KOTA (RAJASTHAN) CA CAC A - allen.ac.in (Main... · ... 25 (3) 37.5 (4) 53.5 8. Equal masses of methane and oxygen are mixed in an empty container at 25ºC. The fraction of the total

$Page 1: KOTA (RAJASTHAN) CA CAC A - allen.ac.in (Main... · ... 25 (3) 37.5 (4) 53.5 8. Equal masses of methane and oxygen are mixed in an empty container at 25ºC. The fraction of the total$
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Path to Success

ALLENCAREER INSTITUTEKOTA (RAJASTHAN)

T M

FORM NUMBER

CLASSROOM CONTACT PROGRAMME(ACADEMIC SESSION 2013-2014)

LEADER & ENTHUSIAST COURSE

PAPER CODE 0 1 C E 3 1 3 0 5 3

Date : 21 - 03 - 2014 PATTERN : JEE (Main)

SCORE-I TEST # 06

ek/;e % fgUnh

Do not open this Test Booklet until you are asked to do so.

1. Immediately fill in the form number on this page of the Test Bookletwith Blue/Black Ball Point Pen. Use of pencil is strictly prohibited.

2. The candidates should not write their Form Number anywhere else(except in the specified space) on the Test Booklet/Answer Sheet.

3. The test is of 3 hours duration.

4. The Test Booklet consists of 90 questions. The maximum marks are360.

5. There are three parts in the question paper A,B,C consisting ofChemistry, Mathematics and Physics and having 30 questions ineach part of equal weightage. Each question is allotted 4 (four) marksfor correct response.

6. One mark will be deducted for indicated incorrect response of eachquestion. No deduction from the total score will be made if noresponse is indicated for an item in the Answer Sheet.

7. Use Blue/Black Ball Point Pen only for writting particulars/markingresponses on Side–1 and Side–2 of the Answer Sheet.Use of pencil is strictly prohibited.

8. No candidate is allowed to carry any textual material, printed or written,

bits of papers, pager, mobile phone any electronic device etc, except

the Identity Card inside the examination hall/room.

9. Rough work is to be done on the space provided for this purpose inthe Test Booklet only.

10. On completion of the test, the candidate must hand over the AnswerSheet to the invigilator on duty in the Room/Hall. However, thecandidate are allowed to take away this Test Booklet with them.

11. Do not fold or make any stray marks on the Answer Sheet.

bl ijh{kk iqfLrdk dks rc rd u [kk sysa tc rd dgk u tk,A

1. ijh{kk iqfLrdk ds bl i"B ij vko';d fooj.k uhys@dkys ckWy ikbaV isuls rRdky HkjsaA isfUly dk iz;ksx fcYdqy oftZr gSaA

2. ijh{kkFkhZ viuk QkeZ ua- (fu/kkZfjr txg ds vfrfjä) ijh{kk iqfLrdk @ mÙkji= ij dgha vkSj u fy[ksaA

3. ijh{kk dh vof/k 3 ?k aVs gSA

4. bl ijh{kk iqfLrdk esa 90 iz'u gaSA vf/kdre vad 360 gSaA

5. bl ijh{kk iqfLrdk es a rhu Hkkx A, B, C gSa] ftlds izR;sd Hkkx esajlk;u foKku] xf.kr ,oa HkkSfrd foKku ds 30 iz'u gSa vkSj lHkh iz'uksa ds vadleku gSaA izR;sd iz'u ds lgh mÙkj ds fy, 4 (pkj) vad fuèkkZfjr fd;s x;s gSaA

6. izR;sd xyr mÙkj ds fy, ml iz'u ds dqy vad dk ,d vad dkVk tk;sxkAmÙkj iqfLrdk esa dksbZ Hkh mÙkj ugha Hkjus ij dqy izkIrkad esa ls ½.kkRedvadu ugha gksxkA

7. mÙkj i= ds i`"B&1 ,oa i`"B&2 ij okafNr fooj.k ,oa mÙkj vafdr djus gsrqdsoy uhys@dkys ck Wy ikbaV isu dk gh iz;ksx djsaAisfUly dk iz;ksx fcYdqy oftZr gSA

8. ijh{kkFkhZ }kjk ijh{kk d{k @ gkWy esa ifjp; i= ds vykok fdlh Hkhizdkj dh ikB~; lkexzh eqfær ;k gLrfyf[kr dkxt dh ifpZ;ksa] istj]eksckby Qksu ;k fdlh Hkh izdkj ds bysDVªkfud midj.kksa ;k fdlh vU;izdkj dh lkexzh dks ys tkus ;k mi;ksx djus dh vuqefr ugha gSaA

9. jQ dk;Z ijh{kk iqfLrdk esa dsoy fu/kkZfjr txg ij gh dhft;sA

10. ijh{kk lekIr gksus ij] ijh{kkFkhZ d{k@gkWy NksM+us ls iwoZ mÙkj i= d{k fujh{kddks vo'; lkSai nsaA ijh{kkFkhZ vius lkFk bl ijh{kk iqfLrdk dks ys tkldrs gSaA

11. mÙkj i= dks u eksM+sa ,oa u gh ml ij vU; fu'kku yxk,saA

IMPORTANT INSTRUCTIONS egRoiw.kZ lwpuk,¡

SPACE FOR ROUGH WORK / jQ dk;Z ds fy;s txg

21–03–2014

01CE313053 KOTA / H-1/29


1. The molar conductivities of KCl, NaCl and KNO3

are 152, 128 and 111 S cm2 mol–1 respectively.What is the molar conductivity of NaNO3 :-(1) 101 S cm2 mol–1 (2) 87 S cm2 mol–1

(3) –101 S cm2 mol–1 (4) –391 S cm2 mol–1

2. For following cell 23 FeAA Fe++

ll , calculateDG° at 298 K

Given : 3A / AE +

°l l

= – 1.66 V ; 1F = 96500 C

2Fe / FeE +

° = – 0.44 V.

(1) – 700.01 kJ (2) – 706.38 kJ(3) – 965.01 kJ (4) None of these

3. Copper crystallises in ƒcc with a unit cell lengthof 361 pm. What is the radius of copper atom(1) 108 pm (2) 127 pm(3) 157 pm (4) 181 pm

4. The solubility of Sb2S3 in water is1.0 × 10–5 mol/litre at 298 K. What will be itssolubility product :-(1) 108 × 10–25 (2) 1.0 × 10–25

(3) 144 × 10–25 (4) 126 × 10–24

5. In a buffer mixture of a weak acid and its salt,the ratio of concentration of acid to salt isincreased ten-fold. The pH of the solution(1) Decreases by one (2) Increases by one-tenth(3) Increases by one (4) Increases ten-fold

1. KCl, NaCl rFkk KNO3 dh eksyj pkydrk,sa Øe'k% 152,128 rFkk 111 S cm2 mol–1 gSaA rc NaNO3 dh eksyjpkydrk D;k gksxh :-(1) 101 S cm2 mol–1 (2) 87 S cm2 mol–1

(3) –101 S cm2 mol–1 (4) –391 S cm2 mol–1

2. fuEu lsy ds fy;s 23 FeAA Fe++

ll 298 K ij DG°

Kkr dhft;sA

fn;k gS : 3A / AE +

°l l

= – 1.66 V ; 1F = 96500 C

2Fe / FeE +

° = – 0.44 V.

(1) – 700.01 fdyks twy (2) – 706.38 fdyks twy(3) – 965.01 fdyks twy (4) buesa ls dksbZ ugha

3. 361 ;wfuV lsy dh yEckbZ ds lkFk dkWij (rk¡ck) ƒcc esafØLVfyr gksrk gSA dkWij ijek.kq dh f=T;k D;k gS(1) 108 pm (2) 127 pm(3) 157 pm (4) 181 pm

4. 298 K ij Sb2S3 dh ty esa foys;rk

1.0 × 10–5 eksy@yhVj gSA bldk foys;rk xq.kuQy gksxk%&

(1) 108 × 10–25 (2) 1.0 × 10–25

(3) 144 × 10–25 (4) 126 × 10–24

5. nqcZy vEy rFkk mlds yo.k ds cQj feJ.k esa vEy rFkkyo.k ds lkUn zrk dk vuqikr 10 xquk rd c<+ tkrk gSA foy;udh pH gksxh%&(1) ,d de gks tkrh gS (2) 1/10 xquk c<+ tkrh gS(3) ,d c<+ tkrh gS (4) 10 xquk c<+ tkrh gS

HAVE CONTROL ¾® HAVE PATIENCE ¾® HAVE CONFIDENCE Þ 100% SUCCESS

PART A - CHEMISTRYBEWARE OF NEGATIVE MARKING


21–03–2014

01CE313053KOTA / H-2/29

TARGET : JEE (Main) 2014

6. 75.2g of C6H5OH (phenol) is dissolved in a solvent

of Kf = 14. If the depression in freezing point is

7K then find the % of phenol that dimerises :-

(1) 50% (2) 75%

(3) 25% (4) 99%

7. Benzene and toluene form nearly ideal solutions. At

20°C, the vapour pressure of benzene is 75 torr and

that of toluene is 22 torr. The partial vapour pressure

of benzene at 20°C for a solution containing 78g of

benzene and 46g of toluene in torr is :-

(1) 50 (2) 25

(3) 37.5 (4) 53.5

8. Equal masses of methane and oxygen are mixed

in an empty container at 25ºC. The fraction of the

total pressure exerted by oxygen is:-

(1) 23

(2) 1 2733 298

´

(3) 13

(4) 12

9. When 3 moles of A and 1 mole of B are mixed

in 1 litre vessel the following reaction takes place

A(g) + B(g) � 2C(g). 1.5 moles of C are formed.

The equilibrium constant for the reaction is :-

(1) 0.12 (2) 0.25

(3) 0.50 (4) 4.0

6. C6H5OH (fQukWy) ds 75.2 xzke dks ,d foyk;d esa ?kksykx;k ftlds fy, Kf = 14 ;fn fgekad esa voueu 7K gSrks fQukWy dk og % D;k gS ftl ij og fgefjr gksrk gS%&

(1) 50% (2) 75%

(3) 25% (4) 99%

7. cSUthu ,oa VkWyqbZu yxHkx vkn'kZ foy;u cukrs gSA 20°C

ij csUthu dk ok"inkc 75 Vksj gS tcfd VkWyqbZu dk22 Vksj gSA 20°C ij 78 xzke csUthu ,oa 46 xzke VkWyqbZu;qDr foy;u ds fy;s csUthu dk vkaf'kd ok"inkc Vksjesa gS%&

(1) 50 (2) 25

(3) 37.5 (4) 53.5

8. 25ºC ij ,d fjDr ik= esa esFksu vkSj vkWDlhtu ds lekuHkkj feyk;s x;sA dqy nkc ij vkWDlhtu }kjk yxk;s x;snkc dk izHkkt gS :-

(1) 23

(2) 1 2733 298

´

(3) 13

(4) 12

9. A ds 3 eksy vkSj B ds 1 eksy dks tc 1 yhVj ds crZuesa vfHkd`r djok;k tkrk gS rks A(g) + B(g) � 2C(g)

vfHkfØ;k gksrh gS rFkk C ds 1.5 eksy curs gS] vfHkfØ;kds fy, lkE; fLFkjkad gksxk :-

(1) 0.12 (2) 0.25

(3) 0.50 (4) 4.0


21–03–2014

01CE313053 KOTA / H-3/29


10. Which conformation of ethylene glycol is most

stable ?

(1) Staggered (2) Gauche

(3) Eclipsed (4) Anti form

11. Which chloro derivative of benzene undergo

hydrolysis most readily with aq. NaOH ?

(1)

CH –Cl2

Cl

(2)

Cl

Cl

(3)

ClNO2

(4)

ClCH3CH3

CH3

10. ,fFkyhu XykbdkWy ds fuEufyf[kr la:i.k esa dkSulklokZfèkd LFkk;h gS ?

(1) lkarfjr :i (2) xkW'k

(3) xzflr :i (4) ,UVh :i

11. fuEufyf[kr csUthu ds Dyksjks O;qRiUuksa esa tyh; NaOH

}kjk fdldk ty vi?kVu vklkuh ls gksxk ?

(1)

CH –Cl2

Cl

(2)

Cl

Cl

(3)

ClNO2

(4)

ClCH3CH3

CH3

fdlh iz'u ij nsj rd :dks ugha A


21–03–2014

01CE313053KOTA / H-4/29


12. The following reaction gives

CHO

OMeOMe

+ HCHO ¾¾¾¾®(i) Conc. NaOH

(ii) H O3Å ?

(1)

COOH

OMeOMe

+ CH3OH

(2)

COOH

OMeOMe

+ HCOOH

(3)

CH –OH2

OMeOMe

+ CH3OH

(4)

CH –OH2

OMeOMe

+ HCOOH

12. fuEufyf[kr vfHkfØ;k nsxh

CHO

OMeOMe

+ HCHO ¾¾¾¾®(i) Conc. NaOH

(ii) H O3Å ?

(1)

COOH

OMeOMe

+ CH3OH

(2)

COOH

OMeOMe

+ HCOOH

(3)

CH –OH2

OMeOMe

+ CH3OH

(4)

CH –OH2

OMeOMe

+ HCOOH


21–03–2014

01CE313053 KOTA / H-5/29


13.

O

CH2

NaBH4

B H2 6

H O /OH2 21

(A)

(B) A and B respectively are:-

(1)

CH3

O

and

O

CH –OH2

(2)

CH2

and

CH –OH2

O

(3)

OH

CH2

and

O

CH –OH2

(4)

CH2

OH

and

CH3

O

OH

13.

O

CH2

NaBH4

B H2 6

H O /OH2 21

(A)

(B) A rFkk B Øe'k% gS :-

(1)

CH3

O

and

O

CH –OH2

(2)

CH2

and

CH –OH2

O

(3)

OH

CH2

and

O

CH –OH2

(4)

CH2

OH

and

CH3

O

OH


21–03–2014

01CE313053KOTA / H-6/29


14. How many stereo isomers are possible for thefollowing molecule :-

Cl CH=CHBr

(1) 4 (2) 8 (3) 12 (4) 16

15. Which of following solvents would NOT be agood choice for the formation of the grignard from1-chloro butane ?

(1) OH (2) O

(3) (4)

16. What is the major organic product of thefollowing sequence of reaction ?

(CH3)2CHCH2–OH 3PBr¾¾¾® Mg¾¾®

¾¾¾¾¾®CH ––CH2 2

O

3H O+

¾¾¾® ?

(1) (CH ) CH–CHCH CH3 2 2 3

OH

(2) (CH ) CHCH CH–CH3 2 2 3

OH

(3) (CH3)2CHCH2CH2–OH

(4) (CH3)2CHCH2CH2CH2OH

14. fn, x, v.kq ds dqy fdrus f=foe leko;oh gksxsa :-

Cl CH=CHBr

(1) 4 (2) 8

(3) 12 (4) 16

15. fuEufyf[kr esa ls dkSulk foyk;d 1-Dyksjks C;wVsu ls xzhU;kjvfHkdeZd cukus dk vPNk fodYi ugh gS ?

(1) OH (2) O

(3) (4)

16. fuEufyf[kr vfHkfØ;k Øe esa dkcZfud eq[; mRikn D;kgksxk ?

(CH3)2CHCH2–OH 3PBr¾¾¾® Mg¾¾®

¾¾¾¾¾®CH ––CH2 2

O

3H O+

¾¾¾® ?

(1) (CH ) CH–CHCH CH3 2 2 3

OH

(2) (CH ) CHCH CH–CH3 2 2 3

OH

(3) (CH3)2CHCH2CH2–OH

(4) (CH3)2CHCH2CH2CH2OH


21–03–2014

01CE313053 KOTA / H-7/29


17. 3-Pentanol is produced by which of the followingreactions ?

(i) CH3CH2Br MgEther¾¾¾® 3 2CH CH CHO¾¾¾¾¾®

2

HH O¾¾¾®

Å

(ii) HCºCH 3 22

2

CH CH CHONaNH HH O

¾¾¾® ¾¾¾¾¾® ¾¾¾®Å

(iii)(CH3CH2)2CO 4

2

LiAlH HEther H O

¾¾¾® ¾¾¾®Å

(1) (i) only (2) (i) & (ii) only(3) (i) & (iii) only (4) (ii) & (iii) only

18. Which of the following reagents would not carryout the following reaction ?

HO? HO

O

(1) PCC (2) KMnO4

(3) H2Cr2O7 (4) H2CrO4

19. Which gives 2 moles of HCHO, 1 mole ofHCOOH and CO2 by HIO4 :-

(1)

CH –OH2

CH–OHC=OCH –OH2

(2)

CH=OCH–OHCH–OHCH=O

(3)

CH –OH2

(CH–OH)2

CH=O(4) All

17. fuEu esa ls fdlesa 3-isUVsuksy curk gS ?

(i) CH3CH2Br MgEther¾¾¾® 3 2CH CH CHO¾¾¾¾¾®

2

HH O¾¾¾®

Å

(ii) HCºCH 3 22

2

CH CH CHONaNH HH O

¾¾¾® ¾¾¾¾¾® ¾¾¾®Å

(iii)(CH3CH2)2CO 4

2

LiAlH HEther H O

¾¾¾® ¾¾¾®Å

(1) (i) dsoy (2) (i) & (ii) dsoy

(3) (i) & (iii) dsoy (4) (ii) & (iii) dsoy

18. fuEufyf[kr vfHkdeZd esa ls fdldks nh xbZ vfHkfØ;k esaiz;qDr ugha dj ldrsA

HO? HO

O

(1) PCC (2) KMnO4

(3) H2Cr2O7 (4) H2CrO4

19. fuEu esa ls dkSu HIO4 ls fØ;k dj 2 eksy HCHO,1 eksy HCOOH rFkk CO2 nsrk gS :-

(1)

CH –OH2

CH–OHC=OCH –OH2

(2)

CH=OCH–OHCH–OHCH=O

(3)

CH –OH2

(CH–OH)2

CH=O(4) lHkh

Take it Easy and Make it Easy


21–03–2014

01CE313053KOTA / H-8/29


20. Rank the following compounds with respect toincreasing order of hydration with H2O, HÅ.

O

H(A)

O

(B)

O

(C)

HCl

O

(D)

HF

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

21. Three of the following fisher projections areidentical, one is not. Which structure is different?

(1)

O H

HO H

CH3

(2) O

OH

HH

CH3

(3)

O H

H OH

CH3

(4) O

OH

HH

CH3

20. fn;s x;s ;kSfxdksa dk H2O, HÅ ds lkFk tyh;dj.k dk ØegS :-

O

H(A)

O

(B)

O

(C)

HCl

O

(D)

HF

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

21. fuEu esa rhu fQ'kj iz{ksi.k lw= leku gS ysfdu ,d vyxgSA fuEu esa dkSulh lajpuk fHkUu gSA

(1)

O H

HO H

CH3

(2) O

OH

HH

CH3

(3)

O H

H OH

CH3

(4) O

OH

HH

CH3


21–03–2014

01CE313053 KOTA / H-9/29


22. MgBr + Et–O–C–O–Et

O

(Excess)

Product,

producut is ?

(1) CH

OH

(2) C

OH

(3) CH2 OH

(4) C

O

23. Which of the following species exist :-(1) NCl5 (2) PH5

(3) CuI2 (4) H3Å

24. Which of the following is not trigonalbipyramidal:-(1) PF5 (2) PCl5 (3) PCl3 (4) PPh5

22. MgBr + Et–O–C–O–Et

O

( )vkf/kD;

mRikn]

mRikn gS?

(1) CH

OH

(2) C

OH

(3) CH2 OH

(4) C

O

23. dkSulh Lih'kht dk vfLrRo gksxk :-(1) NCl5 (2) PH5

(3) CuI2 (4) H3Å

24. fuEu esa ls dkSulk f=dks.kh; f}ijkehfM; ugha gS :-(1) PF5 (2) PCl5

(3) PCl3 (4) PPh5

LoLFk jgks] eLr jgks rFkk i<+kbZ es a O;Lr jgks A


21–03–2014

01CE313053KOTA / H-10/29


25. Choose correct option

(1) H2Å does not exist is gaseous phase

(2) D2 has more molecular weight than T2

(3) H2 has less critical temperature than T2

(4) When two nuclear spins are parallel in H2 thenit is called orthohydrogen

26. MnO2 Fused with KOH

Oxidised with air or KNO3

[X] anion is formed then

what is the oxidation state of Mn in it :-

(1) + 6 (2) + 5 (3) + 7 (4) + 2

27. Permanent hardness of H2O is removed by :-

(1) Treatment with washing soda

(2) Calgon's method

(3) Clark's method

(4) (1) & (2) both

28. Which of the following oxide is neutral :-

(1) SO2 (2) Cl2O7

(3) N2O5 (4) N2O

29. Which of the following species do not showdisproportion reaction :-

(1) ClO1 (2) ClO21

(3) ClO31 (4) ClO4

1

30. CCl4 + H2O Room temperature what happened

(1) CO2 gas is evolved

(2) COCl2 gas is evolved

(3) HOCl is formed

(4) No reaction

25. lgh dFku pqfu,(1) H2

Å xSlh; voLFkk esa ugha gksrk gSA(2) D2 dk v.kqHkkj T2 ls T;knk gksrk gSA(3) H2 dk ØkfUrd rki T2 ls de gksrk gSA(4) tc H2 esa nks ukfHkdh; ?kw.kZu lekukUrj gksrs gSa rc bldks

vkFkksZgkbMªkstu dgk tkrk gSA

26. MnO2 Fused with KOH

Oxidised with air or KNO3

[X] ½.kk;u curk gS rks Mn

dh vkWDlhdj.k voLFkk D;k gksxh :-

(1) + 6 (2) + 5 (3) + 7 (4) + 2

27. H2O dh LFkk;h dBksjrk dks vyx fd;k tk ldrk gS :-(1) /kkou lksMs ls vfHkÏr djokdj(2) dsyxksu (Calgon's) i¼fr }kjk(3) DykdZ (Clark's) i¼fr }kjk(4) (1) o (2) nksuksa

28. fuEu esa ls dkSulk vkWDlkbM mnklhu izÏfr dk gS :-(1) SO2 (2) Cl2O7

(3) N2O5 (4) N2O

29. dkSulh Lih'kht fo"kekuqikrh (disproportion) vfHkfØ;kugha n'kkZrh gS :-(1) ClO1 (2) ClO2

1

(3) ClO31 (4) ClO4

1

30. CCl4 + H2O dejs ds rki ij

D;k gksxk(1) CO2 xSl fudysxh(2) COCl2 xSl fudysxh(3) HOCl cusxk(4) dksbZ vfHkfØ;k ugha gksxh


21–03–2014

01CE313053 KOTA / H-11/29


31. If a, b are the roots of equation ax2 + bx + c = 0,and a + h, b + h are the root of px2 + qx + r = 0,and D1, D2 the respective discriminants of theseequations, then D1 : D2 =

(1) 2

2

ap (2)

2

2

bq (3)

2

2

cr

(4) None

32. If a(p + q)2 + 2bpq + c = 0 anda(p + r)2 + 2bpr + c = 0 then qr =

(1) 2 cpa

+ (2) 2 apc

+

(3) 2 apb

+ (4) 2 bpa

+

33. If a,b,c ÎR and x3 – 3b2x + 2c3 is divisible by(x – a) and (x – b), then(1) a = –b = –c(2) a = 2b = 2c(3) a = b = c or a = –2b = –2c(4) None

34. If the value of determinants

a 1 11 b 11 1 c

is positive,

then :-(1) abc > 1 (2) abc > –8(3) abc < –8 (4) abc > –2

31. ;fn lehdj.k ax2 + bx + c = 0, ds ewy a, b gks rFkklehdj.k px2 + qx + r = 0, ds ewy a + h, b + h gksrFkk D1, D2 muds Øe'k% fofofÙkdj gks rks D1 : D2 cjkcjgksxk&

(1) 2

2

ap (2)

2

2

bq (3)

2

2

cr

(4) None

32. ;fn a(p + q)2 + 2bpq + c = 0 rFkk

a(p + r)2 + 2bpr + c = 0 gks rks qr =

(1) 2 cpa

+ (2) 2 apc

+

(3) 2 apb

+ (4) 2 bpa

+

33. ;fn a,b,c ÎR gk s rFkk x3 – 3b2x + 2c3 O;atd(x – a) rFkk (x – b), ls foHkkftr gks&(1) a = –b = –c(2) a = 2b = 2c(3) a = b = c or a = –2b = –2c(4) None

34. ;fn lkjf.kd a 1 11 b 11 1 c

dk eku /kukRed gk s

rks&(1) abc > 1 (2) abc > –8(3) abc < –8 (4) abc > –2

PART B - MATHEMATICS


21–03–2014

01CE313053KOTA / H-12/29


35. ;fn A = 1 2 12 0 13 2 1

- rFkk A3 + xA2 + yA + zI = 0

rks x + y + z cjkcj gksxk&(1) –2 (2) –0 (3) 6 (4) None

36. ;fn U ,d lkoZf=d leqPp; gS ftlesa 700 vo;o gS rFkk

A, B leqPp; U ds mileqPp; gS ftues a Øe'k% n(A) = 200, n(B) = 300 vo;o gS rFkk n(AÇB) = 100

gks rks n(A'ÇB') cjkcj gksxk&

(1) 400 (2) 600 (3) 300 (4) 0

37. ;fn a,b,c,dÎ{0, 1} rFkk ax + by = 0, cx + dy = 0

rks :-

dFku–1 : mijksDr fudk; ds vuUr gy gksus dh izkf;drk

38

gksxh&

dFku–2 : mijksDr lehdj.k fudk; ds dksbZ gy ugha gksus

dh izkf;drk 'kwU; gksxhA(1) dFku-1 lgh gS vkSj dFku-2 lgh gSA dFku-2, dFku-1

dk lgh Li"Vhdj.k ugha gSA(2) dFku-1 xyr gS vkSj dFku-2 lgh gSA

(3) dFku-1 lgh vkSj dFku-2 xyr gSA

(4) dFku-1 lgh gS vkSj dFku-2 lgh gSA dFku-2, dFku-1 dk lgh Li"Vhdj.k gSA

35. If A = 1 2 12 0 13 2 1

- and A3 + xA2 + yA + zI = 0

then x + y + z is equal to :-(1) –2 (2) –0 (3) 6 (4) None

36. Let U be the universal set containing 700

elements. If A and B are subsets of U such that

n(A) = 200, n(B) = 300 and n(AÇB) = 100 then

n(A'ÇB')=

(1) 400 (2) 600 (3) 300 (4) 0

37. If a,b,c,dÎ{0, 1} and ax + by = 0, cx + dy = 0then:-Statement–1 : The probability that above system

of linear equation has infinite solution 38

Statement–2 : The probability that above systemof linear equation have no solution will be zero.(1) Statement–1 is true, Statement–2 is true;

Statement–2 is not the correct explanationof Statement–1.

(2) Statement–1 is false, Statement–2 is true.(3) Statement–1 is true, Statement–2 is false.(4) Statement–1 is true, Statement–2 is true;

Statement–2 is the correct explanation ofStatement–1.

J ges'kk eqLdjkrs jgs a A


21–03–2014

01CE313053 KOTA / H-13/29


38. Consider the family of lines x(a + b) + y = 1,where a, b and c are the roots of the equationx3 – 3x2 + x + l= 0 such that c Î [1,2]. If the givenfamily of lines makes triangle of area 'A' withcoordinate axis, then maximum value of 'A' (insq. units) will be -

(1) 14

(2) 1 (3) 18

(4) 12

39. A vector ˆ ˆ ˆai bj cka = + +r is said to be rational

vector if a, b, c are all rational. If this vector ar

having magnitude as positive integer, makes an

angle 4p

with vector ˆ ˆ ˆ2i 3 2 j 4kb = + +r

, then

ar

always -(1) lies in xy plane(2) lies in xz plane(3) lies in yz plane(4) lies on x-axis

40. If two tangents drawn from a point P to the parabolay2 = 4x be such that the slope of one tangent is doubleof the other, then P lies on the curve. :-(1) 9y = 2x2 (2) 9x = 2y2

(3) 2x = 9y2 (4) None of these41. The relation R in the R of real numbers; defined

as R = {(a, b) : a £ b2} then R is(1) Reflexive(2) Symmetric(3) Transitive(4) neighter reflexive nor symmetric nor transitive

38. ekuk js[kkvksa dk fudk; x(a + b) + y = 1 gS] tgk¡ a, b

rFkk c lehdj.k x3 – 3x2 + x + l = 0 ds ewy gS rFkk

c Î [1,2] gSA ;fn fn;k x;k js[kk fudk; funsZ'kh v{kksa ds

lkFk {ks=Qy 'A' dk f=Hkqt cukrk gS] rks 'A' dk vfèkdre

eku (oxZ bdkbZ esa) gksxk -

(1) 14

(2) 1 (3) 18

(4) 12

39. ,d lfn'k ˆ ˆ ˆai bj cka = + +r

dks ifjes; lfn'k dgsaxs] ;fna, b, c lHkh ifjes; gksA ;fn lfn'k a

r , ftldk ifjek.k

/kukRed iw.kk±d gS] lfn'k ˆ ˆ ˆ2i 3 2 j 4kb = + +r

ds lkFk

4p

dk dks.k cukrk gS] rks ar

lnSo -

(1) xy lery esa fLFkr gksxk(2) xz lery esa fLFkr gksxk(3) yz lery esa fLFkr gksxk(4) x-v{k ij fLFkr gksxk

40. ;fn ,d fcUnq P ls ijoy; y2 = 4x ij [khaph xbZ Li'kZjs[kkvksa esa ,d dh izo.krk] vU; dh izo.krk dh nksxquhgS] rks P ftl oØ ij gS] og gS :-(1) 9y = 2x2 (2) 9x = 2y2

(3) 2x = 9y2 (4) buesa ls dksbZ ugha41. ,d lEcU/k R, okLrfod la[;kvksa ds leqPp; R esa bl

izdkj ifjHkkf"kr gS R = {(a, b) : a £ b2} rks R gSA(1) LorqY;(2) lefer(3) laØked(4) u LorqY;] u lefer rFkk u laØked


21–03–2014

01CE313053KOTA / H-14/29


42. The length of the shortest path that be gives atthe point (2,5), touches the x-axis and then endsat a point on the circlex2 + y2 + 12x – 20 y + 120 = 0

(1) 13 (2) 4 10

(3) 15 (4) 6 + 89

43. In the figure shown, radius of circle C1 be r and

that of C2 be r2

, where 1

r3

= PQ, then length of

AB is (where P and Q being centres of C1 & C2

respectively)

P

Q

A

B

C1

C2

(1) 2 3 r (2) 3 3 r

4

(3) 3 3 r (4) 3 3 r

2

44. The equation of the tangents to the hyperbola4x2 – y2 = 12 are y = 4x+ c1 & y = 4x + c2, then|c1 – c2| is equal to -(1) 1 (2) 4(3) 6 (4) 12

42. fcUnq (2,5) dh o`Ùk x2 + y2 + 12x – 20 y + 120 = 0 dsfdlh fcUnq ls x-v{k dks Li'kZ djrh gq;h U;wure nwjh Kkr

djsa

(1) 13 (2) 4 10

(3) 15 (4) 6 + 89

43. fn;s x;s fp= esa] o`Ùk C1 rFkk o`Ùk C2 dh f=T;k Øe'k%

r rFkk r2

gS] tgk¡ 1

r3

= PQ, rks AB dh yEckbZ gksxk

(tgk¡ P rFkk Q Øe'k% C1 rFkk C2 ds dsUæ gS)

P

Q

A

B

C1

C2

(1) 2 3 r (2) 3 3 r

4

(3) 3 3 r (4) 3 3 r

2

44. vfrijoy; 4x2 – y2 = 12 dh Li'kZ js[kk dk lehdj.ky = 4x+ c1 rFkk y = 4x + c2 gks] rks |c1 – c2| dk ekugksxk -(1) 1 (2) 4(3) 6 (4) 12


21–03–2014

01CE313053 KOTA / H-15/29


45. The locus of P satisfying the condition

|PF1| + |PF2| = 10 ; F1 = (1, 2) ; F2 = (12, 2) is:-

(1) an ellipse (2) a straight line

(3) a parabola (4) an empty set

46. Let the points P, Q and R have position vectors

1r 3i 2 j k,= - -r

2r i 3j 4k= + +r

and

3r 2i j – 2k= +r

respectively relative to an origin O.

Then the distance of P from the plane OQR is :-

(1) 2 (2) 3 (3) 1 (4) 5

47. The locus of a point equidistant from two given points

whose position vectors are ar

and br

is equal to:-

(1) 1

r (a b) .(a b) 02

é ù- + + =ê úë û

r rr r r

(2) 1

r (a b) .(a b) 02

é ù- + - =ê úë û

r rr r r

(3) 1

r (a b) .a 02

é ù- + =ê úë û

rr r r

(4) None

48. If a, b and c are unit vectors satisfying

| a b- |2 + | b c- |2 | c a- |2 = 9, then

| 2a 3b 3c+ + | is :-(1) 3 (2) 2

(3) 1 (4) 22

45. fcanq P dk fcUnqiFk tks fd |PF1| + |PF2| = 10 ;

F1 = (1, 2) ; F2 = (12, 2) dks larq"V djrk gS] gksxk :-

(1) ,d nh?kZo`Ùk (2) ,d ljy js[kk

(3) ,d ijoy; (4) ,d fjDr leqPp;

46. ekuk rhu fcUnq P, Q rFkk R ftuds fLFkfr lfn'k

1r 3i 2 j k,= - -r

2r i 3j 4k= + +r rFkk

3r 2i j – 2k= +r

ewy fcUnq ds lkis{k gks] rks P dh lery

OQR ls nwjh gksxh :-

(1) 2 (2) 3 (3) 1 (4) 5

47. ml fcUnq dk fcUnqiFk tks nks fcUnqvksa ftuds fLFkfr lfn'k

ar

rFkk br

nks ls leku nwjh ij fLFkr gks] gksxk ?

(1) 1

r (a b) .(a b) 02

é ù- + + =ê úë û

r rr r r

(2) 1

r (a b) .(a b) 02

é ù- + - =ê úë û

r rr r r

(3) 1

r (a b) .a 02

é ù- + =ê úë û

rr r r

(4) dksbZ ugha

48. ;fn a, b rFkk c bdkbZ lfn'k gks tks fd larq"V djrs gks

| a b- |2 + | b c- |2 | c a- |2 = 9, rks

| 2a 3b 3c+ + | gksxk :-(1) 3 (2) 2

(3) 1 (4) 22


21–03–2014

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49. Consider four points A, B, C & D with position

vectors a, b, crr r

and dr

w.r.t. origin O. Also a, b, crr r

are non-coplanar and line OD intersects plane

ABC at some point M such that OM m=uuuur r

.

Statement-1 : [a b c]

m d[d b c] [d c a] [d a b]

=+ +

rr rrr

r r r r rr r r r .

Statement-2 : Four points A, B, C & M are

coplanar if [a b c] [m a b] [m b c] [m c a]= + +r r rr r r r r r r r r

.(1) Statement–1 is true, Statement–2 is true;




50. Consider the planes 3x – 6y – 2z = 15 and2x + y – 2z = 5Statement–1 : The parametric equation of the lineof intersection of the given planes arex = 3 + 14t, y = 1 + 2t, z = 15t

Statement–2 : The vector ˆ ˆ ˆ14i 2 j 15k+ + is

parallel to the line of intersection of given planes.

(1) Statement–1 is true, Statement–2 is true;Statement–2 is not the correct explanationof Statement–1.

(2) Statement–1 is false, Statement–2 is true.

(3) Statement–1 is true, Statement–2 is false.

(4) Statement–1 is true, Statement–2 is true;Statement–2 is the correct explanation ofStatement–1.

49. ekuk pkj fcUnq A, B, C rFkk D ftuds ewy fcUnq O ds

lkis{k fLFkfr lfn'k Øe'k% a, b, crr r

rFkk dr

gS rFkk a, b, crr r

vleryh; gSa vkSj js[kk OD lery ABC dks fdlh fcUnqM ij bl izdkj izfrPNsn djrh gS fd OM m=

uuuur r gksA

dFku–1 : [a b c]

m d[d b c] [d c a] [d a b]

=+ +

rr rrr

r r r r rr r r r .

dFku–2 : pkj fcUnq A, B, C rFkk M leryh; gksaxs

;fn [a b c] [m a b] [m b c] [m c a]= + +r r rr r r r r r r r r

gksA(1) dFku-1 lgh gS vkSj dFku-2 lgh gSA dFku-2, dFku-1

dk lgh Li"Vhdj.k ugha gSA(2) dFku-1 xyr gS vkSj dFku-2 lgh gSA(3) dFku-1 lgh vkSj dFku-2 xyr gSA(4) dFku-1 lgh gS vk Sj dFku-2 lgh gSA dFku-2,

dFku-1 dk lgh Li"Vhdj.k gSA50. ;fn nks lery 3x – 6y – 2z = 15 rFkk 2x + y–2z=5

gks] rksdFku–1 : fn;s x;s leryksa ds izfrPNsnu ls feyusokyh js[kk dk izkpfyd lehdj.k x = 3 + 14t,y = 1 + 2t, z = 15t gSdFku–2 : fn;s x;s leryksa ds izfrPNsnu ls feyus okyh

js[kk ds lekUrj lfn'k ˆ ˆ ˆ14i 2 j 15k+ + gSaA

(1) dFku-1 lgh gS vkSj dFku-2 lgh gSAdFku-2, dFku-1 dk

lgh Li"Vhdj.k ugha gSA

(2) dFku-1 xyr gS vkSj dFku-2 lgh gSA


(4) dFku-1 lgh gS vk Sj dFku-2 lgh gSA dFku-2,

dFku-1 dk lgh Li"Vhdj.k gSA


21–03–2014

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51. The area of the smaller portion between the circlex2 + y2 = 9 and the line x = 1 is

(1) 9 sec–1 3 – 8

(2) 9 cosec–1 3 – 8

(3) sec–1 3 – 8

(4) none of these

52. The area between the parabola x2 = 4y and linex = 4y – 2 is

(1) 9/4 (2) 9/8

(3) 9/2 (2) 9

53. The order and degree of the differential equation

é ùæ ö æ öæ öê ú+ =ç ÷ ç ÷ç ÷+ê ú è øè ø è øë û

4 / 532 3

2 3

d y m d y1

dx m 1 dx are

(1) 3, 5 (2) 5, 3

(3) 3, 2 (4) 3, 3

54. Let a = cos a + cos b – cos (a + b) and

b = 4 sin sin cos2 2 2a b a + b

. Then value of a – b

is

(1) 0 (2) 1

(3) –1 (4) None of these

51. o`Ùk x2 + y2 = 9 rFkk js[kk x = 1 ds e/; NksVs Hkkx dk{ks=Qy gS&

(1) 9 sec–1 3 – 8

(2) 9 cosec–1 3 – 8

(3) sec–1 3 – 8

(4) buesa ls dksbZ ugha

52. ijoy; x2 = 4y rFkk js[kk x = 4y – 2 ds e/; dk{k s=Qy gS&

(1) 9/4 (2) 9/8

(3) 9/2 (2) 9

53. vody lehdj.k

é ùæ ö æ öæ öê ú+ =ç ÷ ç ÷ç ÷+ê ú è øè ø è øë û

4 /532 3

2 3

d y m d y1

dx m 1 dx dh dksfV ,oa ?kkr gSA

(1) 3, 5 (2) 5, 3

(3) 3, 2 (4) 3, 3

54. ;fn a = cos a + cos b – cos (a + b) o

b = 4 sin sin cos2 2 2a b a + b

. rks a – b dk eku gksxk&

(1) 0 (2) 1

(3) –1 (4) buesa ls dksbZ ugha

viuh {kerk dks iwjk olwyus dk iz;kl djs a A


21–03–2014

01CE313053KOTA / H-18/29


55. The number of solutions of

sin2x cos2x = 1 + cos2x sin4x in the interval[0, 2p] is :-

(1) 0 (2) 1

(3) 2 (4) 3

56. If the variance of 1, 2, 3, 4, 5,..., 10 is 9912

, then

the standard deviation of 3, 6, 9, 12,.....30 is :-

(1)2974

(2) 3 332

(3) 3 992

(4) 9912

57. The only statement among the followings that isa tautology is :-

(1) A Ù (A Ú B)

(2) A Ú (A Ù B)

(3) [A Ù (A ® B)]®B

(4) B ® [A Ù (A ® B)]

55. lehdj.k

sin2x cos2x = 1 + cos2x sin4x dh [0, 2p] esa gyksa dhla[;k gksxh&

(1) 0 (2) 1

(3) 2 (4) 3

56. ;fn 1, 2, 3, 4, 5,..., 10 dk izlkj 9912

, gS rks 3, 6, 9,

12,.....30 dk ekud fopyu gksxk&

(1)2974

(2) 3 332

(3) 3 992

(4) 9912

57. fuEu esa ls dkSulk dFku iqu:fDr gS&

(1) A Ù (A Ú B)

(2) A Ú (A Ù B)

(3) [A Ù (A ® B)]®B

(4) B ® [A Ù (A ® B)]


21–03–2014

01CE313053 KOTA / H-19/29


58. Statement-1: A pole standing in the centre of arectangular field of area 2500 sq. units subtendsangle a and b respectively at the mid-points oftwo adjacent sides of the field such thata + b = p/2, the height of the pole is 25 units.

Stament-2: Area of a rectangle is equal to theproduct of the length of the adjacent sides.

(1) Statement–1 is true, Statement–2 is true;Statement–2 is not the correct explanationof Statement–1.

(2) Statement–1 is false, Statement–2 is true.

(3) Statement–1 is true, Statement–2 is false.

(4) Statement–1 is true, Statement–2 is true;Statement–2 is the correct explanation ofStatement–1.

59. If f(x) = x3 – x2 + 100x + 1001, then

(1) f(2000) > f(2001)

(2)1 1f f

1999 2000æ ö æ ö>ç ÷ ç ÷è ø è ø

(3) f(x + 1) < f(x – 1)

(4) f(3x – 5) > f(3x)

60. Maximum value of x2ln1x

is

(1) 2e (2) e

(3) 1e

(4) 12e

58. dFku-1: ,d vk;rkdkj [ksr ftldk {ks=Qy 2500 oxZ

bdkbZ gS ds dsUnz ij [kM+k ,d [kEck nks Øekxr Hkqtkvksa

ds e/; fcUnq ij a o b dks.k bl izdkj cukrk gS fd

a + b = p/2, rks [kEcs dh Å¡pkbZ 25 bdkbZ gksxhA

Stament-2: vk;r dk {ks=Qy mudh Øekxr Hkqtkvksa

dh yEckbZ ds xq.kuQy ds cjkcj gksrk gSA

(1) dFku-1 lgh gS vkSj dFku-2 lgh gSA dFku-2, dFku-1dk lgh Li"Vhdj.k ugha gSA



(4) dFku-1 lgh gS vkSj dFku-2 lgh gSA dFku-2, dFku-1 dk lgh Li"Vhdj.k gSA

59. ;fn f(x) = x3 – x2 + 100x + 1001, rks

(1) f(2000) > f(2001)

(2)1 1f f

1999 2000æ ö æ ö>ç ÷ ç ÷è ø è ø

(3) f(x + 1) < f(x – 1)

(4) f(3x – 5) > f(3x)

60. x2ln1x

dk egÙke eku gSA

(1) 2e (2) e

(3) 1e

(4) 12e


21–03–2014

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61. In the measurement of physical quantity2

1/ 3

A BX

C D= , the percentage error introduced in the

measurements of the quantities A, B, C and D are2%, 2%, 4% and 5%, respectively. Then theminimum amount of percentage error in themeasurement of X is contributed by :-(1) A (2) B (3) C (4) D

62. Vector which is perpendicular toˆ ˆ(a cos )i (bsin ) jq + q is given by :-

(1) ˆ ˆ(bsin )i – (a cos ) jq q

(2) 1 1ˆ ˆsin i cos ja b

æ ö æ öq - qç ÷ ç ÷è ø è ø

(3) ˆ5k(4) all of these

63. A very large number of balls are thrown verticallyupwards in quick succession in such a way thatthe next ball is thrown when the previous one isat the maximum height. If the maximum heightis 5m, then the number of balls thrown per minuteis : (g = 10 ms–2)(1) 120 (2) 80 (3) 60 (4) 40

64. From a balloon rising vertically upwards at 5 ms–1,a stone is thrown up at 10 ms–1 relative to theballoon. Its velocity with respect to the groundafter 2 sec is : (assume g = 10 ms–2)(1) 0 (2) 20 ms–1

(3) 10 ms–1 (4) 5 ms–1

61. ,d HkkSfrd jkf'k 2

1/ 3

A BX

C D= ds ekiu esa A, B, C rFkk

D esa mRiUu izfr'kr =qfV;k¡ Øe'k% 2%, 2%, 4% rFkk 5%

gSA rc X ds ekiu esa U;wure =qfV dk va'k nsxk :-

(1) A (2) B

(3) C (4) D

62. og osDVj] tks ˆ ˆ(a cos )i (bsin ) jq + q ds vfHkyEcor~ gS]gksxk :-(1) ˆ ˆ(bsin )i – (a cos ) jq q

(2) 1 1ˆ ˆsin i cos ja b

æ ö æ öq - qç ÷ ç ÷è ø è ø

(3) ˆ5k(4) ;s lHkh

63. ,d ds ckn ,d vusd xsans rsth ls Å/okZ/kj Åij dh vksjbl izdkj Qsadh tkrh gSa fd tc igyh xsan viuh vf/kdreÅ¡pkbZ ij ig¡qp tkrh gS] rc nwljh xsan Qsadh tkrh gSA ;fnizR;sd xsan dh vf/kdre Å¡pkbZ 5 ehVj gS] rks izfr feuVQsadh tkus okyh xsanksa dh la[;k :(g = 10 ehVj-lsd.M–2)(1) 120 (2) 80 (3) 60 (4) 40

64. 5 ehVj-lsd.M–1 ls Å/okZ/kj Åij dh vksj mB jgs ,d xqCckjsls ,d iRFkj Å/okZ/kj Åij dh vksj xqCckjs ds lkis{k10 ehVj-lsd.M–1 ds osx ls Qsadk tkrk gSA 2 lsd.M i'pkr~i`Foh ds lkis{k bldk osx gS : (g = 10 ehVj-lsd.M–2)(1) 0 (2) 20 ehVj-lsd.M–1

(3) 10 ehVj-lsd.M–1 (4) 5 ehVj-lsd.M–1

PART C - PHYSICS

izR;sd iz'u dks vtqZu cudj djksA


21–03–2014

01CE313053 KOTA / H-21/29


65. Displacement (x) of a particle is related to timet as x = at2 + bt2 – ct3, where a, b and c are constantsof motion. The velocity of the particle when itsacceleration is zero is given by :-

(1) 2b

ac

+ (2) 2b

a2c

+

(3) 2b

a3c

+ (4) 2b

a4c

+

66. The horizontal range of a projectile is 4 3 times

the maximum height. Its angle of projection willbe :-

(1) 45º (2) 60º (3) 90º (4) 30º

67. A body of mass M kg is on the top point of asmooth hemisphere of radius 5 m. It is releasedto slide down the surface of the hemisphere. Itleaves the surface when its velocity is 5 m/s. Atthis instant the angle made by the radius vectorof the body with the vertical is :

(Acceleration due to gravity = 10 ms–2)

(1) 30º (2) 45º (3) 60º (4) 90º

68. An aeroplane is flying horizontally at a height of490 m with a velocity of 150 m/s. A bagcontaining food is to be dropped to the jawans onthe ground. How far from them should the bag bedropped so that it directly reaches them ?

(1) 1000 m (2) 1500 m

(3) 750 m (4) 2000 m

65. ,d d. k d s f oLF k k iu (x) rF k k le; ( t ) e s alEca/k x = at2 + bt2 – ct3 gS, tgk¡ a, b rFkk c xfr

ds fu;rkad gSaA tc d.k dk Roj.k 'k w U; gks] rks bldk

osx gk sxk :-

(1) 2b

ac

+ (2) 2b

a2c

+

(3) 2b

a3c

+ (4) 2b

a4c

+

66. ,d iz{ksI; dh {kSfrt ijkl mldh vf/kdre Å¡pkbZ dh

4 3 xquh gSA bldk iz{ksi.k dks.k gksxk :-

(1) 45º (2) 60º(3) 90º (4) 30º

67. M fdxzk dk ,d fi.M 5 ehVj f=T;k ds fpdus v¼Zxksysdh pksVh ij fLFkr gSA ;g v¼Zxksys dh lrg ij fQlyusds fy;s NksM+k tkrk gSA bldk osx 5 eh/ls gks tkus ij ;gv¼Zxksys dh lrg ls NwV tkrk gSA bl {k.k fi.M dk f=T;h;osDVj Å/okZ/kj ls dks.k cukrk gS :

(xq:Roh; tfur Roj.k = 10 eh/ls2)

(1) 30º (2) 45º

(3) 60º (4) 90º

68. ,d ok;q;ku 490 ehVj dh Å¡pkbZ ls ijUrq {kSfrt fn'kkesa 150 eh/ls ds osx ls tk jgk gSA Hkkstu dk ,d FkSyk tokuksads fy;s ok;q;ku ls Hkwry ij fxjk;k tkrk gSA tokuksa lsfdruh nwjh ij ok;q;ku ls FkSyk fxjkuk pkfg;s ftlls fd;g tokuksa rd ig¡qp tk;s ?

(1) 1000 eh (2) 1500 eh(3) 750 eh (4) 2000 eh


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69. The maximum height reached by projectile is 4m.The horizontal range is 12 m. Velocity ofprojection in ms–1 is : (g is acceleration due togravity)

(1) 5 g / 2 (2) 3 g / 2

(3) 1

g / 23

(4) 1

g / 25

70. Two blocks are placed on a horizontal surface asshown in the figure. The coefficient of frictionbetween the blocks is 0.5 and between block of5 kg and the surface is 0.7. The maximumhorizontal force that can be applied to the blockof mass 5 kg so that the two blocks move withoutslipping is : (Take g = 10 m/s2)

5kg

3kg

(1) 14 N (2) 24 N (3) 56 N (4) 96 N71. A small particle of mass m attached with a light

inextensible thread of length 'l' is moving in avertical circle of radius 'l'. If the particle is movingin complete vertical circle and the ratio of itsmaximum to minimum velocity is 2 : 1, then theminimum velocity of the particle is :-

(1) g3l

(2) 2g3l

(3) g

23l

(4) gl

69. ,d iz{ksI; }kjk izkIr dh x;h vf/kdre Å¡pkbZ 4 ehVj gSA

{kSfrt ijkl 12 ehVj gSA iz{ksI; osx ehVj/lsd.M esa gS :

(g xq:Ro tfur Roj.k gS)

(1) 5 g / 2 (2) 3 g / 2

(3) 1

g / 23

(4) 1

g / 25

70. tSlk fd fp= esa n'kkZ;k x;k gS] nks xqVds ,d {kSfrt lrgij j[ks gq, gSaA xqVdksa ds chp ?k"kZ.k xq.kkad 0.5 gS rFkk5 fdxzk æO;eku ds xqVds vkSj lrg ds chp ?k"kZ.k xq.kkad0.7 gSA 5 fdxzk ds xqVds ij yxk;k x;k og vf/kdre{kSfrt cy ftlls nksuksa xqVds lrg ij fcuk fQlys xfrekugksa: (g = 10 eh/ls2)

5kg

3kg

(1) 14 U;wVu (2) 24 U;wVu(3) 56 U;wVu (4) 96 U;wVu

71. 'l' yEckbZ ds u QSyus okys /kkxs ls m æO;eku ds ,d d.kdks ck¡/kdj mls 'l' f=T;k ds Å/okZ/kj o`Ùk esa ?kqek;k tkrkgSA ;fn d.k Å/okZ/kj o`Ùk esa iwjk pDdj yxkrk gS vkSj bldsvf/kdre o U;wure osxksa dh fu"ifÙk 2 : 1 gS] rks d.k dkU;wure osx gS :-

(1) g3l

(2) 2g3l

(3) g

23l

(4) gl

dksbZ Hkh iz'u Key Filling ls xyr ugha gksuk pkfg,A


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72. Sand falls vertically at the rate of 2 kg/s on aconveyer belt moving horizontally with velocityof 0.2 m/s, the extra power required to keep thebelt moving is :-(1) 0.04 W (2) 0.4 W(3) 0.08 W (4) 0.8 W

73. A shell of mass 2 kg and moving at a rate of4 m/s suddenly explodes into two equal fragments.The fragments go in directions inclined with theoriginal line of motion with equal velocities. Ifthe explosion imparts 48 J of translational kineticenergy to the fragments divided equally, thevelocity (magnitude and direction both) of eachfragment :-(1) 2 m/s, 45º (2) 8 m/s, 45º(3) 8 m/s, 60º (4) 2 m/s, 60º

74. A body of mass 3.0 kg moves under the influenceof some external force such that its position S asa function of time t is given by S = 6t3 – t2 + 1where S is in metres and t is in seconds. The workdone (in joule) by the force in first three secondsis :-(1) 18 J (2) 1800 J(3) 3660 J (4) 36504 J

75. Two springs P and Q are stretched by applyingforces of equal magnitudes at the four ends. If thespring constant of P is 2 times greater than thatof Q and the energy stored in P is E, then theenergy stored in Q is :-

(1) E4

(2) E2

(3) E (4) 2E

72. {kSfrt fn'kk esa 0.2 eh/ls ds osx ls xfreku ,d Hkkj <ksusokyh cSYV ij jsr Å/okZ/kj 2 fdxzk/lsd.M dh nj ls Mkyktk jgk gSA cSYV dks xfreku j[kus ds fy;s nh x;h vfrfjDr'kfDr gksxh :-(1) 0.04 okWV (2) 0.4 okWV(3) 0.08 okWV (4) 0.8 okWV

73. ce dk 2 fdxzk dk ,d xksyk 4 eh/ls ds osx ls pyrs gq,

vpkud nks cjkcj VqdM+ksa esa QV tkrk gSA nksuksa VqdM+s cjkcj

osx ls viuh xfr dh izkjfEHkd fn'kk ls fo{ksfir gksdj tkrs

gSaA ;fn foLQksV esa dqy 48 twy xfrt ÅtkZ dk vknku

iznku gqvk gS tks nksuksa VqdM+ksa esa leku :i ls forfjr gqbZ

gS] rks izR;sd VqdM+s ds osx dk ifjek.k o fn'kk :-

(1) 2 eh/ls, 45º (2) 8 eh/ls, 45º

(3) 8 eh/ls, 60º (4) 2 eh/ls, 60º

74. 3.0 fdxzk æO;eku dk ,d fi.M fdlh cká cy ds izHkko

esa bl izdkj xfr djrk gS fd bldh fLFkfr S le; t lslehdj.k S = 6t3 – t2 + 1 ds vuqlkj O;Dr dh tkrhgSA ;fn S ehVj esa rFkk le; t lsd.M esa gS rks cy }kjk

igys rhu lsd.M esa fd;k x;k dk;Z (twy esa) gS :-

(1) 18 twy (2) 1800 twy

(3) 3660 twy (4) 36504 twy75. nks fLizaxksa P rFkk Q ds pkjksa fljksa ij ,dleku ifjek.k ds

cy yxkdj [khapk tkrk gSA ;fn P dk fLizax fu;rkad Q

ds fLizax fu;rkad ls 2 xquk cM+k gS rFkk P esa lafpr ÅtkZ

E gS rc Q esa lafpr ÅtkZ gS :-

(1) E4

(2) E2

(3) E (4) 2E


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76. Three identical blocks A, B and C are placed ona fixed horizontal surface which is perfectlysmooth. The block B and C are at rest while theblock A is approaching towards B with a uniformvelocity of 10 m/s. The coefficient of restitutionfor all the collisions is 0.5. The block C just afterthe collision moves with the speed of :-

A B C

(1) 3.75 m/s (2) 5.6 m/s(3) 7.5 m/s (4) zero

77. A sphere is rolling without slipping on a fixedhorizontal plane surface. In the figure, A is thepoint of contact, B is the centre of the sphere andC is its topmost point. Then :-

C

B

A

(1) C A B CV V 2(V V )- = -ur ur ur ur

(2) C B B AV V V V- = -ur ur ur ur

(3) C A B C| V V | 2 | V V |- = -ur ur ur ur

(4) C A B| V V | 4 | V |- =ur ur ur

78. The maximum and minimum distances of a cometfrom the sun are 1.6 × 1012 m and 8 × 1010 m. Ifits velocity when nearest to the sun is6 × 104 m/s, what is its velocity when farthest ?Assume in both positions the comet is moving incircular orbit :-(1) 103 m/s (2) 2 × 103 m/s(3) 3 × 103 m/s (4) 4 × 103 m/s

76. rhu ,d tSls xqVds A, B rFkk C ,d iw.kZr% fpduh {kSfrt

lrg ij fLFkr gSaA xqVds B rFkk C fojkekoLFkk esa gSa tcfd

xqVdk A 10 ehVj/lsd.M ds fu;r osx ls B dh vksj vk

jgk gSA lHkh VDdjksa ds fy;s mixeu xq.kkad 0.5 gSA la?kê

ds Bhd i'pkr~ xqVds C dk osx gS :-

A B C

(1) 3.75 eh/ls (2) 5.6 eh/ls

(3) 7.5 eh/ls (4) 'kwU;77. ,d xksyk fdlh n`<+ {kSfrt lery lrg ij fcuk fQlys

gq, fp=kuqlkj] yq<+d jgk gSA A xksys rFkk lrg dk Li'kZfcUnq gS, B xksys dk dsUæ gS rFkk C bldk mPpre fcUnqgSA rc :-

C

B

A

(1) C A B CV V 2(V V )- = -ur ur ur ur

(2) C B B AV V V V- = -ur ur ur ur

(3) C A B C| V V | 2 | V V |- = -ur ur ur ur

(4) C A B| V V | 4 | V |- =ur ur ur

78. lw;Z ds ,d iqPNy rkjs dh vf/kdre o U;wure nwfj;k¡ Øe'k%1.6 × 1012 ehVj rFkk 8 × 1010 ehVj gSaA lw;Z ds lehiLFkvkus ij ;fn bldk osx 6 × 104 ehVj/lsd.M gS rks lw;Zls nwjLFk tkus ij bldk osx D;k gksxk ? ;g eku ysa fdnksuksa fLFkfr;ksa esa iqPNy rkjk o`Ùkkdkj d{kk esa ifjØe.k djjgk gS :-(1) 103 ehVj/lsd.M (2) 2 × 103 ehVj/lsd.M(3) 3 × 103 ehVj/lsd.M (4) 4 × 103 ehVj/lsd.M


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79. A uniform disc of radius R and

F

mass M can rotate withoutfriction on an axle passingthrough its centre andperpendicular to its plane face.The cord is wound over the rimof the disc and a uniform forceF is applied on the cord as shownin the figure. Then the angular acceleration in thedisc is proportional to :-

(1) R0 (2) R (3) R2 (4) 1R

80. Water rises to a height h in a capillary tube heldvertically in a beaker containing water. If thecapillary tube is inclined at an angle 30º with thevertical, the height to which water rises willbe :-

(1) h (2) h2

(3) 2h (4) 2h

381. Statement-1 : In a simple harmonic motion,

velocity does not change uniformly withdisplacement.Statement-2 : The graph between velocity &displacement for a harmonic oscillator is aparabola.(1) Statement–1 is true, Statement–2 is true;

Statement–2 is not the correct explanation ofStatement–1

(2) Statement–1 is false, Statement–2 is true(3) Statement–1 is true, Statement–2 is false(4) Statement–1 is true, Statement–2 is true;

Statement–2 is the correct explanation ofStatement–1

79. M æO;eku rFkk R f=T;k dh ,dleku

F

pdrh dks vius dsUæd ls xqtjus okyh

o ry ds vfHkyEcor~ /kqjh ds ifjr%

fcuk ?k"kZ.k ds ?kqek;h tk ldrh gSA ,d

Mksjh pdrh ds fje ij yisVdj ml ij

fp=kuqlkj F cy yxk;k tkrk gSA rc

pdrh e s a mRiUu dk s.kh; Roj.k

lekuqikrh gS :-

(1) R0 (2) R (3) R2 (4) 1R

80. ty ls Hkjs chdj esa ,d ds'kuyh dks Å/okZ/kj Mqcksus ijds'kuyh esa ty h Å¡pkbZ rd Åij p<+rk gSA ;fn ds'kuyh

dks Å/okZ/kj ls 30º >qdk;k tkrk gS] rks og Å¡pkbZ tgk¡ rd

ty ds'kuyh esa Åij p<+sxk :-

(1) h (2) h2

(3) 2h (4) 2h

381. dFku-1 : ljy vkorZ xfr esa osx] foLFkkiu ds lkFk

,dleku :i esa ifjofrZr ugha gksrk gSA

dFku-2 : vkorZ nksfy= ds fy, osx rFkk foLFkkiu ds chp

[khapk x;k vkjs[k ijoy;kdkj gSA

(1) dFku-1 lgh gS vkSj dFku-2 lgh gSAdFku-2, dFku-1dk




(4) dFku-1 lgh g S vk Sj dFku-2 lgh gSAdFku -2,

dFku-1 dk lgh Li"Vhdj.k gSA


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82. Statement-1 : In an elastic collision between twobodies, the relative speed of the bodies aftercollision is equal to the relative speed before thecollision.Statement-2 : In an elastic collision, the linearmomentum of the system is conserved.(1) Statement–1 is true, Statement–2 is true;

Statement–2 is not the correct explanation ofStatement–1

(2) Statement–1 is false, Statement–2 is true(3) Statement–1 is true, Statement–2 is false(4) Statement–1 is true, Statement–2 is true;

Statement–2 is the correct explanation ofStatement–1

83. Two sources of sound A and B produces the waveof 350 Hz, they vibrate in the same phase. Theparticle P is vibrating under the influence of these twowaves, if the amplitudes at the point P produced bythe two waves is 0.3 mm and 0.4 mm, then theresultant amplitude of the point P will be whenAP–BP = 25 cm and the velocity of sound is 350 m/second :-(1) 0.7 mm (2) 0.1 mm(3) 0.2 mm (4) 0.5 mm

84. A simple harmonic progressive wave isrepresented by the equation :y = 8 sin 2p(0.1x – 2t) where x and y are in cmand t is in seconds. At any instant the phasedifference between two particles separated by 2.0cm in the x-direction is :-(1) 18° (2) 36° (3) 54° (4) 72°

82. dFku-1 : nks fi.Mksa ds chp iw.kZ izR;kLFk la?kê esa] la?kê

ds i'pkr~ fi.Mksa dh lkis{k pky la?kê ls iwoZ dh lkis{k

pky ds cjkcj gksrh gSA

dFku-2 : iw.kZ izR;kLFk la?kê esa fudk; dk js[kh; laosx

lajf{kr jgrk gSA

(1) dFku-1 lgh gS vkSj dFku-2 lgh gSAdFku-2, dFku-1dk




(4) dFku-1 lgh g S vk Sj dFku-2 lgh gSAdFku -2,

dFku-1 dk lgh Li"Vhdj.k gSA83. 350 Hz vko`fÙk ds nks /ofu L=ksr A rFkk B leku dyk

esa dEiUu dj jgs gSa rFkk muds }kjk Hksth xbZ rjaxksa ds

vUrxrZ d.k P dEiUu dj jgk gSA ;fn nksuksa rjaxksa ds }kjk

d.k P ds vk;ke Øe'k% 0.3 mm rFkk 0.4 mm gksa] rks

AP–BP = 25 cm ds fy, d.k dk vk;ke D;k gksxkA

ok;q esa /ofu dh pky 350 eh/ls gS :-

(1) 0.7 mm (2) 0.1 mm(3) 0.2 mm (4) 0.5 mm

84. ,d ljy vkorZ izxkeh rjax dk lehdj.k :

y = 8 sin 2p(0.1x – 2t) gS] tgk¡ x vkSj y lseh esa rFkk

t lsd.M esa gSA fdlh {k.k ,d nwljs ls x fn'kk esa 2.0 lseh

dh nwjh ij fLFkr nks d.kksa ds chp dykUrj gksxk :-

(1) 18° (2) 36° (3) 54° (4) 72°

Use stop, look and go method in reading the question


21–03–2014

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85. Consider the following :-I. Waves created on the surface of a water pond

by a vibrating sources.II. Wave created by an oscillating electric field in airIII. Sound waves travelling under water.Which of these can be polarized :-(1) I and II (2) II only(3) II and III (4) I, II and III

86. The maximum intensity in Young's double slitepperiment is I0. Distance between the slits isd = 5l , where l is is the wavelength ofmonochromic light used in the experiment. Whatwill be the intensity of light in front of one of theslits on a screen at a distance D = 10d -

(1) 0I2

(2) 0

3I

4(3) I0 (4) 0I

487. A 100 Hz sinusoidal wave is travelling in the

positive x-direction along a string with a linearmass density of 3.5 × 10–3kgm–1 and a tension of35 N. At time t = 0, the point x = 0 has zerodisplacement and the slope of the string is p/20.Then select the wrong alternative :-(1) Velocity of wave is 100 m/s(2) Angular velocity is (200p) rad/s(3) Amplitude of wave is 0.025m(4) None of the above

88. The central fringe of the interference patternproduced by light of wavelength 6000 Å is foundto shift to the position of 4th bright fringe aftera glass plate of refractive index 1.5 is introduced.The thickness of the glass plate would be :-(1) 4.8 µm (2) 8.23 µm(3) 14.98 µm (4) 3.78 µm

85. fuEu dFkuksa ij fopkj djsa :-I. dEik;eku L=ksr }kjk rkykc esa ikuh ds i"B ij mRiUu rjaxsII. nksyuh fo|qr {ks= }kjk ok;q esa mRiUu rjaxsaIII. ikuh ds vUnj xfreku /ofu rjaxsamijksDr esa ls fdls /kzqfor fd;k tk ldrk gS :-(1) I vkSj II (2) dsoy II

(3) II vkSj III (4) I, II rFkk III

86. ;ax ds f}&fLyV iz;ksx esa vf/kdre rhozrk I0 gSA fLyVksa

ds chp dh nwjh d = 5l ;gk¡ l iz;qDr ,d o.khZ izdk'k

dh rjaxnS/;Z gSA fdlh Hkh ,d fLyV ds Bhd lkeus nwjh

D = 10d ij izdk'k dh rhozrk gksxh -

(1) 0I2

(2) 0

3I

4(3) I0 (4) 0I

487. ,d 100 Hz vko`fÙk fd T;k ofØ; rjax /kukRed

x-fn'kk esa ,d Mksjh] ftldk js[kh; ?kuRo 3.5 × 10–3kgm–1

rFkk ruko 35 N gS esa lapfjr gks jgh gSA le; t = 0 ij]fcUnq x = 0 dk foLFkkiu 'kwU; gS rFkk ;gk¡ Mksjh fd izo.krkp/20 gS rks vlR; dFku gksxk :-(1) Velocity of wave is 100 m/s(2) Angular velocity is (200p) rad/s(3) Amplitude of wave is 0.025m(4) None of the above

88. 6000 Å rjaxnS/;Z okys izdk'k }kjk mRiUu O;frdj.k uewus

esa dsfUnz; fÝUt] 4 oha pedhyh fÝUt dh fLFkfr ij

f[kld tkrk gS] ;fn 1.5 viorZukad okyh dk¡p dh iêh

j[k nh tkrh gSA dk¡p dh iêh dh eksVkbZ gksuh pkfg;s :-

(1) 4.8 µm (2) 8.23 µm

(3) 14.98 µm (4) 3.78 µm


21–03–2014

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89. dFku–1 : U;wVu us /ofu dk osx Kkr djus dk lw=

izfrikfnr fd;k ftlds fy, mlus ek/;e ds rki dks fu;r

ekukA

dFku–2 : ykIykl us U;wVu lw= dks la'kksf/kr fd;k rFkk

mlds vuqlkj /ofu dk lapj.k ,d lenkch izØe gSA

(1) dFku-1 lgh gS vkSj dFku-2 lgh gSAdFku-2, dFku-1 dk




(4) dFku-1 lgh gS vk Sj dFku-2 lgh gSAdFku -2,

dFku-1 dk lgh Li"Vhdj.k gSA90. ,d Mksjh ij ,d rjax dk vuqizLFk foLFkkiu y(x, t) fn;k

tkrk gS

( ) ( )2 2– ax +bt +2 abxty x, t = e

;g n'kkZrk gS ,d %

(1) vkofÙk b dh ,d vizxkeh rjax

(2) vkofÙk 1

b dh ,d vizxkeh rjax

(3) +x fn'kk esa xfr'khy pky a

b dh ,d rjax

(4) –x fn'kk esa xfr'khy pky b

adh ,d rjax

89. Statement–1 : Newton gave the formula forvelocity of sound for which he assumed thetemperature of medium to remain constant.Statement–2 : Laplace modified the Newton'sformula assuming that propagation of sound is anisoboric process.(1) Statement–1 is true, Statement–2 is true;




90. The transverse displacement y(x, t) of a wave ona string is given by

( ) ( )2 2– ax +bt +2 abxty x, t = e

This represents a :-

(1) standing wave of frequency b

(2) standing wave of frequency 1

b

(3) wave moving in +x direction with speed a

b

(4) wave moving in –x direction with speed b

a

Your moral dutyis to prove that ALLEN is ALLEN


21–03–2014

01CE313053 KOTA / H-29/29



CORRECTIONScore - I : Leader & Enthusiast Course

Q.No. 24 39 40 41 49Ans. Bonus 1 Bonus 4 1

Q.No. 6 17 58Ans. 2 or 4 1 or 4 4

Q.No. 35 36 44 48 56 74Ans. 3 1 4 4 Bonus 3 or 4 (H)

Q.No. 8 20 58 62 77Ans. Bonus 2 or 3 1 or 2 4 2

Test Date : 08-03-2014

Test # 01 (JEE-Main)


Test Date : 01-03-2014

Test Date : 05-03-2014

Test Date : 13-03-2014Test # 04 (JEE-Main)


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KOTA (RAJASTHAN) CA CAC A - allen.ac.in (Main... · ... 25 (3) 37.5 (4) 53.5 8. Equal masses of methane and oxygen are mixed in an empty container at 25ºC. The fraction of the total