NAME
Solutions for MECEE BL Modal Entrance Exam Set –XIV B (2079-04-7)
MECEE BL
Model Entrance Exam 2079
Day Shift
(Set-XIV B)
Date: 2079/04/7
Hints and Solutions
1
Solutions for MECEE BL Modal Entrance Exam Set –XIV B (2079-04-7)
NAME
Zoology
1.
c)
2.
3.
4.
5.
6.
7.
d)
c)
c)
a)
d)
b)
8.
a)
9.
10.
d)
a)
11.
c)
12.
13.
d)
d)
14.
15.
16.
c)
b)
d)
17.
c)
18.
19.
20.
21.
c)
d)
c)
d)
22.
c)
23.
24.
d)
b)
25.
26.
c)
d)
27.
b)
28.
b)
Felidae – is the family of cat. There are 36 species of cats in this family. They have
sharp canine teeth that help them kill their prey.
Absence of paired fins and true jaws.
Sarcoplasmic reticulum release Ca++ ions during muscle contraction.
Common bile duct-it carry bile juice from gall bladder to Duodenum.
Squamos- it is flat cell. Gaseous exchange take place through this layer.
All of the aboveAlbumin- Albumin is a protein made by your liver. Albumin enters your
bloodstream and helps keep fluid from leaking out of your blood vessels into
other.
P-wave- The P wave is the first positive deflection on the ECG. It represents atrial
depolarisation.
RenninAmmonia- Kidneys do not play any significant role in its removal, because
Ammonia produced by metabolism is converted into urea in the liver.
Cochlea-The cochlea is the part of the inner ear involved in hearing. It is a spiralshaped cavity in the bony labyrinth, in humans making 2.75 turns around it.
Both a and b
Antibody- Thymosin is the hormone of the thymus, and it stimulates the
development of B cells produce antibodies to fight bacteria and viruses.
Mammary ampulla
Malignant tumor
Thymus- The thymus is a specialized primary lymphoid organ of the immune
system. The T-cells are produced from the bone marrow. Within the thymus, T
cells mature. T cells are critical to the adaptive immune system, where the body
adapts specifically to foreign invaders. The thymus is the largest and most active
during the neonatal and pre-adolescent periods.
IgE - If you have an allergy, your immune system overreacts to an allergen by
producing antibodies called Immunoglobulin E (IgE).
Hepatic vein
Conusarteriosus.
Erythtocytes of man
Circulation of water through their body. In some of these sponges water enters
through a canal lined by many cells. This body plan provides more circulation to
deliver more oxygen and nutrients.
Polyp- Hydra is a small, freshwater animal belonging to the phylum Cnidaria.
Polyps have a cylindrical body which is elongated at the axis.
All of these
Ascaris- The body of Ascaris is made up of 3 layer- (1) Cuticle (2) Epidermis (3)
Longitudinal muscles lining the body cavity. The nature of Ascaris epidermis is
syncytial. The syncytial epidermis is present in Ascaris.
But pharyngeal nephridia poured into buccal cavity and pharynx.
Platyhelmintes and Annelida. Solenocyte and Nephridia are excretory organs of
Platyhelminthes and earthworm ( annelida).
Lens. Earthworm is small transparent L-shaped lens or phaosome, made up of a
hyaline substance. The lens focuses the light from all directions on neurofibrils.
Anapsida- it means temporal fossae are absent in skull.
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Solutions for MECEE BL Modal Entrance Exam Set –XIV B (2079-04-7)
NAME
29.
30.
31.
32.
b)
d)
d)
b)
33.
a)
34.
c)
35.
36.
d)
c)
37.
38.
a)
b)
39.
c)
40.
c)
Amino acid and simple sugar
Stratified squamous epithelium
Epinephrine
Instinct- Whether behavior is a result of nature (instinct) or nurture (learning) has
been controversial.
Antigen- Any substance that induces the immune system to produce antibodies
against it is called an antigen. Any foreign invaders, such as pathogens (bacteria
and other substance).
Inoviduct- jelly is secreted by oviduct, while passing eggs through it, jelly is
deposited around it.
1600 cc
Re-Capitulation theory- This tutorial explains the recapitulation theory and it
demonstrates the "ontogeny repeats phylogeny" concept regarding evolution.
Urochordata. Notochord present in larval stage but absent in adult stage.
Adrenalin. Adrenaline hormone, also called epinephrine, is part of the body’s fight
or flight response to stress.
CNS. Brain and Spinal cord are part of CNS. Spinal cord is the center for reflex
action.
Proteins. Rennin is an enzyme secreted by the fundic glands to digest milk protein
whereas Renin is an enzyme secreted by the kidneys in response to hypotension.
Botany
41.
d)
42
b)
43
44
c)
a)
45
46
a)
c)
47
b)
48
d)
49
d)
50
51
52
c)
d)
d)
53
a)
54
c)
Cotton blue and lactophenol are two common stains used in staining cell wall of
fungi.
Bacterial cell wall is made from mucopeptide/peptidoglycan which is absent in
mycoplasm (lack cell wall).
Phycophage is viral group attacking on algae.
Lichen is symbiotic relationship between mycobiont (fungi belonging to
ascomycetes) and phycobiont(algae belonging to green algae).
The red rust of coffee is caused by parasitic green algae (Cephaleurosvirscens).
Extra-cellular digestion (digestion outside cell) and osmotrophic mode of nutrition
are common to fungi like yeast.
Leafy gametophyte of moss is gametophore and gametophytic thread is
protonema.
The rachis of Dryopteris bears dictyostele with horse show shape arrangement of
meristeles.
Male cone and female cones are aggregation of microsporophyll and
megasporophyll respectively.
Offset is modified horizontal stem as found in Agave, Pistiaand Trapa.
Cypsellafruits with pappus are common in members of family Compositae.
Myrmecophily is pollinating relationship between flowering plants and ants. It is
beneficial for both (symbiosis) ants (food and shelter) and plants (pollination and
security).
Ultimate source of energy for earth is sun light which is trapped by producer
through photosynthesis.
Pyramid is graphical representation of different trophic levels in an ecoystem.
3
Solutions for MECEE BL Modal Entrance Exam Set –XIV B (2079-04-7)
NAME
55
a)
56
57
58
c)
b)
b)
59
60
c)
a)
61
c)
62
b)
63
64
c)
a)
65
66
c)
a)
67
a)
68
a)
69
c)
70
a)
71
d)
72
b)
73
d)
74
b)
75
76
b)
c)
77
78
a)
d)
79
c)
80
c)
Population is total individuals of single species found in ecological system as
specific time.
Vitality index is the ratio between natality and mortality in a unit of ecosystem.
Among 20 amino acids, glycine is the simplest amino acids with 2 carbons.
The structural and functional unit of chloroplast are thylakoids and quantasomes
respectively.
Palade discovered ribosome and also called as Palade particle.
Emil Heitz discovered dark and light (stainable with acetocarmine) parts of
chromatin and called heterochromatin and euchromatin.
According to Chargaff rule of molecular data of DNA, the molecular amount of
Purine (A+G) is equivalent with Pyrimidine (G+C), i.e. A + G = T + C.
Transfer of characters through sex chromosomes is called sex linked inheritance,
e.g. Hypertrichosis, Myopia, haemophilia, colourblindness, etc.
The similar genotype and phenotype ratio as 1:2:1 found in incomplete dominance.
Multiple effects of single gene is pleiotropism. It gives major and minor expression
as in sickle cell anaemia.
Gene mutation is sudden drastic change in DNA or nucleotides sequences.
Genetic code of codons are triplet set of nucleotides which carries message for
single amino acid.
Allele is a pair of traits which may be similar (homozygous allele) or dissimilar
(heterozygous allele).
Vexillary aestivation (butterfly like) is common character of members of family
Papilionaceae.
Well developed aerenchyma with reduced mechanical tissues is common character
of hydrophytes.
Under natural condition, the osmotic pressure is more than turgor pressure, i.e. OP
> TP.
C4 and C3 plants need 30 ATP and 18 ATP for the synthesis of single molecule of
glucose.
The atmospheric CO2 is considered as limiting factor of photosynthesis.
Photosynthesis is reduction of CO2 into glucose and oxidation of H2O into
molecular oxygen.
Protoplasmic respiration starts from protein molecules while floating respiration
from carbohydrates, lipids and fat.
Respiratory quotient (RQ) depends on respiratory substrates. It becomes one for
carbohydrates and less than one for protein/lipids.
Mimosa (touch me not) show seismonastic movement, response due to vibration.
Pollen chamber, nucellus and filiform apparatus are associated with female
reproductive structure while ubisch bodies are produced from tapetum in male
reproductive structure.
Monocot seeds have edible endosperms as in maize.
At the time of pollination, pollen tube gets nucleus from vegetative cell and
generative divides into two identical male gametes.
Callus is undifferentiated mass of cell produced during cell division or tissue
culture.
Lactic acid, ethanol and curds are produced through fermentation of microbes.
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Solutions for MECEE BL Modal Entrance Exam Set –XIV B (2079-04-7)
NAME
Chemistry
81.
b)
82.
a)
83.
b)
84.
c)
85.
d)
86.
a)
87.
c)
88.
89.
d)
b)
90.
d)
91.
92.
93.
a)
c)
b)
94.
95.
b)
d)
96.
c)
97.
b)
98.
c)
1023 molecules
6.02 ×
of CO =1mole of CO
6.02 × 1024 CO molecules = 10 moles CO
= 10 g atoms of O = 5 g molecules of O2
Empirical formula of compound = CH2
Molecular mass of the compound = 42
∴n = 42/14 = 3
Hence molecular formula = C3H6
2H2 + O2 ⟶ 2H2O
4g
32g
36g
Energy of photon obtained from the transition n = 6 to
n = 5 will have least energy.
1
1
∆E = 13.6Z2 n 2 – n 2
2 
 1
Noble gases have positive values of electron gain enthalpy because the anion is
higher in energy than the isolated atom and electron.
The correct order of dipole moments of HF, H2S and H2O is:
HF < H2S < H2O
PV m
n = RT = M
MPV 34 × 2 × 100
m = RT = 0.082 × 293 = 282.68gm
When work is done by the system, ∆U = q – W
The most favourable conditions are :
(i) High pressure (∆n < 0)
(ii) Low temperature (Exothermic reaction)
(iii) Catalyst Fe is presence of Mo.
In (i) and (ii) both P and S are in highest oxidation state. In (iii) and (iv) ; P has
oxidation state of +4 which can be oxidized to +5 state, while in case of NH3
nitrogen has oxidation state of –3 which can be oxidised.
Smog is caused by oxides of sulphur and nitrogen.
Diamond is covalent or network solid.
∵ 10g glucose is dissolved in 100ml solution
∴ 180g (g mole) is dissolved in
100
= 10 × 180 = 1800ml = 1.8L
Follow ECS, A will replace B.
Ag+ + e– ⟶ Ag
96500 coulombs deposit = 108g of Ag
108
∴ 965 coulombs deposit = 96500 × 965 = 1.08g Ag
k = (mol lit–1)1–ntime–1. For given reaction
n = 2. ∴k = mol–1lit sec–1
∵ r = k[A]n if n = 0
r = k[A]0
or, r = k thus for zero order reactions rate is equal to the rate constant.
Cheese is a liquid dispersed in solid phase.
5
Solutions for MECEE BL Modal Entrance Exam Set –XIV B (2079-04-7)
NAME
99.
c)
100.
101.
a)
d)
102.
c)
103.
b)
104.
c)
105.
106.
b)
a)
107.
108.
109.
a)
a)
d)
110.
111.
112.
a)
b)
d)
113.
c)
114.
c)
115.
d)
116.
b)
117.
118.
c)
c)
In this process sulphide ores are converted into oxide ores.
2ZnS + 3O2 → 2ZnO + 2SO2 ↑
It is lined with haematite (Fe2O3).
Bi forms basic oxides whereas N and P form acidic and As and Sb form amphoteric
oxides.
Ca3P2 + 6H2O3Ca(OH)2 + 2PH3 ; i.e. 2 moles of phosphine are produced from one
mole of calcium phosphide.
Oleum is H2S2O7 (H2SO4 + SO3) which is obtained by dissolving SO3 in H2SO4 and
is called fuming sulphuric acid.
Fluorine is most electronegative element.
Ionisation energy decreases as we move away from nucleus due to less electrostatic
attraction between electrons and nucleus.
4LiNO3 →2Li2O +4NO2 + O2
(CaSO4)2.H2O – Plaster of paris is used for plastering the broken bones.
The order of strength of Lewis acid character for boron halides is, BF3 < BCl3 < BBr3
< BI3 (due to back bonding).
Producer gas is a fuel gas and is mixture of CO and N2.
It is due to surface tension.
Oxide Mn2O7 : Oxidation state of metal + 7
Oxide V2O3 : Oxidation state of metal + 3
Oxide V2O5 : Oxidation state of metal + 5
Oxide CrO : Oxidation state of metal + 2
Oxide Cr2O3 : Oxidation state of metal + 5
The structure of neopentane is
It has 1 quaternary and 4 primary carbons.
The organic reaction which proceed through heterolytic bond cleavage are called
ionic or heteropolar or just polar reactions.
In elimination reactions one or two molecules are lost from the substrate to form a
multiple bond. Dehydration of ethanol is an example of elimination reaction.
Metamerism is shown among compounds of the same functional group.
By adding bromine water to a solution, if the colour of bromine water decolourise
then the compound is unsaturated. This is a confirmatory test for unsaturation
6
Solutions for MECEE BL Modal Entrance Exam Set –XIV B (2079-04-7)
NAME
119.
120.
a)
d)
121.
d)
122.
c)
123.
d)
124.
a)
Thionyl chloride is preferred because the other two products formed in the
reaction are escapable gases. Hence the reaction gives pure alkyl halides
ROH + SOCl2 → R– Cl + SO2 + HCl
Solubility of alcohol in water decreases with increase in molecular mass due to
increase in water repelling alkyl part in alcohol.
Ethers are readily cleaved by HI as follows :
125.
a)
1° Alcohols on catalytic dehydrogenation give Aldehydes.
126.
127.
d)
b)
It is Gatterman-Koch reaction.
The overall reaction involved is
128.
129.
c)
d)
130.
d)
This is an example of Friedel - Craft alkylation.
More the stability of the carbocation, higher will be the reactivity of the parent
chloride. Allyl chloride > Vinyl chloride > Chlorobenzene
On reduction cyanides yield 1 amines. They do not undergo decarboxylation or
electrolysis.
Cl2CHCOOH is most acidic because it has two chlorine at α-position.
For the preparation of Me3CNH2, the required alkyl halide is Me3CX which will
react with potassium phthalimide, a strong base, to form alkene rather than
substituted product. For preparing C6H5NH2, C6H5Cl will be the starting halide in
which Cl is non-reactive.
Orlon is a trade name of polyacrylonitrile.
Physics
131.
b)
coefficient of ^i 2
tanθ =
=
coefficient of ^j 3
2
θ = tan–1 3
 
132.
d)
133.
d)
Two particles are moving with velocities v1 and v2. Their relative velocity is the
maximum, when the angle between their velocities is π.
N = mg – Fsin30º
fl = µsN = Fcos30º
1
3
= 2 × 10 – 10 × 2
⇒ µs × 15 = 10 × 2
= 15
µs =
1
3
7
Solutions for MECEE BL Modal Entrance Exam Set –XIV B (2079-04-7)
NAME
134.
c)
When a force of constant magnitude which is perpendicular to the velocity of
particle acts on a particle, work done is zero and hence change in kinetic energy is
zero.
135.
b)
136.
a)
137.
a)
138.
c)
139.
b)
TL – TH = 6mg
g g
g' = 3 = n
n=3
2R
n–1
3–1
d= n R= 3 R= 3




1
1
K = K0 cos2 ωt
EK = 2 mω2 r2 cos2 ωt ; on comparing 2 mω2 r2 = K0
1
EKmax = 2 mω2 r2 = K0
When a body rolls down on inclined plane, the total potential energy of the body
changes into both rotational and translational KE.
A water proofing agent changes the angle of contact from an acute to obtuse value.
140.
b)
Due to elastic fatigue its elastic property decreases.
141.
b)
γ r = γa + γv ; where γr = coefficient of real expansion,
γa = coefficient of apparent expansion and γv = coefficient of expansion of vessel
For copper γr = C + 3αCu = C + 3A
For silver γr = S + 3αAg
⇒ C + 3A = S + 3αAg ⇒ αAg =
Li
θW - C
142.
a)
θmix =
143.
c)
Vrms ∝
∴
144.
c)
1
=
2
w
2
80
100 – 1
=
= 10ºC
2
1
v1
⇒v =
2
M
M2
M1
M2
32 ⇒ M2 = 16. Hence the gas is CH4
R–L
U – L = constant
R – (–170)
340 – 273
– 33.5 – (–170) = 373 – 273
145.
146.
a)
c)
147.
b)
C – S + 3A
3
R = – 91.9ºX
The first operation involved in a Carnot cycle is isothermal expansion .
The amount of radiation emitted by a perfectly black body is proportional to
Fourth power of temperature on ideal gas scale.
1
C
sinC =
µ=V
µ
1
3 × 108
sin30º =
2= V
µ
µ=2
V = 1.5 × 108 m/s
8
Solutions for MECEE BL Modal Entrance Exam Set –XIV B (2079-04-7)
NAME
148.
c)
Field of view is maximum for convex mirror, hence it is used in vehicles for rear
viewing.
For minimum deviation inside the prism, e = i.
Concave portion will have larger radius since it is less curved.
1
1 1
f = (µ – 1) R1 + R2
1
1 1
f = (1.5 – 1) 20 – 40
f = 80cm
In Young's double slit experiment, a glass plate is placed before a slit which
absorbs half the intensity of light. Under this case, the bright fringes become fainter
and the dark fringes have finite light intensity.
149.
150.
b)
c)
151.
d)
152.
a)
Only transverse waves can be polarised.
153.
c)
y1 = asinωt
π
y2 = acosωt = asin ωt + 2


π
π
Phase difference = ωt + 2 – ωt = 2


154.
b)
2
Beat frequency = 0.4 = 5Hz
155.
156.
b)
d)
I ∝ P02
Sweetness of a sound depends upon its periodicity and reqularity.
157.
a)
Fe = W ⇒ qE = mg
mg 10–6 × 10
E = q = 10–6 = 10V/m upward
158.
d)
σsmall q R2 q (n1/3 r)2
1
= ×
=
×
= n–1/3 = (64)–1/3 = 4
r2
σbig Q r2 nq
159.
b)
R
Resistance of parallel group = 2
160.
b)
161.
c)
162.
c)
163.
a)
164.
b)
165.
166.
d)
a)
167.
c)
R
Total equivalent resistance = 4 × 2 = 2R
The tangent galvanometer, when connected in series with a standard resistance can
be used as a volt meter.
In charging V > E
µ0I
B = 2r
B∝I
Two straight conductors carry current in same direction, then attractive force acts
between them.
Cθ
i = NAB ⇒ i ∝ θ
From the characteristic of B-H curve.
In transformer, core is made of soft iron to reduce hysteresis losses.
200
2π ×
×1
2π
XL 2πvL
tanφ = R = R =
= 1 ⇒ φ = 45º
200
9
Solutions for MECEE BL Modal Entrance Exam Set –XIV B (2079-04-7)
NAME
106
168.
b)
E 6.6 ×
V = B = 1.2
169.
d)
Kmax of photoelectrons doesn’t depends upon intensity of incident light
170.
c)
hc
λmain = eV(energy) ; when KE (or eV) increases, λ decreases.
171.
c)
Lyman series lies in the UV region
172.
a)
Absorb excess neutrons.
173.
b)
N 1n
 1  1n 14 1n
N0 = 2 ⇒ 16 = 2 ⇒ 2 = 2
= 5.5 × 106 m/s
n=4
t 2
1
T1/2 = n = 4 hours = 2 hour = 30 minutes
174.
175.
c)
a)
176.
b)
177.
178.
b)
b)
179.
180.
d)
d)
In P-type semiconductors, holes are the majority charge carriers
In forward biased PN-junction, external voltage decreases the potential barrier, so
current is maximum. While in reversed biased PN-junction, external voltage
increases the potential barrier, so the current is very small.
Due to the large concentration of electrons in N-side and holes in P-side, they
diffuses from their own side to other side. Hence depletion region produces
During reverse biasing, thickness of depletion layer increases.
Acid rain happens because burning of fossil fuels releases oxides of carbon,
nitrogen and sulphur in the atmosphere.
udd form a neutron.
Milky way is one of the enormous galaxies of the universe.
MAT
181.
a)
182.
d)
183.
d)
184.
c)
Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
5x = 20
x = 4.
∴ Age of the youngest child = x = 4 years.
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds (2 minutes).
30
In 30 minutes, they will toll together 2 + 1 = 16 times
Here, S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
n(E)
9
∴ P(E) = n(S) = 20
The figure may be labelled as shown.
10
Solutions for MECEE BL Modal Entrance Exam Set –XIV B (2079-04-7)
NAME
185.
b)
186.
c)
187.
188.
189.
d)
a)
d)
190.
c)
191.
192.
c)
d)
193.
c)
194.
b)
The simplest triangles are AFJ, FJK, FKB, BKG, JKG, JGC, HJC, HIJ, DIH, DEI, EIJ
and AEJ i.e. 12 in number.
The triangles composed of two components each are JFB, FBG, BJG, JFG, DEJ, EJH,
DJH and DEH i.e. 8 in number.
The triangles composed of three components each are AJB, JBC, DJC and ADJ i.e. 4
in number.
The triangles composed of six components each are DAB, ABC, BCD and ADC i.e.
4 in number.
Thus, there are 12 + 8 + 4 + 4 = 28 triangles in the figure.
In fig. (X), one of the dots lies in the region common to the circle and the triangle
only, another dot lies in the region common to the circle, the square and the
triangle only and the third dot lies in the region common to the circle, the square
and the rectangle only. In each of the figures (1) and (3) there is no region common
to the circle and the triangle only. In fig. (4) there is no region common to the circle,
the square and the rectangle only. Only fig. (2) consists of all the three types of
regions.
The third figure in each row comprises of parts which are not common to the first
two figures.
In each step, the dot moves one space CW and the arrow moves two spaces CW.
Given: D is the brother of B.
From statement 1, we can detect that D is son of C (son of D is the grandson of C).
From statement 2, we can detect that B is 'Female' (sister of D).
Therefore, B is daughter of C.
Hospital consists of nurse and patient but nurse and patient are of two different
nature.
As Glove is worn in Hands similarly Socks are worn on feet.
How the number is obtained?
2+4=6
5 + 9 = 14
Similarly,
3+5=8
Therefore, the answer is K8.
(6 + 5) - (7 + 4) = 0
and (7 + 6) - (8 + 4) = 1
Therefore (11 + 2) - (2 + 0) = 11.
Let the number of cows be x and the number of hens be y.
Then, 4x + 2y = 2 (x + y) + 14
4x + 2y = 2x + 2y + 14
2x = 14
11
x = 7.
Solutions for MECEE BL Modal Entrance Exam Set –XIV B (2079-04-7)
NAME
195.
196.
c)
c)
197.
198.
c)
c)
199.
b)
200.
a)
The series is abc/aabc/aabbc/aabbcc/a.
Let the number of boys and girls participating in sports be 3x and 2x respectively.
Then, 3x = 15 or x = 5.
So, number of girls participating in sports = 2x = 10.
Number of students not participating in sports = 60 - (15 + 10) = 35.
Let number of boys not participating in sports be y.
Then, number of girls not participating in sports = (35 -y).
2y
30
y = 15.
Therefore (35 - y) = y + 5
So, number of girls not participating in sports = (35 - 15) = 20.
Hence, total number of girls in the class = (10 + 20) = 30.
Justification:
In the first and second statements, the common code digit is '4' and the common
word is 'good'.
So, '4' stands for 'good'.
In the second and third statements, the common code digit is '7' and the common
word is 'pictures'.
So, '7' means 'pictures'.
Thus, in the second statements, '8' means 'see'.
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