chemistry, mathematics & physics - career point(a) higher colligative properties (b) lower...

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TARGET – IIT JEE CHEMISTRY, MATHEMATICS & PHYSICS Time : 3 : 00 Hrs. MAX MARKS: 246 Name : _____________________________________ Roll No. : __________________________ Date : _____________ 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. 4 marks will be awarded for each correct answer and No negative marking for wrong answer. 6. Reason and Assertion type questions with only one correct answer in each. 3 marks will be awarded for each correct answer and –1 mark for each wrong answer. 7. Passage based single correct type questions. 4 marks will be awarded for each correct answer and –1 mark for each wrong answer. C.FILLING THE OMR : 8. Fill your Name, Roll No., Batch, Course and Centre of Examination in the blocks of OMR sheet and darken circle properly. 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 T A B C D Ensure that all columns are filled. Answers, having blank column will be treated as incorrect. Insert leading zeros (s) 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 '6' should be filled as 0006 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 '86' should be filled as 0086 0 0 0 0 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 '1857' should be filled as 1857 CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000, Fax (0744) 3040050 email : [email protected]; Website : www.careerpointgroup.com 7 RS -11- I -19 SEAL

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Page 1: CHEMISTRY, MATHEMATICS & PHYSICS - Career Point(A) higher colligative properties (B) lower colligative properties (C) same colligative properties (D) none of these Q.5 The solubilities

1

TARGET – IIT JEE

CHEMISTRY, MATHEMATICS & PHYSICS

Time : 3 : 00 Hrs. MAX MARKS: 246

Name : _____________________________________ Roll No. : __________________________ Date : _____________

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. 4 marks will be awarded for each correct answer and No negative

marking for wrong answer. 6. Reason and Assertion type questions with only one correct answer in each. 3 marks will be awarded for each correct

answer and –1 mark for each wrong answer.

7. Passage based single correct type questions. 4 marks will be awarded for each correct answer and –1 mark for each wrong answer.

C. FILLING THE OMR :

8. Fill your Name, Roll No., Batch, Course and Centre of Examination in the blocks of OMR sheet and darken circle properly. 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

3

4

56

7

8

9

012

3

4

56

7

8

9

012

3

4

56

7

8

9

'6' should be filled as 0006

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

'86' should be filled as 0086

0 0 0 00 1 2

3

4

5 6

7

8

9

012

3

4

56

7

8

9

012

3

4

56

7

8

9

012

3

4

56

7

8

9

'1857' should be filled as 1857

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000, Fax (0744) 3040050 email : [email protected]; Website : www.careerpointgroup.com 7

RS -11- I -19

SEA

L

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CHEMISTRY

Section - I Questions 1 to 6 are multiple choice questions. Eachquestion has four choices (A), (B), (C) and (D), out ofwhich ONLY ONE is correct. Mark your response in OMR sheet against the question number of thatquestion. + 3 marks will be given for each correct answerand – 1 mark for each wrong answer.

Q.1 The end product (A) of the reaction is

O

O (i) I2 + NaOH, ∆

(ii) H3O+, ∆ CHI3 + A ; A is

(A)

COOH

O (B)

CHO

O

(C) COOH

COOH (D)

O

Q.2 The end product in the reaction sequence is

HBr Benzoyl peroxide

Mg ether A B CO2

H3O+ C ; C is

(A)

COOH (B) COOH

(C) CH2COOH

(D) None of these

Q.3 Among the following acids which has the lowestpKa value ?

(A) CH3COOH (B) HCOOH (C) (CH3)2CH – COOH (D) CH3CH2COOH

[k.M - I iz'u 1 ls 6 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 lek viukmÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy, + 3 vad fn;stk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA

Q.1 fuEu vfHkfØ;k dk vfUre mRikn A gS

O

O(i) I2 + NaOH, ∆

(ii) H3O+, ∆ CHI3 + A

(A)

COOH

O (B)

CHO

O

(C) COOH

COOH (D)

O

Q.2 fuEu vfHkfØ;k vuqØe esa vfUre mRikn C gS &

HBrBenzoylperoxide

Mg ether A B CO2

H3O+ C

(A)

COOH (B) COOH

(C)

CH2COOH (D) buesa ls dksbZ ugha

Q.3 fuEu esa ls fdldk pKa eku U;wure gS ? (A) CH3COOH (B) HCOOH (C) (CH3)2CH – COOH (D) CH3CH2COOH

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Q.4 Compared to common colloidal sols micelles have : (A) higher colligative properties (B) lower colligative properties (C) same colligative properties (D) none of these

Q.5 The solubilities of MX, MX2, and M3X at 25ºC are 10–6

mol dm–3, 10–4 mol dm–3 and 10–3 mol dm–3 respec-tively. Ksp

values of the salt at 25ºC are in the order. (A) MX > MX2 > M3X (B) M3X > MX2 > MX (C) MX2 > M3X > MX (D) MX > M3X > MX2 Q.6 Electrolysis of water was carried out by passing

0.965 A current for 2000 seconds. Amount of O2

liberated is (in milli moles) (A) 2.5 (B) 5.0 (C) 10 (D) 20

Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and NO NEGATIVE marks for wrong answer. Q.7 When a mixture of calcium acetate and calcium

formate is subjected to dry distillation, it yields. (A) acetone (B) acetaldehyde (C) formaldehyde (D) acetic acid

Q.8 The reagent(s) used to distinguish but-1-yne from but-2-yne is/are

(A) Br2/CCl4 (B) KMnO4/OH– (C) Cu2Cl2 + NH4OH (D) [Ag(NH3)2]OH

Q.4 lkekU; dksykbMh lksyksa dh rqyuk esa felsy ds & (A) v.kqla[;d xq.k/keZ mPp gksrs gS (B) v.kqla[;d xq.k/keZ de gksrs gSa (C) v.kqla[;d xq.k/keZ leku gksrs gS (D) buesa ls dksbZ ugha

Q.5 25ºC ij MX, MX2, rFkk M3X ds foys;rk xq.kuQyØe'k% 10–6 mol dm–3, 10–4 mol dm–3 rFkk 10–3 moldm–3 gSA 25ºC ij yo.k ds Ksp

ekuksa dk Øe gS (A) MX > MX2 > M3X (B) M3X > MX2 > MX (C) MX2 > M3X > MX (D) MX > M3X > MX2

Q.6 0.965 A /kkjk dks 2000 lsd.M rd izokfgr dj tydk fo|qqr vi?kVu fd;k x;kA eqDr gqbZ O2 dh ek=kk(feyh eksy esa) gS

(A) 2.5 (B) 5.0 (C) 10 (D) 20

iz'u 7 ls 10 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkjfodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k vf/kdfodYi lgh gaSA OMR 'khV esa iz'u dh iz'u la[;k ds lekvius mÙkj vafdr dhft,A izR;sd lgh mÙkj ds fy, + 4 vadfn;s tk;saxs rFkk xyr mÙkj ds fy, dksbZ _.kkRed vaduugha gSA Q.7 tc dsfY'k;e ,flVsV rFkk dsfY'k;e QkWesZV ds feJ.k dk

'kq"d vklou fd;k tkrk gSa] rks mRikn izkIr gksaxs & (A) ,flVksu (B) ,flVsfYMgkbM (C) QkWesZfYMgkbM (D) ,flfVd vEy

Q.8 C;wV-1-vkbu rFkk C;wV-2-vkbu esa foHksn djus ds fy,fdu vfHkdeZdksa dk mi;ksx fd;k tkrk gSa

(A) Br2/CCl4 (B) KMnO4/OH– (C) Cu2Cl2 + NH4OH (D) [Ag(NH3)2]OH

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Q.9 A solution of colourless salt H on boiling withexcess NaOH produces a non-flammable gas. Thegas evolution stop after some time. Upon additionof Zn dust to the same solution, the gas evolutionrestarts. The colourless salt(s) H is (are)

(A) NH4NO3 (B) NH4NO2 (C) NH4Cl (D) (NH4)2SO4

Q.10 A gas described by van der waals equation (A) behave similar to an ideal gas in the limit of

large molar volumes (B) behaves similar to an ideal gas is in limit of

large pressures (C) is characterized by van der waals coefficients

that are dependent on the identity of the gas butare independent of the temperature.

(D) has the pressure that is lower than the pressureexerted by the same gas behaving ideally

This section contains 4 questions numbered 11 to 14,(Reason and Assertion type question). Each questioncontains Assertion and Reason. Each question has 4choices (A), (B), (C) and (D) out of which ONLY ONE iscorrect. Mark your response in OMR sheet against thequestion number of that question. + 3 marks will begiven for each correct answer and – 1 mark for eachwrong answer. The following questions given below consist of

an "Assertion" (A) and "Reason" (R) Typequestions. Use the following Key to choose theappropriate answer.

Q.9 jaxghu yo.k H ds foy;u dks vf/kD; NaOH ds lkFk mckyus ij vToyu'khy xSl mRiUu gksrh gSA dqN le; i'pkr~ xSl mRiUu gksuk :d tkrk gSA bl foy;u esa Zn pw.kZ Mkyus ij xSl iqu% mRiUugksrh gSA jaxghu yo.k H gSaA

(A) NH4NO3 (B) NH4NO2 (C) NH4Cl (D) (NH4)2SO4

Q.10 ok.Mjoky lehdj.k ds kjk xSl dks le>krs gS & (A) cgqr vf/kd eksyj vk;ru dh lhek esa vkn'kZ

xSl dh rjg O;ogkj djrh gSA (B) cgqr vf/kd nkc dh lhek esa vkn'kZ xSl dh rjg

O;ogkj djrh gSA (C) ;g ok.Mjoky fu;rkadkssa kjk le>k;h tkrh gSa

tks fd xSl dh fo'ks"krk ij fuHkZj djrs gS ysfdu rki ij fuHkZj ugha djrs gSa

(D) bldk nkc] vkn'kZ O;ogkj dh leku xSl kjk vkjksfir fd;k tkrk gS] ls de gksrk gS

bl [k.M esa 11 ls 14 rd 4 iz'u gSa, (dFku rFkk dkj.k izdkj ds iz'u)A izR;sd iz'u esa dFku rFkk dkj.k fn;s x;s gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) fn;s x;s gSa] ftuesa ls dsoy ,d lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy;s + 3 vad fn;s tk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA

uhps fn;s x;s fuEufyf[kr iz'u "dFku" (A) rFkk "dkj.k" (R) izdkj ds iz'u gSaA vr% mfpr mÙkj dk p;u djus ds fy;s fuEu rkfydk dk mi;ksx dhft;sA

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(A) If both (A) and (R) are true, and (R) is thecorrect explanation of (A).

(B) If both (A) and (R) are true but (R) is not thecorrect explanation of (A).

(C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true.

Q. 11 Assertion : The pKa value of acetic acid is lowerthan that of phenol.

Reason : Phenoxide is more resonance stabilizedthan carboxylate ion.

Q.12 Assertion : Upon heating an amide with Br2 and KOH, primary amine with one carbon atom less is formed.

Reason : The reaction occurs through carbene intermediate.

Q.13 Assertion : Alkali metals dissolve in liquidammonia to give blue solutions.

Reason : Alkali metals in liquid ammonia givesolvated species of the type [M(NH3)n]+

(M = alkali metals)

Q.14 Assertion : Band gap in germanium is small. Reason : The energy spread of each germanium

atomic energy level is infinitesimally small

This section contains 3 paragraphs, each has 3 multiplechoice questions. (Questions 15 to 23) Each question has4 choices (A), (B), (C) and (D) out of which ONLY ONEis correct. Mark your response in OMR sheet against thequestion number of that question. + 4 marks will begiven for each correct answer and – 1 mark for eachwrong answer.

(A) ;fn dFku rFkk dkj.k nksuksa lR; gSa rFkk dkj.k, dFku dk lgh Li"Vhdj.k gSA

(B) ;fn dFku rFkk dkj.k nksuksa lR; gSa rFkk dkj.k, dFku dk lgh Li"Vhdj.k ugha gSA

(C) ;fn dFku lR; gS ysfdu dkj.k vlR; gSA (D) ;fn dFku vlR; gS ysfdu dkj.k lR; gSA

Q. 11 dFku : ,flfVd vEy dk pKa eku fQukWy ls de gksrk gSA

dkj.k : fQukWDlkbM] dkcksZDlhysV vk;u dh viskk vf/kd vuquknh LFkk;h gksrk gSA

Q.12 dFku : ,ekbM dks Br2 rFkk KOH ds lkFk xeZ djus ij ,d dkcZu de okyh izkFkfed ,ehu curh gSA

dkj.k : vfHkfØ;k dkchZu e/;orhZ kjk gksrh gSA Q.13 dFku : nzo veksfu;k esa kkj /kkrqvksa dks ?kksyus ij

uhyk foy;u izkIr gksrk gS dkj.k : nzo veksfu;k esa kkj /kkrq,sa [M(NH3)n]+ izdkj

dh foyk;dhÑr Lih'kht nsrh gSA (M = kkj /kkrq,sa)

Q.14 dFku : tesZfe;e eas cS.M pkSM+kbZ de gksrh gSA dkj.k : izR;sd tesZfu;e ijek.kq ds ÅtkZ Lrj dk

ÅtkZ foLrkj vuUr lwe:i ls de gksrk gSA bl [k.M esa 3 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih iz'u gSaA (iz'u 15 ls 23) izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy;s + 4 vad fn;s tk,saxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA

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vuqPNsn # 1 (iz- 15 ls 17) gkWQeku cksesekbM vfHkfØ;k esa NaOH + Br2 ds mi;ksx

kjk isrd ,ekbM ls ,d dkcZu de okyh ,ehu cuk;h tk ldrh gSA lkekU; vfHkfØ;k bl rjg gS &

RCONH2 + Br2 + 4NaOH → R–NH2 + 2NaBr + Na2CO3 +2H2O fØ;kfof/k : Br2+ 2NaOH → NaBr + NaOBr + H2O

R – C – NH2

O+ OBr– →

−−OH

R – C – N – Br

O

H → −− OH,OH 2

R –– C –– NO

Nitrene

Alkylshift R – N = C = O →

−OH2 R – NH2 + CO3–2

Isocyanate ,ekbM dks LiAlH4 ds mi;ksx ls leku dkcZu okyh

,ehu esa ifjofrZr fd;k tk ldrk gSA

Q.15 tc ,flVkekbM dks lksfM;e gkbiksczksekbV ds lkFk xeZ fd;k tkrk gS rks mRikn cusxk

(A) CH3CN (B) CH3OH (C) CH3NH2 (D) CH3COBr

Q.16 fuEu esa ls dkSulk gkWQeku iwufoZU;kl esa e/;orhZ ds :i esa dk;Z ugha djrk gS ?

(A) RNCO (B) ••

NRCO

(C) RCO••

N Br (D) RNC

Passage # 1 (Ques. 15 to 17) Amines with one carbon less than the parent amide

can be prepared by Hofmann's bromamide reactionusing NaOH + Br2. The general reaction is

RCONH2 + Br2 + 4NaOH → R–NH2 + 2NaBr + Na2CO3 +2H2OMechanism : Br2+ 2NaOH → NaBr + NaOBr + H2O

R – C – NH2

O + OBr– →

−−OH

R – C – N – Br

O

H → −− OH,OH 2

R –– C –– NO

Nitrene

Alkyl shift R – N = C = O →

−OH2 R – NH2 + CO3–2

Isocyanate Amides can be converted into amines with the

same number of carbons using LiAlH4.

Q.15 When acetamide is treated with sodiumhypobromite, the product formed is

(A) CH3CN (B) CH3OH (C) CH3NH2 (D) CH3COBr

Q.16 Among the following which one does not act as anintermediate in Hoffmann rearrangement ?

(A) RNCO (B) ••

NRCO

(C) RCO••

N Br (D) RNC

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Q.17 Identify the product of the reaction C2H5 – C – C

CH3

H NH2

O

(R)-2-methylbutanamide

Br2, OH–

(A) (R)-s-butylamine (B) (S)-s-butylamine (C) (R)-s- pentylamine (D) (S)-s-pentylamine Passage # 2 (Ques. 18 to 20)

Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential (Eº) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their Eº (V with respect to normal hydrogen electrode) values. Using this data obtain the correct explanations to questions 18 to 20.

I2+ 2e– → 2I– Eº = 0.54 Cl2 + 2e– → 2Cl– Eº = 1.36 Mn3+ + e– → Mn2+ Eº = 1.50 Fe3+ + e– → Fe2+ Eº = 0.77 O2 + 4H+ + 4e– → 2H2O Eº = 1.23

Q.18 Among the following, identify the correct statement :

(A) Chloride ion is oxidised by O2 (B) Fe2+ is oxidized by iodine (C) Iodide ion is oxidized by chlorine (D) Mn2+ is oxidized by chlorine

Q.17 fuEu vfHkfØ;k ds mRikn dh igpku dhft, C2H5 – C – C

CH3

H NH2

O

(R)-2-esfFky C;wVsusekbM

Br2, OH–

(A) (R)-s-C;wfVy,ehu (B) (S)-s-C;wfVy,ehu (C) (R)-s-isfUVy,ehu (D) (S)-s-isfUVy,ehu

vuqPNsn # 2 (iz- 18 ls 20) jlk;u foKku rFkk tho foKku esa jsMkWDl vfHkfØ;k,sa

vk/kkjHkwr Hkwfedk fuHkkrh gSA nks v)Zlsy vfHkfØ;kvksa ds ekud jsMkWDl foHko (Eº) ds eku ;g fuf'pr djrs gS fd vfHkfØ;k fdl iFk ij vxzlj gSA ljy mnkgj.k gS Msfu;y lSy ftlesa ftad foy;u esa fey tkrk gS rFkk dkWij tek gksrk gSA uhpsvZlsy vfHkfØ;kvksa ¼vEyh; ek/;e esa½ ds lsV buds Eº ¼V lkekU; gkbMªkstu bysDVªkWM ds lanHkZ esa½ eku ds lkFk fn;sx;s gSA bu vkdM+ksa dk mi;ksx iz'u 18 ls 20 esas dhft,

I2+ 2e– → 2I– Eº = 0.54 Cl2 + 2e– → 2Cl– Eº = 1.36 Mn3+ + e– → Mn2+ Eº = 1.50 Fe3+ + e– → Fe2+ Eº = 0.77 O2 + 4H+ + 4e– → 2H2O Eº = 1.23 Q.18 fuEu esa lgh dFku dh igpku dhft,A (A) Dyksjhu vk;u] O2 kjk vkWDlhÑr gksrk gS (B) Fe2+, vk;ksMhu kjk vkWDlhÑr gksrk gS (C) vk;ksMkbM vk;u] Dyksjhu kjk vkWDlhÑr gksrk gS (D) Mn2+, Dyksjhu kjk vkWDlhÑr gksrk gS

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Q.19 While Fe3+ is stable, Mn3+ is not stable in acidsolution because :

(A) O2 oxidises Mn2+ to Mn3+ (B) O2 oxidises both Mn2+ to Mn3+ and Fe2+ to Fe3+

(C) Fe3+ oxidises H2O to O2 (D) Mn3+ oxidises H2O to O2

Q.20 Sodium fusion extract, obtained from aniline, ontreatment with iron (II) sulphate and H2SO4 in presence of air gives a Prussian blue precipitate.The blue colour is due to the formation of :

(A) Fe4[Fe(CN)6]3 (B) Fe3[Fe(CN)6]2 (C) Fe4[Fe(CN)6]2 (D) Fe3[Fe(CN)6]3

Passage # 3 (Ques. 21 to 23) NiCl2 HCN

KCN → complex A

NiCl2 excessHCl → complex B

A & B complexes have the co-ordination number 4.Q.21 The IUPAC name of complexes ‘A’ & ‘B’ are

respectively : (A) Potassium tetracyanonicklate (II) and Potassium tetrachloronicklate (II) (B) Potassium tetracyanonickel (II) and Potassium tetrachloronickel (II) (C) Potassium cyanonicklate (II) and Potassium chloronicklate (II) (D) Potassium cyanonickel (II) and Potassium chloronickel (II)

Q.19 vEyh; foy;u esa Fe3+ LFkk;h gS tcfd Mn3+ LFkk;h ugha gksrk gS D;ksafd

(A) O2, Mn2+ dks Mn3+ esa vkWDlhÑr djrk gS (B) O2, Mn2+ dks Mn3+ esa rFkk Fe2+ dks Fe3+ esa

vkWDlhÑr djrk gS (C) Fe3+, H2O dks O2 esa vkWDlhÑr djrk gS (D) Mn3+, H2O dks O2 esa vkWDlhÑr djrk gS

Q.20 ,fuyhu ls izkIr lksfM;e laxfyr fu"d"kZ dks ok;q dh mifLFkfr esa vk;ju (II) lYQsV rFkk H2SO4 ds lkFk xeZ djus ij izqf'k;u uhyk voksi izkIr gksrk gSA uhyk jax fdlds cuus ds dkj.k vkrk gS :

(A) Fe4[Fe(CN)6]3 (B) Fe3[Fe(CN)6]2 (C) Fe4[Fe(CN)6]2 (D) Fe3[Fe(CN)6]3

vuqPNsn # 3 (iz- 21 ls 23)

NiCl2 HCNKCN → ladqy A

NiCl2 excessHCl → ladqy B

A o B ladqyksa dh milgla;kstu la[;k 4 gSA

Q.21 ladqy ‘A’ o ‘B’ dk IUPAC uke Øe'k% gS (A) iksVsf'k;e VsVªklk;uksfudysV (II) rFkk iksVsf'k;e VsVªkDyksjksfudysV (II) (B) iksVsf'k;e VsVªklk;uksfudy (II) rFkk iksVsf'k;e VsVªklk;uksfudy (II) (C) iksVsf'k;e lkbuksfudysV (II) rFkk iksVsf'k;e DyksjksfudysV (II) (D) iksVsf'k;e lkbuksfudy (II) rFkk iksVsf'k;e Dyksjksfudy (II)

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Q.22 The hybridization of both complexes are : (A) dsp2 (B) sp2 & dsp2 (C) dsp2 & sp3 (D) both sp3 Q.23 What are the magnetic nature of ‘A’ & ‘B’ ? (A) Both diamagnetic (B) ‘A’ is diamagnetic & ‘B’ is paramagnetic with

one unpaired electrons (C) ‘A’ is diamagnetic & ‘B’ is paramagnetic with

two unpaired electrons (D) Both are paramagnetic

Q.22 nksuksa ladqyksa dk ladj.k gSa (A) dsp2 (B) sp2 o dsp2 (C) dsp2 o sp3 (D) nksuksa dk sp3

Q.23 ‘A’ rFkk ‘B’ dh pqEcdh; izÑfr D;k gS ? (A) nksuksa izfrpqEcdh; (B) ‘A’ izfrpqEcdh; rFkk ‘B’ ,d v;qfXer bysDVªkWu

;qDr vuqpqEcdh; (C) ‘A’ izfrpqEcdh; rFkk ‘B’ nks v;qfXer bysDVªkWuksas

;qDr vuqpqEcdh; (D) nksuksa vuqpqEcdh; gSa

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MATHEMATICS

Section - I Questions 1 to 6 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 mark for each wrong answer.

Q. 1 The polynomial f(x) = x4 + ax3 + bx2 + cx + d hasreal coefficients and f (2i) = f(2 + i) = 0. The value of (a + b + c + d) equals to-

(A) 1 (B) 4 (C) 9 (D) 10

Q. 2 If the sum ∑∞

= +++1k 2kkk)2k(1 =

cba +

where a, b, c ∈ N and lie in [1, 15], then a + b + cequals to-

(A) 6 (B) 8 (C) 10 (D) 11 Q. 3 Triangle ABC is isosceles with AB = AC and BC =

65 cm. P is a point on BC such that theperpendicular distances from P to AB and AC are24 cm and 36 cm, respectively. The area oftriangle ABC in sq cm is-

(A) 1254 (B) 1950

(C) 2535 (D) 5070

[k.M - I iz'u 1 ls 6 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 lek viukmÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy, + 3 vad fn;stk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA

Q. 1 cgqin f(x) = x4 + ax3 + bx2 + cx + d ds xq.kk¡dokLrfod gksa rFkk f (2i) = f(2 + i) = 0 rks(a + b + c + d) dk eku gS -

(A) 1 (B) 4 (C) 9 (D) 10

Q. 2 ;fn ∑∞

= +++1k 2kkk)2k(1 =

cba + tgk¡

a, b, c ∈ N ,oa varjky [1, 15] esa fLFkr gaS, rksa + b + c dk eku gS -

(A) 6 (B) 8 (C) 10 (D) 11 Q. 3 f=kHkqt ABC lefckgq gS] ftlesa AB = AC ,oa

BC = 65 cm gSA ;fn fcUnq P Hkqtk BC ij bl izdkj lsgS fd fcUnq P dh AB o AC ls yEcor~ nwfj;k¡ Øe'k% 24cm o 36 cm gks rks f=kHkqt ABC dk kS=kQy oxZ lseh esagksxk -

(A) 1254 (B) 1950

(C) 2535 (D) 5070

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Q. 4 Three boxes are labelled A, B and C and each box contains four balls numbered 1, 2, 3 and 4. The balls in each box are well mixed. A child chooses one ball at random from each of the three boxes. If a, b and c are the numbers on the balls chosen from the boxes A, B and C resepctively, the child wins a toy helicopter when a = b + c. The odds in favour of the child to receive the toy helicopter are-

(A) 3 : 32 (B) 3 : 29 (C) 1 : 15 (D) 5 : 59 Q. 5 If N = 7p+4.5q.23 is a perfect cube, where 'p' and 'q'

are positive integers, the smallest possible value of 'p + q' is-

(A) 5 (B) 2 (C) 8 (D) 6

Q. 6 The radius of the circle inscribed in a triangle with sides 12, 35 and 37, is-

(A) 4 (B) 5 (C) 6 (D) 7

Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and NO NEGATIVE marks for wrong answer. Q. 7 Let f(x) be a function defined by

f(x) = ∫ +−x

1

2 dx)2x3x(x , 1 ≤ x ≤ 4. Then the

(A) maximum value of f(x) is 53/4 (B) maximum value of f(x) is 63/4 (C) minimum value of f(x) is –1/2 (D) minimum value of f(x) is –1/4

Q. 4 rhu lanwd ftu ij A, B o C vafdr gS rFkk izR;sd lanwd esa pkj xsansa ftu ij 1, 2, 3 o 4 vad n'kkZ;s x;s gSa] j[kh xbZ gSA izR;sd lanwd dh xsnsa Hkyh Hkk¡fr lqesfyr gSA ,d cPpk bu rhuksa lanwdksa esa ls izR;sd ls ;knPN;k ,d xsan dk p;u djrk gS] ;fn a, b o c Øe'k% lanwd A, B o C ls p;u dh xbZ xsanksa ij vk;s vad gksaA cPpk ,d gSyhdkWIVj f[kykSuk thrrk gS ;fn a = b + c rks cPps ds gSyhdkWIVj f[kykSuk thrus ds ik esa la;ksxkuqikr gS -

(A) 3 : 32 (B) 3 : 29 (C) 1 : 15 (D) 5 : 59

Q. 5 ;fn N = 7p+4.5q.23 ,d iw.kZ ?ku gS] tgk¡ 'p' o 'q' /kukRed iw.kkZd gS] rks 'p + q' dk U;wure laHkkfor eku gksxk-

(A) 5 (B) 2 (C) 8 (D) 6

Q. 6 ,d f=kHkqt ftldh Hkqtk,a 12, 35 rFkk 37 gS] ds vUnj cuk, x, oÙk dh f=kT;k gS -

(A) 4 (B) 5 (C) 6 (D) 7

iz'u 7 ls 10 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj fodYi(A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k vf/kd fodYi lgh gaSAOMR 'khV esa iz'u dh iz'u la[;k ds lek vius mÙkj vafdr dhft,A izR;sd lgh mÙkj ds fy, + 4 vad fn;s tk;saxs rFkk xyr mÙkj ds fy, dksbZ _.kkRed vadu ugha gSA Q. 7 ekukfd f(x) ,d Qyu gS tks bl izdkj ifjHkkf"kr gS

fd f(x) = ∫ +−x

1

2 dx)2x3x(x , 1 ≤ x ≤ 4 rks

(A) f(x) dk vf/kdre eku 53/4 gS (B) f(x) dk vf/kdre eku 63/4 gS (C) f(x) dk U;wure eku –1/2 gS (D) f(x) dk U;wure eku –1/4 gS

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Q. 8 ∫π

0

dx)x(sinfx is equal to-

(A) ∫π

π

0

dx)x(sinf2

(B) ∫π

π2/

0

dx)x(sinf

(C) ∫π

π2/

0

dx)x(sinf2 (D) None of these

Q. 9 Let the algebraic sum of perpendicular distancesfrom the points A(2, 0), B(0, 2), C(1, 1) to avariable line be zero, then all such lines

(A) are concurrent (B) passes through fixed point (C) touches a given circle (D) passes through orthocenter of triangle ABC

Q. 10 Value of

π−

451tan2tan 1 is-

(A) 177

− (B) 173

− (C) 3

17− (D) less than

31

This section contains 4 questions numbered 11 to 14,(Reason and Assertion type question). Each questioncontains Assertion and Reason. Each question has 4choices (A), (B), (C) and (D) out of which ONLY ONE iscorrect. Mark your response in OMR sheet against thequestion number of that question. + 3 marks will begiven for each correct answer and – 1 mark for eachwrong answer.

Q. 8 ∫π

0

dx)x(sinfx cjkcj gS -

(A) ∫π

π

0

dx)x(sinf2

(B) ∫π

π2/

0

dx)x(sinf

(C) ∫π

π2/

0

dx)x(sinf2 (D) buesa ls dksbZ ugh

Q. 9 ekuk fcUnqvksa A(2, 0), B(0, 2), C(1, 1) ls fdlh pj js[kk ij Mkys x, yEcksa dk chth; ;ksx 'kwU; gS rc bl izdkj dh js[kk,a

(A) laxkeh gS (B) vpj fcUnq ls xqtjsxh (C) ,d fn, x, oÙk dks Li'kZ djsaxh (D) f=kHkqt ABC ds yEc dsUnz ls xqtjsaxh

Q. 10

π−

451tan2tan 1 dk eku gS-

(A) 177

− (B) 173

− (C) 3

17− (D)

31

− ls de

bl [k.M esa 11 ls 14 rd 4 iz'u gSa, (dFku rFkk dkj.k izdkj ds iz'u)A izR;sd iz'u esa dFku rFkk dkj.k fn;s x;s gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) fn;s x;s gSa] ftuesa ls dsoy ,d lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy;s + 3 vad fn;s tk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA

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The following questions given below consist ofan "Assertion" (A) and "Reason" (R) Typequestions. Use the following Key to choose theappropriate answer.

(A) If both (A) and (R) are true, and (R) is thecorrect explanation of (A).

(B) If both (A) and (R) are true but (R) is not thecorrect explanation of (A).

(C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true. Q. 11 Consider the parabola y2 – 8y + 6x – 9 = 0 and chord of it as λ (x – 3) + y = 4 Assertion (A) : Minimum length of this chord is 3. Reason (R) : Given line passes through the focus.

Q. 12 Assertion (A): If two odd numbers are selectedfrom the set of first 30 natural numbers, then

probability that their sum is even is 297 .

Reason (R): Let A and B are two events, thenprobability of occurrence of event B when event A

has already happened is )A(P

)BA(P ∩ .

Q. 13 Assertion (A): a × b = a × c then b = λ a + c

Reason (R): If a × b = 0 ⇒ a is collinear to b

uhps fn;s x;s fuEufyf[kr iz'u "dFku" (A) rFkk "dkj.k" (R) izdkj ds iz'u gSaA vr% mfpr mÙkj dk p;u djus ds fy;s fuEu rkfydk dk mi;ksx dhft;sA

(A) ;fn (A) rFkk (R) nksuksa lR; gSa rFkk (R), (A) dk lgh Li"Vhdj.k gSA

(B) ;fn (A) rFkk (R) nksuksa lR; gSa rFkk (R), (A) dk lgh Li"Vhdj.k ugha gSA

(C) ;fn (A) lR; gS ysfdu (R) vlR; gSA (D) ;fn (A) vlR; gS ysfdu (R) lR; gSA

Q. 11 ijoy; y2 – 8y + 6x – 9 = 0 ,oa bldh thok λ (x – 3) + y = 4 ij fopkj dhft,A dFku (A) : bl thok dh U;wure yEckbZ 3 gSA dkj.k (R) : nh xbZ js[kk ukfHk ls xqtjrh gSA Q. 12 dFku (A): ;fn izFke 30 izkdr la[;kvksa ds leqPp; esa

ls nks fo"ke la[;k,sa pquh tk, rks mudk ;ksx le gksus

dh izkf;drk 297 gSA

dkj.k (R): ekuk A rFkk B nks ?kVuk,a gS rc B ds ?kVus

dh izkf;drk tcfd A ?kfVr gks pqdh gks )A(P

)BA(P ∩ gSA

Q. 13 dFku (A): a × b = a × c rc b = λ a + c

dkj.k (R): ;fn a × b = 0 ⇒ a o b lajs[kh; gS

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Q. 14 dFku (A) : ∫ + dx)1x(e)xe(F xx = lnx – x + c

fn;k gS fd ∫ dx)x(F = lnx

dkj.k (R) : ∫ ′+ dx)]x(f)x(f[ex = exf(x) + c

bl [k.M esa 3 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih iz'u gSaA (iz'u 15 ls 23) izR;sd iz'u ds pkj fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy;s + 4 vad fn;s tk,saxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA

x|ka'k # 1 (Q. 15 ls 17) ;fn ,d f=kHkqt ABC dh Hkqtkvksa AB, BC, CA

ds lehdj.k Øe'k% 2x + y = 0, x + py = q, x – y = 3 gksa rFkk fcUnq P ds funsZ'kkad (2, 3) gSA rc

Q.15 ;fn P dsUnzd gks rks p + q =

(A) 47 (B) 50 (C) 65 (D) 74

Q.16 ;fn P yEcdsUnz gks rks p + q =

(A) 47 (B) 50 (C) 65 (D) 74

Q.17 ;fn P ifjdsUnz gks, rks p + q =

(A) 47 (B) 50 (C) 65 (D) 74

Q. 14 Assertion (A) : ∫ + dx)1x(e)xe(F xx = lnx – x + c

given that ∫ dx)x(F = lnx

Reason (R) : ∫ ′+ dx)]x(f)x(f[ex = exf(x) + c

This section contains 3 paragraphs, each has 3 multiplechoice questions. (Questions 15 to 23) Each question has4 choices (A), (B), (C) and (D) out of which ONLY ONEis correct. Mark your response in OMR sheet againstthe question number of that question. + 4 marks will begiven for each correct answer and – 1 mark for eachwrong answer.

Passage # 1 (Q. 15 to 17) The equations of the sides AB, BC, CA

of a triangle ABC are 2x + y = 0, x + py = q,

x – y = 3 respectively. The point P is (2, 3).

Q.15 If P is the centroid, then p + q =

(A) 47 (B) 50 (C) 65 (D) 74

Q.16 If P is the orthocentre, then p + q =

(A) 47 (B) 50 (C) 65 (D) 74

Q.17 If P is the circumcentre, then p + q =

(A) 47 (B) 50 (C) 65 (D) 74

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Passage # 2 (Q.18 to 20) An acute angled triangle ABC is inscribed in a

circle of radius R. The altitudes from the vertices

A, B, C meet the circle at D, E, F respectively. Let

s', r' and r'1 be the semiperimeter, inradius and

exradius opposite to D of triangle DEF. Then'

Q.18 R's =

(A) Σ sin 2A (B) Σ sinA (C) Σ cosA (D) Σ cos2A

Q.19 R'r =

(A) 1 – Σ cos2A (B) –1 – Σ cos2A

(C) 1 + Σ sinA (D) –1 + Σ cos2A

Q.20 R'r 1 =

(A) 1 + cos2A – cos2B – cos2C

(B) – cos2A + cos2B + cos2C

(C) 1 + sin2A – sin2B – sin 2C

(D) 2sinA – sinB – sinC

x|ka'k # 2 (Q. 18 ls 20)

,d U;wu dks.k f=kHkqt ABC, R f=kT;k ds oÙk esa

cuk;k x;k gSA 'kh"kksZ A, B, C ls 'kh"kZ yEc oÙk ij

Øe'k% D, E, F ij feyrs gSaA ekuk s', r' rFkk r'1 Øe'k%

v/kZifjeki] vUr%f=kT;k rFkk f=kHkqt DEF ds 'kh"kZ D

ds lEeq[k cká f=kT;k gS rc

Q.18 R's =

(A) Σ sin 2A (B) Σ sinA (C) Σ cosA (D) Σ cos2A

Q.19 R'r =

(A) 1 – Σ cos2A (B) –1 – Σ cos2A (C) 1 + Σ sinA (D) –1 + Σ cos2A

Q.20 R'r 1 =

(A) 1 + cos2A – cos2B – cos2C

(B) – cos2A + cos2B + cos2C

(C) 1 + sin2A – sin2B – sin 2C

(D) 2sinA – sinB – sinC

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Passage # 3 (Q. 21 to 23) Two vectors a

r and br

having unit modulus and angle between them is θ.

Now φ(θ) = ∫×

2

2

)ba(

)b.a(

2 dx)x(f

rr

rr

and f satisfies the

condition f(y) + f(x) = xy

yx + for all x, y ∈ R –0

and h(θ) = – φ(θ) + 21

21 |b.a||ba|

rrrr× , where b2b1

rr= .

Q.21 Fundamental period of φ(θ) is-

(A) 2π (B) π (C) 2π (D) 6π

Q.22 The volume of the parallelopiped formed by a

r , br

and barr

× , where angle between ar and b

ris taken

for which h(θ) is minimum is-

(A) 1 (B) 21 (C)

41 (D)

81

Q.23 If a

r and br

are non collinear vectors, then number of solution of the equation )('3)(4 θφ+θφ = 0 in [0, 2π] are-

(A) 2 (B) 4 (C) 6 (D) 8

x|ka'k # 3 (Q. 21 ls 23)

nks bdkbZ ifjek.k ds lfn'k ar rFkk b

r gS rFkk muds

e/; dks.k θ gS

vc φ(θ) = ∫×

2

2

)ba(

)b.a(

2 dx)x(f

rr

rr

rFkk lHkh x, y ∈ R –0 ds

fy,] f izfrcU/k f(y) + f(x) = xy

yx + dks lUrq"V djrk gS

rFkk h(θ) = – φ(θ) + 21

21 |b.a||ba|

rrrr× tgk¡ b2b1

rr= .

Q.21 φ(θ) dk ewyHkwr vkorZukad gS-

(A) 2π (B) π (C) 2π (D) 6π

Q.22 ar , b

r rFkk ba

rr× kjk cuk, x, lekUrj "kV~Qyd

dk vk;ru] tgk¡ ar rFkk b

r ds e/; dks.k og gS

ftlds fy, h(θ) U;wure gS] gksxk -

(A) 1 (B) 21 (C)

41 (D)

81

Q.23 ;fn ar rFkk b

rvlajs[kh; lfn'k gS rc lehdj.k

)('3)(4 θφ+θφ = 0 ds [0, 2π] esa gyksa dh la[;k gS -

(A) 2 (B) 4 (C) 6 (D) 8

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PHYSICS

Section - I Questions 1 to 6 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 mark for each wrong answer.

Q.1 A cubical sealed vessel with edge L is placed on a

cart, which is moving horizontally with an

acceleration ‘a’ as shown in figure. The cube is

filled with an ideal fluid having density d. The

gauge pressure at the centre of the cubical vessel is –

L

L

a

(A) 2

Ldg (B) 2

Ld (g + a)

(C) 2

Lda (D) 2

Ld (g – a)

[k.M - I iz'u 1 ls 6 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 lek viukmÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy, + 3 vad fn;stk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA

Q.1 L Hkqtk ds ,d ?kukdkj lhy can ik=k dks ,d xkM+h

ij j[kk tkrk gS tks fd fp=kkuqlkj ‘a’ Roj.k ls

kSfrt #i ls xfr dj jgh gSA bl ?ku dks d ?kuRo

ds ,d vkn'kZ nzO; ls Hkjk tkrk gSA bl ?kukdkj ik=k

ds dsUnz ij xst nkc gksxk&

L

L

a

(A) 2

Ldg (B) 2

Ld (g + a)

(C) 2

Lda (D) 2

Ld (g – a)

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Q.2 A U-tube of cross section A and 2A as shown

contains liquid of density d. Initially, the liquid in

the two arms are held with a level difference h.

After being released the levels equalize after some

time. The work done by gravity forces on the liquid

in the process is –

A 2A

h

(A) 3

Adgh2 (B)

2Adgh2

(C) 6

Adgh2 (D)

8Adgh2

Q.3 The equation of stationary wave is

y = 4sin

π

15x cos (96πt). The distance between a

node and its next antinode is -

(A) 7.5 units (B) 1.5 units

(C) 22.5 units (D) 30 units

Q.2 A rFkk 2A vuqizLFk dkV dh ,d U-uyh esa d ?kuRo

dk ,d nzo fp=kuqlkj Hkj gqvk gSA izkjEHk esa nksuksa

Hkqtkvksa eas nzo dks h varj ds Lrj ij j[kk tkrk gSA

eqDr gksus ds i'pkr~ dqN le; ckn nksuksa Lrj leku

gks tkrs gSaA bl izfØ;k esa nzo ij xq#Roh; cy kjk

fd;k x;k dk;Z gksxk& A 2A

h

(A) 3

Adgh2 (B)

2Adgh2

(C) 6

Adgh2 (D)

8Adgh2

Q.3 vizxkeh rjax dk lehdj.k y = 4sin

π

15x cos (96πt)

gSA ,d fuLian rFkk blds vxys izLian ds e/; nwjh gksxh&

(A) 7.5 bdkbZ (B) 1.5 bdkbZ

(C) 22.5 bdkbZ (D) 30 bdkbZ

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Q. 4 Four charged particles (A, B, C, D) of mass 'm' andcharge 'q' each, are connected by light silk threadsof length 'd' forming a tetrahedron floating in outerspace. The thread connecting particles A and Bsuddenly snaps. Find the maximum speed ofparticle A after that –

C

B

D A q q

q

q

(A)

πε1

33

md8q

0

2 (B)

πε1

33

md8q

0

2

(C)

πε1

33

md4q

0

2 (D)

πε1

33

md4q

0

2

Q.5 The ratio of potential difference between 1 µF and 5 µF capacitors –

3 µF

2 µF

1 µF

2 µF

5 µF

10 V (A) 1 : 2 (B) 3 : 1 (C) 1 : 5 (D) 10 : 1

Q. 4 pkj vkosf'kr d.kksa (A, B, C, D) ftuesa izR;sd dk nzO;eku'm' rFkk vkos'k 'q' gS] dks cká vkdk'k esa ,d prq"Qyd cukrs gq, 'd' yEckbZ ds gYds js'ke ds /kkxksa ls tksM+k tkrk gSA A rFkk B d.kksa dks tksM+us okyk /kkxk vpkud dkV fn;k tkrk gS] rks blds i'pkr~ d.k A dh vf/kdre pky Kkr dhft, –

C

B

D Aq q

q

q

(A)

πε1

33

md8q

0

2 (B)

πε1

33

md8q

0

2

(C)

πε1

33

md4q

0

2 (D)

πε1

33

md4q

0

2

Q.5 1 µF rFkk 5 µF ds la/kkfj=kksa ds e/; foHkokarj dk vuqikr gksxk –

3 µF

2 µF

1 µF

2 µF

5 µF

10 V (A) 1 : 2 (B) 3 : 1 (C) 1 : 5 (D) 10 : 1

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Q.6 In a meter bridge circuit as shown in figure, the

bridge is balanced when AJ = 20 cm. On

interchanging P and Q the balance length shifts by –

V

P

A B

G

Q

J

(A) zero (B) 80 cm

(C) 40 cm (D) 60 cm

Questions 7 to 10 are multiple choice questions. Each

question has four choices (A), (B), (C) and (D), out of

which MULTIPLE (ONE OR MORE) is correct. Mark

your response in OMR sheet against the question

number of that question. + 4 marks will be given for

each correct answer and NO NEGATIVE marks for

wrong answer.

Q.7 A tank is filled upto a height h with a liquid and is

placed on a platform of height h from the ground.

To get maximum range xm a small hole is punched

at a distance of y from the free surface of the liquid.

Then :

Q.6 fp=k eas n'kkZ, x, ehVj fczt ifjiFk esa] tc AJ = 20 cm

gksrk gS] rks fczt larqfyr gksrk gSA P rFkk Q dks

vr%ifjofrZr djus ij larqyu yEckbZ foLFkkfir gksxh&

V

P

A B

G

Q

J

(A) 'kwU; (B) 80 cm

(C) 40 cm (D) 60 cm

iz'u 7 ls 10 rd cgqfodYih iz'u gaSA izR;sd iz'u ds pkj

fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls ,d ;k vf/kd

fodYi lgh gaSA OMR 'khV esa iz'u dh iz'u la[;k ds lek

vius mÙkj vafdr dhft,A izR;sd lgh mÙkj ds fy, + 4 vad

fn;s tk;saxs rFkk xyr mÙkj ds fy, dksbZ _.kkRed vadu

ugha gSA

Q.7 ,d VSad dks h Å¡pkbZ rd nzo ls Hkjk tkrk gS rFkk bls

eSnku ls h Å¡pkbZ ij fLFkr ,d IysVQkWeZ ij j[kk tkrk

gSA vf/kdre ijkl xm izkIr djus ds fy, nzo dh eqDr

lrg ls y nwjh ij ,d y?kq fNnz cuk;k tkrk gSA rc :

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h

xm

h

y

(A) xm = 2h (B) xm = 1.5 h

(C) y = h (D) y = 0.75 h

Q.8 ,d flys.Mj nks nzoksa esa rSj jgk gS tSlk fd fp=k esa n'kkZ;k x;k gSA lgh fodYi pqfu,%

2

1

(A) nzo 1 kjk flys.Mj ij vkjksfir dqy cy 'kwU; gS (B) nzo 1 kjk flys.Mj ij vkjksfir dqy cy v'kwU; gS(C) nzo 2 kjk flys.Mj ij vkjksfir cy mRIykou cy ds rqY; gS (D) nzo 2 kjk flys.Mj ij vkjksfir cy mRIykou cy ls vf/kd gS

h

xm

h

y

(A) xm = 2h (B) xm = 1.5 h

(C) y = h (D) y = 0.75 h

Q.8 A cylinder is floating in two liquids as shown infigure. Choose the correct options :

2

1

(A) net force on cylinder by liquid 1 is zero (B) net force on cylinder by liquid 1 is non-zero

(C) net force on cylinder by liquid 2 is equal to the upthrust

(D) net force on cylinder by liquid 2 is more than the upthrust

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Q.9 A string of length L is stretched along the x-axis and is rigidly clamped at its two ends. It undergoestransverse vibration. If n is an integer, which of thefollowing relations may represent the shape of thestring at any time t ?

(A) y = A sin

π

Lxn cos ωt

(B) y = A sin

π

Lxn sin ωt

(C) y = A cos

π

Lxn cos ωt

(D) y = A cos

π

Lxn sin ωt

Q. 10 Two identical conducting balls have positivecharges q1 & q2 respectively. The balls are broughttogether so that they touch and then put back intheir original positions. The force between the ballsmay be -

(A) remain same as it was before the balls touched. (B) greater than before the balls touched (C) less than before the balls touched (D) zero

This section contains 4 questions numbered 11 to 14,(Reason and Assertion type question). Each questioncontains Assertion and Reason. Each question has 4choices (A), (B), (C) and (D) out of which ONLY ONE iscorrect. Mark your response in OMR sheet against thequestion number of that question. + 3 marks will begiven for each correct answer and – 1 mark for eachwrong answer.

Q.9 L yEckbZ dh ,d jLlh dks x-vk ds vuqfn'k [khapk tkrk gS rFkk bls nksuksa fljksa ij [kawVksa ls n<+rk ls ck¡/kk tkrk gSA ;g vuqizLFk dEiUu djrk gSA ;fn n ,d iw.kkZad gks rks fdlh le; t ij fuEu lac/kksa esa ls dkSulk jLlh ds vkdkj dks iznf'kZr djrk gS \

(A) y = A sin

π

Lxn cos ωt

(B) y = A sin

π

Lxn sin ωt

(C) y = A cos

π

Lxn cos ωt

(D) y = A cos

π

Lxn sin ωt

Q. 10 nks le#i pkyd xsanksa ij /kukRed vkos'k Øe'k% q1 rFkkq2 gSA bu nksuksa xasankas dks lkFk&lkFk yk;k tkrk gS ftlls fd ;s Li'kZ djsa rFkk fQj bUgsa okil budh izkjfEHkd fLFkfr eas j[kk tkrk gSA bu xsanksa ds e/; cy&

(A) leku jgsxk tSlk fd;s xsanksa ds Li'kZ djus ls iwoZ Fkk (B) xsanksa ds Li'kZ djus ls iwoZ ds cy ls vf/kd gksxk (C) xsankas ds Li'kZ djus ls iwoZ ds cy ls de gksxk (D) 'kwU;

bl [k.M esa 11 ls 14 rd 4 iz'u gSa, (dFku rFkk dkj.k izdkj ds iz'u)A izR;sd iz'u esa dFku rFkk dkj.k fn;s x;s gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) fn;s x;s gSa] ftuesa ls dsoy ,d lgh gSA OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj vafdr dhft;sA izR;sd lgh mÙkj ds fy;s + 3 vad fn;s tk;asxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA

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The following questions given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the following Key to choose the appropriate answer.

(A) If both (A) and (R) are true, and (R) is the correct explanation of (A).

(B) If both (A) and (R) are true but (R) is not the correct explanation of (A).

(C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true. Q. 11 Assertion : A capillary tube is dipped in water

vertically. It is long enough for the water to rise to the maximum height 'h' in the tube. The length of the immersed portion in water is l(l < h). The lower end of the tube is closed and the tube is taken out of the water and the lower end is opened again. In opening, the water will not flow out of the tube. Reason : When the tube is taken out, a meniscus with concavity upward is formed in the lower part of tube and force due to surface tension acts in upward direction which keeps water in tube.

Q.12 Assertion : In open organ pipe position of pressure node is a little distance out from the opening of tube.

Reason : Sound waves are reaching at the opening of tube are only partially reflected.

uhps fn;s x;s fuEufyf[kr iz'u "dFku" (A) rFkk "dkj.k" (R) izdkj ds iz'u gSaA vr% mfpr mÙkj dk p;u djus ds fy;s fuEu rkfydk dk mi;ksx dhft;sA

(A) ;fn (A) rFkk (R) nksuksa lR; gSa rFkk (R), (A) dk lgh Li"Vhdj.k gSA

(B) ;fn (A) rFkk (R) nksuksa lR; gSa rFkk (R), (A) dk lgh Li"Vhdj.k ugha gSA

(C) ;fn (A) lR; gS ysfdu (R) vlR; gSA (D) ;fn (A) vlR; gS ysfdu (R) lR; gSA g

Q. 11 dFku : ,d dsf'kdk uyh dks ty esa Å/okZ/kj #i ls Mwcks;k tkrk gSA uyh esa ty dks vf/kdre Å¡pkbZ 'h' rd mBkus ds fy, ;g i;kZIr yEch gSA ty esa Mwcs gq, Hkkx dh yEckbZ l(l < h) gSA uyh dk fupyk fljk cUn fd;k tkrk gS rFkk uyh dks ty ls ckgj yk;k tkrk gS rFkk uyh ds fupys fljs dks iqu% [kksyk tkrk gSA [kksyus ij] ty uyh ls ckgj ugha fudysxkA

dkj.k : tc uyh dks ckgj yk;k tkrk gS rks uyh ds fupys Hkkx esa Åij dh vksj vorykdkj esfuLdl curk gS rFkk i"B ruko ds dkj.k Åij dh fn'kk esa dk;Zjr cy uyh esa ty dks cuk, j[krk gSA

Q.12 dFku : [kqys vkWxZu ikbi esa nkch; fuLian dh fLFkfr

uyh ds [kqys fljs ls ckgj dqN nwjh ij gksrh gSA

dkj.k : uyh ds [kqys fljs ij igq¡pus okyh /ofu rjaxsa

dsoy vkaf'kd #i ls ijkofrZr gksrh gSA

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Q.13 Assertion : For a non-uniformly charged thin circular ring with net charge zero, the electric field at any point on axis of the ring is zero.

Reason : For a non-uniformly charged thin circular ring with net charge zero, the electric potential at each point on axis of the ring is zero.

Q.14 Assertion : When an external resistor of resistance R (connected across a cell of internal resistance r) is varied. Power consumed by resistance R is maximum when R = r

Reason : Power consumed by a resistor of constant resistance R is maximum when current through it is maximum

This section contains 3 paragraphs, each has 3

multiple choice questions. (Questions 15 to 23) Each

question has 4 choices (A), (B), (C) and (D) out of

which ONLY ONE is correct. Mark your response in

OMR sheet against the question number of that

question. + 4 marks will be given for each correct

answer and – 1 mark for each wrong answer.

Passage – 1 (Q. 15 to 17) When a source of sound is moving toward a

stationary detection the frequency of sound perceived by the detector is not the same as the emitted by the source. In order to see this quantitatively, consider the time T elapsed

Q.13 dFku : vleku #i ls vkosf'kr iryh oÙkkdkj oy; ftl ij dqy vkos'k 'kwU; gS] ds fy, oy; dh vk ij fdlh fcUnq ij fo|qr ks=k 'kwU; gksrk gSA

dkj.k : vleku #i ls vkosf'kr iryh oÙkkdkj oy; ftl ij dqy vkos'k 'kwU; gS] ds fy, oy; dh vk ij izR;sd fcUnq ij fo|qr foHko 'kwU; gksrk gSA

Q.14 dFku : tc R izfrjks/k dk ,d cká izfrjks/kd (r vkUrfjd izfrjks/k ds ,d lsy ds fljksa ds e/; tqM+k) ifjofrZr fd;k tkrk gS rks tc R = r gksrk gS] rks R izfrjks/k kjk O;fDr O;f;r vf/kdre gksrh gSA

dkj.k : fu;r izfrjks/k R ds ,d izfrjks/kd kjk O;f;r 'kfDr vf/kdre gksrh gS tc blesa ls izokfgr /kkjk vf/kdre gksrh gSA

bl [k.M esa 3 vuqPNsn fn;s x;s gSa] izR;sd esa 3 cgqfodYih

iz'u gSaA (iz'u 15 ls 23) izR;sd iz'u ds pkj fodYi (A),

(B), (C) rFkk (D) gSa, ftuesa ls dsoy ,d fodYi lgh gSA

OMR 'khV esa iz'u dh iz'u la[;k ds lek viuk mÙkj

vafdr dhft;sA izR;sd lgh mÙkj ds fy;s + 4 vad fn;s

tk,saxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k

tk;sxkA

x|ka'k # 1 (Q. 15 ls 17) tc /ofu ds L=kksr dks fLFkj lalwpd dh rjQ xfr

djk;h tkrh gS] rks lalwpd kjk Kkr /ofu dh vkofÙk L=kksr kjk ÅRlftZr /ofu dh vkofÙk ds leku ugha gksrh gSA bls ek=kkRed #i ls ns[kus ds Øe esa] xksykdkj rjaxkxzkas ds ,d Øekxr ;qXe ds ÅRltZuksa

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between the emissions of a successive pair of spherical wave fronts (see figure 1). A Source moving with speed vs emits the first wave front. At this point, the first wave front has already traveled a distance vT, where v = 340 m/s is the speed of sound. Therefore, the wavelength detected along the direction of motio is given by the difference λ′ = vT – vsT. Using the fact that λ′f′ = λφ = v and T = 1/f, we find that the shifted frequency f′ perceived by the detector is f′ = fv/(v – vs).

vs

λ′vsT

vT '

Fig. (1)

The previous equation does not describe situations where vs ≥ v. When the source travels faster than the speed of sound, a shock wave is produced by the spherical wave fronts. Figure 2 shows the spherical wave fronts produced by such a source at equally spaced positions over an arbitrary time t. During this time, the source travels a distance vst, and the first wave front travels a distance vt. However, in this case, the

ds e/; yxs le; T ij fopkj dhft, (fp=k 1 ns[ksa) vs pky ls xfr'khy ,d L=kksr izFke rjaxkxz ÅRlftZr djrk gSA bl fcUnq ij] izFke rjaxkxz vT nwjh igys gh r; dj pqdk gksrk gS] tgk¡ v = 340 m/s /ofu dh pky gSA blfy,] xfr dh fn'kk ds vuqfn'kk lalwfpr rjaxnS/;Z vUrj λ′ = vT – vsT kjk nh tkrh gSA λ′f′ = λφ = v rFkk T = 1/f dk iz;ksx djrs gq, ge ikrs gSa fd lalwpd kjk Kkr foLFkkfir vkofÙk f ′, f ′ = fv/(v – vs) gSA

vs

λ′vsT

vT '

Fig. (1)

iwoZ lehdj.k mu fLFkfr;ksa dk o.kZu ugha djrh tgk¡ vs ≥ v gksrk gSA tc L=kksr /ofu dh pky ls vf/kd pky ls xfr djrs gaS] rks xksykdkj rjaxkxzksa kjk ,d iz?kkrh rjax mRiUu gksrh gSA fp=k 2 fdlh le; t ij leku fLFkfr;ksa ij bl izdkj ds L=kksr kjk mRiUu xksykdkj rjaxkxzksa dks n'kkZrk gSA bl le; ds nkSjku] L=kksr vst nwjh r; djrk gS rFkk izFke rjxkaxz vt nwjh r; djrk gSA ;|fi] bl fLFkfr esa iwoZ esa ÅRlftZr rjaxksa ds rjaxkxzksa ds ihNs xfr djus ds i'pkr~ ;g

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source emits each new wave after traveling beyond the front of the previously emitted wave. The wave fronts bunch along the surface of a cone called the match cone. The resulting rise and fall in air pressure as the surface of the cone passes through a point in space produces a shock wave.

vsθ vst

vt

Fig. (2)

Q.15 A roller skater carrying a portable stereo skates at constant speed an observer at rest. Which of the following accurately represents how the frequency perceived by the observer changes with time ?

L=kksr izR;sd ubZ rjax ÅRlftZr djrk gSA ,d 'kadq dh lrg ds vuqfn'kk rjxkaxzksa dk lewg feyku 'kadq dgykrk gSA tc 'kadq dh lrg vkdk'k esa fdlh fcUnq ls xqtjrh gS rks ok;q nkc esa p<+o rFkk mrkj ds ifj.kkeLo#i ,d iz?kkrh rjax mRiUu gksrh gSA

vsθ vst

vt

Fig. (2)

Q.15 ,d jksyj LdsVj ,d de Hkkj ds LVhfj;ks LdsV~l dks fojkekoLFkk esa fLFkr izskd dh vksj fu;r pky ls yk jgk gSA fuEu esa ls dkSulk ;FkkFkZ #i ls ;g iznf'kZr djrk gS fd izskd kjk izkIr vkofÙk le; ds lkFk dSls ifjofrZr gksrh gS \

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(A)

freq

uenc

y

time

(B)

freq

uenc

y

time

(C)

freq

uenc

y

time

(D)

freq

uenc

y

time

Q.16 A police car moving toward a stationary

pedestrian at a speed of 10 m/s operates its siren.

If the pedestrian perceives the frequency of the

siren to be 1030 Hz. what is the frequency emitted

by the siren ? (Speed of sound = 340 m/s)

(A) 10 Hz

(B) 100 Hz

(C) 1,000 Hz

(D) 10,000 Hz

(A)

vkofÙk

le;

(B)

vkofÙk

le;

(C)

vkofÙk

le;

(D)

vkofÙk

le;

Q.16 ,d iqfyl dh dkj ,d LFkk;h iSny pyus okys O;fDr

dh vksj 10 m/s dh pky ls lk;ju ctkrs gq, xfr dj

jgh gSA ;fn iSny pyus okys O;fDr kjk izkIr lk;ju

dh vkofÙk 1030 Hz gks] rks lk;ju kjk ÅRlftZr vkofÙk

D;k gksxh \ (/ofu dh pky = 340 m/s)

(A) 10 Hz

(B) 100 Hz

(C) 1,000 Hz

(D) 10,000 Hz

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Q.17 A bat flies towards a stationary wall with speed vb. If the bat emits a signal at frequency f, what is the correct expression for the frequency of the reflected signal that the bat hears ?

(A) fbvv

v−

(B) fb

b

vvvv

+−

(C) fvvv b+ (D) f

b

b

vvvv

−+

Passage # 2 (Q. 18 to 20) One end of iron chain is fixed to sphere of mass M

and of diameter D. The volume of sphere is V. The other end is free. The length of chain is l and mass is m. The sphere with the chain is in the reservoir whose depth is H. Density of water is ρw and density of iron is ρF.

M

h

D H

Q.17 ,d pexknM+ ,d fLFkj nhokj dh vksj vb pky ls

mM+rh gSA ;fn pexknM+ f vkofÙk ij ,d ladsr

ÅRlftZr djrh gS] rks ijkofrZr ladsr dh vkofÙk ds

fy, lgh O;atd D;k gksxk tks fd pexknM+ lqurk gS \

(A) fbvv

v−

(B) fb

b

vvvv

+−

(C) fvvv b+ (D) f

b

b

vvvv

−+

x|ka'k # 2 (Q. 18 ls 20) yksgs dh tathj ds ,d fljs dks M nzO;eku rFkk D O;kl

ds ,d xksys ls tksM+k tkrk gSA xksys dks vk;ru V gSA bldk nwljk fljk eqDr gSA tathj dh yEckbZ l rFkk nzO;eku m gSA tathj ds lkFk xksyk tyk'k; esa gS ftldh xgjkbZ H gSA ty dk ?kuRo ρw rFkk yksgs dk ?kuRo ρF gSA

M

h

DH

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Q.18 Total force of gravity acting on the system of sphere and hanging chain is –

(A)

+l

)2/Dh(mM g

(B) (m + M) g

(C)

+ 2

mhMl

g

(D)

+

hmM l g

Q.19 The force of buoyancy acting on the sphere and the hanging chain is –

(A) ρw Vg

(B) ρw

−×

ρ+

l

)2/Dh(mVF

g

(C) ρw

−×

ρ l

)2/Dh(m

F g

(D) F

w Mρ

ρ ×

l

2/Dh × g

Q.20 The value of 'h' i.e. height of sphere above the

bottom of reservoir is –

(A)2D +

)/1(m)MV(

Fw

w

ρρ−−ρ

l (B) 2D +

2l

(C) 2D +

mVw lρ (D)

]/1[m)MV(

Fw

w

ρρ−−ρ l

Q.18 fudk; rFkk yVdh gqbZ tathj ds fudk; ij dk;Zjr dqy xq#Roh; cy gksxk –

(A)

+l

)2/Dh(mM g

(B) (m + M) g

(C)

+ 2

mhMl

g

(D)

+

hmM l g

Q.19 xksys rFkk yVdh gqbZ tathj ij dk;Zjr mRIykou cy gksxk&

(A) ρw Vg

(B) ρw

−×

ρ+

l

)2/Dh(mVF

g

(C) ρw

−×

ρ l

)2/Dh(m

F g

(D) F

w Mρ

ρ ×

l

2/Dh × g

Q.20 'h' vFkkZr~ tyk'k; ds ry ds Åij xksys dh Å¡pkbZ

dk eku gksxk&

(A)2D +

)/1(m)MV(

Fw

w

ρρ−−ρ

l (B) 2D +

2l

(C) 2D +

mVw lρ (D)

]/1[m)MV(

Fw

w

ρρ−−ρ l

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Passage # 3 (Q. 21 to 23) Each component in the infinite network shown

in Fig. has a resistance R = 4 Ω. A battery of emf 1V and negligible internal resistance is connected between any two neighbouring points, say X and Y.

R R R R

R

R

R

R

R R R R R

R

R

R

R

R

R

R

R

R

R

R

R

R

R

R

R R R R

R R R R

R R R R

A 1V

X Y

Ideal ammeter

Answer the following questions

Q.21 Current shown by the ammeter will be -

(A) 0.25A (B) 0.5A

(C) 0.75A (D) 1A

x|ka'k # 3 (Q. 21 ls 23)

fp=k esa n'kkZ, x, vuUr tky esa izR;sd ?kVd dk

izfrjks/k R = 4 Ω gSA fdUgha Hkh nks lehiorhZ fcUnqvksa]

ekuk X rFkk Y ds e/; 1V fo-ok- cy rFkk ux.;

vkUrfjd izfrjks/k dh ,d cSVjh dks tksM+k tkrk gSA

R R R R

R

R

R

R

R R R R R

R

R

R

R

R

R

R

R

R

R

R

R

R

R

R

R R R R

R R R R

R R R R

A 1V

X Y

vkn'kZ vehVj

fuEu iz'uksa ds mÙkj nhft,& Q.21 vehVj kjk iznf'kZr /kkjk gksxh&

(A) 0.25A (B) 0.5A

(C) 0.75A (D) 1A

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Q.22 If the resistance R between X and Y is removed, current shown by ammeter will be -

(A) 0.25A (B) 0.5A (C) 0.75A (D) 1A

Q.23 Equivalent capacitance, if the given network consists of capacitors each C (instead of resistances), and with none of the components removed, will be -

(A) C/4 (B) C/2 (C) 2C (D) 4C

Q.22 ;fn X rFkk Y ds e/; ds izfrjks/k R dks gVk fn;k

tk,] rks vehVj kjk iznf'kZr /kkjk gksxh& (A) 0.25A (B) 0.5A (C) 0.75A (D) 1A

Q.23 ;fn fn, x, tky ds la/kkfj=kksa esa izR;sd dh /kkfjrk

C (izfrjks/kksa ds vfrfjDr) gks rFkk fdlh Hkh ?kVd dks

ugha gVk;k tk,] rks rqY; /kkfjrk gksxh& (A) C/4 (B) C/2 (C) 2C (D) 4C

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Time : 3 : 00 Hrs. MAX MARKS: 246

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 lek lgh mÙkj fpfUgr dhft,A

2. mRrj ds fy,] OMR vyx ls nh tk jgh gSA 3. ifjohkdksa 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, 4 vad fn, tk;saxs rFkk xyr mÙkj ds fy, dksbZ _.kkRed vadu ugha gSA

6. dFku rFkk dkj.k izdkj ds iz'u] ftuesa ls izR;sd esa dsoy ,d mÙkj lgh gSA izR;sd lgh mÙkj ds fy, 3 vad fn;s tk;saxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk,xkA

7. x|ka'k ij vk/kkfjr cgqfodYih izdkj ds iz'u ftuesa ls dsoy ,d fodYi lgh gSA izR;sd lgh mÙkj ds fy, 4 vad fn, tk;saaxs rFkk izR;sd xyr mÙkj ds fy, 1 vad ?kVk;k tk;sxkA

C. OMR dh iwfrZ :

8. OMR 'khV ds CykWdksa esa viuk uke] vuqØek¡d] cSp] dkslZ rFkk ijhkk 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

'6' should be filled as 0006

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

'86' should be filled as 0086

0 0 0 00 1 2

3

4

5 6

7

8

9

012

3

4

56

7

8

9

012

3

4

56

7

8

9

012

3

4

56

7

8

9

'1857' should be filled as 1857

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7

SEA

L