TARGET COURSE FOR IIT-JEE 2011
PHASE- ALL
CHEMISTRY, MATHEMATICS & PHYSICS
TEST NO. 3 [TR-2(I)] (TAKE HOME)
PAPER – I
Date : 13/2/2011
MAX MARKS: 243
Time : 3 : 00 Hrs.
Name : _________________________________________________________ Roll No. : __________________________
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. 3 marks will be awarded for each correct answer and No negative
marking.
6. Passage based single correct type questions. 3 marks will be awarded for each correct answer and –1 mark for each wrong
answer.
Section - III
7. Numerical response (single digit integer answer) questions. 3 marks will be awarded for each correct answer and No
negative marking for wrong answer. Answers to this Section are to be given in the form of single integer only (0 to 9)
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.
9. Use only HB pencil or blue/black pen (avoid gel pen) for darking 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
the wrong method for filling the
bubble are
The answer of the questions in
wrong or any other manner will be
treated as wrong.
1
A
B
C
D
P
Q
R
S
T
'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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SEAL
A.GENERAL :
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 # 1
CHEMISTRY
Section – I
[k.M - I
Questions 1 to 8 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 8 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 dkVk tk;sxkA
Q.1
If in the fermentation of sugar in an enzymatic
solution that is 0.12 M, the concentration of the
sugar is reduced to 0.06 M in 10 h and to 0.03 M in
20 h. What is the order of the reaction ?
(A) 1
(B) 2
(C) 3
(D) 0
Q.1
;fn 'kdZjk ds fd.ou esa ,Utkbeh; foy;u esa tksfd 0.12
M gSA 'kdZjk dh lkUnzrk 10 ?k.Vs esa 0.06 M rFkk 20 ?k.Vs esa
0.03 M cp tkrh gSA vfHkfØ;k dh dksfV D;k gS ?
(A) 1
(B) 2
(C) 3
(D) 0
Q.2
Some graph are sketch for the reaction A → B
(assuming different orders). Where 'α' represent the
degree of dissociation -
Q.2
A → B vfHkfØ;k dks fHkUu dksfV;ks dk ekurs gq, dqN
vkjs[k [khpsa tkrs gSA tgk¡ 'α' fo;kstu dh ek=kk dks
(B) [a]
(A)[a]
t
(A)[a]
(B) [a]
t
t
1
(C) [a]
n'kkZrk gS
t
1
(C) [a]
1
1
t
t
The order of reaction are respectively (A) 0, 1, 2
(B) 1, 0, 2
(C) 2, 0, 1
(D) 1, 2, 0
vfHkfØ;k dh dksfV Øe'k% gS &
(A) 0, 1, 2
(C) 2, 0, 1
(B) 1, 0, 2
(D) 1, 2, 0
Space for rough work
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Page # 1
Q.3
Q.4
A crystal is made of particle X, Y, & Z. X forms fcc
packing. Y occupies all octahedral voids of X and Z
occupies all tetrahedral voids of X, if all the
pariticles along on body diagonal are removed then
the formula of the crystal would be (A) XYZ2
(B) X2YZ2
(D) X5Y4Z8
(C) X8Y4Z5
Q.3
If the rms velocity of nitrogen and oxygen
molecule are same at two different temperature and
same pressure then pick the wrong statement (A) Average speed of molecules is also same
Q.4
;fn ukbVªkstu rFkk vkWDlhtu v.kq dk oxZ ek/; ewy
osx] nks fHkUu rki rFkk leku nkc ij leku gks] rks
xyr dFku dk pquko dhft, &
(A) v.kqvksa dh vkSlr pky Hkh leku jgrh gS
(B) ukbVªkstu rFkk vkWDlhtu dk ?kuRo (gm/lt) Hkh
leku jgrk gS
(C) izR;sd xSl ds eksyks dh la[;k Hkh leku jgrh gSA
(D) v.kqvksa dk lokZf/kd laHkkfor osx Hkh leku jgrk gSA
Q5
,fFky DyksjkbM (C2H5Cl), ,fFkyhu dh gkbMªkstu
DyksjkbM ds lkFk vfHkfØ;k }kjk fufeZr fd;k tkrk gS
prq"Qydh; fjfDr;ksa dks xzg.k djrk gSA ;fn dk;
fod.kZ ds vuqfn'k lHkh d.kksa dks gVk fn;k tk, rks
fØLVy dk lw=k gksxk &
(A) XYZ2
(C) X8Y4Z5
(B) Density (gm/lt) of nitrogen and oxygen is also
equal
(C) Number of moles of each gas is also equal
(D) most probable velocity of molecules is also
equal
Q.5
Ethyl chloride (C2H5Cl), is prepared by reaction of
ethylene with hydrogen chloride
X, Y, rFkk Z d.kksa ls feydj ,d fØLVy fufeZr gksrk
gSA X, fcc ladqy cukrk gSA Y, X dh lHkh v"VQydh;
fjfDr;ksa dks xzg.k djrk gS rFkk Z, X dh lHkh
C2H4(g) + HCl (g) → C2H5Cl(g) ; ∆H = – 72.3 kJ
What is the value of ∆E (in kJ), if 70 g of ethylene
and 73 g of HCl are allowed to react at 300 K (A) – 69.8
(B) –180.75
(C) –174.5
(D) –139.6
(B) X2YZ2
(D) X5Y4Z8
C2H4(g) + HCl (g) → C2H5Cl(g) ; ∆H = – 72.3 kJ
∆E (kJ esa) dk eku D;k gksxk \ ;fn 300 K ij
,fFkyhu ds 70 g rFkk HCl ds 73 g fØ;k djrs gS (A) – 69.8
(C) –174.5
(B) –180.75
(D) –139.6
Space for rough work
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Page # 2
Q.6
Q.7
What is the equivalent weight of H2SO4 in the
reaction ?
H2SO4 + NaI → Na2SO4 + I2 + H2S + H2O
(A) 12.25
(B) 49
(C) 61.25
(D) None of these
Q.6
For the reaction
Q.7
H+
+
H2SO4 + NaI → Na2SO4 + I2 + H2S + H2O
(A) 12.25
(B) 49
(C) 61.25
Select correct statements :
(A) From the following reaction
∆H = q1
CO(g) + O2(g) → CO2(g),
heat of formation of CO2 (g) is q1
(B) From the following reaction
1
C(graphite) + O2(g) → CO(g) ∆H = q2
2
heat of combustion of carbon is q2
(C) From the above reactions, heat of combustion
CO2 is q1 + q2
(D) From the above reactions, heat of combustion
of CO(g) is q1 and that of carbon is q1 + q2
(D) buesa ls dksbZ ugha
vfHkfØ;k
+
–
CH3COCH3 + Br2 H→ CH3COCH2Br + H+ + Br–
ds fy, fuEu vkadM+s ,df=kr fd, x, [,flVksu]
[Br2]
[H+] vfHkfØ;k dh nj
0.15
0.025
0.025 6 × 10–4 (Ms–1)
0.15
0.050
0.025 6 × 10–4 (Ms–1)
0.20
0.025
0.025 8.0 × 10–4 (Ms–1)
0.15
0.025
0.050 12 × 10–4 (Ms–1)
CH3COCH3 rFkk Br2 ds lUnHkZ eas vfHkfØ;k dh dksfV gS (A) 0, 1
(B) 1, 0 (C) 1, 1
(D) 1, 2
CH3COCH3 + Br2 → CH3COCH2Br + H + Br
the following data was collected [Acetone] [Br2]
[H+] Rate of reaction
0.15
0.025
0.025 6 × 10–4 (Ms–1)
0.15
0.050
0.025 6 × 10–4 (Ms–1)
0.20
0.025
0.025 8.0 × 10–4 (Ms–1)
0.15
0.025
0.050 12 × 10–4 (Ms–1)
The order of the reaction with respect to
CH3COCH3 and Br2 respectively are (A) 0, 1
(B) 1, 0 (C) 1, 1
(D) 1, 2
Q.8
fuEu vfHkfØ;k esa H2SO4 dk rqY;kadh Hkkj D;k gS ?
Q.8
lgh dFku gS :
(A) fuEu vfHkfØ;k ls
∆H = q1
CO(g) + O2(g) → CO2(g),
CO2 (g) ds fuekZ.k dh Å"ek q1 gSA
(B) fuEu vfHkfØ;k esa]
1
C(xzsQkbV) + O2(g) → CO(g) ∆H = q2
2
dkcZu ds ngu dh Å"ek q2 gS
(C) mijksDr vfHkfØ;k esa CO2 ds ngu dh Å"ek
q1 + q2 gSA
(D) mijksDr vfHkfØ;kvksa esa CO(g) ds ngu dh Å"ek
q1 rFkk dkcZu dh q1 + q2 gSA
Space for rough work
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Page # 3
Questions 9 to 13 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. + 3 marks will be given for each
correct answer and no negative marks.
iz'u 9 ls 13 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj
fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k ,d ls
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,
+ 3 vad fn;s tk;saxs rFkk dksbZ _.kkRed vadu ugha gSA
Q.9
Q.9
0.1 M solution of KI reacts with excess of H2SO4
and KIO3 solution, according to equation
+
5I– + IO3– + 6H → 3I2 + 3H2O ;
vf/kD; ls fØ;k djrk gS] fuEu lehdj.k ds vuqlkj
+
5I– + IO3– + 6H → 3I2 + 3H2O ;
which of the following statement is correct (A) 200 ml of the KI solution react with 0.004 mole
KIO3
(B) 100 ml of the KI solution reacts with 0.006
mole of H2SO4
(C) 0.5 litre of the KI solution produced 0.005 mole
of I2
(D) Equivalent weight of KIO3 is equal to
Q.10
 molecular Weight 


5


3 moles of the gas C2H6 is mixed with 60 gm of this
gas and 2.4 × 1024 molecules of the gas is removed.
The left over gas is combusted in the presence of
excess oxygen then
(NA = 6 × 1023) (Density of water = 1 gm/ml)
(A) 2 moles of C2H6 left for combustion
(B) Volume of CO2 at S.T.P. produced after
combustion 44.8 litre
(C) Volume of water produced is 54 ml
(D) None
KI dk 0.1 M foy;u] H2SO4 rFkk KIO3 foy;u ds
fuEu esa dkSulk dFku lgh gS &
(A) KI foy;u dk 200 ml, KIO3 ds 0.004 eksy ls
fØ;k djrk gS
(B) KI foy;u dk 100 ml , H2SO4 ds 0.006 eksy ls
fØ;k djrk gS
(C) KI foy;u dk 0.5 yhVj, I 2 ds 0.005 eksy
cukrk gS
 v.kqHkkj 
(D) KIO3 dk rqY;kadh Hkkj; 
 ds cjkcj gksrk gS
 5 
Q.10
C2H6 xSl ds 3 eksyks dks blh xSl ds 60 gm esa
feyk;k tkrk gS rFkk bl xSl ds 2.4 × 1024 v.kqvksa
dks fudky fn;k tkrk gSA 'ks"k cph xSl dk vkWDlhtu
ds vf/kD; esa ngu fd;k tkrk gS] rks &
(NA = 6 × 1023) (ty dk /kuRo = 1 gm/ml)
(A) ngu gsrq C2H6 ds 2 eksy cprs gS
(B) S.T.P ij 44.8 yhVj ds ngu ds i'pkr CO2 dk
vk;ru curk gSA
(C) fufeZr ty dk vk;ru 54 ml gS
(D) dksbZ ugha
Space for rough work
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Page # 4
Q.11
One mole of ideal diatomic gas (CV = 5 cal) was
Q.11
calories as unit of energy and Kelvin for temp)
izkjEHk esa 25°C rFkk 1 L dh voLFkk ls ,d eksy
vkn'kZ f)ijekf.od xSl (CV = 5 cal) dks 100°C rFkk
vk;ru 10 L dh voLFkk esa :ikUrfjr fd;k x;kA vc
bl izØe ds fy, (R = 2 calories/mol/K) (ÅtkZ dh
bdkbZ ds :i esa dSyksjh rFkk rki ds fy, dsfYou yhft,)
(A) ∆H = 525
(A) ∆H = 525
transformed from initial 25°C and 1 L to the state
when temperature is 100°C and volume 10 L. Then
for this process (R = 2 calories/mol/K) (take
(B) ∆S = 5 ln
373
+ 2ln 10
298
(B) ∆S = 5 ln
(C) ∆E = 525
(D) ∆G of the process can not be calculated using
given information
Q.12
(C) ∆E = 525
(D) nh
Q.12
Choose the correct statements :
(A) temperature, enthalpy and entropy are state
functions
(B) for reversible and irreversible both isothermal
expansion of an ideal gas, change in internal
energy and enthalpy is zero
(C) state function are exact differential
Which is correct match for no. of radial node (A) 3 s, 2
(B) 2 p, 0
(C) 4 d, 1
(D) 4 p, 2
xbZ tkudkjh ls ∆G Kkr ugha fd;k tk ldrk
lR; dFku pqfu, :
(A) rki, ,UFkSYih o ,.VªkWih voLFkk Qyu gksrs gS
(B) vkn'kZ xSl ds mRØe.kh; o vuqRØe.kh; çlkj nksuksa
ds fy,] vkUrfjd ÅtkZ] ,UFkSYih esa ifjorZu 'kwU;
gksrk gS
(C) voLFkk Qyu ;FkkFkZ vodyt gksrs gS
(D) pØh;
(D) for cyclic process ∆G = 0
Q.13
373
+ 2ln10
298
Q.13
çØe ds fy, ∆G = 0 gksrk gS
fuEu esa ls fdlesa jsfM;y uksMksa dh la[;k lgh
lqesfyr gS (A) 3 s, 2
(C) 4 d, 1
(B) 2 p, 0
(D) 4 p, 2
Space for rough work
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Page # 5
This section contains 2 paragraphs; passage- I has 2
bl [k.M esa 2 x|ka'k fn;s x;s gSa] x|ka'k -I esa 2 cgqfodYih
multiple choice questions (No. 14 & 15) and passage- II
iz'u (iz'u 14 o 15) gSa rFkk x|ka'k-II esa 3 cgqfodYih iz'u
has 3 multiple (No. 16 to 18). 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
(iz'u 16 ls 18) gSaA izR;sd iz'u ds pkj fodYi (A), (B), (C)
rFkk (D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA OMR 'khV
number of that question. + 3 marks will be given for
esa iz'u dh iz'u la[;k ds le{k viuk mÙkj vafdr dhft;sA
each correct answer and – 1 mark for each wrong
izR;sd lgh mÙkj ds fy;s + 3 vad fn;s tk,saxs rFkk izR;sd
answer.
xyr mÙkj ds fy, 1 vad dkVk tk;sxkA
Passage # 1 (Ques. 14 to 15)
x|ka'k # 1 (iz- 14 ,oa 15)
Some amount of ''20V'' H2O2 is mixed with excess
KI
of acidified solution of KI. The iodine so liberated
H2O2
Q.15
The volume of H2O2 solution is (A) 11.2 ml
(B) 37.2 ml
(C) 5.6 ml
(D) 22.4 ml
dh dqN ek=kk feyk;h tkrh gSA vuqekiu ds fy,
bl izdkj fu"dkflr vk;ksMhu dks 0.1 N Na2S2O3 ds
required 200 ml of 0.1 N Na2S2O3 for titration.
Q.14
ds vEyh;dr̀ foy;u ds vkf/kD; ds lkFk ''20V''
200 ml vko';d
Q.14
H2O2 foy;u
gksrs gSA
dk vk;ru gS -
(A) 11.2 ml
(C) 5.6 ml
The mass of K2Cr2O7 needed to oxidise the above
Q.15
H2O2 ds
(B) 37.2 ml
(D) 22.4 ml
mijksDr foy;u ds vk;ru dks vkWDlhdr̀ djus
volume of H2O2 solution-
gsrq K2Cr2O7 dk vko';d Hkkj gS -
(Atomic mass of Cr = 52)
(Cr dk
(A) 3.6 g (B) 0.8 g (C) 4.2 g (D) 0.98 g
(A) 3.6 g (B) 0.8 g (C) 4.2 g (D) 0.98 g
ijek.kq Hkkj 52)
Space for rough work
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Page # 6
Passage # 2 (Ques. 16 & 18)
x|ka'k # 2 (iz- 16 ,oa 18)
Grahm's Law of Diffusion : The phenomenons of
spontaneous intermixing of gases against the law of
gravitation is known as diffusion. If diffusion
occurs through small orifice of the container, then it
is known as effusion.
The rate of effusion is defined as, r =
PA
2πRTM
folj.k ds fy, xzkg~e dk fu;e: xq:Rokd"kZ.k ds
fo:) og izfØ;k ftlesa xSlksa dk lrr~ :i ls feJ.k
gksrk jgrk gS] folj.k dgykrh gSA ;fn ik=k ds NksVs
fNnz ls folj.k laiUu gksrk gS] rks ;g fu%lj.k
dgykrk gSA
,
fu%lj.k dh nj dks] r =
where P is partial pressure of the gas
A is area of cross-section of the orifice of the
container and M is the molar mass of the container.
Rate of diffusion =
=
Q.16
2πRTM
}kjk ifjHkkf"kr
fd;k tkrk gS] tgk¡ P, xSl dk vkaf'kd nkc gSA
A ik=k ds fNnz dk vuqizLFk dkV dk {ks=kQy gS rFkk
M ik=k dk eksyj Hkkj gSA
folj.k dh nj = fu%lfjr vk;ru = fo%lfjr eksy
le;
le;
nkc voueu r; nwjh
=
=
t
le;
Vol. effused Moles effused
=
Time
Time
Pr essure drop dis tance travelled
=
t
time
1 mole of gas A and 4 moles of gas O2 is taken
inside the vessel, which effuse through the small
orifice of the vessel having same area of crosssection and at the same temperature, then which the
correct % of effused volume of gas A and O2
initially respectively ? (Assume that the gas A does
not react with O2 gas and molar mass of gas A is 2 g)
(A) 50%, 50%
(B) 60%, 40%
(C) 30%, 70%
(D) 10%, 90%
PA
Q.16
,d ik=k esa A xSl ds 1 eksy rFkk O2 xSl ds 4 eksy
fy, tkrs gS] ;fn vuqizLFk dkV ds leku {ks=kQy rFkk
leku rki ij ik=k ds NksVs fNnz ls xSlsa fu%lfjr gksrh
gS] rks izkjEHk eas xSl A rFkk O2 ds fu%lfjr vk;ru dk
lgh izfr'kr D;k gS ? (dYiuk djsa fd xSl A rFkk xSl O2
ijLij fØ;k ugha djrh rFkk A dks eksyj Hkkj 2 g gS)
(A) 50%, 50%
(C) 30%, 70%
(B) 60%, 40%
(D) 10%, 90%
Space for rough work
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Page # 7
Q.17
Which of the following is correct for diffusion ?
Q.17
(A) ∆G must be positive
(A) ∆G /kukRed
(B) ∆H must be negative
(C) ∆S /kukRed
(D) ∆S must be negative
Q.18
from the end "P" and "Q" of the cylinder
respectively. The cylinder has the area of crosssection A, shown as under
gksuk pkfg,
,d csyukdkj ik=k ds "P" rFkk "Q" fljksa ls xSl "X"
ds 1 eksy rFkk xSl "Y" ds 2 eksy izos'k djrs gSA
csyukdkj ik=k ds vuqizLFk dkV dk {ks=kQy A gS] tSls
fuEu fp=k esa n'kkZ;k gS
A
X
gksuk pkfg,
gksuk pkfg,
(D) ∆S _.kkRed
1 mole of gas "X" and 2 moles of gas "Y" enters
P
gksuk pkfg,
(B) ∆H _.kkRed
(C) ∆S must be positive
Q.18
fuEu esa dkSulk folj.k ds fy, gS ?
Q
Y
A
P
Q
X
Y
The length of the cylinder is 150 cms. The gas "X"
intermixes with gas "Y" at the point A. If the
bl csyukdkj ik=k dh yackbZ 150 cms gSA fcUnw A ij
molecular weight of the gases X and Y is 20 and 80
xSl "X" rFkk xSl "Y" ijLij vifefJr gksrh gSA
respectively, then what will distance of point A
from Q ?
;fn xSls X rFkk Y dk v.kqHkkj Øe'k% 20 o 80 gks] rks
(A) 75cms
(B) 50 cms
A dh Q ls
nwjh D;k gksxh ?
(C) 100 cms
(D) 90 cms
(A) 75cms
(B) 50 cms
(C) 100 cms
(D) 90 cms
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Page # 8
Section - III
[k.M - III
This section contains 9 questions (Q.1 to 9).
+3 marks will be given for each correct answer and no
negative marking. The answer to each of the questions is
a SINGLE-DIGIT INTEGER, ranging from 0 to 9. The
appropriate bubbles below the respective question
numbers in the OMR have to be darkened. For example,
if the correct answers to question numbers X, Y, Z and
W (say) are 6, 0, 9 and 2, respectively, then the correct
darkening of bubbles will look like the following :
X Y Z W
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
bl [k.M esa 9 (iz-1 ls 9) iz'u gaSA izR;sd lgh mÙkj ds fy;s
+3 vad fn;s tk,saxs rFkk dksbZ _.kkRed vadu ugha gSA bl
[k.M esa izR;sd iz'u dk mÙkj 0 ls 9 rd bdkbZ ds ,d iw.kk±d
gSaA OMR esa iz'u la[;k ds laxr uhps fn;s x;s cqYyksa esa ls
lgh mÙkj okys cqYyksa dks dkyk fd;k tkuk gSA mnkgj.k ds
fy, ;fn iz'u la[;k ¼ekusa½ X, Y, Z rFkk W ds mÙkj 6, 0, 9
rFkk 2 gkas, rks lgh fof/k ls dkys fd;s x;s cqYys ,sls fn[krs gSa
tks fuEufyf[kr gSA
Q.1
Q.1
A near ultraviolet photon of 300 nm is absorbed by
X
0
1
2
3
4
5
6
7
8
9
300 nm
Y
0
1
2
3
4
5
6
7
8
9
Z
0
1
2
3
4
5
6
7
8
9
W
0
1
2
3
4
5
6
7
8
9
dk ,d fudVre ijkcsxa yh QksVkWu ,d xSl
a gas and then re-emitted as two photons. One
}kjk vo'kksf"kr gksrk gS rFkk rc ;g nks QksVkWuksa ds :i
photon is red with wavelength 760 nm. What would
esa iquZ mRlftZr gksrk gSA 760 nm rjaxnS/;Z ds lkFk ,d
be
QksVkWu yky gSA f}rh; QksVkWu dh rjaxnS/;Z (nm esa) gksxh ?
the
wavelength
of
the
second
photon
(in nm) ? Given the answer by multiply 10–2
vius mÙkj dks 10–2 ls xq.kk djds nhft,A
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Page # 9
Q.2
The density of steam at 100.0ºC and 1 × 105 pascal
Q.2
is 0.6 kg m–3. Calculate the compressibility factor
100.0ºC o 1 × 105 ikLdy
ij] Hkki dk ?kuRo 0.6 kg m–3
gSA bu vkdM+ksa ls lEihM~;rk xq.kkad (Z) Kkr dhft,A
(Z) from these data.
Q.3
The neutralisation of a solution of 1.2 g of a
Q.3
o mnklhu yo.k
v'kqf) ds feJ.k ;qDr inkFkZ ds 1.2 g foy;u 0.5 N
NaOH dk 18.9 mL iz;qDr (consumed) djrk gSA
KMnO4 foy;u ds lkFk mipkfjr djus ij inkFkZ ds
0.4 g inkFkZ dks 0.25 N KMnO4 foy;u ds 21.55 mL
foy;u dh vko';drk
gksrh gSA feJ.k esa
KHC2O4.H2O dh izfr'krrk gSA
vius mÙkj dks 2×10–1ls xq.kk djds nhft,A
Q.4
ty ds ,d uewus esa bldh dBksjrk dsoy CaSO4 ds
dkj.k gksrh gSA tc ty dks _.kk;u fofue; jsftu
ls xqtkjk tkrk gSA SO42– vk;u] OH– ls izfrLFkkfir
gks tkrk gSA bl uewus ds 25.00 mL uewus dks bl
rjg mipkfjr fd;k tkrk gS ftlds QyLo:i blds
vuqekiu ds fy, 1.00 × 10–3 M H2SO4 ds 21.58 mL
dh vko';drk gksA ty ds uewus dh dBksjrk dk
CaSO4 dh ppm esa O;Dr dhft,A ekuk fd ty dk
?kuRo 1 g mL–1 gSA
vius mÙkj dks 10–2 ls xq.kk djds nhft,A
substance containing a mixture of H2C2O4. 2H2O,
KHC2O4.H2O and impurity of a neutral salt
consumed 18.9 mL of 0.5 N NaOH. On treatment
with KMnO4 solution, 0.4 g of the substance
needed 21.55 mL of 0.25 N KMnO4 solution. What
is percentage of KHC2O4.H2O in the mixture ?
Given the answer by multiply 2×10–1.
Q.4
A sample of water has its hardness due only to
CaSO4. When the water is passed through an anion
exchange resin, SO42– ions are replaced by OH–. A
25.00 mL sample of this water sample so treated
requires 21.58 mL of 1.00 × 10–3 M H2SO4 for its
titration. What is hardness of the water sample
expressed in ppm of CaSO4 ? Assume that density
of water is 1 g mL–1.
Given the answer by multiply 10–2
H2C2O4. 2H2O, KHC2O4.H2O
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Page # 10
Q.5
For the reaction aA → x P when [A] = 2.2 mM the
Q.5
–1
rate was found to be 2.4 mMs . On reducing
concentration of A to half, the rate changes to 0.6m
Ms–1. The order of reaction with respect to A is
Q.6
A reaction P → Q is completed 25% in 25 min 50%
aA → x P ,d vfHkfØ;k ds fy, tc [A] = 2.2 mM gks]
rks osx 2.4 mMs–1 ik;k x;kA A dh lkUnzrk vk/kh djus
ij osx ifjofrZr gksdj 0.6m Ms–1 gks tkrk gSA A ds
lkis{k vfHkfØ;k dh dksfV gS
Q.6
completed in 25 min, if [P] is halved, 25%
completed in 50 min if [P] is doubled. The order of
reaction is
P → Q vfHkfØ;k 25 fefuV 25% iw.kZ gksrh gS] ;fn
[P] dks vk/kk djsa rksa 25 fefuV esa 50% iw.kZ gks tkrh
gS] ;fn [P] dks nqxquk djs] rks 50 fefuV esa 25% gks
tkrh gSA vfHkfØ;k dh dksfV gS
Q.7
RbI crystallizes in b.c.c. structure in which each
Rb+ is surrounded by eight iodide ions each of
radius 2.17 Å. Find the length of one side of RbI
unit cell. (in Å)
Q.7
RbI, b.c.c. lajpuk esa fØLVyhdr̀ gksrk gS ftlesa 2.17
Å dh f=kT;k okys 8 vk;ksMkbM vk;uksa }kjk izR;sd
Rb+ vk;u f?kjk jgrk gSA RbI bdkbZ dksf"Vdk ds ,d
Qyd dh yEckbZ Kkr dhft,A (Å esa)
Q.8
An oxybromate compound, KBrOx, where x is
unknown, is analysed and found to contain 52.92 %
Br. what is the value of x ?
Q.8
,d vkWDlhczksesV ;kSfxd KBrOx, tgk¡ x vKkr gS] dks
fo'ys"k.k fd;k tkrk gS rFkk blesa 52.92 % Br ik;k
tkrk gSA x dk eku D;k gS?
Q.9
Potassium chlorate (KClO4) is made in the
following sequence of reactions
Cl2(g) + KOH → KCl + KClO + H2O
KClO → KCl + KClO3
KClO3 → KClO4 + KCl
What mass of Cl2 is needed to produce 1.0 kg of
KClO4 ?
(Give your answer in Kg and in the form of nearest
whole number)
Q.9
fuEufyf[kr vfHkfØ;k Øe }kjk ikSVsf'k;e DyksjsV
(KClO4) fufeZr fd;k tkrk gS
Cl2(g) + KOH → KCl + KClO + H2O
KClO → KCl + KClO3
KClO3 → KClO4 + KCl
KClO4
ds 1 kg dk fuekZ.k djus gsrq Cl2 dk fdruk
Hkkj vko';d gS
(viuk
mÙkj Kg esa rFkk fudVre iw.kZ la[;k esa ns)
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Page # 11
MATHEMATICS
Section – I
[k.M - I
Questions 1 to 8 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 marks for each wrong answer.
iz'u 1 ls 8 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 dkVk tk;sxkA
Q.1
ekuk x ≤ 2 ds fy, f ′ (x) ≥ 5 gS rFkk f(2) = 10 rc-
Q.1
Q.2
Let f ′ (x) ≥ 5 for x ≤ 2 and f(2) = 10. Then(A) f(0) ≤ –40
(B) f(–1) ≤ –5
(A) f(0) ≤ –40
(B) f(–1) ≤ –5
(C) f(x) ≥ 5x
(D) f(1) ≥ 20
(C) f(x) ≥ 5x
(D) f(1) ≥ 20
Which of the following functions satisfy conditions
of Rolle's Theorem-
Q.2
(A) f(x) = |x –1| + |x –3|, x ∈ [1, 2]
(B) g(x) = 1 –
3
(A) f(x) = |x –1| + |x –3|, x ∈ [1, 2]
x 2 , x ∈ [–1, 1]

1
 2 − sin  | x |, x ≠ 0
(C) h(x) = 
,x∈
x

0,
x=0
(B) g(x) = 1 –
3
x 2 , x ∈ [–1, 1]

1
 2 − sin  | x |, x ≠ 0
(C) h(x) = 
,x∈
x

0,
x=0
 1 1
− π , π 


(D) None of these
Q.3
fuEu esa ls dkSulk Qyu jksy izes; dh 'krks± dks
larq"V djrk gS-
 1 1
− π , π 


(D) buesa ls dksbZ ugha
If f(x) = x3 + ax2 + ax + x(tan θ + cot θ) is increasing
Q.3
;fn f(x) = x3 + ax2 + ax + x(tan θ + cot θ) lHkh okLrfod
3π
for all real x and if θ ∈  π,  , then 2 
3π
x ds fy, o/kZeku gS rFkk ;fn θ ∈  π,  gS, rc 2 
(A) a2 –3a – 6 < 0
(A) a2 –3a – 6 < 0
(C) a2 –3a – 6 ≤ 0
2
(C) a –3a – 6 ≤ 0
(B) a2 –3a – 6 > 0
2
(D) a –3a – 6 ≥ 0
(B) a2 –3a – 6 > 0
(D) a2 –3a – 6 ≥ 0
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Page # 12
Q.4
For what values of a, m and b LMVT is applicable
Q.4
to the function f(x) for x ∈ [0, 2];
f(x) ; x ∈ [0, 2] ij ysx
a kz t
a ek/;eku izes; ykxw gksrh gS;
x=0
 3

f(x) = − x 2 + a 0 < x < 1  mx + b 1 ≤ x ≤ 2

x=0
 3

f(x) = − x 2 + a 0 < x < 1
 mx + b 1 ≤ x ≤ 2

(A) a = 3, m = –2, b = 0
(B) a = 3, m = –2, b = 4
(C) a = 3, m = 2, b = 0
(D) No such a, m, b exist
Q.5
 ωr
ω r +1
 r −1
If A(r) = ω
ωr
 ω2 r ω2 r + 2

(A) a = 3, m = –2, b = 0
(B) a = 3, m = –2, b = 4
(C) a = 3, m = 2, b = 0
(D) bl izdkj ds dksbZ a, m, b fo|eku ugha
ωr + 2 

ωr +1  , where ω is
ω2 r + 4 

Q.5
complex cube root of unity, then(A) A(r) is singular only if r is even
(B) A(r) is singular only if r is odd
(C) A(r) is singular
(D) A(r) is non singular
Q.6
a, m ,oa b ds fdu ekuksa ds fy, fuEu Qyu
 2 − 2 − 4
If A = − 1 3
4  , then A is 1 − 2 − 3
Q.6
(A) Idempotent matrix
(B) Involuntary matrix
(C) Nilpotent matrix of index 2
(D) Nilpotent matrix of index 3
 ωr
ωr +1 ωr + 2 
 r −1

ωr
ωr +1  , tgk¡ ω bdkbZ dk
;fn A(r) = ω
 ω2 r ω2 r + 2 ω2 r + 4 


lfEeJ ?kuewy gS, rc(A) A(r) vO;qRØe.kh; gS dsoy ;fn r le gS
(B) A(r) vO;qRØe.kh; gS dsoy ;fn r fo"ke gS
(C) A(r) vO;qRØe.kh; gS
(D) A(r) O;qRØe.kh; gS
 2 − 2 − 4
;fn A = − 1 3 4  gS, rc A gS 1 − 2 − 3
(A) oxZle vkO;wg
(B) vUrZoyuh; vkO;wg
(C) ?kkr 2 dk 'kwU;Hkkoh vkO;wg
(D) ?kkr 3 dk 'kwU;Hkkoh vkO;wg
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Page # 13
Q.7
The number of solution of the equation
log (–2x) = 2 log (x + 1) is(A) zero
(B) 1
(C) 2
(D) None of these
Q.7
lehdj.k log (–2x) =2log (x + 1) ds gyksa dh la[;k gksxh(A) 'kwU;
(B) 1
(C) 2
(D) buesa ls dksbZ ugha
Q.8
If a > 2, roots of the equation (2 – a)x2 + 3ax –1 = 0
are-
Q.8
;fn a > 2 gS, rks lehdj.k (2 – a)x2 + 3ax –1 = 0 ds ewy
gksaxs(A) ,d /kukRed ,oa ,d _.kkRed
(B) nksuksa _.kkRed
(C) nksuksa /kukRed
(D) nksuksa dkYifud
(A) one positive and one negative
(B) both negative
(C) both positive
(D) Both imaginary
Questions 9 to 13 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. + 3 marks will be given for each
correct answer and no negative marks.
iz'u 9 ls 13 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj
fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k ,d ls
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,
+3 vad fn;s tk;saxs rFkk dksbZ _.kkRed vadu ugha gSA
Q.9
Q.9
Which of the following when simplified reduces to
unity(A) log3log27log464
(B) 2 log18( 2 + 8 )
(C) log 2
(D) – log (
(A) log3log27log464
(B) 2 log18( 2 + 8 )
 2 

10 + log 2 
 5
2 −1)
fuEu esa ls dkSulk ljy djus ij bdkbZ esa ifjofrZr gks
tkrk gS-
 2 

(C) log 2 10 + log 2 
 5
( 2 + 1)
(D) – log (
2 −1)
( 2 + 1)
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Page # 14
Q.10
If the quadratic equation
(ab –bc)x2 + (bc –ca)x + ca – ab = 0, a, b, c ∈ R has
both the roots equal, then(A) both roots are equal to zero
(B) both roots are equal to 1
(C) a, c, b are in H.P.
(D) ab2c2, b2a2c, a2c2b are in A.P.
Q.11
πe
eπ
ππ + ee
+
+
= 0 hasx−π−e
x−e
x−π
Q.10
(ab –bc)x2 + (bc –ca)x + ca – ab = 0, a, b, c ∈ R ds
nksuksa ewy leku gSa] rc (A) nksuksa ewy 'kwU; ds cjkcj gksaxs
(B) nksuksa ewy 1 ds cjkcj gksaxs
(C) a, c, b g-Js- esa gksaxs
(D) ab2c2, b2a2c, a2c2b l-Js- esa gksaxs
Q.11
(A) one real root in (e, π) and other in (π – e, e)
(B) ,d okLrfod ewy (e, π) esa rFkk nwljk (π,π + e) esa
(C) two real roots in (π – e, π + e)
(D) No real roots
(C) (π – e, π + e) esa nks okLrfod ewy
(D) dksbZ okLrfod ewy ugha
For the curve represented parametrically by the
Q.12
equations x = 2ln cot t + 1, and y = tan t + cot t
(A) tangent at t = π/4 is parallel to x-axis
(B) normal at t = π/4 is parallel to y-axis
(C) tangent at t = π/4 is parallel to line y = x
(D) tangent and normal intersect at the point (2, 1)
Q.13
πe
eπ
ππ + ee
+
+
= 0 j[krk gSx−π−e
x−e
x−π
(A) ,d okLrfod ewy (e, π) esa rFkk nwljk (π – e, e) esa
(B) one real root in (e, π) and other in (π,π + e)
Q.12
;fn f}?kkr lehdj.k
Let A, B be square matrices of same order with
trace of A2 = 5, trace of B2 = 125 and trace of
BA = 25. If C = AB, D = A + B, E = A – B, then(A) trace of C = 25
(B) trace of D2 = 180
(D) trace of DE = –120
(C) trace of E2 = 80
Q.13
izkpfyd lehdj.kksa x = 2ln cot t + 1 rFkk
y = tan t + cot t }kjk iznf'kZr oØ ds fy,
(A) t = π/4 ij Li'khZ x-v{k ds lekUrj gksxh
(B) t = π/4 ij vfHkyEc y-v{k ds lekUrj gksxk
(C) t = π/4 ij Li'khZ] js[kk y = x ds lekUrj gksxh
(D) Li'khZ ,oa vfHkyEc] fcUnq (2, 1) ij izfrPNsn djrs gSa
ekuk A, B leku dksfV ds oxZ vkO;wg gSa ftlesa
A2 dk Vsªl = 5, B2 dk Vsl
ª = 125 ,oa BA dk Vsl
ª = 25.
;fn C = AB, D = A + B, E = A – B gS, rc(A) C dk Vsªl = 25
(B) D2 dk Vsªl = 180
2
(C) E dk Vsªl = 80
(D) DE dk Vsªl = –120
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Page # 15
This section contains 2 paragraphs; passage- I has 2
multiple choice questions (No. 14 & 15) and passage- II
has 3 multiple (No. 16 to 18). 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 marks for each wrong answer.
Passage # 1 (Ques. 14 & 15)
If roots of the equation x4 –12x3 + bx2 + cx + 81 = 0
are positive.
Q.14
Q.15
The value of b is(A) –54
(B) 54
(C) 27
(D) –27
Roots of equation 2bx + c = 0 is
(A) –
1
2
(B)
1
2
(C) 1
bl [k.M esa 2 x|ka'k fn;s x;s gSa] x|ka'k -I esa 2 cgqfodYih
iz'u (iz'u 14 o 15) gSa rFkk x|ka'k-II esa 3 cgqfodYih iz'u
(iz'u 16 ls 18) 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;s + 3 vad fn;s tk,saxs rFkk izR;sd
xyr mÙkj ds fy, 1 vad dkVk tk;sxkA
x|ka'k # 1 (iz- 14 ls 15)
;fn lehdj.k x4 –12x3 + bx2 + cx + 81 = 0 ds ewy
/kukRed gSA
Q.14
b dk eku gksxk(A) –54
Q.15
(D) –1
(B) 54
(C) 27
(D) –27
lehdj.k 2bx + c = 0 ds ewy gksaxs(A) –
1
2
(B)
1
2
(C) 1
(D) –1
x|ka'k # 2 (iz- 16 ls 18)
Passage # 2 (Ques. 16 - 18)
v( x)
v( x)
dy
in a different
If y = ∫ f (t ) dt , let us define
dx
u ( x)
dy
= v′(x) f 2(v(x)) – u′ (x) f 2(u(x)) and
dx
the equation of tangent at (a, b) as
manner as
;fn y =
∫ f (t ) dt
gS, ekuk
u ( x)
ifjHkkf"kr djrs gSa tSls
dy
dks fofHkUu rjhds ls
dx
dy
= v′(x) f 2(v(x)) – u′ (x) f 2(u(x))
dx
rFkk (a, b) ij Li'khZ dk lehdj.k
dy
y – b =  
(x – a)
 dx  ( a ,b )
dy
y – b =  
(x – a)
 dx  ( a ,b )
Space for rough work
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Page # 16
x2
Q.16
If y = ∫ t dt , then equation of tangent at x = 1 is 2
x2
Q.16
x
(A) y = x + 1
(C) y = x –1
x
Q.17
If f(x) = ∫ e t
2
/2
x
(B) x + y = 1
(D) y = x
(A) y = x + 1
(C) y = x –1
(1 − t 2 ) dt , then f ′(x) at x = 1 is-
x
Q.17
1
(A) 0
If y =
∫
x
(A) 0
;fn f(x) = ∫ e t
(B) x + y = 1
(D) y = x
2
/2
(1 − t 2 ) dt gS, rks x = 1 ij f ′(x) gS-
1
(B) 1
(C) 2
x4
Q.18
;fn y= ∫ t 2 dt gS, rks x = 1 ij Li'kZ js[kk dk lehdj.k gS
ln t dt then lim+
x →0
3
(B) 1
(D) –1
(A) 0
(C) 2
x4
dy
isdx
(C) 2
(B) 1
Q.18
;fn y =
∫ ln t dt
x3
(D) –1
(A) 0
(B) 1
gS] rks lim+
x →0
(D) –1
dy
gSdx
(C) 2
(D) –1
[k.M - III
Section - III
This section contains 9 questions (Q.1 to 9).
+3 marks will be given for each correct answer and no
negative marking. The answer to each of the questions is
a SINGLE-DIGIT INTEGER, ranging from 0 to 9. The
appropriate bubbles below the respective question
numbers in the OMR have to be darkened. For example,
if the correct answers to question numbers X, Y, Z and
W (say) are 6, 0, 9 and 2, respectively, then the correct
darkening of bubbles will look like the following :
bl [k.M esa 9 (iz-1 ls 9) iz'u gaSA izR;sd lgh mÙkj ds fy;s
+3 vad fn;s tk,saxs rFkk dksbZ _.kkRed vadu ugha gSA bl
[k.M esa izR;sd iz'u dk mÙkj 0 ls 9 rd bdkbZ ds ,d iw.kk±d
gSaA OMR esa iz'u la[;k ds laxr uhps fn;s x;s cqYyksa esa ls
lgh mÙkj okys cqYyksa dks dkyk fd;k tkuk gSA mnkgj.k ds
fy, ;fn iz'u la[;k ¼ekusa½ X, Y, Z rFkk W ds mÙkj 6, 0, 9
rFkk 2 gkas, rks lgh fof/k ls dkys fd;s x;s cqYys ,sls fn[krs gSa
tks fuEufyf[kr gSA
Space for rough work
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Page # 17
X
0
1
2
3
4
5
6
7
8
9
Q.1
Y
0
1
2
3
4
5
6
7
8
9
Z
0
1
2
3
4
5
6
7
8
9
W
X
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
If p and q are the roots of the quadratic equation
Q.1
x2 – (α – 2)x – α –1 = 0, then minimum value of
Y
0
1
2
3
4
5
6
7
8
9
Z
0
1
2
3
4
5
6
7
8
9
W
0
1
2
3
4
5
6
7
8
9
;fn p ,oa q f}?kkr lehdj.k x2 – (α – 2)x – α –1 = 0,
ds ewy gSa] rks p2 + q2 dk U;wure eku gksxkA
p2 + q2 is equal to
Q.2
The equation x3 – 6x2 + 9x + λ = 0 have exactly one
Q.2
root in (1, 3) then λ ∈ (α, β) then (β – α) is
Q.3
esa gS rFkk λ ∈ (α, β) gS] rks (β – α) dk eku gksxk A
If x2 + λx + 1 = 0 and (b – c)x2+ (c–a)x + (a –b) = 0
Q.3
have both the roots common then [λ + 7] is (where
Let f(x) =
;fn x2 + λx + 1 = 0 ,oa (b – c)x2+ (c–a)x + (a –b) = 0
ds nksuksa ewy mHk;fu"B gSa] rks [λ + 7] gS (tgk¡ [.] egÙke
[.] denotes the greatest integer function)
Q.4
lehdj.k x3 – 6x2 + 9x + λ = 0 dk Bhd ,d ewy (1, 3)
iw.kk±d Qyu dks iznf'kZr djrk gS)
x2 + 2
, 1 ≤ x ≤ 3, where [.] represents
[ x]
Q.4
greatest integer function then least value of f(x) is
x2 + 2
, 1 ≤ x ≤ 3, (tgk¡ [.] egÙke iw.kk±d
ekuk f(x) =
[ x]
Qyu dks iznf'kZr djrk gS) rc f(x) dk U;wure eku
gksxkA
Space for rough work
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Page # 18
Q.5
Q.6
Let α be the angle in radians between
x2
y2
+
=1 and the circle x2 + y2 = 12 at their
36
4
k
, then
points of intersection. If α = tan–1
2 3
find the value of 'k'
cos ( x + α) cos ( x + β) cos ( x + γ )
If f(x) = sin ( x + α) sin ( x + β) sin ( x + γ )
sin (β − γ )
and f(2) = 3 then find
x
Q.7
Q.5
x2 + y2 = 12
1
Q.6
15
r =1
x
1
Q.7
If f(x) satisfies the equation
f ( x + 1) f ( x + 8) f ( x + 1)
= 0 for all real x.
1
2
−5
2
3
λ
Q.8
 1 0
2
If A = 
 and A = 8A + kI2, then find the
−
1
7


value of |k|
25
∑ f (r ) Kkr dhft,A
r =1
1
3
3x
1
2
4
1
1
;fn f(x) fo"ke Qyu
gS rFkk blds fo"ke eku g(x) ds cjkcj gSa] rks λ dk
eku Kkr dhft,, ;fn λf(1) g(1) = 4 gksA
;fn lHkh okLrfod x ds fy,] f(x) lehdj.k
f ( x + 1)
f ( x + 8)
f ( x + 1)
1
2
2
3
−5
λ
= 0 dks
larq"V
djrk gSA ;fn f vkorZukad 7 okyk vkorhZ Qyu gS]
rks |λ| dk eku gSA
If f is periodic with period 7 then value of |λ| is.
Q.9
gS, rks 'k' dk eku Kkr
2 3
ekuk f(x) = sin 2πx 2 x
x
function and its odd values is equal to g(x) then
find the value of λ, if λf(1) g(1) = 4
Q.8
k
cos ( x + α) cos ( x + β) cos ( x + γ )
f(x)
=
;fn
sin ( x + α) sin ( x + β) sin ( x + γ )
sin (β − γ ) sin ( γ − α) sin (α − β)
1
rFkk f(2) = 3 gS] rks
f (r )
Let f(x) = sin 2πx 2 x 2 1 . If f(x) be an odd
x3
3x 4 1
rFkk oÙ̀k
dhft,A
25
∑
y2
=1
4
ds e/; muds izfrPNsn fcUnq ij fLFkr
dks.k gS ;fn α = tan–1
sin ( γ − α) sin (α − β)
1
15
x2
+
36
ekuk α (jsfM;u esa)
Q.9
1
0
2
;fn A = 
 rFkk A = 8A + kI2 gS, rks |k| dk
− 1 7
eku Kkr dhft,A
Space for rough work
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Page # 19
PHYSICS
Section – I
[k.M - I
Questions 1 to 8 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 8 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 dkVk tk;sxkA
Q.1
,d le:i pdrh ftldk nzO;eku 2 kg o f=kT;k 4m
ls ,d vU; pdrh ftldh f=kT;k 1m o dsUnz O' gS]
dkVk x;k gSA 'ks"k cps Hkkx dk pdrh ds yEcor o
O ls xqtjus okyh ,d v{k ds ifjr% u;k tM+Ro
vk?kw.kZ gS– (O iwjh pdrh dk dsUnz gS )
Q.1
From a uniform disc of mass 2 kg and radius 4m a
small disc of radius 1m with centre O' is extracted.
The moment of inertia of remaining portion about
an axis passing through O perpendicular to plane of
disc is (O is the centre at whole disc)
2m
O'
O
4m
(A) 16 kg m2
(C)
Q.2
255
kg m 2
16
2m
O'
O
4m
(A) 16 kg m2
255
(C)
kg m 2
16
(B) 12 kg m2
(D)
247
kg m 2
16
A ball of mass m is projected from a point P on the
ground as shown in the figure. It hits a fixed smooth
vertical wall at a distance l from P. Choose the most
appropriate option.
Q.2
(B) 12 kg m2
247
(D)
kg m 2
16
m nzO;eku dh ,d xsan dks tehu ls fcUnq P ls
iz{ksfir fd;k x;k gS] fp=k esa n'kkZ;s vuqlkj ;g fcUnq
P ls l nwjh ij fLFkr ,d ?k"kZ.kghu m/oZ nhokj ij
Vdjkrk gSA fuEu esa ls lokZf/kd lgh fodYi pqfu,s&
Space for rough work
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Page # 20
u
u
P
θ
l
(A) xsan P ij ykSVsxh ;fn l = {kSfrt ijkl dk vk/kk
(B) xsan fcUnq P ij ykSVsxh ;fn l ≤ {kSfrt ijkl dk
l
(A) the ball will return to the point P if l = half of
the horizontal range
(B) the ball will return to the point P if l ≤ half of
the horizontal range.
(C) the ball can not return to the initial point if
l > half of the horizontal range
(D) the ball will return to the initial point, if the
collision elastic and l < half of the range
Q.3
θ
P
Block A of mass M = 2kg is connected to another
block B of mass 1 kg with a string and a spring of
force constant k = 600 N/m as shown in the figure.
Initially spring is compressed to 10cm and whole
system is moving on a smooth surface with a
velocity v = 1 m/s. At any time thread is burnt, the
velocity of block A, when B, is having maximum
velocity w.r.t. ground, is (all the surfaces are
frictionless)
vk/kk
(C) xsan izkjfEHkd fcUnq ij ugha ykSV ldrh ;fn
l > {kSfrt ijkl dk vk/kk
(D) xsan izkjfEHkd fcUnq ij ykSVsxh ;fn VDdj izR;kLFk
gks rFkk l < ijkl ds vk/ks ls
Q.3
M = 2kg nzO;eku ds ,d CykWd A dks 1 kg nzO;eku ds
vU; CykWd B ls Mksjh rFkk k = 600 N/m cy fu;rkad
dh fLizax ls fp=kkuqlkj tksM+k x;k gSA izkjEHk esa fLizax
dks 10cm ladwfpr fd;k tkrk gS rFkk iwjk fudk;
fpduh lrg ij osx v = 1 m/s ls xfr djrk gSA
fdlh Hkh le; Mksjh dks tyk;k tk;s] rks CykWd A dk
osx] tc B /kjkry ds lkis{k vf/kdre osx j[krk gS]
gS ¼lHkh lrg ?k"kZ.kjfgr gS½
v
v
m
M
(A) zero
(B) 1 m/s
m
B
M
A
(C) 3 m/s
B
(D) none
(A) 'kwU;
(B) 1 m/s
A
(C) 3 m/s
(D) dksbZ ugha
Space for rough work
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Page # 21
Q.4
Two identical vessels are filled with equal amounts
of ice. The vessels are made of different materials.
If the ice melts in the two vessels in times t1 and t2
respectively then their thermal conductivities are in
the ratio : (A) t2 : t1
Q.5
(B) t 22 : t12
(C) t1 : t2
Q.5
(C) t1 : t2
(D) t12 : t 22
,d df̀".kdk P nj ls fofdj.k mRlftZr djrh gS
mldk rki T gSA bl rki ij rjaxnS/;Z ftl
fofdj.k vf/kdre rhozrk j[krh gS λ0 gSA ;fn vU;
T' ij fofdfjr 'kfDr P' rFkk vf/kdre rhozrk
λ0
gS rc 2
(A) P'T' = 32 PT
(C) P'T' = 8 PT
(B) P'T' = 16 PT
(D) P'T' = 4 PT
The potential energy of a particle of mass 1 kg is ,
U = 10 + (x – 2)2. Here, U is in joule and x in
metres. On the positive x-axis particle travels upto
x = + 6 m. Choose the wrong statement :
(A) On negative x-axis particle travels upto x = –2m
(B) The maximum kinetic energy of the particle is 16 J
(C) The period of oscillation of the particle is
2 π second
(D) None of the above
(B) t 22 : t12
tc
ij
rki
ij
rjaxnS/;Z
intensity is
Q.6
nks le:i ik=kksa dks cQZ dh leku ek=kk ls Hkjk x;k
gSA ik=k fHkUu-fHkUu inkFkksZ ds cus gq, gSA ;fn nks
ik=kksa esa cQZ Øe'k% t1 o t2 le; esa fi?kyrh gS rc
mudh Å"eh; pkydrkvksa dk vuqikr gS (A) t2 : t1
(D) t12 : t 22
A black body emits radiation at the rate P when its
temperature is T. At this temperature the
wavelength at which the radiation has maximum
intensity is λ0. If at another temperature T' the
power radiated is P' & wavelength at maximum
λ0
then 2
(A) P'T' = 32 PT
(C) P'T' = 8 PT
Q.4
Q.6
(B) P'T' = 16 PT
(D) P'T' = 4 PT
1 kg æO;eku ds ,d d.k dh fLFkfrt ÅtkZ
U = 10 + (x – 2)2 gSA ;gk¡ U twy esa rFkk x ehVj esa
gSA /kukRed x-v{k ij d.k x = + 6 m rd pyrk gSA
xyr dFku pqfu;s &
(A) _.kkRed x-v{k ij d.k x = –2m rd pyrk gS
(B) d.k dh vf/kdre xfrt ÅtkZ 16 J gS
(C) d.k dk nksyudky 2 π lsd.M gS
(D) mijksDr esa ls dksbZ ugha
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Page # 22
Q.7
On a cold winter day the temperature of
atmosphere is – TºC. The cylindrical diagram
shown is made of insulating material and it
contains water at 0ºC. If L is latent heat of fusion of
ice, ρ is density of ice and KR is thermal
conductivity of ice, the time taken for total mass of
water to freeze is -
Q.7
,d lnhZ ds 'khry fnu] ok;qe.My dk rki – TºC gSA
fp=k esa iznf'kZr csyukdkj ik=k ,d dqpkyd inkFkZ ls
cuk gS rFkk blesa 0ºC ij ikuh Hkjk gSA ;fn cQZ ds
xyu dh xqIr Å"ek L gS] cQZ dk ?kuRo ρ rFkk cQZ
dh Å"eh; pkydrk KR gks] rks ikuh ds dqy nzO;eku
dks teus esa yxk le; gS &
H
H
2R
2R
ρLH 2
ρLH 2
H2
(A) 2
(B)
(C)
(D)
2kT
kT
ρLkT
H kT
ρL
Q. 8
Two blocks of masses m & M are moving with
speeds v1 & v2 (v1 > v2) in the same direction on
frictionless surface respectively. M being ahead of
m. An ideal spring of force constant K is attached
to back side of M (as shown). The maximum
compression of spring is
v2
v1
m
(A) v1
m
K
(C) (v1 – v2)
(A)
Q. 8
ρL
H 2 kT
?k"kZ.kghu lrg ij leku fn'kk esa xfr djrs gSA
M, m ls vkxs gSA K cy fu;rkad dh ,d vkn'kZ
fLçax (n'kkZ;s vuqlkj) M ds ihNs dh rjQ tksM+h xbZ
gSA fLçax dk vf/kdre ladwpu gS
mM
(D) None of these
K (M + m)
v2
v1
m
(A) v1
M
K
H2
ρLH 2
ρLH 2
(C)
(D)
2kT
kT
ρLkT
m o M æO;eku ds nks CykWd v1 o v2 (v1 > v2) pky ls
M
(B) v2
(B)
m
K
(C) (v1 – v2)
M
(B) v2
mM
(D) buesa
K ( M + m)
M
K
ls dksbZ ugha
Space for rough work
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Page # 23
Questions 9 to 13 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. + 3 marks will be given for each
correct answer and no negative marks.
iz'u 9 ls 13 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj
fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k ,d ls
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,
+ 3 vad fn;s tk;saxs rFkk dksbZ _.kkRed vadu ugha gSA
Q.9
In which of the following cases the centre of mass
of a rod is certainly not at its centre?
(A) the density continuously increases from left to
right
(B) the density continuously decreases from left to
right
(C) the density decreases from left to right up to the
centre and then increases
(D) the density increases from left to right up to the
centre and then decreases
Q.9
Q.10
A disc is given an initial angular velocity ω0 and
placed on a rough horizontal surface as shown in
figure. The quantities which will not depend on the
coefficient of friction is/are
Q.10
buesa ls fdu fLFkfr;ksa esa NM+ dk nzO;eku dsUnz
fuf'pr :i ls blds dsUnz ij ugha gksxk?
(A) ?kuRo ckbZa vksj ls nkbZa vksj lrr :i ls c<+rk
tk;s
(B) ?kuRo ckbZa vksj ls nkbZa vksj lrr :i ls de
gksrk tk;s
(C) ?kuRo ckbZa ls nkbZa vksj dsUnz rd ?kVrk tk;s vksj
blds i'pkr c<+rk tk;s
(D) ?kuRo nkbZa vksj dsUnz rd c<+rk tk;s vkSj blds
i'pkr ?kVrk tk;sA
,d pdrh dks izkjfEHkd dks.kh; Roj.k ω0 nsdj
fp=kkuqlkj ,d [kqjnjh {kSfrt lrg ij j[kk tkrk gSA
jkf'k;k¡ tks ?k"kZ.k xq.kkad ij fuHkZj ugha djsxh] gS
ω0
ω0
(A) The time until rolling begins
(B) The displacement of the disc until rolling
begins
(C) The velocity when rolling begins
(D) The work done by the force of friction
(A) yq<+dus ls igys rd dk le;
(B) yq<+dus ls igys rd dk pdrh dk foLFkkiu
(C) osx tc yq<+duk izkjEHk gksrk gS
(D) ?k"kZ.k cy }kjk fd;k x;k dk;Z
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Page # 24
Q.11
In an elastic collision between two particles
(A) the total kinetic energy of the system is always
constant
(B) the kinetic energy of the system before
collision is equal to the kinetic energy of the
system after collision
(C) the linear momentum of the system is
conserved
(D) none of these
Q.11
nks d.kksa ds e/; izR;kLFk VDdj esa–
(A) fudk; dh dqy xfrt ÅtkZ ges'kk fu;r
jgrh gS
(B) VDdj ds igys fudk; dh xfrt ÅtkZ
VDdj ds ckn fudk; dh xfrt ÅtkZ ds cjkcj
gksrh gS
(C) fudk; dk js[kh; laosx lajf{kr jgrk gS
(D) buesa ls dksbZ ugha
Q.12
A gas is enclosed in a cylinderical vessel with
initial volume V0, temperature T0 and pressure P0.
Initially spring is in its natural length. Now gas is
heated such that its new volume is 2V0. Piston and
surface of cylinder is perfectly insulating. Now
choose the correct one -
Q.12
,d xSl ,d csyukdkj ik=k esa ifjc) gS rFkk mldk
izkjfEHkd vk;ru V0, rki T0 rFkk nkc P0 gSA izkjEHk esa
fLizax viuh okLrfod yEckbZ esa gSA vc xSl dks bl
izdkj xeZ fd;k tkrk gS fd mldk u;k vk;ru 2V0
gSA fiLVu rFkk csyu dh lrg iw.kZr;k dqpkyd gSA
vc lgh dFku dk p;u dhft;s -
Area = A
A
Area = A
k
A
Atmospheric
Pressure (P0)
(A) Final pressure of the gas is P0 +
k
Atmospheric
Pressure (P0)
kV0
A2
(B) work done by atmospheric pressure is – P0V0
(C) work done by the gas is P0V0
(D) temperature of gas remains constant
(A) xSl dk vafre nkc P0 +
kV0
A2
gS
(B) ok;qe.Myh; nkc }kjk fd;k x;k dk;Z – P0V0 gS
(C) xSl }kjk fd;k x;k dk;Z P0V0 gS
(D) xSl dk rki fu;r jgrk gS
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Page # 25
Q.13
The rate of fall of temperature of two identical
solid spheres of different materials are equal at a
certain temperature then (A) Their specific heat capacities are equal
(B) Their heat capacities are equal
(C) Their specific heat capacities are proportional
to their densities
(D) Their specific heat capacities are inversely
proportional to their densities
This section contains 2 paragraphs; passage- I has 2
multiple choice questions (No. 14 & 15) and passage- II
has 3 multiple (No. 16 to 18). 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.
Passage # 1 (Ques. 14 & 15)
In the figure shown a disc A of mass m and radius
r is fixed with the help of nail in a smooth
horizontal (x-y) plane with its plane horizontal.
The co-ordinates of the centre of the disc A are
 r
r 

 another identical disc B of having
,
 2 2
mass and radius same as that of A moving along
−r
with its plane horizontal in x-y
the line x =
2
Q.13
fHkUu-fHkUu inkFkkZs ds nks le:i Bksl xksyksa ds rki de
gksus dh nj fdlh fuf'pr rki ij leku gksrh gS] rks &
(A) mudh fof'k"V Å"ek /kkfjrk,sa cjkcj gSA
(B) mudh Å"ek /kkfjrk,sa cjkcj gSA
(C) mudh fof'k"V Å"ek /kkfjrk,sa muds ?kuRoksa ds
lekuqikrh gSA
(D) mudh fof'k"V Å"ek /kkfjrk,sa muds ?kuRoksa ds
O;qRØekuqikrh gSA
bl [k.M esa 2 x|ka'k fn;s x;s gSa] x|ka'k -I esa 2 cgqfodYih
iz'u (iz'u 14 o 15) gSa rFkk x|ka'k-II esa 3 cgqfodYih iz'u
(iz'u 16 ls 18) 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;s + 3 vad fn;s tk,saxs rFkk izR;sd
xyr mÙkj ds fy, 1 vad dkVk tk;sxkA
x|ka'k # 1 (iz- 14 ,oa 15)
uhps n'kkZ;s fp=k esa ,d pdrh A ftldk nzO;eku m
o f=kT;k r gS] ,d ?k"kZ.kghu {kSfrt ry esa bldk ry
{kSfrt eas jgrs gq, ,d dhy }kjk fQDl gSA pdrh A
 r
r 
ds dsUnz dk funsZ'kkad  ,  gSA
 2
2
,d vU;
pdrh B ftldk nzO;eku o f=kT;k A ds leku gS]
x=
−r
2
js[kk ds vuqfn'k xfr'khy gS] bldk ry
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Page # 26
plane with speed v0, makes elastic impact with A.
The time of impact is ∆t. In elastic impact kinetic
energy of the system is conserved. All surface are
frictionless.
{kSfrt ry esa x-y ry esa jgrs gq, v0 pky ls xfr
djrs gq, A ds lkFk izR;kLFk :i ls VDdj djrh gSA
VDdj dk le; ∆t gSA izR;kLFk VDdj esa fudk; dh
xfrt ÅtkZ lajf{kr jgrh gSA lHkh lrgsa ?k"kZ.kghu gSa&
y
y
A
A
x
v0
B
Q.14
(− î + ĵ)
2
v0
(C) − v 0 î
Q.15
B
Q.14
The velocity of the disc B after collision is
(A)
(B) −
(D)
x
v0
v0
2
(
VDdj ds i'pkr pdrh B dk osx gS –
(A)
î
)
v0
− î + ĵ
2
(− î + ĵ)
2
v0
(B) −
(C) − v 0 î
Net average force exerted by the surface and nail
on the disc A during impact under the assumption
mg is very small compared to impulsive force
Q.15
(D)
v0
2
(
î
)
v0
− î + ĵ
2
VDdj ds nkSjku pdrh A ds lrg o dhy ij
vkjksfir usV vkSlr cy D;k gksxk (;g ekurs g,q fd)
mg vkosx cy dh rqyuk esa cgqr de gS
(A) mv 0 (î + ĵ) / ∆t
(A) mv 0 (î + ˆj) / ∆t
(B) − mv 0 (î + ˆj) / ∆t
(B) − mv 0 (î + ĵ) / ∆t
(C)
2 mv 0 (î + ĵ) / ∆t
(C)
2 mv 0 (î + ĵ) / ∆t
(D)
2 mv 0 (î − ĵ) / ∆t
(D)
2 mv 0 (î − ĵ) / ∆t
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Page # 27
Passage # 2 (Ques. 16 to 18)
This question concern two balloons and two identical
cylinders of gas. One cylinder contains helium,
monoatomic gas of molecular mass 4g mol–1. The
other contain nitrogen, a diatomic gas of molecular
mass 28 g mol–1. The balloons are identical and each
is connected to one of the cylinders. Both gas may
assumed to ideal and both cylinder weigh the same
amount of gas. Initially cylinder was closed.
Temperature of both cylinder is same.
x|ka'k # 2 (iz- 16 ls 18)
bl ç'u esa nks xqCckjs rFkk XkSl ds nks le:i csyu gSA
,d csyu esa 4g mol–1 vkf.od æO;eku dh ,dyijek.kq
ghfy;e xSl Hkjh gSA nwljs esa 28 g mol–1 vkf.od
æO;eku dh f}ijek.kqd ukbVªkstu xSl Hkjh gSA xqCckjs
,dleku gS rFkk çR;sd ,d csyu ls tqMk+ gqvk gSA
nksuksa xSl dks vkn'kZ eku ldrs gS rFkk nksuksa csyuksa esa
leku ek=kk dh xSl Hkjh gSA çkjEHk esa csyu cUn FksA
nksuksa csyu dk rki leku gS &
Q.16
When the cylinders are opened, then (A) Helium balloon will be inflated faster
(B) Nitrogen balloon will be inflated faster
(C) Both balloon will be inflated simultaneously
(D) Data are not sufficient to say any thing about
rate of inflation
Q.16
tc csyu [kqys gS] rc &
s k
(A) ghfy;e okyk xqCckjk rsth ls Qwyx
s u okyk xqCckjk rsth ls Qwyxs k
(B) ukbVªkt
(C) nksuksa xqCckjs ,d lkFk Qwyasxs
(D) Qwyus dh nj ds ckjs esa dqN Hkh dgus ds fy;s
vk¡dM+s vi;kZIr gS
Q.17
If both balloons are filled with same number of
moles of gas n. If both the balloons are now heated
at constant pressure by supplying the same quantity
H of Heat (thermal energy) to each. Due to heating,
temperature of gases in balloon increase. ∆T is
temperature difference between initial and final
temperature then ∆THe : ∆TN 2 is equal to :
Q.17
;fn nksuksa xqCckjksa esa Hkjh xbZ xSlksa ds eksyks dh la[;k
n leku gSA ;fn nksuksa xqCckjksa dks fu;r nkc ij
çR;sd dks leku ek=kk H dh Å"ek (Å"eh; ÅtkZ)
nsdj xeZ fd;k tkrk gSA rks xeZ djus ds dkj.k] xqCckjs
esa xSlksa dk rki c<+rk gSaA çkjfEHkd rFkk vfUre rki ds
e/; rkikUrj ∆T gS rc ∆THe : ∆TN 2 cjkcj gS &
(A)
5
7
(B)
7
5
(C)
3
5
(D)
5
3
(A)
5
7
(B)
7
5
(C)
3
5
(D)
5
3
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Page # 28
Q.18
Find the difference between the volumes of the
two balloon after heating (where P is the initial
pressure of balloon) 4H
3H
2H
6H
(B)
(D)
(A)
(C)
35P
35P
35P
35P
Q.18
xeZ djus ds ckn nksuksa xqCckjksa ds vk;ruksa ds e/;
vUrj Kkr dhft;s (tgk¡ P xqCckjs dk çkjfEHkd nkc gS) (A)
4H
35P
(B)
3H
35P
(C)
2H
35P
(D)
6H
35P
[k.M - III
Section - III
This section contains 9 questions (Q.1 to 9).
+3 marks will be given for each correct answer and no
negative marking. The answer to each of the questions is
a SINGLE-DIGIT INTEGER, ranging from 0 to 9. The
appropriate bubbles below the respective question
numbers in the OMR have to be darkened. For example,
if the correct answers to question numbers X, Y, Z and
W (say) are 6, 0, 9 and 2, respectively, then the correct
darkening of bubbles will look like the following :
X Y Z W
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
bl [k.M esa 9 (iz-1 ls 9) iz'u gaSA izR;sd lgh mÙkj ds fy;s
+3 vad fn;s tk,saxs rFkk dksbZ _.kkRed vadu ugha gSA bl
[k.M esa izR;sd iz'u dk mÙkj 0 ls 9 rd bdkbZ ds ,d iw.kk±d
gSaA OMR esa iz'u la[;k ds laxr uhps fn;s x;s cqYyksa esa ls
lgh mÙkj okys cqYyksa dks dkyk fd;k tkuk gSA mnkgj.k ds
fy, ;fn iz'u la[;k ¼ekusa½ X, Y, Z rFkk W ds mÙkj 6, 0, 9
rFkk 2 gkas, rks lgh fof/k ls dkys fd;s x;s cqYys ,sls fn[krs gSa
tks fuEufyf[kr gSA
X
0
1
2
3
4
5
6
7
8
9
Y
0
1
2
3
4
5
6
7
8
9
Z
0
1
2
3
4
5
6
7
8
9
W
0
1
2
3
4
5
6
7
8
9
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Page # 29
Q.1
A weightless inextensible string which first runs
over a fixed weightless pulley D and then coils on
a spool B of outer radius R and inner radius R/2
tightly. The smaller pulley of spool can roll
without sliding along a horizontal fixed rail, as
shown. The total mass of the spool is M. The axis
O of the spool is perpendicular to the plane of the
drawing and moment of inertia relative to O is 1/2
MR2. If the end A of the string is pulled downward
with constant acceleration g/2, then tension in the
string is nMg/2. Find n. (String does not slip on
spool.)
Q.1
,d Hkkjghu vfoLrkfjr Mksjh tks igys ,d n`<+
Hkkjghu f?kjuh D ds Åij rFkk fQj ckg~; f=kT;k R
rFkk vkUrfjd f=kT;k R/2 ds Liwy B dh dq.Myh ij
dls gq;s xfr djrh gSA Liwy dh NksVh f?kjuh ,d
{kSfrt n`<+ iVjh ij n'kkZ;s vuqlkj fcuk fQlys
yq<+d ldrh gSA Liwy dk dqy nzO;eku M gSA Liwy
dh v{k O fp=k ds ry ds yEcor~ gS rFkk O ds
lkis{k tM+Ro vk?kw.kZ 1/2 MR2 gSA ;fn Mksjh dks g/2
Roj.k ls uhps dh vksj [khapk tkrk gS] rc Mksjh esa
ruko nMg/2 gSA n Kkr dhft,A ¼Mksjh Liwy ij
fQlyrh ugha gSA½
B
C R
O
R/2
B
C R
O
R/2
D
A
Q.2
D
A
g/2
A solid ball of mass m and radius r spinning with
angular velocity ω falls on a horizontal slab of
mass M with rough upper surface (coefficient of
friction µ) and smooth lower surface. Immediately
after collision the normal component of velocity of
the ball remains half of its value just before
collision and it stops spinning. Find the velocity of
the sphere in horizontal direction immediately after
impact (given : Rω = 5)
Q.2
g/2
m nzO;eku rFkk r f=kT;k dh ,d Bksl xsn
a ω dks.kh; osx
ls pØ.k djrh gqbZ M nzO;eku dh ,d {kSfrt iV~Vhdk
ftldh Åijh lrg [kqjnjh (?k"kZ.k xq.kkad µ) rFkk
fupyh lrg fpduh gS] ij fxjrh gSA VDdj ds rqjUr
ckn xsan ds osx dk yEcor~ ?kVd mlds VDdj ds Bdh
igs ds eku dk vk/kk gks tkrk gS rFkk pØ.k djuk can
dj nsrh gSA VDdj ds rqjUr ckn xksys dk {kSfrt fn'kk
esa osx Kkr dhft, (fn;k gS: Rω = 5)
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Page # 30
ω
ω
m
m
v
v
M
M
Q.3
A small ball is projected from point P towards a
vertical wall as shown in figure. It hits the wall
when its velocity is horizontal. Ball reaches point P
after one bounce on the floor. The coefficient of
restitution assuming it to be same for two
collisions is n/2. All surfaces are smooth. Find the
value of n.
Q.3
,d NksVh xsan dks P fcUnq ls ,d Å/okZ/kj nhokj dh
vksj fp=kkuqlkj iz{ksfir fd;k tkrk gSA og nhokj ij
tc Vdjkrh gS mldk osx {kSfrt gSA xsan fcUnq P ij
Q'kZ ij ,d mNky ds ckn igq¡prh gSA ;g ekfu;s fd
nks VDdjksa ds fy;s izR;koLFkku xq.kkad leku n/2 gSA
lHkh lrg fpduh gSA n dk eku Kkr dhft,A
P
P
Q.4
The room heater can provide only 16ºC in the
room when the temperature outside is – 20ºC. It is
not warm and comfortable, that is why the electric
stove with power of 1 kW is also plugged in
together these two devices maintain the room
temperature of 22ºC. Determine the thermal power
of the heater in kW.
Q.4
dejs okyk ghVj dejs esa dsoy 16ºC rki miyC/k
djk ldrk gS tc ckgj dk rki – 20ºC gSA ;g xeZ
rFkk vkjkenk;d ugha gS] bl dkj.k 1 kW dh 'kfDr
dk ,d fo|qr LVkso Hkh pkyw fd;k tkrk gSA nks
;qfDr;ka ,d lkFk dejs dk rki 22ºC cuk;s j[krh gSA
ghVj dh Å"eh; 'kfDr kW esa Kkr djsA
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Page # 31
Q.5
Q.6
In a bowl (with lid) we put together some amount
of water (of mass m) and the same mass of ice both
being at 0ºC. After 160 minutes the entire ice was
melted. After how much time (in minutes)
(approx.) melting of ice, the temperature of water
increases by 1ºC. (Newton law of cooling is
applicable here), temperature of air is 25ºC and
latent heat of ice is 80 cal/gm is given.
Q.5
Identical six rods are used to form given figure.
Rods AB, BC & AC form equilateral triangle. The
temperature of point B is (in °C) can be written as
10x. Find x.
Q.6
[50°C]
E
U;wVu dk 'khryu dk fu;e ykxw gksrk gS) ok;q dk rki
25ºC rFkk cQZ dh xqIr Å"ek 80 cal/gm nh xbZ gSA
,d leku N% NM+sa fn;s x;s fp=k dks cukus esa iz;qä
dh xbZ gSA NM+s AB, BC rFkk AC ,d leckgq f=kHkqt
cukrh gSA fcUnq B dk rki (°C esa) 10x :i esa fy[kk
tk ldrk gSA x Kkr djsaA
[50°C]
E
B
[100°C]
D
<+Ddu;qä ,d ckmy esa ge ikuh (m æO;eku dk) dh
dqN ek=kk rFkk leku æO;eku dh cQZ dks 0ºC ij
lkFk-lkFk j[krs gSA 160 feuV ckn iwjh cQZ fi?ky
tkrh gSA cQZ ds fi?kyus ds fdrus le; ckn
(feuV esa), ikuh dk rki 1ºC ls c<+ tkrk gSA (;gk¡
A
B
[100°C]
D
C
F
[0°C]
A
C
F
[0°C]
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Page # 32
Q.7
The density of the core of a planet is ρ1 and that of
the outer shell is ρ2. The radii of the core and that
of the planet are R and 2R respectively. The
acceleration due to gravity at the surface of the
planet is same as at a depth R. The ρ1/ρ2 can be
written as n/3. Find n.
R
Q.8
Q.9
Q.7
,d xzg dh dksj ¼vkUrfjd dks'k½ dk ?kuRo ρ1 rFkk
ckgjh dks'k dk ?kuRo ρ2 gSA dksj rFkk xzg dh f=kT;k,sa
Øe'k% R rFkk 2R gSA xzg dh lrg ij xq:Roh; Roj.k]
R xgjkbZ ij xq:Roh; Roj.k ds cjkcj gSA ρ1/ρ2 dks
n/3 ds :i esa fy[k ldrs gSAa rks n Kkr djsaA
R
2R
ρ1
ρ1
ρ2
ρ2
A satellite is revolving round the earth in a circular
orbit of radius ‘a’ with velocity v0. A particle is
projected from satellite in a forward direction with
 5 
relative velocity v = 
– 1 v . Then it is found
 4  0


that the maximum distance of particle from earth's
na
centre is
. Then the value of n is
3
Q.8
A disc of radius '5cm' rolls on a horizontal surface
with linear velocity v = 1 î m/s and angular
velocity 50 rad/sec. Height of particle from ground
on rim of disc which has velocity in vertical
direction is (in cm) y
ω
v
x
Q.9
2R
,d mixzg iF̀oh ds pkjksa vksj ‘a’ f=kT;k dh oÙ̀kh; d{kk
esa v0 osx ls pDdj yxk jgk gSA mixzg ls ,d d.k dks
 5

vkxs dh fn'kk esa vkisf{kd osx v =  – 1 v0 ls
 4 
iz{ksfir fd;k tkrk gSA rc ;g ik;k tkrk gS fd
iF̀oh ds dsUnz ls d.k dh vf/kdre nwjh
na
gSA rc
3
n dk eku gS
'5cm' f=kT;k dh ,d pdrh ,d {kSfrt lrg ij
v = 1 î m/s js[kh; osx rFkk 50 rad/sec dks.kh; osx ls
yq<d
+ rh gSA pdrh dh fje ij /kjkry ls d.k dh
Å¡pkbZ tks Å/okZ/kj fn'kk esa osx j[krk gS] gS (cm esa) y
ω
v
x
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Page # 33
Space for Rough Work (jQ+ dk;Z gsrq LFkku)
Space for rough work
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Page # 34
MAX MARKS: 243
Time : 3 : 00 Hrs.
Date : 13/2/2011
Name : _________________________________________________________ Roll No. : __________________________
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, 3 vad fn, tk;saxs rFkk dksbZ
_.kkRed vadu ugha gSA
6. x|ka'k ij vk/kkfjr cgqfodYih izdkj ds iz'u ftuesa ls dsoy ,d gh fodYi lgh gSaA izR;sd lgh mÙkj ds fy, 3 vad fn, tk;saxs
rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA
[k.M – III
7. x.kukRed izdkj ds iz'u gSaA izR;sd lgh mÙkj ds fy, 3 vad fn;s tk;saxs rFkk xyr mÙkj ds fy, dksbZ _.kkRed vadu ugha gSA bl
[k.M esa mÙkj bdkbZ iw.kk±d esa nhft, (tSls 0 ls 9)A
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]
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
rough
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
the wrong method for filling the
bubble are
The answer of the questions in
wrong or any other manner will be
treated as wrong.
A
B
C
D
P
Q
R
S
T
'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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1
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