TARGET COURSE FOR IIT-JEE 2010
PHASE- ALL (REVISION TEST-AT HOME)
MATHEMATICS, PHYSICS, CHEMISTRY
TEST # 5 (TR-2(1))
PAPER - 1
Date : 14/02/2010
Time : 3 : 00 Hrs.
MAX MARKS: 249
Name : _________________________________________________________ Roll No. : __________________________
INSTRUCTIONS TO CANDIDATE
A. GENERAL :
SEAL
1. Please read the instructions given for each question carefully and mark the correct answers against the question
numbers on the answer sheet in the respective subjects.
2. Write your Name & Roll No. in the space provided on this cover page of question paper.
3. The Question paper contains blank space for your rough work. No additional sheet will be provided for rough work.
4. The answer sheet, a machine readable Optical Mark Recognition (OMR) is provided separately.
5. Do not break the seal of the question-paper booklet before being instructed to do so by the invigilators.
6. Blank papers, Clipboards, Log Tables, Slide rule, Calculators, Cellular phones, Pagers and Electronic
gadgets in any form are Not allowed to be carried inside the examination hall.
B.
FILLING THE OMR :
7. Fill your Name, Roll No., Batch, Course and Centre of Examination in the blocks of OMR sheet and
darken circle properly.
8. DO NOT TAMPER WITH/ MUTILATE THE OMR.
C. MARKING SCHEME :
Space for rough work
Each subject in this paper consists of following types of questions:Section - I
9. Multiple choice questions with only one correct answer. 3 marks will be awarded for each correct answer and –1 mark
for each wrong answer.
10. Multiple choice questions with multiple correct option. 4 marks will be awarded for each correct answer and –1 mark for
each wrong answer.
11. Passage based single correct type questions. 4 marks will be awarded for each correct answer and –1 mark for each
wrong answer.
Section –II
12. Column matching type questions. 6 marks will be awarded for the complete correctly matched answer and No
Negative marking for wrong answer. However, 1 mark will be given for a correctly marked answer in any row.
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Page # 1
Important Data (egRoiw.kZ vk¡dM+s)
Atomic Masses: H = 1, C = 12, K = 39, O = 16, Fe= 56, N= 14, I = 127, Ca = 40, Mg = 24, Al = 27, F = 19, Cl = 35.5, S = 32,
(ijek.kq nzO;eku) Na = 23
Constants
: R = 8.314 Jk–1mol–1, h = 6.63 × 10–34 Js , C = 3 × 108 m/s, e = 1.6 × 10–19 Cb,me = 9.1 × 10–31Kg,
(fu;rkad)
: RH = 1.1 × 107 m–1, log 2 = 0.3010, log 3 = 0.4771, log(5.05) = 0.7032; ln2 = 0.693; ln 1.5 = 0.405; ln3 = 1.098
Space for Rough Work (jQ+ dk;Z gsrq LFkku)
Space for rough work
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Page # 2
MATHEMATICS
Section – I
[k.M - I
Questions 1 to 9 are multiple choice questions. Each
iz'u 1 ls 9 rd cgqfodYih iz'u gSaA izR;sd iz'u ds pkj fodYi
question has four choices (A), (B), (C) and (D), out of
(A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 3 marks will be given for each correct answer
Q.1
The value of tan 3 α cot α cannot lie in
(A) ] 0, 2/3 [
(C) ] 4/3, 4 [
vafdr dhft;sA izR;sd lgh mÙkj ds fy, + 3 vad fn;s tk;asxs rFkk
izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA
and – 1 mark for each wrong answer.
Q.1
OMR 'khV esa iz'u dh iz'u la[;k ds le{k viuk mÙkj
tan 3 α cot α dk eku fuEu esa ugha gks ldrk gS
(A) ] 0, 2/3 [
(C) ] 4/3, 4 [
(B) ] 1/3, 3 [
(D) ] 2, 10/3 [
(B) ] 1/3, 3 [
(D) ] 2, 10/3 [
Q. 2
If 2 sin2 ((π/2) cos2x) = 1 – cos (π sin 2x),
x ≠ (2n + 1) π/2, n ∈ I, then cos 2x is equal to
;fn 2 sin2 ((π/2) cos2x) = 1 – cos (π sin 2x),
x ≠ (2n + 1) π/2, n ∈ I, rc cos 2x =
Q. 3
(A) 1/5
(B) 3/5
(C) 4/5
(D) 1
r
r
If u & v are non-zero vectors such that one of
(C) 4/5
(D) 1
r
r
;fn 'kwU; lfn'k u o v bl izdkj gS fd ,d dks
Q. 2
(A) 1/5
Q. 3
(B) 3/5
them can be expressed as a scalar multiple of other
r
r
& satisfy (2cosθ +1) u + ( 3 cot θ + 1) v = 0
nwljs ds vfn'k x.kqu ds :i esa fy[k ldrs gS rFkk
then most general value of θ is
gS rc θ dk O;kid eku gksxk
2π
; n∈I
(A) nπ +
3
(A) nπ +
(C) 2nπ +
2π
; n∈I
3
5π
(B) nπ +
; n∈I
6
(D) 2nπ +
5π
; n∈I
6
r
r
(2cosθ +1) u + ( 3 cot θ + 1) v = 0 dks lUrq"V djrs
(C) 2nπ +
2π
; n∈I
3
2π
; n∈I
3
(B) nπ +
5π
; n∈I
6
(D) 2nπ +
5π
; n∈I
6
Space for rough work
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Page # 1
Q.4

1
1
If z = sec–1  x +  +sec–1  y +  , where xy > 0,
x
y


Q.4
If x + y + z = 12 & x2 + y2 + z2 = 96
1 1 1
& + + = 36 then value of x3 + y3 + z3 is
x y z
(A) 862
Q.6
Q.7
Q. 8
(B) 863
(C) 865
x

y

rc z dk og eku (fn;s x;s ekukas esa ls) tks lEHko
ugha gS] gksxk -
then the value of z (among the given values) is not
possible5π
7π
9π
5π
(A)
(B)
(C)
(D)
6
10
10
3
Q.5

1
1
;fn z = sec–1  x +  + sec–1  y +  , tgk¡ xy > 0,
(A)
Q.5
5π
6
(B)
7π
10
(C)
9π
10
(D)
5π
3
;fn x + y + z = 12 o x2 + y2 + z2 = 96
1 1 1
+ + = 36 rc x3 + y3 + z3 =
x y z
(A) 862 (B) 863
(C) 865
(D) 866
o
(D) 866
The number of solutions of equation
log6 (x + 3) = 7 – x is
(A) 0
(B) 2
(C) 1
(D) None of these
Q.6
If p, q, r, s, t are numbers such that p + q < r + s; q +
r < s + t; r + s < t + p; s + t < p + q; then the largest
and the smallest numbers are respectively(A) p and q
(B) r and t
(C) r and p
(D) q and p
Q.7
In a third order determinant, each element of the
first column consists of sum of two terms, each
element of the second column consists of sum of
three terms and each element of third column
consists of sum of four terms. Then, it can be
decomposed in n determinants, where n has the
value
(A) 24
(B) 16
(C) 9
(D) 1
Q. 8
lehdj.k log6(x + 3) = 7 – x ds gyksa dh la[;k gksxh
(A) 0
(B) 2
(C) 1
(D) bueas
ls dksbZ ugha
;fn la[;k,sa p, q, r, s, t bl izdkj gS fd p + q < r + s ;
q + r < s + t; r + s < t + p; s + t < p + q; rc egÙke o
U;wure la[;k,sa Øe'k% gksxh
(A) p rFkk q
(C) r rFkk p
(B) r rFkk t
(D) q rFkk p
,d rhu Øe ds lkjf.kd esa] izFke LrEHk dk izR;sd
vo;o nks inksa dk ;ksx gS] f}rh; LrEHk dk izR;sd
vo;o rhu inksa dk ;ksx gS rFkk rhljs LrEHk dk
izR;sd vo;o pkj inksa dk ;ksx gS rc bls
n lkjf.kd esa rksM+k tk ldrk gS] rks n dk eku gksxk
(A) 24
(B) 16
(C) 9
(D) 1
Space for rough work
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Page # 2
Q. 9
r − 1
 r
If the matrix Mr is given by Mr = 
,
r − 1 r 
r = 1, 2, 3 .... Then the value of det (M1) + det(M2)
+ .... + det (M2009) is
(A) 2008
(B) 2009
(C) (2009)2
Which of the following set of values of x satisfies
the equation
2( 2 sin
2
x − 3 sin x +1)
+ 2( 2 − 2 sin
2
can lie in
–π π
(A) 
, 
 8 8
3π 5π
(C)  , 
 8 8 
(B) 2009
(D) (2008)2
(A), (B), (C) rFkk (D) gS]a ftuesa ls ,d ;k ,d ls vf/kd fodYi
lgh gaAS OMR 'khV esa iz'u dh iz'u la[;k ds le{k vius mÙkj
vafdr dhft,A izR;sd lgh mÙkj ds fy, + 4 vad fn;s tk;saxs
rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA
Q. 10
x
2
x + 3 sin x )
The solution of the inequality sin6x + cos6x >
r − 1
iz'u 10 ls 14 rd cgqfodYih iz'u gaAS izR;sd iz'u ds pkj fodYi
=9
π
π
(A) x = nπ ± , n ∈ I (B) x = nπ ± , n ∈ I
6
3
π
(C) x = nπ, n ∈ I
(D) x = 2nπ + , n ∈ I
2
Q. 11
 r
;fn eSfVªDl Mr = 
, tgk¡ r = 1, 2, 3 .... gS]
r 
r − 1
rc det (M1) + det(M2) + .... + det (M2009) dk eku
gksxk
(A) 2008
(C) (2009)2
(D) (2008)2
Questions 10 to 14 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE THAN ONE) is
correct. Mark your response in OMR sheet against the
question number of that question. + 4 marks will be
given for each correct answer and –1 mark for each
wrong answer.
Q. 10
Q. 9
ds
ekuksa
( 2 sin 2 x − 3 sin x +1)
dk
+2
og
leqPp;
( 2 − 2 sin 2 x + 3 sin x )
tks
lehdj.k
= 9 dks larq"V
djrk gS] gksxk
π
π
, n ∈ I (B) x = nπ ± , n ∈ I
6
3
π
(C) x = nπ, n ∈ I
(D) x = 2nπ + , n ∈ I
2
5
6
6
vlfedk sin x + cos x > dk gy fuEu esa gksxk
8
(A) x = nπ ±
5
8
Q. 11
π 3π
(B)  , 
4 4 
(D) All are correct
–π π
(A) 
, 
 8 8
π 3π
(B)  , 
4 4 
3π 5π
(C)  , 
 8 8 
(D) lHkh lR;
Space for rough work
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Page # 3
Q.12
Q. 13
2 tan–1 (–3) is equal to
Q.12
(A) – cos–1 (–4/5)
(A) – cos–1 (–4/5)
(B) – π + cos–1 (4/5)
(B) – π + cos–1 (4/5)
(C) – π/2 + tan–1 (–4/3)
(C) – π/2 + tan–1 (–4/3)
(D) cot–1 (4/3)
(D) cot–1 (4/3)
8ax 
If x = 9 is solution ofln (x2 +15a2) –ln (a –2) = ln 

 a −2
Q. 13
then
(A) a =
Q. 14
2 tan–1 (–3) =
;fn x = 9, ln (x2 +15a2) –ln (a – 2) = ln 
8ax 
 dk
 a −2
gy
gks] rc
3
(B) a = 3
5
(A) a =
(C) x = 15 (D) x = 2
Q. 14
If p, q, r, s are in A.P. and
p + sin x q + sin x p − r + sin x
f(x) = q + sin x r + sin x − 1 + sin x such that
r + sin x s + sin x s − q + sin x
p + sin x q + sin x p − r + sin x
f(x) = q + sin x r + sin x − 1 + sin x bl iz d kj
r + sin x s + sin x s − q + sin x
2
∫ f (x)dx = – 4 then the common difference of the
gS fd
A.P. can be –
gks x k
1
2
(C) 1
(D) none of these
∫ f (x)dx =
– 4, rc l-Js - dk lkoZ v Urj
0
0
(B)
(C) x = 15 (D) x = 2
;fn p, q, r, s l-Js - es a gS o
2
(A) – 1
3
(B) a = 3
5
1
2
(A) – 1
(B)
(C) 1
(D) buesa ls dksbZ ugha
Space for rough work
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Page # 4
This section contains 2 paragraphs; each has
3 multiple choice questions. (Questions 15 to 20) Each
question has 4 choices (A), (B), (C) and (D) out of which
ONLY ONE is correct. Mark your response in OMR
sheet against the question number of that question. + 4
marks will be given for each correct answer and –1 mark
for each wrong answer.
Passage # 1 (Ques. 15 to 17)
Let A and B two square matrices such that
B = A–1 BA. For the matrices A and B solve each
of the following.
Q.15
Q.16
Q.17
(A – B)2 is equal to (A) 0
(C) A2 – 2AB + B2
bl [k.M esa 2 vuqPNsn fn;s x;s gS]a izR;sd esa 3 cgqfodYih iz'u
gSAa (iz'u 15 ls 20) izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk
(D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA OMR 'khV esa iz'u
dh iz'u la[;k ds le{k viuk mÙkj vafdr dhft;sA izR;sd lgh
mÙkj ds fy;s + 4 vad fn;s tk,sx
a s rFkk izR;sd xyr mÙkj ds
fy, 1 vad ?kVk;k tk;sxkA
x|ka'k # 1 (iz- 15 ls 17)
ekuk nks oxZ eSfVªDl A rFkk B bl izdkj gS fd
B = A–1 BA eSfVªDl A o B ds fy, fuEu dks gy
djksA
Q.15
2
2
(B) A + B
(D) A – B
If AB = – BA also, then (A + B)2 is equal to (A) 0
(B) A2
(C) A + B
(D) A + B + AB + BA
Q.16
If AB = – BA also, then (A + B)3 is equal to (A) 0
(B) A3 + A2 + A
3
(C) A
(D) A2 + AB + BA + B2
Q17
Passage # 2 (Ques. 18 to 20)
Least positive integral solution (α) of inequality
5x + 8
> – 2 is a root of equation f(x) = 0
x−4
where f(x) =
(A – B)2 =
(A) 0
(C) A2 – 2AB + B2
(B) A2 + B2
(D) A – B
;fn AB = – BA, rc (A + B)2 =
(A) 0
(C) A + B
(B) A2
(D) A + B + AB + BA
;fn AB = – BA, rc (A + B)3 =
(B) A3 + A2 + A
(D) A2 + AB + BA + B2
(A) 0
(C) A3
x|ka'k # 2 (iz- 18 ls 20)
a 2 ( x 2 − 1)
x–1, a∈[0, 1], then
x −
5
5
vlfedk
5x + 8
>–2
x−4
dk U;wure /kukRed iw.kk±d gy
(α) lehdj.k f(x) = 0, tgk¡ f(x) =
a 2 ( x 2 − 1)
x–1,
x −
5
5
a∈[0, 1] dk ,d ewy gS] rc
Space for rough work
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Page # 5
Q.18
Possible values of α is(A) 0
Q.19
(C) 4
(B) 1
Q.19
(C) 1/2
Q.20
x
–1 (B) 5x + 1 (C) 5x + 5 (D) 5x – 1
5
(B) 5
(B) 1
(A)
R
R
R
R
(D) 0, 5
x
–1 (B) 5x + 1 (C) 5x + 5 (D) 5x – 1
5
bl [k.M esa 2 iz'u (iz'u 1, 2) gSaA izR;sd iz'u esa nks LrEHkksa
esa dFku fn;s x;s gSa] ftUgsa lqesfyr djuk gSA LrEHk-I
(Column I ) esa fn;s x;s dFkuksa (A, B, C, D) dks LrEHk-II
(Column II) esa fn;s x;s dFkuksa (P, Q, R, S) ls lqesy djuk
gSA bu iz'uksa ds mÙkj uhps fn;s x;s mnkgj.k ds vuqlkj
mfpr xksyksa dks dkyk djds n'kkZuk gSA ;fn lgh lqesy A-P,
A-S, B-Q, B-R, C-P, C-Q rFkk D-S gS, rks lgh fof/k ls dkys
fd;s x;s xksyksa dk 4 × 4 eSfVªDl uhps n'kkZ;s vuqlkj gksxk :
P Q R S
P Q R S
Q
Q
Q
Q
(C) 1/2
[k.M - II
This section contains 2 questions (Questions 1, 2). Each
question contains statements given in two columns which
have to be matched. Statements (A, B, C, D) in Column I
have to be matched with statements (P, Q, R, S) in
Column II. The answers to these questions have to be
appropriately bubbled as illustrated in the following
example. If the correct matches are A-P, A-S, B-Q, B-R,
C-P, C-Q and D-S, then the correctly bubbled 4 × 4
matrix should be as follows :
P
P
P
P
(D) dksbZ ugha
f(x) dk izfrykse gksxk
Section – II
A
B
C
D
(C) 4
a dk lEHko eku gksxk
(A) 0
(D) 0, 5
The inverse of f(x) is(A)
α dk lEHko eku gksxk
(A) 0
(D) None
Possible values of a is(A) 0
Q.20
(B) 5
Q.18
A
B
C
D
S
S
S
S
Mark your response in OMR sheet against the question
number of that question in section-II. + 6 marks will be
P
P
P
P
Q
Q
Q
Q
R
R
R
R
S
S
S
S
vr% OMR 'khV esa iz'u dh iz'u la[;k ds le{k viuk mÙkj
[k.M-II esa vafdr dhft;sA izR;sd iw.kZ lgh mÙkj ds fy;s + 6
Space for rough work
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Page # 6
given for complete correct answer and No Negative
for a correctly marked answer in any row.
vad fn;s tk;sx
a s rFkk xyr mÙkj ds fy;s dksbZ _
+ .kkRed vadu
ugha gS (vFkkZr~ dksbZ vad ugha ?kVk;k tk;sxk)A fdUrq] fdlh iafDr
esa lgh :i ls fpfUgr mÙkj ds fy, 1 vad fn;k tk;sxkA
Q.5
Q.5
marks for wrong answer. However, 1 mark will be given
Match of the column.
Column-I
(A) Fundamental period of f(x) = [x] + [2x] + [3x] +… …
n (n + 1)
x is (where [.] is represent
2
greatest integer function)
LrEHkksa dks lqesfyr dhft,
LrEHk -I
n (n + 1)
x
2
(tgk¡ [.] egÙke iw.kk±d Qyu dks fu:fir djrk gS)
(A) f(x) = [x] + [2x] + [3x] +… … + [nx] –
+ [nx] –
(B) lim
sin(π cos 2 x )
2
is
(B) lim
πx
(C) Let A and B are square matrices of order 3 × 3
which satisfy AB = A and BA = B.
If (A – B)5 = k(A –B) then k =
x →0
x
x 
0
(D) If A = 2 y y − y  and AA' = I,


 z − z z 
dk ewyHkwr vkorZukad gksxk
x →0
sin(π cos 2 x )
πx 2
(C) ekuk A o B, 3 × 3 Øe ds oxZ eSfVªDl gS tks
AB = A o BA = B dks lUrq"V djrs gS
;fn (A – B)5 = k(A –B) rc k =
x
x 
0
(D) ;fn A = 2 y y − y  o AA' = I, rc x2 + y2 + z2 =


 z − z z 
then x2 + y2 + z2 =
Column-II
(P) sin2 x – cos2 x when x = π/4
=
LrEHk-II
(P) sin2 x – cos2 x tcfd x = π/4
(Q) sin2x + cos2 x when x = π/4
(Q) sin2x + cos2 x tcfd x = π/4
sin x
x →0
x
(S) HCF of (3, 4)
(R) lim
(R) lim
x →0
sin x
x
(S) (3, 4) dk e-l-i-
Space for rough work
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Page # 7
Q.2
Column-I
(A) Number of points of type
Column-II
(P) 4
( cos θ, sin θ ) satisfying
Q.2
LrEHk-I
(A) ( cos θ, sin θ ) izdkj ds
(P) 4
x2 + y2 – 2x – 2y + 1 = 0
x2 + y2 – 2x – 2y + 1 = 0
dks lUrq"V djus okys fcUnqvksa
is/are
dh la[;k gS@gksxh
(B) Minimum value of
4
LrEHk-II
(Q) – 2
2
cos x – 6cos x + 5 is
(C) No. of solution of
(R) 2
tan 2x = tan6x in (0, 2π] is
(D) The value of k for which
the equation
(B) cos4x – 6cos2x + 5 dk
(Q) – 2
U;wure eku gksxk
(C) (0, 2π] esa tan 2x = tan6x ds
(R) 2
gyksa dh la[;k gksxh
(S) 0
(D) k dk og eku ftlds fy,
4
(S) 0
4
2 cos4x – sin4x + k = 0
2 cos x – sin x + k = 0
has at least one solution
dk de ls de ,d gy gks
can be
ldrk gS
Space for rough work
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Page # 8
PHYSICS
Section – I
[k.M - I
Questions 1 to 9 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 3 marks will be given for each correct answer
and – 1 mark for each wrong answer.
iz'u 1 ls 9 rd cgqfodYih iz'u gSaA izR;sd iz'u ds pkj
Q.1
Q.1
A ball falls on an inclined plane as shown in figure.
The ball is dropped from height h. Coefficient of
restitution for collision is e and the surface is
frictionless. If h1, h2…..hn are heights of n
projectiles and t1, t2,…..tn are their corresponding
time of flights, then-
fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYi
lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds le{k viuk
mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy, + 3 vad fn;s
tk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA
,d xsan ,d vkur ry ij fp=k esa n'kkZ;s vuqlkj
fxj jgh gSA xsan dks h Å¡pkbZ ij ls fxjk;k x;k gSA
VDdj dk izR;koLFkku xq.kkad e gS rFkk lrg
?k"kZ.kghu gSA ;fn h1, h2…..hn, n iz{ksI;ksa dh Å¡pkb;k¡ gS
rFkk t1, t2,…..tn buds lEcfU/kr mM~M;u dky gSa] rksh
h
h1
h1
h2
h2
θ
i.
θ
i.
t1, t2,….., tn form a geometric progression of
common ratio e.
ii. h1 > h2 > h3 ……> hn
iiii. t1, t2, ……., tn form a geometric progression of
common ratio e2.
iv. h1, h2, ….. hn form a geometric progression of
common ratio e.
t1, t2,….., tn ,d xq.kksÙkj vuqØe cukrs gSa
ftldk mHk;fu"V vuqikr e gSA
ii. h1 > h2 > h3 ……> hn
iiii. t1, t2, ……., tn ,d xq.kksÙkj vuqØe cukrs gSa]
ftldk mHk;fu"V vuqikr e2 gSA
iv. h1, h2, ….. hn ,d xq.kksÙkj vuqØe cukrs gSa
ftldk mHk;fu"B vuqikr e gSA
Space for rough work
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Page # 9
Evaluate the above statements and choose the
correct option from the following.
(A) Statements i, ii are true and iii, iv are false
(B) Statements i, ii are false and iii, iv are true
(C) All statements are true
(D) All statements are false
Q.2
Q. 3
A particle of mass 1 kg is moving along the line
y = x + 2 (here x and y are in metres) with speed 2
m/s. The magnitude of angular momentum of
particle about origin is (A) 4 kg - m2/s
(B) 2 2 kg - m2/s
(C) 4 2 kg - m2/s
(D) 2 kg - m2/s
As shown in figure, there are two blocks of the
same mass M, one on top of other, lying on a
frictionless horizontal surface. Both the blocks are
at rest. The upper block is much smaller than the
lower block. A force F is applied on the lower
block and both the blocks start moving together
without any relative motion. Suddenly, the lower
block hits a fixed obstacle and come to rest. The
upper block continues to slide on the lower block.
The upper block just manages to reach the opposite
end of the lower block. What is the coefficient of
friction between the two blocks?
mijksDr dFkuksa dks tkafp;s rFkk fuEUk esa ls lgh
fodYi pqfu;s &
(A) dFku i, ii lgh gSa o iii, iv xyr gSa
(B) dFku i, ii xyr gS o iii, iv lgh gSa
(C) lHkh dFku lgh gSa
(D) lgh dFku xyr gSa
Q.2
1 kg nzO;eku dk ,d d.k js[kk y = x + 2 (;gk¡ x rFkk
y ehVj esa gS) ds vuqfn'k 2 m/s pky ls xfr'khy gSA
ewy fcUnq ds lkis{k d.k ds dks.kh; laosx dk ifjek.k gS -
Q. 3
(A) 4 kg - m2/s
(B) 2 2 kg - m2/s
(C) 4 2 kg - m2/s
(D) 2 kg - m2/s
uhps fp=k esa n'kkZ;s vuqlkj] leku nzO;eku M ds nks
CykWd ,d nwljs ij fLFkr gSa o ,d ?k"kZ.kghu {kSfrt
lrg ij fLFkr gaSA nksuksa CykWd fojke esa gSaA Åijh
CykWd fupys CykWd ls cgqr NksVk gSA ,d cy F
fupys CykWd ij vkjksfir fd;k tkrk gS o nksuksa CykWd
fcuk fdlh vkisf{kd xfr ds ,d lkFk xfr djus
yxrs gSaA fupyk CykWd ,dk,d ,d vojks/k ls
Vdjkrk gS o fojke esa vk tkrk gSA Åijh CykWd dk
fupys CykWd ij fQlyuk tkjh jgrk gSA Åijh CykWd]
fupys CykWd ds Bhd vUr fljs rd igqapuk O;ofLFkr
dj ysrk gSA nksuksa CykWdksa ds e/; ?k"kZ.k xq.kkad D;k
gS?
Space for rough work
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Page # 10
Obstacle
Obstacle
F
F
Q.4
l/2
(A) F/Mg
l
(B) F.Mg
l/2
(A) F/Mg
l
(B) F.Mg
(C) F/2Mg
(D) None of these
(C) F/2Mg
(D) buesa ls dksbZ ugha
In figure given below, the velocities are in ground
Q.4
frame and the cylinder is performing pure rolling
rFkk csyu] r[rs ij 'kq) yksVuh xfr dj jgk gS] fcUnq
on the plank, velocity of point ‘A’ would be-
‘A’ dk osx gksxk -
A
C
uhps fn;s x;s fp=k esa osx] tehu rU=k ds lkis{k gSa
A
VC
C
VP
Q.5
(A) 2VC
(B) 2VC + VP
(C) 2VC – VP
(D) 2(VC – VP)
VP
(A) 2VC
(C) 2VC – VP
A disc is rotated about its axis with a certain
Q.5
angular velocity and lowered gently on a rough
inclined plane as shown in figure, thenµ=
30º
VC
(B) 2VC + VP
(D) 2(VC – VP)
,d pdrh dks bldh v{k ds ikfjr% ,d fu;r
dks.kh; osx ls ?kw.kZu djok dj ,d [kqjnqjs vkur ry
ij /khjs ls j[k fn;k tkrk gS] n'kkZ;s fp=kkuqlkj rks -
1
µ=
3
30º
1
3
Space for rough work
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Page # 11
(A) It will rotate at the position where it was placed
(A) ;g ml fLFkfr esa ftlesa bls j[kk x;k gS] igys
and then will move downward
?kw.kZu djsxk fQj uhps dh vksj xfr djsxkA
(B) bldks vkur ij j[ks tkus ds rRdky Ik'pkr~ ;g
uhps xfr djsxkA
(C) ;g igys uhps dh vksj tk;sxk o blds Ik'pkr~
Åij p<s+xk
s k blds Ik'pkr~ uhps xfr djsxk
(D) ;g igys Åij p<+x
(B) It will go downward just after it is lowered
(C) It will go downward first and then climb up
(D) It will climb upwards and then move downwards
Q.6
A conducting rod PQ of length l = 2 m is moving
Q.6
–1
at a speed of 2ms making an angle of 30º with its
length. A uniform magnetic field B = 2 T exists in a
direction perpendicular to the plane of motion.
⊗
P
⊗
v
⊗
30º
⊗
30º dks.k cukrs gq;s 2ms–1 dh pky ls xfr dj jgh
gSA ,d le:Ik pqEcdh; {ks=k B = 2 T, xfr ds ry ds
yEcor~ ,d fn'kk esa fo|eku gS] rks -
Then-
⊗
l = 2 m yEckbZ dh ,d pkyd NM+ bldh yEckbZ ls
⊗
⊗
⊗
⊗
⊗
Q
⊗
P
⊗
v
⊗
30º
⊗
⊗
⊗
⊗
⊗
Q
(A) VP – VQ = 8V
(B) VP – VQ = 4V
(A) VP – VQ = 8V
(B) VP – VQ = 4V
(C) VQ – VP = 8V
(D) VQ – VP = 4V
(C) VQ – VP = 8V
(D) VQ – VP = 4V
Space for rough work
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Page # 12
Q.7
In the given circuit, the current through the 5mH
Q.7
inductor in steady state is
uhps n'kkZ;s ifjiFk esa] fu;r voLFkk (steady state) esa
5mH izsjd ls gksdj xqtjus okyh /kkjk gS-
5mH
5mH
10mH
10mH
20V
20V
5Ω
(A)
Q.8
2
A
3
(B)
8
A
3
5Ω
(C)
1
A
3
(D)
4
A
3
In an elastic collision between two particles–
(A)
Q.8
(i) the kinetic energy of the system before collision
2
A
3
(B)
8
A
3
(C)
1
A
3
(D)
4
A
3
nks d.kksa ds e/; izR;kLFk VDdj esa–
(i) VDdj ds igys fudk; dh xfrt ÅtkZ VDdj ds
is equal to the kinetic energy of the system
ckn fudk; dh xfrt ÅtkZ ds cjkcj gksrh gS
after collision
(ii) the linear momentum of the system is
conserved
(ii) fudk; dk js[kh; laosx lajf{kr jgrk gS
fuEu esa ls lgh dFku pqfu,&
Select the correct statements –
(A) dsoy dFku (i) lgh gS
(A) only statement (i) is correct
(B) dsoy dFku (ii) lgh gS
(B) only statement (ii) is correct
(C) (i) o (ii) nksuksa dFku lgh gS
(C) Both statement (i) & (ii) are correct
(D) Neither statement (i) nor statement (ii) are correct
(D) uk rks dFku (i) lgh gS uk gh dFku (ii) lgh gS
Space for rough work
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Page # 13
Q. 9
Two circular coils X and Y, having equal number
Q. 9
of turns and carrying equal currents in the same
sense, subtend same solid angle at point O. If the
smaller coil X is midway between O and Y and if
we represent the magnetic induction due to bigger
coil Y at O as BY and that due to smaller coil X at
O as BX, then-
nks oÙ̀kh; dq.Mfy;k¡ X o Y leku la[;k esa ?ksjs rFkk
leku fn'kk esa cjkcj /kkjk izokfgr j[krh gaS] fcUnq O
ij leku Bksl (Solid) dks.k vUrfjr djrh gSaA ;fn
NksVh dq.Myh X; O rFkk Y ds e/; fLFkr gS rFkk ;fn
cM+h dq.Myh Y }kjk O ij mRiUu pqEcdh; iszj.k dks
BY }kjk rFkk NksVh dq.Myh X }kjk O ij mRiUu
pqEcdh; izsj.k dks BX }kjk n'kkZ;k tk;s rksY
Y
X
X
d
O
d
O
(A)
BY
=1
BX
(B)
BY
=2
BX
(A)
BY
=1
BX
(B)
BY
=2
BX
(C)
BY
1
=
BX
2
(D)
BY 1
=
BX 4
(C)
BY
1
=
BX
2
(D)
BY 1
=
BX 4
Questions 10 to 14 are multiple choice questions. Each
iz'u 10 ls 14 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj
question has four choices (A), (B), (C) and (D), out of
fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k ,d ls
which MULTIPLE (ONE OR MORE THAN ONE) is
correct. Mark your response in OMR sheet against the
vf/kd fodYi lgh gaSA OMR 'khV esa iz'u dh iz'u la[;k ds
question number of that question. + 4 marks will be
le{k vius mÙkj vafdr dhft,A izR;sd lgh mÙkj ds fy,
given for each correct answer and –1 mark for each
+ 4 vad fn;s tk;saxs rFkk izR;sd xyr mÙkj ds fy, 1 vad
wrong answer.
?kVk;k tk;sxkA
Space for rough work
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Page # 14
Q. 10
Q. 11
Which of the following statements are correct?
(A) If a moving charged particle enters into a
region of magnetic field from outside, It does
not complete a circular path
(B) If a moving charged particle traces a helical
path in a uniform magnetic field, the axis of
the helix is parallel to the magnetic field
(C) The power associated with the force exerted by
a magnetic field on moving charged particle is
always equal to zero
(D) If in a region a uniform magnetic field and a
uniform electric field both exist, a charged
particle moving in this region cannot trace a
circular path
Q. 10
A conductor ABCDE, shaped as shown, carries
current I. It is placed in the x-y plane with the ends
A and E on the x-axis. A uniform magnetic field of
magnitude B exists in the region. The force acting
on it will bey
B
I
a
C
E x
A
a
Q. 11
z
λ
fuEu esa ls dkSulk dFku lR; gS?
(A) ;fn ,d xfr'khy vkosf'kr d.k ,d pqEcdh; {ks=k
ds {ks=k eas ckgj ls izos'k djrk gS] ;g oÙ̀kh; iFk
iw.kZ ugha djrkA
(B) ;fn ,d xfr'khy vkosf'kr d.k ,d le:Ik
pqEcdh; {ks=k esa ,d gSfydy iFk dks vuqlfjr
djrk gS] rks gSfyDl dh v{k pqEcdh; {ks=k ds
lekUrj gksrh gSA
(C) ,d xfr'khy vkosf'kr d.k ij ,d pqEcdh; {ks=k }kjk
yxk;s x, cy ls lEcfU/kr 'kfDr lnSo 'kwU; gksrh gSA
(D) ;fn ,d {ks=k esa ,d le:Ik pqEcdh; {ks=k rFkk
le:Ik fo|qr {ks=k nksuksa fo|;eku gks rks bl {ks=k
esa xfr'khy d.k ,d oÙ̀kh; iFk dks vuqlfjr ugha
dj ldrkA
,d pkyd ABCDE, ftldk vkdkj uhps n'kkZ;s
vuqlkj gS] o /kkjk I j[krk gSA bls blds vUr fljs
x v{k ij A rFkk E ij jgrs gq, x-y ry esa fLFkr
fd;k x;k gSA B ifjek.k dk ,d le:Ik pqEcdh; {ks=k
bl {ks=k esa fLFkr gSA bl ij dk;Zjr cy gksxk y
I
A
a
C
E
x
a
D
z
(A) zero, if B is in the x-direction
(B) λBI in the z-direction, if B is in the y-direction
(C) λBI in the negative y-direction, if B is in the
z-direction
(D) λaBI, if B is in the x-direction
B
λ
D
(A) 'kwU;, ;fn B, x-fn'kk esa gks
(B) λBI, z-fn'kk esa] ;fn B, y-fn'kk esa gS
(C) λBI _.kkRed y-fn'kk esa] ;fn B, z-fn'kk esa gS
(D) λaBI, ;fn B, x-fn'kk esa gS
Space for rough work
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Page # 15
Q.12
In an RLC series circuit shown in figure, the
readings of voltmeters V1 and V2 are 100V and
120V, respectively. The source voltage is 130V.
For this situation mark out the correct statement(s)
Q.12
uhps fp=k esa n'kkZ;s ,d RLC Js.kh ifjiFk eas V1 o
V2 oksYVehVjksa ds ikB~;kad Øe'k% 100V o 120V gSaA
L=kksr oksYVst 130V gSA bl fLFkfr ds lEcU/k esa lgh
dFku pqfu;sV2
V2
V1
V1
V
V
(A) izfrjks/k] izsjd o la?kkfj=k ds fljksa ij oksYVst
Øe'k% 50V, 86.65V rFkk 186.65V gSaA
(A) Voltage across resistor, inductor and capacitor
are 50V, 86.65V and 186.65V respectively
(B) Voltage across resistor, inductor and capacitor
are 10V, 90V and 30V, respectively
(B) izfrjks/k] iszjd o la?kkfj=k ds fljksa ij oksYVst
Øe'k% 10V, 90V rFkk 30V gSaA
5
gSA
(C) ifjiFk dk 'kfDr xq.kkad
13
(D) ifjiFk izdf̀r esa /kkjrh; gSA
5
13
(D) The circuit is capacitance in nature
(C) Power factor of the circuit is
Q. 13
A rod AC of length l and mass m is kept on a
horizontal smooth plane. It is free to rotate and
move. A particle of some mass m moving on the
plane with velocity v strikes the rod at point B
making angle 37º with the rod. The collision is
elastic. After collision-
Q. 13
,d NM+ AC, ftldh yEckbZ l o nzO;eku m gS] dh ,d
{kSfrt ?k"kZ.kghu ry ij fLFkr gSA ;g ?kw.kZu rFkk xfr ds
fy;s Lora=k gSA m nzO;eku dk ,d d.k] ry esa v osx ls
xfr djrk gqvk NM+ ls fcUnq B ij NM+ ls 37º dks.k
ij Vdjkrk gSA VDdj izR;kLFk gSA VDdj ds Ik'pkr~ -
B l/4
A
37º
B l/4
C
A
v
37º
C
v
Space for rough work
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Page # 16
(A) The angular velocity of the rod will be 75/55 v/l
(A) NM+ dk dks.kh; osx 75/55 v/l gSA
(B) ml le;kUrjky esa ftlesa NM+ vk/kk ?kw.kZu iwjk
(B) The centre of the rod will travel a distance πl/3
djrh gSA NM+ dk dsUnz πl/3 nwjh r; djsxkA
(C) vkosxh; cy dk vkosx 24m v/55 gSA
(D) buesa ls dksbZ ugha
in the time in which it makes half rotation
(C) Impulse of the impact force is 24m v/55
(D) None of these
Q. 14
A rigid rod of mass M slides along a fixed
semicircular track (in vertical plane) followed by a
flat rack. At the given instant velocity of end B is v
along the horizontal plane. Then at the given instantr O
A
r
C
B v
Q. 14
M nzO;eku dh n`<+ NM ,d fQDl v)ZoÙ̀kh; Vªsd
¼m/oZry esa½ o blds vkxs lery Vªsd gS] ds vuqfn'k
fQlyrh gSA ,d fn;s x;s {k.k ij {kSfrt ry ds
vuqfn'k B fljs dk osx v gSA rks fn;s x;s {k.k ij r
A
C
(A) Angular speed of the rod is v/r
O
r
B v
(A) NM+ dh dks.kh; pky v/r gS
(B) Velocity of the centre of mass is v/ 2
(B) nzO;eku dsUnz dk osx v/ 2 gS
(C) O ds ikfjr% NM+ dk dks.kh; laosx 2/3 mvr gS
(D) ?kw.kZu o LFkkukUrj.k xfr;ksa dh xfrt ÅtkZvksa
dk vuqikr 1 : 2 gS
(C) Angular momentum of rod about O is 2/3 mvr
(D) The ratio of rotational to translational kinetic
energy is 1 : 2
This section contains 2 paragraphs; each has 3 multiple
choice questions. (Questions 15 to 20) Each question has
4 choices (A), (B), (C) and (D) out of which ONLY ONE
is correct. Mark your response in OMR sheet against the
question number of that question. + 4 marks will be
given for each correct answer and –1 mark for each
wrong answer.
bl [k.M esa 2 vuqPNsn fn;s x;s gS]a izR;sd esa 3 cgqfodYih iz'u
gSAa (iz'u 15 ls 20) izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk
(D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA OMR 'khV esa iz'u
dh iz'u la[;k ds le{k viuk mÙkj vafdr dhft;sA izR;sd lgh
mÙkj ds fy;s + 4 vad fn;s tk,sx
a s rFkk izR;sd xyr mÙkj ds
fy, 1 vad ?kVk;k tk;sxkA
Space for rough work
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Page # 17
Passage # 1 (Ques. 15 to 17)
A ball A of mass m is suspended by a thread of
length r = 1.2m. Another ball B of mass 2m is
projected from the ground with velocity u = 9m/s
such that at the highest point of its trajectory it
collides head on elastically with ball A. It is
observed that during subsequent motion, tension in
the thread at highest point is equal to mg.
9 m/s
x|ka'k # 1 (iz- 15 ls 17)
m nzO;eku dh ,d xsan A, r = 1.2m yEckbZ ds ,d
/kkxs ls yVdh gSA 2m nzO;eku dh ,d vU; xsan B
tehu ls u = 9m/s ds osx ls bl izdkj iz{ksfir dh
tkrh gS rkfd blds xfr iFk ds mPpre fcUnq ij
;g xsan A ls 'kh"kZ :Ik ls izR;kLFk :Ik ls Vdjk,sA
;g izsf{kr fd;k tkrk gS fd Øfed xfr esa] mPpre
fcUnq ij /kkxs esa ruko mg ds cjkcj gSA
A (m)
9 m/s
θ
B (2m)
θ
B (2m)
Q.15
Q.16
Q.17
Q.15
At highest point, velocity of ball A is(A) 6 2 m/s
(B) 2 6 m/s
(C) 3 2 m/s
A (m)
(D) 3 6 m/s
Angle of projection (θ) of ball B is(A) 30º
(B) 60º
(C) 45º
(D) 75º
Q.16
The height of the point suspension of ball A from
the ground is81
129
m
m
(A)
(B)
40
40
81
101
m
m
(C)
(D)
20
40
Q.17
mPpre fcUnq ij A dk osx gS (A) 6 2 m/s
(B) 2 6 m/s
(C) 3 2 m/s
(D) 3 6 m/s
xsan B dk iz{ksI; dks.k (θ) gS (A) 30º
(C) 45º
(B) 60º
(D) 75º
xsan A ds fuyEcu fcUnq dh tehu ls Å¡pkabZ gS 81
m
40
81
m
(C)
20
(A)
129
m
40
101
m
(D)
40
(B)
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Page # 18
Passage # 2 (Ques. 18 to 20)
In a certain region of space, there exists a uniform
and constant electric field of magnitude E along the
positive y-axis of a co-ordinate system. A charged
particle of mass ‘m’ and charge ‘–q’ (q > 0) is
projected from the origin with speed 2v at an angle
of 60º with the positive x-axis in x-y plane. When
x|ka'k # 2 (iz- 18 ls 20)
lrf"V (space) ds ,d fuf'pr {ks=k esa ,d le:Ik o
fu;r ifjek.k dk ,d fo|qr {ks=k E, funs'Z kkad i)fr ds
/kukRed y-v{k ds vuqfn'k fo|eku gSA ‘m’ nzO;eku dk
,d vkosf'kr d.k vkos'k ‘–q’ (q > 0) dks x-y ry esa ewy
fcUnq ls /kUkkRed x-v{k ls 60º dks.k cukrs gq,] 2v pky
ls iz{ksfir fd;k x;k gSA tc d.k dk x-funs'Z kkad
3mv 2
, a
qE
3mv 2
gks tkrk gS] ,d le:Ik o fu;r ifjek.k dk
qE
B izkcY; dk pqEcdh; {ks=k tks fd /kukRed y-fn'kk ds
the x-coordinate of particle becomes
Q.18
uniform and constant magnetic field of strength B is
also switched on along positive y-axis.
Velocity of the particle just before the magnetic
field is switched on is(B) vî +
(A) vî
Q.18
3v ˆ
j
2
3v ˆ
3v ˆ
(D) 2 vî –
j
j
2
2
x-coordinate of the particle as a function of time
after the magnetic field is switched on is (time t = 0
when magentic field is start)-
(C) vî –
Q.19
qB 
3mv 2
– R sin 
t
qE
m 
(A)
(B)
qB 
3mv 2
+ R sin 
t
qE
m 
(B)
(C)
qB 
3mv 2
– R cos 
t
qE
m 
(C)
3mv 2
qB 
+ R cos 
t
qE
m 
3v ˆ
j
2
pqEcdh; {ks=k dk fLop vkWu djus ds Ik'pkr~ d.k dk
x-funsZ'kkad le; ds Qyu ds :Ik esa D;k gksxk
(t = 0 ij pqEcdh; {ks=k 'kq: fd;k tkrk gS)-
(A)
(D)
3v ˆ
j
2
3v ˆ
(D) 2 vî –
j
2
(B) vî +
(A) vî
(C) vî –
Q.19
vuqfn'k fLFkr gS] dk fLop vkWu dj fn;k tkrk gSA
pqEcdh; {ks=k dk fLop vkWu djus ls Bhd iwoZ d.k
dk osx gksxk -
(D)
qB 
3mv 2
– R sin 
t
qE
m 
qB 
3mv 2
+ R sin 
t
qE
m 
qB 
3mv 2
– R cos 
t
qE
m 
qB 
3mv 2
+ R cos 
t
qE
m 
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Page # 19
Q.20
z-coordinate of the particle as a function of time
after the magnetic field is switched on is-
Q.20


 qB 
 qB 
(A) R 1 + cos
t  (B) R 1 + sin 
t 
 m 
 m 


pqEcdh; {ks=k dk fLop vkWu djus ds Ik'pkr~ d.k dk
z-funsZ'kkad le; ds Qyu ds :Ik esa D;k gksxk

 qB 
 qB 
(A) R 1 + cos
t  (B) R 1 + sin 
t 
m


 m 




 qB 
 qB 
(C) –R 1 − cos
t  (D) –R 1 + cos
t 
 m 
 m 




 qB 
 qB 
t  (D) –R 1 + cos
t 
(C) –R 1 − cos
m


 m 


[k.M - II
Section – II
This section contains 2 questions (Questions 1, 2). Each
question contains statements given in two columns which
have to be matched. Statements (A, B, C, D) in Column I
have to be matched with statements (P, Q, R, S) in
Column II. The answers to these questions have to be
appropriately bubbled as illustrated in the following
example. If the correct matches are A-P, A-S, B-Q, B-R,
C-P, C-Q and D-S, then the correctly bubbled 4 × 4
matrix should be as follows :
bl [k.M esa 2 iz'u (iz'u 1, 2) gSaA izR;sd iz'u esa nks LrEHkksa
esa dFku fn;s x;s gSa] ftUgsa lqesfyr djuk gSA LrEHk-I
(Column I ) esa fn;s x;s dFkuksa (A, B, C, D) dks LrEHk-II
(Column II) esa fn;s x;s dFkuksa (P, Q, R, S) ls lqesy djuk
gSA bu iz'uksa ds mÙkj uhps fn;s x;s mnkgj.k ds vuqlkj
mfpr xksyksa dks dkyk djds n'kkZuk gSA ;fn lgh lqesy A-P,
A-S, B-Q, B-R, C-P, C-Q rFkk D-S gS, rks lgh fof/k ls dkys
fd;s x;s xksyksa dk 4 × 4 eSfVªDl uhps n'kkZ;s vuqlkj gksxk :
P Q R S
A
B
C
D
P Q R S
A
B
C
D
P
P
P
P
Q
Q
Q
Q
R
R
R
R
S
S
S
S
P
P
P
P
Q
Q
Q
Q
R
R
R
R
S
S
S
S
vr% OMR 'khV esa iz'u dh iz'u la[;k ds le{k viuk
mÙkj [k.M-II esa vafdr dhft;sA izR;sd iw.kZ lgh mÙkj ds
fy;s + 6 vad fn;s tk;saxs rFkk xyr mÙkj ds fy;s dksbZ
+_.kkRed vadu ugha gS (vFkkZr~ dksbZ vad ugha ?kVk;k
tk;sxk)A fdUrq] fdlh iafDr esa lgh :i ls fpfUgr mÙkj ds
fy, 1 vad fn;k tk;sxkA
Mark your response in OMR sheet against the question
number of that question in section-II. + 6 marks will be
given for complete correct answer and No Negative
marks for wrong answer. However, 1 mark will be given
for a correctly marked answer in any row.
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Page # 20
Assume sufficient friction to prevent slipping. Body
rolls without slipping
Q.1
Column –I
F
(A)
Q.1
;g ekfu;s fd fQlyu jksdus ds fy, ?k"kZ.k i;kZIr
gSA fi.M fcuk fQlys yq<+drh gSa
LrEHk –I
Column-II
(P) Body accelerates
forward
F
(A)
oy;
(B)
F
(Q) Rotation about the
centre of mass is
clockwise
(B)
F
Disc
F
(R) Friction force acts
backward
(C)
lkis{k ?kw.kZu
nf{k.kkorZ gksrk gS
(R) ?k"kZ.k cy ihNs dh
R/2
Solid Cylinder
F
vksj dk;Zjr gksrk gS
Bksl csyu
(D)
(S) No friction acts
Solid Cylinder
(Q) nzO;eku dsUnz ds
pdrh
(C)
R/2
(P) fi.M vkxs dh vksj
Rofjr gksrk gS
Ring
R/2
LrEHk-II
(D)
(S) dksbZ ?k"kZ.k ugha yxrk
R/2
F
F
Bksl csyu
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Page # 21
Q.2
In column I some circuits are given. In all the
circuits except in (A), switch S remains closed for
long time and then it is opened at t = 0 while for
(A), the situation is reversed. Column II tells
something about the circuit quantities. Match the
entries of column I with the entries of column II.
LrEHk I esa dqN ifjiFk n'kkZ;s x;s gSaA ifjiFk (A) esa
NksM+dj 'ks"k lHkh ifjiFkksa esa fLop S dks yEcs le;
rd can j[kdj bls t = 0 ij [kksy fn;k tkrk gS
tcfd (A) ds fy;s fLFkfr O;qRØe gSA LrEHk II, ifjiFk
dh daqN jkf'k;ksa ds ckjs esa tkudkjh nsrk gSA LrEHk-I
dk LrEHk-II ls lgh feyku dhft,&
Column-II
Column -I
(A)
Q.2
L
R
S
(P) Induced e.m.f.
can be greater
than E.
LrEHk-I
(A)
L
(Q) Induced e.m.f.
would be less
than E.
R
S
(R) Finally, energy
stored in
inductor is
zero.
(C)
(R) vfUre :Ik ls]
L
R
S
E
R
R
E
L
L
(Q) izsfjr fo-ok-cE ls de gksxk
L
S
R
(D)
ldrk gS
(B)
E
(C)
(P) iszfjr fo-ok-c-]
E ls vf/kd gks
E
L
S
R
S
E
(B)
LrEHk -II
izsjd eas laxzfgr
ÅtkZ 'kwU; gS
E
S
(S) Finally, energy
stored in
inductor is
non-zero
E
(D)
L
S
R
(S) vafre :Ik ls
izsjd esa laxzfgr
ÅtkZ v'kwU; gS
E
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Page # 22
CHEMISTRY
Section – I
[k.M - I
Questions 1 to 9 are multiple choice questions. Each
iz'u 1 ls 9 rd cgqfodYih iz'u gSaA izR;sd iz'u ds pkj fodYi
question has four choices (A), (B), (C) and (D), out of
(A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA
which ONLY ONE is correct. Mark your response in
OMR sheet against the question number of that
question. + 3 marks will be given for each correct answer
OMR 'khV esa iz'u dh iz'u la[;k ds le{k viuk mÙkj
vafdr dhft;sA izR;sd lgh mÙkj ds fy, + 3 vad fn;s tk;asxs rFkk
and – 1 mark for each wrong answer.
izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA
Q.1
Q.1
Which is not true statement about KMnO4 ?
(A) vEyh; ek/;e esa bldk foy;u vLFkk;h gS
(B) bldh FkksM+h lh ek=kk lkUnz H2SO4 esa feyku ij
(A) Its solution is unstable in acidic medium
(B) Its small quantity added to conc. H2SO4, a
green coloured solution containing
,d gjs jax dk foy;u ftlesa MnO 3+ vk;u gS]
curk gS
(C) {kkjh; foy;u esa MnO −4 dk ifjorZu Mn 2 + esa
gks tkrk gS
(D) Fe 2 + ;k C 2 O 24 − vuqekiu esa ;g Lo;a lwpd gksrk gS
MnO 3+
ion is formed
(C) MnO −4 changes to Mn 2 + in basic solution
(D) It is self-indicator in Fe 2 + or C 2 O 24 − titration
Q. 2
Aluminium hydroxide forms a positively charged
sol, which of the following ionic substances should
be the most effective in coagulating the sol?
(A) NaCl
(B) Fe2(SO4)3
(D) K3PO4
(C) CaCl2
KMnO4 ds fy, dkSulk lgh ugha gS ?
Q. 2
,Y;qfefu;e gkbMªksDlkbM /kukosf'kr lksy cukrk gS]
fuEu esa ls dkSulk vk;fud inkFkZ LdUnhr lksy ds
fy, lokZf/kd izHkko'kkyh gksxk ?
(A) NaCl
(C) CaCl2
(B) Fe2(SO4)3
(D) K3PO4
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Page # 23
Q. 3
Q.4
Q.5
If Ksp of PbBr2 is 8 × 10–5 and it is 80% dissociated
in solution. Its solubility is:
(A) 0.027 mol/L
(B) 0.058 mol/L
(C) 0.017 mol/L
(D) 0.034 mol/L
Q. 3
A buffer solution has equal volume of 0.2 M
NH4OH and 0.02 M NH4Cl. The pKb of the base
is 5. The pH is
(A) 4
(B) 10
(C) 7
(D) 9
Q.4
NaCl is doped with 10–4 mol% of SrCl2, the
Q.5
(A) 0.027 mol/L
(C) 0.017 mol/L
0.2 M NH4OH rFkk 0.02 M NH4Cl ds leku
(A) 4
(B) 10
(C) 7
(D) 9
NaCl dks SrCl2 ds 10–4 eksy% ls Mksihr fd;k x;k]
/kuk;u fjfDr;ksa dh lkUnzrk gksxh -
(A) 6.02 × 10 mol
–1
(B) 6.02 × 10 mol
15
–1
(A) 6.02 × 10 mol
(B) 6.02 × 1016 mol–1
(C) 6.02 × 1017 mol–1
(D) 6.02 × 1014 mol–1
(C) 6.02 × 1017 mol–1
(D) 6.02 × 1014 mol–1
15
–1
16
Total volume of atoms present in a face centred
cubic unit volume of a metal is, (r is atomic
radius)
(A)
Q.7
(B) 0.058 mol/L
(D) 0.034 mol/L
vk;ru dk cQj foy;u gSA {kkj dk pKb 5 gSA
pH gksxk &
concentration of cation vacancies will be-
Q.6
;fn PbBr2 dk Ksp 8 × 10–5 rFkk foy;u esa bldk
fo;kstu 80% gksrk gS rks bldh foys;rk gksxh &
Q.6
20 3
24 3
12 3
16 3
πr (B)
πr (C)
πr (D)
πr
3
3
3
3
The correct order of increasing bond angles is
,d /kkrq ds Qyd dsUnzh; /kuh; bdkbZ vk;ru esa
mifLFkr ijek.kqvksa dk dqy vk;ru gS, (r ijekf.od
f=kT;k gS)
(A)
Q.7
20 3
24 3
12 3
16 3
πr (B)
πr (C)
πr (D)
πr
3
3
3
3
cU/kdks.k dk lgh vkjksgh Øe gS
(A) OF2 < H2O < Cl2O < ClO2
(A) OF2 < H2O < Cl2O < ClO2
(B) ClO2 < OF2 < Cl2O < H2O
(B) ClO2 < OF2 < Cl2O < H2O
(C) ClO2 < Cl2O < H2O < OF2
(C) ClO2 < Cl2O < H2O < OF2
(D) OF2 < Cl2O < H2O < ClO2
(D) OF2 < Cl2O < H2O < ClO2
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Page # 24
Q. 8
Q. 9
The dipole moments of methane and its halogen
derivatives are in the order
(A) CH4 < CH2Cl2 > CHCl3 < CH3Cl
(B) CH3Cl < CH2Cl2 < CHCl3 < CH4
(C) CH4 < CHCl3 < CH2Cl2 < CH3Cl
(D) CH4 < CH3Cl < CH2Cl2 > CHCl3
Q. 8
The correct decreasing order of bond angles is-
Q. 9
esFksu rFkk blds gSykstu O;qRiUuksa ds f}/kzqo vk?kq.kZ dk
lgh Øe gS &
(A) CH4 < CH2Cl2 > CHCl3 < CH3Cl
(B) CH3Cl < CH2Cl2 < CHCl3 < CH4
(C) CH4 < CHCl3 < CH2Cl2 < CH3Cl
(D) CH4 < CH3Cl < CH2Cl2 > CHCl3
(A) ClF3 > PF3 > NF3 > BF3
(B) BF3 > PF3 > NF3 > ClF3
(C) BF3 > NF3 > PF3 > ClF3
(D) BF3 > ClF3 > PF3 > NF3
cU/k dks.k dk lgh vojksgh Øe gS
(A) ClF3 > PF3 > NF3 > BF3
(B) BF3 > PF3 > NF3 > ClF3
(C) BF3 > NF3 > PF3 > ClF3
(D) BF3 > ClF3 > PF3 > NF3
Questions 10 to 14 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE THAN ONE) is
correct. Mark your response in OMR sheet against the
question number of that question. + 4 marks will be
given for each correct answer and –1 mark for each
wrong answer.
iz'u 10 ls 14 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj fodYi
(A), (B), (C) rFkk (D) gS]a ftuesa ls ,d ;k ,d ls vf/kd fodYi
lgh gaAS OMR 'khV esa iz'u dh iz'u la[;k ds le{k vius mÙkj
vafdr dhft,A izR;sd lgh mÙkj ds fy, + 4 vad fn;s tk;sx
a s
rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA
Q. 10
Q. 10
Which of the following are incorrect statements?
(A) Equivalent weight of KMnO4 in acidic medium
is M/5
(B) In acidic medium, MnO 24 − disproportionates to
MnO2 and MnO −4
(C) KMnO 4 spot can be bleached by H2O2
(D) Alkaline KMnO4 can
unsaturation in benzene
be
used
to
test
fuEu esa ls dkSulk@dkSuls vlR; dFku gS ?
(A) vEyh; ek/;e esa KMnO4 dk rqY;kadh Hkkj M/5 gS
(B) vEyh; ek/;e esa] MnO 24 − dk MnO2 o
MnO −4 esa fo"kekuqikru gksrk gS
(C) KMnO 4 ds nkx dks H2O2 }kjk fojaftr dj
ldrs gS
(D) {kkjh; KMnO4 dk mi;ksx csUthu esa vlarÌrrk
ds ifj{k.k esa fd;k tk ldrk gS
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Page # 25
Q. 11
Q.12
On adding AgNO3 solution into KI solution a
negatively charged colloidal sol is obtained when
they are in
(A) 100 mL of 0.1 M AgNO3 + 100 mL of 0.1 M KI
(B) 100 mL of 0.1 M AgNO3 + 100 mL of 0.2 M KI
(C) 100 mL of 0.2 M AgNO3 + 100 mL of 0.1 M KI
(D) 100 mL of 0.15 M AgNO3 + 100 mL of 0.25 M KI
Q. 11
Which of the following statements are true ?
(A) pH + pOH = 14 for all aqueous solutions
(B) pH of 1 × 10–8M HCl is 8
Q.12
fuEu esa l dkSulk dFku lR; gS ?
(A) tyh; foy;uksa ds fy, pH + pOH = 14
(B) 1 × 10–8M HCl dk pH 8 gS
(C) H 2 PO −4 dk la;qXeh {kkj HPO 24 − gS
(D) tc CuSO2 foy;u esa 96,500 dqyke fo|qr
izokfgr dh tkrh gS] rks dSFkksM ij 1g-rqY;kad
dkWij fu{ksfir gksrk gS
Q. 13
fuEu esa ls dkSulk dFku lgh gS ?
dksykbMh lksy izkIr gksrk gS] tc ;g fuEu gks &
(A) 100 mL, 0.1 M AgNO3 + 100 mL, 0.1 M KI
(B) 100 mL, 0.1 M AgNO3 + 100 mL, 0.2 M KI
(C) 100 mL, 0.2 M AgNO3 + 100 mL, 0.1 M KI
(D) 100 mL, 0.15 M AgNO3 + 100 mL, 0.25 M KI
(C) Conjugate base of H 2 PO −4 is HPO 24 −
(D) 96,500 coulombs of electricity when passed
solution
deposits
through
a
CuSO4
1g-equivalent of copper at the cathode
Q. 13
Which of the following statements are correct ?
(A) For NaCl unit cell (edge length = 1),
1
2
(B) For CsCl unit cell (edge – length
AgNO3 dks KI foy;u esa feykus ij _.kkosf'kr
(A) NaCl bdkbZ lsy ds fy, (fdukjk yEckbZ = 1),
r⊕ + rΘ =
r⊕ + rΘ =
= a)
3
a
r⊕ + rΘ =
2
(C) The void space in a bcc unit cell is 68 %
(D) The void space in a fcc unit cell is 26 %
1
2
(B) CsCl bdkbZ lsy ds fy, (fdukjk yEckbZ = a)
r⊕ + rΘ =
3
a
2
(C) bcc bdkbZ lsy esa fjDr LFkku 68 %
(D) fcc bdkbZ lsy esa fjDr LFkku 26 %
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Page # 26
Q. 14
Which of the following statements are correct ?
Q. 14
(A) The crystal lattice of ice is mostly formed by
fuEu esa ls dkSulk dFku lgh gS ?
(A) cQZ dk fØLVy tkyd vf/kdka'kr% lgla;kstd
the covalent as well as hydrogen bonds
rFkk gkbMªkstu cU/kksa }kjk curk gS
(B) The density of water increases when heated
(B) tc ty dks 0°C ls 4°C rd xeZ fd;k tkrk gS
from 0°C to 4°C due to the change in the
rks ty v.kqvksa ds iqat dh lajpuk esa ifjorZu ds
structure of the cluster of water molecules
dkj.k ty dk ?kuRo c<+rk gS
(C) The density of water increase from 0°C to a
(C) ty dk ?kuRo 0°C ls vf/kdre 4°C rd c<+rk
maximum at 4°C because the entropy of the
gS D;ksafd rU=k dh ,.Vªksih c<+rh gS
system increases
This
(D) Above 4°C, the thermal agitation of water
(D) 4°C ds Åij ty v.kqvksa dh rkih; xfr'khyrk
molecules increases. Therefore intermolecular
c<+rh gSA blfy, vUrj vkf.od nwjh c<+rh gS
distance increases and water starts expanding
rFkk ty izlkfjr gksuk 'kq: gks tkrk gS
section
contains
2
paragraphs;
each
has
3 multiple choice questions. (Questions 15 to 20) Each
question has 4 choices (A), (B), (C) and (D) out of which
ONLY ONE is correct. Mark your response in OMR
bl [k.M esa 2 vuqPNsn fn;s x;s gS]a izR;sd esa 3 cgqfodYih iz'u
gSAa (iz'u 15 ls 20) izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk
(D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA OMR 'khV esa iz'u
sheet against the question number of that question. + 4
dh iz'u la[;k ds le{k viuk mÙkj vafdr dhft;sA izR;sd lgh
marks will be given for each correct answer and –1 mark
a s rFkk izR;sd xyr mÙkj ds
mÙkj ds fy;s + 4 vad fn;s tk,sx
for each wrong answer.
fy, 1 vad ?kVk;k tk;sxkA
Space for rough work
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Page # 27
Passage # 1 (Ques. 15 to 17)
A number of molecules and polyatomic ions cannot
be described accurately by a single lewis structure
and a number of descriptions based on the same
skeletal structure are written and these taken
together represent the molecule or ion. These
structures have almost similar energies, same
arrangement of atoms and have same number of
bonding and non-bonding pair of electrons. These
contributing structures or canonical forms taken
together constitute the resonance hybrid which
represents the molecule or ion.
Q.15
The value of bond order of C–O bond for CO 32 −
ion is
(A) 1.25
Q.16
x|ka'k # 1 (iz- 15 ls 17)
vusd v.kq rFkk cgqijek.koh; vk;uksa dh ,d vdsyh
yqbZl lajpuk mls iw.kZ :i ls ugha le>k ldrh gS o
vusd lajpukvksa dks ,d leku lajpuk ds :i esa
fy[kk tkrk gS rFkk blesa v.kq ;k vk;u dks ,d lkFk
n'kkZ;k tkrk gSA ;s lajpuk,sa yxHkx leku ÅtkZ]
ijek.kqvkas dh leku O;oLFkk rFkk cU/khr o vcU/khr
bysDVªkWu ;qXeksa dh leku la[;k dh gksrh gSA ;s
forfjr lajpuk,sa ;k dSukWuhdy :i laxBhr vuquknh
ladj gksrk gS ftls v.kq ;k vk;u ls iznf'kZr djrs gSA
CO 32− vk;u
Q.15
ds fy, C – O cU/k dk cU/k Øe eku
gS
(B) 1.33 (C) 1.5
(A) 1.25
(D) 1.0
Which of the following is not correct about
resonance ?
(A) It averages the bond features as a whole
Q.16
(B) It stabilizes the molecule since energy of the
resonance hybrid is lower than that of any of
the single canonical structure
(C) There is no equilibrium between these
canonical structures
(D) These canonical structures have real existence
also
(B) 1.33 (C) 1.5
(D) 1.0
fuEu esa ls dkSulk vuqukn ds fy, lgh ugha gS ?
(A) blesa cU/k ds vkSlr xq.k lEiw.kZ :i esa gksrs gaS
(B) blesa vuquknh ladj dh ÅtkZ dk eku dksbZ Hkh
vdsyh dsukWuhdy lajpuk ls de gksus ls v.kq
LFkk;h gksrk gS
(C) blesa bldh dsukWuhdy lajpukvksa ds e/; dksbZ
lkE; ugha gksrk gS
(D) budh dsukWuhdy lajpuk,sa Hkh okLrfod LFkkf;Ro
j[krh gS
Space for rough work
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Page # 28
Q.17
Due to resonance in benzene, the value of C – C
Q.17
(A) 115 pm and 1.5
(B) 1.39 Å and 1.33
csUthu eas vuqukn ds dkj.k C – C cU/k yEckbZ
rFkk cU/k Øe dk eku Øe'k% gS
(A) 115 pm rFkk 1.5 (B) 1.39 Å rFkk 1.33
(C) 1.39 Å and 1.5
(D) 115 pm and 2.0
(C) 1.39 Å rFkk 1.5
bond length and bond-order are respectively
Passage # 2 (Ques. 18 to 20)
x|ka'k # 2 (iz- 18 ls 20)
pH scale was introduced by Sorenson to measure
pH iSekus dk ekiu lksjsUlu }kjk ekih x;h
acidity or basicily of a solution. pH value of a
foy;u dh vEyrk vFkok {kkjdrk ls gksrk gSA
foy;u ds pH eku dk fu/kkZj.k foy;u visf{kr
lkeF;Z ij fopkj djds rRdky ugha fn;k tk
ldrk gSA cQj foy;u og foy;u gksrk gS ftlesa
FkksM+h lh ek=kk esa izcy vEy ;k {kkj feykus ij
pH ifjofrZr dqN [kkl ugha gksrk gSA
solution does not instantaneously give us an idea
of the relative strength of the solution. Buffer
solution is the solution whose pH, on addition of
a small amount strong acid or a base, does not
change much.
Q.18
The degree of dissociation of a weak monoprotic
Q.18
acid can be given as
(A) α =
(B) α =
(C) α =
(D) α =
nqcZy eksuksizksVhd vEy ds fo;kstu dh ek=kk bl
rjg nh tkrh gS &
1
1 + 10 ( pK a + pH )
(A) α =
1
1 + 10 ( pK a − pH )
(B) α =
1
1 + 10 ( pH − pK a )
(C) α =
1
1 + 10
(D) 115 pm rFkk 2.0
(D) α =
( pK a / pH )
1
1 + 10 ( pK a + pH )
1
1 + 10
( pK a − pH )
1
1 + 10 ( pH − pK a )
1
1 + 10
( pK a / pH )
Space for rough work
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Page # 29
Q.19
Given a solution of acetic acid. How many times
of the acid concentration, acetate salt should be
added to obtain a solution with pH = 7.0 ?
Q.19
(A) 170 times
(B) 137 times
,flfVd vEy dk foy;u fn;k x;k gSA ,flVsV
yo.k feykus ij foy;u dh pH = 7.0 izkIr gksrk
gS rks vEy lkUnzrk fdrus xquk gks tk,sxh ?
[CH3COOH dk Ka = 1.8 × 10–5]
(B) 137 xquk
(A) 170 xquk
(C) 173 times
(D) 172 times
(C) 173 xquk
[Ka of CH3COOH = 1.8 × 10–5]
Q.20
pH = 7.40, K1 of H2CO3 = 4.5 × 10–7. What will
Q.20
pH = 7.40, H2CO3 dk K1 = 4.5 × 10–7 gSA
be the ratio of [ HCO 3− ] to [H 2 CO 3 ] ?
[ HCO 3− ] o [H 2 CO 3 ] dk vuqikr D;k gksxk ?
(A) 11.3
(B) 1.13
(A) 11.3
(B) 1.13
(C) 1.23
(D) 1.31
(C) 1.23
(D) 1.31
Section – II
[k.M - II
This section contains 2 questions (Questions 1,
2).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with statements
(P, Q, R, S) in Column II. The answers to these
questions have to be appropriately bubbled as
illustrated in the following example. If the correct
matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S,
then the correctly bubbled 4 × 4 matrix should be as
follows :
(D) 172 xquk
bl [k.M esa 2 iz'u (iz'u 1, 2) gSaA izR;sd iz'u esa nks
LrEHkksa esa dFku fn;s x;s gSa] ftUgsa lqesfyr djuk gSA
LrEHk-I
(Column I ) esa fn;s x;s dFkuksa (A, B, C, D) dks LrEHk-II
(Column II) esa fn;s x;s dFkuksa (P, Q, R, S) ls lqesy
djuk gSA bu iz'uksa ds mÙkj uhps fn;s x;s mnkgj.k ds
vuqlkj mfpr xksyksa dks dkyk djds n'kkZuk gSA ;fn lgh
lqesy A-P, A-S, B-Q, B-R, C-P, C-Q rFkk D-S gS, rks lgh
fof/k ls dkys fd;s x;s xksyksa dk 4 × 4 eSfVªDl uhps n'kkZ;s
vuqlkj gksxk :
Space for rough work
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|| IIT Target ||
Page # 30
P Q R S
P Q R S
A
B
C
D
P
P
P
P
Q
Q
Q
Q
R
R
R
R
S
S
S
S
A
B
C
D
Mark your response in OMR sheet against the
P
P
P
P
Q
Q
Q
Q
R
R
R
R
S
S
S
S
any row.
vr% OMR 'khV esa iz'u dh iz'u la[;k ds le{k viuk
mÙkj [k.M-II esa vafdr dhft;sA izR;sd iw.kZ lgh mÙkj ds
fy;s + 6 vad fn;s tk;saxs rFkk xyr mÙkj ds fy;s dksbZ
+_.kkRed vadu ugha gS (vFkkZr~ dksbZ vad ugha ?kVk;k
tk;sxk)A fdUrq] fdlh iafDr esa lgh :i ls fpfUgr mÙkj
ds fy, 1 vad fn;k tk;sxkA
Q.1
Q.1
question number of that question in section-II. + 6
marks will be given for complete correct answer and
No Negative marks for wrong answer. However, 1
mark will be given for a correctly marked answer in
Match the following :
Column -I
LrEHk lqesy dhft;s :
Column-II
(A) Coloured ion
(P) Cu+
(B) µ = 1.73 B.M.
(Q) Cu2+
(C) d10 configuration
(D) more than 3 unpaired electrons
LrEHk -I
LrEHk -II
(A) jaxhu vk;u
(P) Cu+
(B) µ = 1.73 B.M.
(Q) Cu2+
(R) Fe2+
(C) d10 foU;kl
(R) Fe2+
(S) Mn2+
(D) 3 ls vf/kd v;qXehr bysDVªkWu
(S) Mn2+
Space for rough work
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|| IIT Target ||
Page # 31
Q.2
Match the following :
Column -I
(A) Fluorite structure
Q.2
Column-II
LrEHk -I
(P) Distance of
nearest neighbour
=
LrEHk lqesy dhft;s :
(A) ¶yksjkbV lajpuk
LrEHk-II
(P) fudVre
iM+kSlh dh nwjh
3a
4
=
3a
4
(B) Zinc blende structure
(Q) Packing
fraction 73 %
(B) ftad Cys.M lajpuk
(Q) ladqyu fHkUu
73 %
(C) CsCl structure
(R) Packing
fraction 74 %
(C) CsCl lajpuk
(R) ladqyu fHkUu
74 %
(D) Rock salt structure
(S) Packing fraction
79.4%
(D) jksd lkYV lajpuk
(S) ladqyu fHkUu
79.4%
Space for rough work
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|| IIT Target ||
Page # 32
Date : 14/02/2010
Time : 3 : 00 Hrs.
MAX MARKS: 249
Name : _________________________________________________________ Roll No. : __________________________
ijh{kkfFkZ;ksa ds fy, funsZ'k %
A.
lkekU; :
1. Ñi;k izR;sd iz'u ds fy, fn, x, funsZ'kksa dks lko/kkuhiwoZd if<+;s rFkk lEcfU/kr fo"k;kas esa mÙkj&iqfLrdk ij iz'u
SEAL
la[;k ds le{k lgh mÙkj fpfUgr dhft,A
2. iz'u&i=k ds bl eq[k i`"B ij fn, x;s [kkyh LFkku esa viuk uke] vuqØek¡d fyf[k;sA
3. iz'ui=k esa jQ dk;Z gsrq [kkyh LFkku fn;s x, gSaA jQ dk;Z gsrq dksbZ vfrfjDr iqfLrdk ugha nh tk,sxhA
4. mRrj ds fy,] OMR vyx ls nh tk jgh gSA
5. ifjoh{kdksa }kjk funsZ'k fn;s tkus ls iwoZ iz'u&i=k iqfLrdk dh lhy dks ugha rksMsa+A
6. [kkyh dkxt+] fDyicksMZ] y?kqx.kd lkj.kh] LykbM :y] dsYdqysVj] lsY;qyj Qksu] ist+j rFkk bysDVªkWfud midj.kksa dh
fdlh Hkh voLFkk esa ijh{kk d{k esa vanj ys tkus dh vuqefr ugha nh tk,sxhA
B.
OMR dh iwfrZ :
7. OMR 'khV ds CykWdksa esa viuk uke] vuqØek¡d] cSp] dkslZ rFkk ijh{kk dk dsUnz Hkjsa o xksyksa dks mi;qDr :i ls dkyk djsAa
8. OMR 'khV dks xUnk@eksMsa+ ughaA
C.
vadu i)fr:
bl iz'ui=k esa izR;sd fo"k; esa fuEu izdkj ds iz'u gSa:[k.M – I
Space for rough work
9.
cgqfodYih izdkj ds iz'u ftuesa ls dsoy ,d fodYi lgh gSA izR;sd lgh mÙkj ds fy, 3 vad fn;s tk;saxs o izR;sd
xyr mÙkj ds fy, 1 vad ?kVk;k tk,xkA
10. cgqfodYih izdkj ds iz'u ftuesa ls ,d ;k ,d ls vf/kd fodYi lgh gSaA izR;sd lgh mÙkj ds fy, 4 vad fn,
tk;saxas rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk,xkA
11. x|ka'k ij vk/kkfjr cgqfodYih izdkj ds iz'u ftuesa ls dsoy ,d fodYi lgh gSA izR;sd lgh mÙkj ds fy,
4 vad fn;s tk;saxs rFkk xyr mÙkj ds fy, 1 vad ?kVk;k tk,xkA
[k.M –II
12.
LrEHkksa dks lqesfyr djus okys iz'u gSaA iw.kZ :Ik ls lgh lqesfyr mÙkj ds fy, 6 vad fn;s tk;saxs rFkk xyr mÙkj
ds fy, dksbZ _.kkRed vadu ugha gSA fdUrq] fdlh iafDr esa lgh :Ik ls fpfUgr mÙkj ds fy, 1 vad fn;k tk;sxkA
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Page # 33
Space for Rough Work (jQ+ dk;Z gsrq LFkku)
Space for rough work
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Page # 1