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SECTION - I
Single Correct Choice Type
This section contains 6 multiple choice questions. Each
question has 4 choices (A), (B), (C) and (D) for its an-
swer, out of which ONLY ONE is correct.
1. In the reaction
3H C C
C2NH
Cl
O
O
2(1) /NaOH Br
(2)
T
the structure of the Product T is
(a)
3H C CO C−
OO
(b) C
NH
O
3CH
(c)
3H C
C
NH
O
(d)
3H C CNH C−
OO
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PAPER - II
[ CHEMISTRY]
R
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1.Sol.(c)
3H C CO
2NH
21. /NaOH Br→
CO
Cl2.
Step I is Hofmann’s bromamide reaction ∴
3H C CO
2NH
21. /NaOH Br→3H C
2NH
Cl C− −
O
NPSR 3H C NH C− −
O
2. Assuming that Hund’s rule is violated, the bond
order and magnetic nature of the diatomic
molecule 2B is
(a) 1 and diamagnetic
(b) 0 and diamagnetic
(c) 1 and paramagnetic
(d) 0 and paramagnetic
2.Sol.(a) 2 * 2 2 * 2 2
2 1 1 2 2 2 yB s s s s pσ σ σ σ π→
B.O.[ ]6 4
12
−=
No unpaired electron is hence diamagnetic
3. The compounds ,P Q and S
HO 3H C
COOH 3OCH
P Q
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2
C
O
O
S
were separately subjected to nitration using
3 2 4/HNO H SO mixture. The major product
formed in each case respectively, s
(a) HO 3H C
COOH 3OCH
2NO 2NO
2O N
C
O
O
(b) HO 3H C
COOH 3OCH
2NO
2NO
2NO
C
O
O
(c) HO 3H C
COOH 3OCH
2NO2NO
2NO
C
O
O
(d) HO 3H C
COOH 3OCH
2NO2NO
2NO
C
O
O
3.Sol.(c)
COOH
HO(P)
3
2 4
HNO
H SO
+→COOH
HO
2NO
Because OH− is more powerful group
3OCH
HO(Q)
3
2 4
HNO
H SO
+→3OCH
3H C 2NO
Because 3OCH− is more powerful group
CO
O
3
2 4
HNO
H SO
+→
S
CO
O 2NO
because O− is more powerful group & P-is
unoccupied.
Ans =c
4. The species having pyramidal shape is
(a) 3SO (b) 3BrF
(c) 2
3SiO −(d) 2OSF
4.Sol.(d) 2SOF
2 4 03 3 3S s p d→Ground state
S-
Exited state
S-
O FF
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3
O
S
FF
5. The complex showing a spin-only magnetic
moment of 2.82 . .B M is
(a) 4( )Ni CO (b) [ ]24NiCl
−
(c) 3 4( )Ni PPh (d) [ ]24( )Ni CN
−
5.Sol..(b) ( )2n nµ = +
( )2.82 2n n= +
( )7.9 2n n= +
28 2n n= +2 2 8 0n n+ − =
2n = +
( )4
0Ni CO = ; ( )2
4 2Ni Cl−=
( )3 40Ni PPh = ; ( )
2
40Ni CN
− =
2 8 03 4Ni d s+ =
2unpaird
e-
xx x
Cl Cl Cl Cl
sp3
6. The packing efficiency of the two-dimensional
square unit cell shown below is :
L
(a) 39.27% (b) 68.02%
(c) 74.05% (d) 78.54%6. (d) Given diagram is a bcc arrangement
∴volume of two sphere =34
23
rπ×
and 3
4
ar =
3
2
42
3.4
23
r
P Fr
r
π× ×=
×
3
2
2 4 3
3 16 2
r
r r
π× ×=× ×
4
π=
3.14
4=
0.785=78.5%=
SECTION - II
Integer Answer Type
This section contains a group of 5 questions. The answer
to each of the questions is a single digit integer, ranging
from 0 to 9. The correct digit below the question no. in the
ORS is to be bubbled.
7. One mole of an ideal gas is taken from a to b
along two paths denoted y the solid and the
dashed lines as shown in he graph below. If the
work done along the solid line path is sw and
that along the dotted line path is dw , then the
integer closest to the ratio /d sw w is :
4.54.03.53.02.52.01.51.00.50.0
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.5 5.0 6.0
( .)V lit
P( .)atm
b
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7.Sol. Since process is the isothermal because
1 1 2 2PV PV=
i.e of point “a” 1 1 4 0.5 2PV = × =
and at 2V lit= , 1P atm=
∴ 2 2 2 1 2PV = × =
2
1
2.30 logs
Vw RT
V
= −
5.52.303 log
0.5PV= − ×
2.303 2log11= − × ........(1)
4.7sw ≈
[ ]4 2.5 1 1 2.5 0.6dw = − × + × + ×
10= −
∴10
24.6
d
s
w
w
−= =−
Ans =2
8. Among the following, the number of elements
showing only one non-zero oxidation state is
, , , , , , , ,O Cl F N P Sn Tl Na Ti
8.Sol. F→Only -1 ∴ three elements
Na→Only+1
Tl→Only +1
Ans= 3
9. Silver (atomic weight 1108 g mol−= ) has a den-
sity of 310.5 g cm− . The number of silver atoms
on a surface of area 12 210 m− can be expressed
in scientific notation as 10xy× . The value of x
is
9 Sol. 31 10.5cm gm=
13 310.5 46 10
108 3rπ= × × ×
8 1010 10r cm m− −∴ = =Volume corresponding to 12 2 1210 10 2m r− −= ×
∴22
6 3
10.53 10
10
gmwt
m
−−= × ×
∴no. of atom=
22 23
6
2 10 10.5 6.023 10
108 10
−
−
× × × ××
71.17 10= ×∴Ans =7
10. Total number of geometrical isomers for the
complex 3 3[ ( )( )( )]RhCl CO PPh NH is
10.Sol. [ ]Mabcd type of complexes exist in 3 isomeric
form
3NH
CO
Rh
3PPh
Cl
3NH
CORh
3PPh
Cl
3NHCO
Rh
3PPhCl
Ans=3
11. The total number of diprotic acids among the
following is :
3 4H PO 2 4H SO 3 3H PO
2 3H CO 2 2 7H S O 3 3H BO
3 2H PO 2 4H CrO 2 3H SO
11.Sol 2 4 3 3 2 3 2 2 7, , ,H SO H PO H CO H S O
2 4 2 3,H CrO H SO
Ans= 6
SECTION - III
Paragraph Type
This section contains 2 paragraphs. Based upon each of
the paragraphs 3 multiple choice questions have to e
answered. Each question has 4 choices (A), (B), (C) and
(D) for its answer, out of which ONLY ONE is correct.
Paragraph for questions 12 to 14
Two aliphatic aldehydes P and Q react in the presence of
aqueous 2 3K CO to give compound R, which upon
treatment with HCN provides compound S. On acidification
and heating, S gives the product shown below :
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3H C
3H C
OH
O O
12. The compounds P and Q respectively are
(a) 3H C C C
O O
CH
3CH
3andH H C H
(b) 3H C C C
O O
CH
3CH
andH H H
(c) CH C C
O3CH O
2CH3H C3andH H C H
(d) CH C C
O3CH O
2CH3H C andH H H
13. The compound R is
(a) C
O
2CH
CH
OH
3H C
3H C
(b) C
O
CH
CH
OH
3H C
3H C
3H C
(c)
3CH
2CH
O
CH CH
OH
CH3H C
(d)
3CH
CH
O
CH CH
OH
CH3H C
3H C
14. The compound S is
(a)
3CH
2CH
O
CH CH
CN
CH3H C
(b)
2CHCN
C3H C
3H CO
CH
(c)
3CH
2CH
CN
CH CHOH
OH
CH3H C
(d)
2CH
CN
CHOH
OH
C3H C
3H C
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12 to 14 Sol.
C
3H C
3H C
CH
C N≡
OH2CHOH
S
HCN
Nu−←addition
C
3H C
3H C
C
H
O
2CHOH
R
2 3.Aq K CO←
C H C+ −
H
O
H3CH3H C
C
O
H
P Q
or
H C C− −
H
OH
3CH
3CH
C
O
H
3H C
3H C
C
2CH
CH
OH
C
O
OH
OH
Acidificationfollowed by
S
2H O
∆−→
3H C
3H C
OHO
Desired product
∴Ans ( ) ( ) ( )12 ;13 ;14b a d→ → →
Paragraph for questions 15 to 17
The hydrogen-like species 2Li + is in a spherically
symmetric state 1S with one radial node. Upon absorbing
light the ion undergoes transition to a state 2S . The state
2S has one radial node and its energy is equal to the
ground state energy of the hydrogen atom.
15. The state 1S is
(a) 1s (b) 2s
(c) 2p (d) 3s
16. Energy of the state 1S in units of the hydrogen
atom ground state energy is
(a) 0.75 (b) 1.50
(c) 2.25 (d) 4.50
17. The orbital angular momentum quantum number
of the state 2S is
(a) 0 (b) 1
(c) 2 (d) 3
15 - 17 Solution :
15.(b) Ground state energy of hydrogen atom
s 13.6eV= −
∴2
213.6 13.6
Z
n− = − ×
2
2
313.6
n= − ×
∴ 3n = ∴ 2 3S = hence 1 2S S<
Since 1S is spherically symmetric an one radial
node therefore it will be 2s .
16.(c) Energy of 2s = 1E of H-atom
2
2
Z
n×
1E of H-atom
2
2
3
2×
12.25 E= × of H-atom
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17.(b) Since state 2S has also one radial node. Hence
it will be 3p .
∴ 1l =
SECTION - IV
Matrix - Match Type
This section contains 2 questions. Each question has
four statements (A,B,C and D) given in Column-I and five
statements (p, q, r, s and t) in Column-II. Any given
statement in Column-I can have correct matching with
ONE OR MORE statement(s) given in Column-II. For
example, if for a given questions, statement B matches
with the statements given in q and r, then for that particular
questions, against statement B, darken the bubbles
coresponding to q and r in the ORS.
18. Match the reactions in Column-I with appropriate
options in Column-II.
COLUMN - I
(A)
2N Cl + OH 2/NaOH H O00 C
N N= OH
(B)
3 3H C C C CH− − −
OH
3CH
OH
3CH
2 4H SO3H C
3CH3CH
3CHC
C
O
(C) C CH
3CH 3CH
O OH41.LiAlH
32.H O+
(D) HS ClBase S
COLUMN - II
(p) Racemic mixture
(q) Addition reaction
(r) Substitution reaction
(s) Coupling reaction
(t) Carbocation intermediate
18.Sol. (A)
OHElectrophilic
substitution←N N=
2N Cl+
+ OH
O & P-directing groupElectrophile
∴ In electrophilic substitution reaction
carbocation i.e. cycloarenium is formed
Ans = ,r s
(B) 3 3CH C C CH− − −
OHOH
3CH 3CH
2 4H SO→
Pinacol- pinacolone rearrangement in which first
we get carbocation
Ans= t
(C) C
O
3CH
4
3
1.
2.
LiAlH
H O+→
First is addition reaction & then we get optically
active compound ∴Racemic mixture
Ans= ,p q∴
(D) ClSH
NGP
EffectHCl→ + S
Ans= r∴ Ans=
( ) ( ) ( ) ( ) ( ) ( ), ; , ;A r s B t C p q D r→ → → →
19. All the compounds listed in Column-I react with
water. Match the result of the respective reactions
with the appropriate options listed in Column-II.
COLUMN - I COLUMN - II
(A) ( )3 22CH SiCl (p) Hydrogen halide
formation
(B) 4XeF (q) Redox reaction
(C) 2Cl (r) Reacts with glass
(D) 5VCl (s) Polymerization
(t) 2O formation
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19.Sol.
( ) ( ) ( ) ( ), ; , , ; , , ;A p s B p q t C p q t D s→ → → →(A)
3CH
3CH
SiCl
Cl
2H O→3 3CH Si CH− −
OH
OH
2HCl+
HO Si OH− −3CH
3CH
+HO Si OH− − →3CH
3CH
O Si O Si O− − − − − −(( )3CH
3CH 3CH
3CH
Polymersilicone
n
(B)
4 2 3 2
13 6 2 12 1
2XeF H O Xe XeO HF O+ → + + +
(C) 2 2Cl H O HCl HOCl+ → ++1-1
(Redox hydrogen halide)
2 2 3 2
13 2
2Cl H O H O Cl O+ −+ → + +
(Oxygen formation)
(D) ( )I V in five oxidation state on hydrolysis form
5 3
2 4V VO VO+ + −→ ++5 +5
No change in oxidation state when solution of
vanatate ion is concentrated they are polymerised
to form [ ]42 7V O−and higher vanadates i.e.
[ ]610 28V O
− [ ]412 32V O
−
Section - I
Single Correct Choice Type
20. Two adjacent sides of a parallelogram ABCD
are given by ˆˆ ˆ2 10 11AB i j k= + +
and
ˆˆ ˆ2 2AD i j k= − + +
The side AD is rotated by an acute angle α in
the plane of the parallelogram so that ADbecomes AD′ . If AD′ makes a right angle withthe side AB , then the cosine of the angle α is
given by
(a) 8
9(b)
17
9
(c) 1
9(d)
4 5
9
Sol.(b) 090 ADBα = −∠
A B
CDD'
α
( )0cos cos 90 ADBα = −∠
sin ADB= ∠
cosAB AD
ADBAB AD
⋅∠ =
408 / 9
15 3= =
⋅
( )2 17cos sin 1 8 / 9
9ADB∴ α = ∠ = − =
21. Let f be a real-valued function defined on the
interval ( )1, 1− such that
( ) 4
0
2 1
x
xe f x t dt− = + +∫ , for al l
( )1, 1x∈ − , and let 1f − be the inverse function
of f . Then ( )1f −′ (2) is equal to
PAPER - II
[ MATHEMATICS]
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(a) 1 (b) 1
3
(c) 1
2(d)
1
e
Sol.(b) ( )( )1f f x x− =
( )( ) ( )( )1 1 1f f x f x− − ′′⇒ =
( )( )( )( )
1
1
12
2f
f f
−
−
′⇒ =′
Now, ( )0 2f =
( )( ) ( )1 1 12
0 3f
f
− ′⇒ = =′
22. For 0, 1,....., 10r = , let 2 , rA B and rC denote,
respectively, the coefficient of rx in the
expansions of ( ) ( )10 201 , 1x x+ + and
( )301 x+ . Then ( )10
10 10
1
r r r
r
A B B C A=
−∑ is
equal to
(a) 10 10B C−
(b) ( )2
10 10 10 10A B C A−
(c) 0 (d) 10 10C B−
Sol.(d)10 20 30, ,
r r r r r rA C B C C C= = =
( )10 10r r rA B B C A−∑
( )210 10
10 20 30 10
10 10
1 1
r r r
r r
B C C C C= =
= −∑ ∑ ...(1)
( ) ( )20 30 30 20
10 20 10 101 1C C C C= − − −
30 20
10 10C C= −
23. A signal which can be green or red with probability
4
5 and
1
5 respectivley, is received by station A
and then transmitted to station B . The probability
of each station receiving the signal correctly is
3
4. If the signal received at station B is green
then the probability that the original signal was
green is
(a) 3
5(b)
6
7
(c) 20
23(d)
9
20Sol.(c)
24. Let 1, 2,3, 4S − . The total number of unordered
pairs of disjoint subsets of S is equal to
(a) 25 (b) 34
(c) 42 (d) 41
Sol.(d)
25. If the distance of the point ( )1, 2,1P − from the
plane 2 2x y z α+ − = , where 0α > , is 5 ,
then the foot of the perpendicular from P to the
plane is
(a) 8 4 7, ,
3 3 3
−
(b) 4 4 1, ,
3 3 3
−
(c) 1 2 10, ,
3 3 3
(d) 2 1 5, ,
3 3 2
−
Sol.(a) The equation of r⊥ from P
5P on plane is
1 2 1
1 2 2
x y z− + −= =
−its distance form is
1 2 1
1/ 3 2 / 3 2 / 3
x y zd
− + −= = =
−
A point at distance d can be taken
2 21, 2, 1
3 3 3
d d d− + − +
When 5d = when 5d = −
8 4 7 2 16 13, , , ,
3 3 3 3 3 3
− − −
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Section - II
Integer Type
26. Let 1 2 3 11, , , ...,a a a a be real numbers satisfying
1 215, 27 2 0,a a= − > and 1 22k k ka a a− −= −
for 3, 4,....,11k = .
If
2 2 2
1 2 11...90
11
a a a+ + += , then the value of
1 2 11...
11
a a a+ + + is equal to
Sol.(0) 1 2 3 11, , ..........a a a a are in A. P..
Let common difference ' 'd=
2 1 3 1 11 1, 2 .........., 10a a d a a d a a d= + = + = +
Given ( )10
2
1
0
90 11r
a rd=
+ = ×∑
10 10 102 2 2
1 1
0 0 0
1 2 990r r r
a d r a d r= = =
⇒ + + =∑ ∑ ∑
2 2
1 1
1 1111 10 11 21 2 10
6 2a d a d
⇒ + × × × + ×
990=2 2
1 135 10 90a d a d⇒ + + =
2225 35 150 90d d⇒ + + =235 150 135 0d d⇒ + + =
27 21 9 27 0d d d⇒ + + + =
( ) ( )7 3 9 3 0d d d⇒ + + + =
93,
7d
−⇒ = −
33
2d d
−< ⇒ = −∵
27. Two parallel chords of a circle of radius 2 are at a
distance 3 1+ apart. If the chords subtend at
the centre, angles of k
π and
2
k
π, where 0k > .
then the value of [ ]k is
[Note : [ ]k denotes the largest integer less than
or equal to k ]
Sol.(3) Angle subtended by a chord at the centre
12cosp
r
− θ =
O
A
B
cos2
p rθ
∴ =
Given 2 1 3 1, 2p p r± = + =
22 cos cos 3 1
2 2k k
π π + = +
2 3 1cos cos
2 2 2 2k k
π π + = +
3k = satisfies
3 1cos cos
6 3 2 2
π π+ = +
28. Let k be a positive real number and let
2 1 2 2
2 1 2
2 2 1
k k k
A k k
k k
−
= − − −
and
0 2 1
1 2 0 2
2 0
k k
B k k
k k
−
= − − −
If det ( ) ( ) 6det 10adj A adj B+ = , then [ ]k is
equal to
[Note : adj M denotes the adjoint of a square
matrix M and [ ]k denotes the largest integer
less than or equal to k ]
Sol.(4) ( ) ( ) 1det det
nadjA A
−=
Where n is order
( ) ( )2det ) detadjA A∴ =
( ) ( )2det det 0adjB B= =
B∵ is skew symmetric matrix of odd order
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Given, ( ) ( ) 6det det 10adjA adjB+ =
( )2 6det 10A⇒ = .......(1)
Expanding det A , we get
( )3det 2 1A k= +
∴ ( )( )23 62 1 10k + =
9 / 2k⇒ =
[ ] 4k∴ =
29. Consider a triangle ABC and let ,a b and c
denote the lengths of the sides opposite to vertices
,A B and C respectively. Suppose
6, 10a b= = and the area of the triangle is
15 3 . If ACB∠ is obtuse and if r denotes the
radius of the incircle of the triangle, then 2r is
equal to
Sol.(3)
B
CAθ
6
10
Area 1
sin2ab C=
115 3 6 10 sin
2C= × × ×
3
2SinC =
0120C∠ =2 2 2
cos2
a b cC
ab
+ −=∵
2 2 21 6 10
2 2 6 10
c+ −− =
⋅ ⋅
14c =
15 3
15r
s
∆= = 2 3r⇒ =
30. Let f be a function defined on R (the set of all
real numbers) such that
( ) ( )2010 2009f x x′ = −
( ) ( )2 32010 2011x x− − ( )42012x − , for all
x R∈ .
If g is a function defined on R with values in the
interval ( )0, ∞ such that ( ) ( )( )f x n g x= ,
for all x R∈ , then the number of points in R at
which g has a local maximum is
Sol.(1)
Section - III
Paragraph Type
Paragraph -1
Tangents are drawn from the point ( )3, 4P to
the ellipse
2 2
19 4
x y+ = touching the ellipse at
points A and B .
31. The coordinates of A and B are
(a) ( )3,0 and ( )0,2
(b) 8 2 161,
5 15
−
and 9 8,
5 5
−
(c) 8 2 161,
5 15
−
and ( )0, 2
(d) ( )3,0 and 9 8,
5 5
−
Sol.(d)
32. The orthocenter of the triangle PAB is
(a) 8
5,7
(b) 7 25,
5 8
(c) 11 8
,5 5
(d) 8 7
,25 5
Sol.(c)
33. The equation of the locus of the point whose
distances from the point P and the line AB are
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equal, is
(a) 2 29 6 54 62 241 0x y xy x y+ − − − + =
(b) 2 29 6 54 62 241 0x y xy x y+ + − + − =
(c) 2 29 9 6 54 62 241 0x y xy x y+ − − − − =
(d) 2 2 2 27 31 120 0x y xy x y+ − + + − =Sol.(a)
Paragraph -2
Consider the polynomial
( ) 2 31 2 3 4f x x x x= + + +
Let s be the sum of all distinct real roots of ( )f x
and let t s= .
34. The real number s lies in the interval
(a) 1,0
4
−
(b) 3
11,4
− −
(c) 3 1,
4 2
− −
(d) 1
0,4
Sol.(c)
35. The area bounded by the curve ( )y f x= and
the lines 0, 0x y= = and x t= , lies in the
interval
(a) 3,3
4
(b) 21 11
,64 16
(c) ( )9,10 (d) 21
0,64
Sol.(a)
36. The function ( )f x′ is
(a) increasing in 1
,4
t − −
and decreasing in
1,
4t
−
(b) decreasing in 1
,4
t − −
and increasing in
1,
4t
−
(c) increasing in ( ),t t−
(d) decreasing in ( ),t t−
Sol.(b)
Section- IV
Matrix -Match Type
37. Match the statements in Column-I with those in
Column - II.
Column - I
(A) A l ine f rom the origin meets the l ines
2 1 1
1 2 1
x y z− − += =−
8
3 13
2 1 1
xy z
− + −= =−
at P and Q
respectively. If length PQ d= , then 2d is
(B) The values of x satisfying
( ) ( )1 1tan 3 tan 3x x− −+ − −1 3
sin5
− =
are
(C) None-zero vectors ,a b and c
satisfy
0a b⋅ =
, ( ) ( ) 0b a b c− ⋅ + =
and
2 b c b a+ = −
.
If 4a b cµ= +
, then the possible values of µare
(D) Let f be the function on [ ],π π− given by
( )0 9f = and ( )
9sin
2
sin2
x
f xx
=
for 0x ≠ .
The value of ( )2f x dx
π
ππ −∫ is
Column - II
(p) 4−(q) 0
(r) 4
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(s) 5
(t) 637.Sol. (A) Let the equation of line be
x y z
a b c= = ......(1)
Given lines are
2 1 1
1 2 1
x y z− − += =−
.....(2)
& 8 / 3 3 1
2 1 1
x y z− + −= =−
......(3)
Equations (1) & (2) are coplaner
2 1 1
0
1 2 1
a b c
−
=
−
3 5 0a b c⇒ + + = ...(4)Equations (1) & (3) are also coplaner
8 / 3 3 1
0
2 1 1
a b c
−
=
−
⇒ 3 5 0a b c+ − = ......(5)
Solving (4) & (5) by cross multiplication
20 20 8
a b c⇒ = =− −
( ) ( ), , 5, 5, 2a b c⇒ = −
Let a point on this line be ( )5 , 5 ,2k k k−If it is on (2)
5 2 5 1 2 11
1 2 1
k k kk
− − − += = ⇒ =
−
( )5, 5,2P∴ −If it is on (3)
5 3 2 13 2
1 1
k kk
− + −= ⇒ =
−
10 10 4, ,
3 3 3Q
− ∴
distance
2 2 2
2 10 10 45 5 2
3 3 3PQ
= − + − + + −
50 46
9 9= + =
( ) ( )A t∴ →
38.(B) ( ) ( )1 1 1 3tan 3 tan 3 sin5
x x− − −+ − − =
( ) ( )( ) ( )
1 13 3 3
tan tan1 3 3 4
x x
x x
− − + − − = + + −
2
2
6 316 4, 4
8 4x x
x⇒ = ⇒ = ⇒ = −
−
( ) ( ) ( ),B p r∴ →
38.(C) 0a b⋅ =
( ) ( ) 0b a b c− ⋅ + =
2 0b a b b c a c⇒ − ⋅ + ⋅ − ⋅ =2 0b b c a c⇒ + ⋅ − ⋅ = ...(1)
Also, ( ) ( )2 222 b c b a+ = −
( )2 2 2 24 2 2b c b c b a a b⇒ + + ⋅ = + − ⋅
2 2 23 4 8 0b c a b c⇒ + − + ⋅ = ......(2)
If 4a b c= µ + then
2 2 2 216 8a b c b c= µ + + µ ⋅24a c b c c⋅ = µ ⋅ +
2 4 0a b b b c⋅ = µ + ⋅ =2 24 0b b c b c c+ ⋅ −µ ⋅ − =
(A) → t (B)→p,r (C)→q (D)→ r
38. Match the statements in Column-I with those in
Column - II.
[Note : Here z takes values in the complex plane
and Im z and Re z denote, respectively, theimaginary part and the real part of z .]
Column - I
(A) the set of points z satisfying
z i z z i z− = + is contained in or equal to
(B) The set of points z satisfying
4 4 10z z+ + − = is contained in or equal to
(C) If 2w = , then the set of points 1
z ww
= − , is
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PAPER - II
[ PHYSICS]QUESTIONS WITH ONLY ONE CORRECT ANSWERS
39. A tiny spherical oil drop carrying a net charge q
is balanced in still air with a vertical uniform
electric field of strength 5 181
107
Vmπ −× . When
the field is switched off, the drop is observed to
fall with terminal velocity 3 12 10 ms− −× . Given
29.8g ms−= , v iscosity of the air
5 21.8 10 Nsm− −= × and the density of oi l
3900mgm−= , the magnitude of q is :
(a) 191.6 10 C−× (b) 193.2 10 C−×
(c) 194.8 10 C−× (d) 198.0 10 C−×Sol. d)
qE mg= .......(1)
346 .
3rV mg r d gπη π= = .......(2)
6 rV qEπη⇒ =
5 3 581
6 1.8 10 2 10 107
r qπ
π − − −⇒ × × × × × = × ×
22q V r dgη =
3 52 109 2 10 1.8 10 18
102 900 9.8 98
r− −
−× × × ×⇒ = = ×
× ×
109
1079
−= ×
53
107
r m−⇒ = ×
9 6Now E rVπη=
5 53
5
6 1.8 10 3 102 10
81 710
7
qππ
− −−× × ×
= × × ××
186 1.8 3 210
81
−× × ×= ×
contained in or equal to
(D) If 1w = , then the set of points 1
z ww
= + is
contained in or equal to
Column - II
(p) an ellipse with eccentricity 4
5
(q) the set of points z satisfying Im 0z =
(r) the set of points z satisfying Im 1z ≤
(s) the set of points z satisfying Re 2z ≤
(t) the set of points z satisfying 3z ≤
38.Sol.
(A). ( )z i z z i z z i z− = + = − −
' 'z⇒ will lie onr⊥ bisector of line joining
&i z i z−
' 'z⇒ will lie on Real axis.
⇒ 0Im z⇒ =
(B). 4 4 10z z+ + − =
' 'z will lie on an ellipse with
2 10a = & 2 8ae = 4 / 5e⇒ =(C) Let 2 iw e θ=
1 12 2
2 2
ii i
i
ez w e e
w e
− θθ θ
θ= − = − = −
3 5cos sin
2 2i= θ+ θ
3 5Re & Im & 5
2 2z z z≤ ≤ ≤
(D) 1ω = ⇒ω= ω
1z = ω+ = ω+ω
ω
Im 0 & Re 1z z⇒ = ≤
(A) → q,r, (B)→p, (C)→s,t, (D)→ q,s,r,t,t
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18 18 194 1.810 0.8 10 8 10
9
− − −×= × = × = ×
40. A biconvex lens of focal length 15 cm is in front
of a plane mirror. The distance between the lens
and the mirror is 10 cm. A small object is kept at
a distance of 30 cm from the lens. The final image
is :
(a) virtual and at a distance of 16 cm from the
mirror
(b) real and at a distance of 16 cm from the mirror
(c) virtual and at a distance of 20 cm from the
mirror
(d) real and at a distance of 20 cm from the mirror
Sol. (b)
30 cm
F = 15 cm
O
10 cm
Lens
1 1 1
30 15v+ =
1 1 1 1
15 30 30v= − =
30v cm=∴ virton object for mirror. Image will be at 20
cm from mirror toward left. i.e. virtnd object for
lens at 10 cm
1 1 1
10 15v− =
1 1 1 3 2
10 15 30v
+= + =
6v cm⇒ =41. A Vernier calipers has 1 mm marks on the main
scale. It has 20 equal divisions on the Vernier
scale which match with 16 main scale divisions.
For this Vernier calipers, the least count is :
(a) 0.02 mm (b) 0.05 mm
(c) 0.1 mm (d) 0.2 mm
Sol. (d)
1 1LC MSD VSD= −
161
20mm mm= −
1 0.8 0.2mm= − =42. A uniformly charged thin spherical shell of radius
R carries uniform surface charge density of σper unit area. It is made of two hemispherical
shells, held together by pressing them with force
F (see figure). F is proportional to
F
F
(a) 2 21
o
Rσε (b)
21
o
Rσε
(c)
21
o R
σε (d)
2
2
1
o R
σε
Sol. (a)
2 2 2 2
1 2
2 2
0 0 0
( 4 )QQ R RF
R R
σ π σα = =∈ ∈ ∈
[check dimensions]
43. A hollow pipe of length 0.8 m is closed at one
end. At its open end a 0.5 m long uniform string
is vibrating in its second harmonic and it
resonates with the fundamental frequency of the
pipe. If the tension in the wire is 50 N and the
speed of sound is 320 1ms− , the mass of the
string is :
(a) 5 grams (b) 10 grams
(c) 20 grams (d) 40 grams
Sol. (b)
320 2 50
4 0.8 2 0.5 µ=
× ×
5050
µ⇒ =
1/
50kg mµ⇒ =
∴ mass of string 1
0.5 10050
gm= × ×
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10 gm=44. A block of mass 2 kg is free to move along the x-
axis. It is at rest and from t = 0 onwards it is
subjected to a time-dependent force F(t) in the x
direction. The force F(t) varies with t as shown in
the figure. The kinetic energy of the block after
4.5 seconds is :
F t( ) N4
O
s3
s4.5
t
(a) 4.50 J (b) 7.50 J
(c) 5.06 J (d) 14.06 J
Sol. (c)1 1
3 4 1.5 22 2
P∆ = × × − × ×
6 1.5 4.5 /kg m s= − =
4.5 /P kg m s∴ =
2 4.5 4.55.06
2 2 2
PK J
m
×= = =
×Intege type
QUESTIONS WITH INTEGER TYPE
45. At time t = 0, a battery of 10 V is connected across
points A and B in the given circuit. If the capacitors
have no charge initially, at what time (in seconds)
does the voltage across them become 4 V ?
[Take : 5 1.6, 3 1.1n n= = ]
2µF
M2 Ω 2µF
M2 Ω
A
B
Sol. (2)
4 Fπ1mΩ
/
0 (1 )tq q e τ−= −
/
0 (1 )tv v e τ−⇒ = −
/ 44 10 1 te− ⇒ = − / 4
0.6t
e−⇒ =
ln 5 ln34
t⇒ = − 0.5=
2t =46. A diatomic ideal gas is compressed adiabatically to
1/32 of its initial volume. In the initial temperature of
the gas is iT (in Kelvin) and the final temperature is
iaT , the value of a is :
Sol. (4)1 1
1 1 2 2T V T Vγ γ− −=
2
5
1 2
1
32T T
⇒ =
2 14T T⇒ =
47. A large glass slab ( 5 / 3)µ = of thickness 8 cm is
placed over a point source of light on a plane
surface. It is seen that light emerges out of the top
surface of the slab from a circular area of radius R
cm. What is the value of R ?
Sol. (6)
R
8 cm c
c
1 3sin
5c
µ= =
3tan
4c⇒ =
3
8 4
R⇒ =
6R cm⇒ =48. To determine the half life of a radioactive element, a
student plots a graph of ( )dN t
ndt
versus t. Here
( )dN t
dt is the rate of radioactive decay at time t. If
the number of radioactive nuclei of this element
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decreases by a factor of p after 4.16 years, the value
of p is :
3 2
4
5
6
1
2
3
4
5
Years
6
7
8
( )dN tn
dtl
Sol. (8)
0 0
t tdNN N e N e
dt
λ λλ− −= ⇒ − =
0ln ln ( )dN
N tdt
λ λ⇒ = −
Slope of the graph is
111/ 2
2yrλ λ −− = − ⇒ =
14.16
2.082
0
tNe e e
N
λ − ×− −∴ = = =
2.08 8P e∴ = =49. Image of an object approaching a convex mirror of
radius of curvature 20 m along its optical axis is
observed to move from 25
3 m to
50
7 m in 30
seconds. What is the speed of the object in km per
hour ?
Sol. (3)
For 25
, 103
v m f= = +
1 1 1
v u f+ =
3 1 1
25 10u+ =
1 1 3 25 30 5 1
10 25 10 25 10 25 50u
− −− − = = = −
× ×
1 50u⇒ = −Simillarly,
1 1 7 7 50 70 20 1
10 50 50 10 50 10 50 25u
− −= − − = = = −
× ×
2 25u⇒ = −
Object distance 25m in 30 sec80 speed of object
will be 25 18
3 /30 5
km h× =
QUESTIONS WITH COMPREHENSION TYPE
Paragraph For Questions 50 to 52
The key feature of Bohr’s theory of spectrum of
hydrogen atom is the quantization of angular
momentum when an electron is revolving around
a proton. We will extend this to a general
rotational motion to find quantized rotational
energy of a diatomic molecule assuming it to be
rigid. The rule to be applied is Bohr’s quantization
condition.
50. A diatomic molecule has moment of inertia I .
By Bohr’s quantization condition its rotational
energy in the nth level ( n = 0 is not allowed) is
(a)
2
2 2
1
8
h
n Iπ
(b)
2
2
1
8
h
n Iπ
(c)
2
28
hn
Iπ
(d)
22
28
hn
Iπ
Sol. (d)
2
hI nω
π=
2
nh
Iω
π=
2 2 2 22
2 2 2
1 1
2 2 2 8
n h n hI I
I Iω
π π∴= = =
51. It is found that the excitation frequency from
ground to the first excited state of rotation for the
CO molecule is close to 114
10 Hzπ× . Then the
moment of inertia of CO molecule about its center
of mass is close to (Take 342 10h J sπ −= × )
(a) 46 22.76 10 kg m−× (b) 46 21.87 10 kg m−×
(c) 47 24.67 10 kg m−× (d) 47 21.17 10 kg m−×
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Sol. (b)
22 2
2(2 1 )
8
hhv
Iπ= −
34
2 11
3 3 2 10
488 10
hI
v
ππ π
π
−× ×⇒ = =
× × ×
45453 10
0.187 1016
−−×
= = ×
46 21.87 10 kgm−= ×46 21.87 10 kgm−= ×
52. In a CO molecule, the distance between C (mass
= 12 a.m.u.) and O (mass = 16 a.m.u.), where 1
a.m.u. 275
103
kg−= × , is close to :
(a) 102.4 10 m−× (b) 101.9 10 m−×
(c) 101.3 10 m−× (d) 104.4 10 m−×Sol. (c)
C.o.M.CO
2r 1r
1
12 3
28 7r r r= =
2
16 7
28 4r r r= =
2 2
0 1 2cm r m r I+ =
2 2
2 27 463 4 516 12 10 1.87 10
7 7 3r − −
× + × × × = ×
192 1.87 49 3 10
5 336r
−× × ×⇒ =
×
19 200.163 10 1.63 10− −= × = ×101.28 10r −⇒ = ×
Paragraph For Questions 53 to 55
When liquid medicine of density ρ is to be put
in the eye, it is done with the help of a dropper.
As the bulb on the top of the dropper is pressed,
a drop forms at the opening of the dropper. We
wish to estimate the size of the drop. We first
assume that the drop formed at the opening is
spherical because that requires a minimum
increase in its surface energy. To determine the
size, we calculate the net vertical force due to
the surface tension T when the radius of the drop
is R. When this force becomes smaller than the
weight of the drop, the drop gets detached from
the dropper.
53. If the radius of the opening of the dropper is r, the
vertical force due to the surface tension on the
drop of radius R (assuming r<<R) is :
(a) 2 rTπ (b) 2 RTπ
(c)
22 r T
R
π(d)
22 R T
r
π
Sol. (c)
θR
r
Fr
Net vertically upward force
222
r r TrT
R R
ππ ⇒ =
54. If 45 10r m−= × , 3 310 kgmρ −= , 210g ms−=10.11T Nm−= , the radius of the drop when it
detaches from the dropper is approximately
(a) 31.4 10 m−× (b) 33.3 10 m−×
(c) 32.0 10 m−× (d) 34.1 10 m−×
Sol. (a)
232 4
.3
r TR g
R
ππ ρ=
2 84
3
3 3 25 10 0.11
2 2 10 10
r TR
gρ
−× × ×= =
× ×
31.42 10R m−⇒ ×
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55. After the drop detaches, its surface energy is
(a) 61.4 10 J−× (b) 62.7 10 J−×
(c) 65.4 10 J−× (d) 68.1 10 J−×Sol. (b)
2 3 24 4 3.14 (1.42 10 ) 0.11E R Tπ −= = × × × ×
62.7 10 J−= ×
QUESTIONS WITH MATRIX MATCH TYPE
56. You are given may resistances, capacitors and
inductors. These are connected to a variable DC
voltage source (the first two circuits) or an AC
voltage source of 50 Hz frequency (the next three
circuits) in different ways as shown in Column
II. When a current I (steady state for DC or rms
for AC) flows through the circuit, the corresponding
voltage 1V and 2V . (indicated in circuits) are
related as shown in Column I. Match the two
Column I Column II
(A) 10,I V≠ (p) V1 V2 mH6
3µF V
is proportional to I
(B) 2 10,I V V≠ > (q)
V1 V2 mH6
2Ω V
(C) 1 20,V V V= = (r)
V1 V2 mH6
2Ω V
~
(D) 20,I V≠ (s)
V1 V2 mH6
V
~
3 Fµ is proportional to I
(t)
V1 V2
V
~
3 Fµ
k1 Ω
Sol.
(a) → r,s,t
(b) →q,r,s,t
(c) →p,q
(d) →q,r,s,t
57. Two transparent media of refractive indices 1µ
and 3µ have a solid lens shaped transparent
material of refractive index 2µ between them as
shown in figures in Column II. A ray traversing
these media is also shown in the figures. In
Column I different relationships between 1µ , 2µ
and 3µ are given. Match them to the ray
diagrams shown in Column II.
Column I Column II
(A) 1 2µ µ< (p)
µ1
µ2
µ3
(B) 1 2µ µ> (q)
µ1
µ2
µ3
(C) 2 3µ µ= (r)
µ1
µ2
µ3
(D) 2 3µ µ> (s)
µ1
µ2
µ3
(t)
µ1
µ2
µ3
Sol.
(a) →p,r,,
(b) →q,s,t
(c) →p,r,t
(d) →q,s
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