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Multiple choice question for engineering

Set 1

1. Under which of the following operation, NFA is not closed?
a) Negation
b) Kleene
c) Concatenation
d) None of the mentioned

View Answer

Answer: d [Reason:] NFA is said to be closed under the following operations: a) Union b) Intersection c) Concatenation d) Kleene e) Negation.

2. It is less complex to prove the closure properties over regular languages using:
a) NFA
b) DFA
c) PDA
d) Can’t be said

View Answer

Answer: a

3. Which of the following is an application of Finite Automaton?
a) Compiler Design
b) Grammar Parsers
c) Text Search
d) All of the mentioned

View Answer

Answer: d [Reason:] There are many applications of finite automata, mainly in the field of Compiler Design and Parsers and Search Engines.

4. John is asked to make an automaton which accepts a given string for all the occurrence of ‘1001’ in it. How many number of transitions would John use such that, the string processing application works?
a) 9
b) 11
c) 12
d) 15

View Answer

Answer: a

5. Which of the following do we use to form an NFA from a regular expression?
a) Subset Construction Method
b) Power Set Construction Method
c) Thompson Construction Method
d) Scott Construction Method

View Answer

Answer: c [Reason:] Thompson Construction method is used to turn a regular expression in an NFA by fragmenting the given regular expression through the operations performed on the input alphabets.

6. Which among the following can be an example of application of finite state machine(FSM)?
a) Communication Link
b) Adder
c) Stack
d) None of the mentioned

View Answer

Answer: a [Reason:] Idle is the state when data in form of packets is send and returns if NAK is received else waits for the NAK to be received.

7. Which among the following is not an application of FSM?
a) Lexical Analyser
b) BOT
c) State charts
d) None of the mentioned

View Answer

Answer: d [Reason:] Finite state automation is used in Lexical Analyser, Computer BOT (used in games), State charts, etc.

8. L1= {w | w does not contain the string tr }
L2= {w | w does contain the string tr}
Given ∑= {t, r}, The difference of the minimum number of states required to form L1 and L2?
a) 0
b) 1
c) 2
d) Cannot be said

View Answer

Answer: a

9. Predict the number of transitions required to automate the following language using only 3 states:
L= {w | w ends with 00}
a) 3
b) 2
c) 4
d) Cannot be said

View Answer

Answer: a

10. The total number of states to build the given language using DFA:
L= {w | w has exactly 2 a’s and at least 2 b’s}
a) 10
b) 11
c) 12
d) 13

View Answer

Answer: a [Reason:] We need to make the number of a as fixed i.e. 2 and b can be 2 or more. Thus, using this condition a finite automata can be created using 1 states.

Set 2

1. What is the length of a motif, in terms of amino acids residue?
a) 30- 60
b) 10- 20
c) 70- 90
d) 1- 10

View Answer

Answer: b [Reason:] A typical motif is 10-20 amino acids long. For e.g. Zn-finger motif. Hence it is also referred to as super secondary structure. This motif is seen in transcription factors.

2. On average, what is the length of a typical domain?
a) About 100 residues
b) About 300 residues
c) About 500 residues
d) About 900 residues

View Answer

Answer: a [Reason:] The predicted optimal number of residues, which corresponds to the maximum free energy of unfolding, is 100. This is in agreement with a statistical analysis derived from their experimental structures of motifs. For too short chain, change in enthalpy of internal interactions is not favorable enough for folding because of the limited number of inter-residue contacts. And a long chain is also unfavorable for a single domain.

3. Which of the following is false about the ‘loop’ structure in proteins?
a) They connect helices and sheets
b) They are more tolerant of mutations
c) They are more flexible and can adopt multiple conformations
d) They are never the components of active sites

View Answer

Answer: d [Reason:] Loops are frequently components of active sites as they are flexible in nature and as they are situated on the surface of the structure. Besides, they vary in length and 3-D configurations which give even more chances to be component of active sites.

4. Which of the common structural motifs are described wrongly?
a) β-hairpin – adjacent antiparallel strands
b) Greek key – 4 adjacent antiparallel strand
c) β-α-β – 2 parallel strands connected by helix
d) β-α-β – 2 antiparallel strands connected by helix

View Answer

Answer: d [Reason:] In motif, two adjacent β parallel strands are connected by an α helix from the C-terminus of strand 1 to the N-terminus of strand. Most protein structures that contain parallel beta-sheets are built up from combinations of such β-α-βmotifs.

5. Which of the following least describes Long Loop β-hairpins?
a) They are Often referred to as a ‘random coil’ conformation
b) Generally they are referred to as the β-meander supersecondary structure
c) Loop looks similar to the Greek Letter Ω
d) Wide-range of conformations with very specific sequence preferences

View Answer

Answer: d [Reason:] They are wide-range of conformations with no particular sequence preferences. As the name suggests ‘meander’ the conformation they possess is also quite unspecified. Addition to that Long loop β-hairpins are special case of Ω loops, that explains a lot about their structural preferences.

6. Motifs that can form α/β horseshoes conformation are rich with which protein residue?
a) Proline
b) Arginine
c) Valine
d) Leucine

View Answer

Answer: d [Reason:] Specific pattern of Leucine residues, strands form a curved sheet with helices on outside. Leucine-rich repeats (LRRs) are 20-29-residue sequence motifs present in a number of proteins with diverse functions. The primary function of these motifs appears to be to provide a versatile structural framework for the formation of protein-protein interactions.

7. Which of the following wrongly describes protein domains?
a) They are made up of one secondary structure
b) Defined as independently foldable units
c) They are stable structures as compared to motifs
d) They are separated by linker regions

View Answer

Answer: a [Reason:] Protein domains are made up of two or more motifs i.e. the secondary structure to form stable and folded 3-D structures. They are conserved part of the protein sequence and can evolve, function, and exist independently of the rest of the protein chain.

8. The protein structural motif domain- helix loop helix are contained by all of the following except________
a) Scleraxis
b) Neurogenins
c) Transcription Factor 4
d) Leucine zipper

View Answer

Answer: d [Reason:] Leucine zipper is associated with gene regulation and contains alpha helix with leucine at every 7th amino acid. While rest of them are under one of the largest families of dimerizing transcription factors.

9. Which of the following is not the function of Short Linear Motifs?
a) Irreversible cleavage of the peptide at the SLiM
b) Reversible cleavage of the peptide at the SLiM
c) Moiety addition at targeted sites on SLiM
d) Structural modifications of the peptide backbone

View Answer

Answer: a [Reason:] Short Linear Motifs are short stretches of protein sequence that mediate protein-protein interaction. SLiMs can act as recognition sites of endo-peptidases resulting in the irreversible cleavage of the peptide at the SLiM.

10. In the zinc finger, which residues in this sequence motif form ligands to a zinc ion?
a) Cysteine and histidine
b) Cysteine and arginine
c) Histidine and proline
d) Histidine and arginine

View Answer

Answer: a [Reason:] In the zinc finger, which is found in a widely varying family of DNA-binding proteins, cysteine and histidine residues in this sequence motif form ligands to a zinc ion whose coordination is essential to stabilize the tertiary structure.

Set 3

1. The design compressive strength of member is given by
a) Aefcd
b) Ae /fcd
c) fcd
d) 0.5Aefcd

View Answer

Answer: a [Reason:] The design compressive strength of member is given by Pd = Aefcd, where Ae is effective sectional area, fcd is design compressive stress.

2. The design compressive stress, fcd of column is given by
a) [fy / γm0]/ [φ – (φ22)2].
b) [fy / γm0] / [φ + (φ22)].
c) [fy / γm0]/[φ – (φ22)0.5].
d) [fy / γm0] / [φ + (φ22)0.5].

View Answer

Answer: d [Reason:] The design compressive stress, fcd of column is given by fcd = [fy / γm0] / [φ + (φ22)0.5], where fy is yield stress of material, φ is dependent on imperfection factor, λ is non dimensional effective slenderness ratio.

3. What is the value of imperfection factor for buckling class a?
a) 0.34
b) 0.75
c) 0.21
d) 0.5

View Answer

Answer: c [Reason:] The value of imperfection factor, α for buckling class a is 0.21. The imperfection factor considers all the relevant defects in real structure when considering buckling, geometric imperfections, eccentricity of applied loads and residual stresses.

4. If imperfection factor α = 0.49, then what is the buckling class?
a) a
b) c
c) b
d) g

View Answer

Answer: b [Reason:] For buckling class c, the value of imperfection factor is 0.49. The imperfection factor takes into account all the relevant defects in real structure when considering buckling, geometric imperfections, eccentricity of applied loads and residual stresses.

5. The value of φ in the equation of design compressive strength is given by
a) φ = 0.5[1-α(λ-0.2)+λ2].
b) φ = 0.5[1-α(λ-0.2)-+λ2].
c) φ = 0.5[1+α(λ+0.2)-λ2].
d) φ = 0.5[1+α(λ-0.2)+λ2].

View Answer

Answer: d [Reason:] The value of φ in the equation of design compressive strength is given by φ = 0.5[1+α(λ-0.2)+λ2], where α is imperfection factor(depends on buckling class) and λ is non-dimensional effective slenderness ratio.

6. Euler buckling stress fcc is given by
a) (π2E)/(KL/r)2
b) (π2E KL/r)2
c) (π2E)/(KL/r)
d) (π2E)/(KLr)2

View Answer

Answer: a [Reason:] Euler buckling stress fcc is given by fcc = (π2E)/(KL/r)2, where E is modulus of elasticity of material and KL/r is effective slenderness ratio i.e. ratio of effective length, KL to appropriate radius of gyration, r.

7. What is the value of non dimensional slenderness ratio λ in the equation of design compressive strength?
a) (fy /fcc)
b) √(fy fcc)
c) √(fy /fcc)
d) (fy fcc)

View Answer

Answer: c [Reason:] The value of non dimensional slenderness ratio λ in the equation of design compressive strength is given by λ = √(fy /fcc) , where fy is yield stress of material and fcc = (π2E)/(KL/r)2, where E is modulus of elasticity of material and KL/r is effective slenderness ratio i.e. ratio of effective length.

8. The design compressive strength in terms of stress reduction factor is given by
a) Xfy
b) Xfy / γm0
c) X /fy γm0
d) Xfy γm0

View Answer

Answer: b [Reason:] The design compressive strength in terms of stress reduction factor is given by fcd = Xfy / γm0 , where X = stress reduction factor for different buckling class, slenderness ratio and yield stress = 1/ [φ + (φ22)0.5], fy is yield stress of material and γm0 is partial safety factor for material strength.

9. The value of design compressive strength is limited to
a) fy + γm0
b) fy
c) fy γm0
d) fy / γm0

View Answer

Answer: d [Reason:] The value of design compressive strength is given by fcd = [fy / γm0] / [φ + (φ22)0.5] ≤ fy / γm0 i.e. fcd should be less than or equal to fy / γm0.

10. The compressive strength for ISMB 400 used as a column for length 5m with both ends hinged is
a) 275 kN
b) 375.4 kN
c) 453 kN
d) 382 kN

View Answer

Answer: b [Reason:] K = 1 for both ends hinged, KL = 1×5000 = 5000, r = 28.2mm (from steel table), Ae = 7846 mm2(from steel table) KL/r = 5000/28.2 = 177.3 h/bf = 400/140 = 2.82, t = 16mm Therefore, buckling class = b From table in IS code, fcd = 47.85MPa Pd = Ae fcd = 7846 x 47.85 = 375.43 kN.

Set 4

1. What is shear lag effect?
a) the phenomenon of non uniform bending stress not due to influence of shear strain induced on bending stresses in flanges
b) the phenomenon of uniform bending stress not due to influence of shear strain induced on bending stresses in flanges
c) the phenomenon of uniform bending stress due to influence of shear strain induced on bending stresses in flanges
d) the phenomenon of non uniform bending stress due to influence of shear strain induced on bending stresses in flanges

View Answer

Answer: d [Reason:] The shear strain induced influences bending stresses in flanges and causes sections to warp. This consequently modifies the bending stresses determined by simple bending theory and results in higher stresses near junction of web to flange elements with stress dropping as distance from beam web increases. The resultant stress distribution across flange is therefore non uniform and this phenomenon is known as shear lag.

2. As per IS 800:2007, shear lag effects in flanges may be disregarded for outstand elements if
a) bo ≥ L0 / 20
b) bo ≤ L0 / 20
c) bo > L0 / 20
d) bo = L0 / 10

View Answer

Answer: b [Reason:] As per IS 800:2007, shear lag effects in flanges may be disregarded for outstand elements if bo ≤ L0 / 20, where bo = width of flange outstand, L0 = length between points of zero moment in the span.

3. As per IS 800:2007, shear lag effects in flanges may be disregarded for internal elements if
a) bi ≤ L0 / 10
b) bi ≤ L0 / 20
c) bi > L0 / 10
d) bi = L0 / 20

View Answer

Answer: a [Reason:] As per IS 800:2007, shear lag effects in flanges may be disregarded for internal elements if bi ≤ L0 / 10, where bi = width of internal element, L0 = length between points of zero moment in the span.

4. Shear lag effect depends on
a) material of beam
b) width of beam only
c) width-to-span ratio
d) cost

View Answer

Answer: c [Reason:] Shear lag effect depends upon width-to-span ratio, beam end restraints, and type of load.

5. Which of the following is true?
a) point load causes less shear lag than uniform load
b) point load causes more shear lag than uniform load
c) point load causes half times the shear lag than uniform load
d) point load causes equal shear lag as uniform load

View Answer

Answer: b [Reason:] The resultant stress distribution across flanges is non-uniform and is called shear lag. Point load causes more shear lag than uniform load.

6. The moment capacity of plastic section for V > 0.6Vd is given by
a) Mdv = Md – β(Md – Mfd)
b) Mdv = Md + β(Md – Mfd)
c) Mdv = Md – β(Md + Mfd)
d) Mdv = Md + β(Md + Mfd)

View Answer

Answer: a [Reason:] The moment capacity of plastic or compact section for V > 0.6Vd is given by Mdv = Md – β(Md – Mfd), where Md = plastic design moment of whole section disregarding high shear force effect but considering web buckling effect, Mfd = plastic design strength of area of cross section excluding shear area, considering partial safety factor γm0, β is constant.

7. The value of β in equation of moment capacity of plastic section for V > 0.6Vd is given by
a) ([Vd/V] -1)2
b) (2[Vd/V] +1)2
c) (2[Vd/V] -1)2
d) (2[Vd/V] -1)

View Answer

Answer: c [Reason:] The value of β in equation of moment capacity of plastic section for V > 0.6Vd is given by β = (2[Vd/V] -1)2, where Vd = design shear strength as governed by web yielding or web buckling, V = factored applied shear force.

8. The check for moment capacity of plastic section for V > 0.6Vd is given by
a) Mdv ≥ 1.2Zefym0
b) Mdv ≤ 1.2Zefym0
c) Mdv > 1.2Zefym0
d) Mdv = 2.2Zefym0

View Answer

Answer: b [Reason:] The check for moment capacity of plastic section for V > 0.6Vd is given by Mdv ≤ 1.2Zefym0, where Ze = elastic section modulus of whole section, fy = yield stress of material, γm0 = partial safety factor.

Set 5

1. Which IS Code is used for design loads for buildings and structures for wind load?
a) IS 456
b) IS 875 Part 3
c) IS 500
d) IS 1280

View Answer

Answer: b [Reason:] For design loads for buildings and structures for wind load, IS 875-Part 3 given by Bureau of Indian Standards is used.

2. IS Code gives basic wind speed averaged over a short interval of ______
a) 10 seconds
b) 20 seconds
c) 5 seconds
d) 3 seconds

View Answer

Answer: d [Reason:] Basic wind speed is based on peak wind gust velocity averaged over a short interval of 3s and having return period of 50 years and corresponds to mean above ground level in open terrain.

3. Positive sign of pressure coefficient indicates ______________
a) pressure acting towards the surface
b) pressure acting away the surface
c) pressure acting above the surface
d) pressure acting below the surface

View Answer

Answer: a [Reason:] Positive sign of pressure coefficient indicates pressure acting towards the surface and negative sign of pressure coefficient indicates pressure acting away the surface.

4. Which of the following relation is correct for pressure coefficient?
Vp = Actual wind speed at any point on structure at height corresponding to Vz (design wind speed)
a) [1+(Vp/Vz)2].
b) [1+(Vz/Vp)2].
c) [1-(Vz/Vp)2].
d) [1-(Vp/Vz)2].

View Answer

Answer: d [Reason:] Pressure Coefficient is the ratio of difference between pressure acting at point on surface and static pressure of incident wind to the design wind pressure.

5. What is return period?
a) number of years, the reciprocal of which gives the probability of extreme wind exceeding given wind speed in any one year
b) number of years, the reciprocal of which gives the probability of extreme wind less than given wind speed in any one year
c) number of years, the reciprocal of which gives the probability of mild wind exceeding given wind speed in any one year
d) number of years, the reciprocal of which gives the probability of mild wind less than given wind speed in any one year

View Answer

Answer: a [Reason:] Wind load acting on structure varies from year to year based on wind speed and maximum that can be expected to occur at a given location only once in so many years. This period is called return period.

6. Wind Pressure at any height of structure does not depend on _______
a) velocity and density of air
b) angle of wind attack
c) topography of ground surface
d) material of structure

View Answer

Answer: d [Reason:] Wind Pressure at any height of structure depend on (i)velocity and density of air, (ii)height above ground level, (iii)shape and aspect ratio of building, (iv) topography of surrounding ground surface, (v)angle of wind attack, (vi)solidity ratio or openings in the structure.

7. Which of the following relation is correct for design wind speed (Vz) and basic wind speed (Vb) ?
a) Vz ∝ Vb2
b) Vz ∝ 1/Vb2
c) Vz ∝ Vb
d) Vz ∝ 1/Vb

View Answer

Answer: c [Reason:] Vz = k1k2k3Vb , where k1=probability factor(risk coefficient), k2=terrain, height and structure size factor, k3=topography factor.

8. Calculate design wind speed for a site in a city with basic wind speed of 50 m/s, risk coefficient =1, topography factor = 1, terrain is with closely spaced buildings and height of building (class A) = 15m.
a) 40 m/s
b) 48.5 m/s
c) 50 m/s
d) 52.5 m/s

View Answer

Answer: b [Reason:] Vb = 50m/s, k1 = 1, k3 = 1, for terrain with closely spaced buildings, height of building=15m, class A : k2=0.97 (from IS 875 Part 3) Vz = k1k2k3Vb = 1×0.97x1x50 = 48.5 m/s.

9. Which of the following relation between design pressure, pz and design wind speed, Vz is correct?
a) pz ∝ Vz2
b) pz ∝ 1/Vz2
c) pz ∝ Vz
d) pz ∝ 1/Vz

View Answer

Answer: a [Reason:] pz = 0.6Vz2, where pz is in N/m2 and Vz is in m/s. 0.6 factor depends on number of factors and mainly on atmospheric pressure and air temperature.

10. Calculate the design wind pressure if the basic wind speed is 44 m/s, risk coefficient is 1, topography factor is 1, terrain is with closely spaced buildings and height of building(class A) = 20m .
a) 1285 N/m2
b) 1580 N/m2
c) 1085 N/m2
d) 1185 N/m2

View Answer

Answer: d [Reason:] Vb = 44m/s, k1 = 1, k3 = 1, for terrain with closely spaced buildings, height of building=20m, class A: k2=1.01 (from IS 875 Part 3) Vz = k1k2k3Vb = 1×1.01x1x44 = 44.44 m/s pz = 0.6Vz2 = 0.6x(44.44)2 = 1184.95 N/m2.

11. What is the partial safety factor for combination of DL+LL for limit state of strength, where DL=Dead load, LL=imposed load?
a) 1.2
b) 1.0
c) 0.8
d) 1.5

View Answer

Answer: d [Reason:] For limit state of strength, the load combination is 1.5(DL+LL), for limit state of serviceability, the load combination is 1.0(DL+LL), where DL=Dead load, LL=imposed load.

12. Which of the following load combination is not possible?
a) Dead load + imposed load + wind load
b) Dead load + imposed load + earthquake load
c) Dead load + wind load + earthquake load
d) Dead load + imposed load

View Answer

Answer: c [Reason:] According to IS code, it is assumed that maximum wind load and earthquake load will not occur simultaneously on a structure. The following combination of loads with appropriate partial safety factors may be considered : (i)Dead load + imposed load, (ii) Dead load + imposed load + earthquake load or wind load, (iii) Dead load + wind load or earthquake load, (iv) Dead load + erection load.

13. What is the partial safety factor for dead load in combination of DL+LL+WL/EL for limit state of serviceability, where DL=Dead load, LL=imposed load , WL=wind load, EL=earthquake load ?
a) 1.0
b) 0.8
c) 1.5
d) 1.2

View Answer

Answer: b [Reason:] For limit state of strength, the load combination is 1.2(DL+LL+WL/EL), for limit state of serviceability, the load combination is 1.0DL+0.8LL+0.8WL/EL, where DL=Dead load, LL=imposed load, WL=wind load, EL=earthquake load.

14. What is the partial safety factor for dead load in combination of DL+ WL/EL for limit state of serviceability, where DL=Dead load, WL=wind load, EL=earthquake load ?
a) 1.0
b) 1.5
c) 1.2
d) 0.8

View Answer

Answer: a [Reason:] For limit state of serviceability, the load combination is 1.0(DL +WL/EL), for limit state of strength, the load combination is 1.5(DL +WL/EL), where DL=Dead load, WL=wind load, EL=earthquake load.

15. What is the partial safety factor for imposed load in combination of DL+LL+AL , where DL=Dead load, WL=wind load, AL=Accidental load ?
a) 1.0
b) 0.5
c) 0.4
d) 0.35

View Answer

Answer: d [Reason:] The load combination is 1.0DL+ 0.35LL+ 1.0AL, where DL=Dead load, WL=wind load, AL=Accidental load for limit state of strength.