class xii bio - synthesis bikaner€¦ · the omr sheet and fill in the particulars carefully. 3....

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

Date : 15/01/2017 Time: 3.00 hrs. Max Marks: 180×4=720

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 180 questions. The

maximum marks are 720.5. There are three sections in the question paper A,

B, C consisting of Physics, Chemistry , Botany &Zoology having 45 questions in each section ofequal weightage. Each question is allotted 4 (four)marks for 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.

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 180 iz'u gSA vf/kdre vad 720 gSA

5. bl ijh{kk iqfLrdk esa rhu Hkkx A, B, C gSa] ftuds izR;sd Hkkxesa HkkSfrd foKku] jlk;u foKku] ouLifr foKku rFkktho foKku ds 45 iz'u gSa vkSj lHkh iz'uksa ds vad leku gSAizR;sd iz'u ds lgh mŸkj ds fy, 4 (pkj) vad fu/kkZfjr fd;sx;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,A11. 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

BIOGroup

"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

1. Three point charges + q, +q and – q are placed atthe corners of an equilateral triangle of sided 'a'.Another charge +Q is kept at the centroid. Forceexerted on Q is :

(1) 20

1 2qQ4 a?? (2) 2

0

1 6qQ4 a??

(3) 20

1 8qQ4 a?? (4) 2

0

1 12qQ4 a??

2. A charged particle of mass 'm' and charge 'q' initiallyat rest is released in an electric field of magnitude E.Its kinetic energy after time 't' will be :

(1) 2 22E t

mq (2) 2 2 2E q t2m

(3) 2

2

Eq m2t

(4) Eqm

t3. The maximum electric field intensity on the axis of

a uniformly charged ring of charge q and radius Rwill be :

(1) 20

1 q4 3 3R?? (2) 2

1 2q4 3R???

(3) 20

1 2q4 3 3R?? (4) 2

0

1 3q4 2 2R??

4. The equivalent capacity of the combination asshown in figure.

(1) C (2) 2C

(3) 3

C2

(4) C2

1. rhu vkos'kksa + q, +q rFkk – q dks Hkqtk 'a' okys leckgqf=Hkqt ds 'kh"kksZa ij j[kk x;k gSA vU; vkos'k +Q dksf=Hkqt ds dsUnzd ij j[kk x;k gSA Q vkos'k ij vkjksfircy gS :

(1) 20

1 2qQ4 a?? (2) 2

0

1 6qQ4 a??

(3) 20

1 8qQ4 a?? (4) 2

0

1 12qQ4 a??

2. nzO;eku 'm' rFkk vkos'k 'q' okys ,d d.k dks ifjek.k Eokys fo|qr {ks= esa fojkekoLFkk ls eqDr fd;k x;k gSAle; 't' i'pkr~ bldh xfrt ÅtkZ gksxh :

(1) 2 22E t

mq (2) 2 2 2E q t2m

(3) 2

2

Eq m2t

(4) Eqm

t3. vkos'k q rFkk f=T;k R okyh le:i vkosf'kr oy; dh

v{k ij vf/kdre fo|qr {ks= gksxk :

(1) 20

1 q4 3 3R?? (2) 2

1 2q4 3R???

(3) 20

1 2q4 3 3R?? (4) 2

0

1 3q4 2 2R??

4. fp= esa n'kkZ;s x;s lek;kstu dh rqY; /kkfjrk gksxh

(1) C (2) 2C

(3) 3

C2

(4) C2



5. A network of four capcitors of capacity equal to C1

= C, C2 = 2C, C3 = 3C and C4 = 4C are connectedto a battery as shown in the fig. The ratio of thecharges on C2 and C4 is :

(1)74

(2) 223

(3) 322

(4) 47

6. Electric current through a conductor varies withtime as I(t) = 50 sin(100t). Here I is in amperesand t in seconds. Total charge that passes any point

from t = 0 to t = 1

200sec is :

(1) 1.2 C (2) 0.36 C(3) 0.16 C (4) 0.02 C

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

test resistor RT, 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)

5. pkj la/kkfj=ksa dh /kkfjrk;sa C1 = C, C2 = 2C, C3 = 3CrFkk C4 = 4C gSA buls cus ifjiFk dks cSVjh ls fp=kuqlkjtksM+k x;k gSA C2 rFkk C4 ij vkos'kksa dk vuqikr gksxk :

(1)74

(2) 223

(3) 322

(4) 47

6. fdlh pkyd esa ls izokfgr /kkjk le; t ds lkis{kI(t) = 50 sin(100t) ds vuqlkj ifjofrZr gksrh gSA tgk¡I ,Eih;j esa ,oa t lSd.M esa gSA fdlh Hkh fcUnq ls le;

t = 0 ls t = 1

200sec ds e/; izokfgr vkos'k gksxk :

(1) 1.2 C (2) 0.36 C(3) 0.16 C (4) 0.02 C

7. 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)

8. The time cosntant of shown RC circuit is :

(1)RC2

(2) RC

(3) 2RC (4) 4RC

(2)

(3)

(4)

8. fn;s x;s RC ifjiFk gsrq le; fu;rkad gksxk :

(1)RC2

(2) RC

(3) 2RC (4) 4RC



9. In the circuit shown in fig, R3 is variable resistance.As the value R3 is changed, current I through thecell varies as shown. Obviously, the variation isasymptotic, i.e., I 6A as R3 . Resistances R1

and R2 are, respectively :

I I

R2

R2

R1

R3

36V

9

6

3

O ? ? ? R ( )3

? ? ? R ( )3

I(A)

(A) (B)

(4) 4, 2 (2) 2, 4(3) 2, 2 (4) 1, 4

10. A metre bridge with resistance R1 and R2 connectedin the two gaps is balanced at 0.4 m from the zeroend. If the smaller resistance is connected in serieswith a 10 resistor, the balance point is shifted to0.4 m from the other end. The value of the smallerresistance is :(1) 40 (2) 60 (3) 20 (4) 8

11. A steel wire of length L has a magnetic moment M.It is then bent into a semicircular arc; the newmagnetic moment will be :(1) M (2) 2M/(3) M/L (4) M × L

12. If the magnetic dipole moment of an atom ofdiamagentic material, paramagnetic material andferromagnetic material are denoted by d, p andƒ respectively, then :(1) p = 0 and ƒ 0(2) d = 0 and P = 0(3) d 0 and ƒ 0(4) d = 0 and p 0

9. n'kkZ;s x;s ifjiFk eas] R3 ,d ifjorhZ izfrjks/k gSA R3 dseku dks ifjofrZr djus ij lsy ls cgus okyh /kkjk eas gksusokyk ifjorZu fp=kuqlkj n'kkZ;k x;k gSA ;g ifjorZuoØ vuUrLi'khZ gS] vFkkZr~ I 6A tc R3 gSAizfrjks/k R1 rFkk R2 Øe'k% gkasxs &

I I

R2

R2

R1

R3

36V

9

6

3

O ? ? ? R ( )3

? ? ? R ( )3

I(A)

(A) (B)

(4) 4, 2 (2) 2, 4(3) 2, 2 (4) 1, 4

10. ,d ehVj lsrq eas R1 rFkk R2 izfrjks/k yxs gq, gS rFkklUrqyu fcUnq 'kwU; fljs ls 0.4 m dh nwjh ij gSA ;fn NksVsokys izfrjks/k ds lkFk 10 ds izfrjks/k dks Js.khØe eastksM+ fn;k tk;s] rks lUrqyu fcUnq nwljs okys fljs ls0.4 m dh nwjh ij izkIr gksrk gSA NksVs okys izfrjks/k dkeku gksxk &(1) 40 (2) 60 (3) 20 (4) 8

11. yEckbZ L ds LVhy rkj dk pqEcdh; vk?kw.kZ M gSA blsv)Zorkdkj pki ds :i esa eksM+k tkrk gSA u;k pqEcdh;vk?kw.kZ gksxk :(1) M (2) 2M/(3) M/L (4) M × L

12. ;fn izfrpqEcdh;] vuqpqEcdh;] ,oa ykSg pqEcdh; inkFkZds ijek.kq dk pqEcdh; f}/kzqo vk?kw.kZ Øe'k% d, p rFkkƒ gks rks &

(1) p = 0 rFkk ƒ 0(2) d = 0 rFkk P = 0(3) d 0 rFkk ƒ 0(4) d = 0 rFkk p 0



13. In short wave communication which of the followingfrequencies will be reflected back by the ionosphericlayer having electron density 1011 per m3 ?(1) 2 MHz (2) 10 MHz(3) 12 MHz (4) 18 MHz

14. A 1 kW signal is transmitted using a communicationchannel which provides attenuation at the rate of–2 dB per km. If the communication channel has atotal length of 5 km, the power of the signal receivedis :

[Gain in dB = 10 log 0

i

PP

? ?? ?? ? ]

(1) 900 W (2) 100 W(3) 990 W (4) 1010 W

15. In amplitude modulated waves, maximumamplitude is 30 mV and minimum is 5 mV, then themodulating index is :(1) 7 (2) 3(3) 0.7 (4) 0.45

16. A concave mirror of focal length ƒ produces animage n times the size of the object. If the image isreal then the distance of the object from the mirroris :

(1) (n – 1) ƒ (2) (n 1)

ƒn?

(3) (n 1)

ƒn?

(4) (n + 1) ƒ

17. A fish at a depth of 12 cm in water is viewed by anobserver on the bank of a lake. To what height theimage of the fish is raised ?(Refractive index of water = 4/3)(1) 9 cm (2) 12 cm(3) 3.8 cm (4) 3 cm

13. vk;u e.My ijr dk bysDVªkWu ?kuRo 1011 /m3 gSA y?kqrjax lapkj O;oLFkk esa fuEu esa ls dkSulh vkofÙkvk;ue.My }kjk ijkofrZr gks tk;sxh ?(1) 2 MHz (2) 10 MHz(3) 12 MHz (4) 18 MHz

14. fdlh lapkj pSuy esa 1 kW dk ladsr lEizsf"kr fd;k tkrkgSA pSuy –2 dB /km dk ákl n'kkZrk gSA ;fn lapkjpSuy dh dqy yEckbZ 5 km gS rks xzkgh ij izkIr 'kfDrgksxh

[Gain in dB = 10 log 0

i

PP

? ?? ?? ? ]

(1) 900 W (2) 100 W(3) 990 W (4) 1010 W

15. ;fn vk;ke ekWMqfyr rjax dk vf/kdre ,oe~ U;wurevk;ke Øe'k% 30 mV rFkk 5 mV gks rks ekWMqys'ku xq.kkadgksxk :(1) 7 (2) 3(3) 0.7 (4) 0.45

16. Qksdl nwjh ƒ dk ,d vory niZ.k fcEc ds n xquas vkdkjds izfrfcEc dk fuekZ.k djrk gSA ;fn izfrfcEc okLrfodgks rks fcac dh niZ.k ls nwjh :

(1) (n – 1) ƒ (2) (n 1)

ƒn?

(3) (n 1)

ƒn?

(4) (n + 1) ƒ

17. ty ds Hkhrj 12 cm dh xbjkbZ ij fLFkr eNyh >hyds fdukjs cSBs izs{kd }kjk ns[kh tkrh gSA eNyh dkizfrfcEc fdruh Å¡pkbZ rd mBk gqvk fn[kkbZ nsxk(ty dk viorZukad = 4/3)(1) 9 cm (2) 12 cm(3) 3.8 cm (4) 3 cm



18. A point object is placed at a distance of 20 cm froma thin plano-convex lens fo focal length 15 cm. Ifthe plane surface is silvered, the image will form at

(1) 60 cm left of AB(2) 30 cm left of AB(3) 12 cm left of AB(4) 60 cm right of AB

19. An astronomical telescope has an angualrmagnification of magnitude 5 for distant objects.The separation between the objective and theeye-piece is 36 cm and the final image is formed atinfinity. The focal length ƒ0 of the objective and feof the eye-piece are :(1) ƒ0 = 45 cm and ƒe = –9 cm(2) ƒ0 = 50 cm and ƒe = 10 cm(3) ƒ0 = 7.2 cm and ƒe = 5 cm(4) ƒ0 = 30 cm and ƒe = 6 cm

20. In a Young's double slit experiment the intensity at

a point where the path difference is 6?

( being the

wavelength of the light used) is I. If I0 denotes themaximum intensity, I/I0 is equal to :

(1)12

(2) 32

(3) 12

(4) 34

21. In the far field diffraction pattern of a single slit underpolychromatic illumination, the first minimum withthe wavelength 1 is found to be coincident withthe third maximum at 2. So :(1) 31 = 0.32 (2) 31 = 2

(3) 1 = 3.5 2 (4) 0.31 = 32

18. Qksdl nwjh 15 cm okys leryksÙky ySal ls 20 cm dhnwjh ij fcUnq vkdkj dh oLrq j[kh gSA ;fn lery lrgdks jtr ls ysfir dj fn;k tk;s rks izfrfcEc dk fuekZ.kgksxk

(1) AB ds cka;ha vksj 60 cm(2) AB ds cka;ha vksj 30 cm(3) AB ds cka;h vksj 12 cm(4) AB ds nka;h vksj 60 cm

19. nwjLFk fi.Mks gsrq [kxksyh; nwjn'khZ ds dks.kh; vko/kZu dkifjek.k 5 gSA vfHkn';d rFkk vfHkus=d ySal ds e/; nwjh36 cm gS rFkk vafre izfrfcEc vuUr ij curk gSAvfHkn';d ySal ,oa vfHkus=d ySal dh Qksdl nwfj;ksa ƒ0

,oe~ fe Øe'k% gksxh :

(1) ƒ0 = 45 cm rFkk ƒe = –9 cm(2) ƒ0 = 50 cm rFkk ƒe = 10 cm(3) ƒ0 = 7.2 cm rFkk ƒe = 5 cm(4) ƒ0 = 30 cm rFkk ƒe = 6 cm

20. ;ax f}fLyV iz;ksx esa ftl fcUnq ij iFkkUrj 6?gS ogk¡

rhozrk I gSA ( = iz;qDr izdk'k dh rjaXknS/;Z ) ;fn I0 vf/kdre rhozrk gS rks I/I0 dk eku gksxk :

(1)12

(2) 32

(3) 12

(4) 34

21. ,dy fLyV ds nwjLFk {ks= foorZu izk:i esa cgqo.khZ izdk'kdk iz;ksx djus ij 1 rjaxnS/;Z ls cuk izFke fufEu"BrjaxnS/;Z 2 ls cuh rrh; mfPp"B ls lEikrh gksrk gS rks:

(1) 31 = 0.32 (2) 31 = 2

(3) 1 = 3.5 2 (4) 0.31 = 32



22. A beam of light is incident on a glass plate at anangle of incidence 60º. The reflected ray is polarised.What is the angle of refraction when angle ofincidence is 45º :

(1) 1sin ( 3 / 2)? (2) 1cos ( 3 / 2)?

(3) 1sin (1 / 6)? (4) 1sin (1 / 3)?

23. A battery is connected between two points A andB on the circumference of a uniform conductingring of radius r and resistance R.One arc AB of thering subtends an angle at the centre. The valueof the magnetic induction at the centre due to thecurrent in the ring is :(1) Proportion to 2(180º – )(2) Inversely proportional to r(3) Zero only if = 180º(4) Zero for all values of

24. If a long cylindrical conductor carries a steadycurrent parallel to its length :(i) The electric field along the axis is zero.(ii) The magnetic field along the axis is zero.(iii) The magnetic field outside the conductor is zero.(iv) The electric field outside the conductor is zero.(1) (i) and (ii) (2) (iii) and (iv)(3) (ii) and (iv) (4) (i) and (iv)

25. A long straight wire along the z-axis carries a currentI in the negative z-direction. The magnitude of

vector field B?

at a point having coordinates (x, y)

in the z = 0 plane is :

(1) 02 2

I (y i x j)2 (x y )

? ?? ?

? ?

(2) 02 2

I (x i y j)2 (x y )

? ?? ?

? ?

(3) 02 2

I (x j y i )2 (x y )

? ?? ?

? ?

(4) 02 2

I (x i y j)2 (x y )

? ?? ?

? ?

22. dkap dh IysV ij 60º ds vkiru dks.k ij izdk'k iaqtvkifrr gksrk gSA ijkofrZr fdj.k /kzqfor gSA tc vkirudks.k 45º gks] rks viorZu dks.k dk eku gksxk :

(1) 1sin ( 3 / 2)? (2) 1cos ( 3 / 2)?

(3) 1sin (1 / 6)? (4) 1sin (1 / 3)?

23. fdlh le:i pkyd oy; dh ifjf/k ds fdUgh nks fcUnqvksaA ,oae~ B ds e/; cSVjh tksM+h tkrh gSA oy; dh f=T;kr ,oe~ izfrjks/k R gSA pki AB dsUnz ij dks.k vkUrfjrdjrk gSA oy; esa /kkjk ds dkj.k dsUnz ij pqEcdh; izsj.kdk eku :

(1) 2(180º – ) ds lekuqikrh(2) r ds O;qRØekuqikrh(3) 'kwU; dsoy ;fn = 180º(4) ds lHkh ekuksa gsrq 'kwU;

24. ,d lhs/ks yEcs csyukdkj /kkjkokgh pkyd esa yEckbZ dslekUrj LFkk;h /kkjk izokfgr gks jgh gS :(i) v{k ds vuqfn'k fo|qr {ks= 'kwU; gS(ii) v{k ds vuqfn'k pqEcdh; {ks= 'kwU; gS(iii) pkyd ds ckgj pqEcdh; {ks= 'kwU; gS(iv) pkyd ds ckgjk fo|qr {ks= 'kwU; gS(1) (i) rFkk (ii) (2) (iii) rFkk (iv)(3) (ii) rFkk (iv) (4) (i) rFkk (iv)

25. z-v{k ds vuqfn'k j[ks lh/ks yEcs rkj esa _.kkRed z-v{kds vuqfn'k /kkjk I izokfgr gks jgh gSA lery z = 0 esa

funsZ'kkad (x, y) okys fcUnq ij pqEcdh; {ks= lfn'k B?

gksxk :

(1) 02 2

I (y i x j)2 (x y )

? ?? ?

? ?

(2) 02 2

I (x i y j)2 (x y )

? ?? ?

? ?

(3) 02 2

I (x j y i )2 (x y )

? ?? ?

? ?

(4) 02 2

I (x i y j)2 (x y )

? ?? ?

? ?



26. An electron moves in a circular orbit with a uniformspeed v. It produces a magnetic field B at the centreof the circle. The radius of the circle is proportionalto :

(1)vB

(2) vB

(3) Bv

(4) Bv

27. In a coil of resistance 10 , the induced currentdeveloped by changing magnetic flux through it,is shown in fig as a function of time. The magnitudeof change in flux through the coil in weber is :

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

28. In order that the discahrge of an LCR circuit beoscillatory it should satisfy the condition :

(1)2

2

1 RLC 4L

? (2) 2

2

1 RLC 4L

?

(3) 2

2

1 RLC 4L

? (4) 2

2

1 R4LLC

?

29. 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

30. Figure (A) and (B) show two R-L circuits. Growth ofcurrent in the two circuits is as shown in fig. It canbe inferred that(assume equal V in both cases) :

(1) R1 > R2 (2) R1 < R2

(3) L1 > L2 (4) L1 < L2

26. oÙkkdkj d{kk esa ,d bysDVªkWu fu;r pky v ls xfr'khygksrk gSA ;g oÙk ds dsUnz ij pqEcdh; {ks= B mRiUu djrkgSA oÙk dh f=T;k lekuqikrh gksxh :

(1)vB

(2) vB

(3) Bv

(4) Bv

27. izfrjks/k 10 dh fdlh dq.Myh esa ¶yDl ifjorZu dsQyLo:i mRiUu izsfjr /kkjk dk le; ds lkis{k fu:i.kfd;k x;k gSA dq.Myh esa ls ¶yDl ds ifjorZu dkifjek.k ¼oscj½ esa gksxk :

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

28. fdlh LCR ifjiFk dk fujkos'k.k nksyukRed gks] bldsfy;s vko';d 'krZ gksxh :

(1)2

2

1 RLC 4L

? (2) 2

2

1 RLC 4L

?

(3) 2

2

1 RLC 4L

? (4) 2

2

1 R4LLC

?

29. 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

30. fp=ksa (A) rFkk (B) esa nks R-L ifjiFk n'kkZ;s x;s gSA nksuksaifjiFkksa esa /kkjk dh of) Hkh fp= esa n'kkZ;h x;h gSA ;gfu"d"kZ fudkyk tk ldrk gS fd (nksuksa voLFkkvksa esa V dksleku ekusa ) :

(1) R1 > R2 (2) R1 < R2

(3) L1 > L2 (4) L1 < L2



31. A choke coil has :(1) High inductance and low resistance(2) High resistance and low inductance(3) High inductance and high resistance(4) Low inductance and low resistance

32. 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

33. In the circuit of fig, the voltmeter reads 75 V. Valueof C is :

(1) 4 F (2) 2 F(3) 6 F (4) 10 F

34. 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

(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

31. pksd dq.Myh esa gksrk gS :(1) mPp izsjdRo rFkk vYi izfrjks/k(2) mPp izfrjks/k rFkk vYi izsjdRo(3) mPp izsjdRo rFkk mPp izfrjks/k(4) vYi izsjdRo rFkk vYi izfrjks/k

32. 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

33. n'kkZ;s x, ifjiFk esa oksYVehVj dk ikB~;kad 75 V gSA Cdk eku gksxk :

(1) 4 F (2) 2 F(3) 6 F (4) 10 F

34. 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 leku

le; rFkk leku LFkku ij mfPp"B vFkok fufEu"Beku izkIr djrs gS

(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



35. The kinetic energy of most energetic electronsemitted from a metallic surface is doubled when thewavelength of the incident radiation is changed from400 nm to 310 nm, the work function of the metal is(1) 0.9 ev (2) 1.7 eV(3) 2.2 eV (4) 3.1 eV

36. The potential difference that must be applied to stopthe fastest photoelectrons emitted by a nickelsurface, having work function 5.01 eV, whenultraviolet light of 200 nm falls on it, must be :(1) 1.2 V (2) 2.4 V(3) 1.6 V (4) 0.2 V

37. X-rays are incident normally on a crystal of latticeconstants 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

38. 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

39. 200 MeV of energy is obtained in the fission of onenucleus of 235U. A reactor is generating 1000 kWof power. The rate of nuclear fission in the reactoris :(1) 1000 (2) 2 × 108

(3) 3.125 × 1016 (4) 93140. In the transformation sequence represented by

A A 4 A 4 A 4Z Z 2 Z 1 Z 1X Y K K? ? ?

? ? ?? ? ?

the decays are in the order :(1) (2) (3) (4)

35. tc fdlh /kkrq ij vkifrr izdk'k dh rjaxnS/;Z 400 nmds LFkku ij 310 nm dj nh tk;s rks mRlftZr QksVksbysDVªkWu ¼lokZf/kd ÅtkZ okyk½ dh xfrt ÅtkZ nks xquhgks tkrh gSA /kkrq dk dk;ZQyu gksxk :(1) 0.9 ev (2) 1.7 eV(3) 2.2 eV (4) 3.1 eV

36. fudy /kkrq dh lrg dk dk;ZQyu 5.01 eV gSA bl ij200 nm rjaxnS/;Z dk ijkcSaxuh izdk'k vkifrr gks rksQksVks bysDVªkWuksa dk mRltZu jksdus gsrq vko';d vkjksfirfoHkokUrj gksxk :(1) 1.2 V (2) 2.4 V(3) 1.6 V (4) 0.2 V

37. tkyd fu;rkad 0.6 nm okys fØLVy ij X fdj.ksa yEcorvkifrr 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

38. 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

39. 235U ds ,d ukfHkd ds fo[k.Mu ds QyLo:i 200 MeVÅtkZ eqDr gksrh gSA ,d la;a= 1000 kW dh 'kfDrmRiUu djrk gSA la;a= esa ukfHkdh; fo[k.Mu dh nj gksxh

(1) 1000 (2) 2 × 108

(3) 3.125 × 1016 (4) 93140. n'kkZ; x;h :IkkUrj.k Ja[kyk

A A 4 A 4 A 4Z Z 2 Z 1 Z 1X Y K K? ? ?

? ? ?? ? ?

esa fo?kVu Øe'k% gksaxs &(1) (2) (3) (4)



41. MX and MY denote the atomic masses of the parentand the daughter nuclei repsectively in a radioactivedecay.. The Q-value for a decay is Q1 and thatfor a + decay is Q2. If me denotes the mass of anelectron, then which of the following statements iscorrect ?(1) Q1 = (Mx – My) c

2 and Q2 = (Mx – My – 2me) c2

(2) Q1 = (Mx – My) c2 and Q2 = (Mx – My) c

2

(3) Q1 = (Mx–My–2me) c2 and Q2 = (Mx–My+2me) c

2

(4) Q1 = (Mx–My+2me) c2 and Q2 = (Mx–My+2me) c

2

42. 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(4) 100 Hz in the dc output of half-wave and 50

Hz in the dc output of full wave rectifier43. The current flowing through the zener diode in

figure is-

500

10V I1

5V 1k

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

44. The diode used in the circuit shown in the figurehas a constant voltage drop of 0.5 V at all currentsand a maximum power rating of 100 milli watts. Whatshould be the value of the resistor R, connected inseries with the diode for obtaining maximum

current

R 0.5V

1.5V

(1) 1.5 (2) 5 (3) 6.67 (4) 200

41. MX rFkk MY fdlh jsfM;ks/kehZ {k; esa] Øe'k% fir ukfHkd,oe~ larfr ukfHkd ds ijek.kq nzO;eku gSA {k; gsrq Qeku Q1 ,oe~ + {k; gsrq Q2 gSA ;fn me bysDVªkWu dknzO;eku gks rks fuEu esa ls dkSulk dFku lR; gS ?

(1) Q1 = (Mx – My) c2 rFkk Q2 = (Mx – My – 2me) c

2

(2) Q1 = (Mx – My) c2 rFkk Q2 = (Mx – My) c

2

(3) Q1 = (Mx–My–2me) c2 rFkk Q2 = (Mx–My+2me) c

2

(4) Q1 = (Mx–My+2me) c2 rFkk Q2 = (Mx–My+2me) c

2

42. 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) v)Zrjax fn"Vdkjh ds dc esa 100 Hz rFkk iw.kZ rjax

fn"Vdkjh ds dc fuxZr esa 50 Hz43. 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

44. fp= esa n’kkZ;s ifjiFk esa mi;ksx esa fy;s Mk;ksM esa lHkh/kkjkvksa ij 0.5 V dk fu;r foHko iru gksrk gS rFkkvf/kdre 'kfDr 100 feyhokWV gSA vf/kdre /kkjk izkIrdjus ds fy;s Mk;ksM ls Js.kh Øe esa tqM+s izfrjks/k R dkeku D;k gksuk pkfg;s-

R 0.5V

1.5V

(1) 1.5 (2) 5 (3) 6.67 (4) 200



45. 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

46. 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–

47. 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

48. The appearance of colour in solid alkali metal halidesis generally due to(1) Schottky defect (2) Frenkel defect(3) Interstitial positions (4) F-centers

49. Which of the following has been arranged in order ofdecreasing freezing point?(1) 0.05 M KNO3 > 0.04 M CaCl2 > 0.140 M sugar

> 0.075 M CuSO4

(2) 0.04 M BaCl2 > 0.140 M sucrose > 0.075 MCuSO4 > 0.05 M KNO3

(3) 0.075 M CuSO4 > 0.140 M sucrose > 0.04 MBaCl2 > 0.05 M KNO3

(4) 0.075 M CuSO4> 0.05 M NaNO3> 0.140 Msucrose > 0.04 M BaCl2

45. 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

46. 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–

47. 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

48. Bksl {kkjh; /kkrq DyksjkbM esa jax dh mifLFkfr dk eq[;dkj.k gksrk gS(1) 'kkWVdh =qfV (2) Ýsady =qfV(3) vUrjdk'kh; LFkku (4) F-dsUnz

49. fuEu esa ls dkSulk fgekad dk ?kVrk gqvk Øe lgh gS

(1) 0.05 M KNO3 > 0.04 M CaCl2 > 0.140 M sugar> 0.075 M CuSO4

(2) 0.04 M BaCl2 > 0.140 M sucrose > 0.075 MCuSO4 > 0.05 M KNO3

(3) 0.075 M CuSO4 > 0.140 M sucrose > 0.04 MBaCl2 > 0.05 M KNO3

(4) 0.075 M CuSO4> 0.05 M NaNO3> 0.140 Msucrose > 0.04 M BaCl2



50. 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)

51. The Henry's law constant for the solubility of N2 gasin water is 1.0 × 105 atm. The mole fraction of N2 inair is 0.8. The amount of N2 from air dissolved in 10mol 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

52.

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

53. The sequence of ionic mobility in the aqueoussolution is

(1) K Na Rb Cs (2) Cs Rb K Na? ? ? ?? ? ?(3) Rb K Cs Na? ? ? ?? ? ?(4) Na K Rb Cs? ? ? ?? ? ?

54. A hydrogen electrodes is immersed in a solutionwith pH = 0 (HCl). By how much will the potential(reduction) change if an equivalent amount of

NaOH 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

50. 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)

51. N2 xSl dh ty esa foys;rk ds fy, gsujh fu;rkad dk eku1.0 × 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

52.

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

53. tyh; foy;u esa vk;fud xfr'khyrk dk Øe gksrk gS

(1) K Na Rb Cs (2) Cs Rb K Na? ? ? ?? ? ?(3) Rb K Cs Na? ? ? ?? ? ?(4) Na K Rb Cs? ? ? ?? ? ?

54. 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



55. 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

56. The initial rates of reaction3A+2B+C Products, at different initialconcentrations are given below :

Initial rate, Ms–1 [A]0, M [B]0, M [C]0, M 5.0 × 10–3 5.0 × 10–3 1.0 × 10–2 1.25 × 10–3

0.010 0.010 0.010 0.005

0.005 0.005 0.010 0.005

0.010 0.015 0.010 0.010

The order with respect to the ractants A, B and Care respectively(1) 3, 1, 0 (2) 3, 2, 1(3) 2, 1, 0 (4) 2, 2, 1

57. 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

58. 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

59. Which of the following is an example of associatedcolloid :(1) Protein + water(2) Soap + water(3) Rubber + benzene (4) As2O3 + Fe(OH)3

55. 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

56. vfHkfØ;k dh izkjfEHkd nj fofHkUu izkjfEHkd lkUnzrkvkasij uhps nh xbZ gSA3A+2B+C mRikn

izkjfEHkd nj, Ms–1 [A]0, M [B]0, M [C]0, M 5.0 × 10–3 5.0 × 10–3 1.0 × 10–2 1.25 × 10–3

0.010 0.010 0.010 0.005

0.005 0.005 0.010 0.005

0.010 0.015 0.010 0.010

vfHkdkjd A, B rFkk C ds lanHkZ esa dksfV Øe'k% gksrh gS

(1) 3, 1, 0 (2) 3, 2, 1(3) 2, 1, 0 (4) 2, 2, 1

57. 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

58. 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

59. fuEu esa ls dkSulk ,d mnkgj.k ,d laxqf.kr dkWyksbM dkgS :(1) izksVhu + ty(2) lkcqu + ty(3) jcj + cSathu(4) As2O3 + Fe(OH)3



60. 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

61. Mac Arthur - Forrest process is used for theextraction of(1) Aluminium (2) Copper(3) Silver (4) Tin

62. 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

63. For which of the following reaction G vs T plot inthe Ellingham's diagram's show negative slope :

(1) Mg + 12

O2 MgO

(2) 2Ag + 12

O2 Ag2

(3) C + 12

O2 CO

(4) CO + 12

O2 CO2

64. SO2 reacts with Cl2 in sunlight to form(1) Sulphuryl chloride (2) Sulphonyl chloride(3) Sulphur dioxide (4) None of these

65. 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

60. 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

61. eSd vkWFkZj - QkWjsLV izØe dk mi;ksx fdlds fu"d"kZ.k dsfy;s fd;k tkrk gS(1) ,Y;qfefu;e (2) dkWij(3) flYoj (4) fVu

62. 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

63. G rFkk T ds e/; ,yha?ke xzkQ dk <ky fuEu esa lsfdl vfHkfØ;k gsrq _.kkRed gksrk gS :

(1) Mg + 12

O2 MgO

(2) 2Ag + 12

O2 Ag2

(3) C + 12

O2 CO

(4) CO + 12

O2 CO2

64. SO2 lw;Z izdk'k esa Cl2 ds lkFk fØ;k djds cukrk gS(1) LkYQ;wfjy DyksjkbM (2) lYQksfuy DyksjkbM(3) lYQj MkbZvkWDlkbM (4) buesa ls dksbZ ugha

65. 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



66. ‘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

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

68. 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

69. 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

70. 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]

66. ‘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 ugha67. fuEu izØeksa }kjk /kkrqvksa dk fu"d"kZ.k ladqy fuekZ.k }kjk

gksrk gS –I : lkbukbM izØeII : ekWaM izØeIII : 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

68. ,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

69. 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

70. 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]



71. 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

72. Trichloroacetaldehyde (CCl3CHO) reacts withchlorobenzene in presence of sulphuric acid andproduces :

(1) Cl Cl

Cl

C

H

(2) Cl ClC

Cl

OH

(3) Cl ClCH

CCl3

(4) Cl ClC

CH Cl2

Cl

71. 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

72. lY¶;wfjd vEy dh mifLFkfr eas VªkbZDyksjks,lhVsfYMgkbM(CCl3CHO) DyksjkscSathu ds lkFk vfHkfØ;k djds D;kcukrk gS :

(1) Cl Cl

Cl

C

H

(2) Cl ClC

Cl

OH

(3) Cl ClCH

CCl3

(4) Cl ClC

CH Cl2

Cl



73.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 isvfHkfØ;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

74. fuEu vfHkfØ;k ds fy;s mRikn crkb;s :

CH3

CH3

OH

H+

?

(1)

CH3

CH3 (2) H C3 CH3

(3)

CH3

CH3

(4) H C3

H C3

75. 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

73.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

(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

74. Find the product of the given reaction :

CH3

CH3

OH

H+

?

(1)

CH3

CH3 (2) H C3 CH3

(3)

CH3

CH3

(4) H C3

H C3

75. 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



76. ? ?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

77. 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

OH

76. ? ?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

77.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



78. X Y + ZO3

Zn,HO2

OH CH=CH–C–O

Compound X in the above reaction is

(1) C=CH–CH3

(2) C=CHCH3

(3) CH–CH=CH2

(4) CH–CH=CH2

79. 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

78. X Y + ZO3

Zn,HO2

OH CH=CH–C–O

mijksDr vfHkfØ;k esa ;kSfxd X gS &

(1) C=CH–CH3

(2) C=CHCH3

(3) CH–CH=CH2

(4) CH–CH=CH2

79. 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



80.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

(1)

(2)

(3)

(4)

81. 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

82. Compound C H O4 10

No H gas evolved2

3-monochloroproducts

–Ve test

(i) Na metal

(ii) Cl2/h

(iii) Lucas reagent

Compound is

(1) O (2)O

(3) O (4) O83. 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

80.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

(1)

(2)

(3)

(4)

81. 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

82. Compound C H O4 10

No H gas evolved2

3-monochloroproducts

–Ve test

(i) Na metal

(ii) Cl2/h

(iii) Lucas reagent

;kSfxd gS &

(1) O (2)O

(3) O (4) O83. 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



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

85. 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

CH OH2

H

OH

H

OH

OH

H

O

H

CH OH2

HH

O

(3)

H

OH

H

H

H

OH

O

OH

CH OH2

OH

H

H

H

H

OH

O

OH

CH OH2HH

O

(4) H

OH

H

H

H

OH

O

OH

CH OH2

H

OH

H

OH

OH

H

O

H

H

O

H

CH OH2

86. Which of the following will show optical activity:

(A) (B)H

H

COOH

COOH

OH

HO

(C)H

H

CH CH2 3

CH CH2 3

OH

OH(D)

H

H

CH CH2 3

CH CH2 3HO

HO

(E) 50/50 mixture of C and D:(1) A, D, and E (2) A and E only(3) B, C and D (4) All except C

84. 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

85. 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

CH OH2

H

OH

H

OH

OH

H

O

H

CH OH2

HH

O

(3)

H

OH

H

H

H

OH

O

OH

CH OH2

OH

H

H

H

H

OH

O

OH

CH OH2HH

O

(4) H

OH

H

H

H

OH

O

OH

CH OH2

H

OH

H

OH

OH

H

O

H

H

O

H

CH OH2

86. fuEu eas ls dkSu izdkf'kd lfØ;rk n'kkZ;sxk :

(A) (B)H

H

COOH

COOH

OH

HO

(C)H

H

CH CH2 3

CH CH2 3

OH

OH(D)

H

H

CH CH2 3

CH CH2 3HO

HO

(E) C rFkk D dk 50/50 feJ.k(1) A, D, rFkk E (2) dsoy A rFkk E(3) B, C rFkk D (4) C dks NksM+dj lHkh



87. 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–

87. ;kSfxd

O 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–



88. The Keq. values in HCN addition to followingaldehydes are in the correct order :

O

H

MeO

O

H

(I) (II)

O

H

Me N2(III)

(1) I > II > III (2) II > III > I(3) III > I > II (4) II > I > III

89. 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)90. The decreasing order of stability of following anion is

:

CH2

NO2

CH2

CN

CH2

CH3

CH2

(P) (Q) (R) (S)

(1) S > R> Q > P

(2) P > Q > R > S

(3) Q > P > R > S

(4) P > Q > S > R

88. fuEufyf[kr ,fYMgkbMksa esa HCN ;ksx esa Keq. eku dklgh Øe gksxk :

O

H

MeO

O

H

(I) (II)

O

H

Me N2(III)

(1) I > II > III (2) II > III > I(3) III > I > II (4) II > I > III

89. 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)90. fuEufyf[kr _.kk;uksa ds LFkkf;Ro dk ?kVrk Øe gS

CH2

NO2

CH2

CN

CH2

CH3

CH2

(P) (Q) (R) (S)

(1) S > R> Q > P

(2) P > Q > R > S

(3) Q > P > R > S

(4) P > Q > S > R



91. When a diploid plant (male) is crossed with atetraploid (female), the ploidy of the endospermcells in the resulting seed is(1) Diploid (2) Triploid(3) Tetraploid (4) Pentaploid

92. Mendel observed that some characters did notassort independently. Later researchers found itto be due to(1) Crossing-over(2) Linkage(3) Dominance of one trait over the other(4) Amitosis

93. What would be the colour of flower in F1 progenyas a result of a cross between homozygous redand homozygous white-flowered 4 O' Clock plant

(1) Red (2) White(3) Red and white (4) Pink

94. Dichogamy favouring cross pollination is a typeof floral mechanism where(1) Anthers and stigma are placed at different levels(2) Stamens and stigma mature at different times

(3) Structure of anther and stigma act as barrier

(4) Pollen is unable to germinate on its own stigma95. In Angiosperms the functional megaspore in linear

tetrad is the(1) First nearest to the micropyle(2) Second from the micropyle(3) Third from the micropyle(4) Fourth from the micropyle

96. Of the following combinations of cell biologicalprocesses which one is associated withembryogenesis :(1) Mitosis and Meiosis(2) Mitosis and Differentiation(3) Meiosis and Differentiation(4) Differentiation and Reprogramming

91. tc f}xqf.kr (uj) ikS/ks dk prqxqZf.kr (eknk) ikS/ks lsladj.k fd;k tkrk gS rc cuus okys cht dh Hkwz.kiks"kdh dksf'kdkvksa esa lw=xq.krk gksxh(1) f}xqf.kr (2) f=xqf.kr(3) prqxqZf.kr (4) iapxqf.kr

92. es.My us ;g ik;k fd dqN y{k.k LorU= viO;wgu ughdjrsA mÙkjkdkyhu 'kks/kdrkZvksa us ik;k fd bldkdkj.k gS&(1) thu fofue;(2) lgyXurk(3) ,d y{k.k dk nwljs ij izHkkoh gksuk(4) vlw=h foHkktu

93. ;fn 4 O' DykWd ikni ds le;qXeth yky jax ds iq"iokys ikS/ks dk ladj.k le;qXeth lQsn jax ds iq"i okysikS/ks ls fd;k tk, rks izFke ih<h (F1) es fdl jax ds iq"igksxsa &(1) yky (2) lQsn(3) yky vkSj lQsn (4) xqykch

94. Mkbdksxseh] tks ijijkx.k dks lgk;rk djrh gS] ,diq"ih; fof/k gS ftlesa(1) ijkxdks'k vkSj ofrZdkxz fHkUu Lrjksa ij gksrs gSaA(2) iqadslj vkSj ofrZdkxz fHkUu le;ksa ij ifjiDo gksrs gSaA(3) ijkxdks'k vkSj ofrZdkxz dh lajpuk] vojks/k dk dke djrh gS(4) ijkxd.k vius gh ofrZdkxz ij vadqfjr ugha gks ikrk

95. vkorchft;ksa esa jSf[kd prq"d dk izdk;Zd xq:chtk.kqgksrk gS(1) chtk.M}kj ls fudV okyk izFke(2) chtk.M}kj ls nwljk(3) chtk.M}kj ls rhljk(4) chtk.M}kj ls pkSFkk

96. fuEufyf[kr dksf'kdk dh tSfod izfØ;kvksa ds leqPp;esa ls dkSulh izfØ;k Hkwz.kksn~Hkou ls lEcfU/kr gS :

(1) lelw=h foHkktu rFkk v)Zlw=h foHkktu(2) lelw=h foHkktu rFkk foHksnu(3) v)Zlw=h foHkktu rFkk foHksnu(4) foHksnu rFkk iquZ;kstu



97. Perisperm is the remains of(1) pericycle (2) perianth(3) endosperm (4) nucellus

98. Which floral features shown below can be puttogether to represent an anemophilous flower ?

(1) 1, 3 and 6 (2) 2, 4 and 6(3) 2, 3 and 5 (4) 1, 4 and 5

99. Test cross in plants or in Drosophilia involvescrossing(1) Crossing the F1 hybrid with a double recessive

genotype(2) Crossing between two genotypes with

dominant trait(3) Crossing between two genotypes with

recessive trait(4) Crossing between two F1 hybrids

100. Identify the correct number of gametes,genotypes and phenotypes respectively,produced in the self pollination of plant withgenotype of AaBBCCDdEE.(1) 3, 8, 3 (2) 4, 9, 4(3) 4, 6, 8 (4) 4, 8, 4

101. In a linear chromosome, map distance (in cM)between four gene loci are as follows :a – b = 10 cM b – c = 4 cMa – d = 3 cM a – c = 6 cMThe expected cross-over frequency between thegenes c and d is :(1) 9% (2) 3%(3) 5% or 7% (4) 3% or 9%

97. ifjHkwz.kiks"k cpk gqvk(1) ifjjaHk (2) ifjnyiaqt(3) Hkwz.kiks"k (4) chtk.Mdk;

98. uhps fn;s x;s iq"ih; y{k.kksa esa ls dkSuls y{k.k ok;qijkfxr iq"iksa }kjk iznf'kZr gksrs gS ?

(1) 1, 3 rFkk 6 (2) 2, 4 rFkk 6(3) 2, 3 rFkk 5 (4) 1, 4 rFkk 5

99. ikniksa esa ;k MªkslksfQyk esa ijh{k.k ØkWl eas lfEefyrgksrk gS(1) f}d vizHkkoh thuksVkbi ds lkFk Fi ladj dh

ØkWflax(2) izHkkoh y{k.kksa okys nks thuksVkbiksa ds e/; ØkWflax

(3) vizHkkoh y{k.kksa okys lkFk nks thuksVkbiksa d e/;ØkWflax

(4) nks Fi ladjks ads e/; Økflax100. Loijkx.k djus okys ikni ftldk thuizk:i

AaBBCCDdEE gS }kjk mRiUu fd;s x;s ;qXed] thuizk:i rFkk y{k.kizk:i dh lgh la[;k Øe'k% igpkfu;s

(1) 3, 8, 3 (2) 4, 9, 4(3) 4, 6, 8 (4) 4, 8, 4

101. ,d js[kh; xq.klw= ij fLFkr pkj thu LFkyksa ds chp nwjh(cM esa) nh xbZ gS :a – b = 10 cM b – c = 4 cMa – d = 3 cM a – c = 6 cMc rFkk d thuksa ds chp vkisf{kr ØkWflx&vksoj vkorhZgksxh :(1) 9% (2) 3%(3) 5% ;k 7% (4) 3% ;k 9%



102. If ‘K’ is the carrying capacity of the habitat, ‘N’ isthe total number of individuals in a populationwith intrinsic rate of reproduction as ‘r’, the growthrate of such a population will be directlyproportional to:I. r II. 1/NIII. 1/K+N IV. (K+N)/K(1) I and II (2) II and III(3) I and IV (4) Only I

103. Which of the following is considered as a recessivecharacter of Mendel(1) Round seed(2) Inflated pod(3) Terminal flower(4) Green pod

104. The best example of polyembryony is(1) Cocos (2) Citrus(3) Capsicum (4) All of these

105. Mathc the items of column I with column II andselect the correct optionColumn - I Column - IIA. ETS 1. Removes gases like

SO2

B. Scrubber 2. Reduces automobileemission

C. Catalytic converter 3. Removes particulatematter

4. Respiratory chain(1) A - 4, B - 3, C -1 (2) A - 3, B - 2, C - 1(3) A - 1, B - 2, C -4 (4) A - 4, B - 1, C - 2

106. Find out the correct order of succession levels inXerarch(1) Lichen, Annual herb stage, Perennial Scrub

Stage, Perennial herb stage, moss stage(2) Annual herb stage, Perennial herb stage,

Lichen, moss stage, Scrub Stage, Forest(3) Lichen, moss stage, Annual herb stage,

Perennial herb stage, Scrub stage, Forest(4) Scrub stage, Forest, Annual herb stage,

Lichen, moss stage, scrub stage

102. ;fn ‘K’ vkokl dh ogu {kerk gS, ‘N’ lef"V esa O;f"V;ksadh dqy la[;k gS rFkk iztuu dh bUVªsfUld nj ‘r’ gS, rkslef"V dh of) nj lh/ks lekuqikrh gksxh :

I. r II. 1/NIII. 1/K+N IV. (K+N)/K(1) I rFkk II (2) II rFkk III(3) I rFkk IV (4) dsoy I

103. fuEu es ls fdls esaMy dk vizHkkoh y{k.k ekuk tkrk gS&

(1) xksy cht(2) Qwyh gqbZ Qyh(3) varLFk Qwy(4) gjh Qyh

104. cgqHkwz.krk dk lcls vPNk mnkgj.k gS(1) dksdksl (2) flVªl(3) dSfIlde (4) mi;qDr lHkh

105. dkWye I rFkk dkWye II dks lqesfyr dj lgh ;qXe dkp;u dhft,Column - I Column - IIA. ETS 1. xSlsa tSls SO2 dks

gVkukB. ektZd 2. vkWVkseksckby mRltZu

dks de djukC. mRizsjd ifjorZd 3. df.kdke; inkFkksZa dks

gVkuk4. 'oluh; Üka[kyk

(1) A - 4, B - 3, C -1 (2) A - 3, B - 2, C - 1(3) A - 1, B - 2, C -4 (4) A - 4, B - 1, C - 2

106. e:Øed esa] vuqØe.k Lrj dk lgh Øe dkSulk gS

(1) ykbdsu] okf"kZd 'kkd voLFkk] cgqo"khZ; >kM+hvoLFkk] cgqo"khZ; 'kkd voLFkk] ekWl voLFkk

(2) okf"kZd 'kkd voLFkk] cgqo"khZ; 'kkd voLFkk] ykbdsu]ekWl voLFkk] >kM+h voLFkk] ou

(3) ykbdsu] ekWl voLFkk] okf"kZd 'kkd voLFkk] cgqo"khZ;'kkd voLFkk] >kM+h voLFkk] ou

(4) >kM+h voLFkk] ou] okf"kZd 'kkd voLFkk] ykbdsu]ekWl voLFkk] >kM+h voLFkk



107 A biologist studied the population of rats in a barn.He found that the average natality was 250,average mortality 240, immigration 30 andemigration 30. The net increase in population is :(1) 05 (2) Zero(3) 10 (4) 15

108. Match the microbes in column I with theircommercial/industrial products in column II andchoose the correct answerColumn - I Column - IIA. Aspergillus niger 1. EthanolB. Clostridium 2. Statins

butylicumC. Saccharomyces 3. Citric acid

cerevisiaeD. Trichoderma 4. Butyric acid

polysporumE. Monascus 5. Cyclosporin A

purpureus(1) A-4, B-5, C-2, D-1, E-3(2) A-5, B-4, C-1, D-2, E-3(3) A-3, B-4, C-1, D-5, E-2(4) A-3, B-4, C-5, D-1, E-2

109. Which of the following enzymes are used formanufacturing detergents(1) Proteases (2) Glucoamylases(3) Amylases (4) Lactases

110. Excessive accumulation of organic matter in waterbodies leads to(1) Decrease in species diversity(2) Increase in species diversity(3) Green house effect(4) No effect on species diversity

111. Curding of milk takes place by(1) Streptococcus lactis(2) Streptococcus thermophilus(3) Lactobacillus lactis(4) All the above

112. Germinating barley seeds are employed in thepreparation of(1) Cheese (2) Wine(3) Beer (4) Toddy

107. ,d tho foKkuh }kjk [kfygku esa pwgksa dh tula[;k ijv/;;u fd;k x;kA mlus ik;k fd vkSlr tUenj250, vkSlr eR;qnj 240 gS, vkizoklu 30 rFkk mRizoklu30 gSA bl tula[;k esa usV of) gS :(1) 05 (2) 'kwU;(3) 10 (4) 15

108. dkWye I esa fn, x, lw{ethoksa dks dkWye II esa fn, x;smuds O;olkf;d@vkS|ksfxd mRiknksa ls feyku dhft,rFkk lgh mrj dk p;u dhft,Column - I Column - IIA. Aspergillus niger 1. EthanolB. Clostridium 2. Statins

butylicumC. Saccharomyces 3. Citric acid

cerevisiaeD. Trichoderma 4. Butyric acid

polysporumE. Monascus 5. Cyclosporin A

purpureus(1) A-4, B-5, C-2, D-1, E-3(2) A-5, B-4, C-1, D-2, E-3(3) A-3, B-4, C-1, D-5, E-2(4) A-3, B-4, C-5, D-1, E-2

109. viektZd ds fuekZ.k ds fy, dkSu ls ,Utkbe iz;qä gksrsgSa(1) izksfV;st (2) Xywdks,ekbystst(3) ,ekbyst (4) ySDVstst

110. ty&lajpukvksa esa dkcZfud inkFkZ ds vR;f/kd tekodk izHkko gksrk gS(1) tkfr fofo/krk esa deh(2) tkfr fofo/krk esa of)(3) gfjr xg izHkko(4) tkfr fofo/krk ij dksbZ izHkko ugha

111. nw/k dk ngh esa ifjorZu fdlds izHkko ls gksrk gS(1) LVªsIVksdkWdl ysfDVl(2) LVªsIVksdkWdl FkeksZfQyl(3) ySDVkscslhyl ysfDVl(4) mijksä lHkh

112. vadqj.k'khy tkS ds chtksa dks fdlds fuekZ.k esa iz;qäfd;k tkrk gS(1) iuhj (2) 'kjkc(3) ch;j (4) rkM+h



113. Biomagnification of DDT caused decline in Birdpopulation by(1) Bringing diturbance in calcium Metabolism(2) Thinning of egg shell(3) Premature breaking of eggs(4) All of the above

114. In the following which statement is/are not truea. The entire collection (of plants/seeds) having

all the diverse alleles for all genes in a givencrop is called germplasm collection.

b. Agriculture accounts for approximately 45 percent of India’s GDP and employs nearly 80per cent of the population.

c. Semi-dwarf rice varieties were derived fromIR-8, (developed at Central Rice ResearchInstitute , Philippines) and Taichung Native-1(from Japan).

d. In mung bean, resistance to yellow mosaicvirus and powdery mildew were induced bymutations.

Options(1) a & b (2) c & d(3) b & c (4) b,c & d

115. In the following which is produced by distillationof the fermented broth(1) Whisky (2) Brandy(3) Rum (4) All

116. The factor governing the structure of earthsurface is(1) Topographic(2) Edaphic(3) Biotic(4) Temperature

117. In ecological succession from pioneer to climaxcommunity, the biomas shall(1) Decrease(2) Increase and then decrease(3) No relation(4) Increase continuously

113. Mh Mh Vh dk tSo vko/kZu if{k;ksa dh la[;k esa deh ykrkgS(1) dsfY'k;e ds mikip; esa O;o/kku }kjk(2) vaMks ds [kksy ds irys gks tkus ds dkj.k(3) vaMks ds ifjiDou ds iwoZ gh VwVus ds dkj.k(4) mijksä lHkh ds dkj.k

114. fuEu esa ls dkSulk dFku lgh ugha gS \a. fdlh Qly esa ik;s tkus okys lHkh thuksa ds fofo/k

vyhy dk leLr laxzg.k (ikniks@chtks) dks mldktuunzO; (teZIykTe) laxzg.k dgrs gSA

b. Hkkjr ds (GDP) dh yxHkx 45 izfr'kr vk; rFkklef"B fd yxHkx 80% izfr'kr turk dks jkstxkjÑf"k ls izkIr gksrk gSSA

c. v)Z okeu /kku fd fdLeks dks I R–8 (lSVªay jkbZlfjlpZ bUlVhV~;wV] fQyhihal esa iSnk) rFkk FkkbpwaxusfVo–1 (tkiku ls) ls O;qRiUu fd;k x;k gS

d. ew¡x (chu) es ihr ekstsd ok;jl rFkk pw.khZy vklhrkds izfr izfrjks/kd {kerk mRifjorZu }kjk izsfjr gksrhFkh

Options(1) a & b (2) c & d(3) b & c (4) b,c & d

115. fuEufyf[kr es ls fdldk mRikn vklfor fd.ou jlls izkIr gkskr gSA(1) OghLdh (2) czkaMh(3) je (4) lHkh

116. iFoh dh lrg dh lajpuk dks fu/kkZfjr djus okykdkjd gS(1) VksiksxzkfQd(2) bMsfQd(3) ck;ksfVd(4) rkieku

117. bdksyksWftdy vuqØe.k esa iqjks/kk ls pje leqnk; esastSoHkkj gksrk gS(1) de(2) vf/kd rFkk ckn esa de(3) dksbZ lEcU/k ugha(4) yxkrkj c<+rk gS



118. Which of the following ecosystem has very littleprimary productivity(1) Forest (2) River(3) Sea (4) Grass Land

119. In the following which crop variety of Okra isresistant for insect pest(1) Pusa Sawami (2) Pusa Gaurav(3) Pusa Snowball K - 1 (4) Pusa Sadabahar

120. Average primary production (g/m2/yr) of thefollowing has been depicted in the accompanyingdiagram.

i. Tropical rain forestsii. Tundraiii. Swampsiv. Taigav. Coral reefMatch these with the columns from 1 to 5.(1) v, i, iii, iv, ii (2) i, v, iv, iii, ii(3) ii, iv, iii, v, i (4) iv, i, v, iii, ii

121. Match the animals given in column A with theirlocation in column B:Column A Column B(i)Dodo (a) Africa(ii) Quagga (b) Russia(iii) Thylacine (c) Mauritius(iv)Stellar's sea cow (d)AustraliaChoose the correct match from the following:(1) i-a, ii-c, iii-b, iv-d(2) i-d, ii-c, iii-a, iv-b(3) i-c, ii-a, iii-b, iv-d(4) i-c, ii-a, iii-d, iv-b

118. fdl bdksflLVe dh izkFkfed mRikndrk lcls degksrh gS(1) ou (2) unh(3) leqnz (4) ?kkl ds eSanku

119. fuEufyf[kr es ls dkSulh vksdjk fdLe Qly dhVksa dsizfrjks/kd gksrh gS&(1) iwlk Lokeh (2) iwjk xkSjo(3) iwlk LuksckWy K-1 (4) iwlk lnkcgkj

120. uhps fn;s x;s fp= esa fuEufyf[kr ds fy;s vkSlr izkFkfedmRiknu (g/m2/yr) esa fn;s x;s gS

i. m".kdfVca/kh; o"kkZouii. Vq.Mªkiii. vuqiiv. Vsxkv. dksjy jhQbudk feyku LrEHk 1 ls 5 ls dhft;sA(1) v, i, iii, iv, ii (2) i, v, iv, iii, ii(3) ii, iv, iii, v, i (4) iv, i, v, iii, ii

121. dkWye A esa fn;s x;s tUrqvksa dh feyku dkWye B esamuds LFkkuksa (fuokl) ls djksa %&Column A Column B(i) MksMks (a) vÝhdk(ii) Dosxk (b) :l(iii) Fkk;ysflu (c) ekWfjf'k;l(iv) LVhyj leqnzh xk; (d) vkLVªsfy;kfuEu esa ls lgh feyku pqusa(1) i-a, ii-c, iii-b, iv-d(2) i-d, ii-c, iii-a, iv-b(3) i-c, ii-a, iii-b, iv-d(4) i-c, ii-a, iii-d, iv-b



122. Given below are a few statements regardingsomatic hybridization.Choose the correct statements.i) Protoplasts of different cells of the same plant

are fusedii) Protoplasts from cells of different species can

be fusediii) Treatment of cells with cellulase and pectinase

is mandatoryiv) The hybrid protoplast contains characters of

only one parental protoplast(1) (ii) and (iv) (2) (i) and (ii)(3) (iii) and (i) (4) (ii) and (iii)

123. Which of the following statements, support theview that elaborate sexual reproductive processappeared much later in the organic evolution.i. Lower groups of organisms have simpler body

designii. Asexual reproduction is common in lower

groupsiii. Asexual reproduction is common in higher

groups of organismsiv. The high incidence of sexual reproduction in

angiosperms and vertebratesChoose the correct answer from the options givenbelow:(1) i and iii (2) i and iii(3) ii and iv (4) ii and iii

124. Choose the incorrect statement.(1) The Montreal protocol is associated with the

control of emission of ozone depletingsubstances

(2) Methane and carbon dioxide are green housegases

(3) Dobson units are used to measure oxygencontent

(4) Use of incinerators is crucial to disposal ofhospital wastes

122. uhps dkf;d ladj.k ls lEcfU/kr dqN dFku fn;s x;sgSA lgh dFkuksa dks pqusa

(i) ,d gh ikni dh fHkUu dksf'kdkvksa ds izksVksIykLVdk lay;u gksrk gSA

(ii) fHkUu tkfr;ksa dh dksf'kdkvksa ls fy, x, izksVksIykLVdks layf;r fd;k tk ldrk gSA

(iii) dksf'kdkvksa dk lsY;qyst rFkk isfDVust ls mipkjvfuok;Z gSA

(iv)ladj izksVksIykLV esa dsoy ,d gh iSrd izksVksIykLVds y{k.k gksrs gSA

(1) (ii) and (iv) (2) (i) and (ii)(3) (iii) and (i) (4) (ii) and (iii)

123. fuEu esa ls dkSulk dFku bl ckr dk leFkZu djrk gSfd ogn~ ySafxd tuu izfØ;k dkcZfud fodkl esa cgqrle; i'pkr~ izdV gqbZ gSA(i) fuEu oxZ ds tUrqvksa esa 'kkjhfjd lajpuk ljy gksrh

gSA(ii) fuEu oxZ ds tUrqvksa esa vySafxd tuu lkekU; gSA

(iii) mPp oxZ ds tUrqvksa esa vySafxd tuu lkekU; gSA

(iv),fUt;ksLieZ rFkk d'ks:fd;ksa esa ySafxd tuu dkvf/kd ik;k tkuk

uhps fn, x, fodYiksa esa ls lgh mrj pqusa %&

(1) i and iii (2) i and iii(3) ii and iv (4) ii and iii

124. xyr dFku pqusa %&(1) ekWfVª;y izksVksdkWy vkstksu vo{k;dkjh inkFkksZ ds

mRltZu ij fu;a=.k ls lEcfU/kr gSA

(2) feFksu rFkk CO2 xzhugkml xSls gSA

(3) Mksclu bdkb;ksa dk mi;ksx vkWDlhtu ek=k Kkrdjus esa fd;k tkrk gSA

(4) vLirky vof'k"Vksa ds fuLrkj.k ds fy, HkLed dkiz;skx vR;ar vko';d gSA



125. What is common to the techniques (i) in vitrofertilisation, (ii) Cryo preservation and (iii) tissueculture ?(1) All are in situ conservation methods.(2) All are ex situ conservation methods.(3) All require ultra modern equipment and large

space.(4) All are methods of conservation of extinct

organisms.126. Which of the following statements is correct ?

(1) Parthenium is an endemic species of ourcountry.

(2) African catfish is not a threat to indigenouscatfishes.

(3) Steller’s sea cow is an extinct animal.(4) Lantana is popularly known as carrot grass.

127. Formation of tropical forests needs mean annualtemperature and mean annual precipitation as :

(1) 18 – 25°C and 1500 – 4000 mm(2) 5 – 15°C and 50 – 100 cm(3) 30 – 50°C and 100 – 150 mm(4) 5 – 15°C and 100 – 200 cm

128. Approximately how much of the solar energy thatfalls on the leaves of a plant is converted tochemical energy by photosynthesis?(1) Less than 1% (2) 2-10%(3) 30% (4) 50%

129. Which of the following have a positive influenceon the rate of speciation?i. Species richness of habitatii. Heterogeneity of habitatsiii. Long generation time(1) (i), (ii) and (iii)(2) Only i and ii(3) Only i and iii(4) Only ii and iii

125. bu rduhdksa esa D;k lekurk gS] (i) bu&foVªks fu"kspu (ii)'khr laj{k.k (iii) mrd lao/kZu

(1) ;s lHkh Lo&LFkkus laj{k.k iz.kkyh;k¡ gSA(2) ;s lHkh mR&LFkkus laj{k.k iz.kkyh;k¡ gSA(3) lHkh ds fy, ijk vk/kqfud midj.k rFkk cM+s LFkku

dh vko';drk gksrh gSA(4) lHkh foyqIr thoks ds laj{k.k dh fof/k;k¡ gSSA

126. fuEu esa ls dkSulk dFku lR; gS \(1) ikfFkZfu;e gekjs ns'k dh LFkkfud tkfr gSA

(2) vÝhdu dsVfQ'k ns'kh dsVfQ'k ds fy, [krjk ughagSA

(3) LVsyj leqnzh xk; ,d foyqIr tUrq gSA(4) ysUVkuk dks izfl) :i ls xktj ?kkl dgk tkrk

gSA127. m".kdfVca/kh; ouks ds fuekZ.k esa fdruh vkSlr okf"kZd

rkieku rFkk vkSlr okf"kZd o"kZ.k dh vko';drk gksrhgS %&(1) 18 – 25°C rFkk 1500 – 4000 feeh(2) 5 – 15°C rFkk 50 – 100 cm lseh(3) 30 – 50°C rFkk 100 – 150 feeh(4) 5 – 15°C rFkk 100 – 200 lseh

128. ikni dh ifr;ksa ij fxjus okyh lw;Z ÅtkZ dk yxHkxfdruk Hkkx izdk'k la'ys"k.k }kjk jlk;fud ÅtkZ esacny tkrh gS \(1) 1% ls de (2) 2-10%

` (3) 30% (4) 50%129. fuEufyf[kr eas ls dkSu tkfr mn~Hkou dh nj dks

/kukRed :i ls izHkkfor djrs gS ?i. vkokl esa tkfr;ksa dh vf/kdrkii. vkokl esa fo"kekaxh; voLFkkiii. yEck ih<+h dky(1) (i), (ii) rFkk (iii)(2) dsoy i rFkk ii(3) dsoy i rFkk iii(4) dsoy ii rFkk iii



130. Two nearby seashores A and B were comparedusing dominance and diversity indices. It wasfound that the diversity index of A was betterthan that of B while the dominance index of Bwas better than that of A. Which of the followingcan be true?(1) Eutrophication has occurred at A.(2) Eutrophication has occurred at B.(3) Habitat loss could be a problem at A.(4) Indicator species are present at B.

131. Given below is a flow chart showing the effect ofsewage discharge on some importantcharacteristics of a river. Read carefully andidentify A, B, C and D

A

B

C

D

(1) A–BOD, B–Dissolved oxygen,C–Concentration, D–Direction of flow

(2) A–Dissolved oxygen, B–BOD, C–Directionof flow, D–Concentration

(3) A–Dissolved oxygen, B–BOD,C–Concentration, D–Direction of flow

(4) A–BOD, B–Diossolved oxygen, C–Directionof flow, D–Concentration

130. nks fudV fLFkr leqanzh rVksa A rFkk B dks izHkkfork rFkkfofo/krk lwpdkad ds vk/kkj ij rqyuk dh tkrh gSA ;gik;k x;k fd A dk fofo/krk lwpdkad B ls csgrj gStcfd B dk izHkkfork lwpdkad A ls csgrj gSA fuEufyf[kreas ls lR; gS ?

(1) A esa lqiks"k.k gksrk gSA(2) B esa lqiks"k.k gksrk gSA(3) A esa vkoklh; {kfr eq[; leL;k gSA(4) lwpd tkfr;k¡ B esa mifLFkr gSA

131. uhps fn;k x;k izokg fp= unh ds dqN egRoiw.kZ y{k.kksaij okfgr ey foltZu dk izHkko n'kkZrk gSA v/;;udjds A, B, C rFkk D dks igpkfu,

A

B

C

D

(1) A–BOD, B–? k qy h g qb Z vk W Dlhtu ,C–lkaUnzrk, D–izokg fn'kk

(2) A–?kqyh gqbZ vkWDlhtu, B–BOD, C–izokg fn'kk,D–lkanzrk

(3) A–? k qy h g qb Z vk W Dlhtu , B–BOD,C–lkUnzrk, D–izokg fn'kk

(4) A–BOD, B–?kqyh gqbZ vkWDlhtu, C–izokg fn'kk,D–lkUnzrk



132. How many of following represent in-situ (I) andex-situ (E) conservation strategies ?Sacred groves, Wildlife sanctuary, Biospherereserve, Home gardens, Seed bank, Genebank, National park, Aquaria, Arboreta,Sacred lakes, Botanical garden, Zoologicalpark.(1) I = 5, E = 5 (2) I = 5, E = 7(3) I = 3, E = 7 (4) I = 8, E = 2

133. Suppose 2000 J of solar energy is incident ongreen vegetation. On the basis of law of LindemanIdentify A, B and C

(1) A–20J, B–2 J, C–0.2 J(2) A–200 J, B–20 J, C–2J(3) A–400 J, B–40J, C–4 J(4) A–40J, B–4J, C–0.4 J

134. In figure find out coleoptile, shoot apex andepiblast

(1) A, B and C (2) B, C and D(3) D, F and G (4) E, F and G

132. buesa ls fdrus LoLFkkus (I) rFkk cfgLFkkus laj{k.k (E)fof/k dks iznf'kZr djrk gS ?ifo= m|ku , oU;tho vH;kj.;, lajf{krtSoe.My, x`g m|ku, cht cSad, thu cSad, jk"Vªh;m|ku, bDosfj;k, vkWcksZjsVk, ifo= >hy, okUkLifrdm|ku, tUrq m|ku

(1) I = 5, E = 5 (2) I = 5, E = 7(3) I = 3, E = 7 (4) I = 8, E = 2

133. ekuk gjs ikniks ij vkifrr lkSj ÅtkZ 2000 J gSAfy.MeSu ds fu;e ds vk/kkj ij A, B rFkk C igpkfu,

(1) A–20J, B–2 J, C–0.2 J(2) A–200 J, B–20 J, C–2J(3) A–400 J, B–40J, C–4 J(4) A–40J, B–4J, C–0.4 J

134. fp= esa izkadqj pksy] izjksg 'kh"kZ rFkk vf/kdksjd crkb,

(1) A, B rFkk C (2) B, C rFkk D(3) D, F rFkk G (4) E, F rFkk G



135. Which one of the following is a correct matchingpair of certain organism(s) and the kind ofassociation(1) Shark and sucker fish – Commensalism(2) Red algae and fungi in lichens – Mutualism(3) Orchids growing on trees – Parasitism(4) Cuscuta growing – Epiphytes on other flowering plants

136. A method of birth control is :(1) GIFT (2) ZIFT(3) IVF-ET (4) IUDs

137. Which one of the following tRNA will carrymethionine :

(1)

A|C|CG

U A C

(2)

A|C|CG

T A C

(3)

A|C|CG

C A U

(4)

A|C|CG

A U G

138. The layer type breeds of fowls are called :(1) White leghorn(2) New Hempshire(3) Plymouth rock(4) Rhodesian Red

139. Ballworm resistant gene of cotton is :(1) cry I Ac(2) cry II Ab(3) cry I Ab(4) cry I Ac & cry II Ab

135. fuEufyf[kr esa ls thoks rFkk muds var% fØ;kvksa dklgh ;qXe gS

(1) lkdZ rFkk pq"kd eNyh – lgHkksftrk(2) ykbdsu eas yky 'kSoky rFkk dod – lgksidkfjrk(3) o{kks ij mxus okys vkfdZM – ijthfork(4) vU; iq"ih; ikniks ij mxus okyk dLdqVk – vf/kikni

136. tUe fu;a=.k dh ,d fof/k gS :(1) GIFT (2) ZIFT(3) IVF-ET (4) IUDs

137. fuEu esa dkSulk tRNA feF;ksfuu dk ogu djsxk :

(1)

A|C|CG

U A C

(2)

A|C|CG

T A C

(3)

A|C|CG

C A U

(4)

A|C|CG

A U G

138. v.Ms nsus okyh izpfyr eqxsZ dh uLy dgykrhs gS :(1) lQsn ysxgkWuZ(2) U;w gsEi'kk;j(3) IykbZekmFk jkWd(4) jksMsf'k;u jsM

139. fuEu es ls dikl dk ckWyoeZ izfrjks/kh thu gS :(1) cry I Ac(2) cry II Ab(3) cry I Ab(4) cry I Ac & cry II Ab



140. Match the columns :Types of Innate

Immunity ExamplesP. Physical barriers (i) Acid in stomachQ. Physiological barriers (ii) urogenital tractsR. Cellular barriers (iii) InterferonsS. cytokine barriers (iv) leuocytes(1) P - (iv), Q - (iii), R - (i), S - (ii)(2) P - (ii), Q - (i), R - (iv), S - (iii)(3) P - (iii), Q - (i), R - (ii), S - (iv)(4) P - (i), Q - (ii), R - (iii), S - (iv)

141. Blood cancer is :(1) Leukemia(2) Cat cry(3) Philadelphia(4) (1) and (3) both

142. Choose correct sequence of evolution of man ?(1) Java man, Neanderthal man, Homo sapians

sapians, Cromagnon man(2) Heidlberg man, Neanderthal man, Cromagnon

man, Homo sapians sapians(3) Java man, Ramapithecus, Cromagnon man,

Homo sapians sapians(4) Ramapithecus, Heidelberg man,

australopithecus, cromagnon man, Homosapians sapians

143. Transfer RNA :(1) Forms hydrogen bonds between its codon and

the anticodon of an m-RNA in the 'A' site of aribosome

(2) Bind specific amino acid by the help of aminoacylt-RNA synthetase

(3) Uses GTP as the energy source to bind itsamino acid

(4) It translated from m-RNA

140. dkWye feyk, :tUetkr izfrj{kkdk izdkj mnkgj.kP. HkkSfrd vojks/k (i) vkek'k; esa vEyQ. dkf;Zdh; vojks/k (ii) ew= tuu ekxZR. dksf'kdh; vojks/k (iii) bUVjQsjkWuS. lkbVksdkbu vojks/k (iv) Y;qdkslkbV(1) P - (iv), Q - (iii), R - (i), S - (ii)(2) P - (ii), Q - (i), R - (iv), S - (iii)(3) P - (iii), Q - (i), R - (ii), S - (iv)(4) P - (i), Q - (ii), R - (iii), S - (iv)

141. jDr dsUlj gS**(1) Y;wdsfe;k(2) etkZj ØUnu(3) fQykMsyfQ;k(4) (1) rFkk (3) nksuksa

142. ekuo ds fodkl ds lgh Øe dk p;u djks :(1) tkok ekuo] fu,UMjFky ekuo] gksekslsfi;Ul lsfi;Ul]

ØksesXuu ekuo(2) ghMycxZ ekuo] fu,UMjFky ekuo] ØksesXuu ekuo]

gksekslsfi;Ul lsfi;Ul(3) tkok ekuo] jkekfifFkdl] Øk sesXuu ekuo]

gksekslksfi;Ul lsfi;Ul(4) jkekfifFkdl] fgMycxZ ekuo] vkfLVªyksfifFkdl]

ØksesXuu ekuo] gksekslsfi;Ul lsfi;Ul

143. LFkkukUrj.k RNA -(1) jkbckslkse dh 'A' site esa vius dksMksu ,oe m-RNA

ds ,UVhdksMksu ds e/; gkbMªkstu ca/k cukrk gS

(2) ,ehuks ,lkby t-RNA flUFksVst dh lgk;rk lsfof'k"V vehuks vEy dks tksM+rk gS

(3) vehuks vEy dks tksM+us ds fy;s ÅtkZ L=ksr ds :i esaGTP dk mi;ksx djrs gS

(4) ;g m-RNA ls vuqokfnr gksrk gS



144. thok.kq esa vkuqokaf'kd vfHk;kaf=d bUlqfyu cukus ds fy,dkSulk izFke pj.k gksxk &(1) ml thok.kq dh tkudkjh tks yxkrkj DNA jsfIyds'ku

djrk gks(2) gkeksZu dk df=e ek/;e esa 'kqf)dj.k

(3) ml thu dh igpku vkSj foyaxu tks bUlqfyu ds fy,dskfMax djrh gks

(4) thokf.o; thu tks bUlqfyu dh dksfMax djrh gks mlseuq"; eas LFkkukarfjr djuk

145. nk=h dksf'kdk vjDrrk ds fo"k; es D;k lgh gS :(1) blds gksus dk dkj.k gheksXykschu dh chVk Xyksfcu

Ja[kyk esa oSyhu ds LFkku ij XywVSfed vEy dk vktkuk gksrk gS

(2) ;g DNA ds ,d ,dy {kkj&;qXe es ifjorZu vkus lsgksrk gSA

(3) blesa fof'k"Vrk ds rkSj ij yEch nk=h ¼gafl;k½ dh'kDy dh dsUnd;qä yky jä dksf'kdk;sa (RBCs)gksrh ik;h tkrh gSA

(4) ;g ,d vkfyaxlw=h lgyXu izHkkoh VsªV ¼fo'ks"kd½ gSA146. U;weksdksdl thok.kqvksa ij fd;k x;k :ikUrj.k iz;ksx

fdl fl)kUr dks iznf'kZr djrk gS \(1) DNA rFkk izksVhu la'ys"k.k ds chp RNA LFkkukUrj.k

e/;LFk gSA(2) xq.klw= DNA ds cus gksrs gSaA(3) DNA ,d vkuqoaf'kd inkFkZ gSA(4) thok.kqvks es lR; ysSafxd tuu ik;k tkrk gsSA

147. ,sMhukslhu Mh,sfeust (ADA) vHkko okyk vkuqokaf'kd nks"kfdlds }kjk LFkk;h rkSj ij mipkfjr fd;k tk ldrk gSA(1) ,sMhukslhu Mh,sehust lfØ;dksa dk lsou djkdj(2) ADA mRiknu djus okyh vfLFk&eTtk dksf'kdkvksa

dks vkjafHkd Hkzw.k voLFkkvksa ij cu jgh dksf'kdkvksads Hkhrj izos'k djk dj

(3) ,atkbe izfrLFkkiu fpfdRlk }kjk(4) dk;Z'khy ADA, cDNA ls ;qDr vkuqokaf'kdrk

bathfu;fjax fyEQkslkbVksa dks le;≤ ij var%izosf'kr djk dj

144. What is the first step to produce geneticallyengineered insulin in bacteria ?(1) Knowledge of bacteria that replicate DNA rapidly

(2) Purification of the hormone in an artificialmedium

(3) Identification and isolation of gene coding forinsulin

(4) Insertion of bacterial gene coding for insulin inhumans

145. Sickle cell anemia is(1) caused by substitution of valine by glutamic

acid in the beta globin chain of haemoglobin

(2) caused by a change in a single base pair of DNA

(3) characterized by elongated sickle like RBCs witha nucleus

(4) an autosomal linked dominant trait146. Transformation experiments using pneumococcus

bacteria led to the hypothesis that(1) RNA is the transfer link between DNA and

protein synthesis.(2) Chromosomes are made up of DNA(3) DNA is a genetic material.(4) Bacteria have true sexual reproduction.

147. The genetic defect adenosine deaminase (ADA)Deficiency may be cured permanently by :-(1) Administering adenosine deaminase activators.(2) Introducing bone marrow cells producing ADA

into cells at early embryonic stage.

(3) enzyme replacement therapy.(4) Periodic infusion of genetically engineered

lymphocytes having functional ADA cDNA.



148. fuEu esa ls lgh fodYi dk p;u djks :

A

B

C

D

pBR322

ampR

(1) D – ampiciline resistant gene(2) A – BamH I(3) B – Ori(4) C – Pvu I

149. ,fYcfuTe ,d vkWVkslksey vizHkkoh mRifjorZu ds dkj.kgksrk gSA ,d lkekU; Ropk o.kZd okys ;qXy dk izFkef'k'kq ,fYcuks Fkk rks muds f}rh; f'k'kq ds ,fYcuks gksus dhD;k izkf;drk gksxh :

(1) 50% (2) 75%(3) 100% (4) 25%

150. DNA v.kq ds [k.M ds fdukjs fpifpis fdlds dkj.k gksrsgS :(1) eqDr feFkkbyhdj.k (2) ,.MksU;qfDytst(3) v;qfXer {kkj (4) dSfY'k;e vk;u

151. fuEufyf[kr esa ls dkSulh ,d vfHkO;fDr letkr lajpukvksadk lgh lgh vFkZ crkrh gS&(1) 'kkjhjh; lekurkvksa ls ;qä vax] ijarq tks fHkUu dk;Z

djrs gksa(2) 'kkjhjh; vlekurkvksa ls ;qä vax] ijarq tks ,d gh

dk;Z djrs gksa(3) vax ftudk vc dksbZ dk;Z ugha gS ijUrq iwoZtksa esa

mudk dksbZ ,d egRoiw.kZ dk;Z gqvk djrk Fkk(4) ,sls vax tks dsoy Hkzw.k voLFkk esa izdV gksrs gS rFkk

ckn esa o;Ld esa foyhu gks tkrs gS152. feFkkby xqvkuksflu VªkbZQkWLQsV fdlls lEcfU/kr gS :

(1) fcUnq mRifjorZu (2) VkWVksesfjTe(3) dsfiax (4) vksdktkdh [k.M

148. Which is following is correctly explained :

A

B

C

D

pBR322

ampR

(1) D – ampiciline resistant gene(2) A – BamH I(3) B – Ori(4) C – Pvu I

149. Albinism is known to be due to an autosomalrecessive mutation. The first child of a couple withnormal skin pigmentation was an albino. What isthe probability that their second child will also be analbino :(1) 50% (2) 75%(3) 100% (4) 25%

150. The end of fragments of DNA molecule are stickydue to :(1) free methylation (2) endonuclease(3) unpaired bases (4) calcium ions

151. Which one of the following describes correctly thehomologous structures(1) Organs with anatomical similarities, but

performing with different functions(2) Organs with anatomical dissimilarities. but

performing same function(3) Organs that have no function now, but had an

important function in ancestors(4) Organs appearing only in embryonic stage and

disappearing later in the adult152. Methyl guanosine triphosphate is associated with :

(1) point mutation (2) tautomerism(3) capping (4) Okazaki fragments



153. Locations or sites in the human DNA where singlebase DNA differences occurs are called :(1) repetitive DNA (2) VNTR(3) SNPs (4) SSCP

154. Use the following information to answer thequestion.Bt toxin protein in its ____i_____ form creates porethat leads to the lysis of cells in ___ii____ .The information in which alternative completes thegiven statement ?(1) i ii inactive bacterium(2) i ii inactive insects(3) i ii active bacterium(4) i ii active insects

155. A normal-visioned man whose father was colour-blind, marries a woman whose father was alsocolour-blind. They have their first child as a daughter.What are the chances that this child would becolour-blind :(1) 100% (2) 0%(3) 25% (4) 50%

156. Seminal plasma in humans is rich in(1) fructose and calcium but has no enzymes

(2) glucose and certain enzymes but has no clacium

(3) fructose and certain enzymes but poor incalcium

(4) fructose, calcium and certain enzymes

157. Adaptive radiation refers to :(1) evolution of different species from a common

ancestor(2) migration of members of a species to different

geographical areas(3) power of adaptation in an individual to a variety

of environments(4) adaptations due to geographical isolation

153. ekuo DNA ds LFky tgka ,dy {kkj esa mifLFkr fofHkUurk,adgykrh gS :(1) iqujkor DNA (2) VNTR(3) SNPs (4) SSCP

154. iz'u dk mŸkj nsus ds fy, fuEufyf[kr lwpukvksa dkmi;ksx djsa &Bt fo"k izksVhu viuh ____i_____ voLFkk esa fNnz mRiédjrk gS tks fd ___ii____ dksf'kdkvksa dk y;u djrhgSA fn;s x;s dFku dks dkSulh lwpuk,sa iw.kZ djrh gS &

(1) i iivfØ; thok.kq

(2) i iivfØ; dhV

(3) i iilfØ; thok.kq

(4) i iilfØ; dhV

155. ,d lkekU; nf"V okys O;fDr ftlds firk o.kkZa/k FksA ,d,slh efgyk ls 'kknh gksrh gS ftlds firk Hkh o.kkZa/k FksAmudh izFke f'k'kq iq=h gS rks ml iq=h f'k'kq ds o.kkZa/k gksusdh D;k laHkkouk gS :

(1) 100% (2) 0%(3) 25% (4) 50%

156. ekuoksa ds 'kqØh; IykTek easa(1) ÝqDVkst vkSj dSfYl;e rks Hkjiwj gksrs gS] exj ,atkbe

ugah gksrsA(2) Xywdkst vkSj dqN ,atkbe rks Hkjiwj gksrs gS exj

dSfYl;e ugah gksrkA(3) ÝqDVkst rFkk dqN ,atkbe rks Hkjiwj gksrs gS exj

dSfYl;e cgqr ghs de gksrs gsSA(4) ÝqDVkst] dSfYl;e rFkk dqN ,atkbe] rhuks gh Hkjiwj

gksrs gSA157. vuqdwyh; fofdj.k dk rkRi;Z gS :

(1) ,d lkekU; iwoZt ls fofHkUu tkfr;ksa dk fodklgksuk

(2) ,d tkfr ds lnL;ksa dk fofHkUu HkkSxksfyd {ks=ksa esaizoklu

(3) ,d O;fDr esa fofHkUu okrkoj.kksa ds izfr vuqdqyu dh{kerk

(4) HkkSxksfyd iFkDdj.k ds dkj.k vuqdwyrk



158. Darwin, in the 'Theory of Natural Selection', statedthat :

(1) characters acquired during the life of anindividual are inherited by its offsprings

(2) environment does not play any role in theevolution

(3) natural selection acts on favourable variations,which appear among individuals of a species

(4) heritable variations arise through changes in thegene complex of a species

159. Matching the column :–

Column – I (Method)

Column – II (Modes of action)

(A) The pill (a) Prevents sperms reaching cervix

(B) Condom (b) Prevents implantation

(C) Vasectomy (c) Prevents ovulation

(D) Copper T (d) Semen contains no sperms

(1) A-b, B-c, C-a, D-d (2) A-c, B-a, C-d, D-b

(3) A-d, B-a, C-b, D-c (4) A-c, B-d, C-a, D-b

160. Read the following statements :

A. During fertilization, a sperm comes in contactwith the egg plasma membrane and inducesthe changes in the membrane that block theentry of additional sperm

B. The milk produced during intial few days oflactation is called Colostrum having antibodies

C. In human beings, menstrual cycle ceasesaround 50 years of age. It is called menopause.

D. With in one month of pregnancy major organsystem are formed.

(1) All are correct (2) All are incorrect

(3) D is correct (4) All correct except D

158. MkWfoZu us ^izkÑfrdoj.k ds fl)kUr** esa crk;k :

(1) ,d O;fDr ds thou ds nkSjku mikftZr y{k.k mldhlarfr;ksa esa oa'kkxr gksrs gS

(2) okrkoj.k mn~fodkl esa fdlh izdkj dh Hkwfedk ughafuHkkrs gS

(3) ,d tkfr ds lnL;ksa ds e/; vuqdqyh fofHkUurkvksaij izkÑfrd p;u dk;Z djrk gS

(4) ,d tkfr ds thu leqPp; esa ifjorZu ls oa'kkxrfofHkUurk,a mRiUu gksrh gS

159. LrEHk feyku dhft, %&

LrEHk – I (fof/k)

LrEHk – II (fØ;k ds izdkj)

(A) xksyh (a) 'kqØk.kq dks xzhok rd igq¡pus ls jksdrh gS

(B) fujks/k (b) vkjk si.k dks jk sdrh gS

(C) oslsDVkseh ¼ujcU/;rk½

(c) v.MksRltZu dks jksdrh gS

(D) dkWij T (d) oh;Z eas 'kqØk.kq ugha gksrs

(1) A-b, B-c, C-a, D-d (2) A-c, B-a, C-d, D-b

(3) A-d, B-a, C-b, D-c (4) A-c, B-d, C-a, D-b

160. fuEu dFkuksa dks if<;s %

A. fu"kspu ds nkSjku ,d 'kqØk.kq v.M IykTek f>Yyh dslEidZ eas vkrk gS vkSj f>Yyh eas ifjorZu izsfjr djrkgS rkfd vfrfjDr 'kqØk.kqvksa dk izos'k :d tk,A

B. ysDVs'ku ds nkSjku izkjfEHkd dqN le; eas nw/k dkfuekZ.k gksrk gS ftls dksyksLVªe dgrs gS blesa izfrj{khgksrs gSA

C. ekuo esa 50 o"kZ dh vk;q ds vanj jt pØ lekIr gkstkrk gS ;g ehuksikWt dgykrk gSA

D. xHkZ/kkj.k ds ,d eghus ds vanj eq[; vax ra= cuukizkjaHk gks tkrs gSA

(1) lHkh lgh gS (2) dksbZ Hkh lgh ugh gS

(3) D lgh gS (4) D ds vykok lHkh lgh gS



161. The genetic code said to be degenerate anduniversal which means that,

1. amino acids may have more than one codon

2. all amino acids have more than one codon

3. codons are common for higher and lowerorganism

4. codons are not found in bacteria

Code:

(1) 1, 2, 3 & 4 are correct

(2) 1 and 2 correct

(3) 1, 2 and 4 are correct

(4) 1 and 3 are correct

162. Consider the statments given below regardingcontraception and answer as directed there after

A. Medical Termination of Pregnancy (MTP) duringfirst trimester is generally safe.

B. Generally chances of contraception are nil untilmother breast feeds the infant upto two years

C. Intrauterine devices like copper-T are effectivecontraceptives

D. Contraceptive pills may be taken upto one weekafter coitus to prevent conception

(1) A and B

(2) B and C

(3) C and D

(4) A and C

161. vkuqokaf'kd dwV dk vigzkflr rFkk lkoZHkkSfed gksus dkvFkZ gS

1. ,d ,ehuks vEy ds fy, ,d ls vf/kd dksM gks ldrsgSaA

2. lHkh ,sehuks vEy ,d ls vf/kd dksM j[k ldrs gS

3. mPp rFkk fuEu thokas ds fy, dksMksUl leku gksrs gS

4. dksMksUl thok.kq esa ugh ik;s tkrs gSA

dwV %

(1) 1, 2, 3 ,oa 4 lgh gS

(2) 1 rFkk 2 lgh gS

(3) 1, 2 rFkk 4 lgh gS

(4) 1 rFkk 3 lgh gS

162. uhps fn;s x;s dFkuksa eas xHkZ fujks/ku djus ds lanHkZ eas lghmŸkj ds fy, funsZ'k gS %&

A. xHkZ dk fpfdRlh; lekiu (MTP) izFke 3 ekg dsnkSjku izk;% lqjf{kr gksrk gSA

B. ekrk ds }kjk f'k'kq ds 2 o"khZ; Lruiku rd izk;% xHkZ/kkj.k dk volj ux.; gksrk gSA

C. bUVj;wVsjkbu ;qfDr tSls dkWij&T ,d izHkkoh xHkZfujks/kd gSA

D. xHkZ fujks/kd xksfy;k¡ xHkZ/kkj.k dks jksdus ds fy, 1lIrkg ckn rd yh tk ldrh gSA

(1) A rFkk B

(2) B rFkk C

(3) C rFkk D

(4) A rFkk C



163. Column – I Column – II

A. Hyaluronidase I. Grafian follicle

B. Corpus luteum II. Mammary gland

C. Colostrum III. Progesterone

D. Antrum IV. Acrosomal reaction

(1) A–II, B–I, C–IV, D–III

(2) A–IV, B–II, C–III, D–I

(3) A–IV, B–III, C–II, D–I

(4) A–IV, B–III, C–I, D–II

164. A good example for recapitulation theory is :

(1) canine teeth of dog

(2) placenta of mammals

(3) tadpole larva of frog

(4) embryonic membranes of reptiles

165. Choose the correct options :

(A) ELISA is based on the principle of Antigen-antibody interaction.

(B) Over 95% of all existing transgenic animals arerabbits.

(C) Human protein (-1-antitrypsin) used to treatcysticfibrosis

(D) The milk of first transgenic Cow, Rosie,contains the human -lactalbimin and it isnutritionally a more balanced product forhuman babies

(1) A and D (2) C and B

(3) B and D (4) A, B and C

163. LrEHk – I LrEHk – II

A. gk;ywjksfuMst I. xzsfQ;u Qksfydy

B. dkWiZl Y;qfV;e II. Lru xzfUFk;k¡

C. dksyksLVªe III. izkstsLVªkWu

D. ,UVªe IV. ,Øslskey fØ;k

(1) A–II, B–I, C–IV, D–III

(2) A–IV, B–II, C–III, D–I

(3) A–IV, B–III, C–II, D–I

(4) A–IV, B–III, C–I, D–II

164. iqujkor laf{kIr fl)kUr dk ,d vPNk mnkgj.k gS :

(1) dqÙks ds dsukbu nkar

(2) Lrfu;ksa dk vijk

(3) es<+d dk VsMiksy ykokZ

(4) ljhliksZ dh Hkwz.kh; f>Yyh;ka

165. lgh fodYiksa dks pqusa %&

(A) ,ykbtk] izfrtu&izfrj{kh ikjLifjd fØ;k ds fl)kUrij dk;Z djrk gSA

(B) 95 izfr'kr ls vf/kd ijkthoh tUrq [kjxks'k gS

(C) ekuo izksVhu (-1-antitrypsin) iqVh; js'kke;rk dsfunku gsrq mi;ksx fy;k x;kA

(D) izFke ijkthoh xk; jksth* ds nqX/k esa -lactalbiminfeyrk gS] tks ekuo f'k'kq gsrq vR;f/kd larqfyr iks"kdrRo gSa

(1) A vkSj D (2) C vkSj B

(3) B vkSj D (4) A, B vkSj C



166. fuEu esa ls dkSulk dFku vlR; gS \

(1) lnuZ CykWfVax dk mi;ksx Mh-,u-,- esa dqN vkuqokaf'kdfodkj dks Kkr djus esa gksrk gSA

(2) lnuZ CykWfVax dk mi;ksx Mh-,u-,- vaxqyhNki dsfuekZ.k esa gksrk gSA

(3) dksbZ Hkh nks O;fDr;ksa dh Mh-,u-,- vaxqyhNkiu lekugks ldrh gSA

(4) Mh-,u-,- vaxqyhNkiu] ,d O;fDr dh] fof'k"V lnuZCykWV gSA

167. fuEu esa ls fdls lgh le>k;k x;k gS

A

B

C

D

(1) A – 'kqØk.kq blesa rSjrs gS

(2) D – 'kqØk.kqvksa dk lap; gksrk gS

(3) B – dkiZl Liath;ksle ls cuk gksrk gS

(4) C – 'kqØh; IykTek dk L=ko.k djrk gS

168. euq"; eknk ds ekfld pØ dh og dkSu lh voLFkk gS tks7-8 fnu esa lekIr gks tkrh gS :–

(1) QkWfydqyj voLFkk (2) v.MksRlxhZ voLFkk

(3) Y;wfV;y voLFkk (4) ekfld L=ko

166. Which of the following statement is incorrect ?

(1) Southern blotting is useful in determiningcertain genetic defect in DNA

(2) Southern blotting is used to generation DNAfinger– prints

(3) DNA finger prints of any two individual may besimilar

(4) DNA finger print of a person is specificsouthern blot

167. Which one of these correctly matched :–

A

B

C

D

(1) A – Sperms swims in it

(2) D – Stores sperms

(3) B – Made by carpus spongiosum

(4) C – Secrete seminal plasma

168. The phase of menstrual cycle in humans that lastfor 7-8 days is :–

(1) Follicular phase (2) Ovulatory phase

(3) Luteal phase (4) Menstruation



169. 100 izkFkfed v.Mdksa ls fdrus v.Mk.kq cusaxs:–

(1) 300 (2) 400

(3) 200 (4) 100

170. fuEu eas ls dkSulk xyr gS :

D

C

B

A

(1) A, B – dkcksZfDly fljk

(2) C, D – ,feu fljk

(3) A, D – Hkkjh Ja[kyk

(4) B, C – izfrtu cU/ku Ja[kyk

171. lfØ; izfrj{k.k fdlls feyrh gS :–

(1) ,UVhckWMht

(2) jksxk.kqvksa ds nqcZy laØe.k ls

(3) izkdfrd izfrjks/kdrk ls

(4) mijksDr esa ls dksbZ ugh

172. dkSulh bE;wuksXyksfcu vkdkj esa lcls cM+h gS :–

(1) IgA (2) IgD

(3) IgE (4) IgM

169. How many eggs will be formed from 100 primaryoocytes :–

(1) 300 (2) 400

(3) 200 (4) 100

170. Which one of the following is wrong :

D

C

B

A

(1) A, B – Carboxil end

(2) C, D – Amine end

(3) A, D – Heavy chain

(4) B, C – Antigen binding chain

171. Active immunity is obtained by :–

(1) Antibodies

(2) Weakened germs infection

(3) Natural resistance

(4) None of these

172. Which immunoglobulin is the largest in size :–

(1) IgA (2) IgD

(3) IgE (4) IgM



173. ELISA is used to detect viruses, where :–

(1) Alkaline phosphatase is the key reagent

(2) Catalase is the key reagent

(3) DNA-probes are required

(4) Southern blotting is done

174. Elephantiasis (Filariasis) in man is caused by :–

(1) Ancylostoma duodenale

(2) Ascaris lumbricoides

(3) Dracunculus medinensis

(4) Wuchereria bancrofti

175. Mating of more closely related individuals within thesame breed for 4–6 generation is ?

(1) Inbreeding

(2) Cross-breeding

(3) Out-crossing

(4) Inter specific breeding

176. A probe which is a molecule used to locate specificsequences in a mixture of DNA or RNA moleculescould be:

(1) A single stranded RNA

(2) A single stranded DNA

(3) Either RNA or DNA

(4) Can be ss DNA but not ss RNA

173. ,yhlk ok;jl ijh{k.k esa iz;ksx gksrk gS tgk¡:–

(1) ,Ydsykbu QkWLQVst izeq[k vfHkdeZd gS

(2) dSVkyst izeq[k mRizsjd gS

(3) DNA-izksCl dh vko';drk gksrh gS

(4) lnZu CyksfVax dh tkrh gS

174. euq"; esa gkFkhikWo ¼Qk;ysfj,fll½ fdlds dkj.k mRiUugksrk gS :–

(1) ,ulk;yksLVksek M~;wMsusYl

(2) ,Ldsfjl yqfEczdkWbfMl

(3) MªSdUdqyl esMhusfUll

(4) oqpjsfj;k cSuØkW¶Vh

175. ,d gh uLy ds vf/kd fudVLFk tUrq ds e/; 4&6 ih<+hrd laxe gksuk D;k dgykrk gS

(1) vUr%iztuu

(2) ladj.k

(3) cfg%ladj.k

(4) vUrj tkfr; ladj.k

176. ,d izksc ,d v.kq gS tks DNA ;k RNA ds feJ.k esa fo'ks"kvuqØe dh fLFkfr fu/kkZj.k esa dke vkrk gS] ;g izksc D;kgS \

(1) ,d flaxy LVªsUMsM RNA A

(2) ,d flaxy LVªsUMsM DNA A

(3) RNA ;k DNA A

(4) ssDNA gks ldrk gS fdUrq ssRNA ugha A



177. Honey mainly consists of:

(1) monosaccharides

(2) disaccharides

(3) polysaccharides

(4) fats

178. A chromosome contain 1000 nucleosome calculatethe lenght of DNA in it :

(1) 6.8 × 103 nm

(2) 68 × 103 nm

(3) 680 × 103 nm

(4) 6800 × 103 nm

179. Pebrine disease is seen in -

(1) Food of fish

(2) Honey bee

(3) Silk moth

(4) Hen

180. Which of the following process is related to cattle-

(1) Rumination

(2) Castration

(3) Dehorning

(4) All of the above

177. 'kgn eq[;r% cuk gksrk gS &

(1) eksukslSdsjkbM dk

(2) MkblSdsjkbM dk

(3) ikWyhlSdsjkbM dk

(4) olk dk

178. 1000 U;wfDy;kslkse ls cus ,d xq.klw= esa DNA dh yEckbZgksxh

(1) 6.8 × 103 nm

(2) 68 × 103 nm

(3) 680 × 103 nm

(4) 6800 × 103 nm

179. isczkbu jksx ik;k tkrk gS %

(1) eNyh ds Hkkstu esa

(2) e/kqefD[k;ksa esa

(3) js'ke dhV esa

(4) eqfxZ;ksa esa

180. fuEu esa ls dkSu lh fØ;k eos'kh ls tqM+h gS &

(1) :ehus'ku

(2) dkLVªs'ku

(3) MhgkWfuZax

(4) mijksDr lHkh



ANSWER-KEY (XII NEET) Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Ans. 2 2 3 2 3 3 3 3 3 4 2 4 1 2 Q. 15 16 17 18 19 20 21 22 23 24 25 26 27 28 Ans. 3 3 4 3 4 4 3 3 4 3 1 1 4 2 Q. 29 30 31 32 33 34 35 36 37 38 39 40 41 42 Ans. 2 3 1 4 4 4 3 1 2 3 3 1 1 3 Q. 43 44 45 Ans. 4 2 3Q. 46 47 48 49 50 51 52 53 54 55 56 57 58Ans. 1 4 4 1 3 1 2 2 3 2 3 2 3Q. 59 60 61 62 63 64 65 66 67 68 69 70 71Ans. 2 2 3 4 3 1 2 1 3 4 1 3 4Q. 72 73 74 75 76 77 78 79 80 81 82 83 84Ans. 3 4 4 1 4 2 2 3 3 2 2 2 2Q. 85 86 87 88 89 90Ans. 4 1 1 3 3 4Q. 91 92 93 94 95 96 97 98 99 100 101 102 103Ans. 4 2 4 2 4 2 4 2 1 2 4 4 3Q. 104 105 106 107 108 109 110 111 112 113 114 115 116Ans. 2 4 3 3 3 1 1 4 3 4 3 4 1Q. 117 118 119 120 121 122 123 124 125 126 127 128 129Ans. 4 2 1 1 4 4 3 3 2 3 1 2 4Q. 130 131 132 133 134 135Ans. 2 2 2 1 2 1Q. 136 137 138 139 140 141 142 143 144 145 146 147 148Ans. 4 3 1 4 2 4 2 2 3 2 3 2 3Q. 149 150 151 152 153 154 155 156 157 158 159 160 161Ans. 4 3 1 3 3 4 2 4 1 3 2 4 4Q. 162 163 164 165 166 167 168 169 170 171 172 173 174Ans. 4 3 3 1 3 3 1 4 4 2 4 1 4Q. 175 176 177 178 179 180Ans. 1 3 1 2 3 4

class xii bio - synthesis bikaner€¦ · the omr sheet and fill in the particulars carefully. 3....

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