ALLEN 0 0 2 1 4 0 0 6 CLAS AMME · A andida writ i / e we i th ORS ee darkenin ropri ubbl wit hel f...

Ïi;k bu fun sZ'kks a dks /;ku ls i<+ sA vkidks 5 feuV fo'ks"k :i ls bl dke ds fy, fn;s x;s gS aAPlease read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

A. lkekU; / General :1. ;g iqfLrdk vkidk iz'u&i= gSA bldh eqgj rc rd u rksM+s tc rd fujh{kd ds }kjk bldk funZs'k u fn;k tk;sA

This booklet is your Question Paper. Do not break the seal of this booklet before being instructed to do so by theinvigilator.

2. iz'u&i= dk dksM (CODE) bl i`"B ds Åijh nk;sa dkSus ij Nik gSAThe question paper CODE is printed on the right hand top corner of this sheet.

3. dPps dk;Z ds fy, [kkyh i`"B vkSj [kkyh LFkku bl iqfLrdk esa gh gSaA dPps dk;Z ds fy, dksbZ vfrfjDr dkxt ugha fn;k tk;sxkABlank spaces and blank pages are provided in the question paper for your rough work. No additional sheets willbe provided for rough work.

4. dksjs dkxt] fDyi cksMZ] ykWx rkfydk] LykbM :y] dSYdqysVj] dSejk] lsyQksu] istj vkSj fdlh Hkh izdkj ds bysDVªkWfud midj.k dhijh{kk d{k esa vuqefr ugha gSaABlank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers and electronicgadgets of any are NOT allowed inside the examination hall.

5. bl iqfLrdk ds fiNys i`"B ij fn, x, LFkku esa viuk uke vkSj QkWeZ uEcj fyf[k,AWrite your name and Form number in the space provided on the back cover of this booklet.

6. mÙkj i=] ,d ; a=&Js.khdj.k ;ksX; i= (ORS) gS tks fd vyx ls fn;s tk;saxsAThe answer sheet, a machine-readable Optical Response Sheet (ORS), is provided separately.

7. vks-vkj-,l-(ORS) ;k bl iqfLrdk esa gsj&Qsj@foÏfr u djsa / DO NOT TAMPER WITH/MUTILATE THIS ORS OR THIS BOOKLET.8. bl iqfLrdk dh eqgj rksM+us ds i'pkr d`i;k tk¡p ysa fd blesa 40 i`"B gSa vkSj izR;sd fo"k; ds lHkh 20 iz'u vkSj muds mÙkj fodYi Bhd

ls i<+ s tk ldrs gSaA lHkh [kaMksa ds izkjEHk esa fn;s gq, funsZ'kksa dks /;ku ls i<+ sAOn breaking the seal of the booklet check that it contains 40 pages and all the 20 questions in each subject andcorresponding answer choices are legible. Read carefully the instructions printed at the beginning of each section.

B. vks-vkj-,l- (ORS) dk Hkjko / Filling the ORS :9. ijh{kkFkhZ dks gy fd;s x;s iz'u dk mÙkj ORS mÙkj iqfLrdk esa lgh LFkku ij dkys ckWy ikbUV dye ls mfpr xksys dks xgjk djds nsuk gSA

A candidate has to write his / her answers in the ORS sheet by darkening the appropriate bubble with the help ofBlack ball point pen as the correct answer(s) of the question attempted.

10. ORS ds (i`"B la[;k 1) ij ekaxh xbZ leLr tkudkjh /;ku iwoZd vo'; Hkjsa vkSj vius gLrk{kj djsaAWrite all information and sign in the box provied on part of the ORS (Page No. 1).

C. iz'ui= dk izk:i / Question Paper Formate :bl iz'u&i= ds rhu Hkkx (HkkSfrd foKku] jlk;u foKku vkSj xf.kr) gSaA gj Hkkx ds nks [kaM gSaAThe question paper consists of 3 parts (Physics, Chemistry and Mathematics). Each part consists of two sections.

11. [kaM–I / SECTION – I(i) Hkkx esa 8 cgqfodYi iz'u gSaA gj iz'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls ,d ;k vf/kd lgh gSaA

Contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONEor MORE are correct.

(ii) Hkkx esa fl¼kUrksa] iz;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okys 2 vuqPNsn gSA nksuksa vuqPNsnksa ls lacfU/kr pkj iz'u gSaA ftuesa ls gjvuqPNsn ij nks iz'u gSaA fdlh Hkh vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,d lgh gSaAContains 2 paragraphs each describing theory, experiment, data etc. Four questions relate to two paragraphswith two questions on each paragraph. Each question of a paragraph has ONLY ONE correct answer amongthe four choices (A), (B), (C) and (D)

(iii) Hkkx esa 4 cgqfodYi iz'u gSaA izR;sd iz'u esa lqesyu lwpha gSA lwfp;ksa ds fy, dksM ds fodYi (A), (B), (C) vkSj (D) gSa ftuesals dsoy ,d lgh gSaAContains 4 multiple choice questions. Each question has matching lists. The codes for the lists have choices(A), (B), (C) and (D) out of which ONLY ONE is correct

12. [kaM–II o III esa ,d Hkh iz'u ugha gSA / There is no questions in SECTION-II & III13. [kaM-IV es a 4 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh; iw.kk±d gSA

Section-IV contains 4 questions The answer to each question is a single digit integer, ranging from0 to 9 (both inclusive)

funs Z'k / INSTRUCTIONS

Ïi;k 'ks"k funs Z'kks a ds fy;s bl iqfLrdk ds vfUre i`"B dks i<+ sA / Please read the last page of this booklet for rest of the instructions

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TO

DO S

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ds v

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ksa ds fc

uk e

qgjsa u

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le; : 3 ?k.Vs egÙke vad : 204Time : 3 Hours Maximum Marks : 204

isij – 2PAPER – 2

Path to Success

ALLENCAREER INSTITUTEKOTA (RAJASTHAN)

CLASSROOM CONTACT PROGRAMME(ACADEMIC SESSION 2014-2015)

PAPER CODE 0 0 C T 2 1 4 0 0 6

PATTERN : JEE (Advanced)TEST TYPE : MAJOR

ENTHUSIAST & LEADER COURSE

Date : 10 - 05 - 2015TARGET : JEE (Advanced) 2015

ALL INDIA OPEN TEST #02

T M

fo"k; [k.M i`"B la[;kSubject Section Page No.

Hkkx-1 HkkSfrd foKku I(i) ,d ;k vf/kd lgh fodYi izdkj 03 - 06Part-1 Physics One or More Option Correct Type

I(ii) vuqPNsn izdkj 07 - 08Paragraph Type

I(iii) lqesyu lwpha izdkj 09 - 12Matching List Type

IV iw.kk±d eku lgh izdkj (0 ls 9) 13 - 14Integer Value Correct Type (0 to 9)

Hkkx-2 jlk;u foKku I(i) ,d ;k vf/kd lgh fodYi izdkj 15 - 18Part-2 Chemistry One or More Option Correct Type




Hkkx-3 xf.kr I(i) ,d ;k vf/kd lgh fodYi izdkj 27 - 29Part-3 Mathematics One or More Option Correct Type




SOME USEFUL CONSTANTSAtomic No. H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16, Cl = 17,

Br = 35, Xe = 54, Ce = 58,Atomic masse s : H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24,

Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127,Xe = 131, Ba=137, Ce = 140,

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· Boltzmann constant k = 1.38 × 10–23 JK–1

· Coulomb's law constant pe9

0

1 = 9×104

· Universal gravitational constant G = 6.67259 × 10–11 N–m2 kg–2

· Speed of light in vacuum c = 3 × 108 ms–1

· Stefan–Boltzmann constant s = 5.67 × 10–8 Wm–2–K–4

· Wien's displacement law constant b = 2.89 × 10–3 m–K· Permeability of vacuum µ0 = 4p × 10–7 NA–2

· Permittivity of vacuum Î0 = 20

1

cm· Planck constant h = 6.63 × 10–34 J–s

Kota/00CT214006

ALL INDIA OPEN TEST/JEE (Advanced)/10-05-2015/PAPER-2

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BEWARE OF NEGATIVE MARKINGHAVE CONTROL ¾® HAVE PATIENCE ¾® HAVE CONFIDENCE Þ 100% SUCCESS

Space for Rough Work / dPps dk;Z ds fy, LFkku

PART-1 : PHYSICS Hkkx-1 : HkkSfrd foKku

SECTION–I : (i) One or more options correct Type [k.M-I : (i) ,d ;k vf/kd lgh fodYi izdkj

This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and

(D) out of which ONE or MORE are correct.

bl [k.M esa 8 cgqfodYi iz'u gSaA izR;sd iz'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa] ftuesa ls ,d ;k

vf/kd lgh gSA

1. AB is a linear object placed along optical axis as shown in figure. Tick the incorrect statement(s)-

(A) The length of image is smaller than the length of object.

(B) The length of image is larger than the length of object.

(C) The length of image is equal to the length of object.

(D) If middle portion of the lens is painted then the length of image is smaller than length of object.

A B

10cm

30cm

f=20cm

AB ,d js[kh; fcEc gS tks fd fp=kuqlkj izdkf'kd v{k ds vuqfn'k fLFkr gSA xyr dFku@dFkuksa dks fpfUgr dhft;s&

(A) izfrfcEc dh yEckbZ fcEc dh yEckbZ ls de gSA

(B) izfrfcEc dh yEckbZ fcEc dh yEckbZ ls vf/kd gSA

(C) izfrfcEc dh yEckbZ fcEc dh yEckbZ ds cjkcj gSA

(D) ;fn ysal ds e/; Hkkx dks iksr fn;k tk;s rks izfrfcEc dh yEckbZ] fcEc dh yEckbZ ls de gksxhA

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2. A rabbit is hidden in grass at a point A near a pond. A dog is drinking water at a point B in the pond. The

rabbit can run on the ground with a velocity (3i 4j)+$ $ ms–1. Dog can run with a speed 189ms- . The

coordinate of points B and A are (0,0) and (5,10) respectively. If the rabbit starts to run immediately thedog starts then dog catches rabbit in minimum time t0.Tick the correct alternative (s)-(A) X-component of velocity of dog will be 5ms–1

(B) Y-component of velocity of dog will be 8ms–1

(C) The value of to will be 52

s.

(D) Data insufficient

,d [kjxks'k rkykc ds fudV ,d fcUnq A ij ?kkl esa fNik gqvk gSA ,d dqÙkk rkykc esa fcUnq B ij ikuh ih jgk gSA [kjxks'k

/kjkry ij (3i 4j)+$ $ ms–1 ds osx ls nkSM + ldrk gSA dqÙkk 189ms- dh pky ls nkSM + ldrk gSA fcUnq B rFkk A ds funsZ'kkad

Øe'k% (0,0) rFkk (5,10) gSaA tSls gh dqÙkk nkSM+uk 'kq: djrk gS oSls gh ;fn [kjxks'k Hkh rqjUr nkSM+uk izkjEHk dj nsrk gS rksdqÙkk [kjxks'k dks U;wure le; t0 esa idM + ysrk gSA lgh dFku@dFkuksa dks pqfu;s&

(A) dqÙks ds osx dk X-?kVd 5ms–1 gSA (B) dqÙks ds osx dk Y-?kVd 8ms–1 gSA

(C) t0 dk eku 5

2s gSA (D) vkadM+s vi;kZIr gSA

3. A solid sphere S1 is connected to a charge reservoir through a heater H as shown in figure. flux througha closed spherical surface around S1 is given by f = at2 where a is a constant and t is time in seconds.If resistance of heater is R then select correct alternative/s-(A) Power consumed by heater will be a2Î0

2Rt4.(B) Electric flux through a closed spherical surface around S2 will be –at2.(C) Rate of change of electric flux through a closed spherical surface around S2 will be –2at.(D) All of the above are correct.

Solidsphere

ChargeresevoirHeater

S1 S2H

fp=kuqlkj ,d Bksl xksyk S1 ghVj H }kjk ,d vkos'k ik= S2 ls tqM+k gqvk gSA S1 ds pkjksa vksj ,d can xksyh; i`"B ls fuxZr¶yDl f = at2 }kjk fn;k tkrk gSA tgk¡ a ,d fu;rkad gS rFkk t lsd.M esa le; gSA ;fn ghVj dk izfrjks/k R gks rks lghdFku@dFkuksa dks pqfu;s&(A) ghVj }kjk 'kfDr O;; a2Î0

2Rt4 gSA

(B) S2 ds pkjksa vksj ,d can xksyh; i`"B ls fuxZr fo|qr ¶yDl –at2 gSA

(C) S2 ds pkjksa vksj ,d can xksyh; i"B ls fuxZr fo|qr ¶yDl esa ifjorZu dh nj –2at gSA

(D) mijksDr lHkh dFku lR; gSA

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4. If the volume elasticity (i.e. bulk modulus) of fresh water and the sea water are assumed to be the

same. It is necessary that for the velocity of the sound to be the same :

(A) Fresh water must be at a higher temperature

(B) Sea water must be at a higher temperature

(C) Both must be at the same temperature

(D) Fresh water must have higher refractive index

;fn rktk ty rFkk leqnzh ty dh vk;ru izR;kLFkrk (vk;ru izR;kLFkrk xq.kkad) leku ekuk tk;s rks /ofu ds osx leku

gksus ds fy;s vko';d gS fd&

(A) rktk ty mPp rki ij gksA (B) leqnzh ty mPp rki ij gksA

(C) nksuksa dk rkieku leku gksA (D) rktk ty dk viorZukad mPp gksuk pkfg;sA

5. A galvanometer has a coil of resistance 100 W showing a full–scale deflection at 50 mA. Then select

appropriate alternative(s).

(A) The resistance needed to use it as a voltmeter of range 50 volt is 106 W.

(B) The resistance needed to use it as a voltmeter of range 50 volt is 105 W(C) The resistance needed to use it as an ammeter of range 10 mA is 0.5 W(D) The resistance needed to use it as an ammeter of range 10 mA is 1.0 W,d xsYosuksehVj dh dq.Myh dk izfrjks/k 100 W gS rFkk 50 mA /kkjk ij ;g iw.kZ fo{ksi n'kkZrk gS] rks mfpr fodYi@fodYiksadks pqfu;s%&(A) bls 50 oksYV ijkl okys oksYVehVj ds :i esa iz;qDr djus ds fy;s 106 W izfrjks/k dh vko';drk gksxhA(B) bls 50 oksYV ijkl okys oksYVehVj ds :i esa iz;qDr djus ds fy;s 105 W izfrjks/k dh vko';drk gksxhA(C) bls 10 mA ijkl okys vehVj ds :i esa iz;qDr djus ds fy;s 0.5 W izfrjks/k dh vko';drk gksxhA

(D) bls 10 mA ijkl okys vehVj ds :i esa iz;qDr djus ds fy;s 1.0 W izfrjks/k dh vko';drk gksxhA

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6. An X–ray tube is operating at 50 kV and 20mA. The target material of the tube has a mass of 1.0 kg andspecific heat 495 J kg–1 °C–1. One percent of the supplied electric power is converted into X–rays andentire remaining energy goes into heating the target. Then :–(A) the average rate of rise of temperature of the target would be 2°C/sec(B) the minimum wavelength of the X–rays emitted is about 0.25 × 10–10 m(C) a suitable target material must have a low melting temperature(D) a suitable target material must have low thermal conductivity,d X–fdj.k uyh 50 kV rFkk 20mA ij lapkfyr gks jgh gSA uyh ds y{; inkFkZ dk æO;eku 1.0 kg rFkk fof'k"VÅ"ek 495 J kg–1 °C–1 gSA nh xbZ fo|qr 'kfä dk ,d izfr'kr Hkkx X–fdj.kksa esa ifjofrZr gks tkrk gS rFkk 'ks"k lEiw.kZ ÅtkZy{; dks xeZ djus esa iz;qä gks tkrh gSA rc %&(A) y{; ds rkieku esa o`f¼ dh vkSlr nj 2°C/sec gksxhA(B) mRlftZr X–fdj.kksa dh U;wure rjaxnS/;Z yxHkx 0.25 × 10–10 m gksxhA(C) ,d mi;qä y{; inkFkZ dk xyukad vYi gksuk pkfg,A(D) ,d mi;qä y{; inkFkZ dh rkih; pkydrk vYi gksuk pkfg,A

7. Which of the following statement(s) is/are correct?(A) Nuclei with small mass number and atomic number are more likely to undergo fusion than fission(B) Nuclei having a small binding energy per nucleon are more likely to undergo fusion than fission(C) Nuclei having large binding energy per nucleon are more likely to undergo fusion than fission(D) Nuclei with large atomic number are more likely to undergo fusion than fissionlgh dFku@dFkuksa dks pqfu, %&(A) vYi æO;eku la[;k rFkk ijek.kq Øekad okys ukfHkdksa dk fo[k.Mu dh rqyuk esa lay;u vf/kd gksrk gSA(B) izfr U;wfDy;ksu vYi ca/ku ÅtkZ okys ukfHkd fo[k.Mu dh rqyuk esa lay;u dh izfØ;k ls vf/kd xqtjrs gSaA(C) izfr U;wfDy;ksu mPp ca/ku ÅtkZ okys ukfHkd fo[k.Mu dh rqyuk esa lay;u dh izfØ;k ls vf/kd xqtjrs gSaA(D) vf/kd ijek.kq Øekad okys ukfHkd fo[k.Mu dh rqyuk esa lay;u dh izfØ;k ls vf/kd xqtjrs gSaA

8. A cylindrical empty vessel is kept vertically, with its open end up, on a horizontal ground when it rainsuniformly and vertically. The rate of filling in the vessel with rain water will :-(A) remain unchanged, if its length is made inclined to the vertical(B) decrease, if its length is made inclined with the vertical(C) remain unchanged if a slow wind blows horizontally(D) decrease, if a slow wind blows horizontally.,d csyukdkj [kkyh ik= dks Å/okZ/kj :i ls {kSfrt /kjkry ij bl izdkj j[kk tkrk gS fd bldk [kqyk fljk Åij dh vksjgksA ;gk¡ o"kkZ ,dleku :i ls Å/okZ/kj fxj jgh gSA o"kkZ ds ty ls ik= ds Hkjus dh nj %&(A) vifjofrZr jgsxh ;fn bldh yEckbZ dks Å/okZ/kj ds lkis{k > qdk fn;k tk,A(B) ?kV tk,xh ;fn bldh yEckbZ dks Å/okZ/kj ds lkis{k > qdk fn;k tk,A(C) vifjofrZr jgsxh ;fn ;gk¡ ,d /kheh ok;q {kSfrt :i ls cgus yxsA(D) ?kV tk,xh ;fn ;gk¡ ,d /kheh ok;q {kSfrt :i ls cgus yxsA


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Kota/00CT214006

(ii) Paragraph Type (ii) vuqPNsn izdkj

This section contains 2 paragraphs each describing theory, experiment, data etc. Four questions relateto two paragraphs with two questions on each paragraph. Each question of a paragraph has only onecorrect answer among the four choices (A), (B), (C) and (D).bl [k.M esa fl¼kUrksa] iz;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okys 2 vuqPNsn gSA nksuksa vuqPNsnksa ls lacaf/kr pkj iz'ugSa] ftuesa ls gj vuqPNsn ij nks iz'u gSaA vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesals dsoy ,d lgh gSA

Paragraph for Questions 9 and 10 iz'u 9 ,oa 10 ds fy;s vuqPNsn

A block resting over a horizontal floor has a symmetric track ABC as shown in Figure. The mass of theblock is M=3.12 kg and length AB = BC = 1m. A block of mass m=2 kg is put on the track at A and thesystem is released from rest. (Neglecting friction and impact at B.)

,d CykWd ,d {kSfrt Q'kZ ij fp=kuqlkj j[kk gqvk gS rFkk blesa ,d lefer iFk ABC cuk gqvk gSA CykWd dk nzO;eku

M = 3.12 kg rFkk yEckbZ AB = BC = 1m gSA ,d m=2 kg nzO;eku ds CykWd dks iFk ij A ij j[kk tkrk gS rFkk fudk;

dks fojkekoLFkk ls NksM+k tkrk gSA (?k"kZ.k rFkk B ij VDdj dks ux.; ekfu;s)

37° 37°M

mA C

B

9. During motion, the centre of mass of the system moves-(A) Left ward (B) Right ward(C) Upward (D) As periodic motion in vertical direction.

xfr ds nkSjku fudk; dk nzO;eku dsUnz xfr djrk gS&

(A) cka;h vksj (B) nka;h vksj

(C) Åij dh vksj (D) Å/okZ/kj fn'kk esa vkorZ xfr ds vuqlkj10. When the smaller block is at point C, bigger block is-

(A) In rest (B) In motion in lefter ward(C) In motion in rightward (D) No sufficient information

tc NksVk CykWd C fcUnq ij gksrk gS rks cM+k CykWd &

(A) fojkekoLFkk esa gksrk gSA (B) cka;h vksj xfr djrk gSA(C) nka;h vksj xfr djrk gSA (D) lwpuk i;kZIr ugha gSA


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Paragraph for Questions 11 and 12 iz'u 11 ,oa 12 ds fy;s vuqPNsn

The collector of the photocell (in photoelectric experiment) is made of tungsten while the emitter is ofPlatinum having work function of 10 eV. Monochromatic radiation of wavelength 124 Å & power 100watt is incident on emitter which emits photo electrons with a quantum efficiency of 1%. The acceleratingvoltage across the photocell is of 10,000 volts (Use : hc = 12400 eV Å),d izdk'k fo|qr iz;ksx esa QksVks lsy dk laxzkgd VaxLVu dk cuk gqvk gS tcfd mRltZd IysfVue dk gS ftldk dk;ZQyu10 eV gSA 124 Å rjaxnS/;Z rFkk 100 okWV 'kfDr dh ,do.khZ; fofdj.k dks mRltZd ij vkifrr fd;k tkrk gS ftlls1% DokUVe n{krk ds lkFk QksVks bysDVªkWu mRlftZr gksrs gSaA QksVks lsy ds fljksa ij Rojd oksYVrk 10,000 oksYV gSA(hc = 12400 eV Å iz;ksx djsa)-

Platinumemitter Tungsten

target

radiation124Å

V =10,000VA

11. The minimum wavelength of radiation coming from the tungsten target (collector) is-

VaxLVu y{; (laxzkgd) ls vkus okyh fofdj.k dh U;wure rjaxnS/; Z gS&(A) 124 Å (B) 1.24 Å (C) 1.23 Å (D) 12.3 Å

12. If the source of monochromatic radiation of wavelength 124 Å has an efficiency of 50%, and the powerof X ray emitted by the tungsten target is 3W, the overall efficiency of the apparatus of X-rayproduction is-

;fn 124 Å rjaxnS/; Z ds ,do.khZ; fofdj.k L=ksr dh n{krk 50% rFkk VaxLVu y{; }kjk mRlftZr X fdj.kksa dh 'kfDr3W, gks rks X-fdj.k mRiknu ds fy;s midj.k dh laiw.kZ n{krk gksxh&(A) 1% (B) 0.1% (C) 1.5% (D) 6%

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(iii) Matching List Type (iii) lqesyu lwpha izdkj

This Section contains 4 multiple choice questions. Each question has matching lists. The codes for the

lists. have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

bl [k.M esa 4 cgqfodYi iz'u gSaA izR;sd iz'u esa lqesyu lwpha gSA lwfp;ksa ds fy, dksM ds fodYi (A), (B), (C) vkSj (D)

gSa ftuesa ls dsoy ,d lgh gSA

13. n versus Z graph for characteristic X-rays is as shown in figure.

X-fdj.k vfHkyk{kf.kd ds fy;s n - Z vkjs[k fp= esa n'kkZ;k x;k gS&

1 2 34

Z

Ön

List- I/lwph-I List- II/lwph-II

(P) Line -1 (1) La

js[kk -1

(Q) Line -2 (2) Lb

js[kk -2

(R) Line -3 (3) Ka

js[kk -3

(S) Line -4 (4) Kb

js[kk -4Codes :

P Q R S(A) 4 3 2 1

(B) 2 1 4 3

(C) 3 2 1 4

(D) 1 2 3 4

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14. Information I : Six point charges each of magnitude q are placed at six corners of a regular hexagon of side a.Information II: One charge is now removed. Match the List-I with List-II for above given information.lwpuk I : izR;sd q ifjek.k ds N% fcUnq vkos'k] a Hkqtk okys le "kV~Hkqt ds N% dksuksa ij fLFkr gSAlwpuk II: vc ,d vkos'k dks gVk;k tkrk gSA mijksDr nh xbZ lwpuk ds fy;s lwpha I dk lwpha II ls feyku dhft,A

List–I/lwph–I List–II/lwph–II(P) Magnitude of Electric field at centre of hexagon in information I (1) Non–zero

lwpuk I esa "kV~Hkqt ds dsUnz ij fo|qr {ks= dk ifjek.k v'kwU;

(Q) Magnitude of Electric field at centre of hexagon in information II (2) zero

lwpuk II esa "kV~Hkqt ds dsUnz ij fo|qr {ks= dk ifjek.k 'kwU;

(R) Electric potential at centre of hexagon in information I (3)0

6q

4 apÎlwpuk I esa "kV~Hkqt ds dsUnz ij fo|qr foHko

(S) Electric potential at centre of hexagon in information II (4) 20

q

4 apÎlwpuk II esa "kV~Hkqt ds dsUnz ij fo|qr foHko

Codes :

P Q R S

(A) 3 1, 4 2 1

(B) 4 3 1 2

(C) 2 1, 4 3 1

(D) 1 4 2 3

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15. Find correct match for the figure in List–I as shown with the items given in List–II :Match the following column :

lwph–I esa iznf'kZr fp=ksa dk lwph–II esa fn;s x;s ifj.kkekas ds lkFk feyku dhft,A

List-I/lwph-I List-II/lwph-II

(P)

B

I

I

(1) F ¹ 0

(Q)

× ×

× ××

×

×

×× ×

× ×

I

I

B (2) F = 0, t ¹ 0

(R) I

I

R 2B

R

R (3) F = 0, t = 0

(S)

2B B

RI

(4) F ¹ 0, t = 0

Codes :P Q R S

(A) 3 2 4 1(B) 2 3 1 4(C) 4 2 1 3(D) 1 3 4 2

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16. Connections made in a post office box are shown in figure. R = 10 W denotes that when R = 10W thepointer in the galvanometer is as .

iksLV vkWfQl ckWDl esa cuk;s x;s la;kstuksa dks fp= esa iznf'kZr fd;k x;k gSA R = 10 W ;g iznf'kZr djrk gS fd tcR = 10W gS rks xsYosuksehVj dk ikWbUVj izdkj ls gS&

S

R

P Q

G

List-I/lwph-I List-II/lwph-II

(P) P = 100 W, Q = 10W, R = 400 W, R = 500 W (1) 46 W < S < 47 W(Q) P = 100 W, Q = 1 W, R = 460 W, R = 470 W (2) 0.46 W < S < 0.47 W(R) P = 100 W, Q = 10W, R = 460 W, R = 470 W (3) 40 W < S < 50 W(S) P = 1000 W, Q = 1W, R = 460 W, R = 470 W (4) 4.6 W < S < 4.7 W

Codes :

P Q R S

(A) 3 4 1 2

(B) 2 1 3 4

(C) 3 2 4 1

(D) 4 3 2 1

SECTION –II / [k.M – II & SECTION –III / [k.M – IIIMatrix-Match Type / eSfVªDl&esy izdkj Integer Value Correct Type / iw.kk±d eku lgh izdkj

No question will be asked in section II and III / [k.M II ,oa III esa dksb Z iz'u ugha gSA

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SECTION-IV : (Integer Value Correct Type)

[k.M-IV : (iw.kk±d eku lgh izdkj)This section contains 4 questions. The answer to each question is a single digit Integer, ranging from

0 to 9 (both inclusive)

bl [k.M esa 4 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh; iw.kk ±d gSA1. In the figure electric field lines of three charges are shown. If q

1= +8mC and q

3 = –12mC what is

magnitude of charge q2 in mC.

fn;s x;s fp= esa rhu vkos'kksa dh fo|qr {ks= js[kkvksa dks iznf'kZr fd;k x;k gSA ;fn q1= +8mC rFkk q

3 = –12mC gks rks

vkos'k q2 dk ifjek.k mC esa Kkr dhft,A

+

+

–

q1

q3

q2

2. A wire of steel of length 0.5 m and density 4.4gm/cm3 is clamped tightly between two rigid supports.The initial temperature of the support are 100ºC. The temperature of one of the supports is suddenlydropped to 20ºC. If, after the steady state has been reached, the fundamental frequency is(f × 100) Hz . Then find the value of f (Young's modulus of steel Y = 2 × 1011 N/m2, a = 11 × 10–6 K–1)

,d 0.5 m yEckbZ o 4.4gm/cm3 ?kuRo okys LVhy ds rkj dks nks n`<+ vk/kkjksa ds e/; dldj cka/kk tkrk gSA bu

vk/kkjksa dk izkjfEHkd rkieku 100ºC gSA vc buesa ls fdlh ,d vk/kkj dk rkieku vpkud 20ºC rd fxjk nsrs gSaA

;fn LFkk;h voLFkk izkIr gksus ds ckn ewyHkwr vkofÙk (f × 100) Hz gks rks f dk eku Kkr dhft;s (LVhy dk ;ax izR;kLFkrk

xq.kkad Y = 2 × 1011 N/m2, a = 11 × 10–6 K–1)

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3. A resistance coil, wired to an external battery, is placed inside a thermally insulated cylinder fitted with

a frictionless piston and containing an ideal gas. A current I = 240 mA flows through the coil, which

has a resistance R = 500 W. At what speed v (in cm/s) must the piston of mass m = 48 kg move upward

in order that the temperature of the gas remains unchanged? [Assume that the cylinder is placed in vacuum]

fdlh ckg~; cSVjh ls tqM+h gqbZ ,d izfrjks/kh dq.Myh dks Å"eh; dqpkyd csyu ds vUnj j[kk x;k gSA bl csyu

esa ,d ?k"kZ.kjfgr fiLVu yxk gS rFkk blesa vkn'kZ xSl Hkjh gSA dq.Myh dk izfrjks/k R = 500 W gS rFkk blls

I = 240 mA /kkjk izokfgr gksrh gSA nzO;eku m = 48 kg ds fiLVu dks fdl pky v(cm/s esa) ls Åij dh vksj xfr

djuh pkfg;s rkfd xSl dk rki vifjofrZr jgsA (ekuk fd csyu fuokZr~ esa j[kk gqvk gS)

v

Gas

R

I

4. The vernier of a circular scale is divided in to 30 divisions, which coincides with 29 main scale divisions.

If each main scale division is (1/2)o, the least count of the instrument in minutes is

o`Ùkkdkj iSekus dk ofuZ;j 30 Hkkxksa esa foHkkftr gS tks fd eq[; iSekus ds 29 Hkkxksa ds lkFk lEikrh gSA ;fn eq[; iSekus dk

izR;sd Hkkx (1/2)o gks rks bl midj.k dk vYirekad (fefuV esa) gksxkA

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PART-2 : CHEMISTRY

Hkkx-2 : jlk;u foKkuSECTION–I : (i) One or more options correct Type

[k.M-I : (i) ,d ;k vf/kd lgh fodYi izdkjThis section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and



vf/kd lgh gSA

1. Which of the following statements is/are correct regarding adsorption -

(A) Adsorption is always an exothermic process accomplised by decrease in residual forcess on the surface

i.e. decrease in surface energy

(B) 1 gm of activated charcoal adsorb more methane (critical temperature 190K) as compare to SO2

(Critical temperature 630K)

(C) Freundlich adsorption isotherm is applicable for both adsorption of gas on solid and liquid on solid

(D) Chemical adsorption increases with increase in temperature

vf/k'kks"k.k ds lUnHkZ esa fuEu dFkuksa esa ls dkSuls lgh gS@gSa -

(A) vf/k'kks"k.k lnSo ,d Å"ek{ksih izØe gksrk gS] tks lrg ij vo'ks"kh cyksa esa deh }kjk iw.kZ gksrs gS vFkkZr lrg

ÅtkZ esa deh gksrh gS

(B) 1 gm lfØ; pkjdksy SO2 (ØkfUrd rki 630K) dh rqyuk esa esFk su (ØkfUrd rki 190K) dks vf/kd

vf/k'kksf"kr djrk gS

(C) Ýs.Myhp vf/k'kks"k.k lerki] Bksl ij xSl ds rFkk Bksl ij nzo ds vf/k'kks"k.k] nksuksa ds fy, ykxw gksrk gS

(D) jlk;fud vf/k'kks"k.k] rki esas o`f¼ ds lkFk c<+rk gS

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2. Using given information :

Temp='T' Temp='T'Temp='T'

1AX2AX 3AX1BX

2BX 3BX2 2A BY Y

2TP 3TP3 3A BY Y

Where XA , X

B = mol fraction of 'A' & 'B' in liquid

YA , Y

B = mol fraction of 'A' & 'B' in vapour

º ºA BP ,P = vapour pressure of pure liquid A and pure liquid B respectively

PT = vapour pressure of solution

Given : XA1

= XB1

º ºA BP P>

Select the correct set/ sets of relation

(A) YA2

> YB2

, PT2

= PT3

(B) YA2

> XA2

, YB2

< XB2

(C) YA2

> YA3

, PT2

> PT3

(D) XA2

< XB2

, PT2

< PT3

fn xbZ tkudkjh dk mi;ksx djh;s

rki='T' rki='T'rki='T'

1AX2AX 3AX1BX

2BX 3BX2 2A BY Y

2TP 3TP3 3A BY Y

tgk¡ XA , X

B = nzo esa 'A' rFkk 'B' dk eksy izHkkt

YA , Y

B = ok"i eas 'A' rFkk 'B' dk eksy izHkkt

º ºA BP ,P = 'kq¼ nzo A RkFkk 'kq¼ nzo B ds ok"i nkc

PT = foy;u dk ok"i nkc

fn;k gS: XA1

= XB1

º ºA BP P>

lEcU/k dk lgh leqPp@leqP;ks dks pqfu;saA

(A) YA2

> YB2

, PT2

= PT3

(B) YA2

> XA2

, YB2

< XB2

(C) YA2

> YA3

, PT2

> PT3

(D) XA2

< XB2

, PT2

< PT3

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3. Phosgene (COCl2(g)) undergoes dissociation as : COCl

2(g) �� CO(g) + Cl

2(g) ; DH > 0. Which

of the following will increase the dissociation of phosgene

(A) Adding Cl2(g) at constant volume

(B) Adding Neon to system at constant pressure

(C) Decreasing the total pressure

(D) Heating the container in which reaction occurs.

QkWLTkhu (COCl2(g)) dk fo;kstu fuEu izdkj gksrk gSA : COCl

2(g) �� CO(g) + Cl

2(g) ; DH > 0.

fuEu esa lsa dkSulk dkjd QkWLthu ds fo;kstu esa o`f/k djsxkA

(A) fu;r vk;ru ij Cl2(g) feykus ls

(B) fu;r nkc ij ra= esa fuvkWu feykus ls

(C) dqy nkc esa deh djus ls

(D) ik= ftlesa vfHkfØ;k gksrh gS] dks xeZ djus ls

4. For the reaction : I2 + NaOH ® NaIO

3 + NaI + H

2O. Identify the correct statements

(A) Reaction is an example of disproportionation

(B) Oxidation number of iodine in I2, NaIO

3 & NaI is 0, +5 and –1 respectively.

(C) 1 mol of I2 reacts with 6 moles of NaOH

(D) NaI and NaIO3 are formed in 5 : 1 molar ratio

vfHkfØ;k ds fy, : I2 + NaOH ® NaIO

3 + NaI + H

2O. lgh dFku dk p;u dhft;sA

(A) vfHkfØ;k fo"kekuqikru dk ,d mnkgj.k gS

(B) I2, NaIO

3 rFkk NaI esa vk;ksfMu dh vkDlhdj.k la[;k Øe'k% 0, +5 rFkk –1 gS

(C) 1 eksy I2, 6 eksy NaOH ds lkFk fØ;k djrk gS

(D) NaI rFkk NaIO3 , 5 : 1 ds eksy vuqikr esa fufeZr gksrk gS

5. Identify reactions in which, on heating diatomic gas is evolved leaving behind metallic residue :

,slh vfHkfØ;k,sa igpkfu,sa ftuesa xeZ fd;s tkus ij f}ijekf.o ; xSl mRlftZr gksrh gS rFkk /kkfRod vo'ks"k 'ks"k

jgrk gS :

(A) NaN3(s) D¾¾® (B) [Ni(CO)

4](g) 250 ºC

D¾¾¾®

(C) KClO3(s) D¾¾® (D) HgO(s) D¾¾®

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6. If a metal ion M+n forms paramegnetic octahedral complex, paramagnetic tetahedral complex and

paramagnetic square planar complex with suitable ligands. What is the correct electronic configuration

of metal ion M+n :

;fn ,d /kkrq vk;u M+n mi;qDr fyxs.Mks a ds lkFk vuqpqEcdh; v"VQydh; ladqy] vuqpqEcdh; prq"Qydh;

ladqy rFkk vuqpqEcdh; oxkZdkj leryh; ladqy cukrk gSA /kkrq vk;u M+n dk lgh bysDVªkWuh; foU;kl D;k gS :

(A) d7 (B) d8 (C) d9 (D) d10

7. Choose the incorrect option(s).

(A) Rate of SN2

I

> I

(B) Ph – CH—CH – Me

OH D

Optically pure ¾¾®

N

HIS 1

racemic mixture

(C) Me–Cl 2

MoistAg O¾¾¾® Me–O–Me (major) (D) MeI (excess) + NH3 ¾® Me3N

xyr fodYi dk p;u dhft;s&

(A) SN2 dh nj

I

>

I(B) Ph – CH—CH – Me

OH D

izdkf'kd 'kq¼¾¾®

N

HIS 1 jslsfed feJ.k

(C) Me–Cl 2

MoistAg O¾¾¾® Me–O–Me (eq[;) (D) MeI (vkf/kD;) + NH3 ¾® Me3N

8. C

N—OH

EtMeH O3

+

Major product (eq[; mRikn gS)

(A) Me – C – OH

O

(B) Et – C – OHO

(C) Me—NH2 (D) Et—NH2

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(ii) Paragraph Type (ii) vuqPNsn izdkj

This section contains 2 paragraphs each describing theory, experiment, data etc. Four questions relateto two paragraphs with two questions on each paragraph. Each question of a paragraph has only onecorrect answer among the four choices (A), (B), (C) and (D).bl [k.M esa fl¼kUrksa ] iz;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okys 2 vuqPNsn gSA nksuksa vuqPNsnksa ls lacaf/kr pkj iz'ugSa] ftuesa ls gj vuqPNsn ij nks iz'u gSaA vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesals dsoy ,d lgh gSA

Paragraph for Questions 9 and 10

iz'u 9 ,oa 10 ds fy;s vuqPNsnThe iron(II) ion form many crystalline salts like Mohr's salt, green vitriol, sodium nitropruside and

potassium ferrocyanide etc. Some of the complex salts are paramagnetic but some others are diamagnetic

also, which depends on type of ligand coordinated with iron(II) ion.

vk;ju(II) vk;u cgqr ls fØLVyh; yo.k cukrk gS tSls eksgj yo.k] gjk fofVªvkWy] lksfM;e ukbVªksizqlkbM rFkk ikSVsf'k;e

Qsjkslk;ukbM vkfnA dqN ladqy yo.k vuqpqEcdh; gSa rFkk dqN vU; izfrpqEcdh; Hkh gSa tks vk;ju (II) vk;u ds

lkFk milgla;ksftr fyxS.M ds izdkj ij fuHkZj djrk gSaA9. Which of the following does not contain Hexaaquairon(II) ion :

(A) Mohr's salt (B) Greenvitriol (C) Ferric alum (D) None of the above

fuEu esa ls fdlesa gSDlk,Dokvk;ju(II) vk;u mifLFkr ugha gS :

(A) eksgj yo.k (B) gjk fofVªvkWy (C) Qsfjd ,sye (D) mijksDr esa ls dksbZ ugha10. Using the spin only magnetic moment, select which complex of iron (Fe+2 / Fe+3) is 1 : 1 type electrolyte

(cation to anion ratio).

(A) Sodium pentacynonitrosoniumferrate(......) ; µ = 0

(B) Potassium tetrachloridoferrate(......) ; µ = 24 B. M.

(C) Hexaaquairon(......) sulphate ; µ = 24 B. M.

(D) Potassium hexacynidoferrate(......) ; µ = 3 B. M.

dsoy pØ.k pqEcdh; vk?kw.kZ dk iz;ksx djrs gq;s crkbZ;sa fd vk;ju (Fe+2 / Fe+3) dk dkSulk ladqy 1 : 1 izdkj

dk oS|qr vi?kV~; gS (/kuk;u dk ½.kk;u ls vuqikr)

(A) lksfM;e isUVklk;uksukbVªkslksfu;eQSjsV(......) ; µ = 0

(B) ikSVsf'k;e VsVªkDyksjkbMksaQSjsV(......) ; µ = 24 B. M.

(C) gSDlk,Dokvk;ju(......) lYQsV ; µ = 24 B. M.

(D) ikSVsf'k;e gSDlklk;ukbMksQSjsV(......) ; µ = 3 B. M.

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iz'u 11 ,oa 12 ds fy;s vuqPNsnAn organic compound (X) shows following reactions

O

O(X)

aq.K CO2 3 PNH —NH22 QH O22

(i) Mg–HgR(ii) H O2

H+

S LAH T H+

U(Major

product)

,d dkcZfud ;kSfxd (X) fuEu vfHkfØ;k,sa nsrk gSA

O

O(X)

aq.K CO2 3 PNH —NH22 QH O22

(i) Mg–Hg R(ii) H O2

H+

S LAH T H+

U(eq[; mRikn)

11. Compound which can show stereoisomerism.

;kSfxd tks f=foe~ leko;ork iznf'kZr dj ldrk gS&

(A) P (B) Q (C) U (D) T

12. Incorrect statement for ' T '.

(A) Evolve H2 with Na metal (B) Tertiary alcohol

(C) Optically active (D) None

' T ' ds fy, xyr dFku gS&

(A) Na /kkrq ds lkFk H2 xSl mRlftZr djrk gSA (B) r`rh;d ,YdksgkWy gSA

(C) izdkf'kd lfØ; gSA (D) buesa ls dksbZ ugha

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This Section contains 4 multiple choice questions. Each question has matching lists. The codes for thelists. have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.


gSa ftuesa ls dsoy ,d lgh gSA13. List-I List-II

(P) Critical pressure for Vander Waal's gas (1)8a

27Rb

(Q) Z for vander waal's gas at low pressure (2)Pb1RT

+

(R) Z for vander waal's gas at high pressure (3) 2a

27b

(S) Critical temperature for Vander Waal's gas (4)m

a1V RT

-

lwph-I lwph-II

(P) okW.Mjoky xSl dk ØkfUrd nkc (1)8a

27Rb

(Q) U;wu nkc ij okW.Mjoky xSl ds fy, Z (2)Pb1RT

+

(R) mPp nkc ij okW.Mjoky xSl ds fy, Z (3) 2a

27b

(S) okW.Mjoky xSl ds fy, ØkfUrd rki (4)m

a1V RT

-

Codes :P Q R S

(A) 1 2 4 1

(B) 1 4 2 3

(C) 3 4 2 1

(D) 3 2 4 1

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14. List-I List-II

(P) Rate vs time for zero order kinetics (1)

(Q) Concentration of reactant vs time for zero order (2)

(R) Half life vs initial concentration for zero order (3)

(S) Concentration of reactant vs time for 1st order kinetics (4)

lwph--I lwph-II

(P) 'kqU; dksfV xfrdh ds fy, nj v/s le; (1)

(Q) 'kqU; dksfV ds fy, vfHkdkjdks dh lkUnzrk v/s le; (2)

(R) 'kqU; dksfV ds fy, v/kZ vk;q v/s izkjfEHkd lkUnzrk (3)

(S) 1st dksfV xfrdh ds fy, fØ;kdkjdksa dh lkUnzrk vs le; (4)

Codes :P Q R S

(A) 2 1 4 3(B) 2 1 3 4(C) 1 2 4 3(D) 2 4 1 3

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15. List - I List - II(Pair of species in Aqueous solution)

(P) Pb2+ , SO42– (1) Can exist together

(Q) KOH, HCO3

–(2) Does not exist together because both

form a insoluble salt

(R) Sn4+ + Hg2+ (3) Does not exist together because both

under goes in acid base reaction

(S) I–

+ Fe3+ (4) Does not exist together because they

undergoes in redox reaction.

lwph–I lwph–II(tyh; foy;u es a Lih'kht dk ; qXe)

(P) Pb2+ , SO42– (1) lkFk&lkFk jg ldrs gSa

(Q) KOH, HCO3

–(2) lkFk&lkFk ugha jg ldrs gS D;ksafd nksuksa ,d

vfoys;'khy yo.k cukrs gS

(R) Sn4+ + Hg2+ (3) lkFk&lkFk ugha jg ldrs gS D;ksafd nksuksa

vEy&{kkj vfHkfØ;k djrs gSa

(S) I–

+ Fe3+ (4) lkFk&lkFk ugha jg ldrs gS D;ksafd osa jsMkWDl

vfHkfØ;k djrs gSaCodes :

P Q R S

(A) 2 1 3 4

(B) 3 1 4 2

(C) 2 3 4 1

(D) 2 3 1 4

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16. Consider the following compounds.

fuEu ;kSfxdksa ij fopkj dhft,A

(i) IF5

(ii) ClF¯4

(iii) XeO2F

2(iv) NH

2(v) BCl

3

(vi) BeCl2

(vii) AsCl5

(viii) BO3

3–(ix) NO

2

List–I List–II

(Species in which central atom uses (Number of species out

valency shell orbital) of given above)

(P) All three p-orbitals are involved in hybridisation of central atom (1) One

(Q) Only two p-orbitals are involved in hybridisation of central atom (2) Five

(R) Only one p-orbital is involved in hybridisation of central atom (3) Three

(S) Atleast one d-orbital is involved in hybridisation of central atom (4) Four

Select correct code :

lwph–I lwph–II

(nh xbZ Lih'kht es a ls og ftles a dsUæh ; ijek.kq] la;k sth (mijk sDr es a ls Lih'kht dh

dks'k d{kd dk iz;k sx djrs gS) la[;k)

(P) dsUæh; ijek.kq ds ladj.k esa lHkh rhuksa p-d{kd lfEefyr gSa (1) ,d

(Q) dsUæh; ijek.kq ds ladj.k esa dsoy nks p-d{kd lfEefyr gSa (2) ik¡p

(R) dsUæh; ijek.kq ds ladj.k esa dsoy ,d p-d{kd lfEefyr gS (3) rhu

(S) dsUæh; ijek.kq ds ladj.k esa de ls de ,d d-d{kd lfEefyr gS (4) pkj

lgh dwV pqfu,sa :

Codes :

P Q R S

(A) 2 3 1 4

(B) 2 4 1 3

(C) 4 3 2 1

(D) 4 2 1 3



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bl [k.M esa 4 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh ; iw.kk ±d gSA

1. If energy of 2p atomic orbital and 2s atomic orbitals in oxygen atom are –15.85 eV and –32.38 eV

respectively.

Find number of bonding electrons in O2 molecules which have higher energy than –32.38 eV.

;fn vkWDlhtu ijek.kq esa 2p ijekf.o; d{kd rFkk 2s ijekf.o; d{kdksa dh ÅtkZ Øe'k % –15.85 eV rFkk–32.38 eV gS rks vkWDlhtu v.kqvksa esa , sls ca/kh bysDVªkWuks a dh la [;k crkbZ;sa tks ] –32.38 eV ls vf/kd ÅtkZj[krs gSaA

2. In hydroboration oxidation reaction which is a syn addition phenomena

gkbMªkscksjhdj.k vkWDlhdj.k vfHkfØ;k tks ,d flu ;ksx izØe gS] esa

(i) BH THF3

(ii) H O /OH2 2–

Total number of trans isomeric product is/are.

VªkUl leko;oh mRiknks a dh dqy la[;k D;k gS

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3.

O O

¾¾¾®2

KOHI

Number of iodoform molecules produced per molecule.

O O

¾¾¾®2

KOHI

izfr v.kq mRikfnr vk;MksQkWeZ v.kqvksa dh dqy la[;k crkb;sA

4. Number of moles of CrO3 required to oxidise completely of 9 moles of 2-heptanol to the corresponding

ketone ?

dhVksu ds lkFk lEcfU/kr 2-gSIVsukWy ds 9 eksyksa dks iw.kZ vkWDlhdr djus ds fy, vko';d CrO3 ds eksyksa dh la[;kcrkb;s&


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PART-3 : MATHEMATICS Hkkx-3 : xf.kr

SECTION–I : (i) One or more options correct Type [k.M-I : (i) ,d ;k vf/kd lgh fodYi izdkj

This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and



vf/kd lgh gSA

1. Every solution of cos2x = –1 will be a solution of which of the following equation(s)-

cos2x = –1 dk izR;sd gy] fuEu esa ls fdl lehdj.k dk gy gksxk -

(A) cosx + 3cos3x + 5cos5x + 7cos7x = 0 (B) sin2x + 2sin4x + 3sin6x = 0

(C) tan2x + 2tan4x + 3tan6x = 0 (D) sin2x + cos4x + tan6x = 0

2. Let ( )10

10

ƒ x dx-

a = - aò where a Î [–9,11] and maximum and minimum values of ƒ(a) are M & m

respectively, then-

ekuk ( )10

10

ƒ x dx-

a = - aò , tgk¡ a Î [–9,11] rFkk ƒ(a) ds vf/kdre rFkk U;wure eku Øe'k% M rFkk m gks] rks -

(A) m = 50 (B) m = 100 (C) M = 250 (D) M = 220

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3. 5 5 5 5P 1234 1235 1236 .... 4321= + + + +

5 5 5 5Q 1235 1236 1237 .... 4322= + + + +4322

5

1234

R x dx= ò . Then

(A) P < R (B) P + Q > 2R (C) Q > R (D) P + Q < 2R

5 5 5 5P 1234 1235 1236 .... 4321= + + + +

5 5 5 5Q 1235 1236 1237 .... 4322= + + + +4322

5

1234

R x dx= ò gSA rc

(A) P < R (B) P + Q > 2R (C) Q > R (D) P + Q < 2R

4. Let ( )1

3 2xƒ x

x 2

-=

+,

ƒn(x ) = ƒn–1(ƒ1(x)) for n > 2, n Î N and fn(x) = ƒ2(x) + ƒ4(x) +........+ƒ2n(x), n Î N then

(A) ( ) ( )n 1 nx x x R+f > f " Î

(B) ( ) ( )n 1 nx x x R++f > f " Î

(C) ( ) ( )n 1 nx x x R++f < f " Î

(D) Equation f4(x) = 5 sinx will have exactly 3 distinct real solutions.

ekuk ( )1

3 2xƒ x

x 2

-=

+,

tgk¡ n > 2, n Î N ds fy, ƒn(x ) = ƒn–1(ƒ1(x)) rFkk fn(x) = ƒ2(x) + ƒ4(x) +........+ƒ2n(x), n Î N gks] rks

(A) ( ) ( )n 1 nx x x R+f > f " Î

(B) ( ) ( )n 1 nx x x R++f > f " Î

(C) ( ) ( )n 1 nx x x R++f < f " Î

(D) lehdj.k f4(x) = 5 sinx ds Bhd 3 fHkUu okLrfod gy gksaxsA

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5. Let ƒ : [–2p,2p] ® [a,b], ƒ(x) = tan–1x + sinx + x5 – 3x3–x be an onto function, then-

ekuk ƒ : [–2p,2p] ® [a,b], ƒ(x) = tan–1x + sinx + x5 – 3x3–x vkPNknd Qyu gks] rks -(A) 2a + b > 0 (B) 3a + b < 0 (C) a + 2b > 0 (D) 3a + 4b > 0

6. Let nth term of sequence 1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,...... is given by tn, then

ekuk Js.kh 1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,...... dk n ok¡ in tn }kjk fn;k x;k gS] rks(A) t100 = 14 (B) t200 = 20 (C) t300 = 24 (D) t400 = 28

7. Let ƒ : R ® R is polynomial satisfying (x – 4)ƒ(x – 1) = (x – 7)ƒ(x), " x Î R. If ƒ(7) = 6, then-

(A) ƒ'(x) will have three distinct positive real roots(B) ƒ'(x) will have two distinct positive real roots(C) ƒ is surjective

(D) ƒ is injective

ekuk cgqinh; Qyu ƒ : R ® R, (x – 4)ƒ(x – 1) = (x – 7)ƒ(x), " x Î R dks lUrq"V djrk gSA ;fn ƒ(7) = 6 gks] rks -

(A) ƒ'(x) ds rhu fHkUu /kukRed okLrfod ewy gksaxs

(B) ƒ'(x) ds nks fHkUu /kukRed okLrfod ewy gksaxs

(C) ƒ vkPNknd gksxkA

(D) ƒ ,dSdh gksxkA

8. Let ƒ : R0 ® R, ( )2x 4

ƒ x n 23 x

æ ö= + -ç ÷

è øl , then which of the following statement is true about function ƒ(x)

(A) ƒ(x) an increasing function(B) Range of ƒ(x) is R

(C) ƒ(x) = 2 posses exactly three distinct real solution(D) ƒ(x) = 0 posses exactly one real solution

ekuk ƒ : R0 ® R, ( )2x 4

ƒ x n 23 x

æ ö= + -ç ÷

è øl gks] rks fuEu esa ls dkSulk dFku Qyu ƒ(x) ds fy, lR; gksxk

(A) ƒ(x) ,d o/kZeku Qyu gSA

(B) ƒ(x) dk ifjlj R gksxkA

(C) ƒ(x) = 2 ds Bhd rhu fHkUu okLrfod gy gksaxsA

(D) ƒ(x) = 0 dk Bhd ,d okLrfod gy gksxkA

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(ii) Paragraph Type

(ii) vuqPNsn izdkj

This section contains 2 paragraphs each describing theory, experiment, data etc. Four questions relateto two paragraphs with two questions on each paragraph. Each question of a paragraph has only onecorrect answer among the four choices (A), (B), (C) and (D).

bl [k.M esa fl¼kUrksa ] iz;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okys 2 vuqPNsn gSA nksuksa vuqPNsnksa ls lacaf/kr pkj iz'ugSa] ftuesa ls gj vuqPNsn ij nks iz'u gSaA vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesals dsoy ,d lgh gSA


iz'u 9 ,oa 10 ds fy;s vuqPNsn

Let P : x + y + z = 3 is a plane and A(1,1,1) is a point on it. Q is a variable point on plane P such that

QA 2 2= . B(2,3,4) is another point then answer the following questions.

ekuk lery P : x + y + z = 3 rFkk A(1,1,1) bl ij ,d fcUnq gSA Q, lery P ij ,d pj fcUnq bl izdkj

gS fd QA 2 2= gSA B(2,3,4) vU; fcUnq gks] rks fuEu iz'uksa ds mÙkj nhft,A

9. Maximum value of QB is -

QB dk vf/kdre eku gksxk

(A) 3 2 (B) 6 (C) 2 6 (D) 30

10. The coordinates of the point on the locus of Q such that QB is minimum is-

Q ds fcUnqiFk ij ml fcUnq ds funsZ'kkad tc QB U;wure gS] gksxk -

(A) (–1,1,3) (B) (0,1,2) (C) (3,1,–1) (D) (1,–1,3)


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iz'u 11 ,oa 12 ds fy;s vuqPNsn

Let ( )1

n 1

0

ƒ n x sin x dx2

- pæ ö= ç ÷è øò

( )1

n 1

0

g n x cos x dx2

- pæ ö= ç ÷è øò , where n = 1,2,3,.......

ekuk ( )1

n 1

0

ƒ n x sin x dx2

- pæ ö= ç ÷è øò

( )1

n 1

0

g n x cos x dx2

- pæ ö= ç ÷è øò , tgk¡ n = 1,2,3,.......

11. ( ) ( )nlim n 1 ƒ n

®¥+ is equal to-

( ) ( )nlim n 1 ƒ n

®¥+ cjkcj gksxk -

(A) 0 (B) 2

p(C) 1 (D)

2

p

12.( ) ( )( ) ( )2n

3n 1 ƒ nlim

2n 1 g n®¥

+

+ is equal to-

( ) ( )( ) ( )2n

3n 1 ƒ nlim

2n 1 g n®¥

+

+ cjkcj gksxk -

(A) 3

2p(B)

3

2

p(C)

3

4(D)

3

2


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This Section contains 4 multiple choice questions. Each question has matching lists. The codes for the

lists. have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.


gSa ftuesa ls dsoy ,d lgh gSA

13. Let A(t) = [aij], is a matrix of order 3 × 3 given by ij

2cos t if i j

a 1 if i j 1

0 otherwise

=ìï= - =íïî

where |A(t)| is determinant value of matrix A(t).Match List-I with List-II and select the correct answer using the code given below the list.

List-I List-II

(P) ( ) ( )t 0lim A t A 4t

®(1) 0

(Q) maximum value of |A(t)| |A(3t)| (2) 1

(R)4

A A17 17

p pæ ö æ öç ÷ ç ÷è ø è ø

(3) 4

(S) ( ) ( )0

A t A 4t dtp

ò (4) 16

ekuk A(t) = [aij], dksfV 3 × 3 ds vkO;wg

=ìï= - =íïî

ij

2cos t i j

a 1 i j 1

0

;fn;fn

vU;Fkk }kjk fn;k x;k gSA

tgk¡ |A(t)|, vkO;wg A(t) dk lkjf.kd eku gSA

lwph-I dks lwph-II ls lqesfyr dhft, rFkk lwfp;ksa ds uhps fn, u, dksM dk iz;ksx djds lgh mÙkj pqfu;s %

lwph-I lwph-II

(P) ( ) ( )t 0lim A t A 4t

®(1) 0

(Q) |A(t)| |A(3t)| dk vf/kdre eku gksxk (2) 1

(R)4

A A17 17

p pæ ö æ öç ÷ ç ÷è ø è ø

(3) 4

(S) ( ) ( )0

A t A 4t dtp

ò (4) 16

Codes :P Q R S

(A) 4 4 2 1(B) 4 3 2 3(C) 1 4 2 1(D) 4 2 3 1

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14. Let (1 + x)62 = C0 + x C1 + x2 C2+..........+ x62 C62 where Cr = 62Cr.Match List-I with List-II and select the correct answer using the code given below the list.

List-I List-II(P) C0 + C4 + C8 +.....+ C60 is divisible by 2n, (1) 61

then maximum value of n is(Q) C1 + C5 + C9 +.....+ C61 is divisible by 2n, (2) 60

then maximum value of n is(R) C3 + C7 + C11 +.....+ C59 is divisible by 2n, (3) 31

then maximum value of n is(S) C2 + C6 + C10 +.....+ C62 is divisible by 2n, (4) 30

then value of n can not be

ekuk (1 + x)62 = C0 + x C1 + x2 C2+..........+ x62 C62 tgk¡ Cr = 62Cr gSA


lwph-I lwph-II(P) C0 + C4 + C8 +.....+ C60, 2n ls foHkkftr gks] rks n dk (1) 61

vf/kdre eku gksxk(Q) C1 + C5 + C9 +.....+ C61, 2n ls foHkkftr gks] rks n dk (2) 60

vf/kdre eku gksxk(R) C3 + C7 + C11 +.....+ C59, 2n ls foHkkftr gks] rks n dk (3) 31

vf/kdre eku gksxk(S) C2 + C6 + C10 +.....+ C62, 2n ls foHkkftr gks] rks n dk (4) 30

eku ugha gks ldrk gSCodes :

P Q R S(A) 2 4 3 1(B) 2 4 4 1(C) 1 4 4 3(D) 2 1 4 3


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15. The set A = {z : z18 = 1}, B = {w : w48 = 1}, C = {zw : z Î A and w Î B} are three sets of complex

roots of unity and D z : 0 arg(z)3pì ü= £ £í ý

î þMatch List-I with List-II and select the correct answer using the code given below the list.

List-I List-II(P) n(A Ç B) (1) 4

(Q) n(C) (2) 6(R) n(A Ç D) (3) 9(S) n(B Ç D) (4) 12

leqPp; A = {z : z18 = 1}, B = {w : w48 = 1}, C = {zw : z Î A rFkk w Î B} bdkbZ ds lfEeJ ewy

ds rhu leqPp; gS rFkk D z : 0 arg(z)3pì ü= £ £í ý

î þ


lwph-I lwph-II(P) n(A Ç B) (1) 4

(Q) n(C) (2) 6(R) n(A Ç D) (3) 9(S) n(B Ç D) (4) 12

Codes :P Q R S

(A) 2 4 2 3(B) 2 1 1 3(C) 3 4 1 2(D) 2 4 1 3


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16. Let ( ) ( )( )x12ƒ x x 3x 2

-= + + and two equations in n are

n

10k 1

log ƒ(k) 1=

=å .........(1)

n

10k 1

log ƒ(k) 2=

=å ..........(2)

Match List-I with List-II and select the correct answer using the code given below the list.List-I List-II

(P) If n is odd in equation (1), then n is equal to (1) 0(Q) If n is even in equation (1), then n is equal to (2) 1(R) Number of even values of n satisfying equation (2) is (3) 3(S) Number of odd values satisfying equation (2) is (4) 18

ekuk ( ) ( )( )x12ƒ x x 3x 2

-= + + rFkk n esa nks lehdj.kas

n

10k 1

log ƒ(k) 1=

=å .........(1)

n

10k 1

log ƒ(k) 2=

=å ..........(2)

gSa


lwph-I lwph-II(P) ;fn lehdj.k (1) esa] n fo"ke gks] rks n cjkcj gksxk (1) 0(Q) ;fn lehdj.k (1) esa] n le gks] rks n cjkcj gksxk (2) 1(R) lehdj.k (2) dks lUrq"V djus okys n ds le ekuksa dh la[;k gksxh (3) 3(S) lehdj.k (2) dks lUrq"V djus okys n ds fo"ke ekuksa dh la[;k gksxh (4) 18

Codes :P Q R S

(A) 1 4 3 2(B) 4 3 2 1(C) 3 4 2 1(D) 3 1 4 2




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bl [k.M esa 4 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd (nksuksa 'kkfey) ds chp dk ,dy vadh ; iw.kk ±d

gSA

1. Suppose ƒ is a function on the interval [1,3] such that –1 < ƒ(x) < 1 for all x and ( )3

1

ƒ x dx 0=ò .

If maximum value of ( )3

1

ƒ xdx

xò is a

nb

æ öç ÷è ø

l (where a,b are coprime numbers), then a + b is equal to

ekuk ƒ, vUrjky [1,3] esa ,d Qyu bl izdkj gS fd lHkh x ds fy, –1 < ƒ(x) < 1 rFkk ( )3

1

ƒ x dx 0=ò gSA ;fn

( )3

1

ƒ xdx

xò dk vf/kdre eku an

bæ öç ÷è ø

l (tgk¡ a,b lg&vHkkT; la[;k;sa gSa), rks a + b cjkcj gksxk

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2. In the given figure, two circles with radii 6 and 8 are drawn with centers (C1 and C2), 12 units apart. At

P, one of the points of intersection, a line QR is drawn in such a way that the chords QP and PR have

equal length (P is the mid point of QR). Then 21 QP26

is equal to

fn;s x;s vkjs[k esa] nks o`Ùk ftudh f=T;k;sa 6 rFkk 8 gS] rFkk ftuds dsUæ (C1 rFkk C2) ,d&nwljs ls 12 bdkbZ dh nwjh ij

fLFkr gSA izfrPNsnh fcUnqvksa esa ls fdlh ,d fcUnq P ij ,d js[kk QR dks bl izdkj [khapha tkrh gS fd thokvksa QP rFkk

PR dh yEckbZ leku jgs (P, QR dk e/; fcUnq gS)A rc 21 QP26

cjkcj gksxk

QP

RC1 C2

3. y = ƒ(x) is a solution of differential equation x

x

dy 2xey

dx 1 ye

-

+ =+

such that ƒ(0) = 1, then ( )ƒ 1

e

é ù-ê úë û

is equal

to {where [.] greatest integer function}

y = ƒ(x) vody lehdj.k x

x

dy 2xey

dx 1 ye

-

+ =+

dk gy bl izdkj gS fd ƒ(0) = 1 gks] rks ( )ƒ 1

e

é ù-ê úë û

gksxk

{tgk¡ [.] egÙke iw.kk±d Qyu dks n'kkZrk gS}

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4. In a DABC, AB = 85, BC = 90 and AC = 102. P is an interior point and line segments are drawn through

P parallel to the sides of the triangle. If these three line segments (as shown in the diagram by RQ, R'Q',

R''Q'') are of equal length d, then 5d34

is equal to

f=Hkqt ABC esa] AB = 85, BC = 90 rFkk AC = 102 gSA ,d vkUrfjd fcUnq P ls Hkqtkvksa ds lekukUrj js[kk[k.M [khaps

tkrs gSaA ;fn ;g rhu js[kk[k.M] (vkjs[k esa RQ, R'Q', R''Q'' }kjk n'kkZ;k x;k gS) leku yEckbZ d ds gks] rks 5d34

cjkcj gksxk

R'' R'

R Q

C

B Q' Q'' A

P


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D. vadu ;kstuk / Marking scheme :14. [kaM-I (i) ds gj iz'u esa dsoy lgh mÙkjksa (mÙkj) okys lHkh cqycqyksa (cqycqys) dks dkyk djus ij 4 vad vkSj dksbZ Hkh cqycqyk dkyk ugha

djus ij 'kwU; (0) vad iznku fd;k tk;sxkA vU; lHkh fLFkfr;ksa esa ½.kkRed ,d (–1) vad iznku fd;k tk;sxkAFor each question in Section-I (i), you will be awarded 4 marks if you darken all the bubble(s) corresponding to only thecorrect answer(s) and zero mark if no bubbles are darkened. In all other cases minus one (–1) mark will be awarded

15. [kaM-I ( ii & iii) ds gj iz'u esa dsoy lgh mÙkjksa (mÙkj) okys lHkh cqycqyksa (cqycqys) dks dkyk djus ij 3 vad vkSj dksbZ Hkh cqycqyk dkyk ughadjus ij 'kwU; (0) vad iznku fd;k tk;sxkA vU; lHkh fLFkfr;ksa esa ½.kkRed ,d (–1) vad iznku fd;k tk;sxkAFor each question in Section-I (ii & iii), you will be awarded 3 marks if you darken all the bubble(s) corresponding toonly the correct answer(s) and zero mark if no bubbles are darkened. In all other cases minus one (–1) mark will beawarded.

16. [kaM-IV esa gj iz'u esa dsoy lgh mÙkj okys cqycqys (BUBBLE) dks dkyk djus ij 3 vad vkSj dksbZ Hkh cqycqyk dkyk ugha djusij 'kwU; (0) vad iznku fd;k tk;sxk bl [ akM ds iz'uksa esa xyr mÙkj nsus ij dksbZ ½.kkRed vad ugha fn;s tk;saxsaAFor each question in Section-IV, you will be awarded 3 marks if you darken the bubble corresponding to the correctanswer and zero mark if no bubbles are darkened No negative marks will be awarded for incorrect answers inthis section.

17. g = 10 m/s2 iz;qDr djsa] tc rd fd vU; dksbZ eku ugha fn;k x;k gksATake g = 10 m/s2 unless otherwise stated.

Name of the Candidate / ijh{kkFkhZ dk uke

I have read all the instructions and shall abide by them.eSusa lHkh vuqns'kksa dks i<+ fy;k gS vkSj eSa mudk vo'; ikyu d:¡xk@d:¡xhA

Signature of the Candidate / ijh{kkFkhZ ds gLrk{kj

Form Number / QkWeZ la[;k

I have verified all the information filled in by the Candidate.ijh{kkFkhZ }kjk Hkjh xbZ tkudkjh dks eSus a tk¡p fy;k gSA

Signature of the Invigilator / fujh{kd ds gLrk{kj


Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005

+91-744-2436001 [email protected]

Appropriate way of darkening the bubble for your answer to be evaluatedvkids mÙkj ds ewY;kadu ds fy, cqycqys dk s dkyk djus dk mi;qDr rjhdk

a

a

a

a

a

a

The one and the only acceptable,d vkSj dsoy ,d Lohdk;Z

Part darkeningvkaf'kd dkyk djuk

Darkening the rimfje dkyk djuk

Cancelling after darkeningdkyk djus ds ckn jn~n djuk

Erasing after darkeningdkyk djus ds ckn feVkuk

Answer will not be evaluated -no marks, no negative marks

mÙkj dk ewY;kadu ugha gksxk&dksbZ vad ugha] dksbZ ½.kkRed vad ugha

Figure-1 : Correct way of bubbling for valid answer and a few examplex of invalid answers fp=&1 % oS/k mÙkj ds fy, cqycqyk Hkjus dk lgh rjhdk vkSj voS/k mÙkjks a ds dqN mnkgj.kAAny other form of partial marking such as ticking or crossing the bubble will be invalidvkaf'kd vadu ds vU; rjhds tSls cqycqys dks fVd djuk ;k ØkWl djuk xyr gksxkA

10

234

6789

4 2 0 0 0 20

2

0

2 2 23 3 3 3 3

0

4 4 4 45 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

1 1 1 1 1

20

456789

1

54

5

3

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Figure-2 : Correct Way of Bubbling your Form Number on the ORS. (Example Form Numebr : 14200022)fp=&2 % vks-vkj-,l (ORS) ij vkids QkWeZ uEcj ds ccy dks Hkjus dk lgh rjhdkA (mnkgj.k QkWeZ uEcj : 14200022)

Kota/00CT214006Your Target is to secure Good Rank in JEE 201540/40

ALLEN 0 0 2 1 4 0 0 6 CLAS AMME · A andida writ i / e we i th ORS ee darkenin ropri ubbl wit hel f...

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