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CHAMP SQUARE, Sunrise Forum, 2nd Floor (Debuka Nursing Home Lane), Circular Road, Ranchi – 834 001, Ph.: 7360016500
PART TEST –II (MAIN)
PPAARRTT –– AA ((PPHHYYSSIICCSS))
1. The adjacent diagram shows a charge +Q held on an
insulating support S and enclosed by a hollow spherical
conductor. The point O represents the centre of the
spherical conductor and P is a point such that OP = x and
Sp = r. The electric field at point P will be
(A) 2
0
Q
4 x
(B) 2
0
Q
4 r
(C) 0
(D) none of the above
1.Sol. A
Using Gauss’s law, we have
in0
1E dA q
2
0
20
1E 4 x Q Q Q
1 QE
4 x
2. If a point charge is placed at vertex of cube then, flux linked to surface
shaded in figure.
(A) 0
q
8 (B)
0
q
3
(C) 0
q
12 (D) zero
2.Sol. C
Flux through face 1 is zero and through each of the two faces 2 and 3 is
0
q
24 . So, total flux through shaded surface,
0 0
q q0 2
24 12
3. A neutral conducting spherical shell is kept near a charge q as shown.
The potential at point P due to the induced charges is
(point C represents the centre of the shell)
(A) kq
r (B)
kq
r
(C) kq kq
r r
(D)
kq
CP
q
1
2
3
Q x
r
SP = r
OP = x
- Q
+ Q
E
Charge +Q on
insulating support
O x
r
P
SP = r
OP = x
S
q
C
P r’
r q
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3.Sol. C
Inside conducting shell, C PV V
or
0 0
1 q q q 1 qV
4 r R R 4 r
0
1 q qV
4 r r
4. Consider the shown uniform solid insulating sphere with a short and
light electric dipole of dipole moment ˆp j (embedded at its centre)
placed at rest on a horizontal surface. An electric field ˆE i is suddenly
switched on in the region such that the sphere instantly begins rolling
without slipping. Speed of the sphere when the dipole moment
becomes horizontal for the first time is (m = Mass of the sphere)
(A) 5pE
m (B)
10pE
7m
(C) 5pE
2m (D) zero
4.Sol. B
W K
2 21 1or PE cos 90 cos0 m I
2 2
22 2
2
1 1 2or PE m mR
2 2 5 R
10 PE
7m
5. In the figure below, what is the potential difference between
the point A and B between B and C respectively in steady
state.
(A) AB BCV V 100 V
(B) AB BCV 75 V, V 25 V
(C) AB BCV 25V, V 75 V
(D) AB BCV V 50 V
5.Sol. C
The equivalent circuit is shown in figure
1 2V V 100 and 1 22V 6 V
On solving above equations, we get V1 = 75 V, V2 = 25 V
100 V 20 Ω
1 µF
1 µF 3 µF
3 µF
B
10 Ω
C A
V2 V1
r
r’
q
P
C
R +
+
+
+
− − − −
1 µF
100 V 20 Ω 10 Ω
1 µF
1 µF 3 µF
3 µF
B
p
x
y
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6. Two capacitors C1 and C2 = 2 C1 are connected in a circuit with a switch
between them as shown in the figure. Initially the switch is open and C1
holds charge Q. The switch is closed. At steady state, the charge on
each capacitor will be
(A) Q, 2Q
(B) Q / 3, 2Q / 3
(C) 3Q / 2, 3Q
(D) 2Q / 3, 4Q / 3
6.Sol. B
Common potential,
1 1 1
Q QV
C 2C 3C
Now charge, 1 1 2 2
Q 2QQ C V and Q C V
3 3
7. In following circuit, key is closed at time t = 0, then what will be
current through battery and charge on capacitor of 2F at that time
(A) 3 A, 12C
(B) 1. 5 A, 6C
(C) 2 A, 3C
(D) 6 A, 6C
7.Sol. A
At t = 0, capacitors offer zero resistance for dc and so effective resistance of the circuit is R 2
6
i 3A2
8. A mass spectrometer is a device which select particle of equal
mass. A particle with electric charge q > 0 starts at rest from a
source S and is accelerated through potential difference V. It
passes through a hole into a region of constant magnetic field
B perpendicular to the plane of the paper as shown in the
figure. The particle is deflected by the magnetic field and
emerges through the bottom hole at a distance d from the top
hole. The mass of the particle is
(A) q Bd
V (B)
2 2q B d
4V (C)
2 2q B d
8 V (D)
q Bd
2V
8.Sol. C
Radius d m
r2 qB
Also 21 2qV
qV m or2 m
2 2
2qVm
md q B dor m
2 qB 8V
Q C1
R
C2 = 2 C1
S
2 Ω
2 Ω
2 Ω K 6 V
2 µF
1 µF
B
V
S
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9. A circular current loop of radius a is placed in a radial field B
as shown. Then net force acting on the loop is
(A) zero
(B) 2Balcos
(C) 2alBsin
(D) none
9.Sol. C
Loop experiences the magnetic force due to radial component of
magnetic field. Thus,
F B sin I perimeter Bsin I 2 a
10. A conducting wire bent in the form of a parabola y2 = x
carrying a current i = 1 A as shown in the figure. This wire is
placed in a magnetic field ˆB 2k tesla. The unit vector in
the direction of force is
(A) ˆ ˆ3i 4 j
5
(B)
ˆ ˆi j
2
(C) ˆ ˆi 2 j
5
(D) none of these
10.Sol. B
ˆ ˆ ˆ ˆ ˆ ˆab Ob Oa 4 i 2 j i j 3 i 3 j
ˆ ˆ ˆ ˆ ˆ ˆF i i B 1 3i 3 j 2k 6 j 6 i
ˆ ˆ ˆ ˆ6 j 6 i i jF̂
6 2 2
11. Consider three quantities 0 0x E/B, y 1/ and zCR
. Here l is the length of a wire, C is
a capacitance and R is a resistance. All other symbols have standard meanings. Then incorrect
option is
(A) x, y have the same dimensions
(B) y, z have the same dimensions
(C) z, x have the same dimensions
(D) none of the three pairs have the same dimensions
11.Sol. D
E / B has dimensions of velocity. CR has dimensions of time and so CR
has dimensions of velocity . So, x,
y and z all the three have dimensions of velocity.
x =1
x =4 x
b
a (1,1)
O
y
(4,-2)
I a
B sin θ
x =1
x =4 x
b
a
O
y
a
I B
θ
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12. A flexible wire bent in the form of a circle is placed in a uniform
magnetic field perpendicular to the plane of the coil. The radius of the
coil changes as shown in figure. The graph of induced emf in the coil
is represented by
(A)
(B)
(C)
(D)
12.Sol. B
The flux through the coil is given by 2BAcos0 B r
Now d dr
e 2 Brdt dt
Initially dr
0dt
and so e = 0. After that dr
dt
is of constant value, also r increases, so e is also
increases.
Finally dr
dt
becomes zero and so e becomes zero.
13. A rectangular loop of wire shown below is coplanar with a long wire
carrying current I. The loop is pulled to the right as indicated. What
are the directions of induced current in the loop and the magnetic
forces on the left and right sides of the loop?
Induced Current Force on left side Force on right side
(A) Counter clockwise to the left to the right
(B) Clockwise to the left to the right
(C) Counter clockwise to the right to the left
(D) Clockwise to the right to the left
13.Sol. B
As loop move s away, the flux in the loop decreases and so current will induce clockwise to compensate this
flux.
I
t O
e
1 2 t
O
e
1 2
t O
e
1 2 t
O
e
1 2
t (s) O
r
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14. A cylindrical region of radius 1m has instantaneous homogenous
magnetic field of 5T and it is increasing at a rate of 2T/s. A regular
hexagonal loop ABCDEFA of side 1 m is being drawn in to the
region with a constant speed of 1 m/s as shown in the figure. What
is the magnitude of emf developed in the loop just after the shown
instant when the corner A of the hexagon is coinciding with the
centre of the circle?
(A) 5/ 3 V (B) 2 /3V
(C) 5 3 2 /3 V (D) 5 3 V
14.Sol. C
The induced emf across the ends B and F due to motion of the loop,
1e B BF 5 1 2 sin60 5 3 V
The induced emf across the loop due to change in magnetic field
22
2
1dB R dB 2e A 2 V
dt 3 dt 3 3
So, 1 2
2e e e 5 3 V
3
15. Key is in position 2 for long time. Thereafter, it is in position
1 at t = 0, Resistances of the bulb and inductance of
inductor are marked in the figure choose the incorrect
alternative
(A) Bulb 2 dies as soon as key is switched into position 1
(B) Time in which brightness of bulb 1 becomes half of its
maximum brightness does not depend on t.
(C) If t = ∞, total heat produced in bulb 1 is 2
22
L
2R
(D) Ratio of maximum power consumption of bulbs is 1 : 1
15.Sol. D
The maximum current in the inductor 02
iR
Energy stored in the inductor
2 220 2
2 2
1 1 LU Li L
2 2 R 2R
When key is in position 1, this energy will convert into heat energy through resistor
16. Consider the circuit shown with respective specifications of
elements marked in the figure. Capacitor – 1 is charged
such that charge on it is Q0 and its left plate is positively
charged. While capacitor – 2 is uncharged. The switch is
closed at t = 0
(A) Frequency of oscillation of charge on left plate of
capacitor – 1 is 1
2 2L C
(B) Frequency of oscillation of charge on left plate of
capacitor – 1 is 1 2
LC
1 m/s A
1 m
1 m
B C
D
E F
key ε
R2
Bulb 2
Bulb 1
R1
1 2
L
Capacitor -2 Capacitor -1
C C
L
Inductor Switch
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(C) Maximum current through the inductor is 0Q
2L C
(D) Maximum current through the inductor is 0Q
LC
16.Sol. C
eq
1 1f
2 LC 2LC
When i is max, di
0dt
Charge on two capacitors is same
2
02
2 0 00 0
Q
Q Q21Li 2 i
2 2C 4C 2LC
17. A metallic wire is folded to form a square loop of side a. It carries a current i and is kept
perpendicular to the region of uniform magnetic field B. If the shape of the loop is changed from
square to an equilateral triangle without changing the length of the wire and current. The amount
of work done in doing so is
(A) 2 4 3
Bia 19
(B)
2 3Bia 1
9
(C)
22Bia
3 (D) Zero
ANS. A 18. A charged particle of mass m & charge q enters a zone of uniform magnetic
field B with a velocity v making an angle θ with the boundary of the zone,
having width d, when the particle penetrates half – way into the zone, the
change in energy of the particle is
(A)
2mvsin
2B (B)
2
Bqdsin
mv
(C) zero (D) mv d
sinqB
18.Sol. C
Work done by magnetic field is zero.
19. A non conducting ring of radius R1 is charged such that the linear charge density is 2
1cos
where θ is the polar angle. If the radius is increased to R2 keeping the charge constant, the linear
charge density is changed to 2
2 cos . The relation connecting 1 2 1 2R , R , and
(A) 1 1 2 2/R /R (B) 1 2 2 1/R /R (C) 1 2 1 2R R (D) 2 21 1 2 2R R
ANS. B
θ
d
d/2
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20. In the circuit shown the cells are ideal and have equal emf, the
capacitance of the capacitor is C and the resistance of the
resistor is R. The switch X is first connected to Y and then to Z.
After a long time, the total heat produced in the resistor will be
(A) equal to the energy finally stored in the capacitor
(B) half the energy finally stored in the capacitor
(C) twice the energy finally stored in the capacitor
(D) four times the energy finally stored in the capacitor
ANS. D
Numerical Value 1. In the given circuit, the potential difference across the capacitor is
12 V. Each resistance is of 3. The cell is ideal. The emf (in V) of
the cell is
Ans. 15
2. If energy stored in the capacitors C1 and C2 are same, then the
value of 1
2
C
C is given by
25
K . Find K.
Ans. 36
2.Sol.
At steady state the current, i3R
In close loop A H D E F A, 1
1
q0
C or 1 1q 2 C
Now in close loop BGCE FB
2
2
q2R i 0
C
G i 2R B
A ε
ε
i
F
H q1
q2
C1
R
C2
C
D
R 2R
E
E
C1
C2
ε
R R R
R
R C = 3µF
C
A
B R
Z
Y X
− +
+ −
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Or 2
2
q2R 0.
3R C
22
5 Cq
3
Given,
2 21 2
1 2
q q
2C 2C
221 1 1 1
22 2 22
C q 2C C 25or
C 5 /3C C 36q
3. Consider a wire of non uniform cross section. If the area of
cross–section at point A is double of the area of cross – section
at point B. What is ratio of heat energy dissipated in a unit
volume at points A and B?
Ans. 0.25
3.Sol. 21 1
22 2
H i R R 1
H 4R 4i R
4. An electron and a positron are projected in a transverse
magnetic field (of strength B) with speed v each. If width of
magnetic field region is kr (where k is a constant, r is radius
of revolution of each of the particle). Time spent by each of
the particles in magnetic field is one–fourth of the time
period of revolution of particles for k equal to (Assume
elastic collision, if particles collide).
Ans. 1.414
4.Sol. krx 2sin or sinr 4 r
k 2
kr / 2 kr / 2
π / 4
4R R
i
B A
Positron
1.5 r
kr
Electron
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5. In figure, there is a four way key at the middle. If key is
thrown from situation BD to AC, then how much charge
(in C) will flow through point O?
Ans. 72
5.Sol. When key is on the position BD, the situation is shown in the figure.
When key is on the position AD, the situation is shown in the figure
The charge flows to each capacitor is 36 C and so total charge flows through point O is 72 C
O
12 C
- 24 C
3 V
6 V
- 24 C
24 C
- 12 C
O
6 V
- 24 C
24 C
- 12 C
12 C
3 V
3 V
4 µF 4 µF
6 V
A B
O D
K
C
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PPAARRTT –– BB ((CCHHEEMMIISSTTRRYY))
1. When an aromatic compound (A) was treated with Br2 / aq. KOH gives compound (B). Compound
(B) upon treatment with alcoholic potash and another compound (Y) gives a foul smelling gas with
formula C6H5NC was formed. Y was formed by reacting a compound (Z) with Cl2 in the presence
of slaked lime. The compound (Z) is
(A) C6H5NH2 (B) C2H5OH (C) CH3OCH3 (D) CHCl3
1.Sol. B
C
O
NH2Br2 / KOH
NH2
CHCl3 / alc. KOHNC
(A)
H3 CH2 OHCCl2 / Ca (OH)2
CHCl3 + HCOO
(Y)(Z)
2. Acid – catalysed hydration of an unknown compound X (C6H12) yielded Y (C6H14O) as the major
product a racemic mixture. Which (if any) of the following is (are) to be X ?
CH3CH3
CH3
(1)
CH3CH3
CH3(2)
CH2CH3
CH3
(3)
CH2CH3
CH3
CH3
(4)
(A) (1) and (3) (B) only (2) (C) only (4) (D) none of above
2. Sol. D
After hydration non of the compound (1) to (4) form an optically active compound.
3. Bakelite is obtained from phenol and ……………
(A) acetaldehyde (B) acetal (C) formaldehyde (D) chlorobenzene
Ans. C 4. Examine the following statements pertaining to a SN
2 reaction.
(a) The rate of reaction is independent of the concentration of the nucleophile
(b) The nucleophile attacks the carbon atom on the side of the molecule opposite to the group
being displaced
(c) The reaction proceeds with simultaneous bond formation and bond rupture amongst the
following which of the above were true?
(A) a, b (B) a, c (C) a, b, c (D) b, c
4.Sol. D
For SN2 reaction, Rate k R x Nu
Me
Et D
Cl
Nu
SN2
Me
Et D
ClNu- Cl
Me
Et Nu
D
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5. Choose the compound which can react with [Ag(NH3)2]+ and on treatment with alk. KMnO4 gives
(CH3)3C – COOH
(A) CH3CH2CH2 – C C – CH3 (B) (CH3)2 CHCH2 – C CH
(C) (CH3)3 C – C CH (D) (CH3)3 C – C C – CH3
Ans. C 6. Methyl alcohol can be distinguished from ethyl alcohol using
(A) Fehling solution (B) Schiff’s reagent
(C) Sodium reagent (D) Sodium hydroxide and iodine
Ans. D 7. CH3
NH2
(CH3CO)2O
What is X?
(A)
CH3
Br
NH2
(B)
(C) (D)
Br2/CH3COOH H+/H2O
X
CH3
NH2
Br
CH3
NH2
COCH3
CH3
COCH3
NH2
7.Sol. B
CH3
NH2
(CH3CO)2O
CH3
NHCOCH3
Br2 / CH3COOH
NHCOCH3
Br
CH3
H3O+
NH2
Br
CH3
8. Maltose is made of
(A) two molecules of glucose (B) two molecules of fructose
(C) glucose and fructose molecules (D) two molecules of sucrose
Ans. A
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9. An ester (A) with molecular formula C9H10O2 was treated with excess of CH3MgBr and the
compound so formed was treated with conc. H2SO4 to form olefin (B). Ozonolysis of B gave
ketone with formula C8H8O which shows positive iodoform test. The ester (A) is
(A) C2H5COOC6H5 (B) C7H7COOCH3
(C) ethyl benzoate (D) both (A) and (C)
Ans. C 10. C N
OCH3
+ CH3MgBr Q PH3O+
The product 'P' in the above reaction is
(A) (B)
COCH3
OH
(C) (D)
COCH3
OCH3
CH3 CH3
OH
OCH3
CH3 CH3
OH
OH
10.Sol. B
CN
OCH3
CH3MgBr
C NMgBr
CH3
OCH3
H3O+
OH
C CH3O
11. Which of the following pairs are correctly matched for the number of mono-chlorinated products
(ignoring stereoisomers) obtained from the listed isomeric hexanes?
Isomer Number of monochlorinated products
I : 2, 3 - dimethyl butane Two
II : 2, 2 - dimethyl butane Three
III : 2 – methyl pentane Five
Select the correct answer using the code given below:
(A) I, II and III (B) I and II only (C) II and III only (D) I and III only
Ans. A
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12. Consider oxidation of the following alcohols
H2 CHCH2OHC PH2 CHCHOCI :
II : (CH3)3COH H3 C
CH3
CH2C
III : CH3CH2OHR
CH3COOH IV : CH3CH CH CH3
SCH3COOH
The oxidising agents A, B, C and D respectively are
P Q R S
(A) MnO2 Cu/ 4KMnO /OH,
H2CrO4
(B) Cu/ MnO2 H2CrO4 4KMnO /OH,
(C) MnO2 Cu/ H2CrO4
4KMnO /OH,
(D) MnO2 H2CrO4 Cu/
4KMnO /OH,
Ans. C 13. The IUPAC name of following compound is
C CH
CH3
CH
CH3
CH
CHO
CH2 CH3H3
(A) 2 – ethyl – 3, 4 – dimethylpentanal (B) 2, 3 dimethyl – 4 – aldohexane
(C) 2, 3, - dimethyl – 4 – ethyl pentanal (D) 1 – ethyl – 2, 3 - dimethyl pentanal
Ans. A
14. (2R, 3R) – 2,3 – pentandiol is
(A)
C
HHO
OHH
CH3
H2 CH3
(B)
C
OHH
HHO
C
H3
H2 CH3
(C)
C
HHO
OHH
C
H3
H2 CH3
(D) Both (A) & (C)
14.Sol. D
C
HHO
OHH
C
H3
H2 CH3
1
2
3
4 5
(2R, 3R)
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15. Which comparison is correct as indicated?
(A) OH > H3 OHC (acidic nature)
(B)
NNH2
< ( kb value)
(C) COOH < HCOOH (pka
value)
(D) All
Ans. A
16. CH3
Cl
H2
H
CH3C
aq. KOH
SN2
product ( 2 - butanol )
d - (dextrorotatory)
(A) product is (laevorotatory) (B) product is racemic mixture
(C) product is d (D) none of the above
16.Sol. D
In SN2 reaction, inversion in configuration occurs but optical rotation may or may not be opposite
17. Most stable free radical in the following is
(A)
CH3
(B)
(C)
(D)
CH2
17.Sol. C
Higher the conjugation more will be stability.
18 Select the correct statement (s)
(A) All natural amino acids which are constituents of proteins, are -amino acids
(B) All -amino acids are optically active and have L – configuration
(C) In proteins -amino acids are connected by amide linkage
(D) Both (A) and (C)
Ans. D
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19. End product of the reaction
Hg (OAc)2
NaBH4
AOH
CH3 CH3
A is
(A) OH
HO
CH3
(B) O
CH3 CH3
(C) OH
HO
CH3
(D) OH
CH3CH3
19.Sol. B
OH
CH2
Hg (OAc)2 OH
CH3CH3
HgOAc
O
CH3
HgOAc
NaBH4 O
CH3 CH3
20. Glycerol can be separated from spent - lye in soap industry by:
(A) Steam distillation (B) fractional distillation
(C) distillation under reduced pressure (D) ordinary distillation
Ans. C Numerical Value 1. Number of chiral center(s) present in hydrolysis product of A (given below) is
O O
Ans. 1
1.Sol.
O O
H3O+
OH OHHO OH- H2O
O HHO OH
2. Me
NMe
EtBu
OH
Major product
n -
The number of possible hyper conjugation form in major product is
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Ans. 0
2.Sol. There are four β – hydrogens in this quaternary ammonium salt, on heating it will give Hoffman eliminated
product by abstraction of most acidic β – H.
3. The sulphur content of cystine is 32%. If cystine contains two sulphur atoms, then molecular
weight of cystine is
Ans. 200 4. How many of the following is/are correctly matched?
(i) Neoprene CH3 CH2 C
Cl
CH CH2 CH3
n
(ii) Nylon – 66 NHCH3 (CH2)6NH C
O
(CH2)4C
O
CH3
n
(iii) Terylene OCH3 CH2CH2 C
O
C
O
CH3
n
(iv) Teflon CCH3 F2 CF2 CH3
n
Ans. 3
4.Sol.
OC COOCH2 CH2 CH3OCH3
n
Terylene
5. An organic compound ‘A’ having molecular formula C2H3N on reduction gave another compound
‘B’. Upon treatment with nitrous acid, ‘B’ gave ethyl alcohol. On warming with chloroform and
alcoholic KOH ‘B’ forms offensive smelling compound ‘C’. The number of lone pairs of electron in
compound ‘C’ is
Ans. 1
5.Sol.
H3 C NCReduction
CH3CH2NH2
HNO2CH3CH2 OH
(A) (B)
CH3CH2 NH2CHCl3 / KOH
CH3CH2NC
(B)
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PPAARRTT –– CC ((MMAATTHHEEMMAATTIICCSS))
1. Let z1, z2 be two complex numbers such that z1 + z2 and z1z2 both are real, then
(A) z1 = - z2 (B) 1 2z z (C) 1 2z z (D) z1 = z2
1. Sol. (B)
Let z1 = a + ib, z2 = c + id, then z1 and z2 is real
(a + c) + i(b + d) is real
b + d = 0 d = - b
z1z2 is real
(ac – bd) + i(ad + bc) is real
ad + bc = 0
a(-b) + bc = 0
a = c
z1 = a + ib = c – id = 2z a c and b d
2. The value of
6
k 1
2 k 2 ksin icos
7 7
is
(A) -1 (B) 0 (C) – i (D) i
2. Sol. (D)
Let 2 2
z cos isin7 7
Then by De Moivre’s theorem, we have
k 2 k 2 k
z cos isin7 7
3. The points representing the complex number z for which |z + 5|2 - |z – 5|2 = 10 lie on
(A) a straight line
(B) a circle
(C) a parabola
(D) the bisector of the line joining (5, 0) and (-5, 0)
3. Sol. (A)
z 5 z 5 z 5 z 5 10
or 5 z z 25 5 z z 25 10
12 2x 2 x (a)
2
4. If z 2
2z 3
represents a circle, then its radius is equal to
(A) 1
3 (B)
3
4 (C)
2
3 (D) 1
4. Sol. (C)
2 2
z 22
z 3
20 32x y x 0
3 3
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Which is a circle with centre 10
,03
and
radius = 100 32 4 2
9 3 9 3
5. How many ways are there to arrange the letters in the word GARDEN with the vowels in
alphabetical order?
(A) 120 (B) 240 (C) 360 (D) 480
5. Sol. (C)
Take A, E as similar letter (eg. BB)
Then six letters can be arranged in 6 !/2 !, i.e., 360 ways.
6. Seven women and seven men are to sit around a circular table such that there is a man on either
side of every women. The number of seating arrangements is
(A) (7 !)2 (B) (6 !)2 (C) 6 ! 7! (D) 7 !
6. Sol. (C)
First we seat seven women, which can be done by (7 – 1) ! = 6 ! ways
In any such arrangement, seven men can be seated in seven places marked in 7P7 = 7 ! ways
Required number of ways = 6 ! 7 !
7. The number of ways in which 10 persons can go in two boats, so that there may be 5 on each
boat, supposing that two particular persons will not go in the same boat is
(A) 105
1C
2 (B) 8
51
C2
(C) 8
42 C (D) 8C4
7. Sol. (C)
First omit two particular persons. Remaining eight persons may be four in each boat. This can be done in 8C4
ways. The two particular persons may be placed in two ways one in each boat
Total number of ways = 2 8C4
8. The number of ways of dividing 52 cards amongst four players so that three players have 17
cards each and the fourth player just one card, is
(A)
3
52! 1
3!17! (B) 52 ! (C)
3
52!
17! (D) none of these
8. Sol. (A)
3
52! 35! 18! 1 ! 52! 1 !
35!17! 18!17! 17!1! 3! 3!17!
W
W
W
W W
W
W
X
X
X
X X
X
X
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9. There are four balls of different colors and four boxes of colors same as those of the balls. The
number of ways in which the balls, one in each box, could be placed such that a ball does not go
to box of its own colour is
(A) 8 (B) 7 (C) 9 (D) none of these
9. Sol. (C)
1 1 1
4! 12 4 1 92! 3! 4!
10. If the second term in the expansion
n
13
1
xx
x
is 14x5/2 then the value of n n3 2C / C is
(A) 12 (B) 6 (C) 4 (D) 3
10. Sol. (C)
n 1
1n 5 / 2132 1T C x x x 14x
or
n 1 35 / 213 2nx 14x
n
rn
r 1
n3
n2
n 1 3 5
13 2 2
n 14
C n r 1Now, for r 3, n 14
rC
C 14 3 14
3C
11. The number of terms which are free from radical signs in the expansion of 55
1/5 1/10x y is
(A) 6 (B) 7 (C) 8 (D) none of these
11. Sol. (A)
55 r5555 r /10 55 r /5 r /105
r 1 r r
r 0
T C x y x C x y
The terms which are free from radical will correspond to those values of r in which the radical signs
disappear, i.e., r r
and5 10
. Clearly from (i) r should be a multiple of 10 of range 0 to 55.
r = 0, 10, 20, 30, 40, 50
Thus there will be six terms free from radical signs.
12. If the 4th term in the expansion of
n1
axx
is
5
2, then the value of a and n are
(A) 1
,62
(B) 1, 3 (C) 1
,32
(D) cannot be found
12. Sol. (A)
It is given that the fourth term in the expansion
n1 5
of ax isx 2
, therefore
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3n n 3
3
n n 3 n 63
1 5C (ax)
x 2
5C a x
2
We must have x0 for which n – 6 = 0. Hence,
n = 6 and we get 6 3
35
C a2
3 1 1
a a8 2
13. If (1 + x + 2x2)20 = a0 + a1x + a2x2 + …. + a40x40 then a1 + a3 + a5 + ….. + a39 equals
(A) 219(220 – 21) (B) 230(219 – 19) (C) 219(220 + 21) (D) none of these
13. Sol.: (D)
Putting x = 1 and -1 and subtracting, we get 420 – 220 = 2[a1 + a3 + a5 + …. + a39]
219(220 – 1) = a1 + a3 + … + a37 + a39
14. The sum
m
i 0
10 20 p, where 0 if p q
i m i q
is maximum when m is
(A) 5 (B) 10 (C) 15 (D) 20
14. Sol. (C)
m10 20 10 20 10 20 10 20 10 20
i m i 0 m 1 m 1 2 m 2 m 0
i 0
C C C C C C C C ... C C
= Coefficient of xm in the expansion of product (1 + x)10 (x + 1)20
= Coefficient of xm in the expansion of (1 + x)30
= 30Cm
Hence the maximum value 30Cm is 30C15.
15. The largest real value for x such that
4 4 k k
k 0
5 x 8
4 k ! k! 3
is
(A) 2 2 5 (B) 2 2 5 (C) 2 2 5 (D) none of these
15. Sol. (A)
4 4 k k
k 0
4 4 k k
k 0
4 4 4 k kk
k 0
4
5 x
4 k ! k!
5 x 4!
4 k ! k! 4!
C 5 x
4!
5 x
4!
16. A man has 3 pairs of black socks and 2 pairs of brown socks kept together in a box. If he dressed
hurriedly in the dark, the probability that after he has put on a black sock, he will then put on
another black sock is
(A) 1
3 (B)
2
3 (C)
3
5 (D)
2
15
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16. Sol. (A)
6
11 10
1
C 6 3P B
10 5C
2 1 2
1 2 1 2 1
5P B / B B Black
9
P B B P B P B / B
3 5 1
5 9 3
17. A and B are two events such that P(A) > 0, P(B) 1, then P A /B is equal to
(A) 1 – P(A/B) (B) 1 P A /B (C)
1 P A B
P B
(D)
P A
P B
17. Sol. (C)
P A B P A BP A / B
P B P B
1 P A B
P B
18. A bag contains 5 white and 3 black balls. Four balls are successively drawn out and not replaced.
The probability that they are alternately of different colors is
(A) 1
196 (B)
2
7 (C)
1
7 (D)
13
56
18. Sol. (C)
Required probability = P(WBWB) + P(BWBW)
5 3 4 2 3 5 2 4
8 7 6 5 8 7 6 5
1 1 1
14 14 7
19. A dice is tossed 5 times. Getting an odd number is considered a success. Then the variance of
distribution of number of successes is
(A) 8
3 (B)
3
8 (C)
4
5 (D)
5
4
19. Sol. (D)
Here n = 5
p = P (an odd number) = 3 1
6 2
1 1q 1 p 1
2 2
1 1 5Variance npq 5
2 2 4
20. Three of the six vertices of a regular hexagon are chosen at random. The probability that the
triangle with three vertices is equilateral equals
(A) 1/2 (B) 1/5 (C) 1/10 (D) 1/20
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20. Sol. (C)
Out of 6 vertices 3 can be chosen in 6C3 ways, will be equilateral if it is ACE or BDF (2 ways)
Required probability = 6
3
2 2 1
20 10C
Numerical Value
1. If 1 sin2 cos2
f2cos2
then the value of 0 02f 11 f 34 equals
Ans. 1
1.
2 2 2
1 sin2 cos2f
2cos2
cos sin cos sin
2 cos sin cos sin
2cos 1
2 cos sin 1 tan
0 0
0 0
1 1f 11 f 34
1 tan11 1 tan34
0 0 0
1 1
1 tan11 1 tan 45 11
00
0
0
0
1 1
1 tan111 tan111
1 tan11
1 1 tan11 1
2 21 tan11
2. Value of 1
2r 0
1tan
1 r r
is equal to
k
, then k equals to
Ans. 2
2. Sol.
1 1
2
1 r 1 rtan tan
1 r r 11 r r
= tan-1 (r + 1) – tan-1(r)
n1 1 1 1
r 0
tan (r 1) tan (r) tan (n 1) tan (0)
= tan-1 (n + 1)
A B
C
D E
F
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n
1 1
2r 0
1tan tan
21 r r
3. A(3, 1), B(6, 5) and C(x, y) are three points such that angle ACB is a right angle and the area of triangle ABC
= 7 sq. unit, then the number of such points C is
ANS. 0
Sol. It is clear that AB = 5, let AC = b, BC = a,
then area of
ab
ABC 72
Thus such a point C is possible if
a2 + b2 = 5, ab = 14
We easily observe that this system does not have real solutions.
4. The area of the triangle formed by joining the origin to the points of intersection of the line 5x 2y 3 5
and circle x2 + y2 = 10 is
ANS. 5
Sol. The equation of lines joining origin and point of intersection of 5x 2y 3 5 and 2 2x y 10,
is
2
2 2 5x 2yx y 10
3 5 or
2 2x y 8 5xy 0
These lines are perpendicular.
Area of triangle 1
base height2
1
radius radius2
1
10 10 52
5. The foci of the ellipse 2 2
2
x y1,
16 b and hyperbola
2 2x y 1
144 81 25 concide. Then the value of b2 is
ANS. 7
Sol. We must have 12
ae a'e' 4e e'5
Where 2 2 281 144b 16 1 e , e' 1
25 25 from last equation
15e'
12 then from, first equation
3e
4
whence from second equation
2 9b 16 1 7
12
C(x, y)
900
A(3, 1) B(6, 5)