CAREER INSTITUTE Pat S KOTA (RAJASTHAN) DISTANCE … · (ACADEMIC SESSION 2014-2015) PAPE CODE...

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ALLENCAREER INSTITUTEKOTA (RAJASTHAN)

T M

FORM NUMBER

(ACADEMIC SESSION 2014-2015)

PAPER CODE

ALLEN JEE (Main) TEST DATE : 29 - 03 - 2015

0 0 D E 3 1 4 0 2 9

TARGET : JEE (Main) 2015

Corporate OfficeALLEN CAREER INSTITUTE

“SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005

+91-744-2436001 [email protected]

DISTANCE LEARNING PROGRAMME

dlp.allen.ac.in, dsat.allen.ac.in

LEADER TEST SERIES / JOINT PACKAGE COURSE

Full Syllabus Test Pattern : JEE (Main)

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 Booklet

with 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 ofPhysics, Chemistry and Mathematics, having 30 questions in eachpart of equal weightage. Each question is allotted 4 (four) marks forcorrect response.

6. One Fourth mark will be deducted for indicated incorrect responseof each question. No deduction from the total score will be madeif no response 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, 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,A1. ijh{kk iqfLrdk ds bl i"̀B ij vko';d fooj.k uhys@dkys ckWy ikbaV isu

ls rRdky HkjsaA isfUly dk iz;ksx fcYdqy oftZr gSaA2. ijh{kkFkhZ viuk QkeZ ua- (fu/kkZfjr txg ds vfrfjä) ijh{kk iqfLrdk @ mÙkj

i= ij dgha vkSj u fy[ksaA3. ijh{kk dh vof/k 3 ?k aVs gSA4. 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 esa

HkkSfrd foKku] jlk;u foKku ,oa xf.kr] ds 30 iz'u gSa vkSj lHkh iz'uksa ds

vad leku gSaA izR;sd iz'u ds lgh mÙkj ds fy, 4 (pkj)vad fuèkkZfjr fd;s x;s gSaA6. izR;sd xyr mÙkj ds fy, ml iz'u ds dqy vad dk ,d pkSFkkbZ vad dkVk

tk;sxkA mÙkj iqfLrdk esa dksbZ Hkh mÙkj ugha Hkjus ij dqy izkIrkad esa ls½.kkRed vadu 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]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

Note : In case of any correction in the test paper, please mail [email protected] within 2 days along with YourForm No. & Complete Test Details.

uk sV % ;fn bl iz'u i= esa dksbZ Correction gks rks Ïi;k vkidsForm No. ,oa iw.k Z Test Detai ls ds lkFk 2 fnu ds [email protected] ij mail djsaA

IMPORTANT INSTRUCTIONS egRoiw.kZ funs Z'k

Leader Test Series/Joint Package Course/JEE(Main)/29-03-2015

LTS-1/35Kota/00DE314029

SPACE FOR ROUGH WORK

1. The diagram shows a force-extension graph fora rubber band. Consider the followingstatements:-

Extension

O Force

I. It will be easier to compress this rubberthan expand it.

II. Rubber does not return to its original lengthafter it is stretched.

III. The rubber band will get heated if it isstretched and released.

Which of these can deduced from the graph-

(1) III only

(2) II and III

(3) I and III

(4) I only

1. fuEu fp= esa fdlh jcj ds fy, cy&izlkj xzkQ iznf'kZr

gSA fuEu dFkuksa ij fopkj djsa-

Extension

O Force

I. bl jcj dks [khapus ds ctk; laihfM+r djuk vklku

gksxk

II. jcj dks [khapus ds ckn] ;g viuh izkjfEHkd yEckbZ

rd ugha ykSVsxh

III. ;fn bls [khap dj NksM+ fn;k tk;s rks jcj xeZ gks

tk;sxh

xzkQ ls mijksDr esa ls dkSulk dFku lR; gS -

(1) dsoy III

(2) II rFkk III

(3) I rFkk III

(4) dsoy I

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

BEWARE OF NEGATIVE MARKING

PART A - PHYSICS

Kota/00DE314029LTS-2/35

Target : JEE(Main) 2015/MAJOR/29-03-2015


2. Figure shows an infinitely long wire carryingan outward current I

1. The current is along Z

axis. There is a curved wire carrying currentI

2. The magnetic force on this wire between

(x1, y

1) and (x

2, y

2) is :-

(x ,y )1 1

(x ,y )2 2

I 2

I 1

Y

X

(1) 2 2

0 1 2 2 22 21 1

( )

4 ( )

x y

x y

mpI I +

+ (2) 2 2

0 1 2 2 22 21 1

( )

4 ( )

x yn

x y

mpI I +

+l

(3) 0 1 2 2 2

1 1

( )

4 ( )

x yn

x y

mpI I +

+l (4)

2 20 1 2 2 2

2 21 1

( )

2 ( )

x yn

x y

mpI I +

+l

3. A particle of mass m moves with a constantvelocity. Which of the following statements isnot correct about its angular momentumabout O :

D E

A C

B1m

1m

O

Y

X

(1) it is zero when it is at A and movingalong OA

(2) is the same at all points along the line DE(3) is of the same magnitude but oppositely

directed at B and D(4) increases as it moves along the line BC

2. fp= esa ,d vuUr yEck rkj n'kkZ;k x;k gS] ftlesa /kkjkI

1 ckgj dh vksj izokfgr gks jgh gSA /kkjk Z-v{k ds vuqfn'k

gSA ;gka ,d oØh; rkj Hkh gS ftlesa I2 /kkjk izokfgr gks

jgh gSA bl rkj ij (x1, y

1) o (x

2, y

2) ds eè; pqEcdh;

cy gksxk %&

(x ,y )1 1

(x ,y )2 2

I 2

I 1

Y

X

(1) 2 2

0 1 2 2 22 21 1

( )

4 ( )

x y

x y

mpI I +

+ (2) 2 2

0 1 2 2 22 21 1

( )

4 ( )

x yn

x y

mpI I +

+l

(3) 0 1 2 2 2

1 1

( )

4 ( )

x yn

x y

mpI I +

+l (4)

2 20 1 2 2 2

2 21 1

( )

2 ( )

x yn

x y

mpI I +

+l

3. m æO;eku dk ,d d.k fu;r osx ls fp=kuqlkj xfrdjrk gSA fcUnq O ds lkis{k d.k ds dks.kh; laosx ds fy,dkSulk dFku lgh ugha gS-

D E

A C

B1m

1m

O

Y

X

(1) tc ;g A ij gS rFkk OA ds vuqfn'k xfr djrk gS];g 'kwU; gS

(2) DE iFk ds vuqfn'k lHkh fcUnqvksa ij leku gS(3) B ,oa D fcUnqvksa ij ifjek.k esa leku gS fdUrq fn'kk

esa foijhr gS(4) tc d.k iFk BC ds vuqfn'k xfr djrk gS rks bldk

eku c<+rk gS


LTS-3/35Kota/00DE314029


J ges'kk eqLdjkrs jgs a A

4. Two resistance are measured in ohm and isgiven as :-

R1 = 3W ± 1%

R2 = 6W ± 2%

When they are connected in parallel, thepercentage error in equivalent resistance is

(1) 3% (2) 4.5%

(3) 0.67% (4) 1.33%

5. A barometer tube contains a mixture of air andsaturated water vapour in the space above themercury column. It reads 70 cm when the acualatmospheric pressure is 76 cm of mercury. Thesaturated vapour pressure at room temperatureis 1.0 cm of mercury. The tube is now loweredin the reservoir till the space above the mercurycolumn is reduced to half its orignal column.Find the reading of the barometer. Assume that

the temperature remains constant.

(1) 65cm (2) 70cm (3) 38cm (4) 30cm

6. A point source of electromagnetic radiation hasan average power output of 800W. Themaximum value of electric field at a distance3.5 m from the source will be :

(1) 63 V/m (2) 36 V/m

(3) 56 V/m (4) 83 V/m

4. fuEu nks izfrjks/k dk ekiu ohm esa fd;k x;k gS %

R1 = 3W ± 1%

R2 = 6W ± 2%

tc bUgsa lekUrj Øe esa tksM+k tkrk gS rks rqY; izfrjks/k esa

izfr'kr =qfV gksxh %&

(1) 3% (2) 4.5%

(3) 0.67% (4) 1.33%

5. ,d cSjksehVj uyh esa ikjs ds LrEHk ds Åij okys LFkku esa

ok;q rFkk lar`Ir ty ok"i dk feJ.k gSA tc okLrfod

ok;qe.Myh; nkc ikjs ds nkc dk 76 cm gS rks bldk

ikB~;kad 70 cm gSA dejs ds rki ij lar`Ir ok"i nkc dk

eku ikjs ds nkc dk 1.0 cm gSA vc ikjs ds tyk'k; esa

uyh dks rc rd uhps foLFkkfir fd;k tkrk gS tc rd

fd ikjs ds LrEHk ds Åijh LFkku dk vk;ru vius izkjfEHkd

vk;ru dk vkèkk u gks tk;sA cSjksehVj dk ikB~;kad Kkr

dhft;sA eku yhft;s fd rki fu;r jgrk gSA

(1) 65cm (2) 70cm (3) 38cm (4) 30cm

6. fo|qr pqEcdh; rjax dss fcUnq lzksr dh vkSlr fuxZe 'kfDr

800W gSA lzksr ls 3.5 ehVj nwjh ij fo|qr {ks= dk

vfèkdre eku gksxk %&

(1) 63 V/m (2) 36 V/m

(3) 56 V/m (4) 83 V/m

Kota/00DE314029LTS-4/35



7. Conducting plates each are placed face to face& equi-spaced at distance d. Area of each plateis half the previous plate. If area of first plateis A. Then the equivalent capacitance of thesystem shown is :-

A A/2 A/4 A/8 A/16

(1) 0Ad

e(2) 0A

10de

(3) 0A20de

(4) 0A30de

8. To calculate the effective length of pendulum,we measure :-

(1) Length of string with metre scale anddiameter of bob with vernier callipers

(2) Directly distance between point ofsuspension and top of bob with meter scale

(3) Distance between point of suspension andbottom of bob with metre scale

(4) Distance between point of suspension andtop of bob by metre scale and add half ofdiameter of bob measured with verniercallipers.

7. fp= esa pkyd IysVksa dks leku nwjh d o ,d nwljs ds

lEeq[k j[kk x;k gSA izR;sd IysV dk {ks=Qy igys okyh

IysV ls vk/kk gSA ;fn igyh IysV dk {ks=Qy A gks rks

iznf'kZr fudk; dh rqY;kadh /kkfjrk gksxh\

A A/2 A/4 A/8 A/16

(1) 0Ad

e(2) 0A

10de

(3) 0A20de

(4) 0A30de

8. yksyd dh izHkkoh yEckbZ Kkr djus ds fy,] ge

ekirs gS :-

(1) Mksjh dh yEckbZ ehVj&iSekus dh lgk;rk ls rFkk

xksyd dk O;kl ouhZ;j dSyhilZ dh lgk;rk ls(2) fuyEcu fcUnq rFkk xksyd ds 'kh"kZ dh nwjh lh/ks

ehVj&iSekus dh lgk;rk ls(3) fuyEcu fcUnq rFkk xksyd ds fuEu fcUnq dh nwjh

lh/ks ehVj&iSekus dh lgk;rk ls(4) fuyEcu fcUnq rFkk xksyd ds 'kh"kZ ds chp dh nwjh

ehVj&iSekus dh lgk;rk ls rFkk blesa ouhZ;j dSyhilZdh lgk;rk ls ekis x;s xksyd ds O;kl dk vk/kktksM+rs gS


LTS-5/35Kota/00DE314029


9. Six stars of equal mass m moving about thecentre of mass of the system such that they arealways on the vertices of a regular hexagon ofside length a. Their common time period ofrevolution around centre will be :-

(1) 3a

4Gm

p (2) ( )34 3a

2Gm 5 3 4

p+

(3) ( )33a

2Gm 5 3 4

p+

(4) ( )33a

2Gm 5 3 4

p-

10. The wattage rating of a light bulb indicates thepower dissipated by the bulb if it is connectedacross 110V DC potential difference. If a 50Wand a 100W bulbs are connected in series to a110V DC source, how much approximatepower will be dissipated in the 50W bulb ?

(1) 50W (2) 100W

(3) 22W (4) 11W

11. A particle excutes simple harmonic motionwhose displacement changes with time t asx = a sinwt. The average speed of that particlebetween the instants when it passes its meanposition and its displacement is a/2 for the firsttime, is :-

(1) awp

(2) 3aw

p(3)

2awp

(4) 0

9. leku æO;eku m ds N% rkjs buls fufeZr fudk; ds æO;eku

dsUæ ds lkis{k bl izdkj xfr'khy gS fd ; s lnk a Hkqtk

yEckbZ okys ,d le"kV~Hkqt ds 'kh"kksZa ij jgrs gSaA dsUæ ds

lkis{k buds pØ.k dk mHk;fu"B vkorZdky gksxk %&

(1) 3a

4Gm

p (2) ( )34 3a

2Gm 5 3 4

p+

(3) ( )33a

2Gm 5 3 4

p+

(4) ( )33a

2Gm 5 3 4

p-

10. ;fn fdlh izdk'k cYc dks 110V DC foHkokUrj ds

fljksa ij tksM+k tkrk gS rks cYc dk vafdr okWVst] cYc

}kjk O;f;r 'kfDr dks bafxr djrk gSA ;fn 50W rFkk

100W ds cYcksa dks 110V DC lzksr ds Js.khØe esa

tksM+k tkrk gS rks 50W cYc esa yxHkx fdruh 'kfDr

O;f;r gksxh\

(1) 50W (2) 100W

(3) 22W (4) 11W

11. ,d d.k ljy vkorZ xfr djrk gS ftldk foLFkkiu

le; t ds lkFk x = a sinwt ds vuqlkj ifjofrZr gksrk

gSA tc d.k igyh ckj bldh ek/; fLFkfr ls xqtjrk gS

rFkk tc d.k dk foLFkkiu a/2 gS rks mu {k.kksa ds e/;

d.k dh vkSlr pky gksxh %&

(1) awp

(2) 3aw

p(3)

2awp

(4) 0

Kota/00DE314029LTS-6/35



12. Relation between permeability m and

magnetising field H for a sample of iron is

m =40.4( 12 10 )

H-+ ´ henery/meter. where unit

of H is A/m. Find value of H for which

magnetic induction of 1.0 Wb/m2 can be

produce.

(1) 100 A/m (2) 500 A/m

(3) 1000 A/m (4) 5000 A/m

13. In given figure, assuming the diodes to be ideal,

–10VA R D1

D2

B

(1) D1 is forward biased and D

2 is reverse biased

and hence current flows from A to B

(2) D2 is forward biased and D

1 is reverse biased

and hence no current flows from B to A

and vice versa.

(3) D1 and D

2 are both forward biased and hence

current flows from A to B.

(4) D1 and D

2 are both reverse biased and hence

no current flows from A to B and vice versa.

12. yksgs ds ,d uewus dh ikjxE;rk m ,oa pqEcdu {ks= H esa

fuEu lEcU/k gS m = 40.4( 12 10 )H

-+ ´ gsujh/ehVj tgk¡

H dk ek=d A/m gSA H dk og eku Kkr djks tks fd

1.0 Wb/m2 dk pqEcdh; izsj.k mRiu djsA

(1) 100 A/m (2) 500 A/m

(3) 1000 A/m (4) 5000 A/m

13. fn, x;s fp= esa Mk;ksMksa dks vkn'kZ ekusa rks &

–10VA R D1

D2

B

(1) D1 vxzck;flr gS vkSj D

2 i'pck;flr gS] vr%

èkkjk A ls B dh vksj izokfgr gksrh gSA

(2) D2 vxzck;flr gS vkSj D

1 i'pck;flr gS] vr% B

ls A dh vksj vFkok A ls B dh vksj dksbZ /kkjk

izokfgr ugha gksrhA

(3) D1 ,oa D

2 nksuksa vxzck;flr gSa] vr% /kkjk A ls B

dh vksj izokfgr gksrh gSA

(4) D1 ,oa D

2 nksuksa i'pck;flr gaS] vr% A ls B dh

vksj vFkok B ls A dh vksj dksbZ /kkjk izokfgr ugha

gksrhA


LTS-7/35Kota/00DE314029


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

14. In an electromagnetic wave the average energydensity is :-

(1) associated with electric field only

(2) associated with magnetic field only

(3) associated equally with electric andmagnetic fields

(4) zero

15. In the circuit shown in figure, A is a slidingcontact which can move over a smooth rod PQ.Resistance per unit length of the rod PQ is1 ohm/m. Initially slider is just left to the pointP and circuit is in the steady state. At t = 0 sliderstarts moving with constant velocity v = 5 m/stowards right. Current in the circuit at t = 2 secis :-

R = 2 0 WL = 2 H

3 0 V

A VP Q

(1) 1 amp

(2) less than 0.5 amp

(3) more than 1 amp

(4) in between 0.5 to 1.0 amp

14. fo?kqr&pqEcdh; rjaxksa es vkSlr ÅtkZ ?kuRo fdlds lkFk

lEcfU/kr gS :–

(1) dsoy fo|qr {ks=

(2) dsoy pqEcdh; {ks=

(3) leku :i ls fo|qr rFkk pqEcdh; {ks= ds lkFk

(4) vkSlr ÅtkZ ?kuRo 'kwU; gS

15. iznf'kZr ifjiFk esa A ,d lihZ laidZ(sliding contact)

gS tks fd fpduh NM + PQ ij xfr dj ldrk gSA NM + PQ

dh izfr ,dkad yEckbZ dk izfrjks/k 1 W/m gSA izkjEHk esa

LykbMj fcUnq P ds Bhd cka;h vksj gS rFkk ifjiFk LFkk;h

voLFkk esa gSA t = 0 ij LykbMj v = 5 m/s ds fu;r osx ls

nka;h vksj xfr djuk izkjEHk dj nsrk gSA t = 2 sec ij

ifjiFk esa /kkjk gksxh %&

R = 2 0 WL = 2 H

3 0 V

A VP Q

(1) 1 ,fEi;j

(2) 0.5 ,fEi;j ls de

(3) 1 ,fEi;j ls vf/kd

(4) 0.5 ls 1.0 ,fEi;j ds chp

Kota/00DE314029LTS-8/35



16. In the given figure we have two solid cylinders.If the linear acceleration of solid cylinder ofmass m

2 is a

2. Then, its angular acceleration a

2

will be (Assuming that there is no slipping) :-\\\\\\\\\\\\\\\\\\\\\\\\\\\

m ,R1

m ,R2

(1) 2aR (2) 2(a g)

R+

(3) 22(g a )R+

(4) 22(g a )R-

17. n moles of a monoatomic gas undergoes acyclic process ABCDA as shown in figure.Process AB is isobaric, BC is adiabatic, CD isisochoric and DA is isothermal. The maximumand minimum temperature in the cycle are 4T0

and T0 respectively. Then :-P

V

A B

CD

(1) TB > TC > TD

(2) heat is absorbed by the gas in the process CD

(3) heat is released by the gas in theprocess AB

(4) total heat supplied to the gas is 2nRT0 ln(2)

16. fp= esa nks Bksl csyu n'kkZ;s x;s gSA m2 nzO;eku ds ,d

Bksl csyu dk js[kh; Roj.k a2 gS] rc bldk dks.kh; Roj.k

a2gksxk (;gk¡ dksbZ fQlyu ugha gSA) %&

\\\\\\\\\\\\\\\\\\\\\\\\\\\

m ,R1

m ,R2

(1) 2aR (2) 2(a g)

R+

(3) 22(g a )R+

(4) 22(g a )R-

17. ,d ijekf.od xSl ds n eksy fp=kuqlkj pØh; izØe

ABCDA ls gksdj xqtjrs gSaA izØe AB lenkch;, BC

:¼ks"e] CD levk;rfud rFkk DA lerkih; gSA pØ

esa vf/kdre rFkk U;wure rki Øe'k% 4T0 rFkk T0 gSa] rksP

V

A B

CD

(1) TB > TC > TD

(2) izØe CD esa xSl }kjk Å"ek yh tkrh gSA

(3) izØe AB esa xSl }kjk Å"ek eqDr dh tkrh gSA

(4) xSl dks nh xbZ dqy Å"ek 2nRT0 ln(2) gSA


LTS-9/35Kota/00DE314029


18. An aeroplane is flying at a constant speed ofu = 300 m/s, along a straight line at an altitudeof h = 5000 m. At the instant when, it is overan anti-aircraft gun, a shot is fired (as shown).The initial speed of the projectile isv0 = 500 m/s. (Neglect the air resistance.) Theflight time t that must be set for the projectileto explode at the point of meeting with the target(aeroplane) is : (take : g = 10 m/s2)

u

h

a

v0

(1) 10 s (2) 11.6 s (3) 50 s (4) 15.5 s

19. Given set-up which is shown in figure,converges parallel beam of light at point P

1. If

the surrounding medium of below set-up isreplaced by transparent fluid of refractive indexm

m = 2, then same parallel beam converge at

point P2 then distance P

1P

2 is :-

f=10cm

m=3/2

air air

(1) 70 cm (2) 20 cm (3) 30 cm (4) 10 cm

18. ,d ok;q;ku u = 300 m/s dh fu;r pky lsh = 5000m dh Å¡pkbZ ij ,d ljy js[kk ds vuqfn'kmM+ jgk gSA fdlh {k.k tc ;g fdlh foekujks/kh xu dsÅij gksrk gS rks ,d xksyh fp=kuqlkj pyk;h tkrh gSAiz{ksI; dh izkjfEHkd pky v0 = 500 m/s gSA ok;q izfrjksèkdks ux.; ekusA iz{ksI; ds fy;s mMM~;u dky t D;k gksukpkfg;s rkfd ;g y{; (ok;q;ku) ls feyus ij foLQksVgks ldsA (g = 10 m/s2)

u

h

a

v0

(1) 10 s (2) 11.6 s (3) 50 s (4) 15.5 s

19. fp= esa iznf'kZr O;oLFkk ls lekUrj izdk'k iqat fcUnq P1

ij vfHklkfjr gksrk gSA ;fn bl O;oLFkk ds pkjksa vksj ds

ek/;e dks mm = 2 viorZukad okys ikjn'khZ nzO; ls

izfrLFkkfir dj fn;k tk;s rks ;g lekUrj iqat fcUnq P2 ij

vfHklkfjr gksrk gS rc nwjh P1P

2 gksxh%&

f=10cm

m=3/2

air air

(1) 70 cm (2) 20 cm (3) 30 cm (4) 10 cm

Kota/00DE314029LTS-10/35



20. Three rods AB, BC and BD of same length land cross-sectional area A are arranged asshown. The end D is immersed in ice whosemass is 440 gm. Heat is being supplied atconstant rate of 200 cal/sec from the end A.Time in which whole ice will melt. Given :k (thermal conductivity) = 100 cal/m/sec/°C,A = 10 cm2, l = 1m, Latent heat of fusion of iceis 80 cal/gm.

D

B,lk–2

A C

k,l 2k,l200 cal/sec100°C

ice

(1) 40/3 min(2) 700 sec(3) 20/3 min(4) indefinitely long time

21. Two tuning forks A and B produce 8 beats/swhen sounded together. A gas column 37.5 cmlong in a pipe closed at one end resonate to itsfundamental mode with fork A whereas acolumn of length 38.5 cm of the same gas in asimilar pipe is required for a similar resonancewith fork B. (Neglect end correction) Thefrequencies of these two tuning forks, are :-

(1) 308 Hz, 300 Hz (2) 208 Hz, 200 Hz(3) 300 Hz, 400 Hz (4) 350 Hz, 500 Hz

20. leku yEckbZ l rFkk leku vuqizLFk dkV {ks=Qy A okyh

rhu NM+ksa AB, BC rFkk BD dks fp=kuqlkj O;ofLFkr

fd;k x;k gSA fljs D dks cQZ esa Mqcks;k x;k gS ftldk

nzO;eku 440 gm gSA fljs A ls 200 cal/sec dh fu;r

nj ls Å"ek iznku dh tkrh gSA fdrus le; esa lkjh cQZ

fi?ky tk;sxh\ fn;k g S : k (Å"eh; pkydrk)

= 100 cal/m/sec/°C, A = 10 cm2, l = 1m, cQZ

ds laxyu dh xqIr Å"ek 80 cal/gm gS)

D

B,lk–2

A C

k,l 2k,l200 cal/sec100°C

ice

(1) 40/3 min(2) 700 sec(3) 20/3 min(4) vfuf'pr yEcs le; rd

21. A vkSj B nks Lofj= f}Hkqt lkFk&lkFk Lofjr gksus ij çfrlsd.M 8 foLian mRiUu djrs gSaA ,d fljs ls cUn ikbZiesa ,d xSl dk 37.5 lseh yEck LrEHk viuh ewy fo/kk esaLofj= A ls vuqukfnr gksrk gS tcfd Lofj= B blh xSlds ,sls gh ikbZi esa 38.5 lseh LrEHk ls vuqukfnr gksrk gSAnksuksa Lofj= f}Hkqtksa dh vko`fÙk;k¡ gksxh (vUR; la'kks/kudks ux.; ekusa) %&(1) 308 Hz, 300 Hz (2) 208 Hz, 200 Hz

(3) 300 Hz, 400 Hz (4) 350 Hz, 500 Hz


LTS-11/35Kota/00DE314029


22. The horizontal part of the U-shaped tube shownin the figure is 10 cm, and its vertical branchesare long enough. The cross section of the tubeis 1 cm2. First, 20 cm3 water is poured into thetube, then20 cm3 oil is poured into the leftbranch. (density of water = 1000 kg/m3, densityof oil = 800 kg/m3). The whole system isaccelerated towards the left, at an accelerationof a = 10 m/sec2. Determine the differencebetween the level of water and the level ofoil (in cm).

10cm

a=10m/sec2

(1) 6 (2) 10 (3) 12 (4) 18

23. An unpolarized light beam is incident on asurface at an angle of incidence equal toBrewster's angle. Then(1) the reflected and the refracted beams are

both partially polarized.(2) the reflected beam is partially polarized and

the refracted beam is completely polarizedand are at right angled to each other.

(3) the reflected beam is completely polarizedand the refracted beam is partially polarizedand are at right angled to each other.

(4) both the reflected and the refracted beamsare completely polarized and are at rightangled to each other.

22. fp= esa iznf'kZr U-vkdkj dh uyh dk {kSfrt Hkkx 10cm

yEck gS rFkk bldh Å/okZ/kj 'kk[kk,sa i;kZIr :i ls yEchgSA uyh dk vuqizLFk dkV {ks=Qy 1 cm2 gSA igys bluyh esa 20 cm3 ty mM+syk tkrk gS rFkk fQj bldh ck¡;hHkqtk esa 20 cm3 rsy Mkyk tkrk gSA (ty dk ?kuRo =1000 kg/m3, rsy dk ?kuRo = 800 kg/m3) bl iwjsfudk; dks ck¡;h vksj a = 10 m/sec2 Roj.k ls Rofjrfd;k tkrk gSA ty rFkk rsy ds Lrjksa ds e/; vUrj(cm esa) Kkr dhft;sA

10cm

a=10m/sec2

(1) 6 (2) 10 (3) 12 (4) 18

23. ,d v/kqzfor izdk'k iqat] czsosLVj dks.k ds cjkcj vkiru

dks.k ij ,d lrg ij vkifrr gS rks(1) ijkofrZr rFkk viofrZr iqat nksuksa vkaf'kd :i ls

èkqzfor gSaA(2) ijkofrZr iqat vkaf'kd :i ls /kqzfor gSa tcfd viofrZr

iqat iw.kZ :i ls /kqzfor gSa rFkk nksuksa ds e/; dks.k]ledks.k gSA

(3) ijkofrZr iqat iw.kZ :i ls /kqzfor gS tcfd viofrZriqat vkaf'kd :i ls /kqzfor gS rFkk nksuksa ds e/; dks.k]ledks.k gSA

(4) ijkofrZr rFkk viofrZr iqat nksuksa iw.kZ :i ls /kqzforgSa rFkk nksuksa ds e/; dks.k] ledks.k gSA

Kota/00DE314029LTS-12/35



24. Lyman alpha, the n = 1 to n = 2 transition inatomic hydrogen occurs at 1215 Å.(1) Radiation of wavelength shorter than 911 Å

can photo–ionize hydrogen atom in groundstate.

(2) Radiation of wavelength longer than 3645Åcan photo–ionize hydrogen atom in firstexcited state.

(3) Radiation of wavelength longer than 228 Åcan photo–ionize He+ ion in ground state.

(4) Radiation of wavelength longer than1215 Å can photo–ionize He+ ion atom infirst excited state.

25. In an experiment on photoelectric effect, theemitter and the collector plates are placed at aseparation of 10 cm and are connected throughan ammeter without any cell (figure). Amagnetic field B exists parallel to the plates.The work function of the emitter is 2.9 eV andthe light incident on it has wavelength between400 nm and 600 nm. Find the minimum valueof B for which the current registered by theammeter is zero. Neglect any effect of spacecharge.(Assume mass of electron = 9 × 10

–27 kg and

hC = 1240 eV nm)

x x xx x x

B A 10cm

(1) 1.0 × 10–3

T (2) 1.5 × 10–3

T

(3) 2.0 × 10–3

T (4) 2.85 × 10–5

T

24. ijekf.od gkbMªkstu esa n = 1 ls n = 2 laØe.k

(ykbeu ,YQk)1215 Å ij gksrk gSA(1) 911 Å ls de rjaxnS/; Z okyh fofdj.k gkbMªkstu

ijek.kq dks ewy voLFkk esa QksVks vk;fur dj ldrh gSA(2) 3645 Å ls vf/kd rjaxnS/;Z okyh fofdj.k gkbM ªkstu

ijek.kq dks izFke mÙksftr voLFkk esa QksVks vk;furdj ldrh gSA

(3) 228 Å ls vf/kd rjaxnS/;Z okyh fofdj.k] He+

vk;u dks ewy voLFkk esa QksVks vk;fur dj ldrh gSA(4) 1215 Å ls vf/kd rjaxnS/; Z okyh fofdj.k]

He+ vk;u ijek.kq dks izFke mÙksftr voLFkk esaQksVks vk;fur dj ldrh gSA

25. izdk'k fo|qr izHkko ds iz;ksx esa mRltZd rFkk lxzkagd

IysVs ,d&nwljs ls 10 cm dh nwjh ij gS rFkk bUgs fcuk

lsy ,d vehVj ds lkFk fp=kuqlkj tksM+ fn;k tkrk gSA

IysVks a ds lekUrj ,d pqEcdh; {ks= B fo|eku gSA

mRltZd dk dk;ZQyu 2.9 eV gS rFkk bl ij vkifrr

izdk'k dh rjaxnS/; Z 400 nm ls 600 nm ds e/; gSA B

dk U;wure eku D;k gksuk pkfg;s rkfd vehVj 'kwU; /kkjk

ntZ djsA vU; fdlh vkos'k ds izHkko dks ux.; ekusA

(ekuk bysDVªkWu dk æO;eku = 9 × 10–27

kg rFkkhC = 1240 eV nm)

x x xx x x

B A 10cm

(1) 1.0 × 10–3

T (2) 1.5 × 10–3

T

(3) 2.0 × 10–3

T (4) 2.85 × 10–5

T


LTS-13/35Kota/00DE314029


26. A uniform chain is placed at rest on a roughsurface of base length l and height h on anirregular surface as shown in figure. Theminimum coefficient of friction between thechain and the surface so that the chain does notslip is :

(1) h2

m =l

(2) h

m =l

(3) 3h2

m =l

(4) 2h3

m =l

27. Two balls A and B with the same mass m aresuspended. A on an inextensible thread and Bon an elastic string. Both balls are held at 90°with vertical line and released as shown, oneafter other. If both balls pass through same pointon vertical line, they have speeds va & vb

respectively.

90°

Bm A

l

(1) va > vb (2) va < vb

(3) va = vb (4) Can't be determined.

26. ,d le:i tathj fp=kuqlkj ,d ÅcM+&[kkcM+ [kqjnjhlrg ij fojkekoLFkk esa j[kh gqbZ gSA lrg ds vk/kkj dhyEckbZ l rFkk Å¡pkbZ h gSA ;g tathj fQlys ugha bldsfy, tathj rFkk lrg ds e/; U;wure ?k"kZ.k xq.kkad dkeku gksxk%&

(1) h2

m =l

(2) h

m =l

(3) 3h2

m =l

(4) 2h3

m =l

27. leku æO;eku m okyh nks xsanksa A o B dks Øe'k% ,dvforkU; /kkxs rFkk ,d izR;kLFk jLlh }kjk yVdk;k x;kgSA nksuksa xsanksa dks fp=kuqlkj Å/okZ/kj js[kk ls 90° ijjksddj j[kk x;k gSA vc bUgs Øekxr :i ls NksM+k tkrkgSA ;fn nksuksa xsansa Å/okZ/kj js[kk ij fdlh fcUnq ls Øe'k%va o vb pky ls xqtjs rks %&

90°

Bm A

l

(1) va > vb (2) va < vb

(3) va = vb (4) Can't be determined.

Kota/00DE314029LTS-14/35



28. Consider a small water drop in air. If T is the

surface tension, then what is the force due to

surface tension acting on the smaller section

ABC ?

O C

AB

q

(1) 2TRp (2) 2pTRsinq

(3) 2pTR sin2q (4) 2pTR sin3q

29. Consider the four different cases of dispersion

of light ray which has all the wave lengths from

l1 to l2 (l1 > l2). The dotted represents the

light ray of wave length lavg. Which ray

diagram is showing maximum dispersive

power?

(1) (2)

(3) (4)

28. ok;q esa ,d NksVh ty dh cwan ij fopkj djsaA ;fn T i`"B

ruko gks rks NksVs [k.M ABC ij bl i`"B ruko ds dkj.k

yxus okys cy dk eku gksxk ?

O C

AB

q

(1) 2TRp (2) 2pTRsinq

(3) 2pTR sin2q (4) 2pTR sin3q

29. ,d izdk'k fdj.k ds fo{ksi.k ds pkj fofHkUu izdj.kksa ij

fopkj dhft;s ftlesa l1 ls l2 (l1 > l2) rd dh lkjh

rjaxnS/;Z fo|eku gSA fcUnq rjaxnS/; Z lavg okyh izdk'k

fdj.k dks n'kkZrs gSA dkSulk fdj.k vkjs[k vf/kdre

fo{ksi.k {kerk dks n'kkZrk gS\

(1) (2)

(3) (4)


LTS-15/35Kota/00DE314029


30. A circular ring of radius R carries a chargedensity l = l

0(x2 + y2) and placed in the

x–y plane with centre at origin. Which of thefollowing graphs represents the variation ofelectric potential (V) along z-axis?

(1)

V

z

(2)

V

z=z=R2

-R2

z

(3)

V

z

(4)

V

zz= R- z=R

30. R f=T;k dh ,d oÙ̀kkdkj oy; ij l = l0(x2 + y2)

vkos'k ?kuRo gS rFkk bls x–y ry esa j[kk x;k gSA bldk

dsUnz ewy fcUnq ij gSA fuEu esa ls dkSulk xzkQ z-v{k ds

vuqfn'k fo|qr foHko (V) ds ifjorZu dks iznf'kZr djrk

gS\

(1)

V

z

(2)

V

z=z=R2

-R2

z

(3)

V

z

(4)

V

zz= R- z=R

Kota/00DE314029LTS-16/35



31. If a solution of stearic acid (C18H36O2;M = 284) in benzene contains 1.42 gm acid perL. When this solution (100 L) is dropped onsurface, C6H6 gets evaporated and acid formsa unimolecular layer on the surface. If it coversan area 6020 cm2 with unimolecular film. Findthe area covered by one molecule of acid.(1) 2 × 10–20 cm2 (2) 4 × 10–20 cm2

(3) 2 × 1020 cm2 (4) 4 × 1020 cm2

32. A container 'X' containing some liquid wateris connected to another container 'Y' havingvolume double of container 'X' through a valveand containing Br2(liquid) as shown :Calculate mole fraction of H2O vapours incontainer Y after valve is opened for sufficienttimeVapour pressure of H2O(l) = 80 mm HgVapour pressure of Br2(l) = 40 mm HgAssume :(A)Volume occupied by liquid H2O and Br2 is

negligible(B) Temperature to remain constant throughout

Container-Y

H O( )2 l Br ( )2 l

Container-X

(1) 2

3(2)

1

3(3) 1 (4) 0

31. ;fn csUthu esa LVsfjd vEy (C18H36O2 ; M = 284)ds ,d foy;u esa 1.42 gm vEy izfr yhVj mifLFkrgS] rks tc bl foy;u (100 L) dks lrg ij fxjk;k tkrkgSa] rc C6H6 ok"ihd`r gks tkrk gS rFkk vEy lrg ij,dy vkf.od ijr dk fuekZ.k djrk gS ;fn ;g6020 cm2 dk {ks=Qy ,dy v.kqd ijr ds lkFk ?ksjrkgS rks vEy ds ,d v.kq }kjk ?ksjk x;k {ks=Qy Kkr dhft;s(1) 2 × 10–20 cm2 (2) 4 × 10–20 cm2

(3) 2 × 1020 cm2 (4) 4 × 1020 cm2

32. ,d ik= 'X' ] ftlesa dqN ek=k nzo ty dh gS] dks ,d

vU; ik= 'Y' ] ftldk vk;ru X ds vk;ru ls nqxquk

gS] ds lkFk ,d okYo }kjk tksMk x;k gSA ik= 'Y' esa

Br2(nzo) mifLFkr gSA

i;kZIr le; ds fy, okYo dks [kksyus ds i'pkr~ ik=

Y esa H2O ok"i ds eksy izHkkt dh x.kuk dhft;s&

H2O(l) dk ok"i nkc = 80 mm Hg

Br2(l) dk ok"i nkc = 40 mm Hg

ekuk yhft, %

(A)H2O rFkk Br2 nzo }kjk ?ksjk x;k vk;ru ux.; gSaA

(B) lEiw.kZ izØe ds nkSjku rki fu;r jgrk gSA

H O( )2 l Br ( )2 l

ik=-X ik=-Y

(1) 2

3(2)

1

3(3) 1 (4) 0

PART B - CHEMISTRY


LTS-17/35Kota/00DE314029


33. A proton accelerated from rest through apotential difference of 'V' volts has a wavelengthl associated with it. An alpha particle in orderto have the same wavelength must be accelerated

from rest through a potential difference of &(1) V volt (2) 4V volt

(3) 2V volt (4) V8

volt

34. Two radioactive material A and B havedisinterigation constants 10l and 2 lrespectively. If initially they have the samenumber of nuclie then the ratio of number of

nuclie of A and B will be 1

e after a time of –

(1) 1

10l(2)

1

11l(3)

11

10l(4)

1

8l35. Phenolphthalein does not act as an indicator for

the titrations between -(1) HCl and NH4OH(2) Ba(OH)2 and HCl(3) NaOH and H2SO4

(4) KOH and CH3COOH36. E0 of Mg2+ | Mg, Zn+2 | Zn, and Fe2+ | Fe are

–2.37 V, –0.76 V, and –0.44 V, respectively.Which of the following is correct ?(1) Mg oxidizes Fe(2) Zn oxidizes Fe(3) Zn reduces Mg2+

(4) Zn reduces Fe2+

33. ,d izksVkWu fLFkj voLFkk ls 'V' oksYV ds foHkokUrj }kjkRofjr fd;k x;k gS] bl izksVkWu dh rjaxnS/; Z l gSAblh izdkj leku rjaxnS/; Z ds ,d vYQk d.k dks fLFkjvoLFkk ls fdl foHkokUrj }kjk Rofjr fd;k tk ldrkgSA(1) V oksYV (2) 4V oksYV

(3) 2V oksYV (4) V8

oksYV

34. nks jsfM;kslfØ; inkFkZ A rFkk B ds fo;kstu fLFkjkadØe'k% 10l rFkk 2 l gSa ;fn izkjEHk esa buds ukfHkdksadh la[;k leku gS rks fdl le; i'pkRk~ A rFkk B ds

ukfHkdksa dh la[;k dk vuqikr 1

e gksxk &

(1) 1

10l(2)

1

11l(3)

11

10l(4)

1

8l35. fdlds e/; vuqekiu ds fy, ,d lwpd ds :i eas

fQukW¶Fksfyu dk;Z ugha djrk gS-(1) HCl rFkk NH4OH(2) Ba(OH)2 rFkk HCl(3) NaOH rFkk H2SO4

(4) KOH rFkk CH3COOH36. Mg2+ | Mg, Zn+2 | Zn, rFkk Fe2+ | Fe ds E0 eku

Øe'k% –2.37 V, –0.76 V rFkk –0.44 V gS , fuEu esals dkSulk dFku lgh gS ?(1) Mg]Fe dks vkWDlhd`r djrk gS(2) Zn , Fe dks vkWDlhd̀r djrk gS(3) Zn , Mg2+ dks vipf;r djrk gS(4) Zn , Fe2+ dks vipf;r djrk gS

Kota/00DE314029LTS-18/35




37. The pressure of an equilibrium mixture of thethree gases NO, Cl2 and NOCl,

2NO(g) + Cl2(g) 2NOCl(g)

is suddenly decreased by doubling the volumeof the container at constant temperature. Whenthe system returns to equilibrium :

(1) The concentration of NOCl will haveincreased

(2) The value of the equilibrium constant Kc

will have increased

(3) The number of moles Cl2 will haveincreased

(4) The number of moles of NOCl will haveincreased

38. How many electrons should X2H4 liberate sothat in the new compound X shows oxidationnumber of –1/2 [E.N. of X > H]

(1) 3.5 (2) 4 (3) 3 (4) 1.5

39. For the reaction : 2SO2(g) + O2(g) � 2SO3(g);DH = –ve. An increase in temperature shows.

(1) More dissociation of SO3 and a decrease in Kc

(2) Less dissociation of SO3 and an increase in Kc

(3) More dissociation of SO3 and an increase in Kc

(4) Less dissociation of SO3 and a decrease in Kc

37. rhu xSlksa NO, Cl2 rFkk NOCl, ds ,d lkE; feJ.k

dk nkc]

2NO(g) + Cl2(g) 2NOCl(g)

fu;r rki ij ik= dk vk;ru nqxuk djus l s

vpkud de gks tkrk gS] tc iqu% lkE; izkIr djrk gS

rc :

(1) NOCl dh lkUærk c<+ tk;sxh

(2) lkE; fu;rkad Kc dk eku c<+ tk;sxk

(3) Cl2 ds eksyksa dh la[;k c<+ tk;sxh

(4) NOCl ds eksyksa dh la[;k c<+ tk;sxh

38. X2H4 ds fdrus bysDVªkWu mRlftZr gksus pkfg, fd u;k

;kSfxd X esa vkWDlhdj.k la[;k –1/2 iznf'kZr gks :

[X dk E.N. > H](1) 3.5 (2) 4 (3) 3 (4) 1.5

39. vfHkfØ;k ds fy, % 2SO2(g) + O2(g) � 2SO3(g);DH = –ve. rki esa o`f¼ iznf'kZr djrh gS&

(1) SO3 dk vf/kd fo;kstu rFkk Kc esa deh(2) SO3 dk U;wu fo;kstu rFkk Kc esa o`f¼

(3) SO3 dk vf/kd fo;kstu rFkk Kc esa o`f¼(4) SO3 dk U;wu fo;kstu rFkk Kc esa deh


LTS-19/35Kota/00DE314029


40. For a first order reaction choose theCORRECT statement :–

(1) The degree of dissociation is equal to(1 – e–kt)

(2) The pre-exponential factor in the arrheniusequation has the dimension of time–1

(3) A plot of reciprocal concentration of thereactant v/s time gives a straight line

(4) (1) & (2) both

41. Which of the following has higher number ofP–O–P linkages

(1) P4O6

(2) P4O10

(3) both have equal number of P–O–P linkage

(4) None of these

42. Identify the CORRECT order in the following

(1) NH3 < PH3 (bond angle)

(2) Cl > F (Electron affinity)

(3) NO2– > NO3

– (·O – N – O angle)

(4) H2O > O(CH3)2 (bond angle)

43. The general formula for amphibole chainsilicate is-

(1)2n

2 5 n(Si O ) -(2) 2n

2 7 n(Si O ) -

(3) 3x

2 5.5 x(Si O ) -(4) None of these

40. izFke dksfV vfHkfØ;k ds fy, lgh dFku dk p;udhft, :–

(1) fo;kstu dh ek=k (1 – e–kt) ds cjkcj gksrhgS

(2) vkjfguh;l lehdj.k esa iwoZ ?kkrkadh xq.kkd dh foekle;–1 gksrh gS

(3) fØ;kdkjdks dh lkUnzrk dk O;qRØe v/s le; dkoØ ,d lh/kh js[kk gksrh gS

(4) (1) rFkk (2) nksuksa

41. fuEu esa ls fdlesa P–O–P ca/kuksa dh la[;k vf/kdgS

(1) P4O6

(2) P4O10

(3) nksuksa esa P–O–P ca/kuksa dh la[;k leku gS(4) buesa ls dksbZ ugha

42. fuEu esa lgh Øe igpkfu, :

(1) NH3 < PH3 (ca/k dks.k)

(2) Cl > F (bysDVªkWu cU/kqrk)

(3) NO2– > NO3

– (·O – N – O dks.k

(4) H2O > O(CH3)2 (ca/k dks.k)43. ,EQhcksys J`a[kyk flfydsV dk lkekU; lw= gS -

(1)2n

2 5 n(Si O ) -(2) 2n

2 7 n(Si O ) -

(3) 3x

2 5.5 x(Si O ) -(4) buesa ls dksbZ ugha

Kota/00DE314029LTS-20/35



44. Choose the INCORRECT statement :-(1) In alkaline medium, the hydrolysis of

BeCl2 produces clear solution consisting of[Be(OH)4]

2– and HCl.(2) In the clear solution of BiCl3, when large

quantity of water is added, the whiteturbidity of BiOCl is obtained.

(3) SiF4 undergoes partial hydrolysis.(4) The final products on hydrolysis of PCl3 and

POCl3 are not identical.

45. Identify the CORRECT order.(1) CsCl < RbCl < KCl < NaCl < LiCl

(Solubility in water)(2) CsCl < RbCl < KCl < NaCl < LiCl (m.p)(3) CsCl > RbCl > KCl > NaCl > LiCl

(% ionic character)(4) CsCl > RbCl > KCl > NaCl > LiCl

(Lattice Energy)

46. Select the CORRECT IUPAC name of[Pt (NH3)3 Cl (C2H4) (NO2)]3 (PO4)2

(1) Triamminechloro(h2- ethene)nitratoplatinum(IV)phosphate

(2) Triamminechloro(h2- ethene)nitroplatinate(IV)phosphate

(3) Triaminechloro(h2- ethene)nitroplatinum(IV)phosphate

(4) Triamminechloro(h2- ethene)nitroplatinum(IV)phosphate

44. xyr dFku pqfu;sa :-(1) {kkjh; foy;u esa] BeCl2 ds tyvi?kVu ls LoPN

foy;u curk gS ftlesa [Be(OH)4]2– rFkk HCl

mifLFkr gksrs gS(2) BiCl3 ds LoPN foy;u esa tc ty dh vR;f/kd ek=k

feykbZ tkrh gS rks BiOCl dk 'osr /kqa/kykiu izkIr gksrkgS

(3) SiF4 dk vkaf'kd tyvi?kVu gksrk gS(4) PCl3 rFkk POCl3 ds tyvi?kVu ds vafre mRikn

leku ugha gS45. lgh Øe pqfu;sa

(1) CsCl < RbCl < KCl < NaCl < LiCl(ty esa foys;rk)

(2) CsCl < RbCl < KCl < NaCl < LiCl (xyukad)(3) CsCl > RbCl > KCl > NaCl > LiCl

(% vk;fud y{k.k)(4) CsCl > RbCl > KCl > NaCl > LiCl

(tkyd ÅtkZ)

46. fuEu ;kSfxd dk lgh IUPAC uke pqfu,[Pt (NH3)3 Cl (C2H4) (NO2)]3 (PO4)2

(1) Vªkb, sEehuDyk sjk s(h2- ,sFk hu)ukbVª sVk sIysfVue(IV)QkWLQsV

(2) Vª kb, sEehuDyk sjk s(h2- ,sFk hu)ukbVª k s Iy sfVu sV(IV)QkWLQsV

(3) Vª kb,sehuDyk sjk s(h2- ,sFk hu)ukbV ª k sIys fVue(IV)QkWLQsV

(4) Vª kb, sEehuDyk sjk s(h2- ,sFk hu)ukbV ª k sIys fVue(IV)QkWLQsV


LTS-21/35Kota/00DE314029


47. Find out the optically inactive complex fromthe following options -(1) [Be(gly)(acac)]0 (2) [Zn(gly)2]

0

(3) [B(OH)4]–

(4) (1) and (3) both48. In the cynide process for the extraction of silver

from argentite ore, sodium cynaide solutionact as(1) leaching agent(2) Reducing agent(3) Oxidizing agent(4) Activator

49. Which of the following statement is CORRECT(1) Roasting is unnecessarily done for

Fe-extraction because there is no sulphide ore(2) In the smelting step of Cu-extraction,

reduction of the ore takes place.(3) Ores may not be mineral(4) Sphalerite is the ore of the zinc

50. The solid laboratory reagent 'A' gives thefollowing reactions -(i) it imparts green colour to the flame(ii) its solution does not give a precipitate on

passing H2S(iii) when it is heated with solid K2Cr2O7 and

concentration H2SO4, a red gas is evolved.When this gas passed into aqueous solutionof NaOH, turns it yellow -

Identify 'A'(1) PbCl2 (2) BaCl2

(3) NaCl (4) None of these

47. fuEu esa ls izdkf'kd vfØ; ladqy pqfu, -

(1) [Be(gly)(acac)]0 (2) [Zn(gly)2]0

(3) [B(OH)4]–

(4) (1) rFkk (3) nksuksa

48. vtsZUVkbV v;Ld ls flYoj fu"d"kZ.k ds lk;ukbM izØeesa lksfM;e lk;ukbM foy;u dk;Z djrk gS -(1) fu{kkyu vfHkdeZd dk(2) vipk;d dk(3) vkWDlhdkjd dk(4) lfØ;d dk

49. fuEu esa ls dkSulk dFku lgh gS -(1) Fe-fu"d"kZ.k ds fy, HktZu vuko';d :i ls fd;k

tkrk gS D;ksafd ogk¡ dksbZ lYQkbM v;Ld ugha gS(2) Cu-fu"d"kZ.k ds izxyu in esa, v;Ld dk vip;u

gksrk gS(3) v;Ld] [kfut ugha gks ldrk(4) LQsysjkbV] ftad dk v;Ld gS

50. Bksl iz;ksx'kkyk vfHkdeZd 'A' fuEu vfHkfØ;k,sa nsrkgS -(i) ;g Tokyk dks gjk jax iznku djrk gS(ii) bldk foy;u] H2S izokfgr fd;s tkus ij vo{ksi

ugha nsrk gS(iii) tc bls Bksl K2Cr2O7 rFkk lkUæ H2SO4 ds lkFk

xeZ fd;k tkrk gS rks yky xSl mRlftZr gksrh gSA tc;g xSl NaOH ds tyh; foy;u esa izokfgr dhtkrh gS rks foy;u dks ihyk dj nsrh gS -

'A' igpkfu,sa(1) PbCl2 (2) BaCl2

(3) NaCl (4) buesa ls dksbZ ugha

Kota/00DE314029LTS-22/35




51. To prepare diphenyl ether the correct methodis

(1) Cl + O–

(2) Cl + O–

(3) Cl + O– ; Se / D

(4) Cl + OH + NaOH

52.OH

H SO2 4¾ ®¾¾

Which carbocation is involved in the abovereaction ?

(1)

+

(2) +

(3) Both (4) None of these

51. MkbZQsfuy bZFkj dks cukus dh lgh fof/k gS&

(1) Cl + O–

(2) Cl + O–

(3) Cl + O– ; Se / D

(4) Cl + OH + NaOH

52.OH

H SO2 4¾ ®¾¾

mijksDr vfHkfØ;k esa dkSulk dkcZ/kuk;u lfEefyrgS&

(1)

+

(2) +

(3) nksuksa (4) buesa ls dksbZ ugha


LTS-23/35Kota/00DE314029


53.

Br

HBr P1 + P2

P1 & P2 are major dibromo products. What is

the relationship between P1 & P2

(1) Diastereomers (2) Enantiomers

(3) Chain isomer (4) Position isomers

54. Consider the reaction

OH

NH2

+

N2+ Cl–

–OH¾¾¾® X ; 'X' is-

(1)

N2+ Cl–

OH

OH

NN (2)

OH

NH2

N

N

(3)

OH

(4) None of these

53.

Br

HBr P1 + P2

P1 rFkk P2 eq[; MkbZczkseks mRikn gS P1 rFkk P2 ds

e/; D;k lEcU/k gS&

(1) foofje leko;oh (2) izfrfcEc:i leko;oh

(3) J`a[kyk leko;oh (4) fLFkfr leko;oh

54. fuEu vfHkfØ;k ij fopkj dhft;s&

OH

NH2

+

N2+ Cl–

–OH¾¾¾® X ; 'X' gS&

(1)

N2+ Cl–

OH

OH

NN (2)

OH

NH2

N

N

(3)

OH

(4) buesa ls dksbZ ugha

Kota/00DE314029LTS-24/35



55. In the given reaction:

CH3

NHAc

HNO3 (A) H O/H2 (B)+

(C)

NaN

O/conc.H

Cl

2

(D) H PO3 2

(4) is :

(1)

CH3

NO2

(2)

CH3

NO2

(3)

CH3

NO2NO2

(4)

CH3

NO2NO2

OH

55. nh x;h vfHkfØ;k esa

CH3

NHAc

HNO3 (A) H O/H2 (B)+

(C)

NaN

O/

.HC

l2 lkUnz

(D) H PO3 2

(4) gS&

(1)

CH3

NO2

(2)

CH3

NO2

(3)

CH3

NO2NO2

(4)

CH3

NO2NO2

OH


LTS-25/35Kota/00DE314029


56. Which show fastest reaction with NaOH

(1)

C l

N O 2

(2)

C l

N O 2

N O 2

(3)

C l

N O 2

(4)

F

N O 2

N O 2

57. Which is not a pair of reducing sugars-

(1) Glucose, sucrose (2) Fructose, lactose

(3) Glucose, maltose (4) Glucose, fructose

58. A non reducing disacahride is obtained by

condensation of X & Y condensation

takes place between X & Y respectively

at -

OH

OH

CH OH2

HO

HO

OH

HOH C2

OH

H

H

OHOH

CH OH2O

12

3

45

6

(X) (Y)

1

2

34

5

6

(1) C-1 & C-1 (2) C-1 & C-5

(3) C-5 & C-1 (4) C-1 & C-2

56. fuEu esa ls dkSu NaOH ds lkFk rhozre vfHkfØ;k iznf'kZrdjrk gS %

(1)

C l

N O 2

(2)

C l

N O 2

N O 2

(3)

C l

N O 2

(4)

F

N O 2

N O 2

57. fuEu esa ls dkSulk ; qXe vipk;d 'kdZjk dk ugha gS -

(1) Xyqdkst] lwØksl (2) ÝDVksl, ysDVksl

(3) Xyqdkst, ekYVksl (4) Xyqdkst, ÝDVksl

58. X rFkk Y ds la?kuu }kjk ,d vu&vipk;d MkblsdsjkbM

izkIr gksrk gSA X rFkk Y ds ftu dkcZu ijek.kqvksa ds eè;

la?kuu gksrk gS os Øe'k% gSa -

OH

OH

CH OH2

HO

HO

OH

HOH C2

OH

H

H

OHOH

CH OH2O

12

3

45

6

(X) (Y)

1

2

34

5

6

(1) C-1 rFkk C-1 (2) C-1 rFkk C-5

(3) C-5 rFkk C-1 (4) C-1 rFkk C-2

Kota/00DE314029LTS-26/35



59. MgBrH C3 +CO2 HÅ P1NaOHCaO/D

P2

Identify P2

(1)

CH3

COOH

(2)

(3)

CH3

CH3

(4)

CH3

60. Major organic product of reaction

C H – M g C l + C H – C H3 2 2

C l O C H 3

(1) C H – — C H2 2

C H 3 O C H 3

(2) C H – — C H2 2

C l C H 3

(3) C H – — C H2 2

C H O3 O C H 3

(4) C H – — C H2 2

O C H 3C H 3

59. MgBrH C3 +CO2 HÅ P1NaOHCaO/D

P2

P2 igpkfu;s&

(1)

CH3

COOH

(2)

(3)

CH3

CH3

(4)

CH3

60. bl vfHkfØ;k C H – M g C l + C H – C H3 2 2

C l O C H 3

dk

eq[; dkcZfud mRikn gS

(1) C H – — C H2 2

C H 3 O C H 3

(2) C H – — C H2 2

C l C H 3

(3) C H – — C H2 2

C H O3 O C H 3

(4) C H – — C H2 2

O C H 3C H 3


LTS-27/35Kota/00DE314029


61. If a and b are the roots of the equation

6x2 – 6x + 1 = 0, then

2 3 2 31 1(a b c d ) (a b c d )

2 2+ a + a + a + + b + b + b

is equal to-

(1) a + b + c + d (2) a b c d

1 2 3 4+ + +

(3) a + b + c – d (4) a + b – c – d

62. If 13 2

r 1

4r 4tan

4r 4r 3r 3

¥-

=

+æ öç ÷+ + +è ø

å = 1cot m4

-p- + ,

then m is -

(1) 1

3- (2)

1

3(3) 3 (4) –3

63. If ƒ(x) = x2 – 2x, then the set of values of k forwhich |ƒ(|x|)| = k has at least three solutions-

(1) [0,1] (2) 1 3

,2 2

æ ùç úè û

(3) 3 5

,2 2

æ ùç úè û

(4) 5,

2é ö¥÷êë ø

64. If the equation x2 + 2|a|x + 4 = 0 has integral

roots, then minimum value of 'a' is -

(1) 5

2- (2) –2 (3)

17

4- (4)

17

4

61. ;fn a rFkk b, lehdj.k 6x2 – 6x + 1 = 0 ds ewygks] rks

2 3 2 31 1(a b c d ) (a b c d )

2 2+ a + a + a + + b + b + b

dk eku gksxk -

(1) a + b + c + d (2) a b c d

1 2 3 4+ + +

(3) a + b + c – d (4) a + b – c – d

62. ;fn 13 2

r 1

4r 4tan

4r 4r 3r 3

¥-

=

+æ öç ÷+ + +è ø

å = 1cot m4

-p- +

gks] rks m gksxk -

(1) 1

3- (2)

1

3(3) 3 (4) –3

63. ;fn ƒ(x) = x2 – 2x gks] rks k ds ekuksa dk leqPp; ftldsfy, |ƒ(|x|)| = k ds de ls de rhu gy gS] gksxk -

(1) [0,1] (2) 1 3

,2 2

æ ùç úè û

(3) 3 5

,2 2

æ ùç úè û

(4) 5

,2

é ö¥÷êë ø

64. ;fn lehdj.k x2 + 2|a|x + 4 = 0 ds iw.kk±d ewy gks] rks'a' dk U;wure eku gksxk -

(1) 5

2- (2) –2 (3)

17

4- (4)

17

4

PART C - MATHEMATICS

Kota/00DE314029LTS-28/35



65. Let ƒ(x) =sin 2x cos2x

4

-, g(x) = x2 – 4. If

g(ƒ(x)) is surjective, then it is invertible for x Î

(1) ,02

pé ù-ê úë û(2) ,

2

pé ù- pê úë û

(3) ,8 8

p pé ù-ê úë û(4) 0,

2

pé ùê úë û

66. Let ( ) ( )ƒ x 5ƒ yx 5y

ƒ6 6

++æ ö =ç ÷è ø

for all real

x & y. If ƒ'(0) exists and equal to

1 & ƒ(0) = –1, then ƒ(2) is equal to-

(1) 2 (2) 1 (3) 3 (4) –3

67. Solution of (1 + x2y2)ydx + (1 – x2y2)xdy = 0is-

(1) ( )2

x 1log c

y 2 xy- =

(2) ( )2

x 1log c

y xy- =

(3) ( )2

y 1log c

x 3 xy+ =

(4) ( )4

y 1log c

x xy+ =

65. ekuk ƒ(x) =sin 2x cos2x

4

-, g(x) = x2 – 4 gSA ;fn

g(ƒ(x)) vkPNknd gks] rks x ds fuEu vUrjky ds fy,;g O;qRØe.kh; gksxk

(1) ,02

pé ù-ê úë û(2) ,

2

pé ù- pê úë û

(3) ,8 8

p pé ù-ê úë û(4) 0,

2

pé ùê úë û

66. ekuk lHkh okLrfod x rFkk y ds fy,

( ) ( )ƒ x 5ƒ yx 5yƒ

6 6

++æ ö =ç ÷è ø

gSA ;fn ƒ'(0) fo|eku

,oa 1 ds cjkcj rFkk ƒ(0) = –1 gks] rks ƒ(2) dk ekugksxk -(1) 2 (2) 1 (3) 3 (4) –3

67. (1 + x2y2)ydx + (1 – x2y2)xdy = 0 dk gygksxk -

(1) ( )2

x 1log c

y 2 xy- =

(2) ( )2

x 1log c

y xy- =

(3) ( )2

y 1log c

x 3 xy+ =

(4) ( )4

y 1log c

x xy+ =


LTS-29/35Kota/00DE314029



68. In [–2, 2], Lagrange's mean value theorem isnot applicable to -

(1) ƒ(x) = (x – 1)|x – 1|

(2)

sin(x 1), x 1

ƒ(x) (x 1)

1 , x 1

-ì ¹ï= -íï =î

(3)

1 1

|x| xxe , x 0ƒ(x)0 , x 0

æ ö- +ç ÷è ø

ìï ¹= íï =î

(4) ƒ(x) = |x + 3| + |x – 3| + |x – 5|

69. If y(x) is solution of dy

xy 4x 3y 12dx

= + + + ,

then y(1) – e6 y(–1) is -

(1) 2(e6 – 1) (2) 4(e6 +1)

(3) 2(e–6 + 1) (4) 4(e6 – 1)

70. 3 / 2n

1 2 3........ nlim

8n 10n®¥

+ + ++

is equal to-

(1) 2

15(2)

1

15(3)

4

15(4)

7

1571. The length of largest continuous interval in

which the function ƒ(x) = tan4x – 8x ismonotonically decreasing is-

(1) 16

p(2)

4

p(3)

2

p(4)

8

p

68. vUrjky [–2, 2] esa] fdlds fy, ykxzkat ek/; eku izes;ykxw ugha gksxh -

(1) ƒ(x) = (x – 1)|x – 1|

(2)

sin(x 1), x 1

ƒ(x) (x 1)

1 , x 1

-ì ¹ï= -íï =î

(3)

1 1

|x| xxe , x 0ƒ(x)0 , x 0

æ ö- +ç ÷è ø

ìï ¹= íï =î

(4) ƒ(x) = |x + 3| + |x – 3| + |x – 5|

69. ;fn y(x), dy

xy 4x 3y 12dx

= + + + dk gy gks] rks

y(1) – e6 y(–1) gksxk -(1) 2(e6 – 1) (2) 4(e6 +1)

(3) 2(e–6 + 1) (4) 4(e6 – 1)

70. 3 / 2n

1 2 3........ nlim

8n 10n®¥

+ + ++

cjkcj gksxk -

(1) 2

15(2)

1

15(3)

4

15(4)

7

15

71. ml egÙke larr~ vUrjky dh yEckbZ] ftlesa Qyuƒ(x) = tan4x – 8x ,dfn"V Îkleku gS] gksxk -

(1) 16

p(2)

4

p(3)

2

p(4)

8

p

Kota/00DE314029LTS-30/35



72. Maximum value of sum of coefficients in theexpansion of (2 + x.sin(sin–1 nCr))n–r,(when expanded in powers of x), is-

(1) 3n (2) 1 (3) 3 (4) n

73. Seven persons are standing in a ticket counterqueue. Three friends A, B and C enters thequeue at random positions such that no two ofA,B and C are consecutive in the queue. Thenumber of ways in which B gets ticket beforeC but after A is -

(1) 56 × 3! (2) 112

(3) 56 (4) 180 × 7!

74. Let r xa yb= +rr r

where x, y are constant and rr

is a fixed vector such that 2| r |

4a.bxy

£r

rr , then

| a |

| b |

r

r is equal to (where xy > 0)

(1) 1 (2) 2

2

x

y(3)

x

y (4) y

x

75. OABC is a tetrahedron. The position vectors

of A,B and C are ˆ ˆ ˆ ˆ ˆi, i j, j k+ + respectively. 'O'

is the origin. The distance of the plane face

ABC from origin O is-

(1) 1

2(2)

1

2(3)

1

2 2(4) 1

72. (2 + x.sin(sin–1 nCr))n–r ds izlkj esa xq.kkadksa ds ;ksxQy

dk vf/kdre eku gksxk (tc bldks x dh ?kkrksa esa izlkjdjrs gS)-

(1) 3n (2) 1 (3) 3 (4) n

73. lkr O;fDr fVfdV dkmUVj dh drkj esa [kM+s gq;s gSaA rhufe= A, B rFkk C drkj esa ;kn`PN;k bl izdkj izos'kdjrs gSa fd A,B rFkk C esa ls dksbZ Hkh nks drkj esa Øekxrugha gSA mu rjhdksa dh la[;k ftuls B dks C ls] ijUrq Ads ckn fVfdV feys] gksxh&

(1) 56 × 3! (2) 112

(3) 56 (4) 180 × 7!

74. ekuk r xa yb= +rr r

tgk ¡ x, y vpj g S rFk k rr

fu;r lfn'k bl izdkj gS fd 2| r |

4a.bxy

£r

rr gS] rks

| a |

| b |

r

r dk eku gksxk (tgk¡ xy > 0) -

(1) 1 (2) 2

2

x

y(3)

x

y (4) y

x

75. ,d prq"Qyd OABC gSA A,B rFkk C ds fLFkfr lfn'k

Øe'k% ˆ ˆ ˆi, i j+ rFkk ˆ ˆj k+ gSA 'O' ewyfcUnq gSA Qyd

ABC dh ewyfcUnq O ls nwjh gksxh&

(1) 1

2(2)

1

2(3)

1

2 2(4) 1


LTS-31/35Kota/00DE314029


76. If the intercept made by the plane r.n q=r r

onx, y, z axis are respectively are a1, a2 & a3respectively, then -

(1) 1 2 3ˆ ˆ ˆn (qa )i (qa ) j (qa )k= + +

r

(2) 31 2 aa aˆ ˆ ˆn i j kq q q

= + +r

(3) 1 2 3

q q qˆ ˆ ˆn i j ka a a

= + +r

(4) 1 2 3

1 1 1ˆ ˆ ˆn i j ka q a q a q

æ öæ ö æ ö= + + ç ÷ç ÷ ç ÷

è ø è ø è ø

r

77. In a DABC, with usual notations if c2 = a2 + b2,then 4s(s – a) (s– b) (s – c) =(1) s4 (2) c2b2 (3) s2c2 (4) a2b2

78. If M and m are the respective maximum andminimum values of |z| satisfying |z – 4| < 2 butnot |z| < 4, where 'z' is complex number, thenthe value of M – m is-(1) 4 (2) 6 (3) 2 (4) 3

79. If it is 8 : 5 against a husband who is 60 yearsold living till he is 75 and 4 : 3 against his wifewho is now 54 living till she is 69, then theprobability that at least one of them will be alive15 years hence, is-

(1) 15

91(2)

20

91(3)

31

91(4)

59

9180. Number of real value(s) of 'a' for which roots

of the equation 1 + cosx – 2a2 = 0 form anarithmetic progression is-(1) 1 (2) 3 (3) 5 (4) 6

76. ;fn lery r.n q=r r

}kjk Øe'k% x, y, z v{k ij cuk;sx;s vUr%[k.M Øe'k% a1, a2 rFkk a3 gS] rks -

(1) 1 2 3ˆ ˆ ˆn (qa )i (qa ) j (qa )k= + +

r

(2) 31 2 aa aˆ ˆ ˆn i j kq q q

= + +r

(3) 1 2 3

q q qˆ ˆ ˆn i j ka a a

= + +r

(4) 1 2 3

1 1 1ˆ ˆ ˆn i j ka q a q a q

æ öæ ö æ ö= + + ç ÷ç ÷ ç ÷

è ø è ø è ø

r

77. f=Hkqt ABC esa lkekU; ladsrksa ds lkFk] ;fn c2=a2+b2

gks] rks 4s(s – a) (s– b) (s – c) cjkcj gksxk&(1) s4 (2) c2b2 (3) s2c2 (4) a2b2

78. ;fn M rFkk m Øe'k% |z| ds vf/kdre rFkk U;wureeku gS tks |z – 4| < 2 dks larq"V ijUrq |z| < 4 dks larq"Vugha djrk gS] tgk¡ 'z' lfEeJ la[;k gS] rks M – m dkeku gksxk&(1) 4 (2) 6 (3) 2 (4) 3

79. ;fn 60 o"kZ vk;q okys ifr ds 75 o"kZ rd ftUnk jgus dsfoi{k esa la;ksxuqikr 8 : 5 gS rFkk 54 o"kZ vk;q dhmldh ifRu ds 69 o"kZ ftUnk jgus ds foi{k esa la;ksxuqikr4 : 3 gks] rks 15 o"kZ ckn buesa ls de ls de ,d dsftUnk jgus dh izkf;drk gksxh&

(1) 15

91(2)

20

91(3)

31

91(4)

59

9180. 'a' ds okLrfod ekuksa dh la[;k] ftlds fy;s lehdj.k

1 + cosx – 2a2 = 0 ds ewy ,d lekUrj Js.kh cukrs gks]gksxh&(1) 1 (2) 3 (3) 5 (4) 6

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81. Area of the triangle formed by tangent atx = a(a > 0), to the curve y = x3 and coordinateaxes, is -

(1) a4

2

3(2) 32

3a

(3) 42

3a (4) 3

2

3a

82. Value of the integral / 2 7 5 3

2/ 2

2x 3x 4x x 1dx

sec x

p

-p

- + - +ò

is -

(1) 2

p(2)

2

p- (3) p (4) -p

83. Locus of 'z' given by the equation

12

1log z 1 0

2æ ö- - ³ç ÷è ø

, is-

(1) interior of a circle with centre (1,0) and

radius 3

2= units

(2) a region of area 2p sq. units containing nonorigin points only.

(3) a bounded region of area 2p sq. unitscontaining origin.

(4) a bounded region of area p sq. units,containing origin.

81. oØ y = x3 ds fcUnq x = a(a > 0) ij [khaph xbZ Li'kZjs[kk rFkk funsZ'kh v{kksa }kjk fufeZr f=Hkqt dk {ks=Qygksxk-

(1) a4

2

3(2) 32

3a

(3) 42

3a (4) 3

2

3a

82. lekdy / 2 7 5 3

2/ 2

2x 3x 4x x 1dx

sec x

p

-p

- + - +ò dk eku gksxk -

(1) 2

p(2)

2

p- (3) p (4) -p

83. lehdj.k 1

2

1log z 1 0

2æ ö- - ³ç ÷è ø

}kjk fn;k x;k 'z'

dk fcUnqiFk gksxk&

(1) o`Ùk dk vkUrfjd Hkkx ftldk dsUnz (1,0) rFkk

f=T;k 3

2= bdkbZ gSA

(2) dsoy ewy fcUnq jfgr] 2p oxZ bdkbZ {ks=Qy

dk {ks=

(3) ewyfcUnq dks j[kus okyk 2p oxZ bdkbZ dk ifjc¼

{ks=

(4) ewyfcUnq dks j[kus okyk p oxZ bdkbZ dk ifjc¼

{ks=


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84. If xy = yx, then wherever exists dy

dx is equal to-

(1) 2

2

xlog

yey x

loge

æ öç ÷è øæ öç ÷è ø

(2) 2

2

xlog

yey x

loge

æ öç ÷è ø-æ öç ÷è ø

(3) 2

2

ylog

yex x

loge


(4) 2

2

ylog

yex x

loge


85. For x > 2, the expression

( ) ( )( )

1 12

2 x 12 tan x 1 sin

1 x 1- - -

- ++ -

is equal to-

(1) 4 tan–1(x–1) (2) p

(3) 0 (4) 2tan–1(p)

86. ( )22x 0

4lim x 2x

x 2x®

é ù+ ê ú+ë û is equal to (where [.]

denotes greatest integer function)-

(1) –4 (2) 0 (3) 4 (4) 1

87. If A and B are non-singular symmetric matrices

of the same order such that AB = BA and

P = (AB–1)T & Q = (A–1B)T then (PQ)–1 is equal

to-(1) AB–1AB (2) A–1B–1AB(3) A–1BA–1B (4) A2B2

84. ;fn xy = yx gks] rks dy

dx (tgk¡ dgha Hkh fo|eku) gksxk-

(1) 2

2

xlog

yey x

loge


(2) 2

2

xlog

yey x

loge


(3) 2

2

ylog

yex x

loge


(4) 2

2

ylog

yex x

loge


85. x > 2 ds fy, O;atd

( ) ( )( )

1 12

2 x 12 tan x 1 sin

1 x 1- - -

- ++ -

dk eku gksxk-

(1) 4 tan–1(x–1) (2) p

(3) 0 (4) 2tan–1(p)

86. ( )22x 0

4lim x 2x

x 2x®

é ù+ ê ú+ë û dk eku gksxk (tgk¡ [.]

egÙke iw.kk±d Qyu dks n'kkZrk gS) -(1) –4 (2) 0 (3) 4 (4) 1

87. ;fn A rFkk B leku dksfV ds O;qRØe.kh; lefer vkO;wg

bl izdkj gS fd AB = BA rFkk P = (AB–1)T ,oa

Q = (A–1B)T gks] rks (PQ)–1 gksxk-

(1) AB–1AB (2) A–1B–1AB

(3) A–1BA–1B (4) A2B2

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88. Let ƒ : , R2 2

p pæ ö- ®ç ÷è ø

be a function defined byy

ƒ(x) = max {sinx, tan2x}, then the number of

points where ƒ(x) is not differentiable is-

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

89. Value of the expression |x – 1| + |x – 2|

+ |x – 22|+..... + |x – 2100| will be minimum at-

(1)

100r

r 0

2x

101==å

(2) x = 250

(3) x = 251 (4) 1

90. Let A(1,1) & B(2,2) are the vertices of a DABC

on xy plane. If C is a variable point such that

the area of DABC is 5 square units, then locus

of C represents -

(1) Two straight lines which are mutually

perpendicular

(2) Two straight lines which are parallel

(3) A circle having centre at origin

(4) A parabola having vertex at origin

88. ekuk ƒ : , R2 2

p pæ ö- ®ç ÷è ø

,

ƒ(x) = vf/kdre {sinx, tan2x} }kjk ifjHkkf"kr Qyugks] rks mu fcUnqvksa dh la[;k tgk¡ ƒ(x) vodyuh; ughagks] gksxh -

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

89. O;atd |x – 1| + |x – 2| + |x – 22|+..... + |x – 2100|

U;wure gksxk ;fn-

(1)

100r

r 0

2x

101==å

(2) x = 250

(3) x = 251 (4) 1

90. ekuk A(1,1) rFkk B(2,2), xy lery ij f=Hkqt ABC

ds 'kh"kZ gSA ;fn C ,d pj fcUnq bl izdkj gS fd f=Hkqt

ABC dk {ks=Qy 5 oxZ bdkbZ gS] rks C dk fcUnqiFk

iznf'kZr djsxk&

(1) nks ljy js[kk,sa tks ijLij yEcor~ gksxhA

(2) nks ljy js[kk,sa tks lekUrj gksxhA

(3) ,d o`Ùk ftldk dsUnz ewy fcUnq gSA

(4) ,d ijoy; ftldk 'kh"kZ ewy fcUnq ij gSA


LTS-35/35Kota/00DE314029


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

CAREER INSTITUTE Pat S KOTA (RAJASTHAN) DISTANCE … · (ACADEMIC SESSION 2014-2015) PAPE CODE...

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