arelative displacement of the melt and atemperature gradient.

4. Which of the following statementsdescribes best the Bridgman method?a) Melt is outside the temperature furnace,solidification begins at the hotter endb) Melt is inside the temperature furnace,solidification occurs at the cooler endc) Melt is inside the temperature furnace,solidification occurs at the hotter endd) Solidification is achieved by passing themelt through a temperature gradient

Answer: bExplanation: In the Bridgman method, themelt is inside a temperature gradient furnaceand the furnace is gradually cooled so that thesolidification begins at the cooler end. In thismethod, it is again advantageous to use a seedcrystal and atmospheric control may benecessary.

5. In the zone melting method_____________ of the charge is melted atany one time. Fill up the correct option for theblank space from the choices given below.a) Large partb) Small part

c) Solid partd) Anionic part

Answer: bExplanation: In zone melting method thethermal profile through the furnace is suchthat only a small part of the charge is meltedat any one time. Initially, that part of thecharge in contact with the seed crystal ismelted. As the boat is pulled through thefurnace, oriented solidification onto the seedoccurs and at the same time, more of thecharge melts.

6. Which one of the following principle hasbeen used in the zone melting method?a) Impurities concentrate in the solid than inliquid phaseb) Impurities concentrate in the liquid phasethan in gaseous phasec) Impurities concentrate in the liquid phasethan in the solid phased) Impurities concentrate in the gaseous phasethan in the solid phase

Answer: cExplanation: Zone melting method make theuse of the principle that impurities usuallyconcentrate in the liquid rather than in thesolid phase. Impurities are therefore ‘sweptout’ of the crystal by the moving moltenzone. The method has been used for thepurification and crystal growth of the highmelting metals such as tungsten.

7. Which one of the following statements iscorrect for precipitation method?a) Growth of the crystal from a solvent ofsame composition to the crystalsb) Growth of the crystal from the solute ofsame composition to the crystalc) Growth of the crystal from the gaseousspecies of different composition to the crystald) Growth of the crystal from a solvent ofdifferent composition to the crystals.

Answer: dExplanation: In contrast to the methods like


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zone melting, Stockbarger etc method inwhich melts solidify to get crystals that havethe same composition as the melt,precipitation methods involves the growth ofthe crystals from a solvent of differentcomposition to the crystals. The solvent maybe one of the constituents of the desiredcrystal example, crystallization of salt hydratecrystals using water as the solvent.

8. In the precipitation method for the growthof the crystal, the solvent melts are oftenknown as___________a) Electrolyteb) α-particlec) β- particled) Fluxes

Answer: dExplanation: In the precipitation methodwhich used for the growth of crystals fromthe solvent of different composition to thecrystals, the solvent melts are often called asfluxes since they effectively reduce themelting point of the crystals by a considerableamount. The method has recently been usedto grow crystals of β- and β” alumina solidelectrolytes using a borate flux.

9. When was the Verneuil flame fusionmethod first used?a) 1207b) 2016c) 1503d) 1904

Answer: dExplanation: The Verneuil flame fusionmethod was used in 1904 for growing crystalsof high melting oxides, including artificialgemstones such as ruby and sapphire.

10. Which of the following method has beenused to prepare the single crystals of CaOusing plasma torch?a) Stockbarger methodb) Zone melting method

c) Verneuil flame fusion methodd) Bridgman method

Answer: cExplanation: Verneuil flame fusion methodhas been recently used to prepare the singlecrystals of CaO with a melting point equal to2600◦C by using a plasma torch to melt theCaO powder. The starting material in theform of a fine powder is passed through anoxyhydrogen flame or some other hightemperature torch or furnace after melting hastaken place, the droplets fall onto the surfaceof the growing crystals where they solidify.

11. What is the advantage of usingCzochralski, Bridgman- Stockbarger andVerneuil method?a) Gives small crystalsb) High tech apparatusc) Rapid growth ratesd) Uses plasma torch

Answer: cExplanation: The above methods are meltgrowth methods which are used for thegrowth of crystals. Advantages of using thesemelt growth methods are, t gives largecrystals, allows rapid growth rates, andrequires very simple apparatus. While thedisadvantage can be in the crystal qualitywhich can be poor with inhomogeneities andlarge defect concentrations.

12. What is the disadvantage of using asolution growth method for the growth of thecrystals?a) Rapid growth ratesb) Simple apparatusc) Slow growth ratesd) Isothermal conditions

Answer: cExplanation: Solution growth methods likethe water crystallization, flux growth,hydrothermal method etc, which are used forthe growth of the crystals have thedisadvantage that it leads to very slow growth


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rates, face problems of contamination bycontainer or flux. However, the advantage ofusing such methods are that it allows

isothermal conditions with slow growth ratesgive quality crystals of low defectconcentration.


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ph8151 - engineering physics dept. of physics multiple

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