course for iit-jee 2011 phase- all chemistry, mathematics & physics test no. 3 [tr-2(i)] (take...

TARGET COURSE FOR IIT-JEE 2011

PHASE- ALL CHEMISTRY, MATHEMATICS & PHYSICS

TEST NO. 3 [TR-2(I)] (TAKE HOME) PAPER – I

Date : 13/2/2011Time : 3 : 00 Hrs. MAX MARKS: 243

Name : _________________________________________________________ Roll No. : __________________________

INSTRUCTIONS TO CANDIDATE

A. GENERAL : 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

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.

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

P Q R S TA BCD

Ensure that all columns are filled. Answers, having blank column will be treated as incorrect. Insert leading zeros (s)

012

3

4

56

7

8

9

012

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56

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9

012

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56

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9

'6' should be filled as 0006

012

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0 1 2

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5 6

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0 1 2

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5 6

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0 1 2

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0 0 0 00 1 2

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5 6

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0 1 2

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5 6

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012

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9


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SEA

L

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Space for rough work

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)



CHEMISTRY

Section – I

Questions 1 to 8 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct. Mark your response inOMR sheet against the question number of thatquestion. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

Q.1 If in the fermentation of sugar in an enzymaticsolution that is 0.12 M, the concentration of thesugar is reduced to 0.06 M in 10 h and to 0.03 M in20 h. What is the order of the reaction ?

(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 thedegree of dissociation -

(A)

t

[a] (B)

t

[a]

(C)

t

[a] 1 1

The order of reaction are respectively - (A) 0, 1, 2 (B) 1, 0, 2 (C) 2, 0, 1 (D) 1, 2, 0

[k.M - I iz'u 1 ls 8 rd cgqfodYih iz'u gSaA izR;sd iz'u ds pkjfodYi (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 viukmÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy, + 3 vad fn;stk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 vad dkVk tk;sxkA

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 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 n'kkZrk gS

(A)

t

[a] (B)

t

[a]

(C)

t

[a] 11

vfHkfØ;k dh dksfV Øe'k% gS & (A) 0, 1, 2 (B) 1, 0, 2 (C) 2, 0, 1 (D) 1, 2, 0



Q.3 A crystal is made of particle X, Y, & Z. X forms fccpacking. Y occupies all octahedral voids of X and Zoccupies 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 (C) X8Y4Z5 (D) X5Y4Z8

Q.4 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 (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

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

Q.3 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 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 (B) X2YZ2 (C) X8Y4Z5 (D) X5Y4Z8

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

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 (B) –180.75 (C) –174.5 (D) –139.6



Q.6 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.7 For the reaction

CH3COCH3 + Br2 →+H 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 Select correct statements : (A) From the following reaction CO(g) + O2(g) → CO2(g), ∆H = q1

heat of formation of CO2 (g) is q1

(B) From the following reaction

C(graphite) +21 O2(g) → CO(g) ∆H = q2

heat of combustion of carbon is q2

(C) From the above reactions, heat of combustionCO2 is q1 + q2

(D) From the above reactions, heat of combustionof CO(g) is q1 and that of carbon is q1 + q2

Q.6 fuEu vfHkfØ;k esa H2SO4 dk rqY;kadh Hkkj D;k gS ? H2SO4 + NaI → Na2SO4 + I2 + H2S + H2O (A) 12.25 (B) 49 (C) 61.25 (D) buesa ls dksbZ ugha Q.7 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

Q.8 lgh dFku gS : (A) fuEu vfHkfØ;k ls CO(g) + O2(g) → CO2(g), ∆H = q1

CO2 (g) ds fuekZ.k dh Å"ek q1 gSA (B) fuEu vfHkfØ;k esa]

C(xzsQkbV) +21 O2(g) → CO(g) ∆H = q2

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



Questions 9 to 13 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct. Markyour response in OMR sheet against the questionnumber of that question. + 3 marks will be given for each correct answer and no negative marks.

Q.9 0.1 M solution of KI reacts with excess of H2SO4

and KIO3 solution, according to equation 5I– + –

3IO + 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

5Weightmolecular

Q.10 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 ofexcess 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

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 KI dk 0.1 M foy;u] H2SO4 rFkk KIO3 foy;u ds vf/kD; ls fØ;k djrk gS] fuEu lehdj.k ds vuqlkj

5I– + –3IO + 6H+ → 3I2 + 3H2O ;

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, I2 ds 0.005 eksy

cukrk gS

(D) KIO3 dk rqY;kadh Hkkj;

5v.kqHkkj

ds cjkcj gksrk gS

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



Q.11 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

(B) ∆S = 5 ln 298373 + 2ln10

(C) ∆E = 525 (D) nh xbZ tkudkjh ls ∆G Kkr ugha fd;k tk ldrk Q.12 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; çØe ds fy, ∆G = 0 gksrk gS

Q.13 fuEu esa ls fdlesa jsfM;y uksMksa dh la[;k lgh

lqesfyr gS - (A) 3 s, 2 (B) 2 p, 0 (C) 4 d, 1 (D) 4 p, 2

Q.11 One mole of ideal diatomic gas (CV = 5 cal) was transformed from initial 25°C and 1 L to the statewhen temperature is 100°C and volume 10 L. Thenfor this process (R = 2 calories/mol/K) (takecalories as unit of energy and Kelvin for temp)

(A) ∆H = 525

(B) ∆S = 5 ln 298373 + 2ln 10

(C) ∆E = 525 (D) ∆G of the process can not be calculated using

given information

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 internalenergy and enthalpy is zero

(C) state function are exact differential

(D) for cyclic process ∆G = 0

Q.13 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



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 to 15)

Some amount of ''20V'' H2O2 is mixed with excess

of acidified solution of KI. The iodine so liberated

required 200 ml of 0.1 N Na2S2O3 for titration.

Q.14 The volume of H2O2 solution is -

(A) 11.2 ml (B) 37.2 ml

(C) 5.6 ml (D) 22.4 ml

Q.15 The mass of K2Cr2O7 needed to oxidise the above

volume of H2O2 solution-

(Atomic mass of Cr = 52)

(A) 3.6 g (B) 0.8 g (C) 4.2 g (D) 0.98 g

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)

KI ds vEyh;dr foy;u ds vkf/kD; ds lkFk ''20V''

H2O2 dh dqN ek=kk feyk;h tkrh gSA vuqekiu ds fy,

bl izdkj fu"dkflr vk;ksMhu dks 0.1 N Na2S2O3 ds

200 ml vko';d gksrs gSA

Q.14 H2O2 foy;u dk vk;ru gS -

(A) 11.2 ml (B) 37.2 ml

(C) 5.6 ml (D) 22.4 ml

Q.15 H2O2 ds mijksDr foy;u ds vk;ru dks vkWDlhdr djus

gsrq K2Cr2O7 dk vko';d Hkkj gS -

(Cr dk ijek.kq Hkkj 52)

(A) 3.6 g (B) 0.8 g (C) 4.2 g (D) 0.98 g



Passage # 2 (Ques. 16 & 18)

Grahm's Law of Diffusion : The phenomenons ofspontaneous intermixing of gases against the law ofgravitation is known as diffusion. If diffusionoccurs through small orifice of the container, then it is known as effusion.

The rate of effusion is defined as, RTM2

PArπ

= ,

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 = Time

effusedMolesTime

effused.Vol=

= time

travelledcetandist

dropessurePr=

Q.16 1 mole of gas A and 4 moles of gas O2 is taken

inside the vessel, which effuse through the smallorifice of the vessel having same area of cross-section and at the same temperature, then which thecorrect % of effused volume of gas A and O2

initially respectively ? (Assume that the gas A doesnot 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%

x|ka'k # 2 (iz- 16 ,oa 18)

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] RTM2

PArπ

= }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 = le;

syfo%lfjr ekle;

;rufu%lfjr vk =

= le;

r; nwjhnkc voueu=

t

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% (B) 60%, 40% (C) 30%, 70% (D) 10%, 90%



Q.17 Which of the following is correct for diffusion ?

(A) ∆G must be positive

(B) ∆H must be negative

(C) ∆S must be positive

(D) ∆S must be negative

Q.18 1 mole of gas "X" and 2 moles of gas "Y" enters

from the end "P" and "Q" of the cylinder

respectively. The cylinder has the area of cross-

section A, shown as under

A P

X Y Q

The length of the cylinder is 150 cms. The gas "X"

intermixes with gas "Y" at the point A. If the

molecular weight of the gases X and Y is 20 and 80

respectively, then what will distance of point A

from Q ?

(A) 75cms (B) 50 cms

(C) 100 cms (D) 90 cms

Q.17 fuEu esa dkSulk folj.k ds fy, gS ?

(A) ∆G /kukRed gksuk pkfg,

(B) ∆H _.kkRed gksuk pkfg,

(C) ∆S /kukRed gksuk pkfg,

(D) ∆S _.kkRed gksuk pkfg,

Q.18 ,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 P

X Y Q

bl csyukdkj ik=k dh yackbZ 150 cms gSA fcUnw A ij

xSl "X" rFkk xSl "Y" ijLij vifefJr gksrh gSA

;fn xSls X rFkk Y dk v.kqHkkj Øe'k% 20 o 80 gks] rks

A dh Q ls nwjh D;k gksxh ?

(A) 75cms (B) 50 cms

(C) 100 cms (D) 90 cms



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 isa SINGLE-DIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective questionnumbers in the OMR have to be darkened. For example,if the correct answers to question numbers X, Y, Z andW (say) are 6, 0, 9 and 2, respectively, then the correctdarkening of bubbles will look like the following :

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

012

3

4

56

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

Q.1 A near ultraviolet photon of 300 nm is absorbed by

a gas and then re-emitted as two photons. One

photon is red with wavelength 760 nm. What would

be the wavelength of the second photon

(in nm) ? Given the answer by multiply 10–2

[k.M - III 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

012

3

4

56

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

X Y Z W

Q.1 300 nm dk ,d fudVre ijkcsaxyh QksVkWu ,d xSl

}kjk vo'kksf"kr gksrk gS rFkk rc ;g nks QksVkWuksa ds :i

esa iquZ mRlftZr gksrk gSA 760 nm rjaxnS/;Z ds lkFk ,d

QksVkWu yky gSA f}rh; QksVkWu dh rjaxnS/;Z (nm esa) gksxh ?

vius mÙkj dks 10–2 ls xq.kk djds nhft,A



Q.2 The density of steam at 100.0ºC and 1 × 105 pascal

is 0.6 kg m–3. Calculate the compressibility factor

(Z) from these data.

Q.3 The neutralisation of a solution of 1.2 g of 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

Q.2 100.0ºC 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

Q.3 H2C2O4. 2H2O, KHC2O4.H2O 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



Q.5 For the reaction aA → x P when [A] = 2.2 mM the rate was found to be 2.4 mMs–1. 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% completed in 25 min, if [P] is halved, 25% completed in 50 min if [P] is doubled. The order of reaction is

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.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.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.5 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 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, b.c.c. lajpuk esa fØLVyhdr 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 ,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 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)



MATHEMATICS

Section – I

Questions 1 to 8 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct. Mark your response inOMR sheet against the question number of thatquestion. +3 marks will be given for each correct answerand – 1 marks for each wrong answer.

Q.1 Let f ′ (x) ≥ 5 for x ≤ 2 and f(2) = 10. Then-

(A) f(0) ≤ –40 (B) f(–1) ≤ –5 (C) f(x) ≥ 5x (D) f(1) ≥ 20

Q.2 Which of the following functions satisfy conditionsof Rolle's Theorem-

(A) f(x) = |x –1| + |x –3|, x ∈ [1, 2]

(B) g(x) = 1 – 3 2x , x ∈ [–1, 1]

(C) h(x) =

=

≠

−

0,0

0|,|1sin2

x

xxx , x ∈

ππ−

1,1

(D) None of these

Q.3 If f(x) = x3 + ax2 + ax + x(tan θ + cot θ) is increasing

for all real x and if θ ∈

ππ

23, , then-

(A) a2 –3a – 6 < 0 (B) a2 –3a – 6 > 0 (C) a2 –3a – 6 ≤ 0 (D) a2 –3a – 6 ≥ 0

[k.M - I iz'u 1 ls 8 rd cgqfodYih iz'u gSaA izR;sd iz'u ds pkjfodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYilgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds le{k viukmÙkj 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-

(A) f(0) ≤ –40 (B) f(–1) ≤ –5 (C) f(x) ≥ 5x (D) f(1) ≥ 20

Q.2 fuEu esa ls dkSulk Qyu jksy izes; dh 'krks± dks

larq"V djrk gS-

(A) f(x) = |x –1| + |x –3|, x ∈ [1, 2]

(B) g(x) = 1 – 3 2x , x ∈ [–1, 1]

(C) h(x) =

=

≠

−

0,0

0|,|1sin2

x

xxx , x ∈

ππ−

1,1

(D) buesa ls dksbZ ugha

Q.3 ;fn f(x) = x3 + ax2 + ax + x(tan θ + cot θ) lHkh okLrfod

x ds fy, o/kZeku gS rFkk ;fn θ ∈

ππ

23, gS, rc-

(A) a2 –3a – 6 < 0 (B) a2 –3a – 6 > 0 (C) a2 –3a – 6 ≤ 0 (D) a2 –3a – 6 ≥ 0



Q.4 For what values of a, m and b LMVT is applicable to the function f(x) for x ∈ [0, 2];

f(x) =

≤≤+<<+−

=

2110

032

xbmxxax

x -

(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 If A(r) =

ωωωωωωωωω

++

+−

++

42222

11

21

rrr

rrr

rrr

, where ω is

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 If A =

−−−

−−

321431422

, then A is-

(A) Idempotent matrix (B) Involuntary matrix (C) Nilpotent matrix of index 2 (D) Nilpotent matrix of index 3

Q.4 a, m ,oa b ds fdu ekuksa ds fy, fuEu Qyu

f(x) ; x ∈ [0, 2] ij ysaxzkat ek/;eku izes; ykxw gksrh gS;

f(x) =

≤≤+<<+−

=

2110

032

xbmxxax

x

(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

Q.5 ;fn A(r) =

ωωωωωωωωω

++

+−

++

42222

11

21

rrr

rrr

rrr

, tgk¡ ω bdkbZ dk

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

Q.6 ;fn A =

−−−

−−

321431422

gS, rc A gS-

(A) oxZle vkO;wg (B) vUrZoyuh; vkO;wg (C) ?kkr 2 dk 'kwU;Hkkoh vkO;wg (D) ?kkr 3 dk 'kwU;Hkkoh vkO;wg



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.8 If a > 2, roots of the equation (2 – a)x2 + 3ax –1 = 0 are-

(A) one positive and one negative (B) both negative (C) both positive (D) Both imaginary

Questions 9 to 13 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct. Markyour response in OMR sheet against the questionnumber of that question. + 3 marks will be given for eachcorrect answer and no negative marks.

Q.9 Which of the following when simplified reduces to unity-

(A) log3log27log464

(B) 2 log18( 2 + 8 )

(C) 10log2 +

52log2

(D) – )12(log )12( +−

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 ;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

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 fuEu esa ls dkSulk ljy djus ij bdkbZ esa ifjofrZr gks

tkrk gS-

(A) log3log27log464

(B) 2 log18( 2 + 8 )

(C) 10log2 +

52log2

(D) – )12(log )12( +−



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 ex

e

−π +

π−

π

xe +

exee

−π−+ππ

= 0 has-

(A) one real root in (e, π) and other in (π – e, e) (B) one real root in (e, π) and other in (π,π + e) (C) two real roots in (π – e, π + e) (D) No real roots

Q.12 For the curve represented parametrically by theequations 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 Let A, B be square matrices of same order withtrace of A2 = 5, trace of B2 = 125 and trace ofBA = 25. If C = AB, D = A + B, E = A – B, then-

(A) trace of C = 25 (B) trace of D2 = 180 (C) trace of E2 = 80 (D) trace of DE = –120

Q.10 ;fn f}?kkr lehdj.k

(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 ex

e

−π +

π−

π

xe +

exee

−π−+ππ

= 0 j[krk gS-

(A) ,d okLrfod ewy (e, π) esa rFkk nwljk (π – e, e) esa (B) ,d okLrfod ewy (e, π) esa rFkk nwljk (π,π + e) esa (C) (π – e, π + e) esa nks okLrfod ewy (D) dksbZ okLrfod ewy ugha

Q.12 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

Q.13 ekuk A, B leku dksfV ds oxZ vkO;wg gSa ftlesa

A2 dk Vsªl = 5, B2 dk Vsªl = 125 ,oa BA dk Vsªl = 25.

;fn C = AB, D = A + B, E = A – B gS, rc-

(A) C dk Vsªl = 25 (B) D2 dk Vsªl = 180 (C) E2 dk Vsªl = 80 (D) DE dk Vsªl = –120



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;sAizR;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 (B) 54 (C) 27 (D) –27

Q.15 lehdj.k 2bx + c = 0 ds ewy gksaxs-

(A) – 21 (B)

21 (C) 1 (D) –1

x|ka'k # 2 (iz- 16 ls 18)

;fn y = ∫)(

)(

)(xv

xu

dttf gS, ekuk dxdy

dks fofHkUu rjhds ls

ifjHkkf"kr djrs gSa tSlsdxdy = v′(x) f 2(v(x)) – u′ (x) f 2(u(x))

rFkk (a, b) ij Li'khZ dk lehdj.k

y – b = ),( badx

dy

(x – a)

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 questionnumber of that question. +3 marks will be given for eachcorrect 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 The value of b is-

(A) –54 (B) 54 (C) 27 (D) –27 Q.15 Roots of equation 2bx + c = 0 is

(A) – 21 (B)

21 (C) 1 (D) –1

Passage # 2 (Ques. 16 - 18)

If y = ∫)(

)(

)(xv

xu

dttf , let us define dxdy in a different

manner as dxdy = v′(x) f 2(v(x)) – u′ (x) f 2(u(x)) and

the equation of tangent at (a, b) as

y – b = ),( badx

dy

(x – a)



Q.16 If y = ∫2

2x

x

dtt , then equation of tangent at x = 1 is -

(A) y = x + 1 (B) x + y = 1 (C) y = x –1 (D) y = x

Q.17 If f(x) = ∫ −x

t dtte1

22/ )1(2 , then f ′(x) at x = 1 is-

(A) 0 (B) 1 (C) 2 (D) –1

Q.18 If y = ∫4

3

x

x

dttnl then dxdy

x +→0lim is-

(A) 0 (B) 1 (C) 2 (D) –1

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 isa SINGLE-DIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective questionnumbers in the OMR have to be darkened. For example,if the correct answers to question numbers X, Y, Z andW (say) are 6, 0, 9 and 2, respectively, then the correctdarkening of bubbles will look like the following :

Q.16 ;fn y= ∫2

2x

x

dtt gS, rks x = 1 ij Li'kZ js[kk dk lehdj.k gS

(A) y = x + 1 (B) x + y = 1 (C) y = x –1 (D) y = x

Q.17 ;fn f(x) = ∫ −x

t dtte1

22/ )1(2 gS, rks x = 1 ij f ′(x) gS-

(A) 0 (B) 1 (C) 2 (D) –1

Q.18 ;fn y = ∫4

3

x

x

dttnl gS] rks dxdy

x +→0lim gS-

(A) 0 (B) 1 (C) 2 (D) –1

[k.M - III

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



0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

012

3

4

56

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

Q.1 If p and q are the roots of the quadratic equation

x2 – (α – 2)x – α –1 = 0, then minimum value of

p2 + q2 is equal to

Q.2 The equation x3 – 6x2 + 9x + λ = 0 have exactly oneroot in (1, 3) then λ ∈ (α, β) then (β – α) is

Q.3 If x2 + λx + 1 = 0 and (b – c)x2+ (c–a)x + (a –b) = 0 have both the roots common then [λ + 7] is (where[.] denotes the greatest integer function)

Q.4 Let f(x) =][22

xx + , 1 ≤ x ≤ 3, where [.] represents

greatest integer function then least value of f(x) is

012

3

4

56

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

X Y Z W

Q.1 ;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

Q.2 lehdj.k x3 – 6x2 + 9x + λ = 0 dk Bhd ,d ewy (1, 3)

esa gS rFkk λ ∈ (α, β) gS] rks (β – α) dk eku gksxk A

Q.3 ;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

iw.kk±d Qyu dks iznf'kZr djrk gS)

Q.4 ekuk f(x) =][22

xx + , 1 ≤ x ≤ 3, (tgk¡ [.] egÙke iw.kk±d

Qyu dks iznf'kZr djrk gS) rc f(x) dk U;wure eku

gksxkA



Q.5 Let α be the angle in radians between

36

2x + 4

2y =1 and the circle x2 + y2 = 12 at their

points of intersection. If α = tan–1

32k , then

find the value of 'k'

Q.6 If f(x) = )(sin)(sin)(sin)(sin)(sin)(sin)(cos)(cos)(cos

β−αα−γγ−βγ+β+α+γ+β+α+

xxxxxx

and f(2) = 3 then find 151 ∑

=

25

1)(

rrf

Q.7 Let f(x) =13122sin11

43

2

xxxx

xπ . If f(x) be an odd

function and its odd values is equal to g(x) then find the value of λ, if λf(1) g(1) = 4

Q.8 If f(x) satisfies the equation

λ−

+++

32521

)1()8()1( xfxfxf = 0 for all real x.

If f is periodic with period 7 then value of |λ| is.

Q.9 If A =

− 71

01 and A2 = 8A + kI2, then find the

value of |k|

Q.5 ekuk α (jsfM;u esa) 36

2x + 4

2y =1 rFkk oÙk

x2 + y2 = 12 ds e/; muds izfrPNsn fcUnq ij fLFkr

dks.k gS ;fn α = tan–1

32k gS, rks 'k' dk eku Kkr

dhft,A

Q.6 ;fn f(x) = )(sin)(sin)(sin)(sin)(sin)(sin)(cos)(cos)(cos

β−αα−γγ−βγ+β+α+γ+β+α+

xxxxxx

rFkk f(2) = 3 gS] rks 151 ∑

=

25

1)(

rrf Kkr dhft,A

Q.7 ekuk f(x) =13122sin11

43

2

xxxx

xπ ;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

Q.8 ;fn lHkh okLrfod x ds fy,] f(x) lehdj.k

λ−

+++

32521

)1()8()1( xfxfxf = 0 dks larq"V

djrk gSA ;fn f vkorZukad 7 okyk vkorhZ Qyu gS] rks |λ| dk eku gSA

Q.9 ;fn A =

− 71

01rFkk A2 = 8A + kI2 gS, rks |k| dk

eku Kkr dhft,A



PHYSICS

Section – I

Questions 1 to 8 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct. Mark your response inOMR 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 From a uniform disc of mass 2 kg and radius 4m asmall disc of radius 1m with centre O' is extracted.The moment of inertia of remaining portion aboutan axis passing through O perpendicular to plane ofdisc is (O is the centre at whole disc)

O' 2m O

4m

(A) 16 kg m2 (B) 12 kg m2

(C) 2mkg16255 (D) 2mkg

16247

Q.2 A ball of mass m is projected from a point P on theground as shown in the figure. It hits a fixed smoothvertical wall at a distance l from P. Choose the mostappropriate option.

[k.M - I 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 )

O' 2m O

4m

(A) 16 kg m2 (B) 12 kg m2

(C) 2mkg16255 (D) 2mkg

16247

Q.2 m nzO;eku dh ,d xsaan 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&



P

u

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

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

m BA M

v

(A) 'kwU; (B) 1 m/s (C) 3 m/s (D) dksbZ ugha

P

u

θ

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 ifl > half of the horizontal range

(D) the ball will return to the initial point, if thecollision elastic and l < half of the range

Q.3 Block A of mass M = 2kg is connected to anotherblock B of mass 1 kg with a string and a spring offorce constant k = 600 N/m as shown in the figure.Initially spring is compressed to 10cm and wholesystem is moving on a smooth surface with avelocity v = 1 m/s. At any time thread is burnt, thevelocity of block A, when B, is having maximumvelocity w.r.t. ground, is (all the surfaces arefrictionless)

m BA M

v

(A) zero (B) 1 m/s (C) 3 m/s (D) none



Q.4 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 (B) 21

22 t:t (C) t1 : t2 (D) 2

221 t:t

Q.5 ,d df".kdk P nj ls fofdj.k mRlftZr djrh gS tc

mldk rki T gSA bl rki ij rjaxnS/;Z ftl ij

fofdj.k vf/kdre rhozrk j[krh gS λ0 gSA ;fn vU; rki

T' ij fofdfjr 'kfDr P' rFkk vf/kdre rhozrk ij

rjaxnS/;Z 20λ gS rc -

(A) P'T' = 32 PT (B) P'T' = 16 PT (C) P'T' = 8 PT (D) P'T' = 4 PT

Q.6 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 lsd.Mπ2 gS

(D) mijksDr esa ls dksbZ ugha

Q.4 Two identical vessels are filled with equal amountsof 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 (B) 21

22 t:t (C) t1 : t2 (D) 2

221 t:t

Q.5 A black body emits radiation at the rate P when its

temperature is T. At this temperature thewavelength at which the radiation has maximumintensity is λ0. If at another temperature T' thepower radiated is P' & wavelength at maximum

intensity is 20λ then -

(A) P'T' = 32 PT (B) P'T' = 16 PT (C) P'T' = 8 PT (D) P'T' = 4 PT

Q.6 The potential energy of a particle of mass 1 kg is ,U = 10 + (x – 2)2. Here, U is in joule and x inmetres. On the positive x-axis particle travels uptox = + 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

ondsec2π (D) None of the above



Q.7 On a cold winter day the temperature ofatmosphere is – TºC. The cylindrical diagramshown is made of insulating material and itcontains water at 0ºC. If L is latent heat of fusion ofice, ρ is density of ice and KR is thermalconductivity of ice, the time taken for total mass ofwater to freeze is -

H

2R

(A) kTHL

2ρ (B)

kT2LH2ρ (C)

kTLH2ρ (D)

LkTH2

ρ

Q. 8 Two blocks of masses m & M are moving withspeeds v1 & v2 (v1 > v2) in the same direction on frictionless surface respectively. M being ahead ofm. An ideal spring of force constant K is attachedto back side of M (as shown). The maximumcompression of spring is

m Mv2 v1

(A) v1Km (B) v2 K

M

(C) (v1 – v2) )mM(KmM

+ (D) None of these

Q.7 ,d lnhZ ds 'khry fnu] ok;qe.My dk rki – TºC gSA fp=k esaa 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

2R

(A) kTHL

2ρ (B)

kT2LH2ρ (C)

kTLH2ρ (D)

LkTH2

ρ

Q. 8 m o M æO;eku ds nks CykWd v1 o v2 (v1 > v2) pky ls

?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

m Mv2 v1

(A) v1Km (B) v2 K

M

(C) (v1 – v2) )mM(KmM

+(D) buesa ls dksbZ ugha



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 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 nkbZaa 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

Q.10 ,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

(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

Questions 9 to 13 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich MULTIPLE (ONE OR MORE) is correct. Markyour response in OMR sheet against the questionnumber of that question. + 3 marks will be given for eachcorrect answer and no negative marks.

Q.9 In which of the following cases the centre of massof a rod is certainly not at its centre?

(A) the density continuously increases from left toright

(B) the density continuously decreases from left toright

(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 thecentre and then decreases

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 thecoefficient of friction is/are

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



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 ,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 -

k

Area = A

A

Atmospheric Pressure (P0)

(A) xSl dk vafre nkc P0 + 20

AkV 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

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 thesystem after collision

(C) the linear momentum of the system isconserved

(D) none of these

Q.12 A gas is enclosed in a cylinderical vessel withinitial volume V0, temperature T0 and pressure P0. Initially spring is in its natural length. Now gas isheated such that its new volume is 2V0. Piston andsurface of cylinder is perfectly insulating. Nowchoose the correct one -

k

Area = A

A

Atmospheric Pressure (P0)

(A) Final pressure of the gas is P0 + 20

AkV

(B) work done by atmospheric pressure is – P0V0

(C) work done by the gas is P0V0 (D) temperature of gas remains constant



Q.13 The rate of fall of temperature of two identicalsolid spheres of different materials are equal at acertain 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 4choices (A), (B), (C) and (D) out of which ONLY ONE iscorrect. Mark your response in OMR sheet against the question number of that question. + 3 marks will begiven for each correct answer and – 1 mark for eachwrong 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 smoothhorizontal (x-y) plane with its plane horizontal.The co-ordinates of the centre of the disc A are

2r,

2r another identical disc B of having

mass and radius same as that of A moving along

the line 2rx −

= with its plane horizontal in x-y

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

ds dsUnz dk funsZ'kkad

2r,

2r gSA ,d vU;

pdrh B ftldk nzO;eku o f=kT;k A ds leku gS]

2rx −

= js[kk ds vuqfn'k xfr'khy gS] bldk ry



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 arefrictionless.

y

v0

A x

B

Q.14 The velocity of the disc B after collision is

(A) ( )ji2

v0 +− (B) i2

v0−

(C) iv0− (D) ( )ji2

v0 +−

Q.15 Net average force exerted by the surface and nailon the disc A during impact under the assumptionmg is very small compared to impulsive force

(A) t/)ji(mv0 ∆+

(B) t/)ji(mv0 ∆+−

(C) t/)ji(mv2 0 ∆+

(D) t/)ji(mv2 0 ∆−

{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

v0

A x

B

Q.14 VDdj ds i'pkr pdrh B dk osx gS –

(A) ( )ji2

v0 +− (B) i2

v0−

(C) iv0− (D) ( )ji2

v0 +−

Q.15 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) t/)ji(mv0 ∆+

(B) t/)ji(mv0 ∆+−

(C) t/)ji(mv2 0 ∆+

(D) t/)ji(mv2 0 ∆−



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.

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.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 : 2NT∆ is equal to :

(A) 75 (B)

57 (C)

53 (D)

35

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 tqM+k 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 tc csyu [kqys gS] rc &

(A) ghfy;e okyk xqCckjk rsth ls Qwysxk

(B) ukbVªkstu okyk xqCckjk rsth ls Qwysxk (C) nksuksa xqCckjs ,d lkFk Qwysaxs (D) Qwyus dh nj ds ckjs esa dqN Hkh dgus ds fy;s

vk¡dM+s vi;kZIr gS

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 : 2NT∆ cjkcj gS &

(A) 75 (B)

57 (C)

53 (D)

35



Q.18 Find the difference between the volumes of thetwo balloon after heating (where P is the initialpressure of balloon) -

(A) P35

H4 (B) P35

H3 (C) P35

H2 (D) P35

H6

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 isa SINGLE-DIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective questionnumbers in the OMR have to be darkened. For example,if the correct answers to question numbers X, Y, Z andW (say) are 6, 0, 9 and 2, respectively, then the correctdarkening of bubbles will look like the following :

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

012

3

4

56

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

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) P35

H4 (B) P35

H3 (C) P35

H2 (D) P35

H6

[k.M - III

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

012

3

4

56

7

8

9

012

3

4

56

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W



Q.1 A weightless inextensible string which first runs over a fixed weightless pulley D and then coils ona spool B of outer radius R and inner radius R/2tightly. The smaller pulley of spool can rollwithout sliding along a horizontal fixed rail, asshown. The total mass of the spool is M. The axisO of the spool is perpendicular to the plane of thedrawing and moment of inertia relative to O is 1/2MR2. If the end A of the string is pulled downwardwith constant acceleration g/2, then tension in thestring is nMg/2. Find n. (String does not slip onspool.)

O R

R/2

C B

D

A g/2

Q.2 A solid ball of mass m and radius r spinning withangular velocity ω falls on a horizontal slab ofmass M with rough upper surface (coefficient offriction µ) and smooth lower surface. Immediatelyafter collision the normal component of velocity ofthe ball remains half of its value just beforecollision and it stops spinning. Find the velocity ofthe sphere in horizontal direction immediately afterimpact (given : Rω = 5)

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/2Roj.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

OR

R/2

CB

D

A g/2

Q.2 m nzO;eku rFkk r f=kT;k dh ,d Bksl xsan 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 candj nsrh gSA VDdj ds rqjUr ckn xksys dk {kSfrt fn'kk esa osx Kkr dhft, (fn;k gS: Rω = 5)



ω

vM

m

Q.3 A small ball is projected from point P towards avertical wall as shown in figure. It hits the wallwhen its velocity is horizontal. Ball reaches point Pafter one bounce on the floor. The coefficient ofrestitution assuming it to be same for twocollisions is n/2. All surfaces are smooth. Find thevalue of n.

P

Q.4 The room heater can provide only 16ºC in the

room when the temperature outside is – 20ºC. It isnot warm and comfortable, that is why the electricstove with power of 1 kW is also plugged intogether these two devices maintain the roomtemperature of 22ºC. Determine the thermal powerof the heater in kW.

ω

v

M

m

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

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



Q.5 In a bowl (with lid) we put together some amountof water (of mass m) and the same mass of ice bothbeing at 0ºC. After 160 minutes the entire ice wasmelted. After how much time (in minutes)(approx.) melting of ice, the temperature of waterincreases by 1ºC. (Newton law of cooling isapplicable here), temperature of air is 25ºC andlatent heat of ice is 80 cal/gm is given.

Q.6 Identical six rods are used to form given figure.Rods AB, BC & AC form equilateral triangle. Thetemperature of point B is (in °C) can be written as10x. Find x.

[100°C]

[50°C]

[0°C]

E

B

C

F

D A

Q.5 <+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¡

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

Q.6 ,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

[100°C]

[50°C]

[0°C]

E

B

C

F

D A



Q.7 The density of the core of a planet is ρ1 and that ofthe outer shell is ρ2. The radii of the core and thatof the planet are R and 2R respectively. Theacceleration due to gravity at the surface of theplanet is same as at a depth R. The ρ1/ρ2 can be written as n/3. Find n.

2R

ρ2

R ρ1

Q.8 A satellite is revolving round the earth in a circularorbit of radius ‘a’ with velocity v0. A particle is projected from satellite in a forward direction with

relative velocity v =

1–

45 v0. Then it is found

that the maximum distance of particle from earth's

centre is 3

na . Then the value of n is

Q.9 A disc of radius '5cm' rolls on a horizontal surface

with linear velocity v = 1 i m/s and angularvelocity 50 rad/sec. Height of particle from groundon rim of disc which has velocity in verticaldirection is (in cm) -

vω

y

x

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

2R

ρ2

R ρ1

Q.8 ,d mixzg iFoh ds pkjksa vksj ‘a’ f=kT;k dh oÙkh; d{kk esa v0 osx ls pDdj yxk jgk gSA mixzg ls ,d d.k dks

vkxs dh fn'kk esa vkisf{kd osx v =

1–

45 v0 ls

iz{ksfir fd;k tkrk gSA rc ;g ik;k tkrk gS fd

iFoh ds dsUnz ls d.k dh vf/kdre nwjh 3

na gSA rc

n dk eku gS

Q.9 '5cm' f=kT;k dh ,d pdrh ,d {kSfrt lrg ij v = 1 i m/s js[kh; osx rFkk 50 rad/sec dks.kh; osx ls yq<+drh 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) -

v ω

y

x



Space for Rough Work (jQ+ dk;Z gsrq LFkku)



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

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. di;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]

cgqp;ukRed iz'u), [k.M –II (LrEHk lqesyu izdkj ds iz'u), [k.M-III (iw.kkZd mÙkj izdkj ds iz'u½]

Section –I Section-II Section-III

For example if only 'A' choice is correct then, the correct method for filling the bubbles is

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.

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

P Q R S TA BCD

Ensure that all columns are filled. Answers, having blank column will be treated as incorrect. Insert leading zeros (s)

012

3

4

56

7

8

9

012

3

4

56

7

8

9

012

3

4

56

7

8

9

012

3

4

56

7

8

9


012

3

4

56

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


0 0 0 00 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

012

3

4

56

7

8

9

012

3

4

56

7

8

9


Corporate Office : CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000 (6 lines), Fax (0744) 3040050 email : [email protected]; Website : www.careerpointgroup.com

Date : 13/2/2011Time : 3 : 00 Hrs. MAX MARKS: 243

SEA

L

1

course for iit-jee 2011 phase- all chemistry, mathematics & physics test no. 3 [tr-2(i)] (take...

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