RS -11- I -14
TARGET – IIT JEE
CHEMISTRY, MATHEMATICS & PHYSICS
MAX MARKS: 246
Time : 3 : 00 Hrs.
Name : _____________________________________ Roll No. : __________________________ Date : _____________
INSTRUCTIONS TO CANDIDATE
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. The answer sheet, a machine readable Optical Mark Recognition (OMR) is provided separately.
3. Do not break the seal of the question-paper booklet before being instructed to do so by the invigilators.
B. MARKING SCHEME :
Each subject in this paper consists of following types of questions:Section - I
4. Multiple choice questions with only one correct answer. 3 marks will be awarded for each correct answer and –1 mark for
each wrong answer.
5. Multiple choice questions with multiple correct option. 4 marks will be awarded for each correct answer and No negative
marking for wrong answer.
6. Reason and Assertion type questions with only one correct answer in each. 3 marks will be awarded for each correct
answer and –1 mark for each wrong answer.
7. Passage based single correct type questions. 4 marks will be awarded for each correct answer and –1 mark for each wrong answer.
C.FILLING THE OMR :
8. Fill your Name, Roll No., Batch, Course and Centre of Examination in the blocks of OMR sheet and darken circle properly.
Space
9. Use only HB pencil or blue/black pen (avoid gel
pen) for
for rough
darkingwork
the bubbles.
10. While filling the bubbles please be careful about SECTIONS [i.e. Section-I (include single correct, reason type, multiple
correct answers), Section –II ( column matching type), Section-III (include integer answer type)]
Section –I
Section-II
Section-III
For example if only 'A' choice is
correct then, the correct method for
filling the bubbles is
For example if Correct match for (A)
is P; for (B) is R, S; for (C) is Q; for
(D) is P, Q, S then the correct method
for filling the bubbles is
Ensure that all columns are filled.
Answers, having blank column will be treated as
incorrect. Insert leading zeros (s)
A
B
C
D
E
For example if only 'A & C' choices
are correct then, the correct method
for filling the bublles is
A
B
C
D
E
A
B
C
D
P
Q
R
S
T
the wrong method for filling the
bubble are
The answer of the questions in
wrong or any other manner will be
treated as wrong.
7
'6' should be
filled as 0006
0
1
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6
7
8
9
0
1
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9
0
1
2
3
4
5
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0
1
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7
8
9
'86' should be
filled as 0086
0
1
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9
0
1
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'1857' should be
filled as 1857
0 0 0 0
0 0 0 0
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
5 5 5 5
6 6 6 6
7 7 7 7
8 8 8 8
9 9 9 9
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Page # 2
SEAL
A.GENERAL :
Section - I
CHEMISTRY
[k.M - I
Questions 1 to 6 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.
Q.1
1 mole mixture of CH4 & air (containing 80% N2,
iz'u 1 ls 6 rd cgqfodYih iz'u gSaA 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, + 3 vad fn;s
tk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA
Q.1
20% O2 by volume) of a composition such that
when underwent combustion gave maximum heat
(assume combustion of only CH4). Then which of
the statements are correct, regarding composition
of initial mixture:
CH4 o ok;q (vk;rukuqlkj 80% N2, 20% O2 ;qDr) ds 1
eksy feJ.k dk laxBu bl izdkj gS fd tc ngu gksrk
gS] rks vf/kdre Å"ek nsrk gSA (;g ekurs gq, fd dsoy
CH4 dk ngu gksrk gS) rks fuEu esa ls dkSulk dFku
izkjfEHkd feJ.k ds laxBu ds lUnHkZ esa lgh gS&
(A) X CH 4 =
1
2
8
, X O2 = , X N2 =
11
11
11
(A) X CH 4 =
1
2
8
, X O2 = , X N2 =
11
11
11
(B) X CH 4 =
3
1
4
, X O2 = , X N2 =
8
8
2
(B) X CH 4 =
4
1
3
, X O2 = , X N2 =
2
8
8
(C) X CH 4 =
1
1
4
, X O2 = , X N2 =
6
6
3
(C) X CH 4 =
1
1
4
, X O2 = , X N2 =
6
6
3
(D) Data insufficient
(D) vkadM+s vi;kZIr gS
Space for rough work
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Page # 2
Q.2
2 moles of an ideal gas (Cv = 5/2 R) was
Q.2
compressed adiabatically against constant pressure
2 eksy vkn'kZ xSl (Cv = 5/2 R) dks 2 atm fu;r nkc ij
:)ks"e :i ls lEihfM+r fd;k x;k A tks izkjEHk eas 350 K
of 2 atm. Which was initially at 350 K & 1 atm
o 1 atm nkc ij FkhA bl izØe eas fd;k x;k dk;Z D;k gS
pressure. The work involve in the process is equal to
Q.3
(A) 250 R
(B) 300 R
(A) 250 R
(B) 300 R
(C) 400 R
(D) 500 R
(C) 400 R
(D) 500 R
The reactions of higher order are rare because:
Q.3
(A) many body collisions involve very high
mPp dksfV dh vfHkfØ;k,sa nqyZHk gS] D;kasfd –
(A) vusd v.kqvksa dh VDdj vf/kd mPp lfØ;.k
ÅtkZ ls lEcfU/kr gksrh gS
activation energy
(B) many body collisions have a very low
(B) vusd v.kqvksa dh VDdj dh cgqr de izkf;drk
gksrh gS
probability
(C) many body collisions are not energetically
(C) vusd v.kqvksa dh VDdj ÅtkZxfrdh :i ls
vuqdwfyr ugha gksrh gS
favoured
(D) many body collisions can take place only in
(D) vusd v.kqvksa dh VDdj dsoy xSlh; voLFkk esa
gksrh gS
gaseous state
Space for rough work
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Page # 3
Q.4
In the reaction sequence
t-BuOK
PhCHCl2
t-BuOK
Q.4
P
PhC
CPh
Q
vfHkfØ;k vuqØe
[ S ] is -
R
t-BuOK
PhCHCl2
t-BuOK
R
Br2
CPh
Q
esa [ S ] gS -
S
(A) PhCH = CHPh
Ph
(A) PhCH = CHPh
Ph
(B)
Ph
PhC
Br2
S
(B)
P
Ph
Br
Ph
Ph
Ph
Ph
Ph
Br
(C)
Br
Ph
Ph
Ph
Br
(C)
(D) PhCH = CH – C ≡ C – Ph
(D) PhCH = CH – C ≡ C – Ph
Space for rough work
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Page # 4
Q.5
O
||
–
CH3–C–CH3 OH

→ P (conjugated system)
∆
Q.5
O
||
–
CH3–C–CH3 OH

→ P (la;qfXer ra=k)
∆
2 H 5ONa
P + CH2(COOC2H5)2 C
 → Q
2 H 5ONa
P + CH2(COOC2H5)2 C
 → Q
2 H 5ONa
Q C
 → R
2 H 5ONa
Q C
 → R
+
+
. H 3O II. Heat
R I

→ S
The final product ‘S’ is -
. H 3O II. Heat
R I

→ S
vfUre mRikn ‘S’ gS -
O
O
(B)
(A)
O
O
O
(C)
(B)
O
O
O
(D)
(D)
(C)
O
Q.6
(A)
O
Q.6
In the reaction sequence :
O
O
vfHkfØ;k vuqØe esa
CH MgBr
CH 3 − C ≡ CH 3 → CH 4 + A
CH 3 MgBr
CH 3 − C ≡ CH   → CH 4 + A
(i ) CO
 
2 →
B
+
i ) CO 2
(

→
B
+
(ii ) H 2 O / H
(ii ) H 2 O / H
B will be :
(A) CH3 – C ≡ C –MgBr
B gksxk :
(A) CH3 – C ≡ C –MgBr
(B) CH3 – C ≡ C – CH3
(B) CH3 – C ≡ C – CH3
(C) CH3 – C ≡ C – COOH
(C) CH3 – C ≡ C – COOH
(D) CH3 – CH = CH –COOH
(D) CH3 – CH = CH –COOH
Space for rough work
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Page # 5
Questions 7 to 10 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) 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 NO NEGATIVE marks for
wrong answer.
iz'u 7 ls 10 rd cgqfodYih iz'u gaAS izR;sd iz'u ds pkj fodYi
(A), (B), (C) rFkk (D) gS]a ftuesa ls ,d ;k 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 xyr mÙkj ds fy, dksbZ _.kkRed vadu ugha gSA
Q.7
Q.7
CO2 fdlds
lkFk lelajpukRed gS –
CO2 is isostructural with:
Q.8
(A) HgCl2
(B) SnCl2
(A) HgCl2
(B) SnCl2
(C) C2H2
(D) NO2
(C) C2H2
(D) NO2
Which of the following have planar structure?
Q.9
fuEe eas ls fdldh leryh; lajpuk gS?
(A) I3–
(B) [Co(DMG)2]
(A) I3–
(B) [Co(DMG)2]
(C) Ni(CO)4
(D) [Ni (CN)4] 2–
(C) Ni(CO)4
(D) [Ni (CN)4] 2–
Select the reactions in which correct products have
been mentioned (A)
Q.8
2H / Pd
2→
ethyl acetate
solvent
O
CH3
H / Pt
2→

(B)
H H3C
CH3
O
Q.9
vfHkfØ;kvksa dk p;u dhft, ftlesa mRikn lgh :i
ls of.kZr gS (A)
2H / Pd
2→
ethyl acetate
solvent
O
CH3
H / Pt
2→

(B)
O
H H3C
CH3
CH3 H
CH3 H
Space for rough work
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Page # 6
OH
CH3
CH3
(i ) PhMgBr / ether
   →
(ii ) H3O⊕
(C) O
(D)
Q.10
H
Mg / ether
Cl
Cl
CH3
(i ) PhMgBr / ether
   →
(ii ) H3O⊕
(C) O
Ph
Q.10
Choose the correct statement (s) (A) The formation of α-hydroxynitrile from
Cl
(A)
 
→
Cl
,fYMgkbM o dhVksu ls α-gkbMªkWDlhukbVªkby
dk fuekZ.k ,d mRØe.kh; izØe gS
(B) α-gkbMªkWDlhukbVªkby
(B) The formation of α-hydroxynitrile is catalysed
mRiszfjr
by OH–. The use of H+ decreases the
nucleophilicity (HCN
H+ + CN–)
gksrk
gSA
dk fuekZ.k OH– }kjk
H+
ds
(C)
ls
(lkbuksgkbfMªu) ugh
(cyanohydrin)
H + CN )
α-gkbMªkWDlh
ukbVªkby
nsrk gS
(D) csUtsfYMgkbM] HCN/OH– ds
give optically active cyanohydrin molecule.
ls
–
,Øksfyu (CH2=CH–CH=O), HCN/OH– ds lkFk
vfHkfØ;k
HCN/OH– do not give α-hydroxy nitrile
mi;ksx
+
ukfHkdLusgrk ?kVrh gS (HCN
(C) Acrolein (CH2=CH–CH=O) on treatment with
(D) Benzaldehyde on treatment with HCN/OH–
Ph
lgh dFkuksa dk p;u dhft, -
aldehyde and ketone is usually a reversible
process.
H
Mg / ether
(D)
 
→
OH
CH3
lkFk fØ;k dj
izdkf'kd lfØ; lk;uksgkbfMªu v.kq nsrk gS
Space for rough work
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Page # 7
This section contains 4 questions numbered 11 to 14,
(Reason and Assertion type question). Each question
contains Assertion and Reason. 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. + 3 marks will be
given for each correct answer and – 1 mark for each
wrong answer.
The following questions given below consist of
an "Assertion" (A) and "Reason" (R) Type
questions. Use the following Key to choose the
appropriate answer.
(A) If both (A) and (R) are true, and (R) is the
correct explanation of (A).
(B) If both (A) and (R) are true but (R) is not the
correct explanation of (A).
(C) If (A) is true but (R) is false.
(D) If (A) is false but (R) is true.
Q. 11
Assertion (A) : [Co( NH 3 ) 3 Cl 3 ] does not give a
bl [k.M esa 11 ls 14 rd 4 iz'u gSa, (dFku rFkk dkj.k izdkj
ds iz'u)A izR;sd iz'u esa dFku rFkk dkj.k fn;s x;s gSaA
izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) fn;s x;s gSa]
ftuesa ls dsoy ,d 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 + 3 vad fn;s tk;asxs rFkk izR;sd xyr mÙkj ds
fy, 1 vad ?kVk;k tk;sxkA
uhps fn;s x;s fuEufyf[kr iz'u "dFku" (A) rFkk
"dkj.k" (R) izdkj ds iz'u gSaA vr% mfpr mÙkj dk
p;u djus ds fy;s fuEu rkfydk dk mi;ksx dhft;sA
(A) ;fn (A) rFkk (R) nksuksa lR; gSa rFkk (R), (A) dk
lgh Li"Vhdj.k gSA
(B) ;fn (A) rFkk (R) nksuksa lR; gSa rFkk (R), (A) dk
lgh Li"Vhdj.k ugha gSA
(C) ;fn (A) lR; gS ysfdu (R) vlR; gSA
(D) ;fn (A) vlR; gS ysfdu (R) lR; gSA
Q. 11
white ppt with AgNO3
vo{ksi ugh nsrk gS
Reason (R) : Chlorine is not present in the
dkj.k (R) : Dyksjhu] [Co( NH 3 ) 3 Cl 3 ] ds vk;uu
ionization sphere of [Co( NH 3 ) 3 Cl 3 ]
Q. 12
dFku (A) : [Co( NH 3 ) 3 Cl 3 ] , AgNO3 ds lkFk 'osr
e.My esa mifLFkr ugha gksrh gS
Q. 12
Assertion (A) : BeCl2 fumes in moist air.
Reason (R) : BeCl2 reacts with moisture to form
HCl gas
dFku (A) : BeCl2 ue ok;q esa /kwez nsrh gS
dkj.k (R) : BeCl2 ueh ds lkFk fØ;k dj HCl xSl
cukrk gS
Space for rough work
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Q.13
Assertion : Changing solvent from CH3OH to
CH3CN the rate of reaction of azide ion with
gksus ij ,stkbM vk;u dh 1-czkseksC;wVsu ds lkFk
1-bromobutane decreases many fold. (SN1 reaction)
vfHkfØ;k dh nj dbZ xquk ?kV tkrh gS (SN1 vfHkfØ;k)
Reason : CH3OH is a polar protic solvent while
dkj.k (R) : CH3OH /kqzoh; izksfVd foyk;d gS tcfd
CH3CN ,d
CH3CN is polar aprotic.
Q.14
dFku (A) : CH3OH ls CH3CN foyk;d ifjofrZr
Q.13
Q.14
Observe the following reaction
/kqzoh; vizksfVd gSA
fuEu vfHkfØ;k dks izsf{kr dhft,
O
O
KMnO4/H2SO4
P
NaOH/CaO/∆
KMnO4/H2SO4
Q + CaCO3
Assertion : Molecular mass of Q is 58 g/mol.
NaOH/CaO/∆
Q + CaCO3
dFku (A) : Q dk vuqHkkj 58 g/mol gS
Reason : Boiling point of P is greater than Q.
This section contains 3 paragraphs, each has 3 multiple
choice questions. (Questions 15 to 23) 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.
P
dkj.k (R) : P dk DoFkukad Q ls vf/kd gS
bl [k.M esa 3 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih
iz'u gSaA (iz'u 15 ls 23) 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,saxs rFkk
izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA
Space for rough work
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Page # 9
Passage # 1 (Ques. 15 to 17)
A white solid A reacts with dilute H2SO4 to
produce a colourless gas B and a colourless
solution C. The reaction between B and acidified
dichromate yields a green solution and a slightly
coloured precipitate D. The substance D, when
burnt in air, gives a gas E which reacts with B to
yield D and a colourless liquid. Anhydrous copper
sulphate turns blue with this colourless liquid. The
addition of aqueous NH3 or NaOH to C produces a
precipitate that dissolves in an excess of the
reagent to form a clear solution.
x|ka'k # 1 (iz- 15 ls 17)
,d lQsn Bksl A ruq H2SO4 ds lkFk fØ;k dj jaxghu
xSl B rFkk ,d jaxghu foy;u C nsrk gSA B o vEyhdr̀
MkbØksesV ds chp vfHkfØ;k ls ,d gjk foy;u rFkk
gYdk jaxhu vo{ksi D izkIr gksrk gSA inkFkZ D tc ok;q esa
tyrk gS] rks xSl E nsrk gS tks B ds lkFk fØ;k dj D
rFkk ,d jaxghu nzo nsrk gSA futZy dkWij lYQsV bl
jaxghu nzo dks uhyk dj nsrk gSA C esa nzo NH3 ;k
NaOH dks feykus ij ,d vo{ksi nsrk gS tks vfHkdeZd
ds vkf/kD; esa ?kqy tkrk gSA rFkk ,d LoPN foy;u
cukrk gSA
Q.15
Which of the following gases are B and E
respectively?
(A) CO2 and SO2
(B) SO2 and H2S
(D) CO2 and H2S
(C) H2S and SO2
Q.15
What would appear if the gas B is passed through
an aqueous solution of Pb(NO3)2?
(A) White precipitate soluble in hot dilute HNO3
(B) A black precipitate soluble in hot dilute HNO3
Q.16
Q.16
(C) A black precipitate insoluble in hot dilute
HNO3
(D) A yellow precipitate
concentrated HNO3
soluble
in
hot
fuEu esa ls dkSulh xSls Øe'k% B o E gS ?
(A) CO2 o SO2
(B) SO2 o H2S
(C) H2S o SO2
(D) CO2 o H2S
;fn xSl B dks Pb(NO3)2 ds tyh; foy;u ls
xqtkjk tkrk gS] rks D;k gksxk?
(A) 'osr
vo{ksi xeZ ruq HNO3 esa foys; gS
(B) ,d
dkyk vo{ksi xeZ ruq HNO3 esa foys; gS
(C) ,d
dkyk vo{ksi xeZ ruq HNO3 esa vfoys; gS
(D)
,d ihyk vo{ksi xeZ lkUnz HNO3 esa foys; gS
Space for rough work
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Page # 10
Q.17
Suppose the solution obtained by the treatement of
the solution C with an excess of NaOH is acidified
with acetic acid and the gas B is passed through it.
Which of the following will be obtained?
(A) Colourless solution (B) Yellow precipitate
(C) Black precipitate (D) White precipitate
Passage # 2 (Ques. 18 to 20)
Equilibrium constants are given (in atm) for the
following reactions at 0°C
SrCl2 ⋅ 2H2O(s) + 4H2O(g)
SrCl2 ⋅ 6H2O(s)
Kp = 5× 10–12
Na2HPO4 ⋅ 12H2O(s)
Na2SO4 ⋅ 10H2O(s)
Q.17
;g ekurs gq, fd foy;u C dks NaOH ds vkf/kD; ds
lkFk vfHkdr̀ djokus ls izkIr foy;u dks ,lhfVd
vEy ds lkFk vEyhdr̀ rFkk xSl B dks blesa izokfgr
fd;k tkrk gSA fuEu esa ls D;k izkIr gksxk?
(A) jaxghu foys;u
(B) ihyk vo{ksi
(D) 'osr vo{ksi
(C) dkyk vo{ksi
x|ka'k # 2 (iz- 18 ls 20)
0°C
ij fuEu vfHkfØ;kvksa ds lkE; fu;rkad (atm esa)
fn;s x;s gSA
SrCl2 ⋅ 6H2O(s)
Na2HPO4 ⋅ 7H2O(s) + 5H2O(g)
Kp = 2.43 × 10–13
Na2HPO4 ⋅ 12H2O(s)
Na2SO4(s) + 10H2O(g)
Kp = 1.024 × 10–27
Na2SO4 ⋅ 10H2O(s)
The vapor pressure of water at 0°C is 4.56 torr.
0°C ij
SrCl2 ⋅ 2H2O(s) + 4H2O(g)
Kp = 5× 10–12
Na2HPO4 ⋅ 7H2O(s) + 5H2O(g)
Kp = 2.43 × 10–13
Na2SO4(s) + 10H2O(g)
Kp = 1.024 × 10–27
ty dk ok"Ik nkc 4.56 VkWj gSA
Q.18
Which is the most effective drying agent at 0°C?
(A) SrCl2 ⋅ 2H2O
(B) Na2HPO4 ⋅ 7H2O
(D) all equally
(C) Na2SO4
Q.18
0°C ij dkSulk lokZf/kd izHkkoh 'kq"ddkjh gS?
(A) SrCl2 ⋅ 2H2O
(B) Na2HPO4 ⋅ 7H2O
(D) lHkh cjkcj
(C) Na2SO4
Q.19
At what relative humidities will Na2SO4. 10H2O be
efflorescent (release moisture) when exposed to air
at 0°C?
(A) above 33.33%
(B) below 33.33 %
Q.19
tc Na2SO4 10H2O dks 0°C ij ok;q esa [kqyk NksM+k
tkrk gS] rks fdl vkisf{kd vknzZrk ij mRQqfYyr
gksxk (ueh NksM+rk gS)
(A) 33.33% ds Åij (B) 33.33 % ds uhps
(C) 66.66 % ds Åij (D) 66.66 % ds uhps
(C) above 66.66 %
(D) below 66.66 %
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Page # 11
Q.20
At what relative humidities will Na2SO4 be
Q.20
deliquescent (i.e. absorb moisture) when exposed
to the air at 0°C ?
(A) above 33.33 %
(B) below 33.33 %
(C) above 66.66 %
(D) below 66.66 %
tc Na2SO4 dks 0°C ij ok;q esa [kqyk NksM+k tkrk gS]
rks fdl vkisf{kd vknzZrk ij izLosndkjh (ueh dk
vo'kks"k.k) gksxk
(A) 33.33% ds Åij (B) 33.33 % ds uhps
(C) 66.66 % ds Åij (D) 66.66 % ds uhps
Passage # 3 (Ques. 21 to 23)
It occasionaly happens that a reaction proceeds
much faster or much slower than expected on basis
of electrical effects alone. It can be shown that
steric effects are influencing the rate also. An
example of such an effect coming into play is
hydrolysis of alkyl halide by SNl mechanism. The
first step (RDS) involves ionization of alkyl halide
to carbo-cation, thus relieving strain due to change
in hybridisation from sp3 to sp2
x|ka'k # 3 (iz- 21 ls 23)
;g dHkh&dHkh gksrk gS fd dsoy fo|qr izHkko ds
vk/kkj ij vis{kk ls ,d vfHkfØ;k vf/kd rhoz ;k
de /kheh gksrh gSA ;g n'kkZ;k tk ldrk gS fd
f=kfoe izHkko nj dks Hkh izHkkfor djrk gSA bl izdkj
ds izHkko dk mnkgj.k SNl fØ;kfof/k }kjk ,fYdy
gSykbM dk tyvi?kVu gSA izFke in (osx fu/kkZfjr
in) esa ,fYdy gSykbM dk dkcZ/kuk;u esa vk;uu
gksrk gS] vr% sp3 ls sp2 ladj.k eas ifjorZu ds
dkj.k fodf̀r esa deh gksrh gSA
Q.21
Q.21
Relative reaction rates of solvolysis of halides (RBr)
with ethanol,
R = Me > Et > n-propyl > iso-Butyl
can be explained on basis of (A) Carbocation stability
(B) Hyperconjugation
(C) Stability of SN2 transition state
(D) None of these
gSykbMksa (RBr) dh ,FksukWy ds lkFk foyk;dhdj.k
dh vkisf{kd vfHkfØ;k nj
R = Me > Et > n-izksfiy > vkblks-C;qfVy
dks fdlls vk/kkj ij le>k;k tk ldrk gS (A) dkcZ/kuk;u LFkkf;Ro
(B) vfrla;qXeu
(C) SN2 laØe.k voLFkk dk LFkkf;Ro
(D) bues ls dksbZ ugha
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Q.22
Reative rates of hydrolysis of alkyl chlorides (RCl)
Q.22
in 80% aqueous ethanol follows, the order
,fYdy DyksjkbMkas (RCl) dk 80% tyh; ,FksukWy esa
tyvi?kVu dh vkisf{kr nj ds Øe
Et3C – Cl > MeEt2CCl > Me2EtCCl > Me3CCl,
dks fdlds vk/kkj ij le>k;k tk ldrk gS -
Et3C – Cl > MeEt2CCl > Me2EtCCl > Me3CCl,
can be explained by -
(A) dkcZ/kuk;u
(A) Carbocation stability
LFkkf;Ro
(B) vfrla;qXeu
(B) Hyperconjugation
(C)
cM+s lewg ls vfHkdkjd esa fodf̀r vf/kd gksus ds
dkj.k ,fYdy gSykbM dk vk;uu rhoz gksrk gS
(D) SNl fØ;kfof/k esa lewg dss vf/kd cM+k gksus ij
dkcZ/kuk;u dk LFkkf;Ro vf/kd gksrk gS
(C) Greater the strain in reactant due to bulky
groups faster is ionization of alkyl halide
(D) In SNl mechanism greater the bulkiness of
groups more is carbo-cation stability
Q.23
The rate of hydrolysis of following in 80%
Q.23
dk Øe gS
aqueous ethanol follows the order
(I)
Me
Cl
(II) t-BuCl (III)
fuEu ds 80% tyh; ,FksukWy esa tyvi?kVu dh nj
Me
Cl
(I)
Me
Cl
Me
Cl
(II) t-BuCl (III)
(A) I > III > II
(B) II > I > III
(A) I > III > II
(B) II > I > III
(C) I > II > III
(D) III > II > I
(C) I > II > III
(D) III > II > I
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Page # 13
MATHEMATICS
Section - I
[k.M - I
Questions 1 to 6 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 6 rd cgqfodYih iz'u gSaA 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, + 3 vad fn;s
tk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA
Q.1
α
α

If fr (α) =  cos 2 + i sin 2
r
r

×


2α
2α 

 cos 2 + i sin 2 
r
r 

Q.1
Q.3

r
r 

r 
r
α
α
…  cos + i sin  rc lim fn(π) cjkcj gSn →∞
r
r

(A) –1
(B) 1
(C) – i
(D) i
α
α

…  cos + i sin  then lim fn(π) equalsn →∞
r
r

(A) –1
(B) 1
(C) – i
(D) i
Q.2
;fn fr(α)=  cos α2 + i sin α2  ×  cos 2α2 + i sin 2α2 
Sum of the non-real roots of
(x2 + x –2) (x2 + x –3) = 12 is
(A) 1
(B) –1
(C) –6
(D) 6
Q.2
If l, m, n are real, l + m ≠ 0, then the roots of the
equation
(l + m)x2 –3(l –m)x –2(l + m) = 0 are
(A) real and equal
(B) complex
(C) real and unequal (D) None of these
Q.3
(x2 + x –2) (x2 + x –3) = 12 ds vokLrfod ewyksa dk
;ksx gS(A) 1
(B) –1
(C) –6
(D) 6
;fn l, m, n okLrfod gSa, l + m ≠ 0, rc lehdj.k
(l + m)x2 –3(l –m)x –2(l + m) = 0 ds ewy gksaxs(A) okLrfod o leku (B) lfEeJ
(C) okLrfod o vleku (D) buesa ls dksbZ ugha
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Page # 14
Q.4
Q.5
Q.6
The set of all solutions of the equation
log3x log4x log5 x = log3x log4x + log4x log5x
+ log5x log3x is(A) {1}
(B) {1, 60}
(C) {1, 5, 10, 60}
(D) None of these
Q.4
Locus of the mid-points of the chords of the circle
x2 + y2 = 4 which subtend a right angle at the centre
is
(A) x + y = 2
(B) x2 + y2 = 1
2
2
(C) x + y = 2
(D) x – y = 0
Q.5
1 − cos 3 x
is equal to
lim
x →0 x sin x cos x
(A) 2/5
(B) 3/5
(C) 3/2
(D) 3/4
Q.6
Questions 7 to 10 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) 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 NO NEGATIVE marks for wrong
answer.
Q.7
If I =
∫ sec
2
x cosec4 x dx
lehdj.k
log3x log4x log5 x = log3x log4x + log4x log5x
+ log5x log3x ds lHkh gyksa dk leqPp; gS(A) {1}
(B) {1, 60}
(C) {1, 5, 10, 60}
(D) buesa ls dksbZ ugha
oÙ̀k x2 + y2 = 4 dh mu thokvksa ds e/; fcUnqvksa dk
fcUnqiFk tks dsUnz ij ledks.k cukrh gS] gksxk(A) x + y = 2
(C) x2 + y2 = 2
(B) x2 + y2 = 1
(D) x – y = 0
1 − cos 3 x
cjkcj gSx →0 x sin x cos x
(A) 2/5
(B) 3/5
(C) 3/2
(D) 3/4
lim
iz'u 7 ls 10 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj
fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k vf/kd
fodYi lgh gaSA 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 xyr mÙkj ds fy, dksbZ _.kkRed vadu
ugha gSA
Q.7
;fn I = ∫ sec 2 x cosec4 x dx
= K cot2 x + L tan x + M cot x + C rc
(A) K = –1/3
(B) L = 2
(C) M = –2
(D) buesa ls dksbZ ugha
= K cot3 x + L tan x + M cot x + C then
(A) K = –1/3
(B) L = 2
(C) M = –2
(D) None of these
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Page # 15
π/ 2
Q.8
The value of
∫
0
x sin x cos x
sin 4 x + cos 4 x
 5π / 4

sin 2x
(A) 
dx 
4
4


 π cos x + sin x 
π/2
dx is
Q.8
0
2
(B) π2/16
(C) 3π2/4
(D) π2/2
The solution of
dy
=
dx
x 2 + y2 +1
satisfying
2 xy
Q.9
(A) a hyperbola
(B) a system of circles
(C) y2 = x(1 + x) –1
(D) (x –2)2 + (y –3)2 = 5
Q.10
− sin θ
dx dk eku gS2
dy
x 2 + y2 +1
dk gy tks y (1) = 1 dks larq"V
=
2 xy
dx
djrk gS(A) ,d vfrijoy;
(B) oÙ̀kksa dk fudk;
y (1) = 1 is given by-
2
x sin x cos x
sin 4 x + cos 4 x
 5π / 4

sin 2 x
(A) 
dx 
∫
4
4


 π cos x + sin x 
(B) π2/16
(C) 3π2/4
(D) π2/2
∫
Q.9
∫
(C) y2 = x(1 + x) –1
(D) (x –2)2 + (y –3)2 = 5
1
If ∆ = − sin θ
2
sin θ then :
2
−1
− sin θ
2
Q.10
;fn ∆ = − sin θ
−1
− sin θ
1
2
sin θ
− sin θ
2
(A) maximum value of ∆ = 12
(A) ∆ dk vf/kdre eku 12 gS
(B) minimum value of ∆ = 10
(B) ∆ dk U;wure eku 10 gS
(C) minimum value of ∆ = 8
(C) ∆ dk U;wure eku 8 gS
(D) maximum value of ∆ = 10
gS] rc :
(D) ∆ dk vf/kdre eku 10 gS
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Page # 16
This section contains 4 questions numbered 11 to 14,
(Reason and Assertion type question). Each question
contains Assertion and Reason. 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. + 3 marks will be
given for each correct answer and – 1 mark for each
wrong answer.
bl [k.M esa 11 ls 14 rd 4 iz'u gSa, (dFku rFkk dkj.k izdkj
ds iz'u)A izR;sd iz'u esa dFku rFkk dkj.k fn;s x;s gSaA
izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) fn;s x;s gSa]
ftuesa ls dsoy ,d 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 + 3 vad fn;s tk;asxs rFkk izR;sd xyr mÙkj ds fy,
1 vad ?kVk;k tk;sxkA
The following questions given below consist of
an "Assertion" (A) and "Reason" (R) Type
questions. Use the following Key to choose the
appropriate answer.
(A) If both (A) and (R) are true, and (R) is the
correct explanation of (A).
(B) If both (A) and (R) are true but (R) is not
the correct explanation of (A).
(C) If (A) is true but (R) is false.
(D) If (A) is false but (R) is true.
uhps fn;s x;s fuEufyf[kr iz'u "dFku" (A) rFkk
"dkj.k" (R) izdkj ds iz'u gSaA vr% mfpr mÙkj dk
p;u djus ds fy;s fuEu rkfydk dk mi;ksx dhft;sA
(A) ;fn (A) rFkk (R) nksuksa lR; gSa rFkk (R), (A) dk
lgh Li"Vhdj.k gSA
(B) ;fn (A) rFkk (R) nksuksa lR; gSa rFkk (R), (A) dk
lgh Li"Vhdj.k ugha gSA
(C) ;fn (A) lR; gS ysfdu (R) vlR; gSA
(D) ;fn (A) vlR; gS ysfdu (R) lR; gSA
dFku -I : fdlh Hkh f=kHkqt ABC esa
Q.11
Q. 12
Assertion-I : In any triangle ABC,
a cos A + b cos B + c cos C ≤ s
Reason-II : In any triangle ABC,
sin (A/2) sin (B/2) sin (C/2) ≤ 1/8
Assertion - I : In any triangle ABC,
if a : b : c = 4 : 5 : 6 then R : r = 16 : 7
Reason-II :.In any triangle ABC
R abc
=
4s
r
Q. 11
a cos A + b cos B + c cos C ≤ s
dkj.k-II : fdlh Hkh f=kHkqt ABC esa
sin (A/2) sin (B/2) sin (C/2) ≤ 1/8
Q. 12
Hkh f=kHkqt ABC
a : b : c = 4 : 5 : 6 rks R : r = 16 : 7
dkj.k-II :. fdlh Hkh f=kHkqt ABC esa]
dFku-I:
fdlh
esa,
;fn
R abc
=
4s
r
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Page # 17
Q.13
Q.14
Assertion-I: cosec–1(3/2) + cos–1(2/3) –2 cot–1 (1/7)
– cot–17 is equal to cot–1 7.
Reason-II: sin–1 x + cos–1 x = π/2, tan–1 x + cot–1 x
= π/2 , cosec–1 x = sin–1 (1/x), cot–1 (x) = tan–1 (1/x)
Q.13
Assertion I : Slope of the line
(cos θ + sin θ)x + sin 2θy = 1 is
3 −1
if θ = π/6
3
Q.14
dFku-I: cosec–1(3/2) + cos–1(2/3) – 2 cot–1(1/7)
– cot–17 is equal to cot–1 7.
dkj.k-II: sin–1x + cos–1x = π/2, tan–1x + cot–1
x = π/2 , cosec–1x = sin–1 (1/x), cot–1 (x) = tan–1 (1/x)
dFku -I: js[kk dk <ky (cos θ + sin θ)x + sin 2θy = 1
gSA ;fn
3 −1
3
if θ = π/6
dkj.k-II : 3 x + y = 30 ,d js[kk dks iznf'kZr djrh
gS tks ewyfcUnq ls 15 bdkbZ nwjh ij x-v{k dh
/kukRed fn'kk ls π/6 dks.k cukrh gSA
Reason-II : 3 x + y = 30 represents the line
making an angle π/6 with the +ve direction of
x-axis at a distance 15 units from the origin.
This section contains 3 paragraphs, each has 3 multiple
choice questions. (Questions 15 to 23) 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 3 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih
iz'u gSaA (iz'u 15 ls 23) 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,saxs rFkk
izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA
Passage # 1 (Ques. 15 to 17)
ABC is a triangle, the incircle touches the sides BC,
x|ka'k # 1 (iz- 15 ls 17)
ABC ,d f=kHkqt gSA ,d vUr%or̀ Hkqtkvksa BC, CA
CA and AB at D, E, F respectively. BD, CE and AF
rFkk AB dks fcUnq Øe'k% D, E, F ij Li'kZ djrk gSA
are consecutive natural numbers. I is the incentre of
BD, CE rFkk AF Øekxr izkdr̀ la[;k,a gSaA I f=kHkqt
the triangles. The radius of the incircle is 4 units.
Q.15
Sides of the triangle ABC are
(A) 11, 12, 13
(B) 12, 13, 14
(C) 13, 14, 15
(D) 14, 15, 16
ds vUr% dsUnz gSA vUr%oÙ̀k dh f=kT;k 4 bdkbZ gS)
Q.15
f=kHkqt ABC dh Hkqtk,sa gksaxh(A) 11, 12, 13
(C) 13, 14, 15
(B) 12, 13, 14
(D) 14, 15, 16
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Page # 18
Q.16
Q.17
Angles of the triangle DEF are
Q.16
(A) π –2A, π –2B, π –2C
(A) π –2A, π –2B, π –2C
(B) π –A, π – B, π – C
(B) π –A, π – B, π – C
(C) A/2, B/2, C/2
(C) A/2, B/2, C/2
π−A π−B π−C
(D)
,
,
2
2
2
Sides of the triangle DEF are
(D)
Q.17
(A) 4 cos (A/2), 4 cos (B/2), 4 cos (C/2)
(B) 8 sin (A/2), 8 sin (B/2), 8 sin (C/2)
(C) 8 cos (A/2), 8 cos (B/2), 8 cos (C/2)
(D) 4 sin A, 4 sin B, 4 sin C
Q.19
f=kHkqt DEF dh Hkqtk,sa gksasxh-
x|ka'k # 2 (iz- 18 ls 20)
tc ,d 'kCn ds v{kjksa dks lHkh lEHko Øe esa fy[kk
tkrk gS vkSj bu 'kCnksa dks 'kCndks"k ds vuqlkj tek;k
tkrk gS] rc og fLFkfr ftl ij fn;k x;k 'kCn vkrk
gS og mldk Øe dgykrk gSA
The letters A, A, P, T, N are arranged in all
possible ways to form the words PATNA &
TAPAN then number of words between them is :
(A) 11
(B) 10
(C) 9
(D) 13
Q.18
Number of words between DANGER & GARDEN
Q.19
is :
(A) 243
(C) 240
π−A π−B π−C
,
,
2
2
2
(A) 4 cos (A/2), 4 cos (B/2), 4 cos (C/2)
(B) 8 sin (A/2), 8 sin (B/2), 8 sin (C/2)
(C) 8 cos (A/2), 8 cos (B/2), 8 cos (C/2)
(D) 4 sin A, 4 sin B , 4 sin C
Passage # 2 (Ques. 18 to 20)
When letters of a word are written in all possible
orders, these words are written as in a dictionary
then the position at which the given word occurs is
called its rank.
Q.18
f=kHkqt DEF ds dks.k gksaxs-
'kCn PATNA ,oa TAPAN cukus ds fy, v{kjksa A, A,
P, T, N dks lHkh lEHko rjhdksa ls O;ofLFkr fd;k
tkrk gS] rc buds e/; vkus okys 'kCnksa dh la[;k gS :
(A) 11
(C) 9
DANGER rFkk GARDEN ds e/; 'kCnksa dh la[;k gS:
(A) 243
(C) 240
(B) 241
(D) 244
(B) 10
(D) 13
(B) 241
(D) 244
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Page # 19
Q.20
The number of words before the word SUCCESS
in the dictionary is :
(A) 331
(C) 332
Q.20
(A) 331
(C) 332
(B) 330
(D) 333
2n loga f(x) = loga g(x), a > 0, a ≠1, n ∈ N ds :Ik dh
2n loga f(x) = loga g(x), a > 0, a ≠1, n ∈ N is
,d lehdj.k] fudk;
f(x) > 0 rFkk f(x)2n = g(x)
equivalent to the system f(x) > 0 & f(x)2n = g(x)
The number of solutions of
log (2x) = 2 log (4x –15) is
(A) 1
(B) 2
(C) 3
Q.22
Q.21
Q.22
log (8 –10x –12x2) = 3 log (2x –1) is
(A) {1}
(B) {3, 2}
Q.23
1 1
(C)  , 
2 3
(B) 2
Q.23
1
(D)  ( 5 + 7 ) 
2

(C) 3
(D) vuUr
lehdj.k log (8 –10x –12x2) = 3 log (2x –1) dk gy
leqPp; gS(A) {1}
(C) {5}
(D) φ
Solution set of the equation
1
log x = log (x + 1) is
2
1
1

(A)  ( 5 − 1)
(B)  ( 5 + 1)
2

2

log (2x) = 2 log (4x –15) ds gyksa dh la[;k gS(A) 1
(D) Infinity
Solution set of the equation
(C) {5}
(B) 330
(D) 333
x|ka'k # 3 (iz- 21 ls 23)
Passage # 3 (Ques. 21 to 23)
An equation of the form
Q.21
'kCn SUCCESS ds igys vkus okys 'kCnksa dh la[;k
gS :
(B) {3, 2}
(D) φ
lehdj.k log x = 1 log (x + 1) dk gy leqPp; gS2
1
(A)  ( 5 − 1) 
2

1
(B)  ( 5 + 1)
2

1
(C)  ,
2
1
(D)  ( 5 + 7 ) 
2

1

3
Space for rough work
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Page # 20
Section - I
PHYSICS
[k.M - I
Questions 1 to 6 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.
Q.1
Q.2
A body is displaced from (0, 0) to (1m, 1m) along
r
the path x = y by a force F = ( x 2 ˆj + yî ) N . The
work done by this force will be –
4
5
J
J
(B) 6
(A) 3
7
3
J
J
(D) 5
(C) 2
A particle is given an initial speed u inside a
smooth spherical shell of radius R = 1 m such that
it is just able to complete the circle. Acceleration
of the particle when its velocity is vertical is -
iz'u 1 ls 6 rd cgqfodYih iz'u gSaA 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, + 3 vad fn;s
tk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA
Q.1
,d fi.M dks (0, 0) ls (1m, 1m) rd iFk x = y ds
r
vuqfn'k cy F = ( x 2 ˆj + yî ) N }kjk foLFkkfir fd;k
tkrk gSA bl cy }kjk fd;k x;k dk;Z gksxk –
4
J
(A) 3
3
J
(C) 2
Q.2
,d d.k dks R = 1 m f=kT;k dh ,d fpduh
xksykdkj dks'k ds vUnj izkjfEHkd osx u bl izdkj
fn;k x;k gS fd og oÙ̀k iwjk djus esa Bhd l{ke gksA
d.k dk Roj.k tc mldk osx Å/okZ/kj gS] gSA
R
R
u
u
(A) g 10
(C) g
2
5
J
(B) 6
7
J
(D) 5
(A) g 10
(C) g 2
(B) g
(D) 3g
(B) g
(D) 3g
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Page # 21
Q. 3
Two blocks A and B of mass m and 2m are
connected by a massless spring of force constant k.
They are placed on a smooth horizontal plane.
Spring is stretched by an amount x and then
released. The relative velocity of the blocks when
the spring comes to its natural length is –
(C)
2kx
m
m o 2m nzO;eku ds nks CykWd A o B dks k cy
fu;rkad dh nzO;ekughu fLizax ls tksM+k x;k gSA os
,d ?k"kZ.kghu {kSfrt ry ij j[ks gq;s gSA fLiazx dks x
ek=kk [khapdj NksM+k tkrk gSA CykWdksa dk lkisf{kd osx
tc fLiazx viuh izkjfEHkd yEckbZ esa vkrh gS] gS–
 3k 

x
 2m 


(A)
 2k 

x
 3m 


(B)
(D)
B
A
B
A
 3k 

x
 2m 


(A)
Q. 3
3km
2x
(C)
2kx
m
 2k 

x
 3m 


(B)
(D)
3km
2x
Q.4
Q.4
Water
Water
The volume of brick is 2.197 litres. The submerged
brick is balanced by a 2.54 kg mass on the beam scale.
The weight of the brick approximately is – (arm length
are equal, g = 10ms–2 , ρw =103 kg/m3 )
(A) 46 N
(B) 50 N
bZV dk vk;ru 2.197 yhVj gSA Mwch gqbZ bZaV che
iSekus ij 2.54 kg nzO;eku ls lUrqfyr gksrh gSA bZV
dk Hkkj yXkHkx gS– (Hkqtk dh yEckbZ leku gS,
(C) 56 N
(C) 56 N
(D) 72 N
g = 10ms–2 , ρw =103 kg/m3 )
(A) 46 N
(B) 50 N
(D) 72 N
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Page # 22
Q.5
Two organ pipes,each closed at one end, gives 5
Q.5
their fundamental frequencies (in Hz) are -
nks vkWxZu ikbi] izR;sd ,d fljs ij cUn gS] izfr
lsd.M 5 foLiUn nsrs gSa tc nksuksa ewy Loj mRlftZr
djrs gSA ;fn mudh yEckbZ;ksa dk vuqikr 50 : 51 gS]
rks ewy vkof̀Ùk;k¡ (Hzesa) gS –
(A) 255, 250
(B) 255, 260
(A) 255, 250
(B) 255, 260
(C) 260, 265
(D) 265, 270
(C) 260, 265
(D) 265, 270
beats per sec, when emitting their fundamental
notes. If their lengths are in the ratio of 50 : 51,
Q.6
If the balance length corresponding to points B and
Q.6
C is 40 cm on the potentiometer wire. The balance
length corresponding to points C and D is –
;fn foHkoekih rkj ij fcUnqvksa B o C dh lacaf/kr
lUrqyu yEckbZ 40 cm gSA rks fcUnqvksa C rFkk D dh
lEcfU/kr lUrqfyr yEckbZ gS –
()
k
()
k
x
10Ω
B
10Ω
x
y
J
G
4Ω
C
10Ω
B
D
10Ω
6V 1Ω
(A) 25 cm
(B) 32 cm
(C) 40 cm
(D) 64 cm
Questions 7 to 10 are multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which MULTIPLE (ONE OR MORE) 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 NO NEGATIVE marks for wrong
answer.
y
J
G
4Ω
C
D
6V 1Ω
(A) 25 cm
(B) 32 cm
(C) 40 cm
(D) 64 cm
iz'u 7 ls 10 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj
fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k vf/kd
fodYi lgh gaSA 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 xyr mÙkj ds fy, dksbZ _.kkRed vadu
ugha gSA
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Page # 23
Q.7
Two particles of same mass and charge are thrown
Q.7
simultaneously in the same direction along the
horizontal with same velocity v from two different
heights h1 and h2 (h1 < h2). Initially they were
izkjEHk esa os /kjkry ds Åij leku Å/okZ/kj js[kk ij
fLFkr gSA xyr dFku@dFkuksa dk p;u dhft,–
(A) /kjkry ij Vdjkus ds igys nksuksa d.k ges'kk
Å/okZ/kj js[kk ij jgsxsa
(B) nksuksa /kjkry rd igqapus esa leku le; ysxsa
(C) h1 ij fLFkr d.k dk {kSfrt foLFkkiu h2 ij fLFkr
d.k ds {kSfrt foLFkkiu ls de gksrk gS
(D) fdlh Hkh {k.k nzO;eku dsUnz dk Roj.k uhps dh
vksj g gksxk
located on the same vertical line above ground.
Now Choose the wrong statement (s) –
(A) Both the particle will always lie on a vertical
line before hitting the ground.
(B) Both will take same time to reach ground
(C) Horizontal displacement of the particle lying at
h1 is less than that of the particle at h2 .
(D) At any moment acceleration of the centre of
mass will be g downwards
Q.8
In the circuit shown in figure C1 = C2 = 2µF. Then
charge stored in (steady state) 1Ω 2Ω 3Ω
C1
leku nzO;eku rFkk vkos'k ds nks d.kksa dks ,dlkFk
{kSfrt ds vuqfn'k leku fn'kk esa leku osx v ls nks
fHkUu Å¡pkbZ;ksa h1 o h2 (h1 < h2) ls QSadk tkrk gSA
Q.8
fp=k esa n'kkZ;s ifjiFk esa C1 = C2 = 2µF gSA rc lafpr
vkos'k (LFkkbZ voLFkk) 1Ω
2Ω 3Ω
C1
C2
C2
2Ω 1Ω 3Ω
2Ω 1Ω 3Ω
120V
(A) C1 la/kkfj=k esa 'kwU; gS
(B) la/kkfj=k C2 esa 'kwU; gS
(C) nksuksa la/kkfj=kksa esa 'kwU; gS
(D) la/kkfj=k C1 esa 40 µC gS
120V
(A) capacitor C1 is zero
(B) capacitor C2 is zero
(C) both capacitors is zero
(D) capacitor C1 is 40 µC
Space for rough work
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Page # 24
Q.9
A rectangular wire frame rotates with a constant
angular velocity around one of its sides parallel to
a current carrying & rectilinear conductor nearly
as shown in diagram –
i
O
Q.9
O
(A) When rectangular wire frame is in the plane
passing through rectilinear conductor flux
linked through rectangular wire frame is
minimum
(B) When rectangular wire frame is in the plane
passing through rectilinear conductor emf
induced in rectangular wire frame is
minimum
(C) When rectangular wire frame is in plane
perpendicular to the plane passing through
conductor then, emf is maximum
(D) When rectangular wire frame is in plane
perpendicular to the plane passing through
conductor then, flux is minimum
,d vk;rkdkj rkj dk Ýse fu;r dks.kh; osx ls ,d
/kkjkokgh rFkk js[kh; pkyd ds fudV bldh fdlh ,d
Hkqtk dks lekUrj j[krs gq, fp=kkuqlkj ?kwerk gS –
i
O
O
(A) tc vk;rkdkj rkj dk Ýse js[kh; pkyd ds
ry esa j[kk gks rks vk;rkdkj rkj ds Ýse ls
lEc) ¶yDl U;wure gksrk gS
(B) tc vk;rkdkj rkj dk Ýse js[kh; pkyd ds
ry esa j[kk gks rks vk;rkdkj rkj ds Ýse esa
iszfjr fo|qr okgd cy U;wure gksrk gS
(C) tc vk;rkdkj rkj dk Ýse pkyd ls xqtjus
okys ry ds yEcor~ ry esa fLFkr gS] rc fo|qr
okgd cy vf/kdre gksrk gS
(D) tc vk;rkdkj rkj dk Ýse pkyd ls xqtjus
okys ry ds yEcor~ ry esa fLFkr gS] rc ¶YkDl
U;wure gksrk gS
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Page # 25
Q.10
In radioactivity decay according to law N = N0e–λt
which of the following is/are true ?
(A) Probability that a nucleus will decay is 1 – e–λt
(B) Probability that a nucleus will decay four half
lives is 15/16
(C) Fraction nuclei that will remain after two half
lives is zero
(D) Fraction of nuclei that will remain after two
half-lives is ¼
This section contains 4 questions numbered 11 to 14,
(Reason and Assertion type question). Each question
contains Assertion and Reason. 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. + 3 marks will be
given for each correct answer and – 1 mark for each
wrong answer.
(A)
(B)
(C)
(D)
The following questions given below consist of
an "Assertion" (A) and "Reason" (R) Type
questions. Use the following Key to choose the
appropriate answer.
If both (A) and (R) are true, and (R) is the
correct explanation of (A).
If both (A) and (R) are true but (R) is not the
correct explanation of (A).
If (A) is true but (R) is false.
If (A) is false but (R) is true.
fu;e N = N0e–λt ds vuqlkj jsfM;kslfØ; {k; esas
fuEu esa ls dkSulk dFku lgh gS ?
(A) ukfHkd ds {k; gksus dh izkf;drk 1 – e–λt gS
(B) izkf;drk ftlls ukfHkd pkj v)Z&vk;q {kf;r
gksxk 15/16 gS
(C) nks v)Z vk;qvksa ds ckn 'ks"k cps ukfHkd ds d.kksa
dk va'k 'kwU; gS
(D) nks v)Z vk;qvksa ds ckn 'ks"k cps ukfHkd ds d.kksa
dk va'k ¼ gS
Q.10
bl [k.M esa 11 ls 14 rd 4 iz'u gSa, (dFku rFkk dkj.k
izdkj ds iz'u)A izR;sd iz'u esa dFku rFkk dkj.k fn;s x;s
gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) fn;s x;s
gSa] ftuesa ls dsoy ,d 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 + 3 vad fn;s tk;asxs rFkk izR;sd xyr
mÙkj ds fy, 1 vad ?kVk;k tk;sxkA
uhps fn;s x;s fuEufyf[kr iz'u "dFku" (A) rFkk
"dkj.k" (R) izdkj ds iz'u gSaA vr% mfpr mÙkj
dk p;u djus ds fy;s fuEu rkfydk dk mi;ksx
dhft;sA
(A) ;fn (A) rFkk (R) nksuksa lR; gSa rFkk (R), (A) dk
lgh Li"Vhdj.k gSA
(B) ;fn (A) rFkk (R) nksuksa lR; gSa rFkk (R), (A) dk
lgh Li"Vhdj.k ugha gSA
(C) ;fn (A) lR; gS ysfdu (R) vlR; gSA
(D) ;fn (A) vlR; gS ysfdu (R) lR; gSA
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Page # 26
Q.11
A homogeneous cylinder is floating in liquid as
shown.
Assertion : The cylinder can perform both
translational and rotational simple harmonic
oscillations, for small displacement.
Q.11
dFku : csyu vYi foLFkkiu ds fy, LFkkukUrfj;
rFkk ?kw.kZu nksuksas ljy vkorhZ nksyu dj ldrk gSA
dkj.k : csyu dh LFkkukUrjh; lkE;koLFkk LFkkbZ rFkk
?kw.kZu lkE;koLFkk vLFkkbZ gksrh gS
Reason : Translatory equilibrium of cylinder in
stable and rotational equilibrium is unstable.
Q.12
Q.13
,d laekxh csyu n'kkZ;s vuqlkj nzo esa rSj jgk gSA
Assertion : At height h from ground and at depth
h below ground, where h is approximately equal to
0.62 R, the value of g acceleration due to gravity is
same.
Reason : Value of g decreases both sides, in going
up and down.
Q.12
Assertion : Light from an object falls on a
concave mirror forming a real image of the object.
If both the object and mirror are immersed in
water, there is no change in position of the image.
Reason : The formation of image by reflection
does not depend on surrounding medium, so there
is no change in position of image.
Q.13
dFku : /kjkry ls h Å¡pkbZ rFkk /kjkry ls h xgjkbZ
ij] tgk¡ h yxHkx 0.62 R ds cjkcj gS] xq:Ro ds
dkj.k Roj.k g dk eku leku gksrk gS
dkj.k : Åij rFkk uhps nksuksa vksj g dk eku ?kVrk gS
dFku : ,d oLrq ls izdk'k vory niZ.k ij fxjdj
oLrq dk okLrfod izfrfcEc cukrk gSA ;fn oLrq rFkk
niZ.k nksuksa dks ikuh esa Mwck;k tk;s] rks izfrfcEc dh
fLFkfr esa dksbZ ifjorZu ugha gksrk gSA
dkj.k : ijkorZu }kjk izfrfcEc dk cuuk ifjos'k
ek/;e ij fuHkZj ugha djrk gS] vr% izfrfcEc dh
fLFkfr esa dksbZ ifjorZu ugha gksrk gS
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Page # 27
Q.14
Assertion : In Young's double slit experiment, we
observe an interference pattern on the screen if
both the slits are illuminated by two bulbs of same
power.
Reason : The interference pattern is observed
when source is monochromatic and coherent.
This section contains 3 paragraphs, each has 3 multiple
choice questions. (Questions 15 to 23) 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)
A stationary truck has its rear door wide open as
shown. At t = 0 the truck starts to accelerate with
constant acceleration a then the door will begin to
close and at any time 't' the door makes angle θ with its
original orientation. Assume that the door has mass m
uniformly distributed along its length L.
t-axis
L Door
Hinge
ω
a
θ
Q.14
dFku : : ;ax ds f}&js[kk fNnz iz;ksx esa] ge inZs ij
O;frdj.k izk:i izsf{kr djrs gSa ;fn nksuksa fLyVksa dks
leku 'kfDr ds cYcksa }kjk iznhIr fd;k tkrk gSA
dkj.k : O;frdj.k izk:i rc izsf{kr gksrk gS tc
L=kksr ,do.khZ rFkk dyklEc) gksrs gSa
bl [k.M esa 3 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih
iz'u gSaA (iz'u 15 ls 23) 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,saxs rFkk
izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA
x|ka'k # 1 (iz- 15 ls 17)
,d fLFkj Vªd dk fiNyk njoktk iwjk [kqyk gS tSlk fd
n'kkZ;k x;k gSA t = 0 ij Vªd fu;r Roj.k a ls Rofjr
gksuk izkjEHk djrk gS rc njoktk cUn gksuk 'kq: djrk gS] rFkk
fdlh le; 't' ij njoktk bldh ewy fLFkfr ls dks.k θ cukrk
gSA ;g ekfu;s fd njokts dk nzO;eku m mldh yEckbZ L ds
vuqfn'k ,d leku :i ls forjhr gSA
t-axis
L Door
ω
θ
Ft
Ft
a
Fn
Hinge
a
a
Fn
n-axis
n-axis
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Page # 28
Q.15
Q.16
Q.17
Q.15
The tangential force Ft on door is equal to –
(A)
1
ma cos θ
4
(B)
1
ma sin θ
4
(C)
1
ma cos θ
8
(D) none of these
Q.16
The angular speed ω is equal to –
(A)
3a sin θ
L
(B)
3a cos θ
L
(C)
6a sin θ
L
(D) none of these
If t is total time elapsed from t = 0 to the closing of
Q.17
door then –
(A) t =
(B) t =
L
3a
L
3a
njokts ij Li'khZ; cy Ft cjkcj gS–
(A)
1
ma cos θ
4
(B)
(C)
1
ma cos θ
8
(D) buesa ls dksbZ ugh
dks.kh; pky ω cjkcj gS –
(A)
3a sin θ
L
(B)
(C)
6a sin θ
L
(D) buesa ls dksbZ ugh
3a cos θ
L
;fn t = 0 ls njokts ds cUn gksus rd dqy O;rhr
le; t gS] rc –
π/2
∫
cosecθ dθ
(A) t =
0
L
3a
π
∫
cosecθ dθ
(B) t =
0
π
L
sec θ dθ
3a 0
(D) none of these
(C) t =
1
ma sin θ
4
∫
(C) t =
L
3a
L
3a
π/2
∫
cosecθ dθ
0
π
∫
cosecθ dθ
0
π
∫
sec θ dθ
0
(D) buesa ls dksbZ ugh
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Page # 29
Passage # 2 (Ques. 18 to 20)
A charged metal sheet is placed into uniform
electric field E, perpendicularly to the electric
field lines. After placing the sheet into the field,
the electric field on the left side of the sheet will
be E1 = 5.6 × 105 V/m and on the right it will be
E2 = 3.1 × 105 V/m.
E1
Q.18
Q.19
Q.20
x|ka'k # 2 (iz- 18 ls 20)
/kkrq dh ,d vkosf'kr iV~Vhdk ,dleku fo|qr {ks=k
E ds vUnj] fo|qr cy js[kkvksa ds yEcor~ :i ls fLFkr
gSA iV~Vhdk dks {ks=k ds vUnj j[kus ds ckn] iV~Vhdk ds
ck;ha vksj ij fo|qr {ks=k E1 = 5.6 × 105 V/m gksxk
rFkk nk;ha vksj E2 = 3.1 × 105 V/m gksxk&
E1
E2
Find the total charge of the sheet if a electric force
of 0.08N is exerted on it (A) – 0.32 µC
(B) – 0.64 µC
(C) 0.64 µC
(D) 0.32 µC
Q.18
Find the area of sheet of one side (A) 0.02m2
(B) 0.03 m2
2
(D) 0.05 m2
(C) 0.04 m
Q.19
Find the value of E (A) 2.5 × 104 V/m
(B) 12.5 × 104 V/m
4
(D) 8.7 × 104 V/m
(C) 3.5 × 10 V/m
Q.20
E2
iV~Vhdk dk dqy vkos'k Kkr dhft, ;fn ml ij
0.08N dk fo|qr cy yxk;k tkrk gS (A) – 0.32 µC
(C) 0.64 µC
(B) – 0.64 µC
(D) 0.32 µC
iV~~Vhdk ds ,d rjQ dk {ks=kQy Kkr dhft, (A) 0.02m2
(C) 0.04 m2
(B) 0.03 m2
(D) 0.05 m2
E dk eku Kkr dht, (A) 2.5 × 104 V/m
(C) 3.5 × 104 V/m
(B) 12.5 × 104 V/m
(D) 8.7 × 104 V/m
Space for rough work
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Page # 30
Passage # 3 (Ques. 21 to 23)
The rectangular box shown is the place of a lens. If
y = 10 cm, x = 5 cm and OA = 1 cm are given.
x|ka'k # 3 (iz- 21 ls 23)
n'kkZ;k x;k vk;rkdkj ckWDl ysUl ds LFkku ij j[kk x;k gSA
;fn y = 10 cm, x = 5 cm rFkk OA = 1 cm fn;k x;k gS] rks
I
I
O
O
A
A
B
x
x
y
y
Q.21
Q.23
(A)
(D) 10 cm , converging
The co-ordinate of image (I) are.... (if optical
centre is taken as origin) -
ysUl dh Qksdl nwjh gS 10
cm, vfHklkjh
3
(C) 10 cm, vilkjh
10
10
cm, diverging
cm, Converging (B)
3
3
(C) 10 cm, diverging
Q.22
Q.21
The focal length of the lens is (A)
B
Q.22
10
cm, vilkjh
3
(D) 10 cm, vfHklkjh
(B)
izfrfcEc (I) ds funsZ'kkad ..... gS (;fn izdk'kh; dsUnz
dks ewy fcUnw ds :i esa ysrs gSa) -
2 10
(A)  , 
3 3 
21 10
(B)  + ,+ 
3
 3
2 10
(A)  , 
3 3 
21 10
(B)  + ,+ 
3
 3
(C) (2, + 10)
(D) (10, 2)
(C) (2, + 10)
(D) (10, 2)
The object and image pair is respectively (A) Real and real
(B) Virtual and real
(C) Virtual and virtual
(D) Real and virtual
Q.23
oLrq rFkk izfrfcEc ;qXe Øe'k% gS (A) okLrfod rFkk okLrfod
(B) vkHkklh rFkk okLrfod
(C) vkHkklh rFkk vkHkklh
(D) okLrfod rFkk vkHkklh
Space for rough work
CAREER POINT : CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
Page # 31
MAX MARKS: 246
Time : 3 : 00 Hrs.
INSTRUCTIONS TO CANDIDATE
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
la[;k ds le{k lgh mÙkj fpfUgr dhft,A
2. mRrj ds fy,] OMR vyx ls nh tk jgh gSA
3. ifjoh{kdksa }kjk funsZ'k fn;s tkus ls iwoZ iz'u&i=k iqfLrdk dh lhy dks ugha [kksysaA
SEAL
B.
vadu i)fr:
bl iz'ui=k esa izR;sd fo"k; esa fuEu izdkj ds iz'u gSa:[k.M – I
4. 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
5. 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;saxs rFkk xyr
mÙkj ds fy, dksbZ _.kkRed vadu ugha gSA
6. dFku rFkk dkj.k izdkj ds iz'u] ftuesa ls izR;sd esa dsoy ,d mÙkj lgh gSA izR;sd lgh mÙkj ds fy, 3 vad fn;s tk;saxs rFkk
izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk,xkA
7. 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, tk;saxs rFkk
izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA
C. OMR dh iwfrZ :
8. OMR 'khV ds CykWdksa esa viuk uke] vuqØek¡d] cSp] dkslZ rFkk ijh{kk dk dsUnz Hkjsa rFkk xksyksa dks mi;qDr :i ls dkyk djsaA
9. xksyks dks dkyk djus ds fy, dsoy HB isfUly ;k uhys/dkys isu (tsy isu iz;ksx u djsa) dk iz;ksx djsaA
10. dì;k xksyks dks Hkjrs le; [k.Mks dks lko/kkuh iwoZd ns[k ysa [vFkkZr [k.M I (,dy p;ukRed iz'u] dFku izdkj ds iz'u]
rough
ds iz'ufor
), [k.M
-III work
(iw.kkZd mÙkj izdkj ds iz'u½]
cgqp;ukRed iz'u), [k.M –II (LrEHk lqesyu izdkj Space
Section –I
Section-II
Section-III
For example if only 'A' choice is
correct then, the correct method for
filling the bubbles is
For example if Correct match for (A)
is P; for (B) is R, S; for (C) is Q; for
(D) is P, Q, S then the correct method
for filling the bubbles is
Ensure that all columns are filled.
Answers, having blank column will be treated as
incorrect. Insert leading zeros (s)
A
B
C
D
E
For example if only 'A & C' choices
are correct then, the correct method
for filling the bublles is
A
B
C
D
E
A
B
C
D
P
Q
R
S
T
the wrong method for filling the
bubble are
The answer of the questions in
wrong or any other manner will be
treated as wrong.
'6' should be
filled as 0006
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
'86' should be
filled as 0086
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
'1857' should be
filled as 1857
0 0 0 0
0 0 0 0
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
5 5 5 5
6 6 6 6
7 7 7 7
8 8 8 8
9 9 9 9
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7CAREER POINT : CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
Page # 32