Download PHYSICS 1. Ans. (D) Lyman series lies in the UV region. 2. Ans. (B

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Answer to Model Test - 1
ANSWER TO MODEL TEST - 1
PHYSICS
1.
Ans. (D)
Lyman series lies in the UV region.
2.
Ans. (B)
v2
r
It acts along the radius and directed towards the centre of the circular path.
Centripetal acceleration a c 
3.
Ans. (D)
4.
Ans. (C)
  0  R  4107  2000  8104
5.
Ans. (D)
Tension in the string T0 = mR  02
In the second case T = m(2R)(4  02 ) = 8 mR  02 = 8T0
6.
Ans. (D)
From x = n2 2 = n1 1 and   
 n2 2 = n1 1
 92  5898
n 2  n1 1 
99
2
5461
7.
Ans. (B)
 = LI = 5 × 2.5 Wb = 12.5 Wb
8.
Ans. (A)
According to kirchhoff’s voltage law only option (A) is correct.
9.
Ans. (A)
3
2
2g
mg  mg  mR2 or mg  mR2 or  
5
5
5R
10.
Ans. (B)
From law of conservation of energy, PE at height nR = (PE + KE) on the earth’s surface
GMm
GM m 1 2


 mv
(R  nR)
R
2
or
v 2 GM 
1 
n GM

1



2
R  (n  1)  R(n  1)
or
v
MH-CET
2n GM
2ng R 2
2ng R


R(n  1)
R(n  1)
(n  1)
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Answer to Model Test - 1
11.
HORIZON Publications
Ans. (B)
R  A1/3
1/3
R Li  7 
1
  
R Fe  56 
2
12.
Ans. (C)
2
MR 2
5
Using the theorem of parallel axes, moment of inertia of the sphere about a parallel axis tangential
to the sphere is
2
7
I '  I  MR 2  MR 2  MR 2  MR 2
5
5
7
I '  MK 2  MR 2

5
 7
K  
 R
5


Given I =
13.
Ans. (C)
2
I max
I min
2
a 
4 
b 2   1
2
  1
(a  b)
49
b 
3 





2
2
2
(a  b)
1
a 
4 
b 2   1
  1
b 
3 
14.
Ans. (A)
A hoop is a circular ring. Applying theorem of parallel axes.
I = I0 + MR2 = MR2 + MR2 = 2MR2
15.
Ans. (D)
1
v
2
2
2
v n   5   v5  v 2  v
n
v2 5
5
5
16.
Ans. (C)
As no torque is applied
I11 = I22 or (M K12 ) 1 = (MK 22 )2

K1
2

K2
1
17.
Ans. (C)
18.
Ans. (D)
2 2
T



19.
Ans. (C)
Vm = A and  
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2
T
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Answer to Model Test - 1
Total distance 4A

time
T
4A
4A
2A  2Vm



A.S. 
T
2 / 


Average speed 

20.
Ans. (D)
Vmax = A  A
2
2A
K

A
T
m
m
2
K
(Vm)P = (Vm)Q
K1
K2
A1
 A2

m
m

A1
K2

A2
K1
21.
Ans. (B)
22.
Ans. (B)
a sin  = 1 , a 

5000 108 cm

 10 105 cm
sin 
sin 30o
23.
Ans. (C)
The Young’s modulus of steel is higher than copper, glass and rubber.
24.
Ans. (A)
25.
Ans. (D)
The electric field inside a conductor is always zero.
26.
Ans. (C)
The coefficient of linear expansion is defined as
increase in length
l


original length  temp. rise L

Increase in length l = L.
FL
Now Y 
Al
Y Al YAL
F

 YA
or
L
L
27.
Ans. (C)
Resultant cohesive force is
than
FC
, if the surface is plane, otherwise if the surface is concave it is less
2
FC
.
2
When the surface is concave, adhesive force is greater than cohesive force, so FA 
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FC
2
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Answer to Model Test - 1
HORIZON Publications
28.
Ans. (A)
Magnetic moment = IA = M and L = mvr
q e
e
ev

But I   
2

r
t T
2r
v
ev
evr
IA 
 r 2 

2r
2
M
evr
e



L 2mvr 2m
29.
Ans. (D)
In one complete turn, the proton crosses the energy gap between dees of cyclotron twice. Energy
gained by proton in one movement across the gap = 100 keV = 0.1 MeV. The energy gained in one
20
 100
turn = 0.2 MeV. Hence, n 
0.2
30.
Ans. (D)
When ma > 1 then carrier is said to be over modulated.
31.
Ans. (A)
4T
4  1.6

 2560 N / m 2
P =
3
r
2.5 10
32.
Ans. (A)
Electric field at the centre of cube is zero. This is because the electric field due to individual charges
cancel in pairs.
33.
Ans. (B)
34.
Ans. (C)
After reflection from rigid support, a wave suffers a phase change of .
35.
Ans. (A)
Comparing with std stationary wave equation, we get
2x x

   30 cm

15

Distance between node and antinodes   7.5 cm
4
36.
Ans. (A)
The electric field due to infinite sheet of charge density  is uniform, that is distance independent.
Thus, at point P by the principle of superposition of electric field,


2 
2 
Ep 
( k) 
( k) 
( k)
20
20
20


37.

4( k)
2 (k)
Ep 
 Ep 
20
0
Ans. (A)
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Answer to Model Test - 1
38.
Ans. (A)
39.
Ans. (B)
B1 
0
2 nI R 2
and
4 [R 2  (R 3) 2 ]3/2
 0 2 nI
4 R
B1
1
1


So,
3/2
B2 (1  3)
8
B2 
40.
Ans. (A)
C
3RT
M
1930 

3  8.31 103  300
M
3  8.31103  300
2
1930 1930
The gas is H2
M


41.
Ans. (D)
d
d
e
  (3t 2  4t  9)  (6t  4)
dt
dt
e = –[6(2) + 4] = – 16  |e| = 16 volt
42.
Ans. (C)
Force 
43.
Change in momentum
 2 mnu
time
Ans. (B)
The given circuit can be redrawn as follows
Eeq
i
R  req
E eq 
E1r2  E 2r1 5 1  2  2 1

 V
r1  r2
2 1
3
req 
r1  r2 1 2 2

 
r1  r2 1  2 3

i
P1
i
Eeq
req
P2
1/ 3
1
 A  0.03A from P2 to P1
10  2 / 3 32
44.
Ans. (A)
r.m.s. speed does not depend upon pressure.
45.
Ans. (A)
Larger intensity means more incident photons which ejects larger number electrons.
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Answer to Model Test - 1
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46.
Ans. (C)
E 2 T2 500 5
 

E1 T1 300 3
47.
Ans. (D)
48.
Ans. (C)
Maximum magnetic field due to wire carrying current is at surface and is given by
 2I
B.R
B 0
or I 
4 R
2  (0 / 4)

 16

5 104   103 
2
  2A
I
7
2 10
49.
Ans. (C)
Maximum value of generated emf e0 = NBA
50.
Ans. (B)
CHEMISTRY
51.
Ans.
(A)
(B)
(C)
(D)
(A)
Maltose gives two units of glucose on hydrolysis.
Sucrose gives glucose and fructose on hydrolysis.
Lactose gives glucose and galactose on hydrolysis.
Galactose is a monosaccharide.
52.
Ans. (C)
CH3 CH3
CH3 O
|
|
|
||
Mg Hg  H 2O
dil.H 2SO4
2(CH3 )2 CO 
 CH3  C  C  CH3 
 CH3  C  C  CH3
PinacolPinacolone
|
|
|
OH OH
CH3
2, 3Dimethyl2, 3butanediol
3, 3Dimethyl–2butanone
(Pinacol)
53.
Ans. (B)
CH3
|
CH3  C  CH3
2–Methylpropan–2–ol.
|
OH
is a tertiary alcohol.
Rate of reaction with Lucas reagent is of the following order:
Tertiary alcohol > Secondary alcohol > Primary alcohol.
54.
Ans. (A)
H = U + PV = U + P(V2  V1)
Increase in volume means, V is positive, hence H is greater than U, i.e., H > U
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Answer to Model Test - 1
55.
Ans. (B)
Order of carbocations in SN1 reactions is 3oC > 2oC > 1oC.
56.
Ans. (A)
1000 K b Wsolute
Tb =
Wsolvent  M solute
New Tb =
Tb =
1
1000 K b (2Wsolute )
1000 K b Wsolute
=

2
(4Wsolvent )  M solute
Wsolvent  M solute
1
Tb
2
57.
Ans. (B)
Quartz glass is an amorphous solid because in it short range order of constituents is found.
58.
Ans. (C)
59.
Ans. (D)
SO3H
OH
OH
OH
Br

H2SO4
350
Benzene
Benzene Sulphonic
acid
X
60.
Ans. (C)
61.
Ans. (B)
62.
Ans. (B)
9
620
r=
=
= 219.20 pm.
2 2 2  1.414
63.
Ans. (C)
(i) R = k [A]
2R = 2k [A]
R  [A]
Rate  [A][B]0[C]2
Order = 1 + 0 + 2 = 3
64.




CS2
Br2
NaOH
Fusion
Phenol
Y
(ii)
Br
p–bromophenol
+
o–bromophenol
R = k [C]
2.25 R = k [1.5 C]2
2.25 R = 2.25 k [C]2
R  [C]2
Ans. (C)
E 0cell = E 0oxi LHS + E 0red RHS
Since it is given E 0cell is positive,
E 0oxi LHS > E 0red RHS and must be positive
Since E 0red =  E 0oxi
 E 0cell  0.236 + 0.34 = 0.576 V.
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65.
Ans. (B)
Tetragonal system has unit cell dimension a = b  c, and  = β =  = 90
66.
Ans. (B)
Extraction of Zn from ZnS is achieved by roasting followed by reduction with carbon
2ZnS + 3O2 
 2ZnO + 2SO2.
ZnO + C 
 Zn + CO
67.
Ans. (B)
Vapour pressure of a solvent decreases by the addition of non-volatile solute in it. On adding water,
fraction of solvent is increased and hence vapour pressure is also increased.
68.
Ans. (B)
Cu2O
In the graph of rGvsT for formation of oxides, the Cu2O line is almost at the top. So, it is quite
easy to reduce oxide ores of copper directly to the metal by heating with coke (both the lines of C,
CO and C, CO2 are at much lower temperature (500-600K).
Cu2O + C  2Cu + CO
69.
Ans. (C)
In aldol condensation, aldehydes or ketones having Hatom undergo a condensation reaction
between two molecules in the presence of a base or acid catalyst to form a hydroxy aldehyde
(aldol) or a hydroxy ketone (ketol) respectively.
O
O
||
||
B:
CH3  C – H 
CH

C
–H

2

 BH
Acetaldehyde
O
||
CH3  C – H
Carbanion
+
O
OH
O
||
|
||
CH2  C – H 
 CH3 – CH – CH2  C – H
Acetaldol
(A) and (B) are hydroxy ketone and aldehyde respectively. (D) is not possible because aldol
condensation gives only one product, which is the , unsaturated aldehyde or ketone.
OH
O
H
|
||
|


CH3 – CH – CH2 – CH 
CH

C
= CH  CHO
3
H O
2
Acetaldol
70.
Ans. (B)
71.
Ans. (A)
72.
Ans. (D)
O
||
H  C  OH ;
Formic acid
MH-CET
Crotonaldehyde
O
||
it has – C  H (aldehyde) group.
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Answer to Model Test - 1
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73.
Ans. (C)
Racemic mixture is optically inactive due to external compensation.
When the enantiomers are mixed together, the rotation caused by one enantiomer is exactly
cancelled by equal and opposite rotation caused by the other enantiomer.
74.
Ans. (A)
(A) Ethane : CH3 – CH3 (No isomers possible) [same type of Hatoms]
(B) propane. : CH3 – CH2 – CH3 (2 isomers possible) [two type of Hatoms]
CH3
|
(C) 2, 2–Dimethyl pentane : CH3  C  CH 2  CH 2  CH3
|
CH3
(4 isomers possible) [4 different type of H-atoms]
H
|
(D) 2 – Methyl propane : CH3  C  CH3 (2 isomers possible) [2 different type of Hatoms]
|
CH3
Number of isomers depends upon the different types of Hatoms present in the reactant alkane.
75.
Ans. (A)
76.
Ans. (B)
77.
Ans. (B)
During discharging of the cell, SO 24  ions (from H2SO4) are consumed by Pb2+ ions to form PbSO4.
78.
Ans. (B)
|
The functional group of secondary amine is  N  H (imino) group. Structures (i) and (ii) are
primary amines as they have the NH2 group
79.
Ans. (C)
80.
Ans. (B)
O
O
||
||
H
CH3CH 2CH 2  C  OH  HO  CH 2CH3 

CH
CH
CH

C
 O  CH 2CH3
3
2
2
 H 2O
Butanoic acid (C4)
Opiton (A) will give ethanoic acid (C2)
Option (C) will give pentanoic acid (C5)
Option (D) will give propanoic acid (C3)
81.
Ans. (B)
Catenation tendency decreases form S to Po
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82. Ans. (A)
HBr
HOH
CH3 – CH2 – CH = CH2 
 CH3 – CH2 – CH2 – CH2Br 

 CH3 – CH2 – CH2 – CH2 –OH
H O
aq. KOH
2 2
1 – Butene
1 – Bromobutane
(X)
1 – Butanol
(Y)
83.
Ans. (D)
84.
Ans. (B)
Property : Electron gain enthalpy kJ/mol–1
Halogens : F (–328), Cl (–349), Br (–325), I (–296)
85.
Ans. (C)
Alcohols and ethers have same general formula CnH2n+2O. Ketones and aldehydes have same
general formula i.e. CnH2nO. Given molecular formula satisfies only (C).
C3H23+2O = C3H8O.
86.
Ans. (C)
In set Cu2+, Fe2+, Co2+, all the ions have unpaired d-electrons. Hence, they are all coloured ions
Cu+2  [Ar] 3d9 4s0
Fe2+  [Ar] 3d6 4s0
Co2+  [Ar] 3d7 4s0
87.
Ans. (B)
Nickel has +4 oxidation state.
88.
Ans. (C)
Proteins are high molecular weight polymers formed by the condensation of a large number of
amino acids. Conjugated proteins have a nonprotein part called the prosthetic group. In
glycoproteins the prosthetic group is carbohydrate. Casein of milk is a conjugated protein
(phosphoprotein).
89.
Ans. (D)
As the atomic number and nuclear charge increase, the effective nuclear charge experienced by each
of 4f-electrons increases. As a result the whole shell containing 4f-electrons contracts at each of the
successive element. This result in gradual decrease in the size with increase in atomic
number(Lanthanoide contraction).
90.
Ans. (B)
HI
273
(H3C)3C – OCH3
(CH3)3CI
Tert. Butyl iodide +
(X)
alc. KOH
conc.
H2SO4
CH3
|
CH3  C  CH 2
Isobutylene
(Z)
MH-CET
CH3OH
Methyl alcohol
(Y)
 CH 3OH
(Y)
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91.
Ans: (C)
92.
Ans. (A)
Combustion of CH4 is represented as – CH4(g) + 2O2(g) 
H = ?
 CO2(g) + 2H2O(g)
H = fH(products)  fH(reactants)
= [(–394.8) + 2(–286)] – [(–76.2) + (0)][f H0(O2)= 0]; enthalpy of an element in its standard state
is zero.
= – 394.8 – 572 + 76.2
= – 890.6 kJ mole–1
1 m3 = 106 cm3 = 1000 litre
1000
No. of moles of CH4 in 1000 litre at NTP =
= 44.643 moles
22.4
Total heat evolved in the combustion of 44.643 moles of methane
= 890.6  44.643 = 39759.0558 kJ  4  104 kJ
93.
Ans. (A)
Chlorophyll contains magnesium metal.
94.
Ans. (A)
Glucose + Tollen’s reagent  Gluconic acid + Ag-mirror.
95.
Ans. (C)
On electrolysis; M2+ + 2e– 
M
at. wt. of M
eq. Wt. =
2
 1 mole of atom = 2 g – eq. of metal
 1 Faraday deposits 1 g –eq. of metal M.
 2 Faraday deposits 2 g – eq. (1 mole of atom) of metal M.
96.
Ans. (A)
97.
Ans. (D)
The energy of the ns-orbital depends upon its occupancy. When empty, the ns-orbital is lower in
energy than the (n – 1)d-orbital as per the energy level diagram. When the first electron enters the
ns-orbital, the (n – 1) d-orbital becomes lower in energy and the electrons starts entering it.
However in higher shells the energy levels are very close to each other so as to be almost equal.
98.
Ans. (C)
CH3
|
H3C  H 2C  N  CH 2  CH3
99.
Ans. (C)
Both enthalpy and intrinsic energy are additive and depend on the quantity of substance in the
system.
100. Ans. (A)
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BIOLOGY
101. Ans. (A)
Mammary gland or breasts are the secondary sexual character of females. At puberty in females
they begin to develop under the influence of oestrogen and progesterone.
102. Ans. (A)
103. Ans. (C)
104. Ans. (C)
105. Ans. (C)
106. Ans. (D)
107. Ans. (B)
108. Ans. (A)
109. Ans. (A)
110. Ans. (B)
The term Single Cell Protein (SCP) was coined at Massachusetts institute of technology (MIT) by a
group of scientist in 1966. It is dried cell of micro-organisms or michrobes (algae, bacteria,
actinomycetes and fungi) used as food.
111. Ans. (A)
Pons varolii is the only structure which is a commisure as well as connective
112. Ans. (C)
113. Ans. (B)
Ornithine cycle occurs in liver in the presence of enzyme arginase. It synthesizes urea. During this
cycle, carbon dioxide and water are removed from blood.
114. Ans. (B)
115. Ans. (A)
The process of childbirth is usually called labour or parturition. Childbrith begins with a long series
of involuntary contractions of the uterus experienced as labour pain.
116. Ans. (D)
117. Ans. (B)
118. Ans. (D)
119. Ans. (A)
Transformation of the blastocyst into the gastrula with primary germ layers by rearrangement of the
cell is called gastrulation.
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120. Ans. (C)
121. Ans. (B)
122. Ans. (D)
123. Ans. (C)
124. Ans. (C)
125. Ans. (B)
126. Ans. (B)
127. Ans. (A)
128. Ans. (B)
129. Ans. (C)
130. Ans. (C)
131. Ans. (B)
132. Ans. (C)
133. Ans. (C)
Isogametes are morphologically same.
134. Ans. (D)
135. Ans. (A)
136. Ans. (D)
137. Ans. (B)
138. Ans. (B)
Glycolysis takes place in the cytoplasm and does not use oxygen.
139. Ans. (A)
140. Ans. (D)
141. Ans. (C)
142. Ans. (D)
The strand of DNA, the enzyme forms DNA fragments in small pieces again in 5’  3’ direction
initiating from RNA primer. The primer is formed with the help of primase enzyme.
143. Ans. (C)
144. Ans. (A)
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145. Ans. (C)
146. Ans. (A)
147. Ans. (B)
Ductless glands are called endocrine glands. Out of these pituitary, thyroid, adrenal, ovary and
testes are endocrine glands.
148. Ans. (C)
149. Ans. (B)
150. Ans. (A)
151. Ans. (A)
152. Ans. (B)
Unwindase or helicase takes part in separating the two DNA strands. In prokaryotes, helicase is
assisted by gyrase in this function.
153. Ans. (B)
154. Ans. (B)
155. Ans. (C)
156. Ans. (D)
157. Ans. (D)
Cretinism is the result of thyroid malfunctioning from a very early stage of development.
158. Ans. (B)
159. Ans. (B)
The cavity of Graafian follicle is known as antrum. This is filled with a liquor follicular fluid.
160. Ans. (B)
161. Ans. (D)
162. Ans. (D)
163. Ans. (B)
164. Ans. (D)
The generative nucleus of generative cell in pollen tube divides to form two haploid, non motile
male gametes.
165. Ans. (B)
166. Ans. (D)
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167. Ans. (B)
168. Ans. (D)
169. Ans. (B)
170. Ans. (B)
In Krebs cycle, pyruvic acid is oxidised in stepwise manner and energy is released at different steps
of this cycle. Through Krebs cycle 24 ATP (12×2 ATP) molecules are produced.
171. Ans. (C)
Facial nerve (VII), glossopharyngeal (IX), vagus (X), accessory (XI) and hypoglossal (XII) are
arised from medulla oblongata.
172. Ans. (A)
173. Ans. (B)
Juxta Glomerular Apparatus (JGA) of kidney releases an enzyme called renin into the blood which
converts angiotensin-II increases the blood pressure by causing arterioles constriction. It plays a key
role in increasing the blood volume.
174. Ans. (C)
175. Ans. (B)
176. Ans. (A)
177. Ans. (C)
178. Ans. (C)
179. Ans. (D)
180. Ans. (B)
181. Ans. (B)
182. Ans. (C)
183. Ans. (A)
PCT is lined by cuboidal epithelial cells provided with microvilli to form brush border. Microvilli
increase the surface area for reabsorption.
184. Ans. (C)
Exine is made up of sporopollenin (derived from carotenoid)
185. Ans. (A)
Vulva comprises a group of structures jointly referred as female external genitalia. These are
urethral orifice, vaginal orifice, mons veneris, labia majora, labia minora, vestibule, clitoris, hymen etc.
186. Ans. (A)
A small amount approximately 1-2% of sunlight absorbed by plants is utilized for photosynthesis.
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187. Ans. (C)
188. Ans. (C)
189. Ans. (A)
190. Ans. (C)
191. Ans. (A)
192. Ans. (C)
193. Ans. (A)
194. Ans. (B)
195. Ans. (D)
During ultrafiltration all constituents of blood get filtered except blood cells and plasma proteins.
Hence, the liquid/filtrate is more or less similar to the plasma of blood except plasma proteins.
196. Ans. (C)
197. Ans. (C)
198. Ans. (C)
Replicatio of DNA is in unidirectional and Bidirectional
199. Ans. (D)
200. Ans. (D)
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