1. Two blocks are connected by a string, as shown infigure. They are released from rest. The coefficientof kinetic friction between the upper block and thesurface is 0.5. Assume that the pulley is massless andfrictionless. Their common speed after they moveda distance 5m will be :-

m =2kg1

m =2kg2

(1) 10 m/s (2) 5 m/s (3) 2.5 m/s (4) 25 m/s2. Two wheels having radii in the ratio 1 : 3 are

connected by a common belt. If the smaller wheelis accelerated from rest at a rate 1.5 rad s–2 for 10s,the angular velocity of bigger wheel will be :-(1) 5 rad s–1 (2) 15 rad s–1

(3) 45 rad s–1 (4) None of these3. Current I flows around the wire frame along the

edges of a cube as shown in figure. Point P is thecentre of the cube. Then,

YZ

P

X

a

a

a

(1) the magnetic field at P is toward +y direction(2) the magnetic field at P is toward –y direction

(3) the unit vector of magnetic field at P is 1 ˆ ˆ(i j)2

- -

(4) the magnitude of magnetic field at P is 04 2 I

3 a

mp

TOPIC : Work, Energy & Power, Circular Motion, Magnetic effect of current and Magnetism.

1. nks CykWd fp= eas n'kkZ;s vuqlkj ,d Mksjh ls tqM+s gq, gSaA budksfojkekoLFkk ls NksM+k tkrk gSA Åij okys CykWd ,oa lrg ds chpxfrd ?k"kZ.k xq.kkad 0.5 gSA f?kjuh dks nzO;ekughu rFkk ?k"kZ.kjfgr eku yhft;sA buds }kjk 5m nwjh r; djus ds i'pkr] budkmHk;fu"B osx gksxk :-

m =2kg1

m =2kg2

(1) 10 m/s (2) 5 m/s (3) 2.5 m/s (4) 25 m/s2. nks ifg;s ftudh f=T;kvksa dk vuqikr 1 : 3 gS] ,d mHk;fu"B

csYV }kjk tqM+s gq, gSaA ;fn NksVs ifg;s dks 10 s ds fy;s fojkekoLFkkls 1.5 rads–2 Roj.k ds lkFk Rofjr fd;k tk;s] cM+s ifg;s dkdks.kh; osx gksxk :-(1) 5 rad s–1 (2) 15 rad s–1

(3) 45 rad s–1 (4) buesa ls dksbZ ugha3. fp= esa n'kkZ;s vuqlkj ,d ?ku ds fdukjksa ds vuqfn'k cus gq, rkj

ds Ýse ls] I /kkjk izokfgr gks jgh gSA fcUnq P ?ku dk dsUnz gSArks

YZ

P

X

a

a

a

(1) P ij pqEcdh; {ks= +y fn'kk esa gSA(2) P ij pqEcdh; {ks= –y fn'kk esa gSA

(3) P ij pqEcdh; {ks= dk ,dkad lfn'k 1 ˆ ˆ(i j)2

- - gSA

(4) pqEcdh; {ks= dk ifjek.k 04 2 I

3 a

mp

gSA

NEET UG - UNIT TEST - 4 QUESTIONS

4. A smooth block of mass 10 kg moves up frombottom to top of a wedge which is moving withan acceleration a0 = 3 m/s2. Find the work doneby the pseudo force measured by the person sittingat the edge of the wedge.

a0

l = 3m

h = 4mm

(1) 120 J (2) 150 J (3) 90 J (4) zero5. An infinitely long current carrying wire carries

current i. A particle of mass m and charge q isprojected with speed v parallel to the direction ofcurrent at a distance r from it. Then, the radius ofcurvature at the point of projection is :-

(1)0

2rmvq im (2)

0

2 rmvq ipm

(3) r (4) cannot be determined6. A small block of mass m is pushed on a smooth

track from position A with a velocity 2

5 times

the minimum velocity required to reach point D.When the block reaches point B, what is thedirection (in terms of angle with horizontal ) ofacceleration of the block ?

B

D

O

C

A

(1)1 1

tan2

- æ öç ÷è ø

(2) tan–1 (2)

(3)1 2

sin3

- æ öç ÷è ø

(4) The block never reaches point B

4. 10 kg nzO;eku okyk ,d fpduk CykWd a0 = 3 m/s2 Roj.kds lkFk xfr'khy Qkukdkj CykWd ij iSans ls mij dh vksj 'kh"kZrd xfr djrk gSA Qku ds fdukjs ij cSBs O;fDr }kjk izsf{krvkHkklh cy }kjk fd;k x;k dk;Z Kkr dhft;s :

a0

l = 3m

h = 4mm

(1) 120 J (2) 150 J(3) 90 J (4) 'kwU;

5. vuUr yEckbZ okys ,d rkj ls i /kkjk izokfgr gks jgh gSA m nzO;eku,oa q vkos'k okys ,d d.k dks v pky ds lkFk rkj ls r nwjh ij/kkjk dh fn'kk ds lekukUrj iz{ksfir fd;k tkrk gSA rks iz{ksi.kfcUnq ij oØrk f=T;k gS :-

(1)0

2rmvq im (2)

0

2 rmvq ipm

(3) r (4) cannot be determined

6. m nzO;eku okys ,d NksVs CykWd dks] fpdus iFk ij fcUnq A ls]

fcUnq D rd igq¡pus ds fy;s vko';d U;wure osx ds 2

5xquk

osx ds lkFk /kDdk fn;k tkrk gSA tc CykWd fcUnq B ij igq¡prkgS] CykWd ds Roj.k dh fn'kk ({ksfrt ds lkFk dks.k ds inksa esa)D;k gksxh ?

B

D

O

C

A

(1)1 1

tan2

- æ öç ÷è ø

(2) tan–1 (2)

(3)1 2

sin3

- æ öç ÷è ø

(4) CykWd dHkh Hkh B ij ugha ig¡qpsxk

7. If the angles of dip at two places are 30° and 45°respectively, then the ratio of horizontal componentsof earth's magnetic field at the two places will be

(1) 3 : 2 (2) 1: 2 (3) 1: 3 (4) 1 : 2

8. A bullet leaving the muzzle of a rifle barrel with avelocity v penetrates a plank and loses one-fifth ofits velocity. It then strikes second plank, which it justpenetrates through. Supposing the average resistanceto the penetration is same in both the cases, the ratioof the thickness of the planks will be :-

(1)9

16(2)

14

(3) 15

(4) 925

9. A car is moving on a circular road of diameter50 m with a speed 5 ms–1. It is suddenlyaccelerated at a rate 1 ms–2. If its mass is 500 kg,net force acting on the car is :-(1) 500 N (2) 1000 N

(3) 500 2N (4) 500 / 2N10. Two long parallel conductors carry currents in the

same direction as shown in figure. Conductor Acarries a current of 150 A and is held firmly inposition. Conductor B carries a current IB and isallowed to slide freely up and down (parallel to A)between a set of non-conducting guides. If the massper unit length of conductor B is 0.100 g cm–1 whatvalue of current IB will result in equilibrium whenthe distance between the two conductors is 2.50 cm?

IA

IB

B

A

(1)2503

(2) 253

1(3)

253

(4) 12.5

3

7. nks LFkkuksa ij ueu dks.kksa dk eku Øe'k% 30° ,oa 45° gSa] rks nksuksaLFkkuksa ij i`Foh ds pqEcdh; {ks= ds {kSfrt ?kVdksa dk vuqikrgksxk

(1) 3 : 2 (2) 1: 2 (3) 1: 3 (4) 1 : 2

8. jkbQy dh uky ls v osx ds lkFk fudyus okyh xksyh ,diV~Vs dks ikj djrh gS vkSj blds osx dk ik¡poka Hkkx [kks nsrhgSA blds i'pkr~ ;g nwljs iV~Vs ls Vdjkrh gS vkSj bldks ikjek= dj ikrh gSA ;g ekudj fd ikj djrs gq, nksuksa fLFkfr;ksaesa izfrjks/k dk eku ,d leku gS] iV~Vksa dh eksVkb;ksa dk vuqikrgS :-

(1)9

16(2)

14

(3) 15

(4) 925

9. ,d dkj 5 ms–1 pky ds lkFk 50 m O;kl okys o`Ùkkdkj iFkij xfr'khy gSA ;g vpkud 1 ms–1 dh nj ls Rofjr gksrh gSA;fn bldk nzO;eku 500 kg gS] dkj ij yx jgk dqycy gS :-(1) 500 N (2) 1000 N

(3) 500 2N (4) 500 / 2N10. nks yEcs] lekukUrj pkydksa ls fp= es n'kkZ;s vuqlkj leku fn'kk

esa /kkjk,a izokfgr gks jgh gSA pkyd A esa] 150A /kkjk gS vkSjbldks bldh fLFkfr esa n`<+rkiwoZd j[kk tkrk gSA pkyd B esa IB

/kkjk izokfgr gks jgh gS vkSj ;g dqpkyd nf.Mdkvksa ds chp eqDr:i ls mij&uhps (A ds lekukUrj) xfr ds fy;s LorU= gSA ;fnpkyd B dh izfr bdkbZ yEckbZ dk nzO;eku 0.100 gcm–1

gS] rks tc pkydksa ds chp dh nwjh 2.50 cm gks] IB ds fdruseku ls lkE;koLFkk izkIr gksxh&

IA

IB

B

A

(1)2503

(2) 253

1(3)

253

(4) 12.5

3

11. A constant power P is applied to a particle of massm. The distance travelled by the particle when itsvelocity increases from v1 to v2 is- (neglect friction)

(1) ( )2 22 1

3P v vm

- (2) 2 1m (v v )3P

-

(3) ( )3 32 1

m v v3P

- (4) ( )2 22 1

m v v3P

-

12. Current I enters at A in a square loop of uniformresistance and leaves at B. The ratio of magneticfield at E, the center of square, due to segment ABto that due to DC is :-

CD

AI I

E

(1) 1 (2) 2 (3) 3 (4) 413. A car moves on a horizontal track of radius r, the

speed is increasing at a constant rate "a", coefficientof friction between road and tyre is m, the speedat which car will skid :-

(1) [(m2g2 + a2)r2]1/4 (2) grm

(3) [(m2g2 – a2)r2]1/4 (4) ar14. The magnetic moment of a magnet of mass

75 gm is 9 × 10–7 A-m2. If the density of thematerial of magnet is 7.5 × 103 kg/m3 then intensityof magnetisation will be :-(1) 0.9 A/m (2) 0.09 A/m(3) 9 A/m (4) 90 A/m

15. A car of mass 500 kg moving with a speed36 km/h in a straight road unidirectionally doublesits speed in 1 min. The power delivered by theengine is :-(1) 1250 W (2) 1250 HP(3) 625 W (4) 625 HP

11. m nzO;eku okys d.k dks fu;r 'kfDr iznku dh tkrh gSA d.kdk osx v1 ls v2 rd c<+us esa blds }kjk r; dh x;h nwjh gS-(?k"kZ.k ux.; gS)

(1) ( )2 22 1

3P v vm

- (2) 2 1m (v v )3P

-

(3) ( )3 32 1

m v v3P

- (4) ( )2 22 1

m v v3P

-

12. ,d leku :i ls izfrjks/k okys oxkZdkj ywi esa /kkjk I, A ij izos'kdjrh gS vkSj B ij ckgj fudyrh gSA Hkkx AB rFkk CD dsdkj.k] oxZ ds dsUnz E ij pqEcdh; {ks= dk vuqikr gksxk :-

CD

AI I

E

(1) 1 (2) 2 (3) 3 (4) 413. ,d dkj r f=T;k okys {kSfrt iFk ij fu;r nj "a" ls c<+rh gqbZ

pky ds lkFk xfr dj jgh gS] lM+d ,oa Vk;j ds chp ?k"kZ.k xq.kkadm gS] fdl pky ij dkj fQly tk;sxh :-

(1) [(m2g2 + a2)r2]1/4 (2) grm

(3) [(m2g2 – a2)r2]1/4 (4) ar

14. ,d pqEcd nzO;eku 75 gm ds pqEcdh; vk?kw.k Z dkeku 9 × 10–7 A-m2 ;fn pqEcd ds inkFk Z dk ?kuRo7.5 × 103 kg/m3 gS rks pqEcdu dh rhozrk gksxh :-

(1) 0.9 A/m (2) 0.09 A/m

(3) 9 A/m (4) 90 A/m

15. 36 km/h pky ds lkFk lh/kh lM+d ij xfr'khy 500 kgnzO;eku okyh dkj mlh fn'kk esa] 1 feuV esa viuh pky nqxuhdj ysrh gSA batu }kjk iznÙk 'kfDr gS :-

(1) 1250 W (2) 1250 HP

(3) 625 W (4) 625 HP

16. Same current i = 2A is flowing in a wire frame asshown in fig. The frame is a combination of twoequilateral triangles ACD and CDE of side 1m. Itis placed in uniform magnetic field B = 4 T actingperpendicular to the plane of frame. The magnitudeof magnetic force acting on the frame is-

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

C

A

D

E

(1) 24 N (2) zero (3) 16 N (4) 8 N17. A motor van weighing 4400 kg rounds a level

curve of radius 1000 m on unbanked road at70 m/s. What should be the minimum value of thecoefficient of friction to prevent skidding(g = 9.8 m/s2)(1) 0.3 (2) 0.4 (3) 0.5 (4) 0.6

18. A block is placed on the top of a inclined planeinclined at 37º with horzontal. The length of theinclined plane is 5m. The block slides down the planeand reaches the bottom, if a coefficient of friction is0.25, then the speed of the block at the bottom is

37º

h5m

(1) 7.67 ms–1 (2) 12.5 ms–1

(3) 6.26 ms–1 (4) 3.13 ms–1

19. A charged particle enters a magnetic field at right anglesto the magnetic field. The field exists for a length equalto 1.5 times the radius of the circular path of the particle.The particle will be deviated from its path by(1) 90° ×

×××

××××

××××

1.5R

Q,m(2) sin–1

23

(3) 30°(4) 180°

16. rkj ds ,d Ýse esa fp=kuqlkj leku /kkjk i = 2A izokfgr gks jghgSA ;g Ýse 1 eh- Hkqtk okys nks leckgq f=Hkqtksa ACD ,oa CDEdk la;kstu gSA bldks Ýse ds ry ds yEcor~ bafxr le:ipqEcdh; {ks= B = 4 T esa j[kk x;k gSA Ýse ij yx jgs pqEcdh;cy dk ifjek.k gS&

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

C

A

D

E

(1) 24 N (2) 'kwU; (3) 16 N (4) 8 N17. 4400 kg Hkkj okyh ,d eksVj osu] 1000 m f=T;k okyh

lery ,oa vcafdr lM+d ij 70 m/s ls ?kqerh gSA fQlyusls cpkus ds fy, ?k"kZ.k xq.kkad dk U;wure eku fdruk gksukpkfg,:-(g = 9.8 m/s2)(1) 0.3 (2) 0.4 (3) 0.5 (4) 0.6

18. ,d CykWd {ksfrt ls 37° vkufr okys urry ds 'kh"kZ ij j[kkgqvk gSA urry dh yEckbZ 5m gSA CykWd fQlyrk gS vkSj iSansrd igq¡prk gS] ;fn ?k"kZ.k xq.kkad 0.25 gS] rks iSans ij CykWd dhpky gS -

37º

h5m

(1) 7.67 ms–1 (2) 12.5 ms–1

(3) 6.26 ms–1 (4) 3.13 ms–1

19. ,d vkosf'kr d.k pqEcdh; {ks= ds yEcor~ izos'k djrk gSA d.kds oÙkkdkj iFk dh f=T;k ds 1.5 xquk yEckbZ eas {ks= fo|ekugSA d.k vius iFk ls fopfyr gksxk&(1) 90°

××××

××××

××××

1.5R

Q,m(2) sin–1

23

(3) 30°(4) 180°

20. Due to the earth's magnetic field, charged cosmicray particles(1) Require greater kinetic energy to reach the

equator than the poles(2) Require less kinetic energy to reach the

equator than the poles(3) Can never reach the equator(4) Can never reach the poles

21. Car moves at a constant speed on a road as shownin figure. The normal force by the road on the car isNA and NB when it is at the points A and B respectively

A

B

(1) NA = NB

(2) NA > NB

(3) NA < NB

(4) insufficient information to decide the relationof NA & NB

22. A circular coil having mass m is kept above theground (x-z plane) at some height. Coil carries acurrent i in the direction shown in fig. In which

direction a uniform magnetic field B®

be appliedso that the magnetic force balances the weight ofthe coil :

y

x

z

(1) positive x-direction(2) negative x-direction(3) positive z-direction(4) none of these

20. i`Foh ds pqEcdh; {ks= ds dkj.k vkosf'kr dkWfLed fdj.k dsd.k(1) dks /kqzo dh rqyuk esa fo"kqor~ js[kk ij igq¡pus esa vf/kd mGtkZ

dh vko';drk gksrh gS(2) dks /kqzo dh rqyuk esa fo"kqor~ js[kk ij igq¡pus esa de mGtkZ

dh vko';drk gksrh gS(3) dHkh Hkh fo"kqor~ js[kk ij ugha igq¡p ldrs gSa(4) dHkh Hkh /kqzo ij ugha igq¡p ldrs gSa

21. fp= esa iznf'kZr dh xbZ lM+d ij ,d dkj fu;r pky ls xfr'khygSA tc ;g fcUnq A rFkk B ij gksrh gS rks dkj ij lM+d dkvfHkyEcor~ izfrfØ;k cy Øe'k% NA rFkk NB gksrk gS&

A

B

(1) NA = NB

(2) NA > NB

(3) NA < NB

(4) NA rFkk NB esa lEcU/k Kkr djus ds fy;s nh x;h lwpukvi;kZIr gSA

22. m nzO;eku okyh ,d o`Ùkkdkj dq.Myh tehu (x-z ry) lsdqN Å¡pkbZ ij j[kh x;h gSA dq.Myh ls fp= esa n'kkZ;h x;h

fn'kk esa i /kkjk izokfgr gks jgh gSA ,d le:i pqEcdh; {ks= B®

fdl fn'kk esa yxk;k tk;s fd ;g dq.Myh ds Hkkj dks larqfyrdj ys:

y

x

z

(1) /kukRed x-fn'kk esa

(2) ½.kkRed x-fn'kk esa

(3) /kukRed z-fn'kk esa

(4) buesa ls dksbZ ugha

23. A rod of length l is pivoted at one of the ends andis made to rotate in a horizontal plane as shown infigure with a constant angular speed. A ball of massm suspended by a string of length l from the otherend of the rod. If the angle made by the string withthe vertical is q, then angular speed of rotation is :-

l

q

m

(1)g

(1 sin )+ ql(2)

gcosql

(3)g tan

(1 sin )q

+ ql(4)

gcot(1 cot )

q+ ql

24. Find the magnetic field at P due to the arrangementshown :

d

P

ii

(1) m

p0i

2 d

11 –

2

æ öç ÷è ø Ä

(2) p02µ i

2 d Ä

(3) m

p0i

2 d Ä

(4) m

p0i

2 d

æ öç ÷è ø

11 +

2 Ä

23. l yEckbZ okyh NM + ds ,d fljs ij /kqjh yxh gqbZ gS vkSj bldksfp= esa n'kkZ;s vuqlkj fu;r dks.kh; pky ds lkFk {ksfrt ry esa?kqek;k tkrk gSA NM+ ds nwljs fljs ij l yEckbZ okyh Mksjh dhlgk;rk ls m nzO;eku dh xsan yVdh gqbZ gSA ;fn Mksjh }kjkÅ/oZ ls q dks.k cuk;k tk jgk gS] rks ?kw.kZu dh dks.kh;pky gS :-

l

q

m

(1)g

(1 sin )+ ql(2)

gcosql

(3)g tan

(1 sin )q

+ ql(4)

gcot(1 cot )

q+ ql

24. iznf'kZr O;oLFkk ds dkj.k P ij pqEcdh; {ks= Kkr dhft;s :

d

P

ii

(1) m

p0i

2 d

11 –

2

æ öç ÷è ø Ä

(2) p02µ i

2 d Ä

(3) m

p0i

2 d Ä

(4) m

p0i

2 d

æ öç ÷è ø

11 +

2 Ä

25. A block slides along a track from one level to a higherlevel by moving through an intermediate valley asshown in figure. The track is friction less until theblock reaches the higher level. Then there is frictionforce which stops the block at a distance d. Theblock's initial speed v0 is 6.0 ms–1; the heightdifference h is 1.1 m, and the coefficient of kineticfriction m is 0.60. The value of d is :-

h d

m=0.60m=0

v0

(1) 1.17 m (2) 3 m (3) 2.34 m (4) 6 m26. A pendulum is released from rest from the point

A as shown in the figure. The string of thependulum is taut. OA makes an angle 30° with thevertical. The acceleration of the pendulum bob atthis instant would be :-

30°O

BA

(1) along AOuuur

(2) along the vertical

(3) in a direction perpendicular to OA

(4) In a direction making an angle less than 30°with the vertical

27. A conducting rod of length l and mass m ismoving down a smooth inclined plane ofinclination q with constant velocity v. A currenti is flowing in the conductor in a directionperpendicular to paper inwards. A vertically

upward magnetic field Br

exists in space. Then

magnitude of magnetic field Br

is -

q

qv F =mi Bl

(1) ql

mgsin

i (2) ql

mgtan

i (3) q

l

mgcosi (4) ql

mgi sin

w

25. ,d CykWd fp= esa n'kkZ;s vuqlkj fQlydj ,d Lrj ls] chpdh ?kkVh esa gksdj mPp Lrj rd igq¡prk gSA tc rd CykWd mPpLrj rd ugha ig¡qprk gS iFk ?k"kZ.k jfgr gSA blds i'pkr~ ogkW ij?k"kZ.k cy gS tks CykWd dks d njh esa jksd nsrk gSA CykWd dh izkjfEHkdpky v0 dk eku 6.0 ms–1] Å¡pkbZ ds vUrj h dk eku1.1 m vkSj xfrd ?k"kZ.k xq.kkad m dk eku 0.60 gSA d dk ekugS :-

h d

m=0.60m=0

v0

(1) 1.17 m (2) 3 m (3) 2.34 m (4) 6 m

26. ,d yksyd dks fp= esa iznf'kZr fcUnq A ls fojkekoLFkk lsNksM+r s g SaA yksyd dh jLlh ruh gqbZ gSA OA Å/okZ/kj ls30° dks.k cukrk gSA bl {k.k ij yksyd ds xksyd dk Roj.kgksxk :-

30°O

BA

(1) AOuuur

ds vuqfn'k

(2) Å/okZ/kj ds vuqfn'k

(3) OA dh yEcor~ fn'kk esa

(4) Å/okZ/kj ls 30° ls de dks.k cuk jgh fn'kk esa

27. yEckbZ l o nzO;eku m okyh ,d pkyd NM+ ?k"kZ.k jfgr urry(ftldk >qdko dks.k q gS) ij fu;r osx v ls fp=kuqlkj uhpsdh rjQ xfr djrh gSA pkyd esa /kkjk i isij ds yEcor~ vUnj

dh rjQ gSA {ks= esa m/okZ/kj Åij dh vksj pqEcdh; {ks= Br

fo|eku gSA pqEcdh; {ks= Br

dk eku gS &

q

qv F =mi Bl

(1) ql

mgsin

i (2) ql

mgtan

i

(3) q

l

mgcosi (4) ql

mgi sin

28. Primary origin of magnetism lies in :-(a) Atomic currents(b) Pauli exclusion principle(c) Polar nature of molecules(d) Intrinsic spin of electron(1) Only d (2) b, d (3) a, d (4) c, d

29. Angular velocity of a particle is w = 2

2q

where q is

angular position of particle. Tangential accelerationof particle when it has covered 1 revolution in a circleof radius 2 m :-(1) 4p3 (2) 8p3 (3) 2p3 (4) 16p3

30. A child’s pogo stick (figure) stores energy in aspring with a force constant of 2.5 × 104 N/m. Atposition (A) (xA = 0.10 m), the spring compressionis a maximum and the child is momentarily at rest.At position (B) (xB = 0), the spring is relaxed andthe child is moving upward. At position (C), thechild is again momentarily at rest at the top of thejump. The combined mass of child and pogo stickis 25 kg. Calculate the total energy of the child-stick-earth system if both gravitational and elasticpotential energies are zero for x = 0.

xC

xA

(A)(B)

(C)

(1) 125 J (2) 100 J (3) 150 J (4) 75 J31. A wire of length l is bent in the form of a circular

coil of some turns. A current i flows through thecoil. The coil is placed in a uniform magnetic fieldB. The maximum torque on the coil can be-

(1) 2iB

4pl

(2) 2iB

pl

(3) 2iB

2pl

(4) 22iB

pl

28. pqEcdRo dk ewy L=ksr gS :-(a) ijekf.od /kkjk,sa(b) ikWmyh dk viotZu fu;e(c) v.kqvksa dh /kqzoh; izofÙk(d) bysDVªkWu dk uSt pØ.k(1) dsoy d (2) b, d (3) a, d (4) c, d

29. ,d d.k dk dks.kh; osx w = 2

2q

gS] tgk¡ q d.k dh dks.kh;

fLFkfr gSA tc d.k 2 m f=T;k okys o`Ùk ij 1 pDdj r; djys bldk Li'kZ js[kh; Roj.k gS :-(1) 4p3 (2) 8p3

(3) 2p3 (4) 16p3

30. ,d cPps dh iksxks fLVd] 2.5×104 N/m cy fu;rkad okyhfLizax esa ÅtkZ lafpr djrh gSA fLFkfr (A) (xA = 0.10 m)esa, fLizax dk ladqpu vf/kdre gS] vkSj cPpk {kf.kd :i lsfojkekoLFkk esa gSA fLFkfr (B) (xB = 0) esa, fLizax eqDr gS vkSjcPpk mij dh vksj xfr'khy gSA fLFkfr (C) esa cPpk iqu% viuhdqnku ds 'kh"kZ ij {kf.kd :i ls fojkekoLFkk esa gSA cPps rFkk^iksxks fLVd* dk la;qDr nzO;eku 25 kg gSA ;fn x = 0 ds fy;sxq:Roh; rFkk izR;kLFk fLFkfr ÅtkZ 'kwU; gS Hkweh&cPpk&fLVdfudk; dh dqy ÅtkZ dh x.kuk dhft;s

xC

xA

(A)(B)

(C)

(1) 125 J (2) 100 J (3) 150 J (4) 75 J31. l yEckbZ ds rkj dks eksM+dj dqN Qsjksa okyh o`Ùkkdkj dq.Myh

cuk;h x;h gSA dq.Myh ls i /kkjk izokfgr gksrh gSA dq.Myh dksle:i pqEcdh; {ks= B esa j[kk x;k gSA dq.Myh ij vf/kdrecyk?kw.kZ gks ldrk gS &

(1) 2iB

4pl

(2) 2iB

pl

(3) 2iB

2pl

(4) 22iB

pl

32. Soft iron has property of :(1) low retentivity(2) High coercivity(3) High retentivity(4) Low susceptibility

33. Superconductors shows(a) Perfect diamagnetism(b) Perfect conductivity(c) Magnetic levitation(d) Meissener effect(1) a,b (2) a,b,d(3) a,b,c (4) a,b,c,d

34. A block of mass m is lying at rest at point P of awedge having a smooth semi-circular track ofradius R. What should be the minimum value ofa0 so that the mass m can just reach point Q.

m

P

Q

a0

(1)g2

(2) g

(3) g (4) Not possible35. A pendulum bob swings along a circular path on

a smooth inclined plane as shown in figure, wherem = 3 kg, l = 0.75 m, q = 37°. At the lowest pointof the circle the tension in the string is 274 N.The speed of the bob at the lowest point is(g = 10 m/s2) :-

mv0

q

l

(1) 9.2 ms–1

(2) 9 ms–1

(3) 6.5 ms–1

(4) 8 ms–1

32. dPps yksgs dk xq.k gS :(1) vYi /kkj.k'khyrk(2) mPp fuxzkfgrk(3) mPp /kkj.k'khyrk(4) vYi pqEcdh; izofÙk

33. vfr pkyd n'kkZrs gSa(a) iw.kZr% izfr pqEcdRo(b) iw.kZr% pkydrk(c) pqEcdh; mÙkksyu(d) ehluj izHkko(1) a,b (2) a,b,d(3) a,b,c (4) a,b,c,d

34. m nzO;eku okyk ,d CykWd] R f=T;k okys v¼ZoÙkkdkj fpdusiFk okys ,d vU; CykWd ds fcUnq P ij fojke voLFkk esa gSAa0 dk U;wure eku fdruk gks fd nzO;eku m fcUnq Q rd ig¡qpek= tk;sA

m

P

Q

a0

(1)g2

(2) g

(3) g (4) lEHko ugha gS35. ,d yksyd dk xksyk] fp= esa n'kkZ;s vuqlkj vkur ry ij

,d o`Ùkkdkj iFk ds vuqfn'k > wyrk gS] tgk¡ m = 3 kg,l = 0.75 m, q = 37° gSA oÙk ds fuEure fcUnq ij Mksjh esa ruko274 N gSA

fuEure fcUnq ij xksyd dh pky gS (g = 10 m/s2) :-

mv0

q

l

(1) 9.2 ms–1

(2) 9 ms–1

(3) 6.5 ms–1

(4) 8 ms–1

36. Figure shows a square current carrying loop

ABCD of side 2m and current 1i A2

= . The

magnetic moment M®

of the loop is-

C

A

B

zD

30° x

i = 12 A

y

(1) ˆ ˆ(i 3 k)- A-m2

(2) ˆ ˆ( j k)- A-m2

(3) ˆ ˆ( 3 i k)+ A-m2

(4) (i + k)ˆ ˆ A-m2

37. Curie law for paramagnetic substances is valid-(1) above curie temperature(2) below curie temperature(3) Above magnetic saturation(4) below magnetic saturation

38. A small ball is rolled with speed u from point Aalong a smooth circular track as shown in figure.If x = 3R, then determine the required speed u sothat the ball returns to A, the point of projectionafter passing through C, the highest point.

A

C

R

u

x B Referencelevel

X

Y

(1)3

gR2

(2) 1

gR2

(3)5

gR3

(4) 5

gR2

36. 2 eh HkqTkk okys oxkZdkj ywi ABCD esa izokfgr /kkjk dk

eku 1i A2

= gSA oy; ds pqEcdh; vk?kw.kZ M®

dk eku

gksxk&

C

A

B

zD

30° x

i = 12 A

y

((1) ˆ ˆ(i 3 k)- A-m2

(2) ˆ ˆ( j k)- A-m2

(3) ˆ ˆ( 3 i k)+ A-m2

(4) (i + k)ˆ ˆ A-m2

37. vuqpqEcdh; inkFkks± ds fy;s D;wjh dk fu;e ykxw gksrk gS -(1) D;wjh rki ds mGij(2) D;wjh rki ls uhps(3) pqEcdh; larIrrk ls mGij(4) pqEcdh; larIrrk ls uhps

38. ,d NksVh xsan u pky ds lkFk fp= esa n'kkZ;s vuqlkj ,d fpduso`Ùkkdkj iFk ds fcUnq A ls yq<+drh gSA ;fn x = 3R gS] rks udk og vko';d eku Kkr dhft;s fd xsan mPpre fcUnq C lsxqtjus ds i'pkr~ iqu% iz{ksi.k fcUnq A ij ykSVdj vk;s :-

A

C

R

u

x B Referencelevel

X

Y

(1)3

gR2

(2) 1

gR2

(3)5

gR3

(4) 5

gR2

39. Same current i is flowing in three infinitely longwires along positive x, y and z directions. Themagnetic field at a point (0, 0, –a) would be-

(1) 0µ i ˆ ˆ( j i)2 a

-p

(2) 0µ i ˆ ˆ(i j)2 a

+p

(3) 0µ i ˆ ˆ(i j)2 a

-p

(4) 0µ i ˆ ˆ ˆ(i j k)2 a

+ +p

40. An aircraft executes a horizontal loop of radius1 km with a steady speed of 900 kmh–1. Itscentripetal acceleration is.(1) 62.5 m/s2 (2) 30 m/s2

(3) 32.5 m/s2 (4) 72.5 m/s2

41. A large straight current carrying conductor is bent

in the form of L shape. Find Br

at P :-

Pa

a

ii

(1) Zero

(2) 0µ i ˆ(2 2) k4 a

+p

(3) 0µ i4 ap

(4) 02µ iap

42. A pump is required to lift 1000 kg of water perminute from a well 20 m deep and eject it at a rateof 20 ms–1. What HP (horsepower) engine isrequired for the purpose of lifting water ?(1) 8.85 (2) 4.42 (3) 17.7 (4) 2.6

39. /kukRed x, y o z fn'kkvksa esa vuqfn'k fLFkr vuar yEckbZ;ksa okysrhu rkjksa ls leku /kkjk i izokfgr gks jgh gSA fcUnq (0, 0, –a)ij pqEcdh; {ks= gksxk –

(1) 0µ i ˆ ˆ( j i)2 a

-p

(2) 0µ i ˆ ˆ(i j)2 a

+p

(3) 0µ i ˆ ˆ(i j)2 a

-p

(4) 0µ i ˆ ˆ ˆ(i j k)2 a

+ +p

40. ,d ok;q;ku 1 km f=T;k okys {ksfrt ywi dks fLFkj pky900 kmh–1 ds lkFk iwjk djrk gSA bldk vfHkdsUnzh; Roj.kgS&(1) 62.5 m/s2 (2) 30 m/s2

(3) 32.5 m/s2 (4) 72.5 m/s2

41. ,d yEcs ,oa lh/ks /kkjkokgh pkyd dks L vkÏfr esa eksM + fn;k

x;k gSA P ij Br

Kkr dhft, :-

Pa

a

ii

(1) 'kwU;

(2) 0µ i ˆ(2 2) k4 a

+p

(3) 0µ i4 ap

(4) 02µ iap

42. ,d 20 m xgjs dq¡, ls 20 ms–1 dh pky ds lkFk] izfr feuV1000 kg ikuh ckgj fudkyus ds fy;s ,d iEi vko';d gSAikuh dks ckgj [khapus ds fy;s vko';d batu fdruh v'o 'kfDrdk gksuk pkfg, ?(1) 8.85 (2) 4.42 (3) 17.7 (4) 2.6

43. A car is moving in a circular horizontal track ofradius 10m with constant speed of 10 m/s. A plumbbob is suspended from the roof of the car by a lightstring length 1m. The angle made by the stringwith track is :-(1) Zero (2) 30°(3) 45° (4) 60°

44. Rank the value of B.dlòurr

Ñ for the closed paths

shown in figure from the smallest to largest.

a

2A×

2A.

3A.

bd

c

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

45. Figure shows a plot of the potential energy as afunction of x for a particle moving along thex-axis.

a b c d

U

Consider the following statements :(A) a, c and d are points of equilibrium(B) a is a point of stable equilibrium(C) b is a stable equilibrium pointCorrect statements are :-(1) only A(2) only A and B(3) only C(4) A, B and C

43. ,d dkj 10m f=T;k okys o`Ùkkdkj {kSfrt iFk ij fu;r pky10 m/s ds lkFk xfr dj jgh gSA dkj dh Nr ls 1m yEchMksjh ls ,d lkoy dk xksyd yVdk;k x;k gSA Mksjh }kjk iFkds lkFk cuk;k x;k dks.k gS :-

(1) 'kwU; (2) 30°

(3) 45° (4) 60°

44. fp=kuqlkj can iFkksa ds fy, B.dlòurr

Ñ ds ekuksa dks NksVs ls cM+s dh

vksj Øe nhft,A

a

2A×

2A.

3A.

bd

c

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

45. x-v{k ds vuqfn'k xfr dj jgs d.k ds fy;s] x ds Qyu :iesa fLFkfrt ÅtkZ dk vkjs[k fp= esa n'kkZ;k x;k gSA

a b c d

U

fuEufyf[kr dFkuksa ij fopkj dhft, &(A) a, c rFkk d lkE;koLFkk okys fcUnq gSa(B) a LFkk;h lkE;koLFkk dk fcUnq gS(C) b LFkk;h lkE;koLFkk dk fcUnq gSlgh dFku gS %&(1) dsoy A

(2) dsoy A o B

(3) dsoy C

(4) A, B rFkk C

46. For the reaction, 2N2O5(g) ® 4 NO2(g) + O2(g), therate of formation of O2 is 0.032g hour–1, then rateof conversion of N2O5 in g hour–1 is :-(1) 0.216 (2) 0.098(3) 0.627 (4) 0.313

47. For a reaction 2A + 2B + C ® productswith the help of the following table, find order withrespect to A, B, C

[A] [B] [C] rate ofreaction

0.010 0.005 0.010 5 × 10–3

0.010 0.005 0.015 5 × 10–3

0.010 0.010 0.010 1 × 10–2

0.005 0.005 0.010 2.5 × 10–3

(1) 3,2,0 (2) 1,1,0 (3) 3,2,1 (4) 2,2,148. Decomposition reaction of N2O5 follows first order

and is written as :-2N2O5(g) ® 4NO2(g) + O2(g); rate = k[N2O5]N2O5(g) ® 2NO2(g) + O2(g); rate = k' [N2O5]which of the following relation is true ?(1) k = k' (2) k > k' (3) k > 2k' (4) 2k = k'

49. The mechanism of the reaction 2NO2 + F2 ® 2NO2Fis :-

(i) NO2 Slow¾¾¾® NO + O

(ii) F2 + O + NO fast¾¾® NO2F + F

(iii) F + NO2 fast¾¾® NO2F.

Select the correct one(1) the reaction is of 3rd order(2) the molecularity of the reaction is sum of

molecularities of all steps(3) reaction is zero order w.r.t. F2

(4) half life of reaction depends upon initialconcentration of NO2.

50. If the slope of a line of the graph between time(in hour) and concentration of formed product forzero order reaction is 0.2 then what will be theinitial concentration of reactant in mol L–1, if after30 minute its concentration is 0.05 mol L–1:(1) 0.01 (2) 0.15 (3) 0.25 (4) 0.50

TOPIC : Chemical Kinetics, Equilibrium, Ionic equilibrium.

46. vfHkfØ;k, 2N2O5(g) ® 4 NO2(g) + O2(g), ds fy, O2 dsfuekZ.k dh nj 0.032g hour–1 gS rks N2O5 ds ifjorZu dhnj g hour–1 esa gS :-(1) 0.216 (2) 0.098(3) 0.627 (4) 0.313

47. vfHkfØ;k 2A + 2B + C ® mRikn] ds fy,fuEu lkj.kh dh lgk;rk ls A, B, C ds lkis{k dksfV gS&

[A] [B] [C] vfHkfØ;k osx

0.010 0.005 0.010 5 × 10–3

0.010 0.005 0.015 5 × 10–3

0.010 0.010 0.010 1 × 10–2

0.005 0.005 0.010 2.5 × 10–3

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

48. N2O5 dk fo?kVu izFke dksfV dk vuqlj.k djrk gS ftls fuEuizdkj fy[kk x;k gS :-2N2O5(g) ® 4NO2(g) + O2(g); nj = k[N2O5]N2O5(g) ® 2NO2(g) + O2(g); nj = k' [N2O5]fuEu esa ls dkSulk lEca/k lgh gS ?(1) k = k' (2) k > k' (3) k > 2k' (4) 2k = k'

49. vfHkfØ;k 2NO2 + F2 ® 2NO2F ds fy, fØ;kfof/k infn, x, gSa&

(i) NO2 ¾¾¾®/khek NO + O

(ii) F2 + O + NO ¾¾®rhoz NO2F + F

(iii) F + NO2 ¾¾®rhoz NO2F.

lgh fodYi dks pqfu,&(1) vfHkfØ;k dh dksfV 3 gSA(2) vfHkfØ;k dh vkf.odrk lHkh inksa dh vkf.odrk dk ;ksx

gksxkA(3) vfHkfØ;k dh dksfV F2 ds lkis{k 'kwU; gSA(4) vfHkfØ;k dk v¼Z vk;qdky NO2 dh izkjfEHkd lkUnzrk

ij fuHkZj djrk gSA50. 'kwU; dksfV vfHkfØ;k ds fy, le; (?kaVs esa) rFkk mRikn dh mRiUu

lkUnzrk ds e/; xzkQ dk <+ky 0.2 gSa rks fØ;kdkjd dh izkjfEHkdlkUnzrk eksy L–1 esa D;k gksxh ;fn 30 feuV i'pkr~ bldh lkUnzrk0.05 eksy L–1 gks &(1) 0.01 (2) 0.15(3) 0.25 (4) 0.50

51. At 300 K the half-life of a sample of a gaseouscompound intially at 1 atm is 100 s. When thepressure is 0.5 atm the half-life is 50 s. The orderof reaction of compound is :-(1) 0 (2) 1 (3) 2 (4) 3

52. A first order reaction is 50% completed in 20 min at27ºC and in 5 min. at 47ºC. The energy of activationof the reaction is :-(1) 43.85 KJ/mol (2) 55.3 KJ/mol(3) 11.97 KJ/mol (4) 6.65 KJ/mol

53. The rate of reaction is expressed in different waysas follows :

1 d[C] 1 d[D] 1 d[A] d[B]2 dt 3 dt 4 dt dt

+ = - = + = -

The reaction is :(1) 4A + B ® 2C + 3D(2) B + 3D ® 4A + 2C(3) A + B ® C + D (4) B + D ® A + C

54. For the reaction 2A ® B + 3C ; if

– 21

d[A]k [A]

dt= ; 2

2

d[B]k [A]

dt= ; 2

3

d[C]k [A]

dt= the

correct relation between k1, k2 and k3 is :(1) k1 = k2 = k3 (2) 2k1 = k2 = 3k2

(3) 4k1 = k2 = 3k2 (4) 1 32

k kk

2 3= =

55. For A(g) + B(g) ¾® C(g) ; rate = k[A]1/2[B]2, if initialconcentration of A and B are increased by factors4 and 2 respectively, then the initial rate is changedby the factor :(1) 4 (2) 6(3) 8 (4) None of these

56. The reaction 2NO(g) + O2(g) ¾¾® 2NO2(g) is offirst order. If volume of reaction vessel is reducedto 1/3, the rate of reaction would be :(1) 1/3 times (2) 2/3 times(3) 3 times (4) 6 times

57. The molecularity of a complex reaction givenbelow is :

2N2O5 (g) ¾® 4NO2(g) + O2(g)(1) 1 (2) 2(3) 3 (4) has no meaning

51. 300 K ij] 1 atm nkc ij j[ks xSlh; ;kSfxd ds izfrn'kZdh v¼Z&vk;q 100 lSd.M gSA tc nkc 0.5 atm gS rksv¼Z&vk;q dky 50 lSd.M gSA ;kSfxd dh vfHkfØ;k dh dksfVgS :-(1) 0 (2) 1 (3) 2 (4) 3

52. ,d izFke dksfV vfHkfØ;k 27ºC ij 20 feuV esa 50% iw.kZ gksrhgS rFkk 47ºC ij 5 feuV esa 50% iw.kZ gksrh gSA vfHkfØ;k dhlfØ;.k ÅtkZ gS&(1) 43.85 KJ/mol (2) 55.3 KJ/mol(3) 11.97 KJ/mol (4) 6.65 KJ/mol

53. fdlh vfHkfØ;k dh nj fuEu izdkj ls iznf'kZr dh tk ldrhgS &

1 d[C] 1 d[D] 1 d[A] d[B]2 dt 3 dt 4 dt dt

+ = - = + = -

rks vfHkfØ;k gS &(1) 4A + B ® 2C + 3D(2) B + 3D ® 4A + 2C(3) A + B ® C + D (4) B + D ® A + C

54. vfHkfØ;k 2A ¾® B + 3C ds fy, ; fn

– 21

d[A]k [A]

dt= ; 2

2

d[B]k [A]

dt= ; 2

3

d[C]k [A]

dt= gSa] rks

k1, k2 rFkk k3 esa lgh lEcU/k gS\(1) k1 = k2 = k3 (2) 2k1 = k2 = 3k2

(3) 4k1 = k2 = 3k2 (4) 1 32

k kk

2 3= =

55. A(g) + B(g) ¾® C(g) ds fy, nj = k[A]1/2[B]2 gSA ;fnA o B dh izkjafHkd lkUnzrk,¡ Øe'k% 4 o 2 xquh dj nha tk;srks izkjafHkd nj esa gksus okyh o`f¼ fdrus xquk gS\

(1) 4 (2) 6

(3) 8 (4) buesa ls dksbZ ugha

56. vfHkfØ;k 2NO(g) + O2(g) ¾¾® 2NO2(g) izFke dksfV dhgSA ;fn vfHkfØ;k ik= dk vk;ru 1/3 dj fn;k tk;s rksvfHkfØ;k dh nj gS &(1) 1/3 xquk (2) 2/3 xquk(3) 3 xquk (4) 6 xquk

57. uhps nh x;h tfVy vfHkfØ;k dh v.kqla[;rk gS &

2N2O5 (g) ¾® 4NO2(g) + O2(g)

(1) 1 (2) 2

(3) 3 (4) dksbZ vFkZ ugha gSA

58. A first order reaction is 75% completed in100 minutes. How long time will it take for it's87.5% completion ?(1) 125 min (2) 150 min(3) 175 min (4) 200 min

59. Rate constant for a chemical reaction takes placeat 500 K is expressed as K = A. e–1000 . Theactivation energy of the reaction is :(1) 100 cal/mol (2) 1000 kcal/mol(3) 104 kcal/mol (4) 106 kcal/mol

60. Which of the following represents the expression

for 3

th4

life of first order reaction ?

(1)k

log 4 / 32.303

(2) 2.303

log 3 / 4k

(3) 2.303

log 4k

(4) 2.303

log 3k

61. The minimum energy for molecules required togive effective collision is called :(1) Kinetic energy (2) Potential energy(3) Threshold energy (4) Activation energy

62. An endothermic reaction A ® B has an activationenergy 15 k cal/mol and the heat of reaction is5 k cal/mol. What will be the threshold energyof the reaction B ® A if potential energy ofproduct is 10 kcal/mol :-(1) 20 k cal/mol (2) 15 k cal/mol(3) 10 k cal/mol (4) Zero

63. For a certain gaseous reaction rise of temperaturefrom 25° C to 35° C doubles the rate of reaction.What is the value of activation energy :–

(1) 10

2.303R 298 308´ ´

(2) 2.303 10298 308R

´´

(3) 0.693R 10290 308

´´

(4) 0.693R 298 308

10´ ´

58. izFke dksfV dh ,d vfHkfØ;k 100 feuV esa 75% iw.kZ gkstkrh gSA vfHkfØ;k ds 87.5% iw.kZ gksus esa yxus okyk le;gksxk&(1) 125 min (2) 150 min(3) 175 min (4) 200 min

59. 500 K ij gk s jgh ,d vfHkfØ;k ds fy, nj fLFkjk adK = A. e–1000 gSA vfHkfØ;k dh lfØ;.k ÅtkZ gS &

(1) 100 dSyksjh@eksy (2) 1000 fdyks dSyksjh@eksy

(3) 104 fdyks dSyksjh@eksy (4) 106 fdyks dSyksjh@eksy

60. fuEufyf[kr esa ls dkSulk O;atd fdlh izFke dksfV dh vfHkfØ;k

ds 34

iw.kZ gksus esa yxus okys le; dks iznf'kZr djrk gS &

(1)k

log 4 / 32.303

(2) 2.303

log 3 / 4k

(3) 2.303

log 4k

(4) 2.303

log 3k

61. v.kqvksa dh izHkkoh VDdjksa ds fy, vko';d U;wure ÅtkZ dgykrhgSA(1) xfrt ÅtkZ (2) fLFkfrt ÅtkZ(3) nsgyh ÅtkZ (4) lfØ;.k ÅtkZ

62. ,d Å"ek'kk s"kh vfHkfØ;k A ® B dh lfØ;.k ÅtkZ15 kcal/mol gS vkSj vfHkfØ;k Å"ek 5 fd- dSyksjh@eksy gSvfHkfØ;k B ® A dh nsgyh ÅtkZ D;k gksxh ;fn mRikn dhfLFkfrt ÅtkZ 10 fd- dSyksjh@eksy gS%

(1) 20 fd- dSyksjh@eksy (2) 15 fd- dSyksjh@eksy

(3) 10 fd- dSyksjh@eksy (4) 'kwU;

63. ,d xSlh; vfHkfØ;k ds fy, rki 25° C ls 35° C rd c<+kusij vfHkfØ;k dh nj nks xquk gks tkrh gSA lfØ;.k ÅtkZ dk ekuD;k gS %&

(1) 10

2.303R 298 308´ ´

(2) 2.303 10298 308R

´´

(3) 0.693R 10290 308

´´

(4) 0.693R 298 308

10´ ´

64. For a reaction

3 2 26H 5Br BrO 3Br 6H O+ - -+ + ¾¾® +

If rate of consumption of 3BrO- is x mol L–1s–1

then calculate rate of formation of Br2:-

(1)x3

(2) 2x3

(3) x4

(4) 3x

65. Arrhenius equation may not be written as :–

(1) K = AeaE-

RT

(2) lnK = lnA aE2.303RT

-

(3) logK = logA aE2.303RT

-

(4) ddT

(ln K) = a2

ERT

66. 2A(g) + B(g) �������� Product, is an elementary reaction,If pressure is increased three times of the initialpressure, the velocity of forward reaction will be------ of the previous velocity:-(1) 9 times (2) 27 times

(3) 19

times (4) 127

times

67. log p

c

K

K + log RT = 0 is true relationship for the

following reaction:-(1) PCl5 � PCl3 + Cl2 (2) 2SO2 + O2 � 2SO3

(3) N2 + 3H2 � 2NH3 (4) (2) and (3) both68. PCl5 (g) � PCl3 (g) + Cl2 (g)

In above reaction, at equilibrium condition molefraction of PCl5 is 0.4 and mole fraction of Cl2 is0.3. Then find out mole fraction of PCl3

(1) 0.3 (2) 0.7 (3) 0.4 (4) 0.669. Given the reaction between 2 gases represented

by A2 and B2 to give the compound AB(g).A2(g) + B2(g) = 2AB(g).At equilibrium, the concentrationof A2 = 3.0 × 10–3 Mof B2 = 4.2 × 10–3 Mof AB = 2.8 × 10–3 MIf the reaction takes place in a sealed vessel at527°C, then the value of KC will be :-(1) 0.62 (2) 4.5 (3) 2.0 (4) 1.9

64. vfHkfØ;k

3 2 26H 5Br BrO 3Br 6H O+ - -+ + ¾¾® +

ds fy, ;fn 3BrO- ds iz;qDr gksus dh nj x eksy L–1s–1 gS rksBr2 ds cuuss dh nj Kkr djks :-

(1)x3

(2) 2x3

(3) x4

(4) 3x

65. vkjsfu;l lehdj.k dks ugha fy[kk tk ldrk gS :–

(1) K = AeaE-

RT

(2) lnK = lnA aE2.303RT

-

(3) logK = logA aE2.303RT

-

(4) ddT

(ln K) = a2

ERT

66. 2A(g) + B(g) �������� mRikn] ,d izkjfEHkd vfHkfØ;k gS

mijksä fØ;k ds fy, ;fn nkc dks izkjfEHkd nkc dk rhu xqukdj nsa rks vxz fØ;k dk osx igys osx dk---------gks tk;sxk%(1) 9 xquk (2) 27 xquk

(3) 19

xquk (4) 127

xquk

67. log p

c

K

K + log RT = 0, fuEu vfHkfØ;k ds fy;s lgh lEcaèk

gksxk%(1) PCl5 � PCl3 + Cl2 (2) 2SO2 + O2 � 2SO3

(3) N2 + 3H2 � 2NH3 (4) (2) rFkk (3) nksuksa

68. vfHkfØ;k PCl5(g) ������ PCl3(g) + Cl2(g) esa lkE; voLFkkij PCl5 dh eksy fHkUu 0.4 rFkk Cl2 dh eksy fHkUu 0.3 gS]rks PCl3 dh eksy fHkUu gksxh%(1) 0.3 (2) 0.7 (3) 0.4 (4) 0.6

69. A2 rFkk B2 }kjk iznf'kZr dh xbZ nks xSlksa ds e/; vfHkfØ;k gksusij AB(g) ;kSfxd curk gSA

A2(g) + B2(g) = 2AB(g).lkE; ij

A2 dh lkUærk 3.0 × 10–3 MB2 dh lkUærk 4.2 × 10–3 MAB dh lkUærk 2.8 × 10–3 M

;fn 527°C ij vfHkfØ;k ,d can ik= esa gksrh gS rks KC dkeku gksxk :-(1) 0.62 (2) 4.5 (3) 2.0 (4) 1.9

70. For the reaction C(s) + CO2(g) � 2CO(g) the partialpressure of CO and CO2 are 2.0 and 4.0 atm.respectively at equilibrium. The KP for the reaction is(1) 0.5 (2) 4.0(3) 8.0 (4) 1

71. For the equilibrium SO2Cl2(g) � SO2(g) + Cl2(g) what

is the temperature at which P

C

KK

= 1 ?

(1) 36.54 K (2) 273 K(3) 12.18 K (4) 0 K

72. At a certain temperature, the equilibrium constantKC is 25 for the reaction

A2(g) + B2(g) � C2(g) + D2(g)

If we take 1 mol of each of the four gases in a1 L container, what would be equilibriumconcentration of A2 ?(1) 0.33 (2) 0.133(3) 0.033 (4) 1.33

73. In equation N2(g) + O2(g) � 2NO(g), KC = 10–40

Initially 0.4 mole of N2 and 0.9 mole of O2 weremixed then moles of N2, O2 and NO at equilibriumare (approximate) :-(1) 0.4, 0.9, 6 × 10–21

(2) 4 × 10–21, 6 × 10–21, 6 × 10–21

(3) 0.4, 0.9, 0.9(4) 6 × 10–21, 0.4, 0.9

74. At 200°C, nitrogen oxide reacts with oxygen toform nitrogen dioxide as follows :

2NO + O2 � 2NO2, KC = 3 × 106

In a mixture of the three species at equilibrium,we can accurately predict that(1) The concentrations of both NO and O2 will be

much larger than the concentration of NO2

(2) The concentrations of both NO and O2 will bemuch smaller than the concentration of NO2

(3) The concentrations of either NO or O2 andpossibly both will be much smaller than theconcentration of NO2

(4) The concentration of O2 will be exactly onehalf the concentration of NO

70. vfHkfØ;k C(s) + CO2(g) � 2CO(g) ds fy;s lkE; ijCO rFkk CO2 ds vkaf'kd nkc Øe'k% 2.0 rFkk 4.0 ok;qe.MygSaA vfHkfØ;k ds fy;s KP gS :-(1) 0.5 (2) 4.0(3) 8.0 (4) 1

71. SO2Cl2(g) � SO2(g) + Cl2(g) vfHkfØ;k esa fdl rki ij

P

C

KK

= 1 gksxk\

(1) 36.54 K (2) 273 K(3) 12.18 K (4) 0 K

72. fdlh rki ij fuEu vfHkfØ;kA2(g) + B2(g) � C2(g) + D2(g)

dk KC dk eku 25 gSA ;fn izR;sd xSl ds 1 eksy 1 L ik=esa fy;k tk, rks A2 dh lkE; ij lkaærk D;k gksxh\(1) 0.33(2) 0.133(3) 0.033(4) 1.33

73. N2(g) + O2(g) � 2NO(g), KC = 10–40

izkjEHk esa 0.4 eksy N2 rFkk 0.9 eksy O2 fefJr fd;s x;sAlkE;koLFkk ij N2, O2 o NO ds eksy gksaxs (yxHkx) :-

(1) 0.4, 0.9, 6 × 10–21

(2) 4 × 10–21, 6 × 10–21, 6 × 10–21

(3) 0.4, 0.9, 0.9

(4) 6 × 10–21, 0.4, 0.9

74. 200°C rki ij ukbVªkstu vkWDlkbM] O2 ds lkFk fØ;k djfuEukuqlkj NO2 cukrk gS :

2NO + O2 � 2NO2, KC = 3 × 106

lkE;koLFkk ij rhuksa Lih'kht ds feJ.k ds ckjs esa dkSulk dFkufcYdqy lgh gS

(1) NO rFkk O2 nksuksa dh lkUærk,¡ NO2 dh lkUærk ls cgqrvf/kd gSaA

(2) NO rFkk O2 nksuksa dh lkUærk,¡ NO2 dh lkUærk ls cgqrde jgrh gSaA

(3) NO ;k O2 dh lkUærk vFkok NO rFkk O2 nksuksa dh lkUnzrkNO2 dh lkUnzrk ls cgqr de gSA

(4) O2 dh lkUærk, NO dh lkUærk dh fcYdqy vk/khgSA

75. For the reaction N2(g) + 3H2(g) � 2NH3(g) initialN2 and H2 were taken in the mole ratio of 1 : 3. Atthe point of equilibrium 40% of each reactant hasbeen reacted. If the total pressure at equilibrium wasP. The partial pressure of NH3 will be -

(1) P3 (2)

P6

(3) P4

(4) P8

76. BOH is a weak monoacidic base (Kb = 1.5 × 10–5).

The molar concentration of BOH that provides éOHùë û

of 1.5 × 10–3 M will be :-(1) 0.15 (2) 0.0015(3) 1.5 × 10–5 (4) 0.015

77. Which of the following salt will give cationichydrolysis ?(1) CH3COONa

(2) 100 ml N10

NH4OH + 100 ml N10

HCl

(3) 100 ml N10

NH4OH + 100 ml N20

HCl

(4) 100 ml N10

NaOH + 100 ml N10

HCl

78. Out of the following statements the correctstatement is :-(a) H+ form H9O4

+ in H2O(b) H+ from 0.1 M CH3COOH is less than 0.1 M HCl(c) 0.2 M NaCN + 0.1 M HCl forms buffer(d) Compounds in which OH– is present is known

as bases only(1) a,b,c only(2) b,c,d only(3) a,b,c,d(4) a,b only

79. The equilibrium constant for the reaction.OCl–(aq) + H2O(g) � HOCl(aq) + OH–

(aq) is 3.6×10–7

hence Ka for HOCl is :-(1) 3.6 × 10–7 (2) 6 × 10–1

(3) 2.8 × 106 (4) 2.8 × 10–8

75. N2(g) + 3H2(g) � 2NH3(g) vfHkfØ;k ds fy;s izkjEHkesa N2 : H2 dk eksy vuqikr 1 : 3 fy;k x;kA lkE; igq¡pusrd izR;sd dk 40% vfHkd`r gks pqdk gksrk gSA ;fn lkE; ijdqy nkc P gks rks NH3 dk vkaf'kd nkc gksxk-

(1) P3 (2)

P6

(3) P4

(4) P8

76. ,dy vEyh; nqcZy {kkj BOH ds fy, Kb = 1.5 × 10–5 gSABOH dh eksyj lkUnzrk Kkr djks tks 1.5 × 10–3

M [OH–]iznku dj ldsA

(1) 0.15 (2) 0.0015

(3) 1.5 × 10–5 (4) 0.015

77. fuEu esa ls dkSulk yo.k /kuk;fud ty vi?kVu nsxk\

(1) CH3COONa

(2) 100 ml N10

NH4OH + 100 ml N10

HCl

(3) 100 ml N10

NH4OH + 100 ml N20

HCl

(4) 100 ml N10

NaOH + 100 ml N10

HCl

78. fuEu dFkuksa esa ls dkSulk dFku lgh gS :-(a) H+ ikuh esa H9O4

+ cukrk gSA(b) 0.1 M HCl dh rqyuk esa] 0.1 M CH3COOH ls H+

de izkIr gksrk gSA(c) 0.2 M NaCN + 0.1 M HCl cQj foy;u cukrs gSaA(d) os ;kSfxd ftuesa OH– mifLFkr gksrs gaS] dsoy os gh {kkj

dgykrs gSaA(1) dsoy a,b,c(2) dsoy b,c,d(3) a,b,c,d(4) dsoy a,b

79. fuEu vfHkfØ;k ds fy, lkE; fu;rkadOCl–

(aq) + H2O(g) � HOCl(aq) + OH–(aq) 3.6 × 10–7

gS vr% HOCl ds fy, Ka gksxk :-(1) 3.6 × 10–7 (2) 6 × 10–1

(3) 2.8 × 106 (4) 2.8 × 10–8

80. The precipitate of Ag2CrO4 (KSP = 1×10–8) isobtained when equal volume of following are mixed?(1) 10–4 M Ag+ + 10–4 M CrO4

2–

(2) 10–2 M Ag+ + 10–3 M CrO42–

(3) 10–5 M Ag+ + 10–3 M CrO42–

(4) 10–2 M Ag+ + 10–5 M CrO42–

81. Which ions will have highest Ksp values of theirsulphides.(1) Zn2+ Cu2+ (2) Pb2+ Hg2+

(3) Zn2+ Mn2+ (4) Cd2+ Bi3+

82. At 25°C, the dissociation constant of CH3COOHand NH4OH in aqueous solution are almost thesame. The pH of a solution 0.01 N CH3COOH is4.0 at 25°C. The pOH of 0.01 N NH4OH solutionat the same temperature would be :-(1) 3.0 (2) 4.0 (3) 10.0 (4) 10.5

83. A weak acid, HA with an ionization constant ofapproximately 10–4, is prepared in a 1.0 molaraqueous solution. What will happen to thepercentage dissociation of this acid as more purewater is added to the solution ?(1) Percentage dissociation remains unchanged(2) Percentage dissociation will increase(3) Percentage dissociation will decrease(4) The ionization constant will increase

84. The [OH–] in 0.01 M aqueous solution of NaOCN(Kb for OCN– = 10–10) will be :-(1) 10–6 M (2) 10–3 M(3) 10–8 M (4) None of these

85. A 100 mL portion of water is added to each of thefollowing two solutions :-(i) 100 mL of 0.02 M KCN(ii) 100 mL of 0.02 M HClWhich of the following statements is correct ?(1) There will be no change in pH of solution (i)

and (ii)(2) The pH of solution (i) will decrease but pH of

solution (ii) will increase(3) The pH of solution (i) will remain same but of

solution (ii) will decrease(4) The pH of solution (ii) will remain same but

of solution (i) will increase

80. Ag2CrO4 (KSP = 1×10–8) dk vo{ksi fuEu vk;uksa ds lekuvk;ruksa dks feykus ij izkIr gksrk gS &(1) 10–4 M Ag+ + 10–4 M CrO4

2–

(2) 10–2 M Ag+ + 10–3 M CrO42–

(3) 10–5 M Ag+ + 10–3 M CrO42–

(4) 10–2 M Ag+ + 10–5 M CrO42–

81. fuEu vk;uksa esa ls fduds lYQkbZM yo.kksa dk Ksp eku lokZfèkdgksxk?(1) Zn2+ Cu2+ (2) Pb2+ Hg2+

(3) Zn2+ Mn2+ (4) Cd2+ Bi3+

82. 25°C ij ,flfVd vEy rFkk NH4OH ds tyh; foy;u esafo;kstu fLFkjkad ds eku yxHkx leku gSaA 25°C rki ij0.01 N CH3COOH foy;u dh pH = 4 gks rks blh rkiij 0.01 N NH4OH dh pOH Kkr djks :-(1) 3.0 (2) 4.0(3) 10.0 (4) 10.5

83. ,d nqcZy vEy HA, ftldk vk;uu fLFkjk ad yxHkx10–4 gS] dk 1.0 eksyj tyh; foy;u cuk;k tkrk gSA blfoy;u esa vfrfjDr 'kq¼ ty feykus ij izfr'kr fo;kstu ij D;kizHkko gksxk ?

(1) izfr'kr fo;kstu vifjofrZr jgrk gSA

(2) izfr'kr fo;kstu c<+sxkA

(3) izfr'kr fo;kstu ?kVsxkA

(4) vk;uu fLFkjkad c<+sxkA

84. 0.01 M, NaOCN ds tyh; foy;u esa [OH–] dh lkUnzrkgksxh (OCN– ds fy, Kb = 10–10) :-(1) 10–6 M (2) 10–3 M(3) 10–8 M (4) buesa ls dksbZ ugha

85. 100 mL ty fuEu nks foy;uksa (izR;sd esa) feyk;k tkrkgS :-(i) 100 mL 0.02 M KCN foy;u(ii) 100 mL 0.02 M HCl foy;ufuEu esa ls dkSulk dFku lgh gS ?(1) foy;u (i) rFkk (ii) dh pH esa dksb Z ifjorZu ugha

gksxkA(2) foy;u (i) dh pH ?kVsxh rFkk foy;u (ii) dh pH

c<+sxhA(3) foy;u (i) dh pH vifjofrZr jgsxh rFkk foy;u (ii) dh

pH ?kVsxhA(4) foy;u (ii) dh pH vifjofrZr jgsxh rFkk foy;u (i) dh

pH c<+ sxhA

86. Which of the following aqueous solutions is abuffer with a pH greater than 7.0 ? For HNO2,Ka = 4.6 × 10–4, and for NH3, Kb = 1.8 × 10–5.

(1) 10 mL of 0.1 M NH3(aq) + 10 mL of 0.1 MHCl (aq)

(2) 10 mL of 0.1 M HNO2(aq) + 5.0 mL of 0.1 MNaOH (aq)

(3) 10 mL of 0.1 M HNO2(aq) + 10 mL of 0.1 MNaOH (aq)

(4) 10 mL of 0.1 M NH3(aq) + 5.0 mL of 0.1 MHCl (aq)

87. In a saturated solution of H2S decreasing the pHof the solution will cause :-

(1) The S2– concentration to decrease

(2) The H2S concentration to decrease

(3) The S2– concentration to increase

(4) No change in either the H2S or S2–concentration

88. The strongest conjugate base is –

(1) NO3– (2) Cl–

(3) SO4–2 (4) CH3COO–

89. 200 mL, N20

HNO3, 100 mL N20

H2SO4, 300 mL

M30

HCl, 50 mL, 0.1M Ba(OH)2 and 250 mL

M10

NaCl, solutions were mixed then value of pH

of resultant solution is :-

(1)1.65 (2) 2.0

(3) 1.78 (4) 0.16

90. 1M NaCl and 1M NaOH are present in an aqueoussolution. The solution is :–

(1) Not a buffer solution with pH < 7

(2) Not a buffer solution with pH > 7

(3) A buffer solution with pH < 7

(4) A buffer solution with pH > 7

86. fuEu esa ls dkSulk tyh; foy;u] pH = 7.0 ls vf/kd dkcQj gSA HNO2 ds fy, Ka = 4.6 × 10–4 rFkk NH3 ds fy,Kb = 1.8 × 10–5.

(1) 10 mL of 0.1 M NH3(aq) + 10 mL of 0.1 MHCl (aq)

(2) 10 mL of 0.1 M HNO2(aq) + 5.0 mL of 0.1 MNaOH (aq)

(3) 10 mL of 0.1 M HNO2(aq) + 10 mL of 0.1 MNaOH (aq)

(4) 10 mL of 0.1 M NH3(aq) + 5.0 mL of 0.1 MHCl (aq)

87. H2S ds larIr foy;u esa] pH de djus ij :-

(1) S2– dh lkUærk ?kVsxh

(2) H2S dh lkUærk ?kVsxh

(3) S2– dh lkUærk c<+sxh

(4) H2S ;k S2– dh lkUærk esa dksbZ ifjorZu ugha gksxkA

88. çcyre la;qXeh {kkj gS –

(1) NO3– (2) Cl–

(3) SO4–2 (4) CH3COO–

89. 200 mL, N20

HNO3, 100 mL N20

H2SO4, 300 mL

M30

HCl, 50 mL, 0.1M Ba(OH)2 rFkk 250 mL,

M10

NaCl, foy;u fefJr fd;s x;s rks ifj.kkeh foy;u dh

pH dk eku gksxk :-

(1)1.65 (2) 2.0

(3) 1.78 (4) 0.16

90. fdlh tyh; foy;u esa 1M NaCl o 1M NaOH gS] rksfoy;u :–

(1) pH < 7 ;qDr cQj foy;u ugha gSA

(2) pH > 7 ;qDr cQj foy;u ugha gSA

(3) pH < 7 ;qDr cQj foy;u gSA

(4) pH > 7 ;qDr cQj foy;u gSA

91. The genotype of a plant showing the dominantphenotype, it can be determined by :(1) Pedigree analysis (2) Back cross(3) Test cross (4) Dihybrid cross

92. Match the terms given in Column I with theirexample in Column II and select the correct option:

Column I Column II

(a) Co-dominance (i) Phenylketonuria inhuman

(b) Pleiotropy (ii) Human blood group

(c) Multipleallelism (iii) Coat colour of cattle

(d) Polygenicinheritance (iv) Skin colour of

human

Code :-a b c d

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

93. If the frequency of an autosomal dominant alleleis 0.6. Calculate the frequency of recessivephenotype in a population of 6000 :-(1) 1200 (2) 4000 (3) 960 (4) 1000

94. Shaded symbol represent black guinea pig whileopen symbol represents white guinea pig. Black skinis dominant over white skin in guinea pig. What isthe possible genotype of II–1 and III–1 respectively.

1 2

1 2 3 4

1 2

I

II

III

(1) Bb and bb (2) BB and bb(3) bb and Bb (4) BB or Bb and bb

TOPIC : GENETICS-I : Principles of Inheritance and variation, Biology in Human Welfare, Microbesin human welfare, Biotechnology : Principles and Process of Biotechnology, Biotechnology and Its Applications.

91. ,d ,sls ikSèks dk thuiz:i ftlesa izHkkoh y{k.kiz:i fn[kk;h nsjgk gks] fdlds }kjk fuèkkZfjr fd;k tk ldrk gS\(1) oa'kkoyh fo'ys"k.k (2) izrhi izladj.k(3) ijh{k.kkFkZ izladj.k (4) f}ladj izladj.k

92. LrEHk I esa fn;s x;s 'kCnksa dks LrEHk II esa muds o.kZu ls lqesfyrdhft, ,oa lR; fodYi pqfu;s :

I

(a) lgizHkkfork (i)ekuo esa Qhukby

dhVksuwfj;k

(b) cgqizHkkfork (ii) ekuo dk jDr lewg

(c) cgq;qXefodYih (iii) eos'kh esa Ropk dk jax

(d) cgqthuh oa'kkxfr (iv) ekuo dh Ropk dk jax

dksM :-a b c d

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

93. ;fn vkVkslksey çHkkoh ; qXefodYih dh vko`fÙk 0.6. rks 6000dh tula[;k esa vçHkkoh y{k.k çk:i okys lnL;ksa dh la[;kD;k gksxh(1) 1200 (2) 4000 (3) 960 (4) 1000

94. uhps nh xbZ oa'kkoyh esa xgjs fpUg dkys xquh;k lqvj ds tcfd[kkyh fpUg lQsn xqfu;k lqvj dks fu:fir dj jgs gSa rFkk dkyhRopk lQsn Ropk ij izHkkoh gSA II–1 ,oa III–1 dk lEHko thuizk:i Øe'k% gksxk ?

1 2

1 2 3 4

1 2

I

II

III

(1) Bb ,oa bb (2) BB ,oa bb(3) bb ,oa Bb (4) BB ;k Bb ,oa bb

95. Genes which are tightly linked on chromosomeshows :-(1) Very low recombination(2) High recombination(3) Very low parental combination(4) Independent assortment

96. In a dihybrid cross between AABB and aabb theratio of AABB, AABb, aaBb, aabb in F2 generationis :-(1) 9:3:3:1 (2) 1:1:1:1(3) 1:2:2:1 (4) 1:1:2:2

97. Barr body absent in(1) Down syndromic female(2) Klinefelter syndrome(3) Turner syndrome(4) Thalasemic female

98. Match the Column-I with Column-II

Column-I Column-II

a Allele i Two alleles of a geneare identical

b Genotype ii An individual havingtwo different alleles ofa gene

c Homozygous iii Genetic constitution ofan organism

d Heterozygous iv Alternative forms of agene

(1) a-i, b-ii, c-iii, d-iv(2) a-iv, b-iii, c-i, d-ii(3) a-iii, b-iv, c-ii, d-i(4) a-iii, b-ii, c-iv, d-i

99. Which of the following characters of garden peais present on 4th chromosome.(1) Plant height, pod colour, cotyledon colour(2) Plant height, flower position, pod shape(3) Pod colour, seed colour, flower colour(4) Pod shape, seed shape, plant height

95. tks thu xq.klw=kas s ij etcwrh ls lgyXu gksrh gS oks n'kkZrhgS :-(1) cgqr de iqu;ksZtu(2) cgqr T;knk iqu;ksZtu(3) cgqr de iSr`d la;ksx(4) Lora= viO;wgu

96. nks thuiz:i AABB ,oa aabb ds chp ladj.k ls mRikfnrAABB, AABb, aaBb ,oa aabb larfr;ksa dk f2 vuqikrD;k gksxk %&(1) 9 : 3 : 3 : 1 (2) 1 : 1 : 1 : 1(3) 1 : 2 : 2 : 1 (4) 1 : 1 : 2 : 2

97. ckj&ckWMh vuqifLFkr gksrh gSA(1) Mkmu flaMªkse okyh eknk(2) dykbuQsUVj flaMªkse esa(3) VuZj flaMªkse eas(4) FkSuklhfe;k okyh eknk esa

98. dkWye-I dks dkWye-II ls feyku djsaA

dkWye-I dkWye-II

a ;qXefodYih(,yhy)

i fdlh thu ds nksuksa ,yhyleku gksa

b thu izk:i(thuksVkbi)

ii fdlh thu ds nksuksa ,yhyfo"ke gks

c le;qXeth(gksekstkbxl)

iii fdlh tho dk vkuqoaf'kdlaxBu

d fo"ke;qXeth(gsVjkstkbxl)

iv fdlh thu dk ,dkUrfjrLo:i

(1) a-i, b-ii, c-iii, d-iv(2) a-iv, b-iii, c-i, d-ii(3) a-iii, b-iv, c-ii, d-i(4) a-iii, b-ii, c-iv, d-i

99. fuEu esa ls dkSuls y{k.k xkMZu eVj ds 4th xq.klw= ij mifLFkrgksrs gSa(1) ikni yEckbZ] Qyh dk jax] chti= dk jax(2) ikni yEckbZ] iq"i dh fLFkfr] Qyh dh vkd`fr(3) Qyh dk jax] cht dk jax] iq"i dk jax(4) Qyh dk vkdkj] cht dk vkdkj] ikni yEckbZ

100. Find out the correct match from column I, II andIII of following table :-

Column I Column II Column III

Cross type No. ofgenotypes

No. ofphenotypes

(A) Monohybrid (I) 9 (X) 8

(B) Dihybrid (II) 27 (Y) 4

(C) Trihybrid (III) 3 (Z) 2

(1) A – III – Z, B – I – Y, C – II – X

(2) A – II – X, B – III – Y, C – I – Z

(3) A – I – Y, B – III – X, C – II – Z

(4) A – III – Z, B – II – X, C – I – Y

101. A trihybrid mullato man marries with a whitewoman, how many different types of phenotypeswill be obtained ?

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

102. If substrate 'S' is transformed into product 'P' withthe help of enzyme 'E'. Enzyme 'E' has beenintroduced from wild Allele 'A'. After mutation 'A'changes into modified forms as follows :

'A' wildAllele

Mutation

'a' Allele Produces normalenzyme

' ' Allelea

' ' Alleleb

' ' Alleleg

Produces less efficientenzymeProduces non functionalenzyme

Produces no enzymeat all

Producesnormalenzyme

modifiedUnmodified

After considering the above findings, which of thefollowing allele pair will produce recessivephenotype ?

(1) a, a (2) a, b

(3) A, a (4) b, g

100. fuEu rkfydk ds dkWyeks I, II rFkk III feyku djrs gq, lghfodYi dk p;u dhft, :-

dkWye I dkWye II dkWye III

ØkWl dk çdkjthuh çk:iks adh la[;k

n`'; çk:iksadh la[;k

(A) ,dladj ØkWl (I) 9 (X) 8

(B) f}ladj ØkWl (II) 27 (Y) 4

(C) f=ladj ØkWl (III) 3 (Z) 2

(1) A – III – Z, B – I – Y, C – II – X

(2) A – II – X, B – III – Y, C – I – Z

(3) A – I – Y, B – III – X, C – II – Z

(4) A – III – Z, B – II – X, C – I – Y

101. ,d f=ladj mullato uj dk fookg lQsn eknk ls gks rks larfr;ksaesa fdrus izdkj ds y{k.kizk:i izkIr gksaxs\

(1) 2 (2) 3

(3) 4 (4) 5

102. ;fn fØ;kdkjd 'S' ,atkbe 'E' dh lgk;rk ls mRikn 'P' esaifjofrZr gksrk gS] ,atkbe 'E' oU; ;qXefodYih 'A' ls mRikfnrfd;k x;k gS] mRifjorZu ds i'pkr~ 'A' fofHkUu :i esa bl izdkjls :ikarfjr gksrk gS :

'A' oU;;qXefodYih

Mutation

'a' ;qXefodYih lekU; ,atkbedk mRiknu

' ' a ;qXefodYih

' ' b ;qXefodYih

' ' g ;qXefodYih

de dk;Z'khy ,atkbedk mRiknuvdk;Z'khy ,atkbedk mRiknufdlh Hkh ,atkbedk mRiknu ugha gksxk

lekU; izdkjds , atkbedk mRiknu

v:ikarfjr

mijksDRk ifj.kkeksa dks /;ku j[krs gq;s ; s crk;s fd dkSulk;qXefodYih tksM+k vizHkkoh y{k.k izk:i dks mRikfnr djsxk

(1) a, a (2) a, b

(3) A, a (4) b, g

103. Phenotype Genotype Phenotype

TT50 cm 50 cm

Tt

tt

35 cm

20 cm

50 cm

20 cm

A B

A B

(1) Polygenicqualitative

Polygenicquantitative

(2) Monogenicquantitative

Monogenicqualitative

(3) Monogenicquantitative

Monogenicquantitative

(4) Monogenicqualitative

Monogenicqualitative

104. A normal man and woman who have phenylketonuria(affected) child, which is autosomal recessive trait. Thewoman becomes pregnant again and carryingmonozygotic twins. What is the probability ofphenylketonuria in twins

(1)1

16(2)

14

(3) 12

(4) 9

16

105. Find out the number of plants produced withgenotype AabbCc out of 256 seeds collected fromF2 progenies of a trihybrid cross :-(1) 32 (2) 16 (3) 12 (4) 8

106. T.H. Morgan experimented on :-(1) Fruit fly (2) Fire fly(3) Sweet pea (4) Garden pea

107. The period for Mendel's hybridization experimentsis ?(1) 1865 - 1884 (2) 1856 - 1863(3) 1822 - 1856 (4) 1884 - 1900

103. Phenotype Genotype Phenotype

TT50 cm 50 cm

Tt

tt

35 cm

20 cm

50 cm

20 cm

A B

A B

(1) cgqthuhxq.kkRed

cgqthuhek=kRed

(2) ,dthuhek=kRed

,dthuhxq.kkRed

(3) ,dthuhek=kRed

,dthuhek=kRed

(4) ,dthuhxq.kkRed

,dthuxq.kkRed

104. ,d lkekU; iq:"k o L=h dh larku QhukbydhVksuwfj;k lsizHkkfor gS tks fd vfyaxh vizHkkoh Vª sV gSA bl efgyk dsnwljh ckj xHkZ/kkj.k fd;k vkSj og ,d;qXeth tqMoka larkusgks axhA bu tqM+okvksa ds QhukbydwVksuwfj;k gksus dh laHkkoukgS&

(1)1

16(2)

14

(3) 12

(4) 9

16

105. ,d f=ladj Øk Wl ls ,df=r 256 chtk s a ls mRikfnrikS/kksa esa AabbCc thuizk:i okys F2 larfra;ks dh la[;k dk irkyxk,¡\(1) 32 (2) 16 (3) 12 (4) 8

106. Vh ,p ekWjxu us iz;ksx fd;k FkkA(1) Qy eD[kh (2) tqxuw(3) ehBh eVj (4) m|ku eVj

107. esaaMy ds ladj.k iz;ksxksa dh vof/k gS ?

(1) 1865 - 1884 (2) 1856 - 1863

(3) 1822 - 1856 (4) 1884 - 1900

108. Which of the following is incorrect for garden pea?

Character Phenotype Genotype

1 Seed colour Yellow Homozygousheterozygous

2 Pod colour Green Homozygousheterozygous

3 Flower position Terminal Homozygousheterozygous

4 Pod shape Constricted Onlyhomozygous

109. Choose correct option based on following statement:-

(i) Emasculation is done in plant which isconsidered as male parent

(ii) Emasculation is done in plant which isconsidered as female parent

(iii) Emasculation prevent underisable cross-pollination

(iv) Emasculation is done to prevent self-pollination.

(1) i, ii (2) ii, iii

(3) i, iii (4) ii, iv

110. Given below is a pair of homologous chromosomes:-

aB

CdEE

DC

BA

How many loci are in heterozygous condition ?

(1) 2 (2) 3

(3) 5 (4) 10

108. m|ku eVj ds lanHkZ esa fuEu esa dkSu vlR; gS ?

1 cht dk jax ihyk le;qXeth ofo"ke;qXeth

2 Qyh dk jax gjk le;qXeth ofo"ke;qXeth

3 iq"i dh fLFkfr 'kh"kZLFk le;qXeth ofo"ke;qXeth

4 Qyh dhvkdfr

fldqM+h dsoy le;qXeth

109. fuEu dFkuksa ij vk/kkfjr lgh fodYi pqfu, :-

(i) foiqlau ml ikS/ksa esa fd;k tkrk gSA ftls uj tud ekuk tkrkgSA

(ii) foiqlau ml ikS/ks esa fd;k tkrk gSA ftls eknk tud ekuktkrk gSA

(iii) foiqlau vokafNr ij&ijkx.k jksdrk gSA

(iv) foiqlau Lo&ijkx.k jksdrk gSA

(1) i, ii (2) ii, iii

(3) i, iii (4) ii, iv

110. uhps letkr xq.klw=ksa dk ,d tksM+k fn;k x;k gS :-

aB

CdEE

DC

BA

fdrus LFkyksa ij fo"ke;qXeth voLFkk gS ?

(1) 2 (2) 3

(3) 5 (4) 10

111. Read the following statements :-

(a) It produces disorder in females more oftenthan in males

(b) All female offsprings will exhibit disorder, iffather possesses the same

(c) Do not transmitted to son if mother does notexhibit disorder.

Which of the following gene will have the abovestated features?

(1) Sex-linked recessive gene

(2) Sex-linked dominant gene

(3) Autosomal dominant gene

(4) Autosomal recessive gene

112. Given below is a representation of kind ofchromosomal mutation. What kind of mutation isrepresented in given diagram ?

A B C D E F G H®

A

D C B

E F G H

(1) Deletion (2) Duplication

(3) inversion (4) Translocation

113. Which of the following is not true for F2 generationof dihybrid cross ?

(1) 9-phenotypic categories, 4 genotypic categories

(2) 9-genotypic categories, 16 zygotic combination

(3) 62.5% parental combination, 37.5% newcombination

(4) 4-types male gametes, 4 type female gemate

114. Multiple alleles are present on :-

(1) Same locus of homologous chromosome

(2) Different locus of different chromesome

(3) Different locus of same chromosome

(4) Same locus of non homologous chromosome

115. The substitution of amino acid in the globin proteinresults due to the single base substitution at thesixth code of the beta globin gene (DNA) from :-

(1) GAG to GUG (2) CTC to CAC

(3) GUG to GAG (4) CAC to CTC

111. fuEufyf[kr dFkuksa dks if<, :-

(a) uj dh rqyuk esa eknk esa jksx mRiUu gksus dh laHkkouk vfèkdgksrh gSA

(b) ;fn firk esa ;g mifLFkr gksrk gS] rks lHkh eknk larfr;ksa esajksx ik;k tkrk gSA

(c) iq=ksa esa ;g LFkkukUrfjr ugha gksrk gS ;fn ekrk esa jksx ughagksA

fuEu esa ls dkSulk thu mijksä fn, x;s y{k.k izn£'kr djrkgS\

(1) fyax-lgyXu vizHkkoh thu

(2) fyax-lgyXu izHkkoh thu

(3) vksVkslksey izHkkoh thu

(4) vksVkslksey vizHkkoh thu

112. uhps fn;s x;s fp= esa xq.klw=h; mRifjorZu dks iznf'kZr fd;kx;k gSA uhps fn;s x;s fp= esa fdl izdkj dk mRifjorZugS ?

A B C D E F G H®

A

D C B

E F G H

(1) deh (2) vuqfyfidj.k

(3) izrhiu (4) LFkkukUrj.k

113. f}ladj ØkWl dh F2 ih<+h ds fy, D;k lR; ugha gS ?

(1) 9-y{k.kizk:i dh Jsf.k;k¡, 4 thuizk:i Jsf.k;k¡

(2) 9-thuizk:i Jsf.k;k¡, 16 tk;xksVhd la;kstu

(3) 62.5% iSr`dh; la;ksx, 37.5% u;s la;ksx

(4) 4-izdkj ds uj ; qXed, 4 izdkj ds eknk ;qXed

114. cgq;qXe fodYih mifLFkr gksrs gSa :-

(1) le;qXeth xq.klw=ksa ds leku LFky ij

(2) fHkUu xq.klw=ksa ds fHkUu LFky ij

(3) leku xq.klw= ds fHkUu LFky ij

(4) vletkr xq.klw= ds leku LFky ij

115. Xyksfcu izksVhu esa vehuksa vEy dk izfrLFkkiu] chVk (B) Xykschuthu (Mh-,u-,-) ds NBs dksMksu ds fdl ,dy {kkj izfrLFkkiuds dkj.k gksrk gS :-

(1) GAG ls GUG (2) CTC ls CAC

(3) GUG ls GAG (4) CAC ls CTC

116. Which of the following statement is correct aboutpolygenic inheritance ?(1) In polygenic inheritance one gene regulates

the expression of many characters(2) Qualitative characters which are regulated by

polygenes(3) Besides the involvement multiple genes,

polygenic inheritance also takes into accountthe influence of environment

(4) Human skin colour is an example of qualitativecharacter

117. In a reciprocal cross results are not same and itseems that the character is most probably inheritedonly through female parents. It represents :-(1) Sex linkage inheritance(2) Autosomal inheritance(3) Cytoplasmic inheritance(4) Polygenic inheritance

118. When a dihybrid (AaBb) is selfed then thegenotypes AABb, AaBb, AABB, aabb would bein a proportion of ?(1) 2 : 4 : 2 : 2 (2) 2 : 4 : 1 : 1(3) 1 : 2 : 1 : 1 (4) 1 : 4 : 1 : 1

119. What is true about sex determination in chickens?(1) ZZ males and ZW females(2) ZW males and ZZ females(3) XO males and XX females(4) XX males and XO females

120. Two genes A and B present on same chromosomeshow 40% cross over value, then what will bepercentage of gamete aB for the plant havinggenotype AaBb (Cis arrangement) :-(1) 60% (2) 20% (3) 80% (4) 30%

121. The most popularly known blood group is the ABOblood group. It's name is ABO but not ABC,because "O" in it refers to :-(1) Antigens A and B are absent on RBCs(2) Besides antigen A and B are present on RBCs(3) Overdominance of this type of genes over

A and B antigens(4) Only one antibody - either anti–A or anti–B

present on the RBCs

116. cgqthuh oa'kkxfr ds lanHkZ esa fuEufyf[kr esa dkSu lk dFku lR;gS ?

(1) cgqthuh oa'kkfxr esa ,d thu dbZ y{k.kksa ds izdVu dksfu;af=r djrh gSA

(2) xq.kkRed y{k.kksa dk fu;a=.k cgqthuksa }kjk gksrk gSA

(3) vusd thuksa ds 'kkfey gksus ds vfrfjDr cgqthuh oa'kkxfresa i;kZoj.k ds izHkko dks Hkh ij[kk tkrk gSA

(4) euq"; esa Ropk dk jax ,d xq.kkRed y{k.k gSA

117. ,d O;qRØe ØkWl ds ifj.kke leku ugha gS rFkk ,slk izrhr gksjgk gS fd y{k.k eknk ds ek/;e ls oa'kkxr gks jgk gSA ;g ,dmnkgj.k gS :-(1) fyax lgyXurk dh oa'kkxfr dks(2) dkf;d oa'kkxfr dks(3) dksf'kdk æO;h; oa'kkxfr dks(4) cgqthoh oa'kkxfr dks

118. ,d f}ladj (AaBb) dk Lo%ijkx.k djus ij thuizk:iksaAABb, AaBb, AABB, aabb dk vuqikr D;kgksxk ?(1) 2 : 4 : 2 : 2 (2) 2 : 4 : 1 : 1(3) 1 : 2 : 1 : 1 (4) 1 : 4 : 1 : 1

119. pwtksa esa fyax fu/kkZj.k ds ckjs esa D;k lR; gS ?(1) ZZ uj rFkk ZW eknk(2) ZW uj rFkk ZZ eknk(3) XO uj rFkk XX eknk(4) XX uj rFkk XO eknk

120. nks thUl A rFkk B leku xq.klw= ij mifLFkr gSa rFkk 40% thufofue; (crossing over) n'kkZrs gSa] rks ;qXed (aB) dk izfr'krD;k gk sxk ;fn ikni dk thuk sVkbi AaBb g S(fll O;oLFkk gS) ?(1) 60% (2) 20% (3) 80% (4) 30%

121. vke yksxksa esa lokZf/kd tkuk&igpkuk jDr lewg ABO jDr oxZgSA bls ABC u dgds ABO dk uke fn;k x;k D;ksafd blesa"O" ;g n'kkZrk gS fd :-(1) RBCs ij A vkSj B nksuksa ,saVhtu (izfrtu) vuqifLFkr gksrs gSa(2) A vkSj B ,saVhtu ds vfrfjDr RBCs ij vU; ,saVhtu gksrs gSa(3) A vkSj B ,saVhtu ds thuksa ij bl iz:i dh vfrizHkkfork

gksrh gS(4) RBCs ij dsoy ,d ,s aVhckWMh mifLFkr gksrh gS ;k rks

,saVh – A ;k ,s aVh – B

122. Represented below is the inheritance pattern of acertain type of traits in humans. Which one of thefollowing conditions could not be an example ofthis pattern?

Female Mother

Male Father

Daughter Son

(1) Haemophilia(2) Colourblindness(3) Phenylketonuria(4) Duchenne muscular dystrophy

123. Given below is the pedigree of myotonic dystrophy.

In this pedigree the genotype of all affectedchildren will be –(1) AA (2) Aa(3) AA or Aa (4) aa

124. The phenotypic ratio in F2 generation for theinheritance of flower colour in Antirrhinum would be:-(1) 3 : 1 (2) 1 : 2 : 1(3) 1 : 1 (4) 2 : 1

125. A child has blood groups 'O'. If father has bloodgroup 'A' and mother has blood group 'B'. Findout the genotypes of the parents.(1) IAIA and IBI0 (2) IAI0 and IBI0

(3) IAI0 and I0I0 (4) I0I0 and IBIB

126. If both parents are carrier for thalassemia, which isan autosomal recessive disorder, what is the chancesof pregenancy resulting in an affected child ?(1) 25% (2) 100%(3) 0 % (4) 50%

122. uhps fn;s tk jgs vkjs[kh; fu:i.k esa ekuoksa esa ik;s tkus okys ,d[kkl izdkj ds fo'ks"kdksa (VsªsVksa) dk oa'kkxfr izfr:i n'kkZ;k x;kgSA crkb, fd fuEufyf[kr esa ls dkSulh ,d n'kk gS tks blhizfr:i dk ,d mnkgj.k ugha gks ldrh gS\

eknk ekrk uj firk

iq=h iq=

(1) gheksfQfy;k(2) o.kkZU/krk(3) QhukbydhVksU;wfj;k(4) Mdu eLdwyj fMLVªksQh

123. fn;k x;k oa'kkoyh pkVZ ,d ek;ksVksfud fMLVªksQh dk gS&

bl oa'kkoyh esa lHkh jksxh larfr;ksa dk thuizk:i D;kgksxk\(1) AA (2) Aa(3) AA or Aa (4) aa

124. F2 ih<+h esa ,aVhjkbue ds Qwyksa ds jax ds oa'kkuqxr dk y{k.k izk:ivuqikr gksxk\(1) 3 : 1 (2) 1 : 2 : 1(3) 1 : 1 (4) 2 : 1

125. ,d cPps dk jDr lewg 'O' gS o mldh ekrk dk 'B' ofirk dk 'A' gSA rks firk o ekrk dk jDr lewg Kkrdhft,\(1) IAIA rFkk IBI0 (2) IAI0 rFkk IBI0

(3) IAI0 rFkk I0I0 (4) I0I0 rFkk IBIB

126. nks isr`d tks dh FkSyslseh;k [vksVkslksey vizHkkoh jksx] ds okgdgSA rks larrh ds jksxh gksus ds D;k laHkkouk gksxh\

(1) 25% (2) 100%

(3) 0% (4) 50%

127. Flower colour of Mirabilis jalapa is due to :-(1) Complete dominance(2) Incomplete dominance(3) Linkage(4) Mutation

128. The distance between genes on chromosomesrepresenting 12% cross over. Find out the distancebetween two genes.(1) 12 mapunit (2) 24 mapunit(3) 21 mapunit (4) 42 mapunit

129. In a typical monohybrid cross, we have crosseda hybrid tall plant with another hybrid tall plant,Which of the following result can not be expected?(1) 25% homozygous tall(2) 25% homozygous dwarf(3) 100% heterozygous tall(4) 75% tall and 25% dwarf

130. If a dwarf variety of a pea plant is treated with GAhormone, it grows as tall as the tall pea plant. Onselfing of this plant, the phenotypic ratio is likelyto be :-(1) All dwarf(2) All tall(3) 50% tall(4) 75% tall and 25 dwarf

131. The term hemizygous means :-(1) only one allele of a gene is present.(2) Many alleles of a gene are present.(3) Pair of contrasting traits are present(4) Two similar alleles are present.

132. Column-A Column-B

I. Gregor Mendel (A) Alkaptonurea

II. T.H. Morgan (B) Laws of inheritance

III. A. Garrod (C) LinkageIV Henking (D) X-body

Find the correct match :(1) I-A, II-C, III-D, IV-B(2) I-B, II-C, III-A, IV-D(3) I-D, II-A, III-C, IV-B(4) I-A, II-D, III-B, IV-C

127. fejkfcfyl tykik ds iq"iksa ds jax dk dkj.k gSA(1) iw.kZ izHkkfork(2) viw.kZ vizHkkfork(3) lgyXurk(4) mRifjorZu

128. Øksekslkse ds Åij ikbZ tkus okyh nks thuksa dks izfrfuf/kRo12% Økl vksoj }kjk gksrk gSA thuksa ds chp nwjh Kkrdhft,A(1) 12 esibdkbZ (2) 24 esibdkbZ(3) 21 esibdkbZ (4) 42 esibdkbZ

129. ,d ,dladj ØkWl esa] ge ,d yEcs ikni dk nwljs yEcs iknils ØkWl djokrs gSa] nksuksa iSr`d bl y{k.k ds fy, fo"ke;qXeh gSAfuEu esa ls dkSulk ifj.kke izkIr ugha fd;k tk ldrk gS %&(1) 25% le;qXeth yEcs(2) 25% le;qXeh ckSus(3) 100% fo"ke;qXeth yEcs(4) 75% yEcs rFkk 25% ckSus

130. ;fn eVj ds ,d ckSus ikni dks GA gkjeksu ls mipkfjr dj mlseVj ds yEcs ikni tSlk yEck dj fn;k tkrk gS] ,oa mls Loijkfxrdjus ij y{k.kizk:i dk vuqikr D;k gksxk%&(1) lHkh ckSus(2) lHkh yEcs(3) 50% yEcs(4) 75% yEcs vkSj 25 ckSus

131. v¼Z ;qXeuth 'kCn dk vFkZ gS %&(1) ,d thu dk dsoy ,d ,yhy mifLFkr gksA(2) ,d thu ds dbZ ,yhy mifLFkr gksrs gksA(3) foi;kZlh y{k.kksa ds tksM+s dh mifLFkrh(4) nks leku ,yhy dh mifLFkrh

132. dkWye-A dkWye-B I. Gregor Mendel (A) ,YdsIVksU;wfj;kII. T.H. Morgan (B) oa'kkxfr ds fu;eIII. A. Garrod (C) lgyXurkIV. Henking (D) X-dk;

lgh feyku dks pqus &(1) I-A, II-C, III-D, IV-B(2) I-B, II-C, III-A, IV-D(3) I-D, II-A, III-C, IV-B(4) I-A, II-D, III-B, IV-C

133. Read the following statements :

(A) Variation always exists among the membersof species

(B) Crossing over causes continuous variation

(C) Variation may or may not required for evolution

(D) Environment also plays its role in variation

How many statement(s) is/are correct ?

(1) One (2) Two

(3) Three (4) Four

134. What type of allele produces its effect only inhomozygous condition ?

(1) Recessive

(2) Dominant

(3) Both recessive and dominant

(4) Incomplete recessive

135. Which of the following statements is generallytrue ?

(1) A dominant allele also determines its recessivetrait

(2) A recessive allele is always beneficial.

(3) A recessive allele always does not produce itsrecessive trait when paired with its dominantallele

(4) A dominant allele is always better for anorganism

136. In a cross EeFfGg × eeffGg, what will be theprobability of organisms, which are homozygousfor first character and heterozygous for second andthird characters ?

(1) 28

(2) 18

(3) 1

16(4)

416

137. A true-breeding line is one that :-

(1) Have undergone continuous self pollination

(2) Have undergone continuous cross pollination

(3) Shows stable trait inheritance and expressionfor several generations.

(4) Both (1) & (3)

133. fuEufyf[kr dFkuksa dks i<+s &(A) fdlh tkfr ds lnL;ksa ds e/; fofHkUurk;sa ges'kk mifLFkr

gksrh gSA(B) thu fofue; ds dkj.k lrr~ fofHkUurk,sa mRiUu gksrh gSa(C) fodkl ds fy, fofHkUurk,sa vko';d gks Hkh ldrh gS vkSj

ugha HkhA(D) fofHkUurkvksa dks mRiUu djus esa okrkoj.k dk Hkh ;ksxnku gksrk gSAfdrus dFku lR; gS \(1) ,d (2) nks (3) rhu (4) pkj

134. fdl izdkj dk vyhy] viuk izHkko dsoy le;qXeh O;oLFkkesa dj ldrk gS &(1) vizHkkoh(2) izHkkoh(3) nksuksa izHkkoh o vizHkkoh(4) viw.kZ vizHkkoh

135. fuEufyf[kr esa dkSulk dFku lkekU;r% lgh gS \

(1) ,d izHkkoh ,yhy vius vizHkkoh foi;kZlh dk Hkhfu/kkZj.k djrk gSa

(2) ,d vizHkkoh ,yhy] ges'kk Qk;nsean gksrk gSA

(3) ,d vizHkkoh ,yhy] ges'kk viuh vizHkkoh foi;kZlh ughcukrk gSa tc bldks izHkkoh ,yhy ds lkFk ; qfXer fd;ktkrk gSA

(4) ,d izHkkoh ,yhy ,d tho ds fy, ges'kk mÙke gksrkgSA

136. ,d ØkWl EeFfGg × eeffGg, esa mu thuksa dh çkf;Drk D;kgksxh] tks igys y{k.k ds le;qXeth o nwljs o rhljs y{k.k dsfy, fo"ke;qXeth gksa\

(1) 28

(2) 18

(3) 1

16(4)

416

137. ,d rnwzi çtuu ykbu (True-breeding line) og gS] tks@ftlesa:-(1) lrr~ Lo%ijkx.k fd;k x;k gksA(2) lrr~ ijijkx.k fd;k x;k gksA(3) dbZ ih<+h;ksa rd] LFkkbZ VªsV oa'kkuqxfr o çn'kZu çnf'kZr djrh

gSA(4) (1) o (3) nksuksa

138. Identify the restriction enzymes used to cut thesite 'X' and 'Y' respectively :

Ori

pBR322ampR tetR

X

Y

ROP

(1) Pvu I, Pst I (2) Sal I, EcoRI(3) Pvu II, Pst I (4) Bam H I, Sal I

139. Consider the following statement and select thecorrect option :-(A)Any protein encoding gene is expressed in a

heterologous host, is called a recombinantprotein.

(B) Stirred tank reactor facilitates even mixing ofreactor content and oxygen availabilitythroughout the bioreactor.

(C) The down stream processing and qualitycontrol testing similar to all products.

Options :-(1) A and C are correct while B is wrong(2) A and B are correct while C is wrong(3) C and B are correct while A is wrong(4) Only A is correct while B and C are wrong

140. In the following figure

4

3

2

1

Which DNA is not digested ?(1) 1 (2) 2 (3) 3 (4) 4

141. Among the following select which is not the toolsof rDNA technology(1) Restriction enzymes(2) Vectors(3) Host (4) Promotor

138. mu jsfLVªD'ku ,Utkbe dks igpkfu, tks Øe'k% 'X' o 'Y' LFkyij dkVrs gS :

Ori

pBR322ampR tetR

X

Y

ROP

(1) Pvu I, Pst I (2) Sal I, EcoRI(3) Pvu II, Pst I (4) Bam H I, Sal I

139. fuEu dFkuksa ij fopkj dhft, rFkk lgh fodYi dk p;udhft, :-(A);fn dksbZ çksVhu dwVys[ku thu fdlh gsVsjksyksxl ijiks"kh

(Host) esa vfHkO;Dr gksrh gSa rks blls cuus okys çksVhu dksiqu;ksZxt çksVhu dgrs gSA

(B) foyksfMt gkSt fj,DVj ds vUnj varZoLrq ds feJ.k eas rFkkvkWDlhtu dh miyC/krk ds fy, mi;ksxh gSA

(C) vuqçokg lalk/ku o xq.koÙkk fu;a=.k ijh{k.k lHkh mRiknksads fy, leku gksrk gSA

fodYi :-

(1) A rFkk C lgh gS tcdh B xyr gSA(2) A rFkk B lgh gS tcdh C xyr gSA(3) C rFkk B lgh gS tcdh A xyr gSA(4) dsoy A lgh gS tcdh B rFkk C xyr gSA

140. fn;s x;s fp= gS

4

3

2

1

dkSuls DNA dk ikpu ugha gqvk gS\(1) 1 (2) 2 (3) 3 (4) 4

141. uhps fn;s x;s esa ls dkSulk ,d rDNA rduhd dk tool ughagSA(1) izfrcU/ku ,atkbe (2) laokgd(3) ijiks"kh (4) mUuk;d

142. Purified DNA ultimately precipitate out after theaddition of chilled ethanol. The DNA that separateout can be removed by :-(1) Electrophoresis(2) Downstream processing(3) PCR(4) Spooling

143. The given below figure shows three steps (A,B, C)in Polymerase Chain Reaction (PCR). Select theoption having correct identification with any onerepresentation.

5'

5'

3'

5'

5'

3'

5'3'

3'

3'

3'

3'

5'

3'

3'

5'

3'5'

5'

5'

A

C

B

dsDNA

Region to be amplified

Options :(1) C-Extension in the presence of heat stable DNA

polymerase(2) A-Annealing with two sets of primers(3) B-Denaturation at a temperature of about 98°C

separating the two DNA strands(4) A-Denaturation at a temperature of about 50°C

144. Find out correct recognisation sequence offollowing restriction endonuclease enzyme :-

(1) Bam HIGGATCCCCTAGG

Eco RIGAATTCCTTAAG

(2) Bam HIGAATCAACTTAGTT

Eco RITTGCAACAACGTTG

(3) Bam HIGCATGGCGTACC

Eco RIAGCTCCTCGAGG

(4) Bam HIGACTAACTGATT

Eco RIGCCTTACGGAAT

142. 'kqf¼d`r DNA, tc vfr'khfrr ,FksukWy dks Mkyk tkrk gS rksPrecipitate ds :i esa vk tkrk gSA ;g DNA tks vyx gksx;k ckn esa gVk;k tkrk gSA(1) bysDVªksQksjsfll }kjk(2) vuqizokg lalk/ku }kjk(3) PCR }kjk(4) Liwfyax }kjk

143. uhps fn;s x;s fp= esa ikSyhejst psu jh,sD'ku (PCR) ds rhu pj.k(A,B, C) fn[kk;s x, gSaA fuEufyf[kr esa ls fdl ,d fodYiesa ,d pj.k dk fu:i.k lgh igpkuk x;k gS\

5'

5'

3'

5'

5'

3'

5'3'

3'

3'

3'

3'

5'

3'

3'

5'

3'5'

5'

5'

A

C

B

dsDNA

izo/kZu fd;k tkus okyk {ks=

fodYi %(1) C-rkiLFkk;h DNA ikSyhejst dh mifLFkfr es foLrkj.k(2) A-izkbejksa ds nks lsVksa ds lkFk rkikuq'khyu(3) B-yxHkx 98°C ds rkieku ij fuf"Ø;dj.k ftlls nks

DNA jTtqd i`Fkd gks x;s(4) A-yxHkx 50°C ds rkieku ij fuf"Ø;dj.k

144. fn;s x;s izfrcaf/kr ,.MksU;wfDy,st ,atkbe ds fy;s lgh igpkuØe dk irk dhft, :-

(1) Bam HIGGATCCCCTAGG

Eco RIGAATTCCTTAAG

(2) Bam HIGAATCAACTTAGTT

Eco RITTGCAACAACGTTG

(3) Bam HIGCATGGCGTACC

Eco RIAGCTCCTCGAGG

(4) Bam HIGACTAACTGATT

Eco RIGCCTTACGGAAT

145. The construction of first r-DNA emerged from thepossibility of linking a gene encoding antibioticresistance with native plasmid of :-(1) Salmonella typhimurium(2) E. coli(3) Yeast(4) Thermus aquaticus

146. The linking of desired gene with the plasmidvector become possible with the special enzyme,which acts on cut DNA molecules and join theirends, this enzyme is called :-(1) Restriction enzyme(2) DNA ligase(3) DNA polymerase(4) Reverse transcriptase

147. Which of the following is a mismatch regardingthe transgenic plant and its application :-(1) Golden rice Þ Vitamin A enriched rice(2) Flavr savr tomato Þ Delay ripening(3) Tobacco Þ Herbicide resistant(4) Bt corn Þ Resistant to nematode

148. The use of bioresorces by multinationalcompanies and other organization without properauthorisation from countries and people conceredwith compensatory payment it is known as :-(1) Biopiracy (2) Biopatent(3) Biowar (4) Bioremediation

149. Microinjection is suitable for :-(1) DNA fingerprinting(2) Transformation of animal cells(3) Transformation of plant cells(4) Transformation of bacteria

150. A permanent remedy, against ADA deficiency inpatients can be :-(1) Periodic infusion of genetically engineered

lymphocytes in patients carrying correct ADAgene

(2) Introduction of ADA gene into the cells atearly embryonic stages

(3) Bone marrow transplantation in early childhood(4) Enzyme replacement therapy in early childhood

145. izFke iqu;kZ sxt Mh,u, dk fuekZ.k fdl thok.kq ds lgtIykfTeM esa izfrtSfod izfrjks/kh dwVys[ku thu ds tM+us ls gksldk Fkk :-(1) lkyeksusyk VkbQhE;wfj;e(2) bZ · dksykbZ(3) ;hLV(4) FkeZl ,DosfVdl

146. okafNr thu ds laokgd ds lkFk tksMus dk dke fo'ks"k ,atkbeds }kjk gksrk gSA tks Mh ,u , ds v.kq ds dVs gq,s Hkkx ij dk;Zdj mlds fdukjs dks tksMus dk dke djrk gSA bl , atkbe dksdgrs gS :-(1) izfrcU/ku ,atkbe(2) DNA ykbxst(3) DNA ikWyhejst(4) fjolZ VªkUlfØIVst

147. fuEu esa ls dkSulk xyr feyku ikjthoh ikS/kk o mlds vuqiz;ksxds laca/k esa gS :-(1) lqugjs pkoy Þ foVkkfeu A dh izpqj ek=k(2) Flavr savr VekVj Þ VekVj idus dh fØ;k esa nsjh(3) rackdw Þ 'kkduk'kh izfrjks/kh(4) Bt eDdk Þ lw=d`fe ls izfrjks/kh

148. cgqjk"Vªh; daifu;ksa o nwljs laxBuksa }kjk fdlh jk"Vª ;k mllslacfèkr yksxks ls fcuk O;ofLFkr vuqeksnu o {kfriwjd Hkqxrkuds tSo lalk/kuks dk mi;ksx djuk dgykrk gS :-

(1) ck;ksikbjslh (2) ck;ksisVsUV

(3) tSfod;q¼ (4) tSfod mipkj

149. ekbØksbatsD'ku fdlds fy, mi;qDr gS :-(1) DNA fQaxjfizafVx(2) tarq dksf'kdkvksa dk #ikUrj.k(3) ikni dksf'kdkvksa dk #ikUrj.k(4) thok.kq dksf'kdk dk #ikUrj.k

150. ADA dh deh okys jksfx;ksa dk LFkk;h mipkj fd;k tk ldrkgS :-

(1) jksxh esa lgh ADA thu okyh vkuqokaf'kd vfHk;kfU=dfyEQkslkbV dks le;&le; ij LFkkukUrfjr djds

(2) ADA thu dks izkjafHkd Hkzq.kh; voLFkk okyh dksf'kdkvksaesa LFkkukUrfjr djdsA

(3) cpiu esa vfLFk&eTtk çR;kjksi.k }kjkA

(4) cpiu esa ,Utkbe izfrLFkkiu mipkj }kjkA

151. Given below diagram showing cloning vectorpBR 322. Mark the incorrect statement statements:-

EcoR I

ampR tetR

ab

(1) Label 'a' represent Rop site, which codes for theproteins involved in the replication of plasmid

(2) Bam H 1 site is located in ampR region(3) Label 'b' repersent 'ori' site(4) Sal I site is located in tetR region

152. Advancement in genetic engineering has beenpossible due to discovery of :-(1) Oncogenes(2) Transposons(3) Restriction endonuclease(4) Exonucleases

153. Pvu I site is present at in pBR322 :-(1) ampR (2) tetR (3) Ori (4) rop

154. I. ROP also controls the copy numbers of thelinked DNA.

II. If a foreign DNA ligates at the Bam HI site oftetracycline resistance gene in the vectorpBR322, the recombinant plasmid loses thetetracycline resistance due to insertion offoreign DNA.

Choose regarding the above statements.(1) I is true, II is false (2) II is true, I is false(3) Both are true (4) Both are false

155. A bacterium modifies its DNA by adding methylgroups to the DNA, it does so to :-(1) Protect its DNA from its own restriction

endonuclease(2) Clone its DNA(3) Be able to transcribe many genes simultaneously.(4) Turn its gene ON

151. uhps fn;k x;k fp= Dyksfuax okgd pBR 322 dk gSA vlR;dFkuksa@dFku dks Nk¡fV;s :-

EcoR I

ampR tetR

ab

(1) fpfUgr 'a' 'Rop' LFky dks n'kkZ jgk gS tks fd IykfTeMf}xq.ku esa lfEefyr izksVhu dks dqV djrk gSA

(2) Bam H 1 LFky] ampR {ks= esa ik;k tkrk gSA(3) fpfUgr 'b', 'ori' LFky dks n'kkZrk gSA(4) Sal I LFky] tetR {ks= esa ik;k tkrk gSA

152. vkuqoaf'kd vfHk;kfU=dh rduhd esa mUufr fdldh [kkst lsgqbZ :-(1) dsUlj thu(2) VªkUlikstksUl(3) izfrcfU/kr ,UMksU;wfDy;st(4) ,DlksU;wfDy;st

153. Pvu I LFky pBR322 esa mifLFkr gksrk gS :-(1) ampR (2) tetR (3) Ori (4) rop

154. I. ROP tksM+s x, DNA dh copy numbers dks Hkh fu;af=rdjrh gSA

II. ;fn ,d fotkrh; DNA dk s okgd pBR322 es aVsVªklkbfDyu izfrjks/kh thu ds Bam HI LFky esa tksM+k tkrkgS] rks iwu;kZsxt plasmid dk VsVªklkbfDyu izfrjks/kh ckgjhDNA ds fuos'ku ls lekIr gks tkrk gSA

mijksDr dFkuksa ds laca/k esa p;u dhft,A

(1) I lgh gS, o II xyr gSA (2) II lgh gS, o I xyr gSA

(3) nksuksa lgh gSaA (4) nksuksa xyr gSaA

155. ,d thok.kq vius DNA esa esfFky lewg tksM+dj :ikUrfjr djnsrk gS ,slk fd;k tkrk gS :-(1) vius DNA dks Loa; ds jsfLVªd'ku ,UMksU;wfDy,t dh fØ;k

ls cpkus ds fy;sA(2) vius DNA dks Dyksu djus gsrq(3) bl dkj.k dbZ thu dk VªkalfØIlu ,d lkFk gks tkrk gSA(4) thu dks lfØ; cukus gsrqA

156. Which of the following is commonly used as avector for introducing a DNA fragment in plants:-(1) Retrovirus (2) Ti plasmid(3) l-phase (4) pBR322

157. GEAC is related to :-(1) The safety of introducing GM organisms for

public service(2) Shows extraordinary courage and dedication

in protecting wild life(3) To control the emission of ozone deplating

substances.(4) For separation and purification of bioreactor

product158. The C-peptide is :-

(1) Not present in proinsulin(2) Present in mature insulin(3) Removed during maturation of insulin(4) Also present in artificial insulin

159. cryIIAb endotoxins obtained from Bacillusthuringiensis are effective against :-(1) Nematodes (2) Bollworms(3) Mosquitoes (4) Flies

160. Denaturation, annealing and extension are thecorrect order of steps in :-(1) RNAi (2) PCR(3) r-DNA (4) BOD

161. Strategy used to prevent nematode infection oftobacco roots is(1) Use of agrochemicals (2) Bt toxin gene(3) Gene mutation (4) RNA interference

162. Which one is not a transgenic product :-(1) Flavr Savr Tomato (2) Herbicide resistant tobacco(3) Golden rice (4) Iron fortified rice

163. Which is not true about PCR (Polymerase chainreaction) ?

(1) PCR can amplify very small amount of DNA

(2) It can be used to detect HIV in suspects

(3) Only RNA can be used as primer

(4) Use of thermostable DNA polymerase

156. ikS/kksa esa Mh.,u.,. ds ,d VqdM+s ds fuos'k ds fy, fuEufyf[kresa ls dkSulk oSDVj mi;qDr fd;k tkrk gSA(1) jsVªksokbjl (2) Ti IykfTeM(3) l-Qkt (4) pBR322

157. th.bZ.,.lh. dk lEcU/k gS&

(1) tu lsokvksa ds fy, GM-thoksa ds lfUuos'k dh lqj{kk vkfnds ckjs esa fu.kZ; ysukA

(2) oU;thoksa dh j{kk ds fy, mn~Hkqr lkgl vkSj leiZ.kfn[kkuk

(3) vkstksu vo{k;dkjh inkFkk Z s ds mRltZu ij fu;a=.kdjuk

(4) ck;ksfj,DVj ds mRikn ds iFkDdj.k o 'kks/ku ds fy,

158. C-isIVkbM :-(1) izkd~balqfyu esa mifLFkr ugha gksrk gSA(2) ifjiDo balqfyu esa mifLFkr gksrk gSA(3) balqfyu ds ifjiDou ds nkSjku gVk fn;k tkrk gSA(4) Ïf=e balqfyu esa Hkh mifLFkr gksrk gSA

159. csflyl FkqfjfUt,afll ls izkIr cryIIAb ,aMksVkWfDlu fdldsizfr dkjxj gksrs gSa \(1) uheSVksM (2) cksyoeZ(3) ePNj (4) efD[k;k¡

160. fod`rhdj.k] vuhyu rFkk foLrj.k vkfn pj.kksa dk lgh Øefeyrk gSA(1) vkj ,u , vkbZ (2) ih.lh.vkj.(3) vkj.-Mh.,u.,. (4) ch.vks.Mh.

161. raEckdw ikni dh tM+ksa esa nematode ds laØe.k dks jksdus dsfy, j{kk ra= gSA(1) ,xzks jlk;uksa ds bLrseky (2) Bt toxin dk thu(3) thu mRifjorZu (4) RNA ck/kk

162. dkSulk ikjthuh mRikn ugha gSA(1) ¶ysoj lkoj VekVj (2) 'kkduk'kh izfrjks/kd rEckdw(3) lqugjs pkoy (4) ykSg rRo tSoiqf"V/kku

163. fuEu esa ls PCR (Polymerase chain reaction) ds fy,D;k lR; ugh gS &(1) DNA dh cgqr de ek=k dks PCR ds mi;ksx ls c<+k;k

tk ldrk gSA(2) bldh enn ls fdlh O;fDr esa HIV laØe.k dk irk yxk;k

tk ldrk gSA(3) blesa dsoy RNA dk mi;ksx izkbej ds :i esa fd;k tk

ldrk gSA(4) bleas rkiLFkkbZ Mh.,u., ikWfyejst dk mi;ksx gksrk gSA

164. In agarose gel electrophoresis :(1) DNA migrates towards negative electrode(2) Larger molecules migrate faster than smaller

molecules(3) Ethidium bromide can be used to visualize the

DNA fragments(4) Blue coloured bands of DNA can be visualize

when gel exposed to UV light165. Choose the incorrect match about pBR322 :

(1) Ori : Responsible for controlling the copynumber of linked DNA

(2) Rop : Codes for the protein involved inreplication of plasmid

(3) Hind III : Restriction site on pBR322(4) Sal-I : Selectable marker on plasmid of

Salmonella typhimurium

166. Column-I Column-II (I) PCR (A) Large scale culture (II) Bioreactor (B) To introduce alien

DNA in host cell (III) Gene gun (C) Restriction

endonuclease enzyme (IV) Eco R1 (D) Amplification of

gene of interest

(1) I-D, II-A, III-B, IV-C (2) I-B, II-A, III-D, IV-C(3) I-C, II-B, III-D, IV-A (4) I-D, II-B, III-A, IV-C

167. Identify the palindromic nucleotide sequence ?(1) GTCTAGAC (2) CTACATC(3) AGCAGCT (4) TTTCAAA

168. Match the following list of microbes and theirimportance :

(a) Saccharomycescerevisiae

(i) Production ofimmunosuppressive agents

(b) Monascuspurpureus

(ii) Ripening of Swiss cheese

(c) Trichodermapolysporum

(iii) Commercial production ofethanol

(d) Propionibacteriumsharmanii

(iv) Production of bloodcholesterol lowering agents

(a) (b) (c) (d)(1) (iii) (i) (iv) (ii)(2) (iii) (iv) (i) (ii)(3) (iv) (iii) (ii) (i)(4) (iv) (ii) (i) (iii)

164. ,xkjkst tSy fo|qr dk lapyu (bysDVªksQksjsfll) esaa &(1) DNA ½.kkRed bysDVªksM dh rjQ LFkkukUrfjr gksrk gSA(2) y?kq v.kqvksa dh rqyuk esa o`gn v.kq rsth ls LFkkukUrfjr gksrs

gSA(3) DNA [k.Mksa dks ns[kus gsrq bfFkfM;e czksekbM + dk mi;ksx

fd;k tk ldrk gSA(4) tSy dks ijkcSaxuh izdk'k esa iznf'kZr djus ij mlesa DNA

ds uhys jax ds cS.M ns[ks tk ldrs gSA165. pBR322 ds ckjs esa xyr feyku dk p;u dhft, &

(1) Ori : vuqØe tksM+s x, DNA ds izfr:iksa dh la[;k dsfu;a=.k ds fy, mÙkjnk;h gSA

(2) Rop : IykfTeM+ ds izfrd`fr esa Hkkx ysus okys izksVhu dkdwV ys[ku djrk gSA

(3) Hind III : pBR322 esa izfrca/ku LFky(4) Sal-I : lkyeksusyk VkbQhewfj;e ds IykfTeM+ ij fLFkr oj.k

;ksX; fpUgd

166. dkWye -I dkWye -II (I) PCR (ihlhvkj) (A) cM+k iSekuk ij lao/kZu (II) ck;ksfj,DVj (B) ijiks"kh dksf'kdk esa ckgjh

DNA dk izos'k (III) thu xu (C) jsLVªhD'ku

,.MksU;wfDy,t ,atkbe (IV) Eco R1 (D) mi;ksxh thu dk vko/kZu

(1) I-D, II-A, III-B, IV-C (2) I-B, II-A, III-D, IV-C(3) I-C, II-B, III-D, IV-A (4) I-D, II-B, III-A, IV-C

167. iSyhUMksfed U;wfDy;ksVkbM vuqØe dks igpkfu, ?(1) GTCTAGAC (2) CTACATC(3) AGCAGCT (4) TTTCAAA

168. lw{ethoksa dh vkSj muds egRo dh fuEufyf[kr lwph dk feykudhft, :

(a) lSdSjksekblht +lfoZflvkbZ

(i) izfrj{kh laned dkjdksadk mRiknu

(b) eksuSLdl iI;wZfj;l (ii) fLol pht dks idkuk

(c) VªkbdksMekZ iksyhLiksje (iii) bZFkSukWy dk O;kolkf;dmRiknu

(d) izksfivkfu cSDVhfj;e'kekZukbZ

(iv) :f/kj esa dksysLVªkWy dedjus dk dkjd

(a) (b) (c) (d)(1) (iii) (i) (iv) (ii)(2) (iii) (iv) (i) (ii)(3) (iv) (iii) (ii) (i)(4) (iv) (ii) (i) (iii)

169. Methanogens grow anaerobically on cellulosicmaterial but do not produce :-(1) Methane (2) Oxygen(3) Carbon dioxide (4) Hydrogen

170. Which of the following is correct ?(1) VAM is not a biofertilizer(2) Azolla and BGA are the not biofertilizers(3) Synthetic fertilizers cause pollution(4) Probiotics are live pathogenic food supplement

171. Identify the blank spaces A, B, C and D in thefollowing given table and select the correct options.

Type ofmicrobe

Scientificname

Commercialproduct

Bacterium A Streptokinase

B Aspergillusniger

Citric acid

Fungus Trichodermapolysporum

C

Bacterium D Butyric acid

(1)A-Streptococcus, B-Fungus, C-Cyclosporin-A,D-Clostridium butylicum

(2) A-Clostridium butylicum, B-Streptococcus,C-Fungus, D-Cyclosporin-A

(3) A-Streptococcus, B-Yeast, C-Cyclosporin-A,D-Lactobacillus

(4) A-Streptococcus, B-Cyclosporin-A, C-Statins,D-Clostridium butylicum

172. Bioactive molecule __ is used as immunosuppressiveagent in organ transplant patients. It is producedby the fungus_____.(1) Antibiotic, Trichoderma polysporum(2) Cyclosporin A, Trichoderma polysporum(3) Trichoderma polysporum, Cyclosporin A(4) Trichoderma polysporum, Antibiotic

173. What role is played by lactic acid bacteria (LAB)in our stomach ?(1) Beneficial (2) Harmful(3) Neutral (4) Rarely beneficial

169. ehFkSukstu lSY;wyksth; inkFkksZ esa vok;oh; :i ls o`f¼ djrs gSaysfdu os D;k mRiék ugha djrs gaSA(1) ehFksu (2) vkWDlhtu(3) dkcZu MkbZ vkDlkbM (4) gkbMªkstu

170. fuEu esa ls dkSulk lR; gSA(1) VAM ,d tSo moZjd ugha gSA(2) ,TkkSyk rFkk BGA tSo moZjd ugha gSA(3) d`f=e moZjdksa ls i;kZoj.k iznw"k.k gksrk gSA(4) izksck;ksfVd thfor jksxtud iwjd vkgkj gSA

171. uhps nh xbZ lkj.kh esa ls A, B, C rFkk D fjDr LFkkuksa dks igpkfu,rFkk lgh fodYi dks pqfu,A

cSDVhfj;e A LVªsIVksdkbust

B ,sLijftylukbxj

flfVªd vEy

dod VªkbdksMekZikWyhLiksje

C

cSDVhfj;e D C;qfVªd vEy

(1)A-LVª s IV k sdk sdl, B-dod , C-lkbDyk sLik sfju-A,D-DyksLVªhfM;e C;wVkbyhde

(2) A-Dyk sLVª h fM;e C; wVkbyhde, B-LVª s IV k sdk sdl,C-dod, D-lkbDyksLiksfju-A

(3) A-LV ª sIVk sdk sdl, B-;hLV, C-lkbDyk sLik sfju-A,D-ysDVksoSlhyl

(4) A-LV ª s IVk sdk sdl, B-lkbDyk sLiksfju-A, C-LVSfVu,D-DyksLVªhfM;e C;wVkbyhde

172. tSo lfØ; v.kq _____ vax izR;kjksi.k jksxh esa izfrj{kk laneudkjd ds :i esa iz;qDr fd;k tkrk gSA ;g ____ dod }kjkmRiUu gksrk gSA(1) izfrtSfod] VªkbZdksMekZ ikWfyLiksje(2) lkbDyksLiksfju A, VªkbZdksMekZ ikWfyLiksje(3) VªkbZdksMekZ ikWfyLiksje] lkbDyksLiksfju A(4) VªkbZdksMekZ ikWfyLiksje] izfrtSfod

173. gekjs vkek'k; esa ySfDVd vEy thok.kq (LAB) dkSu lh HkwfedkfuHkkrs gSa ?(1) ykHknk;d (2) gkfudkjd(3) mnklhu (4) ;nk dnk ykHknk;d

174. Read the following statements :-(A) Cyclosporin A is used as a clot buster(B) Whisky, brandy and rum are produced by

distillation of fermented broth(C) NPV are the major Baculoviruses used as

biological control agents(D) The biogas production technology in India is

developed by ICARWhich of the above statements are not incorrect?(1) A & B (2) B & C(3) C & D (4) A & D

175. Propionibacterium sharmanii is a prokaryotesand it is used for the production of :(1) Toddy (2) Swiss cheese(3) Clot buster (4) Biogas

176. A good producer of lactic acid is :-(1) Aspergillus (2) Lactobacillus(3) Clostridium (4) Monascus

177. Statins commercially used as blood cholesterollowering agents are produced by :-(1) Yeast (2) Algae(3) Bacteria (4) Actinomycetes

178. Which one of the following is an example ofcarrying out biological control of pests/diseasesusing microbes?(1)Bt-Cotton to increase cotton yield(2)Lady bird beetle against aphids in mustard(3)Trichoderma sp. against certain plant pathogens.(4)Nucleopolyhedrovirus against white rust in

Brassica179. Which one of the following forms symbiotic

association with plants and helps them in their nutrition?(1) Glomus (2) Trichoderma(3) Azotobacter (4) Aspergillus

180. Which of the following bacteria is associated withbiofertilizer :(1) Azotobacter (2) Methanobacterium(3) Agrobacterium (4) Trichoderma

174. fuEufyf[kr dFkuksa dks if<, :-(A) lkbDyksLiksfju &, dks FkDdk LQksVu ds :i esa mi;ksx djrs gSa(B) OghLdh] czkaMh o je fdf.or rjy ds vklou ls cukbZ tkrh

gS(C) NPV eq[; cSD;wyksokbjl gS tks tSo fu;a=d dkjd ds :i

esa mi;ksx fd, tkrs gSa(D) Hkkjr esa ck;ksxSl mRiknu rduhd dk fodkl ICAR }kjk

fd;k x;k gSAmijksDr dFkuksa esa ls dkSuls dFku vlR; ugha gS ?(1) A o B (2) B o C(3) C o D (4) A o D

175. izkSfivksfucSDVhfj;e 'kkjekSukbZ ,d izkSdsfj;ksV~l gSa ftldk mi;ksxfuEu ds mRiknu ds fy, gksrk gS &(1) VksMh (2) fLol pht(3) DykV cLVj (4) tSo xSl

176. ysfDVd vEy dk vPNk mRiknd dkSulk gS\(1) ,sLijftyl (2) ySDVkscSlhyl(3) DyksLVªhfM;e (4) ekWukldl

177. LVsfVu tks fd jDr dksyLVªky dks de djus ds fy, vkS/kksfxd:i esa iz;ksx fd;k tkrk gS ;g fdlls mRikfnr gksrk gSA(1) ;hLV ls (2) 'kSoky ls(3) thok.kq ls (4) ,DVhuksekblhfVt ls

178. lw{ethoksa dk mi;ksx djrs gq, ihM+dksa@jksxksa ds tSfodh; fu;a=.kdk] fuEufyf[kr esa ls ,d mnkgj.k dkSu&lk gS ?(1) dikl dh mit esa c<+ksrjh djus ds fy, Bt-dikl cukuk(2) ljlksa esa ,fQMksa ds izfr ^^ysMh cMZ chVy** dk gksuk(3) dqN [kkl ikni jksxtudksa ds fy, VªkbdksMekZ Lih- ds

fo:¼ gksuk(4) czSfldk esa 'osr fdê ds izfr U;wfDvksikSyhgsMªksok;jl ds

fo:¼ gksuk179. fuEufyf[kr esa og dkSulk ,d gS tks ikS/kksa ds lkFk lgthou

LFkkfir djds muds iks"k.k esa lgk;rk djrk gS ?(1) Xyksel (2) VªkbdksMekZ(3) ,T+kksVkscSDVj (4) ,sliftZyl

180. fuEu esa ls dkSu ls thok.kq dk lca/k tSo moZjd ls gS &

(1) ,StksVkscSDVj (2) ehFkSukscSDVhfj;e

(3) ,xzkscSDVhfj;e (4) VªkbdksMekZ

Questions

1. Any problem in subscription of test series:

[2] Not at all [1] Some time [0] Problem faced

2. Test paper start on time:

[2] As per schedule [1] Some time deviate from schedule [0] Always delay

3. Test paper timing :

[2] Comfortable [1] Average [0] Need to be change

4. Location of test center:

[2] Good and approachable [1] Average in terms of approach [0] difficult to reach

5. Are you satisfy with result analysis :

[2] Outstanding [1] Average [0] Below average

6. The level of test paper [meet all the requirement of competitive examination]

[2] Outstanding [1] Average [0] Below average

7. Number of mistake in test papers

[2] Negligible [1] Are very less [0] Maximum

8. Do you think our test series is able to improve speed, accuracy & developing examination temperament?

[2] Yes [1] Partly [0] Not at all

9. Response from ALLEN on email / telephonically

[2] Always good and prompt [1] Some time delay [0] Not satisfactory

Response on test center

[2] Satisfactory [1] Partly Satisfactory [0] Not good