Target : JEE (Main+Advanced) 2019

®

Academic Session: 2018-19

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PART – A

Straight Objective Type This section contains 30 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its answer, out of which ONLY ONE is correct.

1. Air has refractive index 1.0003. The thickness of air column, which will have one more wavelength of yellow light of wavelength 6000Å than in same thickness of vacuum is :

(1) 2 cm (2) 2 mm (3) 2 m (4) 2 km

2. If each resistance is 'r' along the 12 edges of a

cubical skeleton of uniform wires, then find equivalent resistance between X and Y, where X and Y are the mid points of two opposite edges of a face of the cube :

A x

X F Y G

H E

C D

B

(1) 7r

8

(2) r2

(3) 4r5

(4) 7r5

Hkkx– A

lh/ks oLrqfu"B izdkj

bl [k.M esa 30 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4

fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA

1. ok;q dk viorZukad 1.0003 gSA fuokZr dh leku

eksVkbZ dh rqyuk esa ok;q LrEHk dh eksVkbZ esa 6000Å

rjaxnS/;Z ds ihys izdk'k dh rjaxnS/;Z ls ,d

rjaxnS/;Z T;knk gS] rc ok;q LrEHk dh eksVkbZ gksxhA (1) 2 cm

(2) 2 mm (3) 2 m (4) 2 km

2. ;fn ,d leku rkjksa ls cus 12 Hkqtkvksa ds ,d ?ku

dh izR;sd Hkqtk dk izfrjks/k 'r' gSA rc fp=k esa n'kkZ;s

vuqlkj X o Y ds e/; rqY; izfrjks/k Kkr dhft,,

tgkW X rFkk Y ?ku ds ,d Qyd dh nks foifjr

Hkqtkvksa ds e/; fcUnq gSA

A x

X F Y G

H E

C D

B

(1) 7r

8

(2) r2

(3) 4r5

(4) 7r5


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3. A particle is thrown at some angle with horizontal from point O, it falls at C as shown. Two triangles OAC & OBC are shown in the figure. Angle of projection from horizontal is :

A B

C x

y

O 10 cm20 cm

10 cm5 cm

(1) 1 5tan

3

(2) 1 7tan4

(3) 1 6tan5

(4) tan–1 (2) 4. A body of mass m is placed on an inclined

plane, the angle of inclination is = 37° and attached to the top end of the incline with a thread which is parallel to the slope. Then the slope is moved with a constant horizontal acceleration a. Friction is negligible. If the body

pushes the incline plane with a force of 34

mg,

value of a is:

a m

(1) 3g

4

(2) 4g3

(3) g9

(4) g12

3. ,d d.k dks fcUnq O ls {kSfrt ds lkFk dqN dks.k cukrs

gq, fp=kkuqlkj iz{ksfir fd;k tkrk gSaA tks fd fp=kkuqlkj

fcUnq C ij fxjrk gSA nks f=kHkqt OAC rFkk OBC fp=k esa iznf'kZr gSA {kSfrt ls iz{ksi.k dks.k D;k gksxkA

A B

C x

y

O 10 cm20 cm

10 cm5 cm

(1) 1 5tan

3

(2) 1 7tan4

(3) 1 6tan5

(4) tan–1 (2)

4. m nzO;eku dh oLrq = 37° urdks.k okys urry

ij j[kh gqbZ gS rFkk ;g urry ds 'kh"kZ ij ca/kh gqbZ

jLlh ls tqM+h gqbZ gSA jLlh urry ds lekUrj gSA

vc urry {kSfrt fn'kk esa fu;r Roj.k xfr izkjEHk

djrk gSA ?k"kZ.k ux.; gSA ;fn oLrq urry dks

34

mg cy ls nckrh gks rks a dk eku gksxkA

a m

(1) 3g

4

(2) 4g3

(3) g9

(4) g12


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5. A particle moving with constant acceleration on a straight line covers 20 m in 7th sec and 24 m in 9th sec. Distance travelled by the particle in10th sec is: (1) 26 m (2) 28 m (3) 30 m (4) 32 m

6. Three particles with the mass ratio 3:4:5

(the mass of the lightest body is m) are kept at three different points on the inner surface of a smooth hemispherical cup of radius r. The cup is fixed at its lowest point on a horizontal surface. At a certain instant, the bodies are released. All the three bodies make perfectly inelastic collision at the bottom most point of the hemisphere, such that energy loss due to the collision is maximum. Angle between the velocity vector of lightest particle and velocity vector of heaviest particle just before collision will be: (1) 53° (2) 127° (3) 143° (4) 90°

7. Moment of inertia of a disc of mass

M = 2 0.02 kg, inner radius a = 5 ± 0.05 cm, outer radius b = 7 ± 0.07 cm, with respect to an axis passing through geometrical centre of the disc and perpendicular to the plane of disc is given by (A0 ± B0 × 10–2) kg – cm2. The value

of 0

0

BA

is:

(1) 2 (2) 2.5 (3) 3 (4) 3.5

5. lh/kh js[kk ds vuqfn'k fu;r Roj.k ls xfr'khy d.k

7th sec esa 20 m nwjh rFkk 9th sec esa 24 m nwjh r;

djrk gSA 10th sec esa pyh xbZ nwjh gksxh :

(1) 26 m (2) 28 m (3) 30 m (4) 32 m

6. rhu d.kksa ds nzO;ekuksa dk vuqikr 3:4:5 gS] rFkk

lcls gYds d.k dk nzO;eku m gSA budks r f=kT;k ds

fpdus v)Zxksykdkj I;kys ds vUnj rhu fHkUu&fHkUu

fcUnqvksa ij j[kk tkrk gSA I;kys dk uhpyk fcUnq

{kSfrt lrg ij tM+or~ gSA fdlh le; d.kksa dks

NksM+k tkrk gSA lHkh d.k I;kys ds isans ij iw.kZ

vizR;kLFk :i ls bl izdkj Vdjkrs gS fd VDdj ds

dkj.k ÅtkZ gkfu vf/kdre~ gSA VDdj ds Bhd i'pkr~

lcls gYds d.k rFkk lcls Hkkjh d.k ds osx lfn'kksa

ds e/; dks.k D;k gksxkA

(1) 53° (2) 127° (3) 143° (4) 90°

7. ,d pdrh ftldk nzO;eku M = 2 0.02 kg,

vkUrfjd f=kT;k a = 5 ± 0.05 cm, ckg~; f=kT;k

b = 7 ± 0.07 cm gS] dk mlds T;kferh; dsUnz ls

xqtjus okyh rFkk pdrh ds ry ds yEcor~ v{k ds

ifjr% tM+Ro~ vk?kw.kZ (A0 ± B0 × 10–2) kg – cm2

}kjk fn;k x;k gSA 0

0

BA

dk eku Kkr djksaA

(1) 2 (2) 2.5 (3) 3 (4) 3.5


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8. We have a specially designed mirror, part of

which is a plane mirror and part of which is a

spherical mirror as shown.

C

R = 20 cm

Radius of curvature of spherical part is 20 cm,

C is its centre of curvature and the line shown

is its principal axis. Sizes of the mirrors are

small enough, so that paraxial approximation is

valid. A point object is moving on the principal

axis and at certain instant it is at distance

30 cm from the mirror, moving towards the

mirror with velocity 1 cm/s. At the instant

mentioned in the question, relative speed of

approach of one of the images with respect to

another image is:

(1) 34

cm/s

(2) 74

cm/s

(3) 54

cm/s

(4) 57

cm/s

8. fp=kkuqlkj fo'ks"k :Ik ls fufeZr niZ.k dk dqN Hkkx

lery rFkk dqN Hkkx xksfy; gS

C

R = 20 cm

xksfy; Hkkx dh o`Ørk f=kT;k 20 cm, rFkk o`Ørk

dsUnz C gS rFkk iznf'kZr js[kk bldh eq[; v{k gSA

niZ.k dk vkdkj bruk NksVk gS dh bl ij vkifrr

fdj.kks dks lek{kh; fdj.ks eku ldrs gSA niZ.k ls

30 cm nqjh ij fLFkr ,d fcUnqor oLrq niZ.k dh

rjQ eq[; v{k ij 1 cm/s osx ls xfr'khy gSA

iznf'kZr le; ij niZ.k ds ,d Hkkx esa izkIr izfrfcEc

rFkk nqljs Hkkx esa izkIr izfrfcEc ds e/; lkfeI; pky

D;k gksxhA

(1) 34

cm/s

(2) 74

cm/s

(3) 54

cm/s

(4) 57

cm/s


®



9. Consider a hollow sphere of radius R, its

quarter part is cut and kept such that

geometrical centre lies at origin as shown in

figure.

x

y

x-y axis lies in plane of paper and z-axis is

to the plane of the paper. Sphere is kept in a

manner such that plane portions of sphere lies

in x-z and y-z plane. Uniform electric field

0 0 0ˆ ˆ ˆE E i 2E j 2E k

(N/c) lies in the complete

space. Electric flux coming out of the curved

surface of sphere is:

(1) 20

3 R E2

(2) 20

5 R E2

(3) 202 R E

(4) Zero

9. R f=kT;k ds [kks[kys xksys fd dYiuk fdft,A blesa

ls ,d pkSFkkbZ Hkkx bl izdkj dkVk tkrk gS fd

bldk T;kferh dsUnz fp=kkuqlkj eqy fcUnq ij gSA

x

y

x-y v{k dkxt ds ry esa gS rFkk z-v{k dkxt ds

ry ds yEcor gSaA xksyk bl izdkj j[kk gSa fd xksys

ds lery Hkkx x-z rFkk y-z ry esa gSA ,d le:Ik

fo|qr {ks=k 0 0 0ˆ ˆ ˆE E i 2E j 2E k

(N/c) bl ry esa

iq.kZ :Ik ls mifLFkr gSA xksys ds oØ i`"B ls ckgj

vkus okys fo|qr QyDl D;k gksxkA

(1) 20

3 R E2

(2) 20

5 R E2

(3) 202 R E

(4) 'kwU;


®



10. A solenoid of inductance L and resistance R

2is connected in parallel to a resistance R. An ideal battery of emf E is connected across the parallel combination as shown in figure. Switch S is kept closed for long time and it is opened at time t = 0. Total heat generated in the solenoid after opening the switch is:

R

L, R/2

S E

(1) 2

24 LE3 R

(2) 2

21 LE3 R

(3) 2

22 LE3 R

(4) Zero

11. A homogeneous solid ball of specific gravity

2 and volume 1m3 is suspended on a weightless thread from an end of a homogeneous rod of specific gravity 3, volume 1m3. The rod is placed on the edge of a tumbler with water so that half of the ball is submerged in water when the system is in equilibrium. Density of water is 0.. Neglecting the surface tension on the boundaries between the ball and water, the ratio y/x of the parts of the rod to the brim is:

y x

(1) 4

3

(2) 2 (3) 3

(4) 73

10. L izsjdRo rFkk R2

izfrjks/k dh ifjukfydk ,d R izfrjks/k

ls lekUrj Øe esas tqM+h gSaA E fo|qr okgd cy dh ,d

vkn'kZ cSVjh fp=k esa n'kkZ;suqlkj lekarj Øe la;kstu ds

fljksa ls tqM+h gSA fLop S dks yEcs le; ds fy, cUn

j[kk x;k gS rFkk bls t = 0 ij [kksyk x;k gSA fLop

[kksyus ds ckn ifjukfydk esa mRiUu dqy Å"ek gS %

R

L, R/2

S E

(1) 2

24 LE3 R

(2) 2

21 LE3 R

(3) 2

22 LE3 R

(4) 'kwU;

11. ,d lekax Bksl xsan dk fof'k"V xq:Ro 2 rFkk vk;ru

1 m3 gS bldks nzO;ekghu /kkxs }kjk fp=kkuqlkj NM+ ds

,d fljs ls yVdk;k tkrk gSA NM+ dk fof'k"V xq:Ro

3 rFkk vk;ru 1 m3 gSA NM+ dks ikuh ls Hkjs gq, ik=k

dh nhokj ds fdukjs ij fp=kkuqlkj bl izdkj

O;ofLFkr fd;k tkrk gS fd vk/kh xsan ikuh esa Mwch

gqbZ gS rFkk fudk; lkE;koLFkk esa gSA ikuh dk ?kuRo

0 gSA ikuh rFkk xsan dh lEidZ lrg ij i`"B ruko

dk izHkko ux.; ekuksaA NM+ ds nksuksa Hkkxksa dh

yEckbZ;ksa y/x dk vuqikr D;k gksxkA y x

(1) 43

(2) 2 (3) 3

(4) 73


®



12. A uniform solid sphere of density and radius R starts to move with constant velocity v in a gravity free viscous medium. Viscous drag follows exact stokes law, Due to Viscous drag, heat produced is completely absorbed by the sphere and sphere evaporates uniformly. Latent heat of vaporization of material of sphere is L. How long after radius of the sphere will become R/2? Coefficient of viscosity for the medium is

(1) 2

2LR

4 v

(2) 2LR

3 v

(3) 23 LR

v

(4) 22 LR3 v

13. The capacitor is charged by closing the switch

S. The switch is then opened and the capacitor is allowed to discharge. Take R1 = R2 = R3 = R (Battery is ideal and connecting wire has negligible resistance).

The fraction of the total heat generated, lost in

R1 during discharging is:

(1) 16

(2) 13

(3) 12

(4) 23

12. rFkk R f=kT;k dk ,d le:i Bksl xksyk xq:Ro

eqDr ';ku ek/;e esa fu;r osx v ls xfr izkjEHk djrk

gSA vkjksfir ';ku cy LVksd ds fu;e dk ikyu

djrk gSA ';ku cy ds dkj.k mRiUu Å"ek xksys }kjk

iw.kZ :i ls vo'kksf"kr gks tkrh gS rFkk blds dkj.k

xksyk ,d leku :i ls okf"ir gksrk gSA xksys ds

inkFkZ ds ok"iu dh xqIr Å"ek L gSA fdrus le;

i'pkr~ xksys dh f=kT;k R/2 gks tk;sxh ? ek/;e dk

';kurk xq.kkad gSA

(1) 2

2LR

4 v

(2) 2LR

3 v

(3) 23 LR

v

(4) 22 LR3 v

13. la/kkfj=k fLop S dks cUn djds vkosf'kr fd;k tkrk

gSA vc fLop dks [kksyk tkrk gS rFkk la/kkfj=k dks

fujkosf'kr gksus fn;k tkrk gSA R1 = R2 = R3 = Rekfu,A (cSVªh vkn'kZ gS rFkk la;ksftr rkj dk izfrjks/k

ux.; gSA)

fujkos'ku ds nkSjku R1 esa mRiUu dqy Å"ek gkfu dk

fdruk va'k gksxkA

(1) 16

(2) 13

(3) 12

(4) 23


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14. In the situation as shown in figure time period

of small vertical oscillation of block will be –

(String, springs and pulley are ideal)

(1) 2 cos 2mk

(2) 2 sec

2

mk

(3) 2 sin 2mk

(4) 2 cosec 2mk

15. An experiment is performed to determine the

I - V characteristics of a Zener diode, which

has a protective resistance of R =100 , and

maximum power of dissipation rating of 1W.

The minimum voltage range of the DC source

in the circuit is :

(1) 0 – 12V

(2) 0 – 5V

(3) 0 – 24

(4) 0 – 8V

14. fp=k esa iznf'kZr fLFkfr ds fy, fi.M+ ds vYi Å/okZ/kj

nksyuksa dk vkorZdky gksxk– (jLlh, fLiazx rFkk f?kjuh

vkn'kZ gS)

(1) 2 cos 2mk

(2) 2 sec2

mk

(3) 2 sin 2mk

(4) 2 cosec 2mk

15. tsuj Mk;ksM dk I - V vfHkykf{k.kd Kkr djus ds

fy, ,d iz;ksx fd;k tkrk gS] ftlesa R =100 dk

lqj{kkRed izfrjks/k vkSj vf/kdre~ 'kfDr dk viO;;

jsfVax 1W gSA fn"V /kkjk L=kksr (DC source) dk

U;wure~ oksYVst ijkl D;k gksxkA

(1) 0 – 12V

(2) 0 – 5V

(3) 0 – 24

(4) 0 – 8V


®



16. A V shaped ladder is kept on the truck as

shown.

53º 53º

Friction between the truck and the ladder is

sufficient enough to prevent any slipping. A rod

is fixed between two arms (at their midpoints)

of the ladder so that ladder doesn't collapse.

This rod is massless and mass of the ladder is

m, which is uniformly distributed throughout its

length. Truck is initially at rest. Truck can

accelerate with maximum acceleration 20 m/s2

and retard with max retardation 10 m/s2.

Maximum speed of the truck can be 60 m/s.

This truck is to complete a journey of 1500 m

with maintaining the ladder in same position

with respect to the truck as it was before the

truck started moving, starting from rest and

ending at rest. Minimum time taken by the

truck for this task will be:

(1) 30 sec

(2) 29.5 sec

(3) 5 2 sec

(4) 30.5 sec

16. ,d V vkdkj dh lh<h fp=kkuqlkj Vªd ij fLFkr gSA

53º 53º

Vªd rFkk lh<h ds e/; ?k"kZ.k cy dgh Hkh fQlyu

jksdus ds fy, i;kZIr gSA lh<h dh nksuks Hkqtkvksa ds

e/; fcUnqvksa ds chp ,d NM+ tqMh gqbZ gSa ftlds

dkj.k lh<h dh nksuks Hkqtk,s vkil esas ugh feyrh gSaA

NM nzO;ekufgu gS rFkk lh<h dk nzO;eku m gS tks

fd mldh yEckbZ ij ,d leku forfjr gSaSA Vªd

izkjEHk esa fojke voLFkk esa gSA Vªd vf/kdre Roj.k

20 m/s2 rFkk vf/kdre eanu 10 m/s2 ls xfr dj

ldrk gSA Vªd dh vf/kdre pky 60 m/s gks ldrh

gSaA Vªd lh<h dks iznf'kZr fLFkfr esa j[krs gq, 1500

m dh nqjh r; djrk gSaA ;k=kk ds izkjaHk esa rFkk vUr

esa Vªd dk osx 'kqU; gSA Vªd }kjk ;k=kk esa fy;k x;k

U;qure le; D;k gksxkA

(1) 30 sec

(2) 29.5 sec

(3) 5 2 sec

(4) 30.5 sec


®



17. Two uniform solid spheres A and B of same material, painted completely black and placed in free space separately. Their radii are R and 2R respectively and the dominating wavelengths (wavelength corresponding to which spectral emissive power is maximum) in their spectrum are observed to be in the ratio 1 : 2. Which of the following is not correct.

(1) Ratio of their temperatures is 2 : 1 (2) Ratio of their emissive powers is 4 : 1 (3) Ratio of their rates of heat loss is 4 : 1

(4) Ratio of their rates of cooling is 32 : 1 18. A photon is incident upon a hydrogen atom

ejects an electron with a kinetic energy 10.7 eV. If the ejected electron was in the first excited state. Energy of photon incident nearly will be :

(1) 14.5 eV (2) 20.9 eV (3) 24.3 eV (4) 14.1 eV

19. Consider a brass rod and a steel rod (80 cm

longer than the brass rod) at 0°c. It is observed that on increasing temperatures of the two rods by same amount difference in lengths of the two rods does not change. Given that thermal coefficient of linear expansion for steel and brass are 11 × 10–6/°C and 19 × 10–6/°C respectively, the sum of lengths of the two rods at 0°C is

(1) 2 m (2) 4m (3) 3 m (4) 1.5 m

17. nks ,d leku Bksl xksys A rFkk B leku inkFkZ ds

cus gq, gS rFkk budksa iw.kZ:i ls dkyk jax djds

eqDr vkdk'k esa vyx&vyx j[kk tkrk gSA budh

f=kT;k,sa Øe'k% R rFkk 2R gS rFkk buds LisDVªe ds

laxr eq[; izsf{kr rjaxnS/;ks± ¼rjaxnS/;Z tks vf/kdre

mRltZu LiSDVªe {kerk ds laxr gS½ dk vuqikr 1 : 2

gSA fuEu esa ls dkSulk fodYi lgh ugha gSA

(1) buds rkieku dk vuqikr 2 : 1 gS

(2) budh mRltZu {kerk dk vuqikr 4 : 1 gS

(3) budh Å"ek Ðkl dh nj dk vuqikr 4 : 1 gS

(4) buds B.Ms gksus dh nj dk vuqikr 32 : 1 gS

18. ,d QksVkWu gkbMªkstu ijek.kq ij vkifrr gksrk gS] tks

10.7 eV. xfrt ÅtkZ ,d bysDVªkWu mRlftZr djrk

gSA;fn mRlftZr bysDVªkWu izFke mRrsftr voLFkk esa

gSA rc vkifrr QksVkWu dh ÅtkZ yxHkx gksxh

(1) 14.5 eV (2) 20.9 eV (3) 24.3 eV (4) 14.1 eV

19. 0°C ij LVhy dh NM+ ihry dh NM+ ls 80 cm

T;knk yEch gSA ;g izsf{kr fd;k tkrk gS fd nksuksa

NM+ksa dk leku rkieku c<+kus ij nksuksa NM+ksa dh

yEckbZ esa vUrj ifjofrZr ugha gksrk gSA ;fn LVhy

rFkk ihry dh NM+ksa ds rkih; jS[kh; izlkj xq.kkad

Øe'k% 11 × 10–6/°C rFkk 19 × 10–6/°C gks rks 0°C

ij nksuksa NM+ksa dh yEckbZ dk ;ksx gksxk &

(1) 2 m (2) 4m (3) 3 m (4) 1.5 m


®



20. If unit of mass, length and time are doubled

then in this new system the which of the

following, will not have same magnitude : (1) Gravitational constant

(2) Gravitational force

(3) Power of gravitational force

(4) Gravitational energy

21. To get an output of 1 from the circuit shown in

figure the input must be :

a b

c

Y

(1) a =1, b =1, c = 0

(2) a =1, b =0, c = 0

(3) a =0, b = 0, c = 1

(4) a =1, b =0, c = 1

22. Astronomers observe two separate solar

systems, each consisting of a planet orbiting a

sun. The two orbits are circular and have the

same radius R. It is determined that the planets

have angular momenta of the same magnitude

L about their suns, and that the orbital periods

are in the ratio of three to one ; i.e., T1 = 3T2.

The ratio m1/m2 of the masses of the two

planets is

(1) 1

(2) 3

(3) 2

(4) 3

20. ;fn nzO;eku, yEckbZ rFkk le; dh bdkbZ nqxuh gks

tk;s rks bl u;h i)fr esa dkSulh jkf'k dk ifjek.k

leku ugha gksxk (1) xq:Rokd"kZ.k fu;rkad

(2) xq:Rokd"kZ.k cy

(3) xq:Rokd"kZ.k cy dh 'kfDr

(4) xq:Rokd"kZ.k ÅtkZ

21. fp=k esa fn[kk;s x;s ifjiFk esa fuxZr 1 izkIr djus ds

fy, fuos'k D;k gksuk pkfg, :

a b

c

Y

(1) a =1, b =1, c = 0

(2) a =1, b =0, c = 0 (3) a =0, b = 0, c = 1

(4) a =1, b =0, c = 1

22. ,d [kxksy'kkL=kh nks iFkd lksyj fudk; dks izsf{kr djrk

gSA izR;sd fudk; ,d lw;Z ds ifjr% ifjØek djrs gq,

xzg ls fufeZr gS, nksuksa d{kk;sa oÙkkdkj gS rFkk nksuksa dh

f=kT;k R gSA ;g Kkr gS fd xzgksa dk lw;Z ds ifjr%

dks.kh; laosx dk ifjek.k L leku rFkk muds d{kh;

vkoZrdkyksa dk vuqikr 3 : 1 gS] vFkkZr T1 = 3T2 A nksuksa

xzgksa ds nzO;ekuksa dk vuqikr m1/m2 gksxkA

(1) 1

(2) 3

(3) 2

(4) 3


®



23. 5 identical thin conducting plates are kept parallel to each other such that separation between each plate is d. Area of each plate is A. Plates are arranged as shown in figure. All wires are conducting.

Switch S is closed, find the work done by

battery. (A >> > d2)

(1) d5

AV20

(2) d5AV2 2

0

(3) d2AV3 2

0

(4) dAV2 2

0

24. Conductor PQRSTU is carrying a current ‘i’ as

shown and QR and ST are bent in the form of quarter circles. OX, OY and OZ are three orthogonal coordinate axes. i , j and k have usual meaning. The magnetic filed at O is

(1) 0i ˆ ˆ[ j k]

8a

(2) 0i ˆ ˆ[ j 0.25k]8a

(3) 0i ˆ ˆ[ i j]8a

(4) 0i ˆ ˆ[ i 0.25 j]8a

23. 5 le:i iryh pkyd IysVsa lekUrj esa ,d&nwljs ls

d nwjh ij j[kh gqbZ gSA çR;sd IysV dk {ks=kQy A gSA IysVsa fp=k esa n'kkZ,uqlkj O;ofLFkr gSA lHkh rkj

pkyd gSA

fLop S cUn gS, cSVjh }kjk fd;k x;k dk;Z Kkr

dhft,A (A >> > d2)

(1) d5

AV20

(2) d5AV2 2

0

(3) d2AV3 2

0

(4) dAV2 2

0

24. pkyd PQRSTU esa fp=kkuqlkj ‘i’ /kkjk izokfgr gks jgh gS rFkk QR o ST dks prqFkkZ'k oÙk ds :i esa eksM+k tkrk gSA

OX, OY rFkk OZ rhu yEcor~ funsZ'kkad gSA i , j rFkk

k ds vius lkekU; vFkZ gSA O ij pqEcdh; {ks=k gksxkA

(1) 0i ˆ ˆ[ j k]8a

(2) 0i ˆ ˆ[ j 0.25k]8a

(3) 0i ˆ ˆ[ i j]8a

(4) 0i ˆ ˆ[ i 0.25 j]8a


®



25. A transverse sinusoidal wave of amplitude

2 mm is setup in a long uniform string.

Snapshot of string from x = 0 to x = meter is

taken at t = 0, which is shown. Velocity of point

P is in –y direction. Magnitude of relative

velocity of P with respect to Q is 2 cm/s.

Choose the correct option:

(1) wave equation is y = (2 × 10–3) sin

(–5t – 2x + 6 ) (m)


(5t + 2x + 6 ) (m)


(5t + 2x + 56 ) (m)


(5t – 2x + 6 ) (m)

25. 2 mm vk;ke dh ,d vuqizLFk T;kofØ; rajx ,d

yEch ,d leku Mksjh esa lapfjr gSA x = 0 ls

x = ehVj rd Mksjh dk izfr:i t = 0 ij fp=kkuqlkj

fy;k x;k gSA fcUnq P dk osx _.kkRed –y fn'kk esa

gSA Q ds lkis{k P ds osx dk ifjek.k 2 cm/s gSA

lgh fodYi dk p;u dhft,A

(1) rjax lehdj.k y = (2 × 10–3) sin

(–5t – 2x + 6 ) (m) gksxhA


(5t + 2x + 6 ) (m) gksxhA


(5t + 2x + 56 ) (m) gksxhA


(5t – 2x + 6 ) (m) gksxhA


®



26. In the circuit shown, the power factor of the

circuit is 35

. Power factor of only RC circuit is

45

. Source voltage is 100 volt and its angular

frequency = 100 rad/sec. RMS current in circuit is 5A. If inductive reactance is greater then capacitive reactance then the value of self inductance L is.

L C R

~

V = 100 volt = 100 rad/sec

(1) 1H2

(2) 1H3

(3) 1H

(4) H41

27. When a metallic surface is illuminated with

monochromatic light of wavelength , the stopping potential is 5 V0. When the same surface is illuminated with light of wavelength 3, the stopping potential is V0. Then the work function of the metallic surface is :

(1) hc6

(2) hc5

(3) hc4

(4) 2hc4

26. n'kkZ;s x;s ifjiFk esa ifjiFk dk 'kfDr xq.kakd 35

gSA

dsoy RC ifjiFk dk 'kfDr xq.kkad 45

gSA L=kksr

oksYVst 100 oksYV ,oa bldh dks.kh; vko`fÙk

= 100 rad/sec gSA ifjiFk esa oxZ ek/; ewy /kkjk

5A gSA ;fn izsjfd; izfr?kkr /kkjrh; izfr?kkr ls

vf/kd gS] rc LoizsjdRo L dk eku gksxkA

L C R

~

V = 100 oksYV = 100 rad/sec

(1) 1H2

(2) 1H3

(3) 1H

(4) H41

27. tc ,d /kkfRod lrg dks , rjaxnS/;Z ds ,do.khZ;

çdk'k ls çdkf'kr fd;k tkrk gS rks fujks/kh foHko

5 V0 gSA tc bl lrg dks 3 rjaxnS/;Z ds çdk'k ls

çdkf'kr fd;k tkrk gS rks fujks/kh foHko V0 gSA

/kkfRod lrg dk dk;ZQyu gS

(1) hc6

(2) hc5

(3) hc4

(4) 2hc4


®



28. A body of mass 2 kg is moved from A to B by an external agent in a conservative force field. If the speed of the body at the point A and B are 5m/s and 3m/s respectively and the work done by the external agent is –10J then find the change in potential energy in the process.

(1) – 16 J (2) 6 J (3) 10 J (4) – 10 J

29. A small bead of mass 'm' is threaded on a

frictionless circular wire of radius 'a'. The circular wire frame is rotated about its vertical diameter as shown. (Assume acceleration due to gravity is g) The angular speed required if the bead is to be made to move in a horizontal

circle of radius a 32

is :

(1) 1/ 2g

a

(2) 1/ 22g

a

(3) 1/23g

2a

(4) 1/ 23g

a

28. 2 kg dh ,d oLrq dks ckg~; dkjd }kjk laj{kh cy

{ks=k esa A ls B rd xfr djk;h tkrh gSA ;fn fcUnq

A rFkk B ij oLrq dh pky Øe'k% 5m/s rFkk 3m/s gS

rFkk ckg~; dkjd }kjk fd;k x;k dk;Z –10J gSA rc

izØe ds nkSjku fLFkfrt ÅtkZ esa ifjorZu gksxkA (1) – 16 J

(2) 6 J (3) 10 J (4) – 10 J

29. 'm' nzO;eku dk ,d NksVk eksrh a f=kT;k ds o`Ùkh;

?k"kZ.k jfgr rkj ij fLFkr gSA ;g o`Ùkh; rkj dk Ýse

Å/okZ/kj O;kl ds ifjr% fp=kkuqlkj ?kw.kZu dj jgk gSA

(ekuk xq:Ro ds dkj.k Roj.k g gSA) ?k"kZ.k dh

vuqifLFkfr esa eksrh dks a 3

2 f=kT;k ds {kSfrt o`Ùkh;

iFk esa pykus ds fy, vko';d dks.kh; pky D;k

gksxhA

(1) 1/ 2g

a

(2) 1/ 22g

a

(3) 1/23g

2a

(4) 1/ 23g

a


®



30. In Young's double slit experiment, the distance

between slits and the screen is 1.0 m and

monochromatic light of 600 nm is being used.

A person standing near the slits is looking at

the fringe pattern. When the separation

between the slits is varied, the interference

pattern disappears for a particular distance d0

between the slits. If the angular resolution of

the eye is 160 the value of d0 is close to

(1) 2 mm

(2) 1 mm

(3) 3 mm

(4) 4 mm

30. ;ax ds f}fLyV iz;ksx esa] fLyV vkSj insZ ds chp nwjh

1.0 m vksj 600 nm rjaxnS/;Z dh ,do.khZ; izdk'k

dk bLrseky fd;k x;k gSA ,d O;fDr fLyV ds

lehi [kM+k gS] og fÝat izk:i dks ns[krk gSA tc

fLyVksa ds chp dh nwfj;ksa dks cnyk tkrk gS rks

O;frdj.k izk:i] fLyVksa ds chp ,d fo'ks"k nwjh d0 ds

fy, xk;c gks tkrk gSA vxj vkW[kkas dk dks.kh;

ladYi (,asX;yj jst+yw'k~u~) 160 gS rks d0 dk eku

fuEufyf[kr fodYiksa esa ls fdlds lehi gS

(1) 2 mm

(2) 1 mm

(3) 3 mm

(4) 4 mm

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Space for Rough Work / (dPps dk;Z ds fy, LFkku )

PART – B

Atomic masses : [H = 1, D = 2, Li = 7, C = 12,

N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27,

Si = 28, P = 31, S = 32, Cl = 35.5, K = 39, Ca = 40,

Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65,

As = 75, Br = 80, Ag = 108, I = 127, Ba = 137,

Hg = 200, Pb = 207]

SECTION - I

Straight Objective Type

This section contains 30 multiple choice questions.

Each question has 4 choices (1), (2), (3) and (4) for

its answer, out of which ONLY ONE is correct.

31. A solution is prepared by urea and water. If

mole fraction of water is 0.8 in the solution

then find the ratio of mass of urea & water.

(1) 6

5

(2) 4

1

(3) 1

1

(4) 1

4

Hkkx– B

Atomic masses : [H = 1, D = 2, Li = 7, C = 12,

N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27,

Si = 28, P = 31, S = 32, Cl = 35.5, K = 39, Ca = 40,

Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65,

As = 75, Br = 80, Ag = 108, I = 127, Ba = 137,

Hg = 200, Pb = 207]

[k.M - I

lh/ks oLrqfu"B izdkj

bl [k.M esa 30 cgq&fodYih iz'u gSaA izR;sd iz'u ds

4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d

lgh gSA

31. ;wfj;k o ty }kjk ,d foy;u cuk;k x;kA ;fn

foy;u esa ty dh eksy fHkUu 0.8 gS rks ;wfj;k

o ty ds nzO;eku dk vuqikr Kkr dhft,A

(1) 6

5

(2) 4

1

(3) 1

1

(4) 1

4

mailto:[email protected]





32. Decomposition of A follows first order kinetics

by the following equation.

4A(g)

B(g) + 2C(g)

If initially, total pressure was 800 mm of Hg

and after 10 minutes it is found to be 650 mm

of Hg. What is half-life of A ?

(Assume only A is present initially)

(1) 10 minute

(2) 5 minute

(3) 7.5 minute

(4) None of these

33. Which of the following is incorrect statement

regarding A = [Co(en)2Cl

2]+ and

B = [Co(NH3)

4Cl

2]+.

(1) A is thermodinamically more stable than B

(2) A can show optical isomerism

(3) A and B both are diamagnetic

(4) A is outer orbital complex while B is inner

orbital complex

34. For a gaseous reaction A + 3B 2C H

= –90 KJ, S = – 200 JK–1 at 400 K.

For the reaction B2

3A

2

1 C at 400 K,

G (in KJ) is:

(1) –10 KJ

(2) – 5 KJ

(3) 20 KJ

(4) None of these

32. A dk fo;kstu fuEu lehdj.k }kjk izFke dksfV dh

cyxfrdh dk ikyu djrk gSA

4A(g)

B(g) + 2C(g)

;fn izkjEHk esa] dqy nkc 800 mm Hg ik;k x;k

rFkk 10 feuV i'pkr~ ;g 650 mm Hg ik;k x;kA

A dh v)Z&vk;q D;k gS ? ¼ekuk fd izkjEHk esa dsoy

A mifLFkr gSa½

(1) 10 feuV

(2) 5 feuV

(3) 7.5 feuV

(4) buesa ls dksbZ ugha

33. fuEu esa ls dkSulk dFku A = [Co(en)2Cl

2]+ rFkk

B = [Co(NH3)

4Cl

2]+ ds lanHkZ esa vlR; gS \

(1) A Å"ekxfrdh; :i ls, B dh rqyuk esa vf/kd

LFkk;h gSA

(2) A izdkf'kd leko;ork n'kkZ ldrk gSA

(3) A rFkk B nksuksa izfrpqEcdh; gSA

(4) A ,d ckâ; d{kd ladqy gS tcfd

B vkUrfjd d{kd ladqy gSA

34. ,d xSlh; vfHkfØ;k A + 3B 2C ds fy,

400 K ij H = –90 KJ, S = – 200 JK–1

400 K ij vfHkfØ;k B2

3A

2

1 C ds fy,

G (KJ esa) gksxk&

(1) –10 KJ

(2) – 5 KJ

(3) 20 KJ

(4) bues ls dksbZ ugha






35. The incorrect statement amongs the following

is :

(1) beryllium oxide is amphoteric in nature

(2) solubility of sulphates of second group

elements decreases down the group

(3) reducing power of hydride of alkali metal

decreases down the group

(4) Berylium has diagonal relationship with

alumunium

36. The e.m.f. of a cell corresponding to the

reaction :

Zn + 2H+ (aq) Zn2+ (0.1M) + H2(g) (1 bar)

is 0.169 volt at 25ºC. Calculate the pH of the

solution at the hydrogen electrode.

º

/ZnZn +2E = – 0.76 volt and º

/HH 2+E

= 0

(1) 9

(2) 11.5

(3) 8.5

(4) 10.5

37. HSOXOHOSMn 242

282

2

The magnetic moment (spin only) of [X] is :

(1) 0

(2) 3 BM

(3) 15 BM

(4) 24 BM

35. fuEu esa ls xyr dFku gS &

(1) csjhfy;e vkWDlkbM mHk;/kehZ izd`fr dk gksrk gSA

(2) f}rh; oxZ ds rRoksa ds lYQsVksa dh foys;rk oxZ

esa uhps dh vksj ?kVrh gSA

(3) {kkjh; /kkrqvksa ds gkbMªkbM dh vipk;d {kerk

oxZ esa uhps dh vksj ?kVrh gS

(4) csfjfy;e] ,Y;qfefu;e ds lkFk fod.kZ laca/k

j[krk gSA

36. vfHkfØ;k %

Zn + 2H+ (aq) Zn2+ (0.1M) + H2(g) (1 bar)

ls lacaf/kr lsy dk e.m.f. 25°C ij 0.169 oksYV gS

rks gkbMªkstu bysDVªksM ij] foy;u ds pH dh x.kuk

dhft,A

º

/ZnZn +2E = – 0.76 oksYV rFkk º

/HH 2+E

= 0

(1) 9

(2) 11.5

(3) 8.5

(4) 10.5

37. HSOXOHOSMn 242

282

2

[X] dk pqEcdh; vk?kw.kZ ¼dsoy pØ.k½ gS &

(1) 0

(2) 3 BM

(3) 15 BM

(4) 24 BM






38. The complex [Fe(H2O)5(NO)]2+ is formed in

the –3NO radical brown ring test. Select the

incorrect statement regarding this complex :

(1) Colour is due to charge transfer.

(2) It has iron in +1 oxidation state and NO+

ligand.

(3) It has magnetic moment (spin only) of

3.87 BM with three unpaired electrons.

(4) It has d2sp3 hybridisation.

39. (NH4)2Cr2O7 on heating gives a gas / vapour,

which is also produced in significant quantity

by :

(I) heating NH4Cl

(II) heating CuSO4.5H2O

(III) heating Ba(N3)2

(IV) treating NaNO3 with Zn in basic

medium.

(1) Only III

(2) I, IV

(3) Only IV

(4) II, III

40. Total number of nodal planes in *p2 x

molecular orbital is ………

(1) 0

(2) 1

(3) 3

(4) 2

38. ladqy [Fe(H2O)5(NO)]2+, –3NO ewyd ds Hkwjh

oy; ijh{k.k esa curk gSA bl ladqy ds lanHkZ esa

xyr dFku dk p;u dhft,&

(1) jax vkos'k LFkkukUrj.k ds dkj.k gksrk gSA

(2) ;g +1 vkWDlhdj.k voLFkk es vk;ju rFkk

NO+ fyxs.M j[krk gSA

(3) ;g rhu v;qfXer bysDVªkWuksa ;qDr 3.87 BM dk

pqEcdh; vk?kw.kZ ¼dsoy pØ.k½ j[krk gSA

(4) ;g d2sp3 ladj.k j[krk gSa

39. (NH4)2Cr2O7 xeZ djus ij xSl@ok"i nsrk gS] tks

fuEu }kjk Hkh lkFkZd ek=kk es mRiUu gksrh gS &

(I) NH4Cl dks xeZ djus ij

(II) CuSO4.5H2O dks xeZ djus ij

(III) Ba(N3)2 dks xeZ djus ij

(IV) {kkjh; ek/;e esa Zn ds lkFk NaNO3 dks

mipkfjr djus ij

(1) dsoy III

(2) I, IV

(3) dsoy IV

(4) II, III

40. *p2 x

vkf.od d{kd esa dqy uksM+y ry dh la[;k -

---------------- gSA

(1) 0

(2) 1

(3) 3

(4) 2






41. The graph of compressibility factor (Z) v/s P

for 1 mol of a real gas is shown in following

diagram. The graph is plotted at 273 K

temperature. If slope of graph at very high

pressure dZ

dP

is 1

2.8

atm–1 then calculate

volume of 1 mol of real gas molecules in

(L/mol).

[Given NA = 6× 1023 and R = 22.4

273 L atm K–1

mol–1]

–1dZ 1atm

dP 2.8

P

1

Z

(1) 1

(2) 2

(3) 3

(4) 4

42. The total number of orbitals associated with

the principal quantum number 4 is:

(1) 5

(2) 20

(3) 16

(4) 10

43. 50 mL of 0.2 M ammonia solution is treated

with 50 mL of 0.2 M HCl . If pKb of ammonia

solution is 4.7, the pH of the mixture will be :

(1) 4.15

(2) 3.75

(3) 8.85

(4) None of these

41. ,d eksy okLrfod xSl ds fy, lEihM~;rk xq.kkad

(Z) rFkk P ds e/; xzkQ fuEu vkjs[k esa n'kkZ;k x;k

gSA 273 K rki ij xzkQ vkysf[kr fd;k x;k gSA

;fn vfr mPp nkc ij xzkQ dk

<+kydZ

dP

,1

2.8

atm–1 gS] rc 1 eksy okLrfod

xSl v.kqvksa dk vk;ru (L/eksy) ifjdfyr dhft,A

[fn;k x;k gS NA = 6× 1023 rFkk R = 22.4

273 L atm

K–1 mol–1]

–1dZ 1atm

dP 2.8

P

1

Z

(1) 1

(2) 2

(3) 3

(4) 4

42. eq[; DokaVe la[;k 4 ls lEcfU/kr d{kdksa dh dqy

la[;k gS %

(1) 5

(2) 20

(3) 16

(4) 10

43. 0.2 M veksfu;k foy;u dk 50 mL, 0.2 M HCl ds

50 mL vEy ds lkFk fØ;k djrk gSA ;fn veksfu;k

foy;u dk pKb 4.7 gS] rc feJ.k dk pH gksxh %

(1) 4.15

(2) 3.75

(3) 8.85







44. An element has body centered cubic

structure with a cell edge length of 3.0 Å. The

density of the metal is 2 amu/Å3. How many

unit cells are present in 216 amu of this

element ?

(1) 1

(2) 2

(3) 4

(4) None of these

45. Which of the following statements are

correct?

(A) Principle of chromatographic operation is

adsorption.

(B) For adsorption : G < O, S < O, H < O

(C) Zeolites are used as catalyst in

petrochemical industries during cracking.

(D) In chemical adsorption, multilayer may be

formed on adsorbent.

(1) A, B, D

(2) B, C

(3) A, B, C

(4) B, D

44. ,d rRo 3.0 Å dksf"Bdk dksj yEckbZ okyh dk;

dsfUnz; ?kuh; lajpuk j[krk gSA /kkrq dk ?kuRo

2 amu/Å3 gSA bl rRo ds 216 amu esa fdruh

,dd dksf"Bdk;sa mifLFkr gS \

(1) 1

(2) 2

(3) 4


45. fuEu esa ls dkSuls dFku lgh gSa \

(A) o.kZysf[kdh (chromatographic) izØe dk

fl)kUr vf/k'kks"k.k gSA

(B) vf/k'kks"k.k ds fy, : G < O, S < O, H < O

(C) ftvksykbV] isVªksfy;e m|ksxksa esa Hkatu

(cracking) ds nkSjku mRizsjd ds :i esa iz;qDr

gksrk gSA

(D) jklk;fud vf/k'kks"k.k esa] cgqijr vf/k'kks"kd

(adsorbent) ij cu ldrh gSA

(1) A, B, D

(2) B, C

(3) A, B, C

(4) B, D






46. In order to refine ‘blister copper' it is melted in

furnace and is stirred with green logs of

wood. The purpose is:

(1) to expel the dissolved gases in blister

copper

(2) to reduce the metallic oxide impurities

with hydrocarbon gases liberated from

the wood

. (3) to bring the impurities to surface and

oxidize them

(4) to increase the carbon content in the

copper

47. Which of the following statement(s) are

correct about SO2Cl2 ?

(I) ClSClClSOOSO

bond angle

(II) Number of p– p bonds < Number of

p–d bonds

(III) 1 mol of this reacts with 0.1 mol P4

(IV) It's aqueous solution does not give pink

colour with phenolphthalein.

(1) I, II, III

(2) I,II,III, IV

(3) III, IV

(4) I, II

46. ^QQksysnkj dkWij* dks 'kksf/kr djus ds fy, bldks

HkV~Vh esa xyk;k tkrk gSa ,oe~ ydM+h ds gjs yV~Bksa

ds lkFk fgyk;k tkrk gaSA bldk m)s'; fuEu gSa&

(1) QQksysnkj dkWij esa ?kqyh xSlksa dks fudkyukA

(2) ydM+h ds yV~Bksa ls fudyus okyh gkbMªksdkcZu

xSlksa ds }kjk /kkfRod vkWDlkbM v'kqf);ksa dks

vipf;r djuk

(3) lrg ij v'kqf);ksa dks ykdj vkWDlhd`r djukA

(4) dkWij esa dkcZu vo;o dh o`f) djukA

47. SO2Cl2 ds lUnHkZ esa fuEu esa ls dkSuls dFku lgh

gSa\

(I) ClSClClSOOSO

ca/k dks.k

(II) p– p ca/kksa dh la[;k < p–d ca/kksa dh la[;k

(III) blds 1 eksy 0.1 eksy P4 ds lkFk fØ;k djrs

gSA

(IV) bldk tyh; foy;u fQukW¶Fksyhu ds lkFk

xqykch jax ugha nsrk gSA

(1) I, II, IV

(2) I,II,III, IV

(3) III, IV

(4) I, II






48. Let the plasma cell wall of shark is a

semipermeable membrane. When these cells

were kept in a series of NaCl solution of

different concentration at 25°C, these cells

remain intact in 0.7% (w/w) NaCl solution,

shrank in more concentrated solution and

swelled in dilute solution. What is the

osmotic pressure of the cell cytoplasm at

25°C? Given Kf of water = 1.86 kgmol–1K and

water freezes from the 0.7% (w/w) salt

solution at – 0.418°C. Assume solution is

dilute in each case. (R = 0.08 atm L / mol K)

(1) 2.3 atm

(2) 4.7 atm

(3) 5.3 atm

(4) 10.1 atm

49. When grignard reagent (excess) is treated

with isopropyl formate followed by acid

hydrolysis we get:

(1) Aldehyde

(2) 2º alcohol

(3) 3º alcohol

(4) 1º alcohol

48. ekuk 'kkdZ dh IykTek dksf'kdk fHkfÙk ,d

v)ZikjxE; f>Yyh gS] tc bu dksf'kdkvksa dks 25°C

ij fofHkUu lkUnzrk ds NaCl foy;u dh Js.kh esa

j[krs gSa rks ;s dksf'kdk,sa 0.7% (w/w) NaCl foy;u

esa v[kf.Mr jgrh gS] ;s dksf'kdk,¡ vf/kd lkfUnzr

foy;u esa fldqM+ tkrh gS rFkk ruq foy;u esa Qwy

tkrh gSA 25°C ij dksf'kdk ds dksf'kdk nzO;

(cytoplasm) dk ijklj.k nkc D;k gS \ fn;k gS %

ty ds fy, Kf = 1.86 kgmol–1K rFkk – 0.418°C

ij 0.7% (w/w) yo.k ds tyh; foy;u esa ty

terk gSA ekuk foy;u izR;sd fLFkfr esa ruq gSA

(R = 0.08 atm L / mol K)

(1) 2.3 atm

(2) 4.7 atm

(3) 5.3 atm

(4) 10.1 atm

49. fxzU;kj vfHkdeZd ¼vkf/kD;½ dks vkblksizksfiy QkWesZV

ds lkFk vfHkd`r djus ds ckn vEyh; ty vi?kVu

ls izkIr mRikn gksxk %

(1) ,fYMgkbM

(2) 2º ,YdksgkWy

(3) 3º ,YdksgkWy

(4) 1º ,YdksgkWy






50. Which of the following reactions report the

correct major products?

I. + CH3–CH2–CH2–Cl 3AlCl

II. 3AgNO.aq

III. Zn

IV.

H

OH2CH2CH3CH

(1) I, III (2) III, IV

(3) II, III, IV (4) I, III, II

51.

Product P will be :

(1)

O

(2)

OH

(3)

(4)

50. fuEu esa ls dkSulh vfHkfØ;k,¡ lgh eq[; mRikn

crkrh gSa \

I. + CH3–CH2–CH2–Cl 3AlCl

II. 3AgNO.aq

III. Zn

IV.

H

OH2CH2CH3CH

(1) I, III (2) III, IV

(3) II, III, IV (4) I, III, II

51.

eq[;

mRikn P D;k gksxk %

(1)

O

(2)

OH

(3)

(4)






52. An optically active aldotetraose which retains

its optical activity with aqueous Br2 but

looses its optical activity with conc. HNO3

has the following structure :

(i)

H OH

CHO

H OH

CH OH2

(ii) H OH

CHO

CH OH2

(iii)

CHO

CH OH2

(iv)

CHO

CH OH2

(1) i,ii

(2) i,iii

(3) ii, iii

(4) ii,iv

53. Consider the following sequence of reaction:

Ph–COOH A B

C H /Ni2

D.

The final product D is :

(1) Benzonitrile

(2) Benzylamine

(3) Aniline

(4) Benzamide

52. ,d izdkf'kd lfØ; ,YMksVsVªksl ftldh tyh; Br2

ds lkFk vfHkfØ;k djus ij bldh izdkf'kd

lfØ;rk ;FkklEHko jgrh gS ysfdu lkUnz HNO3 ds

lkFk ;g izdkf'kd lfØ;rk [kks nsrk gSA vr%

mijksDr ;kSfxd dh lajpuk gksxh %

(i)

H OH

CHO

H OH

CH OH2

(ii) H OH

CHO

CH OH2

(iii)

CHO

CH OH2

(iv)

CHO

CH OH2

(1) i,ii

(2) i,iii

(3) ii, iii

(4) ii,iv

53. vfHkfØ;k ds fuEu Øe ij fopkj dhft;s %

Ph–COOH A B

C H /Ni2

D.

vfUre mRikn D gksxk %

(1) csUtksukbVªkby

(2) csfUty,ehu

(3) ,fuyhu

(4) csUtkekbM






54. OHEt/ONaEt

/4SO2H [X]

The product [X] is :

(1) (2)

(3) (4)

55. Which of the following acid is a vitamin?

(1) Aspartic acid

(2) Ascorbic acid

(3) Adipic acid

(4) Saccharic acid

56. The correct order of electron density in the

benzene ring in the following compounds :

OH

OC H2 5

(1) I > II > III > IV

(2) II > IV > I > III

(3) IV > III > II > I

(4) I > III > IV > II

54.

OHEt/ONaEt

/4SO2H [X]

mRikn [X] gS :

(1) (2)

(3) (4)

55. fuEu esa ls dkSulk vEy ,d foVkfeu gS\

(1) ,LikfVZd vEy

(2) ,LdkWfcZd vEy

(3) ,fMfid vEy

(4) lsdsfjd vEy

56. fuEufyf[kr ;kSfxdksa esa csUthu oy; ds bysDVªkWu

?kuRo dk lgh Øe dkSulk gS \

OH

OC H2 5

(1) I > II > III > IV

(2) II > IV > I > III

(3) IV > III > II > I

(4) I > III > IV > II






57. Which of the following is not aromatic cation?

(1)

(2)

(3)

(4)

58.

Na metal

excess

In the above reaction find out the number of

moles of the gases formed in the both

reactions respectively ?

H2(g) SO

2(g) CO

2(g) NO

2(g)

(1) 1.5 1 1 0

(2) 3 1 1 1

(3) 1.5 0 3 0

(4) 1.5 0 2 0

57. fuEu esa ls dkSulk ,sjkseSfVd /kuk;u ugha gS\

(1)

(2)

(3)

(4)

58.

Na /kkrq

vkf/kD;

tyh; NaHCO3

vkf/kD;

mijksDr vfHkfØ;k ds lUnHkZ esa] nksuksa vfHkfØ;kvksa esa

cuus okys xSlh; mRiknksa dh eksy la[;k Øe'k% gS \

H2(g) SO

2(g) CO

2(g) NO

2(g)

(1) 1.5 1 1 0

(2) 3 1 1 1

(3) 1.5 0 3 0

(4) 1.5 0 2 0






59. The end product of following reaction is :

(1) (2)

(3) (4)

60. The most probable product of the following

sequence of reactions would be :

1

PCC ‘Y’

(1)

(2)

(3)

(4)

59. fuEu vfHkfØ;k dk vfUre mRikn gS &

NaOH ( )

(2)

tyh; ty vi?kVu

(1) (2)

(3) (4)

60. fuEufyf[kr vfHkfØ;k vuqØe dk lokZf/kd izkf;drk

mRikn dkSulk gS\

1

PCC Y’

(1)

(2)

(3)

(4)


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PART – C Straight Objective Type

This section contains 30 multiple choice questions.Each question has 4 choices (1), (2), (3) and (4) for its answer, out of which ONLY ONE is correct.

61. The equation of plane containing z-axis and

passing through the point (1, 2, 3) is

(1) x = 2y – z

(2) 2x = y

(3) 9x + 6y – 7z = 0

(4) x + 2y – z = 0

62. The value of 1 2

2–1

x – 1x 1 dx + 2

1

0

1– x1 x dx is

equal to

(1) –2

(2) –1

(3) 2

(4) 0

63. If A =

2 1 –10 1 40 0 3

, then tr(adj (adj A)) is equal

to

(1) 18

(2) 24

(3) 36

(4) 48

Hkkx – C lh/ks oLrqfu"B izdkj

bl [k.M esa 30 cgq&fodYih iz'u gSaA izR;sd iz'u ds

4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA

61. z-v{k dks j[kus okys rFkk fcUnq (1, 2, 3) ls xqtjus

okys lery dk lehdj.k gS&

(1) x = 2y – z

(2) 2x = y

(3) 9x + 6y – 7z = 0

(4) x + 2y – z = 0

62. 1 2

2–1

x – 1x 1 dx + 2

1

0

1– x1 x dx dk eku cjkcj gS&

(1) –2

(2) –1

(3) 2

(4) 0

63. ;fn A =

2 1 –10 1 40 0 3

, rc tr(adj (adj A)) cjkcj

gS&

(1) 18

(2) 24

(3) 36

(4) 48




64. Let w1 and w2 be the set of words which can be formed using all the letters of the words "SHREYANSH" and "SANIDHYA" respectively. A set is randomly selected and a word is selected from it. If the probability that it contains two alike letters together is

pq

, p, q N then the least value of (q – 3p)

is (1) 4

(2) 3 (3) 5 (4) 7

65. Let a

, b

, c

be vectors represent three co-terminous edges of a tetrahedron such

that a

b

= b c

= c

a

= 3

and

4(a

. a

) + 3(b

. b

) + 2(c

.c

) = 144. If V is the volume of the tetrahedron, then the maximum value of V is

(1) 4 (2) 6 (3) 8 (4) 10

66. Let Sn = n0

n1

+ n1

n2

+ ………

..… + nn –1

nn

, where n N. If n 1

n

SS

=154

then the sum of all possible values of 'n' is (1) 4

(2) 6 (3) 8 (4) 10

64. ekuk w1 rFkk w2 mu 'kCnksa dk leqPp; gS tks 'kCnksa

"SHREYANSH" rFkk "SANIDHYA" ds lHkh

v{kjksa dh lgk;rk ls cuk;k tk ldrk gSA ,d

leqPp; ;kn`fPNd pquk tkrk gS rFkk blls ,d

'kCn dks ;knfPNd pquk tkrk gSA ;fn blds nks]

,d leku v{kjksa ds lkFk j[kus dh izkf;drk pq

,

p, q N gS] rc (q – 3p) dk U;wure eku gS&

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

65. ekuk a

, b

, c

rhu lfn'k ,d prq"Qyd ds rhu

layXu dksj dks O;Dr djrs gS tcfd

a

b

= b c

= c

a

= 3

rFkk

4( a

.a

) + 3(b

. b

) + 2(c

. c

) = 144. ;fn V prq"Qyd dk vk;ru gS] rc V dk

vf/kdre eku gS& (1) 4

(2) 6 (3) 8 (4) 10

66. ekuk Sn = n0

n1

+ n1

n2

+ ……

……+ nn –1

nn

, tgk¡ n N. ;fn n 1

n

SS

=154

,

rc n ds lHkh laHkkfor ekuksa dk ;ksxQy gS&

(1) 4 (2) 6 (3) 8 (4) 10




67. Let p(x) be a real polynomial of least degree

which has a local maximum at x = 2 and a

local minimum at x = 4. If p(2) = 8, p(4) = 1,

then p(0) is

(1) 1685

(2) 42

(3) 43

(4) 45

68. Let f and g be two real valued differentiable

functions on R If f(x) = g(x) and g'(x) = f(x)

x R and f(3) = 5, f (3) = 4, then the value

of (f2() – g2()) is equal to

(1) 3

(2) 5

(3) 7

(4) 9

69. For each m R, the line y = (m – 1)x + n + 2

always. passes through a fixed point P. If the

ordinate of P is 3, then the value of 'n' is

(1) 1

(2) 2

(3) 3

(4) 4

67. ekuk fd p(x) U;wure ?kkr dk og okLrfod cgqin

(real polynomial) gS ftldk ,d LFkkuh; mPpre

(local maximum) x = 2 ij gS vkSj ,d LFkkuh;

U;wure (local minimum) x = 4 ij gSA ;fn p(2) = 8

vkSj p(4) = 1 gS rc p(0) dk eku gS&

(1) 1685

(2) 42

(3) 43

(4) 45

68. ekuk f vkSj g nks okLrfod vodyuh; Qyu] R ij

ifjHkkf"kr gSA ;fn f(x) = g(x) vkSj g'(x) = f(x)

x R rFkk f(3) = 5, f (3) = 4 gS] rc

(f2() – g2()) dk eku cjkcj gS&

(1) 3

(2) 5

(3) 7

(4) 9

69. izR;sd m R ds fy, js[kk y = (m – 1)x + n + 2

lnSo ,d fLFkj fcUnq P ls xqtjrh gSA ;fn P dh

dksfV 3 gS] rc n dk eku gS&

(1) 1

(2) 2

(3) 3

(4) 4




70. The absolute value of

/ 2sinx

0/ 2

cos x

0

(x cos x 1)e dx

(x sin x 1)e dx

is

equal to

(1) e

(2) e

(3) e2

(4) e

71. The number of possible arrangements of

letters of the word REVISION such that there

are exactly two vowels between E and V is

(1) 4320

(2) 480

(3) 600

(4) 720

72. –x 0lim

2

–1 –12

| x |1– x2 tan x – cos1 x

equal to

(1) –1

(2) 1

(3) 14

(4) –14

70.

/ 2sinx

0/ 2

cos x

0

(x cos x 1)e dx

(x sin x 1)e dx

dk fujis{k eku gS&

(1) e

(2) e

(3) e2

(4) e

71. 'kCn REVISION ds v{kjksa ds mu lHkh laHkkfor

foU;klkas dh la[;k] ftueas E vkSj V ds e/; Bhd nks

Loj gks] gS&

(1) 4320

(2) 480

(3) 600

(4) 720

72. –x 0lim

2

–1 –12

| x |1– x2 tan x – cos1 x

cjkcj gS&

(1) –1

(2) 1

(3) 14

(4) –14




73. If 4a2 + 9b2 + 16c2 = 2(3ab + 6bc + 4ca),then

a, b, c are in (where a, b, c R – {0})

(1) A.P.

(2) G.P.

(3) H.P.

(4) A.G.P.

74. If the variance of the data 2, 4, 5, 6, 8, 17 is

23.33, then variance of 5, 11, 14, 17, 23, 50 is

(1) 46.66

(2) 93.32

(3) 209.97

(4) 213.97

75. Equation of line in the plane x + 3y – z = 9,

which is perpendicular to the line

r

= i + j + k + (2 i + j – k ) and passing

through a point where the plane P meets the

given line, is

(1) x – 3

2 =

y – 21

= z5

(2) x – 3–2

= y – 2

–1 =

z5

(3) x – 3–5

= y – 2

1 =

z5

(4) x – 3

1 =

y – 25

= z2

73. ;fn 4a2 + 9b2 + 16c2 = 2(3ab + 6bc + 4ca), rc

a, b, c esa gksxsa& (tgk¡ a, b, c R – {0})

(1) A.P.

(2) G.P.

(3) H.P.

(4) A.G.P.

74. ;fn 2, 4, 5, 6, 8, 17 dk izlj.k 23.33 gS] rc

5, 11, 14, 17, 23, 50 dk izlj.k gS&

(1) 46.66

(2) 93.32

(3) 209.97

(4) 213.97

75. lery x + 3y – z = 9 esa ml js[kk dk lehdj.k] tks

js[kk r

= i + j + k + (2 i + j – k) ds yEcor~ gS

rFkk ml fcUnq ls xqtjrh gS tgk¡ lery

P, nh xbZ js[kk dks feyrk gS] gS&

(1) x – 3

2 =

y – 21

= z5

(2) x – 3–2

= y – 2

–1 =

z5

(3) x – 3–5

= y – 2

1 =

z5

(4) x – 3

1 =

y – 25

= z2




76. Let f(x) is continuous function and f(x) exists

everywhere. If f(2) = 10 and f '(x) –3 for all,

then the least possible value of f(4) is

(1) 1

(2) 2

(3) 3

(4) 4

77. If the range of values of 'a' for which

the function f : R R defined by

f(x) = 2sin 2x – 3cos2x – (a2 + a – 7) x + 5,

a R is strictly increasing is [p, q], where

p, q , then |p + q| is equal to

(1) 0

(2) 2

(3) 1

(4) 3

78. The area bounded by curve

f(x) = cos–1

]x[sin–2

sin and x-axis

from x = 0 and x = 2 is (where [x] and {x}

denotes greatest integer function and

fractional part function)

(1) sq.unit

(2) 2 sq. unit

(3) 3 sq.unit

(4) 2

sq.unit

76. ekukfd f(x) lrr Qyu gS rFkk loZ=k f (x) fo|eku

gSA ;fn f(2) = 10 vkSj f '(x) –3 lHkh x ds fy,]

rc f(4) dk U;wure laHkkfor eku gS&

(1) 1

(2) 2

(3) 3

(4) 4

77. ;fn a ds ekuksa dk ifjlj ftuds fy, Qyu

f : R R tgk¡

f(x) = 2sin 2x –3cos2x – (a2 + a – 7) x + 5, a R,

fujUrj o/kZeku Qyu] vUrjky [p, q] gS tgk¡

p, q rc |p + q| cjkcj gS&

(1) 0

(2) 2

(3) 1

(4) 3

78. oØ f(x) = cos–1

]x[sin–

2sin rFkk x-v{k]

x = 0 vkSj x = 2 ls ifjc) {ks=kQy gS (tgk¡ [x]

egÙke iw.kk±d Qyu gS rFkk {x} fHkUukRed Hkkx Qyu

dks O;Dr djrk gS&)

(1) oxZ bdkbZ

(2) 2 oxZ bdkbZ

(3) 3 oxZ bdkbZ

(4) 2

oxZ bdkbZ




79. If z lies on the curve arg(z + i) = 4

, then the

minimum value of |z + 4 – 3i| + |z – 4 + 3i| is

(1) 5

(2) 10

(3) 15

(4) 20

80. Let ABC be an acute angled triangle such

that a = 14, sin B = 1312 and c, a, b (in the

order) form an A.P., then the in-radius of

triangle ABC is

(1) 5

(2) 3

(3) 4

(4) 6

81. If , are roots of x2 – 2px + q = 0 and ,

are roots of x2 – 2rx + s = 0 where , , , ,

are in arithmetic progression, then s – q =

(1) r2 – p2

(2) r2 + p2

(3) p2 – r2

(4) pr – p2

79. ;fn z, oØ arg(z + i) = 4

ij fLFkr gS] rc

|z + 4 – 3i| + |z – 4 + 3i| dk U;wure eku gS&

(1) 5

(2) 10

(3) 15

(4) 20

80. ekuk ABC U;wudks.k f=kHkqt bl izdkj gS fd

a =14, sin B = 1312

rFkk c, a, b (Øe esa gS) lekUrj

Js<h cukrs gS] rc f=kHkqt ABC dh vUr%f=kT;k gS&

(1) 5

(2) 3

(3) 4

(4) 6

81. ;fn , lehdj.k x2 – 2px + q = 0 ds ewy gS

rFkk , lehdj.k x2 – 2rx + s = 0 ds ewy gS]

tgk¡ , , , , lekUrj Js<h esa gS] rc s – q =

(1) r2 – p2

(2) r2 + p2

(3) p2 – r2

(4) pr – p2




82. A tangent to the circle x2 + y2 = 4 intersects

the hyperbola x2 – 2y2 = 2 at P and Q. If locus

of mid point of PQ is (x2 – 2y2)2 = (x2 + 4y2),

then equal to

(1) 2

(2) 4

(3) 6

(4) 8

83. An ellipse with major axis 8 and minor axis 6

is drawn such that lines x – 3y + 2 = 0 and

3x + y – 4 = 0 are tangents to it, then locus of

its centre is

(1) x2 + y2 – 2x – 2y – 25 = 0

(2) x2 + y2 – 4x – 4y – 23 = 0

(3) x2 + y2 – 2x – 2y – 23 = 0

(4) x2 + y2 – 4x – 4y – 25 = 0

84. If (x + y3) dxdy = y and y(0) = 2, then sum of all

possible value(s) of y(1) is

(1) –4

(2) 4

(3) 0

(4) 2

82. oÙk x2 + y2 = 4 dh Li'kZ js[kk] vfrijoy;

x2 – 2y2 = 2 dks P vkSj Q ij feyrh gSA ;fn PQ

ds e/; fcUnq dk fcUnqiFk (x2 – 2y2)2 = (x2 + 4y2)

gS] rc cjkcj gS&

(1) 2

(2) 4

(3) 6

(4) 8

83. ,d nh?kZoÙk ftldh nh?kZv{k 8 rFkk y?kqv{k 6 gS]

bl izdkj cuk;k tkrk gS fd js[kk,a x – 3y + 2 = 0

rFkk 3x + y – 4 = 0 bl ij Li'kZ js[kk,a gS blds

dsUnz dk fcUnqiFk gS&

(1) x2 + y2 – 2x – 2y – 25 = 0

(2) x2 + y2 – 4x – 4y – 23 = 0

(3) x2 + y2 – 2x – 2y – 23 = 0

(4) x2 + y2 – 4x – 4y – 25 = 0

84. ;fn (x + y3) dxdy = y vkSj y(0) = 2, rc y(1) ds

lHkh laHkkfor ekuksa dk ;ksxQy gS&

(1) –4

(2) 4

(3) 0

(4) 2




85. ekuk A vkSj B nks Lora=k ?kVuk,a bl izdkj gS fd

P(A) =31 vkSj P(B) =

41

. ekuk fd fuEu dFku gS&

() P(A B) = 21 () P(A B ) =

61

() P

BAA =

52 (V) P( BA ) =

21

fuEu esa ls dkSulk fodYi lgh gS ?

(1) (),(),()

(2) (),()

(3) (),()

(4) mijksDr lHkh

86. ekuk f : R R bl izdkj gS fd

f(x) = x3 + 2x2 + 4x + sin

2x rFkk g(x), f(x)

dk izfrykse Qyu gS] rc g(8) dk eku cjkcj gS&

(1) 21

(2) 9

(3) 111

(4) 11

87. rhu oÙkksa

x2 + y2 – 2x + 3y – 7 = 0, x2 + y2 + 5x – 5y + 9 = 0

,oa x2 + y2 + 7x – 9y + 29 = 0 dks yEcdks.kh;

dkVus okys oÙk dk lehdj.k gS–

(1) x2 + y2 – 16x – 18y – 4 = 0 (2) x2 + y2 – 7x + 11y + 6 = 0

(3) x2 + y2 + 2x – 8y + 9 = 0 (4) x2 + y2 + 16x – 18y – 4 = 0

85. Let A and B are two independent events such

that P(A) = 31 and P(B) =

41

. Let consider

the statement

() P(A B) = 21 () P(A B ) =

61

() P

BAA =

52 (V) P( BA ) =

21

Which of the following option is correct ? (1) (),(),()

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

86. Let f : R R be defined as

f(x) = x3 + 2x2 + 4x + sin

2x and g(x) be

the inverse function of f(x), then g(8) is equal to

(1) 21

(2) 9

(3) 111

(4) 11 87. Equation of the circle cutting orthogonally

the three circles x2 + y2 – 2x + 3y – 7 = 0, x2 + y2 + 5x – 5y + 9 = 0 and x2 + y2 + 7x – 9y + 29 = 0 is

(1) x2 + y2 – 16x – 18y – 4 = 0 (2) x2 + y2 – 7x + 11y + 6 = 0

(3) x2 + y2 + 2x – 8y + 9 = 0 (4) x2 + y2 + 16x – 18y – 4 = 0




88. ekukfd fuEu dFku gS&

p : vki lQy gksuk pkgrs gSA

q : vkidks ,d jkg izkIr gksxh

rc ~(p q) dk udkjkRed dFku gS&

(1) vki lQy gksuk pkgrs gS rFkk vki ,d jkg

izkIr djrs gSA

(2) vki lQy gksuk pkgrs gS rFkk vki ,d jkg

izkIr ugha djrs gSA

(3) ;fn vki lQy gksuk pkgrs gS rc vki ,d

jkg izkIr ugha dj ldrs gSA

(4) ;fn vki lQy gksuk ugha pkgrs gS rc vki ,d

jkg izkIr djrs gSA

89. okLrfod la[;kvksa ds leqPp; esa ,d lEcU/k R bl

izdkj ifjHkkf"kr gS fd x R y tc |x| + |y| 1 rc

lEcU/k R gS&

(1) LorqY; vkSj lefer ijUrq laØked ugha

(2) lefer ijUrq u rks laØked vkSj u gh LorqY;

(3) laØked ijUrq u rks lefer vkSj u gh LorqY;

(4) LorqY;] lefer vkSj laØked esa ls dksbZ ugha

90. p ds iw.kk±d ekuksa dh la[;k gksxh ftlds fy,

lehdj.k 99cos3 – 20sin3 = 20p + 35 dk

,d gy gS&

(1) 8

(2) 9

(3) 10

(4) 11

88. Consider the following statements

p : you want to success

q : you will find a way,

then the negation of ~(p q) is

(1) you want of success and you find a way

(2) you want of success and you do not find

a way

(3) If you want of success then you cannot

find a way

(4) If you do not want to succeed then you

will find a way

89. In a set of real numbers a relation R is

defined as x R y such that |x| + |y| 1 then

relation R is

(1) reflexive and symmetric but not transitive

(2) symmetric but not transitive and reflexive

(3) transitive but not symmetric and reflexive

(4) none of reflexive, symmetric and transitive

90. The number of integral values of p for which

the equation 99cos3 – 20sin3 = 20p + 35

will have a solution is

(1) 8

(2) 9

(3) 10

(4) 11

IIMMPPOORRTTAANNTT IINNSSTTRRUUCCTTIIOONNSS // eeggÙÙooiiww..kk ZZ ffuunnss ZZ''kk A. General % A. lkekU; : 1. Immediately fill the particulars on this page of the Test

Booklet with Blue / Black Ball Point Pen. Use of pencil is strictly prohibited.

1. ijh{kk iqfLrdk ds bl i"B ij vko';d fooj.k uhys@dkys ckWy

IokbaV isu ls rRdky HkjsaA isfUly dk iz;ksx fcYdqy oftZr gSA

2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and fill in the particulars carefully.

2. mÙkj i=k bl ijh{kk iqfLrdk ds vUnj j[kk gSA tc vkidks ijh{kk

iqfLrdk [kksyus dks dgk tk, rks mÙkj i=k fudky dj lko/kkuhiwoZd

fooj.k HkjsaA

3. The Test Booklet consists of 90 questions. The maximum marks are 360.

3. bl ijh{kk iqfLrdk esa 90 iz'u gSA vf/kdre vad 360 gSA

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

4. bl ijh{kk iqfLrdk es rhu Hkkx A, B, C gSA ftlds izR;sd Hkkx esa

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

iz'uksa ds vad leku gSA izR;sd iz'u ds lgh mÙkj ds fy, 4¼pkj½ vad fu/kkZfjr fd;s x;s gSA

5. Candidates will be awarded marks as stated above in Instructions No. 4 for correct response of each question. 1/3 [one third (–1)] marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet.

5. vH;kfFkZ;ksa dks izR;sd lgh mÙkj ds fy, mijksDr funsZ'ku la[;k 4 ds

funsZ'kkuqlkj vad fn;s tk,axsA izR;sd iz'u ds xyr mÙkj ds fy;s

1/3oka Hkkx (–1) dkV fy;k tk;sxkA ;fn mÙkj iqfLrdk esa fdlh

iz'u dk mÙkj ugha fn;k x;k gks] rks dqy izkIrkad ls dksbZ dVkSrh ugha

fd tk;sxhA

6. There is only one correct response for each question. Filling up more than one response in any question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instructions 5 above.

6. çR;sd iz'u dk dsoy ,d gh lgh mÙkj gSA ,d ls vf/kd mÙkj nsus

ij mls xyr mÙkj ekuk tk;sxk vkSj mijksDr funsZ'k 5 ds vuqlkj

vad dkV fy;s tk;saxsA

B. Filling the Top-half of the ORS : Use only Black ball point pen only for filling the ORS. 7. Write your Roll no. in the boxes given at the top left corner of

your ORS with black ball point pen. Also, darken the corresponding bubbles with Black ball point pen only. Also fill your roll no. on the back side of your ORS in the space provided (if the ORS is both side printed).

B. vksvkj,l (ORS) ds Åijh&vk/ks fgLls dk Hkjko : ORS dks Hkjus ds fy, dsoy dkys ck¡y iSu dk mi;ksx dhft,A

7. ORS ds lcls Åij cka;s dksus esa fn, x, ck¡Dl esa viuk jksy uEcj

dkys ck¡y ikbUV ls fyf[k, rFkk laxr xksys Hkh dsoy dkys isu ls

Hkfj;sA ORS ds ihNs dh rjQ Hkh viuk jksy uEcj fyf[k, (;fn ORS nksuksa rjQ Nih gqbZ gSA)

8. Fill your Paper Code as mentioned on the Test Paper and darken the corresponding bubble with Black ball point pen.

8. ORS ij viuk isij dksM fyf[k, rFkk laxr xksyksa dks dkys ck¡y isu

ls dkys dhft,A 9. If student does not fill his/her roll no. and paper code

correctly and properly, then his/her marks will not be displayed and 5 marks will be deducted from the total.

9. ;fn fo|kFkhZ viuk jksy uEcj rFkk isij dksM lgh vkSj mfpr rjhds

ls ugha Hkjrk gS rc mldk ifj.kke jksd fy;k tkosxk rFkk izkIrkad esa

ls 5 vad dkV fy, tkosaxsaA 10. Since it is not possible to erase and correct pen filled bubble,

you are advised to be extremely careful while darken the bubble corresponding to your answer.

10. pawfd isu ls Hkjs x, xksys feVkuk vkSj lq/kkjuk laHko ugha gS blfy,

vki lko/kkuh iwoZd vius mÙkj ds xksyksa dks HkjsaA

11. Neither try to erase / rub / scratch the option nor make the Cross (X) mark on the option once filled. Do not scribble, smudge, cut, tear, or wrinkle the ORS. Do not put any stray marks or whitener anywhere on the ORS.

11. fodYi dks u feVk,a@u Ldzsp djsa vkSj u gh xyr (X) fpUg dks HkjsaA

ORS dks u dkVs u gh QkMs u gh xUnk ugha djsa rFkk dksbZ Hkh

fu'kku ;k lQsnh ORS ij ugha yxk,aA 12. If there is any discrepancy between the written data and the

bubbled data in your ORS, the bubbled data will be taken as final. 12. ;fn ORS esa fdlh izdkj dh fy[ks x, vkadMksa rFkk xksys fd, vkadMksa esa

fojks/kkHkkl gS] rks xksys fd, vkadMksa dks gh lgh ekuk tkosxkA

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AALLLL IINNDDIIAA OOPPEENN TTEESSTT--11 ((AAIIOOTT--11)) ((JJEEEE MMAAIINN PPAATTTTEERRNN)) CCOOUURRSSEE :: AALLLL IINNDDIIAA TTEESSTT SSEERRIIEESS ((VVIIKKAALLPP)) || CCLLAASSSS -- XXIIII//XXIIIIII

DDAATTEE :: 2233--1122--22001188 SSEETT -- 1

Name of the Candidate (ijh{kkFkhZ dk uke) :

I have read all the instructions and shall abide by them eSaus lHkh funsZ'kksa dks i<+ fy;k gS vkSj eSa mudk

vo'; ikyu d:¡xk@d:¡xhA

...................................... Signature of the Candidate

ijh{kkFkhZ ds gLrk{kj

Roll Number (jksy uEcj) :

I have verified all the information filled by the candidate. ijh{kkFkhZ }kjk Hkjh xbZ lkjh tkudkjh dks eSusa

tk¡p fy;k gSA

...................................... Signature of the Invigilator

ijh{kd ds gLrk{kj

P02-18

Target : JEE (Main+Advanced) 2019

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