class xii - synthesissynthesis.ac.in/.../12th_maths_paper_2017.pdf · student name :..... roll no....

Student Name :...................................................................... Roll No. :.................................................

Date : 15/01/2017 Time: 3.00 hrs. Max Marks: 90×4=360

INSTRUCTIONS funsZ'k

1. Immediately fill in the particulars on this page of theTest Booklet with Blue/Black Ball Point Pen.

2. The OMR Sheet is kept inside the Test Booklet. Whenyou are directed to open the Test Booklet, take outthe OMR Sheet and fill in the particulars carefully.

3. The test is of 3.00 hours duration.4. The Test Booklet consists of 90 questions. The

maximum marks are 360.5. There are three sections in the question paper A,

B, C consisting of Physics, Chemistry & Mathshaving 30 questions in each section of equalweightage. Each question is allotted 4 (four) marksfor correct response.

6. Candidates will be awarded marks as stated above ininstruction 5 for correct response of each question.1/4 (one forth) marks will be deducted for indicatingincorrect response of each question. No deductionfrom the total score will be made if no response isindicated for an item in the OMR sheet.

7. There is only one correct response for each question.Filling up more than one response in each questionwill be treated as wrong response and marks forwrong response will be deducted accordingly as perinstruction 6 above.

8. Use Blue/Black Ball Point Pen only for writingparticulars/marking responses on OMR Sheet.

9. No candidate is allowed to carry any textual material,

printed or written bits of papers, pager, mobile phone,

any electronic device etc. inside the examination room.

10. Rough work is to be done on the space provided forthis purpose in the Test Booklet only. This space isgiven at the end of the test booklet.

11. On completion of the test, the candidate must handover the OMR Sheet to the Invigilator on duty in theRoom. However, the candidates are allowed to takeaway this Test Booklet with them.

1. ijh{kk iqfLrdk ds bl i"B ij vko';d fooj.k uhys] dkys ckWyIokbaV isu ls rRdky HkjsaA

2. mŸkj i= (OMR) bl ijh{kk iqfLrdk ds vUnj j[kk gSA tcvkidks ijh{kk iqfLrdk [kksyus dks dgk tk,] rks mŸkj i=fudky dj lko/kkuhiwoZd fooj.k HkjsaA

3. ijh{kk dh vof/k 3.00 ?kaVs gSA4. bl ijh{kk iqfLrdk esa 90 iz'u gSA vf/kdre vad 360 gSA

5. bl ijh{kk iqfLrdk esa rhu Hkkx A, B, C gSa] ftuds izR;sd Hkkxesa HkkSfrd foKku] jlk;u foKku ,oa xf.kr ds 30 iz'ugSa vkSj lHkh iz'uksa ds vad leku gSA izR;sd iz'u ds lgh mŸkjds fy, 4 (pkj) vad fu/kkZfjr fd;s x;s gSaA

6. vH;fFkZ;ksa dks izR;sd lgh mŸkj ds fy, mijksDr funsZ'ku la[;k5 ds vuqlkj ekDlZ fn;s tk;saxsA izR;sd iz'u ds xyr mŸkj dsfy;s 1/4 oka Hkkx dkV fy;k tk;sxkA ;fn mŸkj esa fdlh iz'udk mŸkj ugh fn;k x;k gks rks dqy izkIrkad ls dksbZ dVkSrh ughdh tk;sxhA

7. izR;sd iz'u dk dsoy ,d gh lgh mŸkj gSA ,d ls vf/kd mŸkj

nsus ij mls xyr mŸkj ekuk tk;sxk vkSj mijksDr funsZ'k 6

ds vuqlkj vad dkV fy;s tk;saxsA

8. mŸkj iqfLrdk esa dsoy uhys@dkys isu dk gh iz;ksx djsaA

9. ijh{kkFkhZ }kjk ijh{kk d{k@gkWy esa izos'k dkMZ ds vykok fdlhHkh izdkj dh ikB~; lkexzh] eqfnzr ;k gLrfyf[kr] dkxt dhifpZ;k¡] istj] eksckby] Qksu ;k fdlh Hkh izdkj ds bysDVªkWfudmidj.kksa ;k fdlh vU; izdkj dh lkexzh dks ys tkus ;kmi;ksx djus dh vuqefr ugha gSA

10. jQ dk;Z ijh{kk iqfLRkdk esa dsoy fu/kkZfjr txg ij dhft,A

;g txg izR;sd i"B ij uhps dh vksj nh xbZ gSA11. ijh{kk lekIr gksus ij] ijh{kkFkhZ d{k@gkWy NksM+us ls iwoZ mŸkj

i= d{k fujh{kd dks vo'; lkSai nsaA ijh{kkFkhZ vius lkFk blijh{kk iqfLrdk dks ys tk ldrs gSaA

ClassXII

MATHSGroup

"Gyan Tower" Plot No. 1,2 , Near Shivbari Circle, Old Shivbari Road, Bikaner (0151-2206735, 2233735, Mob : 80030948 - 91/92/93,)

Class : XII Page - 2

Space for Rough Work

1. A graph is shown between object distance u andimage distance v for a lens. The focal length of thelens is :

(1) 10 ± 0.1(2) 10 ± 0.02(3) 10 ± 0.3(4) 10 ± 0.4

2. Two bars of radius '3r' and 'r' are kept in contact asshown. An electric current I is passed through thebars. Which one of the following is correct :

(1) Heat produced in bar (1) is 3 times the heatproduced in bar (2)

(2) Electric field in both halves is equal(3) Current density across AB is thrice that of

across BC(4) Potential difference across BC in 9 times that of

across AB

1. ,d ySUl ds fy, oLrq dh nwjh u rFkk izfrfcEc dh nwjhv ds chp xzkQ uhps n'kkZ;k x;k gSA ySUl dh Qksdl nwjhgS%

(1) 10 ± 0.1(2) 10 ± 0.02(3) 10 ± 0.3(4) 10 ± 0.4

2. '3r' rFkk 'r' f=T;kvksa dh nks NM+sa n'kkZ;s vuqlkj lEidZ esaj[kh gSA bu NM+ksa dh izfrjks/kdrk leku gSA buesa ls I/kkjk cgrh gSA fuEu esa ls dkSulk lgh gS%

(1) NM+ (2) esa mRiUu Å"ek NM+ (1) esa mRiUu Å"ek lsrhu xquh gS

(2) nksuksa Hkkxksa esa fo|qr {ks= cjkcj gS(3) AB esa /kkjk ?kuRo BC ls rhu xquk gS

(4) BC ds fljksa ij foHkokUrj] AB ds fljksa ls ukS xquk gS

SECTION-A (PHYSICS)




3. Find the time constant for the given RC circuits incorrect order :

R1 R2 C1

C2

C2

C1 R1

R2 C1

C2

R3

R1

R2

R1 = 2, R2 = 4, R3 = 6, C1 = 8F, C2 = 4F.(1) 16, 48, 54 (2) 16, 16, 144(3) 16, 24, 48 (4) 48, 27/4, 60

4. A magnetic field 0ˆB B j

exists in the region a < x

< 2a and 0ˆB B j

in the region 2a < x < 3a where

B0 is a positive constant. A positive point charge

moving with a velocity 0ˆv v i

, where v0 is a

positive constant, enters the magnetic field at

x = a. The trajectory of the charge in this region

can be like :

(1)

3. fn, x, RC ifjiFk dk le; fu;rkad lgh Øe esa Kkrdhft,%

R1 R2 C1

C2

C2

C1 R1

R2 C1

C2

R3

R1

R2

R1 = 2, R2 = 4, R3 = 6, C1 = 8F, C2 = 4F.(1) 16, 48, 54 (2) 16, 16, 144(3) 16, 24, 48 (4) 48, 27/4, 60

4. {ks= a < x < 2a esa pqEcdh; {ks= 0ˆB B j

gS rFkk {ks=

2a < x < 3a esa pqEcdh; {ks= 0ˆB B j

gS] tgka B0

èkukRed vpj gSA osx 0ˆv v i

] tgka v0 /kukRed vpj

gS] ls pyrk gq, ,d /kukRed fcUnq vkos'k pqEcdh; {ks=

esa x=a ij izos'k djrk gSA bl {ks= esa vkos'k ds iFk dk

lgh fu:i.k ;g gks ldrk gS%

(1)




(2)

(3)

(4)

5. Electrons with de-Broglie wavelength fall on thetarget in an X-ray tube. The cut off wavelength(0) of the emitted X-rays is :

(1)2

0

2mch

(2) 0

2hmc

(3)2 2 3

0 2

2m ch

(4) 0 =

6. A radioactive sample S1 having an activity of 5Ci

has twice the number of nuclei as another sample

S2 which has an activity of 10Ci. The half lives of

S1 and S2 can be :

(1) 20 years and 5 years, respectively

(2) 20 years and 10 years, respectively

(2)

(3)

(4)

5. Mh&czksXyh rjax nS/;Z ds bysDVªkWu X-fdj.k ufydk dsy{; ij fxjrs gSaA mRlftZr X-fdj.k dh vUrd rjaxnSè;Z(0) gksxh%

(1)2

0

2mch

(2) 0

2hmc

(3)2 2 3

0 2

2m ch

(4) 0 =

6. ,d jsfM;ksdehZ uewuk S1 ¼ftldh lfØ;rk 5Ci) esa

ukfHkdksa dh la[;k ,d&nwljs jsfM;ks/kehZ uewus S2 ¼ftldh

lfØ;rk 10CigS½ ls nks xquh gSA S1 rFkk S2 dh vèkZ&vk;q

fuEu gks ldrh gS%

(1) Øe'k% 20 o"kZ o 5 o"kZ

(2) Øe'k% 20 o"kZ o 10 o"kZ




(3) 10 years each

(4) 5 years each

7. A parallel plate capacitor C with plates of unit area

and separation d is filled with a liquid of dielectric

constant K = 2. The level of liquid is d/3 initially.

Suppose the liquid level decreases at a constant

speed V, the time constant as a function of time t

is:

(1) 06 R5d 3Vt

(2) 02 2 2

(15d 9Vt) R2d 3dVt 9V t

(3) 06 R5d 3Vt

(4) 02 2 2


8. S1 and S2 are two co-herent sources of

monochromatic light of wavelength . The distance

between them is 5 / 3. A circle is drawn taking

their midpoint as centre. What will be the number

of maxima and minimas obtained respectively.

(3) nksuksa dh 10 o"kZ

(4) nksuksa dh 5 o"kZ

7. ,d lekarj ifêdk la/kkfj= C dks] ftldh IysVksa dk

{ks=Qy bdkbZ gS rFkk vUrjky d gS] ,d rjy inkFkZ ls

Hkjk tkrk gS ftldk ijkoS|qrkad K = 2 gSA 'kq: esa rjy

ds Lrj dh ÅapkbZ d/3 gSA eku ysa fd rjy dk Lrj fLFkj

xfr V ls ?kVrk gSA le; t ds Qyu ds lkFk le;&fLFkjkad

gksxk%

(1) 06 R5d 3Vt

(2) 02 2 2


(3) 06 R5d 3Vt

(4) 02 2 2


8. S1 o S2 ,do.khZ rjaxnS/;Z ds dyk lEca) izdk'k L=ksr

gSA vkSj buds chp dh nwjh 5 / 3 gSA buds e/; fcUnq dks

dsUnz ekudj ,d fp=kuqlkj or [khaprs gS rks bl or ij

Øe'k% fdrus mfp"B o fuEuh"B gksaxs




(1) 6, 8 (2) 8, 6(3) 6, 6 (4) 8, 8

9. Photoelectric effect experiments are performed

using three different metal plates p, q and r having

work functions p = 2.0 eV, q = 2.5 eV and

r = 3.0 eV respectively. A light beam containing

wavelengths of 550 nm, 450 nm and 350 nm with

equal intensities illuminates each of the plates. The

correct I-V graph for the experiment is :

[Take hc = 1240 eV nm]

(1) (2)

(3) (4)

10. A steady current I goes through a wire loop PQR

having shape of a right angle triangle with PQ =

3x, PR = 4x and QR = 5x. If the magnitude of the

magnetic field at P due to this loop is 0Ik

48 x

,

find the value of k.

(1) 8 (2) 7

(3) 5 (4) 10

11. A solid sphere of radius R has a charge Q distributed

in its volume with a charge density = kra, where k

(1) 6, 8 (2) 8, 6(3) 6, 6 (4) 8, 8

9. /kkrq dh rhu fHkUu IysVksa p, q rFkk r ftuds dk;Z Qyu

Øe'k% p = 2.0 eV, q = 2.5 eV rFkk r = 3.0 eV gS] dks

ysdj izdk'k oS|qr izHkko iz;ksx fd, x,A rhuksa iz;ksxksa esa

,d izdk'k iqat dk iz;ksx fd;k tkrk gS tks leku rhozrk

okys 550 nm, 450 nm rFkk 350 nm rjaxnS/;Z okys

izdk'kksa ls feykdj cuk;k x;k gSA bu rhuksa ds I-V vkjs[k

dk lgh fp=.k gSA

[hc = 1240 eV nm ysa]

(1) (2)

(3) (4)

10. ,d fLFkj fo|qr /kkjk I ledks.k f=Hkqtkdkj ywi PQR esa

izokfgr gks jgh gSA bl f=Hkqt dh Hkqtkvksa dh yEckbZ]

Øe'k% PQ = 3x, PR = 4x rFkk QR = 5x gSA fcUnq P ij

ywi ls mRiUu pqEcdh; {ks= dk eku 0Ik

48 x

gSA k dk

eku gksxk%

(1) 8 (2) 7

(3) 5 (4) 10

11. f=T;k R ds ,d Bksl xksys ds lEiw.kZ vk;ru esa vkos'k Q

dks bl izdkj ckaVk x;k gS fd vkos'k&?kuRo = kra gSA




and a are constants and r is the distance from its

centre. If the electric field at r = R/2 is 1/8 times

that at r = R, find the value of a :

(1) 1 (2) 3

(3) 4 (4) 2

12. To verify Ohm's law, a student is provided with a

test resistor R1, a high resistance R1, a small

resistance R2, two identical galvanometers G1 and

G2, and a variable voltage source V. The correct

circuit to carry out the experiment is :

(1)

(2)

(3)

;gka k rFkk a fu;rkad gSa rFkk r xksys ds dsUæ ls nwjh gSA

;fn bl xksys esa r = R/2 ij fo|qr {ks= dk eku r = R ij

fo|qr {ks= ds eku dk 1/8 gS] rks fu;rkad a dk eku gksxk%

(1) 1 (2) 3

(3) 4 (4) 2

12. vkse ds fu;e dks lR;kfir djus ds fy, ,d fo|kFkhZ dks

,d VsLV&izfrjks/k RT ] ,d mPp izfrjks/k R1 ] ,d fuEu

izfrjks/k R2] nks vfHkUu xsYosuksehVj G1 rFkk G2 ,oa ,d

ifjorhZ oksYVrk L=ksr fn, x, gSaA iz;ksx dks djus ds fy,

fuEu esa ls lgh ifjiFk pqfu,%

(1)

(2)

(3)




(4)

13. An AC voltage source of variable angular frequency

and fixed amplitude V0 connected in series with a

capacitance C and an electric bulb of resistance R

(inductance zero). When is increeased :

(1) the bulb glows dimmer

(2) the bulb glows brighter

(3) total impedence of the circuit is unchanged

(4) total impedence of the circuit increases

14. A thin flexible wire of length L is connected to two

adjacent fixed points carries a current I in the

clockwise direction, as shown in the figure. When

system is put in a uniform magnetic field of strength

B goint into the plane of paper, the wire takes the

shape of a circle. The tension in the wire is :

(1) IBL (2) IBL/

(3) IBL/2 (4) IBL/4

15. A light ray travelling in glass medium is incident on

glass-air interface at an angle of incidence . The

reflected (R) and transmitted (T) intensities, both

as function of , are plotted. The correct sketch is:

(4)

13. ,d C /kkfjrk okys la/kkfj= rFkk R izfrjks/k okys ,d

fo|qr cYc ¼ftldk izsjdRo 'kwU; gS½ dks ,d ifjorhZ

dks.kh; vkofÙk rFkk fLFkj vk;ke V0 okys AC oksYVrk

L=ksr ls Js.khØe esa tksM+k x;k gSA dk eku c<+kus ij%

(1) cYc dh nhfIr ean gks tkrh gS

(2) cYc dh nhfIr rhoz gks tkrh gS

(3) ifjiFk dh dqy izfrck/kk ugha cnyrh gS

(4) ifjiFk dh dqy izfrck/kk c<+ tkrh gS

14. fn, x, fp=kuqlkj L yEckbZ dk ,d yphyk rkj nks

fudVorhZ fLFkj fcUnqvksa ls tqM+k gqvk gSA blesa nf{k.kkorZ

fn'kk esa I eku dh fo|qr /kkjk izokfgr gks jgh gSA bl

fudk; dks dkxtry ds vUnj tkrs gq, B rhozrk ds ,d

,dleku pqEcdh; {ks= esa j[kk tkrk gS rks ;g oÙkkdkj

:i esa vk tkrk gSA rkj esa ruko dk eku gS%

(1) IBL (2) IBL/

(3) IBL/2 (4) IBL/4

15. ,d izdk'k fdj.k dkap esa pydj dkap&ok;q varjki"B ij

vkiru dks.k ls iM+ jgh gSA ijkofrZr (R) rFkk fuxZfer

(T) rhozrkvksa esa cnyko dks.k ds ln'k [khaps x;s gSaA

lgh Ldsp gS%




(1) (2)

(3) (4)

16. A long insulated copper wire is closely wound as a

spiral of 'N' turns. The spiral has inner radius 'a' and

outer radius 'b'. The spiral lies in the X-Y plane and a

steady current I flows through the wire. The Z-

component of the magnetic field at the center of

the spiral is :

(1)0NI b

ln2(b a) a

(2)0NI b a

ln2(b a) b a

(3)0NI b

ln2b a

(4)0NI b a

ln2b b a

(1) (2)

(3) (4)

16. ,d yEcs fo|qr&jks/kh rkacs ds rkj ls 'N' pDdjksa dh fp=

esa fn[kkbZ xbZ dq.Mfyuh cukbZ x;h gSA bldh vkarfjd

f=T;k 'a' rFkk ckg~; f=T;k 'b' gSA ;g X-Y ry esa j[kh

gS vkSj rkj esa /kkjk I izokfgr gksrh gSA blds dsUæ ij

pqEcdh; {ks= dk Z-?kVd gksxk %

(1)0NI b

ln2(b a) a

(2)0NI b a

ln2(b a) b a

(3)0NI b

ln2b a

(4)0NI b a

ln2b b a




17. In Young's double slit experiment the two light

waves (from the coherent sources, each having

intensity I0) has initial phase difference . The

intensity at the point of central maxima which is

equidistant from the slits is observed with variation

of . The corresponding graph is :

(1)

(2)

(3)

17. ,d ;ax ds f}&f>jh iz;ksx esa nks izdk'k rjaxksa ¼dyk&lEc)

L=ksrksa ls çR;sd dh leku rhozrk I0½ dk izkjafHkd dykUrj

gSA f>fjZ;ksa ls leku nwjh okys dsUæh; nhfIr fcUnq ij

rhozrk dk oØ ds lkFk forj.k vkjs[k gksxk%

(1)

(2)

(3)




(4)

18. An infinitely long hollow conducting cylinder withinner radius R/2 and outer radius R carries a uniformcurrent density along length. The magnitude of the

magnetic field, |B |

as a function of the radial

distance r from the axis is best represented by :

(1)

(2)

(3)

(4)

(4)

18. ,d vlhfer yEckbZ ds [kks[kys pkyd&csyu dh vkarfjdf=T;k R/2 vkSj ckgjh&f=T;k R gSA bldh yEckbZ dhfn'kk esa ,dleku /kkjk ?kuRo gSA blds pqEcdh; {ks= dk

eku |B |

v{k ls f=T;h; nwjh r ds lkFk ftl izdkj

cnyrk gS mldk loksZÙke o.kZu fdl xzkQ esa gS%

(1)

(2)

(3)

(4)




19. A circular wire loop of radius R is placed in thex-y plane centered at the origin O. A square loop ofside a (a << R) having one turn is placed with its

center at 3 R along the axis of the circular wire

loop, as shown in figure. The plane of the squareloop makes an angle of 30° with respect to the z-axis. If the mutual inductance between the loop is

given by 2

0p

a2 R

, then the value of p is :

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

20. A cylindrical cavity of diameter a exists inside a

cylinder of diameter 2a shown in the figure. Both

the cylinder and the cavity are infinitely long. A

uniform current density J flows along the length. If

the magnitude of the magnetic field at the point P

is given by N24

0 aJ, then the value of N is :

19. fp= esa n'kkZ;s vuqlkj R f=T;k dk ,d oÙkkdkj rkj ywi

¼ik'k½ x-y ry esa j[kk gS vkSj bldk dsUæ O ij gSA bl

oÙkkdkj ywi ds v{k ij Hkqtk a(a << R) dh ,d Qsjksa

okyh oxZ dqaMyh j[kh gS ftldk dsUæ 3 R ij gS ¼fp=

nsf[k,½A dq.Myh dk ry z-v{k ls 30° dks.k ij gSA ;fn

ywi vkSj dqaMyh dk vU;ksU; izsdRo 2

0p

a2 R

gS] rc p dk

eku D;k gS%

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

20. O;kl 2a ds ,d csyu esa] O;kl a dk ,d [kks[kyk csyuh;

xqfgdk gS ¼fp= nsf[k,½ vkSj csyu rFkk xqfgdk nksuksa

vifjfer yEcs gSaA budh yEckbZ dh fn'kk esa blesa ,d

leku /kkjk&/kuRo J izokfgr gksrh gSA ;fn fcUnq P ij

pqEcdh; {ks= dk eku N24

0 aJ gS] rc N dk eku gksxk%




(1) 1 (2) 2

(3) 4 (4) 3

21. particle is fired from very far away towards a

nucleus with charge Q = 120 e, where e is the

electronic charge. It makes a closest approach of

10 fm to the nucleus. The de Brogle wavelength

(in units of fm) of the particle at its start is 7

p

fm then find the value of p.

(1) 2 (2) 6

(3) 8 (4) 7

22. A bi-convex lens is formed with two thin plano-

convex lenses as shown in the figure. Refractive

index n of the first lens is 1.5 and that of the second

lens is 2. Both the curved surfaces are of the same

radius of curvature R = 15cm. One surface of this

lens is silvered. For this bi-convex lens, for an object

distance of 6 cm, the image distance will be :

(1) 1 (2) 2

(3) 4 (4) 3

21. ,d d.k dks lh/ks ,d ukfHkd (Q = 120 e, tgka e

bysDVªkWfud vkos'k gS½ dh vksj cgqr nwj ls nkxk tkrk gSA

;g izksVkWu ukfHkd ls 10 fm dh fudVre nwjh rd igqaprk

gSA d.k ds pyuk vkjaHk djrs le; mldh Mh&czksXyh

rjaxnS/;Z (fm esa) 7

p fm gS rc p dk eku gksxk%

(1) 2 (2) 6

(3) 8 (4) 7

22. fp= esa n'kkZ;s vuqlkj nks irys lery&mÙky ySalksa dks

feykdj ,d mHk;ksÙky ySal cuk gSA igys ySal dk viorZukad

(n) 1.5 rFkk nwljs dk 2 gSA nksuksa ySalksa ds xksyh;Qydksa

dh oØrk f=T;k] R = 15cm gSA bl ySal dh ,d lrg

pkanh ls iksfy'k dh xbZ gSA bl mHk;ksÙky ySal ds fy,

;fn fcEc nwjh 6 cm gks] rc izfrfcEc nwjh gksxh%




(1) –3cm (2) –6 cm

(3) +3cm (4) +6 cm23. X-rays are incident normally on a crystal of lattice

constants 0.6 nm. The first order reflection ondiffraction from the crystal occurs at an angle of30º. What is the wavelength of X-rays used ?(1) 0.3 nm (2) 0.6 nm(3) 1.2 nm (4) 2.4 nm

24. An electron of stationary hydrogen atom passesfrom the fifth energy level to the ground level. Thevelocity that the atom acquired as a result of photoemission will be :

(1) 25m24hR (2)

24m25hR

(3) 24hR25m (4)

25hR24m

25. With an ac input from 50 Hz power line,the ripplefrequency is :(1) 50 Hz in the dc output of half-wave as well as

full-wave rectifier(2) 100 Hz in the dc output of half-wave as well as

full wave rectifier(3) 50 hz in the dc output of half wave and 100 Hz

in the dc output of full wave rectifier

(1) –3cm (2) –6 cm

(3) +3cm (4) +6 cm

23. tkyd fu;rkad 0.6 nm okys fØLVy ij X fdj.ksa yEcrorvkifrr gksrh gSA fØLVy ls izFke dksVh ijkorZu dkfoorZu izk:i 30º ds dks.k ij curk gSA iz;qqDr X-fdj.kksadh rjaxnS/;Z D;k gksxh ?(1) 0.3 nm (2) 0.6 nm(3) 1.2 nm (4) 2.4 nm

24. fLFkj gkbMªkstu ijek.kq esa bysDVªkWu ikapoh d{kk ls izFked{kk esa laØe.k djrk gSA mRlftZr fofdj.k ds dkj.kgkbMªkstu ijek.kq }kjk izkIr fd;k x;k osx gksxk :

(1) 25m24hR (2)

24m25hR

(3) 24hR25m (4)

25hR24m

25. 50 Hz vkofÙk dh 'kfDr iznk; js[kk ds fuos'kh ls mfeZdkvkofÙk (Ripple Frequency) gksxh :(1) v)Zrjax rFkk iw.kZ rjax fn"Vdkjh nksuksa ds dc fuxZr

esa 50 Hz(2) v)Zrjax rFkk iw.kZ rjax fn"Vdkjh nksuksa ds dc fuxZr

esa 100 Hz(3) v)Zrjax fn"Vdkjh ds dc fuxZr esa 50 hz rFkk iw.kZ

rjax fn"Vdkjh ds dc fuxZr esa 100 Hz




(4) 100 Hz in the dc output of half-wave and 50Hz in the dc output of full wave rectifier

26. The current flowing through the zener diode infigure is-

500

10V I1

5V 1k

(1) 20 mA (2) 25 mA(3) 15 mA (4) 5 mA

27. The figure shown a logic circuit two inputs A and Band the output C. The voltage wave forms acrossA, B and C are as given. The logic circuit gate is :

(1) AND gate (2) NAND gate(3) OR gate (4) NOR gate

28. Which of the following statements is false for theproperties of electromagnetic waves ?(1) These waves do not require any material

medium for propagation(2) Both electric and magnetic field vectors attain

the maxima and minima at the same place andsame time

(4) v)Zrjax fn"Vdkjh ds dc esa 100 Hz rFkk iw.kZ rjaxfn"Vdkjh ds dc fuxZr esa 50 Hz

26. thuj Mk;ksM ls izokfgr /kkjk I1 dk eku Kkr dhft, -

500

10V I1

5V 1k

(1) 20 mA (2) 25 mA(3) 15 mA (4) 5 mA

27. n'kkZ;s x;s fp= esa rkfdZd ifjiFk esa nks fuos'kh A rFkk B,oe~ ,d fuxZr C gSA A, B rFkk C dh oksYVrk rjaxsa n'kkZbZxbZ gSA rkfdZd ifjiFk }kj gksxk :

(1) AND gate (2) NAND gate(3) OR gate (4) NOR gate

28. fo|qr pqEcdh; rjaxksa ds xq.k/keksZa ds lanHkZ esa dkSulk dFkuvlR; gS ?(1) bu rjaxksa dks lapj.k gsrq inkFkZ ek/;e dh vko';drk

ugha gksrh

(2) fo|qr {ks= lfn'k rFkk pqEcdh; {ks= lfn'k lekule; rFkk leku LFkku ij mfPp"B vFkok fufEu"Beku izkIr djrs gS




(3) The energy in electromagnetic wave is dividedequally between electric and magnetic fieldvectors

(4) Both electric and magnetic field vectors areparallel to each other and perpendicular to thedirection of propgation of wave

29. In the circuit shown X and Y are identical bulbs.After a short while on pressing the switch :

(1) X and Y glow with equal brightness(2) Y glows brighter than X(3) X glows but Y does not glow(4) Y glows but X does not glow

30. A coil of inductance 30 mH and resistance 2 isconnected to a source of voltage 2V. The currentbecome half of its steady state value in :(1) 0.05 s (2) 0.01 s(3) 0.15 s (4) 0.3 s

(3) fo|qr pqEcdh; rajx dh ÅtkZ fo|qr {ks= rFkk pqEcdh;{ks=ksa esa leku :i ls foHkkftr gksrh gSA

(4) fo|qr {ks= lfn'k rFkk pqEcdh; {ks= lfn'k ijLijlekUrj gksrs gS rFkk lapj.k dh fn'kk ds yEcor~ gksrsgS

29. n'kkZ;s x, ifjiFk esa X rFkk Y loZle cYc gSA fLop dksnckus ds rqjUr i'pkr~ :

(1) X rFkk Y leku ped ls txexkrs gS(2) X dh ped Y ls vf/kd gksrh gS(3) X pedrk gS ijUrq Y ugha(4) Y pedrk gS ijUrq X ugha

30. izsjdRo 30 mH ,oe~ izfrjks/k 2 dh dq.Myh dks 2V dsoksYVrk L=ksr ls tksM+k x;k gSA fdrus le; eas /kkjk viusLFkk;h eku dk vk/kk eku izkIr djsxh :(1) 0.05 s (2) 0.01 s(3) 0.15 s (4) 0.3 s




SECTION-B (CHEMISTRY)

31. Ferrous oxide has a cubic structure and each edgeof the unit cell is 5.0 Å. Assuming density of theoxide as 4.0 g cm–3, the number of Fe2+ and O2–

ions present in each unit cell will be(1) four Fe2+ and four O2–

(2) two Fe2+ and four O2–

(3) four Fe2+ and two O2–

(4) three Fe2+ and three O2–

32. Analysis show that nickel oxide consist of nickel ionwith 96% ions having d8 configuration and 4% havingd7 configuration. Which amongst the following bestrepresents the formula of the oxide.(1) Ni1.02O1.00 (2) Ni0.96O1.00

(3) Ni0.96O0.98 (4) Ni0.98O1.00

33. The correct order of magnetic moments(1) [Fe(CN)6]

4– > [MnCl4]2– > [CoCl4]

2–

(2) [MnCl4]4– > [Fe(CN)6]

4– > [CoCl4]2–

(3) [MnCl4]2– > [CoCl4]

2– > [Fe(CN)6]4–

(4) [Fe(CN)6]4– > [CoCl4]

2– > [MnCl4]2–

(Atomic no. : Mn=25 ; Fe=26 ; Co=27 ; Ni = 28)34. The Henry's law constant for the solubility of N2

gas in water is 1.0 × 105 atm. The mole fraction ofN2 in air is 0.8. The amount of N2 from air dissolvedin 10 mol of water at 298 K and 5 atm. pressure is(1) 4.0 × 10–4 mol (2) 4.0 × 10–5 mol(3) 5.0 × 10–4 mol (4) 4.0 × 10–6 mol

31. Qsjl vkWDlkbM dh ?kuh; lajpuk gS vkSj ek=d dksf"Bdk dhizR;sd Hkqtk 5.0 Å gSA vkWDlkbM dk ?kuRo 4.0 g cm–3 gS, rksizR;sd ek=d dksf"Bdk esa Fe2+ rFkk O2– vk;uksa dh la[;kgksxh(1) pkj Fe2+ rFkk pkj O2–

(2) nks Fe2+ rFkk pkj O2–

(3) pkj Fe2+ rFkk nks O2–

(4) rhu Fe2+ rFkk rhu O2–

32. fo'ys"k.k n'kkZrk gS fd fudy vkWDlkbM esa fudy vk;u96%, d8 foU;kl j[krs gS rFkk 4%, d7 foU;kl j[krs gSArks fuEu esa ls dkSulk vkWDlkbM ds lw= dks lgh n'kkZrkgS(1) Ni1.02O1.00 (2) Ni0.96O1.00

(3) Ni0.96O0.98 (4) Ni0.98O1.00

33. pqEcdh; vk?kw.kZ dk lgh Øe gS –(1) [Fe(CN)6]

4– > [MnCl4]2– > [CoCl4]

2–

(2) [MnCl4]4– > [Fe(CN)6]

4– > [CoCl4]2–

(3) [MnCl4]2– > [CoCl4]

2– > [Fe(CN)6]4–

(4) [Fe(CN)6]4– > [CoCl4]

2– > [MnCl4]2–

(ijek.kq Øekad : Mn=25 ; Fe=26 ; Co=27 ; Ni = 28)34. N2 xSl dh ty esa foys;rk ds fy, gsujh fu;rkad dk eku

1.0 × 105 atm gSA N2 dk ok;q esa eksy va'k 0.8 gS] rks298 K o 5 atm nkc ij ok;q ?kqfyr ty ds 10 eksy esaN2 dh fdruh ek=k mifLFkr gksxh :(1) 4.0 × 10–4 mol (2) 4.0 × 10–5 mol(3) 5.0 × 10–4 mol (4) 4.0 × 10–6 mol




35.

Eº

0.71 v

3ClO 0.54VClO– 0.45V 1.07 V

Cl–212Cl

The E° in the given diagram is,(1) 0.5 (2) 0.6(3) 0.7 (4) 0.8

36. A hydrogen electrodes is immersed in a solutionwith pH = 0 (HCl). By how much will the potential(reduction) change if an equivalent amount ofNaOH is added to the solution. (Take

2HP = 1 atm)T = 298 K.(1) increase by 0.41 V (2) increase by 50 mV(3)decrease by 0.41 V (4) decrease by 59 mV

37. For two reaction the value of rate constant k1 and k2

are 1016 e–2000/T and 1015 e–1000/T. At which temperaturek1 = k2

(1) 2000 K (2) 1000

K2.303

(3) 1000 K (4) 2000

K2.303

38. The rate of the reaction

2 5 2 22N O 4NO O? ? can be written in three ways

2 5 2 22 5 2 5 2 5

d N O d NO d Ok N O , k ' N O , k " N O

dt dt dt

The relationship between k and k' and between kand k" becomes(1) k'=2k; k"=k (2) k'=2k; k"=k/2(3) k'=2k; k"=2k (4) k'=k; k"=k

35.

Eº

0.71 v

3ClO 0.54VClO– 0.45V 1.07 V

Cl–212Cl

fn;s x;s fp= esa E°dk eku gS(1) 0.5 (2) 0.6(3) 0.7 (4) 0.8

36. gkbMªkstu bysDVªkWM dks pH = 0 (HCl) foy;u esa Mqcks;kx;kA rks vip;u foHko esa D;k ifjorZu gksxk ;fn blfoy;u esa NaOH ds cjkcj rqY;kad feyk;s tkrs gSA

(fn;k gS 2HP = 1 atm T = 298 K).

(1) 0.41 V ls c<sxk (2) 50 mV ls c<sxk(3) 0.41 V ls ?kVsxk (4) 59 mV ls ?kVsxk

37. nks fHkUu vfHkfØ;kvksa ds fy;s nj fLFkjkad k1 rFkk k2 Øe'k%1016 e–2000/T rFkk 1015 e–1000/T gSA fdl rkieku ij k1 =k2 gksxk

(1) 2000 K (2) 1000

K2.303

(3) 1000 K (4) 2000

K2.303

38. vfHkfØ;k 2 5 2 22N O 4NO O? ? , dh nj dks fuEu rhu

izdkj ls fy[kk x;k gS

2 5 2 22 5 2 5 2 5

d N O d NO d Ok N O , k ' N O , k " N O

dt dt dt

k o k' rFkk k o k" ds chp lEcU/k D;k gksxk

(1) k'=2k; k"=k (2) k'=2k; k"=k/2(3) k'=2k; k"=2k (4) k'=k; k"=k




39. 60 mL of 1M oxalic acid is shaken with 0.5 g ofwood charcoal. The final concentration of thesolution after adsorption is 0.5 M. Amount of oxalicacid adsorbed per gram of charcoal is :(1) 3.15 g (2) 3.45 g(3) 6.3 g (4) None of these

40. When a graph is plotted between log x/m and logp, it is straight line With an angle 45° and intercept0.3010 on y-axis. If initial pressure is 0.3 atm, whatwill be the amount of gas adsorbed per gm ofadsorbent :(1) 0.4 (2) 0.6(3) 0.8 (4) 0.1

41. Excess KI reacts with CuSO4 solution and theNa2S2O3 solution is added to it. Which of thestatements is incorrect for this reaction ?(1) Cu2I2 is formed(2) Evolved I2 is reduced(3) Na2S2O3 is oxidised(4) CuI2 is formed

42. ‘Vortex rings’ is the –(1) White smoke of P2O5 formed on combustion of phosphine in air(2) White smoke formed on burning of P in air(3) White fumes formed due ot hydrolysis of PCl3(4) None of the above

43. Extraction of metals by following processes isthrough the complex formation –I : cyanide processII : Mond's process

39. 1M vksDlsfyd vEy ds 60 mL dks dk"B pkjdksy ds0.5 g ds lkFk fgyk;k tkrk gS rks vo'kks"k.k ds cknfoy;u dh vfUre lkUnzrk 0.5 M gksrh gSA pkjdksy dsizfr xzke }kjk vf/k'kksf"kr vkWDlsfyd vEy dh lkUnzrkD;k gksxh :(1) 3.15 g (2) 3.45 g(3) 6.3 g (4) buesa ls dksbZ ugha

40. tc log x/m rFkk log p ds chp xzkQ [khapk tkrk gS rks;g ,d lh/kh js[kk izkIr gksrh gS] ftldk dks.k 45° gSrFkk vUr% [k.M y v{k ij 0.3010 gSA ;fn izkjafHkd nkc0.3 atm gS rks vf/k'kks"kd ds izfrxzke }kjk vf/k'kksf"kr xSldh ek=k gS(1) 0.4 (2) 0.6(3) 0.8 (4) 0.1

41. CuSO4 foy;u ds lkFk vkf/kD; esa KI fØ;k djrk gS rFkkblesa Na2S2O3 foy;u feykrs gS] rks bl vfHkfØ;k dsfy;s dkSulk dFku vlR; gS ?(1) Cu2I2 curk gS(2) mRlftZr I2 vipf;r gksrh gS(3) Na2S2O3 vkWDlhdr gksrk gSA(4) CuI2 cur~k gSA

42. ‘okWVZsZDl oy;’ gS –(1) ok;q esa QkWLQhu ds tyus ls cuk P2O5 dk lQsn /kqavk

(2) ok;q esa QkWLQksjl ds tyus ls cuk lQsn /kqavk(3) PCl3 ds ty vi?kVu ls cuk lQsn /kqavk

(4) buesa ls dksbZ ugha43. fuEu izØeksa }kjk /kkrqvksa dk fu"d"kZ.k ladqy fuekZ.k }kjk

gksrk gS –I : lkbukbM izØeII : ekWaM izØe




III : Photographic fixing processComplexes formed in these methods are –

I II III(1) [Ag(NH3)2]Cl Ni(CO)4 [Ag(CN)2]

–

(2) [Cd(CN)4]2– Ni(CO)4 [Ag(S2O3)2]

3–

(3) [Ag(CN)2]– Ni(CO)4 [Ag(S2O3)2]

3–

(4) [Ag(CN)2]– [Ag(S2O3)2]

3– Ni(CO)4

44. In a solid AB having the NaCl structure, A atomsoccupy the corners of the cubic unit cell. If all theface centered atoms along one of the axes areremoved, then the resultant stoichiometry of thesolid is -(1) AB2 (2) A2B

(3) A4B3 (4) A3B4

45. For the four successive transition elements (Cr, Mn,Fe and Co), the stability of +2 oxidation state willbe in the following order :(1) Mn>Fe>Cr>Co (2) Fe>Mn>Co>Cr(3) Co>Mn>Fe>Cr (4) Cr>Mn>Co>Fe

46. Which one of the following is expected to exhibitoptical isomerism [en = ethylenediamine] –(1) Trans-[Co(en)2Cl2] (2) Cis-[Pt(NH3)2Cl2](3) Cis-[Co(en)2Cl2] (4) Trans-[Pt(NH3)2Cl2]

47. In diborane(1) 4 bridged hydrogens and two terminal hydrogen are present(2) 2 bridged hydrogens and four terminal hydrogen are present(3) 3 bridged and three terminal hydrogen are present(4) None of the above

III : QksVksxzkQh fLFkj izØebu fof/k;ksa esa cuus okys ladqy gS –

I II III(1) [Ag(NH3)2]Cl Ni(CO)4 [Ag(CN)2]

–

(2) [Cd(CN)4]2– Ni(CO)4 [Ag(S2O3)2]

3–

(3) [Ag(CN)2]– Ni(CO)4 [Ag(S2O3)2]

3–

(4) [Ag(CN)2]– [Ag(S2O3)2]

3– Ni(CO)4

44. ,d Bksl AB, NaCl izdkj dh lajpuk j[krk gSA ftlesa Aijek.kq ?kuh; ,dd dksf"Bdk ds dksuks ij fLFkr gSA ;fn,d v{k ij vkus okys lHkh Qyd dsfUnzr ijek.kqvksa dks gVkfn;k tk;s rks Bksl dh ifj.kkeh jllehdj.kfefr gksxh -(1) AB2 (2) A2B

(3) A4B3 (4) A3B4

45. pkj Øekxr laØe.k rRoksa (Cr, Mn, Fe o Co) ds fy;s +2vkWDlhdj.k voLFkk ds LFkkf;Ro dk fuEu esa ls dkSulkØe gksxk :(1) Mn>Fe>Cr>Co (2) Fe>Mn>Co>Cr(3) Co>Mn>Fe>Cr (4) Cr>Mn>Co>Fe

46. fuEu esa ls dkSuls ladqy ls izdkf'kd leko;ork n'kkZusdh vk'kk dh tk ldrh gS [en = ethylenediamine] –(1) Trans-[Co(en)2Cl2] (2) Cis-[Pt(NH3)2Cl2](3) Cis-[Co(en)2Cl2] (4) Trans-[Pt(NH3)2Cl2]

47. Mkbcksjsu esa -(1) 4 lsrq gkbMªkstu rFkk nks VfeZuy gkbMªkstu mifLFkr gksrs gS(2) 2 lsrq rFkk pkj VfeZuy gkbMªkstu mifLFkr gksrs gS

(3) 3 lsrq gkbMªkstu rFkk 3 VfeZuy gkbMªkstu mifLFkr gksrs gS(4) buesa ls dksbZ ugha




48. In the reaction

A.

OHOH

CH3

CH3 conc. H SO2 4

The product is -

(1) O

CH3

CH3

(2)

CH3

CH3

(3)

CH3

COCH3

(4)

CH3

CH3

O

49.O(x) CH OH2

pyridine,

SOCl2 Mg(C H ) O2 5 2 H

+O

CH – C(CH )2 3 2

Product

Product of the reaction is

(1) CH – C – CH – CH2 2 3

OH

CH3

O

48. vfHkfØ;k esa

A.

OHOH

CH3

CH3 conc. H SO2 4

mRikn gS &

(1) O

CH3

CH3

(2)

CH3

CH3

(3)

CH3

COCH3

(4)

CH3

CH3

O

49.O(x) CH OH2

pyridine,

SOCl2 Mg(C H ) O2 5 2 H+

OCH – C(CH )2 3 2

Product

Product of the reaction is

vfHkfØ;k dk mRikn gS &

(1) CH – C – CH – CH2 2 3

OH

CH3

O




(2) CH– CH – C(CH )3 22

OO

(3) CH – C – H2 C OH2 – O

CH3

CH3

(4) CH – CH2 OH2 – C– O

CH3

CH3

50. Which one of the following reagents will reducediethyl ether to ethane and ethanol ?(1) Na/liquid NH3 (2) cold HI(3) H2SO4/high pressure (4) Al2O3/heat

51. ? ?3 3 2 3

3

CH CCl Cl /FeCl HBrAlCl HeatAnisole X? ? ? ? ? ? ? ? ? ? ? ? ??

The product X in the above series of reactions is

(1)

OCH3

C(CH )3 3

(2)

Br

C(CH )3 3

Cl

(3)

Br

C(CH )3 3

Br

(4)

OH

C(CH )3 3

Cl

(2) CH– CH – C(CH )3 22

OO

(3) CH – C – H2 C OH2 – O

CH3

CH3

(4) CH – CH2 OH2 – C– O

CH3

CH3

50. fuEu esa ls dkSulk ,d vfHkdeZd MkbZ,fFky bZFkj dks,sFksu rFkk ,sFksukWy esa vipf;r dj nsrk gS ?(1) Na/nzo NH3 (2) B.Mk HI(3) H2SO4/mPp nkc (4) Al2O3/Å"ek

51. ? ?3 3 2 3

3

CH CCl Cl /FeCl HBrAlCl Heat X? ? ? ? ? ? ? ? ? ? ? ? ??,fulkys

mijksDr vfHkfØ;k dh Js.kh esa mRikn X gS

(1)

OCH3

C(CH )3 3

(2)

Br

C(CH )3 3

Cl

(3)

Br

C(CH )3 3

Br

(4)

OH

C(CH )3 3

Cl




52. CH3 COCHO P QOH H O3+

2Q R The product R is :H SO2 4

(1) CH3 CH(OH)COOH

(2) CH3 CH CH CH3

O||C

C||O

O

O

(3) CH3 CH = CH – COOH

(4) CH3 CH3CH

O

O

C

OH53. Consider the following reaction sequence

HNO + H SO3 2 4 Sn + HCl NaNO + HCl2

(6H)

Cu (CN) + HCN2 2 H O2 Product

Product is :

(1)

CH OH2

(2)

CN

OH

(3)

COOH

(4)

NH2 CN

52.CH3 COCHO P QOH 3

+

2Q RH SO2 4

mRikn R gS &

(1) CH3 CH(OH)COOH

(2) CH3 CH CH CH3

O||C

C||O

O

O

(3) CH3 CH = CH – COOH

(4) CH3 CH3CH

O

O

C

OH

53. fuEu vfHkfØ;k ds vuqØe dks ns[ksa &

HNO + HSO3 2 4 Sn + HCl NaNO + HCl2

(6H)

Cu(CN) + HCN2 2 HO2 mRikn gS mRikn gS&

(1)

CH OH2

(2)

CN

OH

(3)

COOH

(4)

NH2 CN




54.O C – OC H2 5

O|| (1) HCI

(2) CH OH2

CH OH2

X Y(i) LiAIH /Et O4 2

(ii) H O2

H O3+

Z, Identify X, Y and Z

O

O

O

C–OC H2 5

||

O

O

O

C–OC H2 5

||

O

O

O

C–OC H2 5

||

O

O CH OH2

O

COC H2 5O||

O

OCOOH

O

OCH OH2

HO CH OH2

HO CH OH2

HO CH OH2

O CH OH2

HO COOH

X Y Z

(A)

(B)

(C)

(D)

55. What will be the product of the following reaction

NH2H C – C – (CH ) – C – CH3 2 2 3

O|| ||

OH /

+

(1) CH3 CH3

Ph

N(2) CH3 CH3

PhN

(3) N|Ph

HC3 CH3

(4) N|Ph

HC3

CH3

56. The secondary structure of a protein refers to(1) fixed configuration of the polypeptide backbone(2) -helical backbone(3) hydrophobic interactions(4) sequence of -amino acids

54.O C – OC H2 5

O|| (1) HCI

(2) CH OH2

CH OH2

X Y(i) LiAIH /Et O4 2

(ii) H O2

H O3+

Z, Identify X, Y and Z

O

O

O

C–OC H2 5

||

O

O

O

C–OC H2 5

||

O

O

O

C–OC H2 5

||

O

O CH OH2

O

COC H2 5O||

O

OCOOH

O

OCH OH2

HO CH OH2

HO CH OH2

HO CH OH2

O CH OH2

HO COOH

X Y Z

(A)

(B)

(C)

(D)

55. fuEu vfHkfØ;k dk mRikn D;k gS&

NH2H C – C – (CH ) – C – CH3 2 2 3

O|| ||

OH /

+

(1) CH3 CH3

Ph

N(2) CH3 CH3

PhN

(3) N|Ph

HC3 CH3

(4) N|Ph

HC3

CH3

56. izksVhu dh f}rh; lajpuk ds fy;s funsZ'k gS(1) ikWyhisIVkbM jh<+ dk fuf'pr foU;kl(2) -gSfydy jh<+(3) gkbMªksQksfcd vUrjfØ;k(4) -,ehuks vEyksa dk Øe




57. Which of the following is not correct regardingterylene ?(1) Step - growth polymer(2) Synthetic fibre(3) Thermoplastic polymer(4) It is also called dacron

58. Which of the following is a nonreducing sugar ?

(1) OHCH – C – (CHOH) – CH OH2 3 2

O

(2)

H

OH

H

H

H

OH

O

OH

CHOH2

H

OH

H

OH

OH

H

O

H

CHOH2

HH

O

(3)

H

OH

H

H

H

OH

O

OH

CHOH2

OH

H

H

H

H

OH

O

OH

CHOH2

HH

O

(4) H

OH

H

H

H

OH

O

OH

CHOH2

H

OH

H

OH

OH

H

O

H

H

O

H

CHOH2

57. fuEu esa ls dkSulk Vsjhfyu ds lanHkZ eas lgh ugha gS ?

(1) in - of) cgqyd(2) la'ysf"kr js'ks(3) rkin<+ cgqyd(4) ;g MsØksu Hkh dgykrk gSA

58. fuEu esa ls dkSulh vukip;h 'kdZjk gS ?

(1) OHCH – C – (CHOH) – CH OH2 3 2

O

(2)

H

OH

H

H

H

OH

O

OH

CHOH2

H

OH

H

OH

OH

H

O

H

CHOH2

HH

O

(3)

H

OH

H

H

H

OH

O

OH

CHOH2

OH

H

H

H

H

OH

O

OH

CHOH2

HH

O

(4) H

OH

H

H

H

OH

O

OH

CHOH2

H

OH

H

OH

OH

H

O

H

H

O

H

CHOH2




59. The compound :O N2

HO

C COOH

C

OH

CH3

CH

is treated with 2 mol of NaNH2. The productobtained is :

(1)

O N2

–O

C COO–

C

OH

CH3

CH

(2)

O N2

HO

C COO–

C

O–

CH3

CH

(3)

O N2

HO

C COOH

C

O–

CH3

C–

(4)

O N2

HO

C COO–

C

OH

CH3

C–

59. ;kSfxdO N2

HO

C COOH

C

OH

CH3

CH

NaNH2 ds 2 eksy ds lkFk vfHkdr fd;k tkrk gS] rksmRikn izkIr gksxk

(1)

O N2

–O

C COO–

C

OH

CH3

CH

(2)

O N2

HO

C COO–

C

O–

CH3

CH

(3)

O N2

HO

C COOH

C

O–

CH3

C–

(4)

O N2

HO

C COO–

C

OH

CH3

C–




60. What is/are the products of the following reaction

CH CH O + CH CBr ?3 2 3–

CH3

CH3

(1) CH CH OC—CH3 2 3

CH3

CH3

(2) CH2=CH2

(3) CH C=CH3 2

CH3

(4) (1) and (2)

60. nh xbZ vfHkfØ;k dk eq[; mRikn D;k gksxk@gksaxs ?

CH CH O + CH CBr ?3 2 3–

CH3

CH3

(1) CH CH OC—CH3 2 3

CH3

CH3

(2) CH2=CH2

(3) CH C=CH3 2

CH3

(4) (1) and (2)




SECTION-C (MATHS)

61. The range of the function

f(x) = 222

log 2 – log (16 sin x 1) is

(1) (–,1] (2) (– , 2)(3) (–,1) (4) (–, 2]

62. Let f: {x, y, z} {1, 2, 3} be a one-one mappingsuch that only one of the following three statementsis true and remaining two are false : f(x) 2,f(y) = 2, f(z) 1, then(1) f(x) > f(y) > f(z) (2) f(x) < f(y) < f(z)(3) f(y) < f(x) < f(z) (4) f(y) < f(z) > f(x)

63. Set of all values of x such that nlim

n–1

1

4 tan 2x1

is

non-zero and finite number, where n N, is

(1) 1 1

– ,2 2

(2) 1

0,2

(3) (–1, 1) (4) 1

– ,02

64. If f(x) = maximum 1

cos x, , {sin x}2

, 0 x 2,

where { . } represents fractional part function,then number of points of which f(x) is continousbut not differentiable, is(1) 1 (2) 2(3) 3 (4) 4

61. Qyu

f(x) = 222

log 2 – log (16 sin x 1) dh ijkl gS%

(1) (–,1] (2) (– , 2)(3) (–,1) (4) (–, 2]

62. ekuk f: {x, y, z} {1, 2, 3} ,d ,dSdh izfrfp=.kbl izdkj gS fd vkxs fn, x, rhu dFkuksa esa ls dsoy,d lR; gS] 'ks"k nks vlR; gS% f(x) 2, f(y) = 2,f(z) 1 rks%(1) f(x) > f(y) > f(z) (2) f(x) < f(y) < f(z)(3) f(y) < f(x) < f(z) (4) f(y) < f(z) > f(x)

63. x ds ekuk s a dk leqPp;] bl i zdkj g S fd

n–1

1

4 tan 2x1

v'kwU; rFkk lhfer la[;k gS] tgka

n N gS] gksxk%

(1) 1 1

– ,2 2

(2) 1

0,2

(3) (–1, 1) (4) 1

– ,02

64. ;fn f(x) = vf/kdre 1

cos x, , {sin x}2

, 0 x 2,

tgka { . } Qyu ds vkaf'kd Hkkx dks O;Dr djrk gS]rks mu fcUnqvksa dh la[;k tgka f(x) larr gS ijUrqvodyuh; gS] gksxh%(1) 1 (2) 2(3) 3 (4) 4




65. If f ''(x) = – f(x) and g(x) = f '(x) and

F(x) = 2

xf

2

+ 2

xg

2

and given that F(5),

then F(10) is equal to(1) 5 (2) 10(3) 0 (4) 15

66. If t, n, t', n' are the lenghts of tangent, normal,subtangent & subnormal at a point P (x1,y1) onany curve y = f(x) then

(1) t2 + n2 = t'n' (2) 2

1t

+ 2

1n

= 1

t 'n '(3) t'n' = tn (4) nt' = n't

67. The tangent to the graph of the function y = f(x) atthe point with absciassa x = 1 form an angle of /6 andat the point x = 2, an angle of /3 and at the point x= 3, an angle of /4. The value of

3

1

f '(x)f ''(x) dx + 3

2

f ''(x) dx (f'(x) is supposed to

be continuous) is :

(1) 4 3 – 1

3 3(2)

3 3 – 12

(3) 4 – 3

3(4) None of these

68. If 0an 1

+ 1an

+ 2an – 1

+ ............. + n–1a2

+ an =

0, then the equationa0xn + a1xn–1 + a2xn–2 + ........... + an –1 x + an

= 0 has

65. ;fn f ''(x) = – f(x) rFk k g(x) = f '(x) ,o a

F(x) = 2

xf

2

+ 2

xg

2

vkSj fn;k gS fd F(5)

gS] rks F(10) =(1) 5 (2) 10(3) 0 (4) 15

66. ;fn t, n, t', n' Øe'k% Li'kZjs[kk] vfHkyEc] v/kksLi'kZ js[kkrFkk v/kksvfHkyEc dh yEckbZ;ksa dks fdlh fcUnq P(x1,y1) tks oØ y = f(x) ij fLFkr gS] dks fu:fir djsa]rks%

(1) t2 + n2 = t'n' (2) 2

1t

+ 2

1n

= 1

t 'n '(3) t'n' = tn (4) nt' = n't

67. oØ y = f(x) ds fcUnqvksa x = 1, x = 2 rFkk x = 3 ijØe'k% [khaph xbZ Li'kZjs[kk,a v{k ls /6, /3 rFkk /4

dks.k cukrs gSa] rks 3

1

f '(x)f ''(x) dx + 3

2

f ''(x) dx(f'(x)) =

(1) 4 3 – 1

3 3(2)

3 3 – 12

(3) 4 – 3

3(4) buesa ls dksbZ ugha

68. ;fn 0an 1

+ 1an

+ 2an – 1

+ ......... + n–1a2

+ an =

0, rks lehdj.ka0xn + a1xn–1 + a2xn–2 + ........... + an –1 x + an

= 0 j[krh gS%




(1) exactly one root in (0, 1)(2) at least one root in (0, 1)(3) no root in (0, 1)(4) at the most one root in (0, 1)

69. Lef f(x) be a non-constant twice differentiablefunction defined on (– , ) such that

f(x) = f(1 – x) and f'14

= 0, Then,

(1) f''(x) vanishes at least twice on [0, 1]

(2) f' 12

= 0

(3) 1/2

–1/2

1f x

2 sin x dx = 0

(4) None of these70. The equation x3 – 3x + [a] = 0,where [.] denotes

the greatest integer function, will have three realand distinct roots if(1) a (– , 2) (2) a (0, 2)(3) a (, –2) (4) a [–1, 2]

71. If f(x) = 2x3 – 3(a + 1) x2 + 6ax – 12 has localmaximum at x1 and local minimum at x2 and if 2x1

= x2 then value of a is :

(1) 1 (2) 12

(3) –1 (4) 2

72. 4 cos x6

cos 2x. cos 5

cos x6

dx

(1) sin 4x sin2x

– x4 2

+ c

(2) sin 4x sin2x

– x4 2

+ c

(1) varjky (0, 1) esa Bhd ,d ewy(2) varjky (0, 1) esa de ls de ,d ewy(3) varjky (0, 1) esa dksbZ ewy ugha(4) varjky (0, 1) esa vf/kd ls vf/kd ,d ewy

69. ekuk f(x), (– , ) esa nks ckj vodyuh; gS rFkk vpjQyu ugha gSA

f(x) = f(1 – x) rFkk f'14

= 0 gS] rks %

(1) f''(x), [0, 1] esa de ls de nks ckj 'kwU; gksrk gS

(2) f' 12

= 0

(3) 1/2

–1/2

1f x

2 sin x dx = 0

(4) buesa esa dksbZ ugha70. lehdj.k x3 – 3x + [a] = 0 tgka [.] e-iw-Q- dks O;Dr

djrk gS] rhu okLrfod rFkk fHkUu&fHkUu ewy j[krh gS];fn %(1) a (– , 2) (2) a (0, 2)(3) a (, –2) (4) a [–1, 2]

71. ;fn f(x) = 2x3 – 3(a + 1) x2 + 6ax – 12; x1 ijLFkkuh; mfPp"B rFkk x2 ij LFkkuh; fufEu"B gS ,oa;fn 2x1 = x2 gS] rks a dk eku gS%

(1) 1 (2) 12

(3) –1 (4) 2

72. 4 cos x6

cos 2x. cos 5

cos x6

dx

(1) sin 4x sin2x

– x4 2

+ c

(2) sin 4x sin2x

– x4 2

+ c




(3) sin 4x sin2x

– x –4 2

+ c

(4) sin 4x sin2x

– x –4 2

+ c

73. If x–1

2

e(x – 5x 4) 2x dx = A F(x – 1) + B F(x –

4) + C and F(x) = xe

x dx. then A & B ordered set

is

(1) 2 8

– ,3 3

(2) 32 8e

– ,3 3

(3) 8 2

,3 3

(4) 32 8e

– , –3 3

74. 20

sin x1 cos x

dx = 2

cos1 sin

(1) for no value of (2) for exactly two values of in (0, )

(3) for at least one in ,2

(4) for exactly one in 0,2

75.2

0

x x x ... dx is equal to (x > 0)

(1) 196

(2) 176

(3) 136 (4) Can't determine

76. If Sn = 12n

+ 2

1

4n – 1 + 2

1

4n – 4 + ......... +

(3) sin 4x sin2x

– x –4 2

+ c

(4) sin 4x sin2x

– x –4 2

+ c

73. ;fn x–1

2

e(x – 5x 4) 2x dx = A F(x – 1) + B F(x

– 4) + C rFkk F(x) = xe

x dx rks A o B Øfer

leqPp; gS%

(1) 2 8

– ,3 3

(2) 32 8e

– ,3 3

(3) 8 2

,3 3

(4) 32 8e

– , –3 3

74. 20

sin x1 cos x

dx = 2

cos1 sin

(1) ds fdlh eku ds fy, ugha(2) varjky (0, ) esa ds Bhd nks ekuksa ds fy,

(3) ,2

esa de ls de ,d ds fy,

(4) 0,2

esa Bhd ,d ds fy,

75.2

0

x x x ... dx cjkcj gS% (x > 0)

(1) 196

(2) 176

(3) 136 (4) x.kuk ugha dj ldrs

76. ;fn Sn = 12n

+ 2

1

4n – 1 + 2

1

4n – 4 + ......... +




2

1

3n 2n – 1, 0, N, then

nlim

Sn is equal to

(1) 2

(2) 2

(3) 1 (4) 6

77. Area of the region bounded by x = 0, y = 0, x =2, y ex and y nx , is(1) 6 – 4 n 2 (2) 4 n 2 – 2(3) 2 n 2 – 4 (4) 6 – 4 n 2

78. If gradient of a curve at any point P(x, y) is

x y 12y 2x 1

and it is passes through origin, then curve is

(1) 2 (x + 3y) = n 3x 3y 2

2

(2) x + 3y = n 3x 3y 2

2

(3) 3y + x = n (3x + 2y + 1)

(4) 6y – 3x = n 3x 3y 2

2

79. The solution of differential equation

(x2 – 1) dydx + 2 xy = 2

1x –1

is

(1) y (x2 – 1) = 12 log

x – 1x 1 + C

(2) y (x2 + 1) = 12 log

x – 1x 1 + C

2

1

3n 2n – 1, 0, N rks

nlim

Sn =

(1) 2

(2) 2

(3) 1 (4) 6

77. x = 0, y = 0, x = 2, y ex rFkk y nx ls ifjc){ks= dk {ks=Qy gS%(1) 6 – 4 n 2 (2) 4 n 2 – 2(3) 2 n 2 – 4 (4) 6 – 4 n 2

78. ;fn oØ ds fdlh fcUnq P(x, y) ij izo.krk

x y 12y 2x 1

gks rFkk ;g ewy fcUnq ls xqtjrk gks] rks oØ gS%

(1) 2 (x + 3y) = n 3x 3y 2

2

(2) x + 3y = n 3x 3y 2

2

(3) 3y + x = n (3x + 2y + 1)

(4) 6y – 3x = n 3x 3y 2

2

79. vody lehdj.k

(x2 – 1) dydx + 2 xy = 2

1x –1

dk gy gS%

(1) y (x2 – 1) = 12 log

x – 1x 1 + C

(2) y (x2 + 1) = 12 log

x – 1x 1 + C




(3) y(x2 – 1) = 52 log

x – 1x 1 + C

(4) None of these

80. In a room are 4 students each of which is equallylikely to be a girl or a boy. 2 students have walkedout from the room, first is found to be a boy andthe second a girl. The probablity that the remainingstudents are boys is

(1) 27

(2) 14

(3) 12

(4) 38

81. A bag contains 5 balls all of different colous (one ofwhich is white) three persons A, B and C whose

probabilities of speaking truth are 12

, 23

and 34

respectively assert that a ball drawn from the bagis white, the probability of truth of their assertion,is

(1) 9697

(2) 2425

(3) 1

20(4)

67

82. There are two urns A and B. A contains 5 red, 3blue and 2 white balls, urn B contains 4 red, 3 blueand 3 white balls. An urn is choosen at randomand a ball is drawn. Probability, that the ball drawn

(3) y(x2 – 1) = 52 log

x – 1x 1 + C

(4) buesa ls dksbZ ugha

80. ,d dejs esa 4 fo|kFkhZ gSa] muesa ls izR;sd fo|kFkhZ leku:i ls yM+dk ;k yM+dh gks ldrk gSA nks fo|kFkhZ dejsls ckgj tkrs gSa] muesa igyk yM+dk rFkk nwljh yM+dhik;h tkrh gS rc 'ks"k fo|kfFkZ;ksa ds yM+ds gksus dhizkf;drk gS%

(1) 27

(2) 14

(3) 12

(4) 38

81. ;fn fdlh FkSys esa 5 vyx&vyx jax dh xsansa gS ftuesals ,d xsan lQsn gSA rhu O;fDr A, B o C ds lp cksyus

dh izkf;drk 12

, 23

rFkk 34

gSA ;fn rhuksa O;fDr FkSys

esa fudkyh x;h xsan ds lQsn gksus dks crkrs gSa] rks mudsbl dFku ds lp gksus dh izkf;drk gS%

(1) 9697

(2) 2425

(3) 1

20(4)

67

82. ekukfd nks ik= A o B gSaA ik= A esa 5 yky, 3 uhyhvkSj 2 lQsn xsansa gSa] ik= B esa 4 yky, 3 uhyh rFkk 3lQsn xsansa gSaA ,d ik= ;knfPNd pquk tkrk gS vkSj,d xsan fudkyh tkrh gSA fudkyh x;h xsan ds yky




is red, is equal to

(1) 9

10(2)

12

(3) 1120

(4) 920

83. If A and B are two square matrices of order 3 × 3which satisfy AB = A and BA = B then (A + B)7 is

(1) 7 (A + B) (2) 7.I3 × 3

(3) 64 (A + B) (4) 128 I3 × 3

84. System of equation

x + 3y + 2z = 6

x + y + 2z = 7

x + 3y + 2z = has

(1) unique solution of = 2, m 6(2) infinitely many solution = 4, = 6

(3) no solutioin if = 5, = 7(4) no solution of = 3, = 5

85. If

1a ,

2a ,

3a are non-coplanar vectors and

(x + y – 3)

1a + (2x – y + 2)

2a + (2x +

y + )

3a = 0

holds for some 'x' and 'y' then '' is

(1) 73

(2) 2

gksus dh izkf;drk gS%

(1) 9

10(2)

12

(3) 1120

(4) 920

83. ;fn A o B; 3 × 3 dksfV ds nks oxZ vkO;qg gSa] tksAB = A o BA = B dks larq"V djrs gSa] rks (A + B)7 gS%

(1) 7 (A + B) (2) 7.I3 × 3

(3) 64 (A + B) (4) 128 I3 × 3

84. lehdj.k fudk;

x + 3y + 2z = 6

x + y + 2z = 7

x + 3y + 2z = j[krk gS%

(1) = 2 dk fof'k"V gy, m 6(2) vuar dbZ gy ;fn = 4, = 6

(3) dksbZ gy ugha ;fn = 5, = 7

(4) dksbZ gy ugha ;fn = 3, = 5

85. ;fn

1a ,

2a ,

3a vleryh; lfn'k gSa rfkk

(x + y – 3)

1a + (2x – y + 2)

2a + (2x + y +

)

3a = 0

fdUgha 'x' rFkk 'y' ds fy, lR; gks] rks '' dk eku gS%

(1) 73

(2) 2




(3) 10–3

(4) 53

86. If

(a b) ×

(c d) .

(a d) = 0, then which

of the following is always true

(1) a ,

b ,

c ,

d are necessarily coplanar

(2) either a or

d must lie in the plane of

b &

c

(3) either b or

c must lie in plane of

a and

d

(4) either a or

b must lie in plane of

c and

d

87. If the foot of the perpendicular from the origin to aplane is P(a, b, c), the equation of the plane is

(1) x y za b c

= 3

(2) ax + by + cz = 3

(3) ax + by + cz = a2 + b2 + c2

(4) ax + by + cz = a + b + c

88. Equation of line in the plane 2x – y + z – 4 =0 which is perpendicular to the whose equation is

x – 21

= y – 2

–1 =

z – 3–2

and which passes through

the point of intersection of and is

(1) x – 2

3 =

y – 15

= z – 1–1

(2) x3

= y – 3

5 =

z – 5–1

(3) x 22

= y 1

–1 =

z 11

(3) 10–3

(4) 53

86. ;fn

(a b) ×

(c d) .

(a d) = 0 gS] rks fuEu

esa ls dksulk lnSo lR; gS%

(1) a ,

b ,

c ,

d vko';d :i ls leryh; gSa

(2) ;k rks a ;k

d ; b rFkk

c ds ry esa fLFkr gksuk pkfg, gS

(3) ;k rks b ;k

c ; a rFkk

d ds ry esa fLFkr gksuk pkfg, gS

(4) ;k rks a ;k

b ; c rFkk

d ds ry esa fLFkr gksuk pkfg,

87. ;fn ewy fcUnq ls ,d lery ij Mkys x, yEc ds iknP(a, b, c) gSa] rks ley dk lehdj.k gS%

(1) x y za b c

= 3

(2) ax + by + cz = 3

(3) ax + by + cz = a2 + b2 + c2

(4) ax + by + cz = a + b + c

88. lery 2x – y + z – 4 = 0 esa fLFkr vkSj js[kk

( )x – 2

1 =

y – 2–1

= z – 3–2

gS] ds yEcor~ rFkk

,oa ds izfrPNsn fcUnq ls xqtjus okyh js[kk dklehdj.k gS%

(1) x – 2

3 =

y – 15

= z – 1–1

(2) x3

= y – 3

5 =

z – 5–1

(3) x 22

= y 1

–1 =

z 11




(4) x – 2

2 =

y – 1–1

= z – 1

1

89. Let A (1, 1, 1) and B(2, 3, 5), C(–1, 0, 2) be threepoints, then equation of a plane parallel to theplane ABC which is at a distance 2 is

(1) 2x – 3y + z + 2 14 = 0

(2) 2x – 3y + z – 14 = 0

(3) 2x – 3y + z + 2 = 0

(4) 2x – 3y + z – 2 = 0

90. The solut ion of the inequalitylog1/2 sin–1 x > log1/2 cos–1 x is

(1) x

10,2 (2) x

1 ,12

(3) x

10,2 (4) None of these

(4) x – 2

2 =

y – 1–1

= z – 1

1

89. ekuk A (1, 1, 1) rFkk B(2, 3, 5), C(–1, 0, 2) rhufcUnq gSa] rks lery ABC ds lekUrj vkSj ewyfcUnq ls 2bdkbZ nwjh ij fLFkr lery dk lehdj.k gS%

(1) 2x – 3y + z + 2 14 = 0

(2) 2x – 3y + z – 14 = 0

(3) 2x – 3y + z + 2 = 0

(4) 2x – 3y + z – 2 = 0

90. vlfedk log1/2 sin–1 x > log1/2 cos–1 x dk gy gS%

(1) x

10,2 (2) x

1 ,12

(3) x

10,2 (4) buesa ls dksbZ ugha




FINAL ANSWER KEY OF CLASS XII (JEE)Q. 1 2 3 4 5 6 7 8 9 10A. 1 4 2 1 1 1 1 1 1 2Q. 11 12 13 14 15 16 17 18 19 20A. 4 3 2 3 3 1 3 4 3 4Q. 21 22 23 24 25 26 27 28 29 30A. 3 2 2 3 3 4 3 4 4 2Q. 31 32 33 34 35 36 37 38 39 40A. 1 4 3 1 2 3 2 2 3 2Q. 41 42 43 44 45 46 47 48 49 50A. 4 1 3 4 1 3 2 4 4 1Q. 51 52 53 54 55 56 57 58 59 60A. 4 2 3 3 2 2 2 4 1 3Q. 61 62 63 64 65 66 67 68 69 70A. 4 3 2 4 1 2 4 2 4 4Q. 71 72 73 74 75 76 77 78 79 80A. 2 1 2 4 1 4 1 4 1 2Q. 81 82 83 84 85 86 87 88 89 90A. 1 4 3 1 3 3 3 2 1 3

Note : - B means Bonus

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