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CHEMISTRY
PART – C
96. HBr reacts with CH2 = CH – OCH3 under anhydrous conditions at room temperature to give
(1) CH3CHO and CH3Br (2) BrCH2CHO and CH3OH
(3) BrCH2 – CH2 – OCH3 (4) H3C – CHBr – OCH3
Ans. (4)
Sol. Electrophilic addition reaction more favourable.
H2C CH OCH3
HBr H2C CH
H
OCH3
Br H C 3 CH
Br
OCH3
97. The IUPAC name of the compound shown below is
Cl
Br
(1) 2-bromo-6-chlorocyclohex-1-ene (2) 6-bromo-2-chlorocyclohexene
(3) 3-bromo-1-chlorocyclohexene (4) 1-bromo-3-chlorocyclohexene
Ans. (3)
98. The increasing order of the rate of HCN addition to compounds A – D is
(A) HCHO (B) CH3COCH3
(C) PhCOCH3 (D) PhCOPh
(1) A < B < C < D (2) D < B < C < A
(3) D < C < B < A (4) C < D < B < A
Ans. (3)
99. How many moles of magnesium phosphate, Mg3(PO4)2 will contain 0.25 mole of oxygen atoms?
(1) 0.02 (2) 3.125 × 10–2
(3) 1.25 × 10–2 (4) 2.5 × 10–2
Ans. (2)
Sol. Mg3(PO4)2
‘n’ moles
8n = 0.25
n 0.25
8
=
25 3.125 10 2
8 100
= = × −
×
100. According to Bohr’s theory, the angular momentum of an electron in 5th orbit is
(1) 25 h
π
(2) 1.0 h
π
(3) 10 h
π
(4) 2.5 h
π
Ans. (4)
Sol. mvr nh
2
=
π
5h 2.5 h
2
= =
π π
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101. Which of the following molecules/ions does not contain unpaired electrons?
(1) 22
O − (2) B2
(3) N2+ (4) O2
Ans. (1)
102. Total volume of atoms present in a face-centre cubic unit cell of a metal is (r is atomic radius)
(1) 20 r3
3
π (2) 24 r3
3
π
(3) 12 r3
3
π (4) 16 r3
3
π
Ans. (4)
Sol. V n 4 r3
3
= × π
4 4 r3
3
= × π
16 r3
3
= π
103. A reaction was found to be second order with respect to the concentration of carbon monoxide. If the
concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction
will
(1) remain unchanged (2) triple
(3) increase by a factor of 4 (4) double
Ans. (3)
Sol. R ∝ [W]2
R′ ∝ [2CO]2
R ∝ 4[W]2
R ∝ 4M
104. Which of the following chemical reactions depicts the oxidizing behaviour of H2SO4?
(1) 2HI +H2SO4 →I2 + SO2 + 2H2O (2) Ca(OH)2 +H2SO4 →CaSO4 + 2H2O
(3) NaCl +H2SO4 →NaHSO4 +HCl (4) 2PCl5 +H2SO4 →2POCl3 + 2HCl + SO2Cl2
Ans. (1)
105. The IUPAC name for the complex [Co(NO2)(NH3)5]Cl2 is
(1) nitrito-N-pentaamminecobalt (III) chloride (2) nitrito-N-pentaamminecobalt (II) chloride
(3) pentaammine nitrito-N-cobalt (II) chloride (4) pentaammine nitrito-N-cobalt (III) chloride
Ans. (4)
106. The term anomers of glucose refers to
(1) isomers of glucose that differ in configurations at carbons one and four (C-1 and C-4)
(2) a mixture of (D)-glucose and (L)-glucose
(3) enantiomers of glucose
(4) isomers of glucose that differ in configuration at carbon one (C-1)
Ans. (4)
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107. In the transformation of 238 234
92 U to 92 U, if one emission is an α-particle, what should be the other
emission(s)?
(1) Two β– (2) Two β– and one β+
(3) One β– and one γ (4) One β+ and one β–
Ans. (1)
Sol. 238 234 4 0
92 92 2 1 U U He2e− → + +
108. Phenyl magnesium bromide reacts with methanol to give
(1) a mixture of anisole and Mg(OH)Br (2) a mixture of benzene and Mg(OMe)Br
(3) a mixture of toluene and Mg(OH)Br (4) a mixture of phenol and Mg(Me)Br
Ans. (2)
109. CH3Br +Nu− →CH3 −Nu + Br−
The decreasing order of the rate of the above reaction with nucleophiles (Nu–) A to D is
[Nu– = (A) PhO–, (B) AcO–, (C) HO–, (D) CH3O–]
(1) D > C > A > B (2) D > C > B > A
(3) A > B > C > D (4) B > D > C > A
Ans. (1)
110. The pyrimidine bases present in DNA are
(1) cytosine and adenine (2) cytosine and guanine
(3) cytosine and thymine (4) cytosine and uracil
Ans. (3)
111. Among the following the one that gives positive iodoform test upon reaction with I2 and NaOH is
(1) CH3CH2CH(OH)CH2CH3 (2) C6H5CH2CH2OH
(3)
H3C
CH3
OH
(4) PhCHOHCH3
Ans. (4)
112. The increasing order of stability of the following free radicals is
(1) (CH3 )2 CH (CH3 )3 C (C6H5 )2 CH (C6H5 )3 C
• • • • < < <
(2) (C6H5 )3 C (C6H5 )2 CH (CH3 )3 C (CH3 )2 CH
• • • •
< < <
(3) (C6H5 )2 CH (C6H5 )3 C (CH3 )3 C (CH3 )2 CH
• • • •
< < <
(4) (CH3 )2 CH (CH3 )3 C (C6H5 )3 C (C6H5 )2 CH
• • • •
< < <
Ans. (1)
113. Uncertainty in the position of an electron (mass = 9.1 × 10–31 kg) moving with a velocity 300 ms–1,
accurate upto 0.001%, will be
(1) 19.2 × 10–2 m (2) 5.76 × 10–2 m
(3) 1.92 × 10–2 m (4) 3.84 × 10–2 m
(h = 6.63 × 10–34 Js)
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Ans. (3)
Sol. x. V h
4 m
Δ Δ ≥
π
34
31
x h 6.63 10
4 m V 4 3.14 9.1 10 300 0.001
100
−
−
×
Δ ≥ =
π Δ × × × × ×
34
31 3
6.63 10
4 3.14 9.1 3 10 10
−
− −
×
=
× × × × ×
= 0.01933
= 1.93 × 10–2
114. Phosphorus pentachloride dissociates as follows, in a closed reaction vessel,
PCl5(g)PCl3(g) + Cl2(g)
If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x, the
partial pressure of PCl3 will be
(1) x P
x 1
+
(2) 2x P
1 x
−
(3) x P
x 1
−
(4) x P
1 x
−
Ans. (1)
Sol. PCl5(g) PCl3(g) Cl2(g)
(1 x) x x
+
−
PCl3
P x P
1 x
= × +
115. The standard enthalpy of formation o
(ΔfH ) at 298 K for methane, CH4(g), is –74.8 kJ mol–1. The
additional information required to determine the average energy for C – H bond formation would be
(1) the dissociation energy of H2 and enthalpy of sublimation of carbon
(2) latent heat of vapourization of methane
(3) the first four ionization energies of carbon and electron gain enthalpy of hydrogen
(4) the dissociation energy of hydrogen molecule, H2
Ans. (1)
116. Among the following mixtures, dipole-dipole as the major interaction, is present in
(1) benzene and ethanol (2) acetonitrile and acetone
(3) KCl and water (4) benzene and carbon tetrachloride
Ans. (2)
117. Fluorobenzene (C6H5F) can be synthesized in the laboratory
(1) by heating phenol with HF and KF
(2) from aniline by diazotisation followed by heating the diazonium salt with HBF4
(3) by direct fluorination of benzene with F2 gas
(4) by reacting bromobenzene with NaF solution
Ans. (2)
118. A metal, M forms chlorides in its +2 and +4 oxidation states. Which of the following statements about
these chlorides is correct?
(1) MCl2 is more volatile than MCl4
(2) MCl2 is more soluble in anhydrous ethanol than MCl4
(3) MCl2 is more ionic than MCl4
(4) MCl2 is more easily hydrolysed than MCl4
Ans. (3)
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119. Which of the following statements is true?
(1) H3PO3 is a stronger acid than H2SO3
(2) In aqueous medium HF is a stronger acid than HCl
(3) HClO4 is a weaker acid than HClO3
(4) HNO3 is a stronger acid than HNO2
Ans. (4)
120. The molar conductivities o o
∧NaOAc and ∧HCl at infinite dilution in water at 25oC are 91.0 and
426.2 S cm2/mol respectively. To calculate o
∧HOAc , the additional value required is
(1) o
H2O
∧ (2) o
∧KCl
(3) o
∧NaOH (4) o
∧NaCl
Ans. (4)
Sol. o o o
CH3COONa CH COO Na 3
− + λ =λ +λ ……….. (1)
o o o
HCl H+ Cl− λ = λ +λ ………………….. (2)
o o o
NaCl Na Cl− λ = λ +λ …………………. (3)
o
CH3COOH λ = (1) + (2) − (3)
121. Which one of the following sets of ions represents a collection of isoelectronic species?
(1) K+ , Cl–, Ca2+, Sc3+ (2) Ba2+, Sr2+, K+, S2–
(3) N3–, O2–, F–, S2– (4) Li+, Na+, Mg2+, Ca2+
Ans. (1)
122. The correct order of increasing acid strength of the compounds
(a) CH3CO2H (b) MeOCH2CO2H
(c) CF3CO2H (d)
CO2H
Me
Me
is
(1) b < d < a < c (2) d < a < c < b
(3) d < a < b < c (4) a < d < c < b
Ans. (3)
123. In which of the following molecules/ions are all the bonds not equal?
(1) SF4 (2) SiF4
(3) XeF4 (4) BF4−
Ans. (1)
124. What products are expected from the disproportionation reaction of hypochlorous acid?
(1) HClO3 and Cl2O (2) HClO2 and HClO4
(3) HCl and Cl2O (4) HCl and HClO3
Ans. (4)
125. Nickel (Z = 28) combines with a uninegative monodentate ligand X– to form a paramagnetic complex
2
[NiX4 ] − . The number of unpaired electron(s) in the nickel and geometry of this complex ion are,
respectively
(1) one, tetrahedral (2) two, tetrahedral
(3) one, square planar (4) two, square planar
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Ans. (2)
Sol. 28Ni: ………. 3s2, 3p6, 3d8, 4s2
Ni2+: 3s2, 3p6, 3d8
3d 4s 4p
sp3
Tetrahedral geometry
126. In Fe(CO)5, the Fe – C bond possesses
(1) π-character only (2) both σ and π characters
(3) ionic character (4) σ-character only
Ans. (2)
127. The increasing order of the first ionization enthalpies of the elements B, P, S and F (lowest first) is
(1) F < S < P < B (2) P < S < B < F
(3) B < P < S < F (4) B < S < P < F
Ans. (4)
128. An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If Ti is the
initial temperature and Tf is the final temperature, which of the following statements is correct?
(1) (Tf)irrev > (Tf)rev
(2) Tf > Ti for reversible process but Tf = Ti for irreversible process
(3) (Tf)rev = (Tf)irrev
(4) Tf = Ti for both reversible and irreversible processes
Ans. (1)
129. In Langmuir’s model of adsorption of a gas on a solid surface
(1) the rate of dissociation of adsorbed molecules from the surface does not depend on the surface
covered
(2) the adsorption at a single site on the surface may involve multiple molecules at the same time
(3) the mass of gas striking a given area of surface is proportional to the pressure of the gas
(4) the mass of gas striking a given area of surface is independent of the pressure of the gas
Ans. (3)
130. Rate of a reaction can be expressed by Arrhenius equation as:
k = A e−E /RT
In this equation, E represents
(1) the energy above which all the colliding molecules will react
(2) the energy below which colliding molecules will not react
(3) the total energy of the reacting molecules at a temperature, T
(4) the fraction of molecules with energy greater than the activation energy of the reaction
Ans. (2)
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131. The structure of the major product formed in the following reaction
I
Cl
NaCN
DMF →
is
(1)
CN
CN
(2)
I
Cl
NC
(3)
CN
Cl
(4)
I
CN
Ans. (4)
132. Reaction of trans-2-phenyl-1-bromocyclopentane on reaction with alcoholic KOH produces
(1) 4-phenylcyclopentene (2) 2-phenylcyclopentene
(3) 1-phenylcyclopentene (4) 3-phenylcyclopentene
Ans. (4)
Sol. According to E2 mechanism.
133. Increasing order of stability among the three main conformations (i.e. Eclipse, Anti, Gauche) of
2-fluoroethanol is
(1) Eclipse, Gauche, Anti (2) Gauche, Eclipse, Anti
(3) Eclipse, Anti, Gauche (4) Anti, Gauche, Eclipse
Ans. (3)
134. The structure of the compound that gives a tribromo derivative on treatment with bromine water is
(1) CH3
OH
(2) CH2OH
(3) CH3
OH
(4) CH3
OH
Ans. (1)
135. The decreasing values of bond angles from NH3 (106o) to SbH3 (101o) down group-15 of the periodic
table is due to
(1) increasing bp-bp repulsion (2) increasing p-orbital character in sp3
(3) decreasing lp-bp repulsion (4) decreasing electronegativity
Ans. (4)
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136. Me
N
Et
Me
n-Bu
Δ→
OH
The alkene formed as a major product in the above elimination reaction is
(1)
Me
(2) CH2 = CH2
(3) Me
(4) Me
Ans. (2)
137. The “spin-only” magnetic moment [in units of Bohr magneton, (μB )] of Ni2+ in aqueous solution would
be (Atomic number of Ni = 28)
(1) 2.84 (2) 4.90
(3) 0 (4) 1.73
Ans. (1)
138. The equilibrium constant for the reaction
3 2 2
SO (g) SO (g) 1O (g)
2
+
is Kc = 4.9 × 10–2. The value of Kc for the reaction
2SO2(g) + O2(g)2SO3(g)
will be
(1) 416 (2) 2.40 × 10–3
(3) 9.8 × 10–2 (4) 4.9 × 10–2
Ans. (1)
Sol.
2
c 2
K 1
4.9 10−
′ = ×
104 100 100
4.9 4.9 24.01
×
= =
×
= 4.1649 × 100
= 416.49
139. Following statements regarding the periodic trends of chemical reactivity of the alkali metals and the
halogens are given. Which of these statements gives the correct picture?
(1) The reactivity decreases in the alkali metals but increases in the halogens with increase in atomic
number down the group
(2) In both the alkali metals and the halogens the chemical reactivity decreases with increase in
atomic number down the group
(3) Chemical reactivity increases with increase in atomic number down the group in both the alkali
metals and halogens
(4) In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic
number down the group
Ans. (4)
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140. Given the data at 25oC,
Ag + I− →AgI + e− ; Eo = 0.152 V
Ag→Ag+ + e−; Eo = −0.800 V
What is the value of log Ksp for AgI?
2.303RT 0.059 V
F
=
(1) –8.12 (2) +8.612
(3) –37.83 (4) –16.13
Ans. (4)
Sol. AgI(s) + e− Ag(s) + I− ; Eo = −0.152
Ag(s) Ag e ; Eo 0.8 → + + − = −
AgI(s) Ag I ; Eo 0.952 → + + − = −
o
cell
E 0.059 logK
n
=
sp
0.952 0.059 log K
1
− =
sp
log K 0.952 16.135
0.059
= − = −
141. The following mechanism has been proposed for the reaction of NO with Br2 to form NOBr:
NO(g) + Br2(g)NOBr2(g)
NOBr2(g) +NO(g)→2NOBr(g)
If the second step is the rate determining step, the order of the reaction with respect to NO(g) is
(1) 1 (2) 0
(3) 3 (4) 2
Ans. (4)
Sol. NO(g) + Br2(g)NOBr2(g)
NOBr2(g) +NO(g)→2NOBr(g)
2 R = K[NOBr ] [NO]
2
c 2 c
2
K.K [NO] [Br ][NO], where K [NOBr ]
[NO] [Br ]
= =
2
2 = K′[NO] [Br ]
142. Lanthanoid contraction is caused due to
(1) the appreciable shielding on outer electrons by 4f electrons from the nuclear charge
(2) the appreciable shielding on outer electrons by 5d electrons from the nuclear charge
(3) the same effective nuclear charge from Ce to Lu
(4) the imperfect shielding on outer electrons by 4f electrons from the nuclear charge
Ans. (4)
143. Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 Ω.
The conductivity of this solution is 1.29 S m–1. Resistance of the same cell when filled with 0.2 M of
the same solution is 520 Ω. The molar conductivity of 0.02 M solution of the electrolyte will be
(1) 124 × 10–4 S m2 mol–1 (2) 1240 × 10–4 S m2 mol–1
(3) 1.24 × 10–4 S m2 mol–1 (4) 12.4 × 10–4 S m2 mol–1
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Ans. (4)
Sol. There is one mistake in Question paper.
Assuming concentration of solution is 0.2 M instead of 0.02 M. Since resistance of 0.2 M is 520 Ω.
R = 100 Ω
K 1
R a
=
1.29 1
100 a
=
129 m 1
a
= −
R = 520Ω , C = 0.2 M
K 1 1 (129) 1m 1
R a 520
= = Ω− −
in cm3 μ = K × V
1 129 1000 10 6 m3
520 0.2
= × × × −
129 1000 10 6
520 0.2
= × × −
= 1.24 ×10−3
= 12.4 ×10−4
144. The ionic mobility of alkali metal ions in aqueous solution is maximum for
(1) K+ (2) Rb+
(3) Li+ (4) Na+
Ans. (2)
145. Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is
(1) 1.14 mol kg–1 (2) 3.28 mol kg–1
(3) 2.28 mol kg–1 (4) 0.44 mol kg–1
Ans. (3)
146. The enthalpy changes for the following processes are listed below:
Cl2(g) = 2Cl(g), 242.3 kJ mol–1
I2(g) = 2I(g), 151.0 kJ mol–1
ICl(g) = I(g) + Cl(g), 211.3 kJ mol–1
I2(s) = I2(g), 62.76 kJ mol–1
Given that the standard states for iodine and chlorine are I2(s) and Cl2(g), the standard enthalpy of
formation for ICl(g) is
(1) –14.6 kJ mol–1 (2) –16.8 kJ mol–1
(3) +16.8 kJ mol–1 (4) +244.8 kJ mol–1
Ans. (3)
Sol. 2 2
1I (s) 1Cl ICl(g)
2 2
+ →
[ ] I2(s) I2(g) I I Cl Cl I Cl
H 1 H 1 1
2 2 2 → − − −
Δ = Δ + μ + μ − μ
1 62.76 1 151.0 1 242.3 (211.3)
2 2 2
= × + × + × −
= 228.03 – 211.3
ΔH = 16.73
147. How many EDTA (ethylenediaminetetraacetic acid) molecules are required to make an octahedral
complex with a Ca2+ ion?
(1) Six (2) Three
(3) One (4) Two
Ans. (3)
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148. OH
+ CHCl3 + NaOH →
O Na
CHO
The electrophile involved in the above reaction is
(1)
dichloromethyl cation (CHCl2 )
(2) dichlorocarbene( :CCl2 )
(3)
trichloromethyl anion (CCl3 )
(4)
formyl cation (CHO)
Ans. (2)
149. 18 g of glucose (C6H12O6) is added to 178.2 g of water. The vapour pressure of water for this aqueous
solution at 100oC is
(1) 759.00 Torr (2) 7.60 Torr
(3) 76.00 Torr (4) 752.40 Torr
Ans. (4)
Sol.
o
s
s
P P n
P N
−
=
s
s
18 1
760 P 180 10 0.1
P 178.2 9.9 9.9
18
−
= = =
s s
760 P 1 P
99
− =
760 × 99 – Ps × 99 = Ps
760 × 99 = 100 Ps
s
P 760 99 752.4
100
×
= =
150. (ΔH− ΔU) for the formation of carbon monoxide (CO) from its elements at 298 K is
(R = 8.314 J K–1 mol–1)
(1) –1238.78 J mol–1 (2) 1238.78 J mol–1
(3) –2477.57 J mol–1 (4) 2477.57 J mol–1
Ans. (1)
Sol. g ΔH− ΔU = Δn RT
1 8.314 298
2
= − × ×
= –1238.78
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MATHEMATICS
PART − A
1. ABC is a triangle, right angled at A. The resultant of the forces acting along AB, AC
with magnitudes 1
AB
and 1
AC
respectively is the force along AD
, where D is the
foot of the perpendicular from A onto BC. The magnitude of the resultant is
(1)
2 2
2 2
AB AC
(AB) (AC)
+
(2) (AB)(AC)
AB + AC
(3) 1 1
AB AC
+ (4) 1
AD
Ans. (4)
Sol: Magnitude of resultant
=
1 2 1 2 AB2 AC2
AB AC AB AC
+ + = ⋅
BC BC 1
AB AC AD BC AD
= = =
⋅ ⋅
A B
C
D
2. Suppose a population A has 100 observations 101, 102, … , 200, and another
population B has 100 observations 151, 152, … , 250. If VA and VB represent the
variances of the two populations, respectively, then A
B
V
V
is
(1) 1 (2) 9/4
(3) 4/9 (4) 2/3
Ans. (1)
Sol:
2
i 2x
d
n
σ = Σ . (Here deviations are taken from the mean)
Since A and B both has 100 consecutive integers, therefore both have same
standard deviation and hence the variance.
∴ A
B
V
1
V
= ( 2 )
As Σdi is same in both the cases .
3. If the roots of the quadratic equation x2 + px + q = 0 are tan30° and tan15°,
respectively then the value of 2 + q − p is
(3) 2 (2) 3
(3) 0 (4) 1
Ans. (2)
Sol: x2 + px + q = 0
tan 30° + tan 15° = − p
tan 30° ⋅ tan 15° = q
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tan 45° = tan30 tan15 p 1
1 tan30 tan15 1 q
° + ° −
= =
− ° ° −
⇒ − p = 1 − q
⇒ q − p = 1 ∴ 2 + q − p = 3.
4. The value of the integral,
6
3
x dx
9 − x + x ∫ is
(1) 1/2 (2) 3/2
(3) 2 (4) 1
Ans. (2)
Sol: I =
6
3
x dx
9 − x + x ∫
I =
6
3
9 x dx
9 x x
−
− + ∫
2I =
6
3
∫dx = 3 ⇒ I = 3
2
.
5. The number of values of x in the interval [0, 3π] satisfying the equation
2sin2x + 5sinx − 3 = 0 is
(1) 4 (2) 6
(3) 1 (4) 2
Ans. (1)
Sol: 2 sin2 x + 5 sin x − 3 = 0
⇒ (sin x + 3) (2 sin x − 1) = 0
⇒ sin x = 1
2
∴ In (0, 3π), x has 4 values
6. If (a × b)× c = a × (b × c), where a, b and c are any three vectors such that a ⋅ b ≠ 0 ,
b ⋅ c ≠ 0 , then a and c are
(1) inclined at an angle of π/3 between them
(2) inclined at an angle of π/6 between them
(3) perpendicular
(4) parallel
Ans. (4)
Sol: (a × b) × c = a × (b × c), a ⋅ b ≠ 0, b ⋅ c ≠ 0
⇒ (a ⋅ c) b − (b ⋅ c)a = (a ⋅ c) b − (a ⋅ b)c
(a ⋅ b)c = (b ⋅ c)a
a c
7. Let W denote the words in the English dictionary. Define the relation R by :
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R = {(x, y) ∈ W × W | the words x and y have at least one letter in common}. Then R
is
(1) not reflexive, symmetric and transitive
(2) reflexive, symmetric and not transitive
(3) reflexive, symmetric and transitive
(4) reflexive, not symmetric and transitive
Ans. (2)
Sol: Clearly (x, x) ∈ R ∀ x ∈ W. So, R is reflexive.
Let (x, y) ∈ R, then (y, x) ∈ R as x and y have at least one letter in common. So, R is
symmetric.
But R is not transitive for example
Let x = DELHI, y = DWARKA and z = PARK
then (x, y) ∈ R and (y, z) ∈ R but (x, z) ∉ R.
8. If A and B are square matrices of size n × n such that A2 − B2 = (A − B) (A + B), then
which of the following will be always true ?
(1) A = B
(2) AB = BA
(3) either of A or B is a zero matrix
(4) either of A or B is an identity matrix
Ans. (2)
Sol: A2 − B2 = (A − B) (A + B)
A2 − B2 = A2 + AB − BA − B2
⇒ AB = BA.
9. The value of
10
k 1
sin 2k icos 2k
= 11 11
π π +
Σ is
(1) i (2) 1
(3) −1 (4) −i
Ans. (4)
Sol:
10 10 10
k 1 k 1 k 1
sin 2k icos 2k sin 2k i cos 2k
11 11 11 11 = = =
π π π π + = +
Σ Σ Σ
= 0 + i (− 1) = − i.
10. All the values of m for which both roots of the equations x2 − 2mx + m2 − 1 = 0 are
greater than −2 but less than 4, lie in the interval
(1) −2 < m < 0 (2) m > 3
(3) −1 < m < 3 (4) 1 < m < 4
Ans. (3)
Sol: Equation x2 − 2mx + m2 − 1 = 0
(x − m)2 − 1 = 0
(x − m + 1) (x − m − 1) = 0
x = m − 1, m + 1
− 2 < m − 1 and m + 1 < 4
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m > − 1 and m < 3
− 1 < m < 3.
11. A particle has two velocities of equal magnitude inclined to each other at an angle θ.
If one of them is halved, the angle between the other and the original resultant
velocity is bisected by the new resultant. Then θ is
(1) 90° (2) 120°
(3) 45° (4) 60°
Ans. (2)
Sol:
u sin
tan 2
4 u u cos
2
θ θ
=
+ θ
⇒ sin 1 sin cos 1 sin cos
4 2 4 2 4
θ θ θ
+ θ= θ
∴ 2sin sin 3 3sin 4sin3
4 4 4 4
θ θ θ θ
= = −
∴ sin2 1
4 4
θ
= ⇒ 30
4
θ
= ° or θ = 120°.
u
u
θ/2
θ/4 θ/4
R2 R1
u/2
12. At a telephone enquiry system the number of phone cells regarding relevant enquiry
follow Poisson distribution with an average of 5 phone calls during 10-minute time
intervals. The probability that there is at the most one phone call during a 10-minute
time period is
(1) e
6
5
(2) 5
6
(3) 6
55
(4) 5
6
e
Ans. (4)
Sol: P (X = r) =
e mmr
r !
−
P (X ≤ 1) = P (X = 0) + P (X = 1)
= e−5 + 5 × e−5 = 5
6
e
.
13. A body falling from rest under gravity passes a certain point P. It was at a distance of
400 m from P, 4s prior to passing through P. If g = 10 m/s2, then the height above the
point P from where the body began to fall is
(1) 720 m (2) 900 m
(3) 320 m (4) 680 m
Ans. (1)
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Sol: We have h 1 gt2
2
= and h + 400 = 1 g(t 4)2
2
+ .
Subtracting we get 400 = 8g + 4gt
⇒ t = 8 sec
∴ h 1 10 64 320m
2
= × × =
∴ Desired height = 320 + 400 = 720 m.
h
400m
Q(t)
P(t+4)
14.
0
xf(sin x)dx
π∫
is equal to
(1)
0
f(cos x)dx
π
π∫ (2)
0
f(sinx)dx
π
π∫
(3)
/ 2
0
f(sinx)dx
2
π π∫ (4)
/ 2
0
f(cos x)dx
π
π ∫
Ans. (4)
Sol: I =
0 0
xf(sin x)dx ( x) f(sin x)dx
π π
∫ = ∫ π −
=
0
f(sin x)dx I
π
π∫ −
2I =
0
f(sin x)dx
π
π∫
I =
/ 2
0 0
f(sin x)dx f(sin x)dx
2
π π π
∫ = π ∫
= π
/ 2
0
f(cos x)dx
π
∫ .
15. A straight line through the point A(3, 4) is such that its intercept between the axes is
bisected at A. Its equation is
(1) x + y = 7 (2) 3x − 4y + 7 = 0
(3) 4x + 3y = 24 (4) 3x + 4y = 25
Ans. (3)
Sol: The equation of axes is xy = 0
⇒ the equation of the line is
x 4 y 3 12
2
⋅ + ⋅
= ⇒ 4x + 3y = 24.
16. The two lines x = ay + b, z = cy + d; and x = a′y + b′, z = c′y + d′ are perpendicular to
each other if
(1) aa′ + cc′ = −1 (2) aa′ + cc′ = 1
(3) a c 1
a c
+ = −
′ ′
(4) a c 1
a c
+ =
′ ′
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Ans. (1)
Sol: Equation of lines x b y z d
a c
− −
= =
x b y z d
a c
− ′ − ′
= =
′ ′
Lines are perpendicular ⇒ aa′ + 1 + cc′ = 0.
17. The locus of the vertices of the family of parabolas
a3x2 a2x y 2a
3 2
= + − is
(!) xy 105
64
= (2) xy 3
4
=
(3) xy 35
16
= (4) xy 64
105
=
Ans. (1)
Sol: Parabola: y =
a3x2 a2x 2a
3 2
+ −
Vertex: (α, β)
α =
2
3
a /2 3
2a / 3 4a
−
= − , β =
4 3
4
3 3
a 4 a 2a 1 8 4 3 a 4 3
a 4 a 4 3 3
− + ⋅ ⋅ − +
= −
= 35 a 3 35 a
12 4 16
− × =−
αβ = − 3 35 a 105
4a 16 64
− =
.
18. The values of a, for which the points A, B, C with position vectors
2ˆi − ˆj + kˆ, ˆi − 3ˆj − 5kˆ and aˆi − 3ˆj + kˆ respectively are the vertices of a right-angled
triangle with C
2
π
= are
(1) 2 and 1 (2) −2 and −1
(3) −2 and 1 (4) 2 and −1
Ans. (1)
Sol: ⇒ BA = ˆi − 2ˆj + 6kˆ
CA = (2 − a)ˆi + 2ˆj
CB = (1− a)ˆi − 6kˆ
CA ⋅CB
= 0 ⇒ (2 − a) (1 − a) = 0
⇒ a = 2, 1.
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19. ( ) ( )
/ 2
3 2
3 /2
x cos x 3 dx
−π
− π
+ π + + π ∫ is equal to
(1)
4
32
π
(2)
4
32 2
π π
+
(3)
2
π
(4) 1
4
π
−
Ans. (3)
Sol: I =
/ 2
3 2
3 /2
(x ) cos (x 3 ) dx
−π
− π
+ π + + π ∫
Put x + π = t
I =
/ 2 / 2
3 2 2
/ 2 0
t cos t dt 2 cos tdt
π π
−π
+ = ∫ ∫
=
/ 2
0
(1 cos 2t)dt 0
2
π π
∫ + = + .
20. If x is real, the maximum value of
2
2
3x 9x 17
3x 9x 7
+ +
+ +
is
(1) 1/4 (2) 41
(3) 1 (4) 17/7
Ans. (2)
Sol:
2
2
y 3x 9x 17
3x 9x 7
+ +
=
+ +
3x2(y − 1) + 9x(y − 1) + 7y − 17 = 0
D ≥ 0 ∵ x is real
81(y −1)2 − 4x3(y −1)(7y −17) ≥ 0
⇒ (y − 1) (y − 41) ≤ 0 ⇒ 1 ≤ y ≤ 41.
21. In an ellipse, the distance between its foci is 6 and minor axis is 8. Then its
eccentricity is
(1) 3
5
(B) 1
2
(C) 4
5
(D) 1
5
Ans. (1)
Sol: 2ae = 6 ⇒ ae = 3
2b = 8 ⇒ b = 4
b2 = a2(1 − e2)
16 = a2 − a2e2
a2 = 16 + 9 = 25
a = 5
e 3 3
a 5
∴ = =
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22. Let A =
1 2
3 4
and B =
a 0
0 b
, a , b ∈ N. Then
(1) there cannot exist any B such that AB = BA
(2) there exist more than one but finite number of B’s such that AB = BA
(3) there exists exactly one B such that AB = BA
(4) there exist infinitely many B’s such that AB = BA
Ans. (4)
Sol: 1 2 a 0
A B
3 4 0 b
= =
a 2b
AB
3a 4b
=
a 0 1 2 a 2a
BA
0 b 3 4 3b 4b
= =
AB = BA only when a = b
23. The function f(x) = x 2
2 x
+ has a local minimum at
(1) x = 2 (2) x = −2
(3) x = 0 (4) x = 1
Ans. (1)
Sol: x 2
2 x
+ is of the form x 1 2
x
+ ≥ & equality holds for x = 1
24. Angle between the tangents to the curve y = x2 − 5x + 6 at the points (2, 0) and (3, 0)
is
(1)
2
π
(2)
2
π
(3)
6
π
(4)
4
π
Ans. (2)
Sol: dy 2x 5
dx
= −
∴ m1 = (2x − 5)(2, 0) = −1, m2 = (2x − 5)(3, 0) = 1
⇒ m1m2 = −1
25. Let a1, a2, a3, … be terms of an A.P. If
2
1 2 p
2
1 2 q
a a a p , p q
a a a q
+ + ⋅ ⋅ ⋅
= ≠
+ +⋅⋅⋅+
, then 6
21
a
a
equals
(1) 41
11
(2) 7
2
(3) 2
7
(4) 11
41
Ans. (4)
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Sol:
( )
( )
( )
( )
1 2 1
2
1
1
p 2a p 1 d 2a p 1 d 2 p p
q 2a q 1 d q 2a q 1 d q
2
+ − + −
= ⇒ =
+ − + −
1
1
a p 1 d
2 p
a q 1 d q
2
− +
=
− +
For 6 6
21 21
a a 11 , p 11, q 41
a a 41
= = → =
26. The set of points where f(x) x
1 |x|
=
+
is differentiable is
(1) (−∞, 0) ∪ (0, ∞) (2) (−∞, −1) ∪ (−1, ∞)
(3) (−∞, ∞) (4) (0, ∞)
Ans. (3)
Sol: ( )
( )
2
2
x 1 , x 0 , x 0 (1 x) f x 1 x f (x) x 1 , x 0 , x 0
1 x 1 x
< < − = − ⇒ ′ =
≥ ≥
+ +
∴ f′(x) exist at everywhere.
27. A triangular park is enclosed on two sides by a fence and on the third side by a
straight river bank. The two sides having fence are of same length x. The maximum
area enclosed by the park is
(1) 3 x2
2
(2)
x3
8
(3) 1 x2
2
(4) πx2
Ans. (3)
Sol: Area = 1 x2 sin
2
θ
2
max
A 1 x at sin 1,
2 2
π = θ = θ =
θ
x x
28. At an election, a voter may vote for any number of candidates, not greater than the
number to be elected. There are 10 candidates and 4 are of be elected. If a voter
votes for at least one candidate, then the number of ways in which he can vote is
(1) 5040 (2) 6210
(3) 385 (4) 1110
Ans. (3)
Sol: 10C1 + 10C2 + 10C3 + 10C4
= 10 + 45 + 120 + 210 = 385
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29. If the expansion in powers of x of the function 1
(1− ax)(1− bx)
is
a0 + a1x + a2x2 + a3x3 + … , then an is
(1)
bn an
b a
−
−
(2)
an bn
b a
−
−
(3)
an 1 bn 1
b a
+ − +
−
(4)
bn 1 an 1
b a
+ − +
−
Ans. (4)
Sol: (1 ax) 1 (1 bx) 1 (1 ax a2x2 ......)(1 bx b2x2 ....) − − − − = + + + + + +
∴ coefficient of xn = bn + abn−1 + a2bn−2 + .... + an−1b + an =
bn 1 an 1
b a
+ − +
−
∴
n 1 n 1
n
a b a
b a
+ − +
=
−
30. For natural numbers m, n if (1 − y)m (1 + y)n = 1 + a1y + a2y2 + … , and a1 = a2 = 10,
then (m, n) is
(1) (20, 45) (2) (35, 20)
(3) (45, 35) (4) (35, 45)
Ans. (4)
Sol: ( )m ( )n m m 2 n n 2
1− y 1+ y = 1− C1y + C2y − .... 1+ C1y + C2y + ...
= ( ) ( ) ( ) m m 1 n n 1 2
1 n m mn y .....
2 2
− − + − + + − +
2 2
1 2
a n m 10 and a m n m n 2mn 10
2
+ − − −
∴ = − = = =
So, n − m = 10 and (m − n)2 − (m + n) = 20 ⇒ m + n = 80
∴ m = 35, n = 45
31. The value of
a
1
∫[x] f′(x)dx , a > 1, where [x] denotes the greatest integer not exceeding
x is
(1) af(a) − {f(1) + f(2) + … + f([a])} (2) [a] f(a) − {f(1) + f(2) + … + f([a])}
(3) [a] f([a]) − {f(1) + f(2) + … + f(a)} (4) af([a]) − {f(1) + f(2) + … + f(a)}
Ans. (2)
Sol: Let a = k + h, where [a] = k and 0 ≤ h < 1
[ ] ( ) ( ) ( ) ( ) ( )
a 2 3 k k h
1 1 2 k 1 k
x f ' x dx 1f ' x dx 2f ' x dx ........ k 1 dx kf ' x dx
+
−
∴ ∫ =∫ + ∫ + ∫ − + ∫
{f(2) − f(1)} + 2{f(3) − f(2)} + 3{f(4) − f(3)}+…….+ (k−1) – {f(k) − f(k − 1)}
+ k{f(k + h) − f(k)}
= − f(1) − f(2) − f(3)……. − f(k) + k f(k + h)
= [a] f(a) − {f(1) + f(2) + f(3) + …. + f([a])}
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32. If the lines 3x − 4y − 7 = 0 and 2x − 3y − 5 = 0 are two diameters of a circle of area
49π square units, the equation of the circle is
(1) x2 + y2 + 2x − 2y − 47 = 0 (2) x2 + y2 + 2x − 2y − 62 = 0
(3) x2 + y2 − 2x + 2y − 62 = 0 (4) x2 + y2 − 2x + 2y − 47 = 0
Ans. (4)
Sol: Point of intersection of 3x − 4y − 7 = 0 and 2x − 3y − 5 = 0 is (1 , − 1), which is the
centre of the circle and radius = 7.
∴ Equation is (x − 1)2 + (y + 1)2 = 49 ⇒ x2 + y2 − 2x + 2y − 47 = 0.
33. The differential equation whose solution is Ax2 + By2 = 1, where A and B are arbitrary
constants is of
(1) second order and second degree (2) first order and second degree
(3) first order and first degree (4) second order and first degree
Ans. (4)
Sol: Ax2 +By2 = 1 … (1)
Ax By dy 0
dx
+ = … (2)
2 2
2
A By d y B dy 0
dx dx
+ + =
… (3)
From (2) and (3)
2 2
2
x By d y B dy By dy 0
dx dx dx
− − + =
⇒
2 2
2
xy d y x dy y dy 0
dx dx dx
+ − =
34. Let C be the circle with centre (0, 0) and radius 3 units. The equation of the locus of
the mid points of the chords of the circle C that subtend an angle of 2
3
π
at its centre
is
(1) x2 y2 3
2
+ = (B) x2 + y2 = 1
(3) x2 y2 27
4
+ = (D) x2 y2 9
4
+ =
Ans. (4)
Sol:
2 2
cos h k h2 k2 9
3 3 4
π +
= ⇒ + =
35. If (a, a2) falls inside the angle made by the lines y x , x 0
2
= > and y = 3x, x > 0, then a
belongs to
(1) 0, 1
2
(2) (3, ∞)
(3) 1, 3
2
(4) 3, 1
2
− −
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Ans. (3)
Sol: a2 − 3a < 0 and a2 a 0 1 a 3
2 2
− > ⇒ < <
36. The image of the point (−1, 3, 4) in the plane x − 2y = 0 is
(1) 17 , 19 , 4
3 3
− −
(2) (15, 11, 4)
(3) 17 , 19 ,1
3 3
− −
(4) (8, 4, 4)
Sol: If (α, β, γ) be the image then 1 2 3 0
2 2
α − β + − =
∴ α − 1 − 2β − 6 ⇒ α − 2β = 7 … (1)
and 1 3 4
1 2 0
α + β − γ −
= =
−
… (2)
From (1) and (2)
9, 13, 4
5 5
α = β = − γ =
No option matches.
37. If z2 + z + 1 = 0, where z is a complex number, then the value of
2 2 2 2
2 3 6
2 3 6
z 1 z 1 z 1 z 1
z z z z
+ + + + + + ⋅ ⋅ ⋅ + +
is
(1) 18 (2) 54
(3) 6 (4) 12
Ans. (4)
Sol: z2 + z + 1 = 0 ⇒ z = ω or ω2
so, 2 2 2 3 3 3
2 3
z 1 1, z 1 1, z 1 2
z z z
+ = ω+ω = − + = ω +ω = − + = ω +ω =
4 5 6
4 5 6
z 1 1, z 1 1 and z 1 2
z z z
+ = − + = − + =
∴ The given sum = 1 + 1 + 4 + 1 + 1 + 4 = 12
38. If 0 < x < π and cosx + sinx = 1
2
, then tanx is
(1) (1 7)
4
−
(B) (4 7)
3
−
(3) (4 7)
3
+
− (4) (1 7)
4
+
Ans. (3)
Sol: cos x sinx 1 1 sin2x 1 sin2x 3 ,
2 4 4
+ = ⇒ + = ⇒ =− so x is obtuse
and 2
2
2tanx 3 3tan x 8tanx 3 0
1 tan x 4
= − ⇒ + + =
+
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∴ tanx 8 64 36 4 7
6 3
− ± − − ±
= =
tanx 0 tanx 4 7
3
− −
∵ < ∴ =
39. If a1, a2, … , an are in H.P., then the expression a1a2 + a2a3 + … + an−1an is equal to
(1) n(a1 − an) (2) (n − 1) (a1 − an)
(3) na1an (4) (n − 1)a1an
Ans. (4)
Sol:
2 1 3 2 n n1
1 1 1 1 ..... 1 1 d
a a a a a a−
− = − = = − = (say)
Then 1 2 2 3 n 1 n
1 2 2 3 n 1 n
a a a a a a
a a , a a ,......., a a
d d d
−
−
− − −
= = =
∴ 1 n
1 2 2 3 n 1 n
a a
a a a a ....... a a
d −
−
+ + + = Also, ( )
n 1
1 1 n 1 d
a a
= + −
⇒ 1 n ( )
1 n
a a
n 1aa
d
−
= −
40. If xm ⋅ yn = (x + y)m+n , then dy
dx
is
(1) y
x
(2) x y
xy
+
(3) xy (4) x
y
Ans. (1)
Sol: xm.yn (x y)m n mlnx nlny (m n)ln(x y) + = + ⇒ + = + +
∴ m n dy m n 1 dy m m n m n n dy
x y dx x y dx x x y x y y dx
+ + + + = + ⇒ − = − + + +
⇒ ( ) ( )
my nx my nx dy dy y
x x y y x y dx dx x
− −
= ⇒ = + +
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41. A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity
‘v’ that varies as v = α x . The displacement of the particle varies with time as
(1) t3 (2) t2
(3) t (4) t1/2
Ans: (2)
Sol. dx dx dt
dt x
= α ∫ = ∫α ⇒ x α t2
42. A mass of M kg is suspended by a weightless string. The horizontal force that is required to
displace it until the string makes an angle of 45° with the initial vertical direction is
(1) Mg( 2 −1) (2) Mg( 2 +1)
(3) Mg 2 (4) Mg
2
Ans: (1)
Sol. F sin 45 = Mg ( − cos 45)
F = Mg (√2 − 1)
43. A bomb of mass 16 kg at rest explodes into two pieces of masses of 4 kg and 12 kg. The velocity
of the 12 kg mass is 4 ms−1. The kinetic energy of the other mass is
(1) 96 J (2) 144 J
(3) 288 J (4) 192 J
Ans: (3)
Sol. m1v1 = m2v2
2
2 2
KE 1m v 1 4 144 288J
2 2
= = × × =
44. A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. the work done by the
force of gravity during the time the particle goes up is
(1) 0.5 J (2) −0.5 J
(3) −1.25 J (4) 1.25 J
Ans: (3)
Sol. − mgh = − mg
v2 1.25J
2g
= −
.
45. A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary
person with speed v ms−1. The velocity of sound in air is 300 ms−1. If the person can hear
frequencies upto a maximum of 10,000 Hz, the maximum value of v upto which he can hear the
whistle is
(1) 30 ms−1 (2) 15 2 ms−1
(3) 15 / 2 ms−1 (4) 15 ms−1
Ans: (4)
Sol.
( )
app
f 300
f v 15m/ s
300 v
= ⇒ =
−
46. A electric dipole is placed at an angle of 30° to a non-uniform electric field. The dipole will
experience
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(1) a torque only
(2) a translational force only in the direction of the field
(3) a translational force only in a direction normal to the direction of the field
(4) a torque as well as a translational force
Ans: (4)
Sol. A torque as well as a translational force
47. A material ‘B’ has twice the specific resistance of ‘A’. A circular wire made of ‘B’ has twice the
diameter of a wire made of ‘A’. Then for the two wires to have the same resistance, the ratio A /B
of their respective lengths must be
(1) 2 (2) 1
(3) 1
2
(4) 1
4
Ans: (1)
Sol. A A
1 2
A
R
R
ρ
=
π
B B
2 2B
R
R
ρ
=
π
2 2
A B A A A
2 2
B A B A A
R 2 R
R 4R
ρ ρ
= =
ρ ρ ⋅
⇒ B
A
= 2
48. The Kirchhoff’s first law (Σi = 0) and second law (ΣiR =ΣE) , where the symbols have their
usual meanings, are respectively based on
(1) conservation of charge, conservation of energy
(2) conservation of charge, conservation of momentum
(3) conservation of energy, conservation of charge
(4) conservation of momentum, conservation of charge
Ans: (1)
Sol. Conservation of charge, conservation of energy
49. In a region, steady and uniform electric and magnetic fields are present. These two fields are
parallel to each other. A charged particle is released from rest in this region. The path of the
particle will be a
(1) circle (2) helix
(3) straight line (4) ellipse
Ans: (3)
Sol. Straight line
50. Needles N1, N2 and N3 are made of a ferromagnetic, a paramagnetic and a diamagnetic
substance respectively. A magnet when brought close to them will
(1) attract all three of them
(2) attract N1 and N2 strongly but repel N3
(3) attract N1 strongly, N2 weakly and repel N3 weakly
(4) attract N1 strongly, but repel N2 and N3 weakly
Ans: (3)
Sol. attracts N1 strongly, N2 weakly and Repel N3 weakly
51. Which of the following units denotes the dimensions ML2/Q2, where Q denotes the electric
charge?
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(1) Weber (Wb) (2) Wb/m2
(3) Henry (H) (4) H/m2
Ans: (3)
Sol. Henry (H)
52. A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process is
completed in 0.1 s, the force of the blow exerted by the ball on the hand of the player is equal to
(1) 300 N (2) 150 N
(3) 3 N (4) 30 N
Ans: (4)
Sol. (mv − 0)⇒0.15× 20
F 3 30N
0.1
= =
53. A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves
0.2 m which applying the force and the ball goes upto 2 m height further, find the magnitude of the
force. Consider g = 10 m/s2
(1) 22 N (2) 4 N
(3) 16 N (4) 20 N
Ans: (4)
Sol. mgh = Fs
F = 20 N
54. Consider a two particle system with particles having masses m1 and m2. If the first particle is
pushed towards the centre of mass through a distance d, by what distance should the second
particle be moved, so as to keep the centre of mass at the same position?
(1) d (2) 2
1
m
m
d
(3) 1
1 2
m
m +m
d (4) 1
2
m
d
m
Ans: (4)
Sol. m1 d + m2 x = 0
1
2
m d
x
m
=
55. Starting from the origin, a body oscillates simple harmonically with a period of 2 s. After what time
will its kinetic energy be 75% of the total energy?
(1) 1 s
12
(2) 1 s
6
(3) 1 s
4
(4) 1 s
3
Ans: (2)
Sol. 2 2
max
1mv 3 1mv
2 42
=
A2 2 cos2 t 3 A2 2
4
ω ω ⇒ ω
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cos t 3
2
ω =
t t 1 sec
6 6
π
ω = ⇒ =
56. The maximum velocity of a particle, executing simple harmonic motion with an amplitude 7 mm, is
4.4 m/s. The period of oscillation is
(1) 100 s (2) 0.01 s
(3) 10 s (4) 0.1 s
Ans: (2)
Sol. Aω = vmax
max
T 2 2 A 0.01sec
v
π π
= = =
ω
57. A string is stretched between fixed points separated by 75 cm. It is observed to have resonant
frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two.
Then, the lowest resonant frequency for this string is
(1) 10.5 Hz (2) 105 Hz
(3) 1.05 Hz (4) 1050 Hz
Ans: (2)
Sol. ( ) ( ) n n 1 v 315, v 420
2 2
+
= =
Solving v 105
2
=
58. Assuming the sun to be a spherical body of radius R at a temperature of T K, evaluate the total
radiant power, incident on Earth, at a distance r from the Sun.
(1)
2 4
2
R T
r
σ
(2)
2 2 4
0
2
4 r R T
r
π σ
(3)
2 2 4
0
2
r R T
r
π σ
(4)
2 2 4
0
2
r R T
4 r
σ
π
where r0 is the radius of the Earth and σ is Stefan’s constant.
Ans: (3)
Sol.
2
0 4 2
2
r
( T 4 R )
4 r
π
σ ⋅ π
π
4 2 2
0
2
T R r
r
σπ
=
59. The refractive index of glass is 1.520 for red light and 1.525 for blue light. Let D1 and D2 be an of
minimum deviation for red and blue light respectively in a prism of this glass. Then
(1) D1 > D2
(2) D1 < D2
(3) D1 = D2
(4) D1 can be less than or greater than depending upon the angle of prism
Ans: (2)
Sol. D = (μ − 1)A
D2 > D1
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60. In a Wheatstone’s bridge, there resistances P, Q and R connected in the three arms and the
fourth arm is formed by two resistances S1 and S2 connected in parallel. The condition for bridge
to be balanced will be
(1)
1 2
P R
Q S S
=
+
(2)
1 2
P 2R
Q S S
=
+
(3) 1 2
1 2
P R(S S )
Q SS
+
= (4) 1 2
1 2
P R(S S )
Q 2SS
+
=
Ans: (3)
Sol. 1 2
1 2
P R(S S )
Q SS
+
=
61. The current I drawn from the 5 volt source will be
(1) 0.17 A (2) 0.33 A
(3) 0.5 A (4) 0.67 A
Ans: (3)
Sol. i 5 0.5
10
= =
5Ω 10Ω 5Ω
5 Volt
10Ω
I 10Ω
+ −
62. In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 kΩ with C = 2μF. The
resonant frequency ω is 200 rad/s. At resonance the voltage across L is
(1) 4 × 10−3 V (2) 2.5 × 10−2 V
(3) 40 V (4) 250 V
Ans: (4)
Sol. i 100 0.1A
1000
= =
L C 6
V V 0.1 250
200 2 10− = = =
× ×
V
63 Two insulating plates are both uniformly charged in such a way that
the potential difference between them is V2 −V1 = 20 V. (i.e. plate 2
is at a higher potential). The plates are separated by d = 0.1 m and
can be treated as infinitely large. An electron is released from rest on
the inner surface of plate 1. What is its speed when it hits plate 2?
(e = 1.6 × 10−19 C, me = 9.11 × 10−31 kg)
(1) 32 × 10−19 m/s (2) 2.65 × 106 m/s
(3) 7.02 × 1012 m/s (4) 1.87 × 106 m/s
0.1 m
Y
X
1 2
Ans: (2)
Sol. 1mv2 eV
2
=
v 2eV
m
= = 2.65 × 106 m/s
64. The resistance of a bulb filament is 100 Ω at a temperature of 100°C. If its temperature coefficient
of resistance be 0.005 per °C, its resistance will become 200 Ω at a temperature of
(1) 200°C (2) 300°C
(3) 400°C (4) 500°C
Ans: (2)
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Sol. 200 = 100[1 +(0.005 × Δt)]
T− 100 = 200
T = 300° C
65. In an AC generator, a coil with N turns, all of the same area A and total resistance R, rotates with
frequency ω in a magnetic field B. The maximum value of emf generated in the coil is ‘
(1) N.A.B.ω (2) N.A.B.R.ω
(3) N.A.B (4) N.A.B.R
Ans: (1)
Sol. NBAω
66. The flux linked with a coil at any instant ‘t’ is given by
φ = 10t2 − 50t + 250
The induced emf at t = 3 s is
(1) 190 V (2) −190 V
(3) −10 V (4) 10 V
Ans: (3)
Sol. e d (20 t 50) 10
dt
φ
= − = − − = − volt
67. A thermocouple is made from two metals, Antimony and Bismuth. If one junction of the couple is
kept hot and the other is kept cold then, an electric current will
(1) flow from Antimony to Bismuth at the cold junction
(2) flow from Antimony to Bismuth at the hot junction
(3) flow from Bismuth to Antimony at the cold junction
(4) not flow through the thermocouple
Ans: (1)
Sol. Flow from Antimony to Bismuth at cold junction
68. The time by a photoelectron to come out after the photon strikes is approximately
(1) 10−1 s (2) 10−4 s
(3) 10−10 s (4) 10−16 s
Ans: (3)
Sol. 10−10 sec.
69. An alpha nucleus of energy 1mv2
2
bombards a heavy nuclear target of charge Ze. Then the
distance of closest approach for the alpha nucleus will be proportional to
(1) 1
Ze
(2) v2
(3) 1
m
(4) 4
1
v
Ans: (3)
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Sol. 1
m
70. The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV, and the
stopping potential for a radiation incident on this surface 5 V. The incident radiation lies in
(1) X-ray region (2) ultra-violet region
(3) infra-red region (4) visible region
Ans: (2)
Sol. 1242eVnm 1100
11.2
λ = ≈ Å
Ultraviolet region
71. The energy spectrum of β-particles [number N(E) as a function of β-energy E] emitted from a
radioactive source is
(1)
N(E)
E0
E
(2)
N(E)
E0
E
(3)
N(E)
E0
E
(4)
N(E)
E0
E
Ans: (4)
Sol.
N(E)
E0
E
72. When 3Li7 nuclei are bombarded by protons, and the resultant nuclei are 4Be8, the emitted
particles will be
(1) neutrons (2) alpha particles
(3) beta particles (4) gamma photons
Ans: (4)
Sol. Gamma-photon
73. A solid which is transparent to visible light and whose conductivity increases with temperature is
formed by
(1) Metallic binding (2) Ionic binding
(3) Covalent binding (4) Van der Waals binding
Ans: (3)
Sol. Covalent binding
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74. If the ratio of the concentration of electrons that of holes in a semiconductor is 7
5
and the ratio of
currents is 7 ,
4
then what is the ratio of their drift velocities?
(1) 4
7
(2) 5
8
(3) 4
5
(4) 5
4
Ans: (4)
Sol. e
n
n 7
n 5
= e
n
I 7
I 4
=
d e
d n
(V )
(V )
⇒ e n
n e
I n 5
I n 4
× =
75. In a common base mode of a transistor, t collector current is 5.488 mA for an emit current of 5.60
mA. The value of the base current amplification factor (β) will be
(1) 48 (2) 49
(3) 50 (4) 51
Ans: (2)
Sol. Ib = Ie − Ic
β = c
b
I
49
I
=
76. The potential energy of a 1 kg particle free move along the x-axis is given by
x4 x2 V(x) J
4 2
= −
The total mechanical energy of the particle 2 J. Then, the maximum speed (in m/s) is
(1) 2 (2) 3 / 2
(3) 2 (4) 1/ 2
Ans: (2)
Sol. k Emax = ET − Umin
Umin (±1) = −1/4 J
KEmax = 9/4 J ⇒ U = 3
2
J
77. A force of −Fkˆ acts on O, the origin of the coordinate system. The torque about the point (1, −1)
is
(1) −F(ˆi − ˆj) (2) F(ˆi − ˆj)
(3) −F(ˆi + ˆj) (4) F(ˆi + ˆj)
Ans: (3)
Sol. τ = (−ˆi + ˆj)×(−Fkˆ)
= – F( ˆi + ˆj)
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78. A thin circular ring of mass m and radius R is rotating about its axis with a constant angular
velocity ω. Two objects each of mass M are attached gently to the opposite ends of a diameter of
the ring. The ring now rotates with an angular velocity ω′ =
(1) m
(m 2M)
ω
+
(2) (m 2M)
m
ω +
(3) (m 2M)
(m 2M)
ω −
+
(4) m
(m M)
ω
+
Ans: (1)
Sol. Li = Lf
mR2ω = (mR2 + 2MR2)ω′
ω′ = m
m 2M
ω
+
79. If the terminal speed of a sphere of gold (density = 19.5 kg/m3) is 0.2 m/s in a viscous liquid
(density = 1.5 kg/m3) of the same size in the same liquid.
(1) 0.2 m/s (2) 0.4 m/s
(3) 0.133 m/s (4) 0.1 m/s
Ans: (4)
Sol. s s
g g
v ( )
v ( )
ρ − ρ
=
ρ −ρ
vs = 0.1 m/s
80. The work of 146 kJ is performed in order to compress one kilo mole of gas adiabatically and in
this process the temperature of the gas increases by 7° C. The gas is
(R = 8.3 J mol−1 K−1)
(1) monoatomic (2) diatomic
(3) triatomic (4) a mixture of monoatomic and diatomic
Ans: (2)
Sol. 146 = CvΔT
⇒ Cv = 21 J/mol K
81. The rms value of the electric field of the light coming from the Sun is 720 N/C. The average total
energy density of the electromagnetic wave is
(1) 3.3 × 10−3 J/m3 (2) 4.58 × 10−6 J/m3
(3) 6.37 × 10−9 J/m3 (4) 81.35 × 10−12 J/m3
Ans: (2)
Sol. Uav = 2 6
0Erms 4.58 10ε = × − J/m3
2 6 3
rms E = 4.58 ×10− J/m
82. A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of
angular frequency ω. The amplitude of oscillation is gradually increased. The coin will leave
contact with the platform for the first time
(1) at the highest position of the platform (2) at the mean position of the platform
(3) for an amplitude of 2
g
ω
(4) for an amplitude of
2
2
g
ω
Ans: (3)
Sol. Aω2 = g
⇒ A = g/ω2
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83. An electric bulb is rated 220 volt − 100 watt. The power consumed by it when operated on 110
volt will be
(1) 50 watt (2) 75 watt
(3) 40 watt (4) 25 watt
Ans: (4)
Sol.
2 2
1 2
1 2
V V
Resistance
P P
= =
⇒ P2 = 25 W
84. The anode voltage of a photocell is kept fixed. The wavelength λ of the light falling on the cathode
is gradually changed. The plate current I of the photocell varies as follows :
(1)
O λ
I
(2)
O λ
I
(3)
O λ
I
(4)
O λ
I
Ans: (3)
85. The ‘rad’ is the correct unit used to report the measurement of
(1) the rate of decay of radioactive source
(2) the ability of a beam of gamma ray photons to produce ions in a target
(3) the energy delivered by radiation to a target.
(4) the biological effect of radiation
Ans: (4)
86. If the binding energy per nucleon in 73
Li and 4
2He nuclei are 5.60 MeV and 7.06 MeV
respectively, then in the reaction
7 4
p +3 Li→ 2 2He
energy of proton must be
(1) 39.2 MeV (2) 28.24 MeV
(3) 17.28 MeV (4) 1.46 MeV
Ans: (3)
Sol. EP
= (8 × 7.06 – 7 × 5.60) MeV = 17.28 MeV
87. If the lattice constant of this
semiconductor is decreased, then which
of the following is correct?
EC
EV
Eg
Condutton band width
Band gap
Valence badn width
(1) All Ec, Eg, Ev decrease
(2) All Ec, Eg, Ev increase
(3) Ec, and Ev increase but Eg decreases
(4) Ec, and Ev, decrease Eg increases
Ans: (4)
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88. In the following, which one of the diodes is reverse biased?
(1)
R
+ 5 V
(2)
R
+ 10 V
+ 5 V
(3)
− 10 V R
− 5 V
(4)
R
−10 V
Ans: (1)
89.
The circuit has two oppositely connect ideal diodes
in parallel. What is the current following in the
circuit?
4 Ω
3 Ω 2 Ω
D1
12 V
D2
(1) 1.33 A (2) 1.71 A
(3) 2.00 A (4) 2.31 A
Ans: (3)
Sol. D1 is reverse biased therefore it will act like an open circuit.
i 12 2.00 A
6
= =
90. A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its centre is
6.28 × 10−2 Weber/m2. Another long solenoid has 100 turns per cm and it carries a current i/3. The
value of the magnetic field at its centre is
(1) 1.05 × 10−4 Weber/m2 (2) 1.05 × 10−2 Weber/m2
(3) 1.05 × 10−5 Weber/m2 (4) 1.05 × 10−3 Weber/m2
Ans: (2)
Sol. B2 =
2
1 2 2 2
1 1
B n i (6.28 10 )(100 i / 3) 1.05 10
n i 200(i)
−
× × −
= = × W/m2
91. Four point masses, each of value m, are placed at the corners of a square ABCD of side . The
moment of inertia through A and parallel to BD is
(1) m2 (2) 2 m2
(3) 3m2 (4) 3 m2
Ans: (4)
Sol. I = 2m (/ 2 )2 = 3 m2
92. A wire elongates by mm when a load W is hanged from it. If the wire goes over a pulley and two
weights W each are hung at the two ends, the elongation of the wire will be (in mm)
(1) /2 (2)
(3) 2 (4) zero
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Ans: (2)
93. Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole
of nitrogen at temperature T0, while Box B contains one mole of helium at temperature (7/3) T0.
The boxes are then put into thermal contact with each other and heat flows between them until
the gases reach a common final temperature. (Ignore the heat capacity of boxes). Then, the final
temperature of the gases, Tf, in terms of T0 is
(1)f 0
T 5 T
2
= (2) f 0
T 3 T
7
=
(3)f 0
T 7 T
3
= (4) f 0
T 3 T
2
=
Ans: (4)
Sol. ΔU = 0
⇒ f 0 f 0
3R(T T ) 1 5R(T 7 T ) 0
2 2 3
− + × − =
Tf
= 0
3 T
2
94. Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm
and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium
condition, the ratio of the magnitude of the electric fields at the surface of spheres A and B is
(1) 1 : 4 (2) 4 : 1
(3) 1 : 2 (4) 2 : 1
Ans: (4)
Sol. A B
B A
E r 2
E r 1
= =
95. An inductor (L = 100 mH), a resistor (R = 100 Ω) and a
battery (E = 100 V) are initially connected in series as
shown in the figure. After a long time the battery is
disconnected after short circuiting the points A and B. The
current in the circuit 1 mm after the circuit is
E
A B
R
L
(1) 1 A (2) 1/e A
(3) e A (4) 0.1 A
Ans: (2)
Sol. I = I0 e−Rt /L = 1 A
e
* * * * * *