Main Full 2.pdf

®

Academic Session: 2018-19

SET - 1

MAIN MAJOR TEST (MMT) (JEE MAIN PATTERN)

Target : JEE (Main+Advanced) 2019 Date: 30-12-2018 | Duration : 3 Hours | Max. Marks: 360 COURSE : ALL INDIA TEST SERIES (VIKALP) | CLASS - XII/XIII

Please read the last page of this booklet for the instructions. (Ñi;k

funsZ'kksa ds fy;s bl iqf Lrdk ds vfUre i`"B dks i<+sA)

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80034 44888

DO NOT BREAK THE SEAL WITHOUT BEING INSTRUCTED TO DO SO BY THE INVIGILATOR tc rd ifjos"kd funsZ'k ugha nsa rc rd iz'u i=k dh lhy dks ugha [kksaysA

P02-18

PART – A

Hkkx– A

Straight Objective Type

lh/ks oLrqfu"B izdkj

This section contains 30 multiple choice questions.

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

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

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

its answer, out of which ONLY ONE is correct.

1.

For the given combination of gates, if the logic

1.

states of inputs A, B, C are as follows

fn;s x;s rkfdZd }kjks ds fudk; ds fy, ;fn fuos'kh (inputs) A, B, C dh fLFkfr A = B = C = 0 rFkk

A = B = C = 0 and A = B = 1, C = 0 then the

A = B = 1, C = 0 gks] rks fuxZr (output) D dh

logic states of output D are

fLFkfr gksxh :

(1) 0, 0 (1) 0, 0

(2) 0, 1

(2) 0, 1

(3) 1, 0

(3) 1, 0

(4) 1, 1

2.

(4) 1, 1

A particle performs simple harmonic motion

2.

,d d.k A vk;ke ls ljy vkoZr xfr djrk gSA tc

with amplitude A. If its speed is doubled at the instant when it is at distance

A 3

;g ek/; fLFkfr ls from

equilibrium position. The new amplitude of the motion is (1)

nqxuk dj fn;k tkrk gS] d.k dk u;k vk;ke gksxkA (1)

11A

(2)

33A 2

(3)

33A 3

11A

(2)

33A 2

(3)

33A 3

A nwjh ij gS rc bldk osx 3

(4) mijksDr esa ls dksbZ ugha

(4) None of these Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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PSET101JRMMT2301218-1

3.

A disc of mass ‘m’ and radius ‘r’ is attached to

3.

a spring of stiffness ‘k’. During its motion, the

‘m’ nzO;eku vkSj ‘r’ f=kT;k dh ,d pdrh ,d fLizax

fu;rkad ‘k’ ds fLizax ls tqM+h gSA xfr ds nkSjku

disc rolls on the ground. When released from

pdrh yksVuh xfr djrh gSaA tc f[kpko dh fLFkfr

some stretched position, the centre of the disc will execute harmonic motion with a time period

ls NksM+k tkrk gS] rc blds nzO;eku dsUnz ds ljy

of

vkorhZ xfr dk vkorZdky gksxkA m

m k

k

r

4.

r

(1) 2

m 2k

(1) 2

m 2k

(2) 2

m k

(2) 2

m k

(3) 2

3m 2 k

(3) 2

3m 2 k

(4) 2

2m k

(4) 2

2m k

A uniform rope of mass M = 0.1 kg and length L = 10 m hangs from the ceiling. [g = 10 m/s ]

4.

M = 0.1kg vkSj L = 10m dh ,d le:Ik Mksjh Nr

2

(1) Speed of the transverse wave in the rope increases linearly from bottom to the top with

ls yVdh gSa [g = 10 m/s2] (1) Mksjh esa vuqizLFk rjax dh pky fups ls Åij tkus

distance.

ij nqjh ds lkFk js[kh; :Ik ls c
(2) Speed of the transverse wave in the rope

(2) Mksjh esa vuqizLFk rjax dh pky fups ls Åij tkus

decreases linearly from bottom to the top with

ij nqjh ds lkFk js[kh; :Ik ls ?kVrh gSA

distance.

(3) Mksjh esa vuqizLFk rjax dh pky fups ls Åij tkus

(3) Speed of the transverse wave in the rope remain constant along the length of the rope

fu;r jgrh gSA

(4) Time taken by the transverse wave to travel

(4) fups ls Åij tkus esa rjax dks 2 sec dk le;

the full length of the rope is 2 sec

yxrk gSA

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

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PSET101JRMMT2301218-2

5.

In the system shown, the wire connecting two

5.

n'kkZ;s

x;s fudk;

esa] jsf[k, nzO;eku

?kuRo

1 kg/m. 20

1 kg/m dk ,d rkj nks nzO;ekuksa ds e/; ca/kk gSA 20

A tuning fork of 50 Hz is found to be in

f?kjuh o CykWd A ds e/; rkj dk {kSfrt Hkkx 50 Hz

resonance with the horizontal part of wire

ds Lofj=k f}Hkqt ds lkFk vuqukn esa ik;k tkrk gSA

masses has linear mass density of

between pulley and block A. (Assuming nodes

¼ekfu, dh f?kjuh o CykWd A ij fuLiUn curs gS½A

at block A and pulley). Now at t = 0, system is

vc t = 0 ij] fudk; dks fojkekoLFkk ls NksMk tkrk

released from rest. The ratio of time gap

gSA t = 0 ls izkjEHk gksus ds i'pkr ~leku

between successive resonance with the same

f}Hkqt ds lkFk Øekxr vuqukn ds e/; le; vUrjky

tuning fork starting from t = 0. (take g = 10 m/s2) 50Hz 4kg A

Lofj=k

dk vuqikr gksxk (fyft, g = 10 m/s2)

60cm

50Hz 4kg A

60cm

B 4kg B 4kg

(1) 2 : 1 (1) 2 : 1

(2) 1 : 2 (3) 1 :

(2) 1 : 2

2 1

(3) 1 : (4) 1: 2

6.

2 1

(4) 1: 2

Initially two stretched strings are in resonance with each other. If tension in one of the string is

6.

çkjEHk esa nks ruh gqbZ Mksfj;ka ,d&nwljs ds lkFk vuqukn esa gSA ;fn ,d Mksjh esa ruko 2% ls c<+k

increases by 2%, then they produce 6 beats per second. Then original frequency of two

fn;k tkrk gS] rc os 6 foLiUn çfr lSd.M mRiUu

strings is approximately : (Assume their original

djrh gSA rc nks Mksfj;ksa dh ewy vko`fÙk yxHkx gksxh:

frequency is very-very greater than beat

(ekfu, fd mudh ewy vko`fÙk foLiUn vko`fÙk ls cgqr

frequency)

T;knk vf/kd gS)

(1) 600 Hz

(1) 600 Hz

(2) 300 Hz

(2) 300 Hz

(3) 1200 Hz

(3) 1200 Hz

(4) 900 Hz

(4) 900 Hz Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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PSET101JRMMT2301218-3

7.

Two sources S1 and S2 of same frequency f emits sound. The sources are moving as shown with speed u each. A stationary observer hears that sound. The beat frequency is (v = velocity of sound)

(1) (2) (3)

7.

2 u2 f

(1)

v 2  u2 2 v2 f

(2)

v 2  u2 2 u v f

(3)

9.

Two light waves are given by, E 1 = 2 sin (100  t - k x + 30º) and E 2 = 3 cos (200  t - k x + 60º) The ratio of intensity of first wave to that of second wave is : 2 (1) 3 4 (2) 9 1 (3) 9 1 (4) 3

8.

In the circuit, if the forward voltage drop for the diode is 0.5V, the current in the circuit will be (approximately)– 0.5 V

9.

8V

2 u2 f v 2  u2 2 v2 f v 2  u2 2 u v f

v 2  u2 2u (4) f v

v 2  u2 2u (4) f v

8.

nks L=kksr S1 rFkk S2 leku vko`fÙk f ds gS o /ofu mRiUu djrs gSA L=kksrksa dh xfr;k fp=kkuqlkj gSA ftlesa izR;sd dh pky u gSA ,d fojke esa :dk gqvk izs{kd /ofu dks lqurk gSA foLiUn (beat) vko`fÙk gksxhA (v = /ofu dk osx)

2.2 k

(1) 3.4 mA (2) 2 mA (3) 2.5 mA (4) 3 mA

nks izdk'k rjaxs E1 = 2 sin (100  t - k x + 30º) rFkk E2 = 3 cos (200  t - k x + 60º) ls nh tkrh gSaA rks igyh o nwljh rjaxksa dh rhozrk esa vuqikr gSA 2 3 4 (2) 9 1 (3) 9 1 (4) 3

(1)

ifjiFk esa] ;fn Mk;ksM ds fy, vxz foHko iru 0.5V gS, rc ifjiFk esa /kkjk yxHkx gksxhA 0.5 V 8V

2.2 k

(1) 3.4 mA (2) 2 mA (3) 2.5mA (4) 3 mA Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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PSET101JRMMT2301218-4

10.

If the magnetic field at 'P' can be written as K

10.

 tan   then K is : 2

11.

;fn 'P' ij pqEcdh; {ks=k K (tan(/2) ls fy[kk tk;sa] rks K gSA

(1)

0  4d

(1)

0  4d

(2)

0  2d

(2)

0  2d

(3)

0  d

(3)

0  d

(4)

20  d

(4)

20  d

Find current through the battery just after switch S is closed. Initially all the capacitors are uncharged :

11.

dqath S cUn djus ds Bhd ckn cSVjh ls xqtjus okyh /kkjk Kkr dhft,A çkjEHk lHkh la/kkfj=k vukosf'kr gSA

(1)

4 A 3

(1)

4 A 3

(2)

5 A 3

(2)

5 A 3

(3)

3 A 4

(3)

3 A 4

(4)

3 A 5

(4)

3 A 5

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

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PSET101JRMMT2301218-5

12.

A 50 Hz ac source is connected to a capacitor

12.

C in series with a resistance 1 K  . The rms

50 Hz vko`fr dk ,d izR;korhZ L=kksr la?kkfj=k C ,oa

izfrjks/k 1 K  ls Js.khØe esa tqM+k gqvk gSA muds

voltage measured across them are 5 volt and

lkis{k ekis x;s foHko dk oxZek/; ewy eku Øe'k%

2 volt respectively. Assume the capacitor to be ideal. The peak value of the source voltage

5 volt vkSj 2 volt gSA ekfu;s fd la?kkfj=k vkn'kZ gSA

and the capacitance are respectively.

L=kksr foHko dk f'k[kj eku ,oa /kkfjrk Øe'k% gSa µ

(1) 7V, 1.27 F

(1) 7V, 1.27 F

(2) 5.3 V, 2.3 F

(2) 5.3 V, 2.3 F

(3) 7.62 V, 1.27 F

(3) 7.62 V, 1.27 F

(4) 3 V, 2.3 F 13.

(4) 3 V, 2.3 F

In an atom, two electrons move around the

13.

nucleus in circular orbits of radii R & 4R. The

,d ijek.kq esa ukfHkd ds pkjkas vkSj nks bysDVªksu R o 4R f=kT;kvksa dh d{kkvksa esa xfr djrs gSaA nksuksa ds

ratio of the time taken by them to complete one

}kjk ,d pØ iwjk djus esa yxs le;ksa dk vuqikr gS&

revolution is: (assume that electrons exert

¼ekuk fd nksuksa bysDVªksu ,d nwljs ij ux.; cy

negligible force on each other) (1) 1: 4

yxkrs gSa &

(2) 4: 1

(1) 1: 4

(3) 1: 8

(2) 4: 1 (3) 1: 8

(4) 8: 1

(4) 8: 1

 14.



Two radioactive materials A and B have decay constants 5 and  respectively. Initially both A and B have the same number of nuclei. The

14.

nks jsfM;ks ,fDVo inkFkZ A vkSj B ds vi{k; fu;rkad 5 vkSj  Øe'k% gSA izkjEHk esa nksuksa ds ukfHkdks dh la[;k cjkcj gSA A ds ukfHkdks dh la[;k vkSj B ds

ratio of the number of nuclei of A to that of B will be

ukfHkdks dh la[;k dk vuqikr

1 after a time e

brus le; ckn gks

tk;sxkA

(1)

1 5

(1)

1 5

(2)

1 4

(2)

1 4

(3)

5 4

(3)

5 4

(4)

1 e

4 5

(4)

4 5

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

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PSET101JRMMT2301218-6

15.

A large open tank is filled with water upto a

15.

height H. A small hole is made at the base of

,d foLr`r [kqyk Vsad H Å¡pkbZ rd ikuh ls Hkjk gqvk gSA Vsad ds ryh esa ,d NksVk Nsn fd;k tkrk gSA

the tank. It takes T1 time to decrease the height of water to

blesa ikuh dk Lrj

H (n > 1) and it takes T2 time to n

yxrk gS rFkk ;g 'ks"k ikuh [kkyh djus esa T2 le;

take out the remaining water. If T1 = T2, then

ysrk gSA ;fn T1 = T2 gS rc n dk eku gS :

the value of n is :

16.

H (n > 1) rd ?kVus esa T1 le; n

(1) 2

(1) 2

(2) 3

(2) 3

(3) 4

(3) 4

(4) 2 2

(4) 2 2

Surface tension of a liquid is S = 0.01N/m. Density of the liquid is 1g/cm3. A capillary tube contain 12cm the liquid column. If R1 and R2

16.

,d nzo dk i`"B ruko S = 0.01N/m] ?kuRo 1g/cm3 gSA ,d dS'k uyh esa bl nzo LrEHk dh

Å¡pkbZ 12cm gSA ;fn nks lrgksa dh oØrk f=kT;k

are radii of curvature of two surfaces then

Øe'k% R1 rFkk R2 gks rks

R1  R2 is : R1R2

R1  R2 R1R2

dk eku gksxkA

h

h

(1) 3 × 104 m–1

(1) 3 × 104 m–1

(2) 4 × 104 m–1

(2) 4 × 104 m–1

(3) 12 × 103 m–1

(3) 12 × 103 m–1

(4) 6 × 104 m–1

(4) 6 × 104 m–1

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

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PSET101JRMMT2301218-7

17.

In determination of young modulus of elasticity

17.

of wire, a force is applied and extension is

,d rkj ds ;ax xq.kkad ds irk yxkus es ,d cy vkjksfir fd;k tkrk gS ,oa foLrkj vkysf[kr fd;k

recorded. Initial length of wire is ‘ 1 m ’. The

tkrk gSA rkj dh izkjfEHkd yEckbZ ‘ 1 m ’ gSA foLrkj o

curve between extension and stress is depicted

izfrcy ds e/; oØ fn;k x;k gS rks rkj dk ;ax

then young modulus of wire will be:

xq.kkad gksxk & 4mm 4mm

Extension 2mm

foLrkj 2mm

8000KN/m2

4000KN/m2

Stress

4000KN/m2

KN/m2

(1) 2 × 10 N/m 9

çfrcy

KN/m2

2

(1) 2 × 109 N/m2 (2) 1 × 109 N/m2 (3) 2 × 1010 N/m2 (4) 1 × 1010 N/m2

(2) 1 × 109 N/m2 (3) 2 × 1010 N/m2 (4) 1 × 1010 N/m2 18.

Two moles of hydrogen are mixed with n moles

18.

of helium. The root mean square speed of gas molecules in the mixture is

ek/; oxZ pky feJ.k esa /ofu dh pky ls (1) 3 (2) 2 (3) 1.5 (4) 2.5

(2) 2 (3) 1.5 (4) 2.5 Two uniform solid spheres A and B of same material, painted completely black and placed in free space separately. Their radii are R and respectively

2 xq.kk

gS] rks n gS

(1) 3

2R

gkbMªkstu ds nks eksyksa dks ghfy;e ds n eksyksa ds lkFk feyk;k tkrk gSA feJ.k esa xSl v.kqvksa dh ewy

2 times the

speed of sound in the mixture. Then n is.

19.

8000KN/m2

and

the

dominating

wavelengths 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

19.

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± dk vuqikr 1 : 2 gSA fuEu esa ls dkSulk fodYi xyr 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

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

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PSET101JRMMT2301218-8

20.

A semi circular disk of mass ‘M’ and radius ‘R’

20.

shown in figure OA is initially vertical. Disk is

‘M’ nzO;eku rFkk ‘R’ f=kT;k dh ,d v)Z o`Ùkkdkj

pdrh fp=kkuqlkj fLFkr gS] izkjEHk esa OA Å/okZ/kj gSA

released from this position when OA becomes

pdrh dks bl fLFkfr ls NksM+k tkrk gSA tc OA

horizontal, then angular velocity of disk is :

A

{kSfrt gks tkrk gS] rc pdrh dk dks.kh; osx gksxkA

M, R

A

O Hing

21.

M, R

O Hing ¼fuyfEcr½

(1)

4g  4  1   3R  3 

(1)

4g  4  1   3R  3 

(2)

g R

(2)

g R

(3)

4g 3R

(3)

4g 3R

(4)

g 4  1   R  3 

(4)

g 4  1   R  3 

A plank P is placed on a solid cylinder S, which rolls on a horizontal surface. The two are of

21.

CykWd P dks {kSfrt lrg ij ?kw.kZu xfr dj jgs csyu S ij j[kk tkrk gSA nksauks ds nzO;eku leku gSA

equal mass. There is no slipping at any of the surfaces in contact. The ratio of the kinetic

lEidZ fcUnq ds chp lkis{k xfr ugha gS rks P dh

energy of P to the kinetic energy of S is:

xfrt ÅtkZ o S dh xfr ÅtkZ dk vuqikr gksxk :

(1) 1: 1

(1) 1: 1

(2) 2: 1

(2) 2: 1

(3) 8: 3

(3) 8: 3

(4) 1: 4

(4) 1: 4 Space for Rough Work / (dPps dk;Z ds fy, LFkku)

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PSET101JRMMT2301218-9

22.

A particle of mass m0, travelling at speed v 0,

22.

strikes a stationary particle of mass 2m0. As a

2m0 nzO;eku ds d.k ls Vdjkrk gSA ifj.kkeLo:i

result, the particle of mass m0 is deflected through 45º and has a final speed of

v0 2

m0 nzO;eku dk d.k çkjfEHkd fn'kk ls 45º dks.k ij

.

xfr djus yxrk gS rFkk bldh vfUre pky

Then the speed of the particle of mass 2m0

(4)

23.

(1)

v0 2 2

(2)

2v0

gks

v0 2 v0 2 2

(3)

v0

(4)

2

A pump is required to lift 800 kg of water per

23.

minute from a 10 m deep well and eject it with

2v0

v0 2

,d iEi }kjk ,d feuV esa 800 kg ikuh dks 10 ehVj xgjs dqa,sa ls fudky dj 20 m/s dh pky

a speed of 20 m/s. The minimum required

}kjk Qsadk tkrk gSA iEi dh vko';d U;qure 'kfDr

horse power of the pump will be:

¼v'o 'kfDr esa½ Kkr djsaA.

(1) 6

(1) 6

(2) 1.87

(2) 1.87

(3) 5.36

(3) 5.36

(4) 5.5 24.

2

dh pky gksxhA

v (1) 0 2

(3)

v0

tkrh gS rks VDdj ds i'pkr~ 2m0 nzO;eku ds d.k

after this collision is :

(2)

m0 nzO;eku dk ,d d.k v 0 pky ls pyrk gqvk fLFkj

(4) 5.5

A block of mass 5 kg slides down an inclined plane which makes an angle  with the horizontal. The co-efficient of friction between

24.

5 kg nzO;eku dk xqVdk ,d ur ry tks {kSfrt ls  dk dks.k cukrk gS ij uhps fQlyrk gSA

the block and the plane is =3/4. The force

xqVds o ry ds e/; ?k"kZ.k xq.kkad =3/4gSA xqVds

exerted

}kjk ry ij vkjksfir cy gS (g = 10 m/s2):

by the

block

on

the

plane is

2

(g = 10 m/s ): (1) 25 N

(1) 25 N (2) 125/4 N





(2) 125/4 N

(3) 15 N

(3) 15 N

(4) 75/4 N 

(4) 75/4 N 





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

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PSET101JRMMT2301218-10

25.

A perfect smooth sphere A of mass 2kg is in contact with a rectangular block B of mass 4kg and vertical wall as shown in the figure. All surfaces are smooth. Find normal reaction by vertcial wall on sphere A.

(1) 20 N (3) 30 N 26.

25.

lkFk fp=k esa iznf'kZr gSA lHkh lrg ?k"kZ.kjfgr gSA xksys A ij m/okZ/kj nhokj }kjk yxk vfHkyEc cy Kkr djksaA

(2) 25 N (4) 20 3 N

At an instant particle-A is at origin and moving with constant velocity (3iˆ  4ˆj) m / s and

2kg nzO;eku dk ,d iq.kZ ?k"kZ.kjfgr xksyk A, 4kg ds nzO;eku ds vk;rkdkj CykWd B o m/okZ/kj nhokj ds

(1) 20 N (3) 30 N 26.

fdlh {k.k ij d.k A ewy fcUnq ij rFkk fu;r osx (3iˆ  4ˆj) m / s

particle-B is at (4,4)m and moving with constant velocity (4iˆ  3ˆj) m / s . Then at this instant which of the following options is incorrect: (1) relative velocity of B w.r.t. A is (iˆ  7ˆj) m / s

(2) 25 N (4) 20 3 N

ls xfr'khy gS] rFkk d.k B fcUnq

(4,4)m ij gS] rFkk fu;r osx (4iˆ  3ˆj) m / s . ls

xfr'khy gS] rks bl le; dkSulk fodYi vlR; gS : (1) A ds lkis{k B dk lkis{k osx (iˆ  7ˆj) m / s gSA

(2) approach velocity of A and B is 3 2 m/s (2) A rFkk B dk lkehI; osx 3 2 m/s gSA

(3) relative velocity of B w.r.t. A remains constant (4) approach velocity of A and B remains constant 27.

a 2 directly above the centre of a horizontal square of side 'a' while another point charge –q is placed at a distance 'a' directly below one of its (the square's) corners. The electric flux through the square is q (1) 60

A point charge q is placed at a distance

(3) A ds lkis{k B dk lkis{k osx fu;r jgrk gSA (4) A rFkk B dk lkehI; osx fu;r jgrk gSA 27.

Hkqtk yEckbZ a ds {kSfrt oxkZdkj ds dsUnz ls Bhd a nwjh ij fcUnq vkos'k q fLFkr gS] tcfd ,d vU; 2 fcUnq vkos'k –q, bl oxZ ds fdlh 'kh"kZ ds Bhd uhps a

Åij

nwjh ij fLFkr gSA oxZ ls ikfjr oS|qr ¶yDl gksxkA (1)

q 60

(2)

q 80

(2)

q 80

(3)

q 240

(3)

q 240

(4)

5q 240

(4)

5q 240

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

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PSET101JRMMT2301218-11

28.

Three identical conducting large plates each of area A are placed perpendicular to the plane of the paper and parallel to each other as shown. The outer plates are given charge Q and –2Q. The electrostatic force on the middle plate is :

28.

gS] isij ds yEcor~ lrg esa ,d&nwljs ds lekUrj j[kh gSA ckgjh IysVksa dks Q rFkk–2Q vkos'k fn;k x;k gSA e/; IysV ij fo|qr cy gksxkA

(1) is directed towards left (2) is directed towards right (3) is either directed towards left or towards right (4) is neither directed towards left nor towards right 29.

In the given circuit the power generated in 5  resistance will be maximum for ' x ' equal to:

(1) ckW;h vksj (2) nkW;h vksj (3) ckW;h ;k nkW;h vksj es ls dksbZ ,d (4) ckW;h ;k nkW;h vksj es ls dksbZ ughaA 29.

fn;s x;s ifjiFk esa 5 ds izfrjks/k esa mRiUu gksus okyh 'kfDr vf/kdre gksxh tc 'x' dk eku gksxk

(1) 1 

(1) 1  (2) 7  (3) 2/3  (4) 0  30.

rhu ,d leku cM+h pkyd IysVsa ftudk {ks=kQy A

(2) 7  (3) 2/3  (4) 0 

A beam of light from a distant axial point source is incident on the plane surface of a thin planoconvex lens ; a real image is formed at a distance of 40 cm. Now if the curved surface is silvered, the real image is formed at a distance of 7.5 cm. The radius of curvature of the curved surface of the lens and the refractive index of the material of the lens respectively, are : (1) 40 cm, 1.5 (2) 24 cm, 1.6 (3) 20 cm, 1.6 (4) 7.5 cm, 1.5

30.

,d leryksÙky ySal ds lery lrg ij nwjLFk v{kh; fcUnq L=kksr ls izdk'k dh iqat fxjrh gS] vkSj ,d okLrfod izfrfcEc 40 cm nwjh ij curk gSA vc ;fn oØ lrg ij flYoj dj fn;k tk;s] rc okLrfod izfrfcEc 7.5 cm nwjh ij curk gSaA oØlrg dh oØrk f=kT;k vkSj ysal ds inkFkZ dk viorZukad gksxkA (1) 40 cm, 1.5 (2) 24 cm, 1.6 (3) 20 cm, 1.6 (4) 7.5 cm, 1.5

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

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PSET101JRMMT2301218-12

PART – B

Hkkx– B

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

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

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

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,

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

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

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

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

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

Ba = 137,

Hg = 200, Pb = 207]

Hg = 200, Pb = 207] SECTION - I

[k.M - I

Straight Objective Type

lh/ks oLrqfu"B izdkj

This section contains 30 multiple choice questions.

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

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

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

its answer, out of which ONLY ONE is correct.

lgh gSA

31.

31.

The correct statement regarding defects in

Bksl esa =kqfV;ksa ds lEcU/k esa lgh dFku gS &

solid is : (1) Schottky defect is usually favoured by

(1)

large difference in the sizes of cation and

/kuk;u o _.kk;u ds vkdkj esa cM+k vUrj lkekU;r% 'kkWV~dh =kqfV esa lgk;d gksrk gSA

anion. (2)

Compounds

having

F-centres

are

diamagnetic.

(2) F-dsUnz

j[kus okys ;kSfxd izfrpqEcdh; gksrs gSaA

(3) 'kkWV~dh

=kqfV Bksl dk ?kuRo c<+krh gSA

(3) Schottky defect increases the density of (4)

solid. (4) Due to Frenkel defect density of solid

Ýsady =kqfV ds dkj.k Bksl dk ?kuRo leku jgrk gSA

remains same.

Space for Rough Work / (dPps

dk;Z ds fy, LFkku )

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CSET101JRMMT2301218-13

32.

If urea is added in the sugar solution then which

of

the

following

statements

32.

are

rks fuEu esa ls dkSuls dFku lgh gSa\

correct? (a) Vapour pressure of solution decreases (b) Relative lowering in vapour pressure increases (c) Freezing point decreases (d) Depression in freezing point increases

(b) ok"inkc

esa vkisf{kd voueu c<+rk gSA

(c) fgekad

?kVrk gSA

(d) fgekad

esa voueu c<+rk gSA

DoFkukad esa mUu;u rFkk foy;u dk DoFkukad

(f) Osmotic pressure decreases

(f) ijklj.k

(1) a, b, c, d, e, f

(1) a, b, c, d, e, f

(2) a, b, c, f

(2) a, b, c, f

(3) a, b, c, d

(3) a, b, c, d

(4) a, b, c, d, e

(4) a, b, c, d, e

In a compound XY, ionic radii of X+ and

33.

nkc ?kVrk gSA

,d ;kSfxd XY esa] X+ rFkk Y– dh vk;fud f=kT;k

Y– are 90 pm and 200 pm respectively then

Øe'k% 90 pm rFkk 200 pm gSa] rc X+ dh leUo;

what is coordination number of X+ ?

la[;k D;k gS ?

(1) 4

(1) 4

(2) 6

(2) 6

(3) 2

(3) 2

(4) 8

(4) 8

Calculate

the

temperature the

temperature, given

above

reaction

this

34.

become

C(s) + H2O(g)  CO(g) + H2(g) Hº = +131.3 KJ/mole ;

ml rkieku dh x.kuk dhft,] ftl rkieku ds Åij fn x;h vfHkfØ;k LOkr% gksxhA

spontaneous 

dk ok"inkc ?kVrk gSA

c<+rk gSA

point of solution increases.

34.

(a) foy;u

(e)

(e) Elevation in boiling point as well as boiling

33.

;fn ;wfj;k dks 'kdZjk foy;u esa feyk;k tkrk gS]

C(s) + H2O(g)  CO(g) + H2(g) 

Hº = +131.3 KJ/mole ; Sº = +0.1336 KJ/mole K

Sº = +0.1336 KJ/mole K

(1) 98.8 K

(1) 98.8 K

(2) 709.8ºC

(2) 709.8ºC

(3) 491.4 K

(3) 491.4 K

(4) 354.9ºC

(4) 354.9ºC Space for Rough Work / (dPps

dk;Z ds fy, LFkku )

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CSET101JRMMT2301218-14

35.

36.

When borax is dissolved in water, then :

35.

tc cksjsDl dks ty esa ?kksyk tkrk gS] rc&

(1) B(OH)3 is formed only

(1) dsoy B(OH)3 curk

(2) [B(OH)4]– is formed only

(2) dsoy [B(OH)4]– curk

(3) both B(OH)3 and [B(OH)4]– are formed

(3) nksuksa B(OH)3 rFkk [B(OH)4]– curs

(4) [B3O3(OH)4]– is formed only

(4) dsoy [B3O3(OH)4]– curk

A 1.5 m solution of acetic acid () is mixed

36.

with 3 m solution of the acetic acid () to prepare 2 m solution (m is molality of

2 m

gSA gSA gSaA

gSA

foy;u (m foy;u dh eksyyrk gS) cukus ds

fy,] ,lhfVd vEy ()ds 1.5 m foy;u dks ,lhfVd vEy () ds 3 m foy;u ds lkFk fefJr djrs gSA

solution). Select the correct statement(s)

1   (I) Mass ratio of solvents mixed   is 2   

lgh dFku@dFkuksa dk p;u dhft, & 

fefJr foyk;dksa   dk nzO;eku vuqikr   

(I)

2   (II) Mass ratio of solvents mixed   is 1    (III) Mass ratio of solutions mixed

     is  

1 gSA 2 (II) fefJr

(IV) Mass ratio of solutions mixed

     is  

1



109 59 (IV)

59 109

  

fefJr foy;uksa   dk nzO;eku vuqikr   

(III)

109 59

 2 foyk;dks   dk nzO;eku vuqikr gSA

gSA

 fefJr foy;uksa   dk nzO;eku vuqikr  

59 109



gSA

(1) I, IV (1) I, IV

(2) II, III

(2) II, III (3) I, III

(3) I, III

(4) II, IV

(4) II, IV

Space for Rough Work / (dPps

dk;Z ds fy, LFkku )

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CSET101JRMMT2301218-15

37.

37.

The value of Kp for the reaction

CO2 (g) + C(s)

ij vfHkfØ;k

CO2 (g) + C(s)

2 CO(g)

2 CO(g) ds

fy, Kp dk

is 3.0 at 1000 K. If initially PCO2 = 0.48 bar,

eku 3.0 gSA ;fn izkjEHk es PCO2 = 0.48 ckj,

PCO = 0 bar and pure graphite is present

PCO = 0 ckj rFkk 'kq) xzsQkbV mifLFkr gS] rc CO

then determine equilibrium partial pressure

rFkk CO2 dk lkE; vkaf'kd nkc fu/kkZfjr dhft,A

of CO and CO2.

(1) PCO = 0.15 ckj, PCO2 = 0.66 ckj

(1) PCO = 0.15 bar, PCO2 = 0.66 bar

(2) PCO = 0.66 ckj, PCO2 = 0.15 ckj

(2) PCO = 0.66 bar, PCO2 = 0.15 bar

(3) PCO = 0.33 ckj, PCO2 = 0.66 ckj

(3) PCO = 0.33 bar, PCO2 = 0.66 bar

(4) PCO = 0.66 ckj, PCO2 = 0.33 ckj

(4) PCO = 0.66 bar, PCO2 = 0.33 bar 38.

1000 K

When freshly precipitated Fe(OH)3 is shaken

38.

with small amount of aqueous solution of

tc rktk vo{ksfir Fe(OH)3 dks FeCl3 ds tyh; foy;u dh lw{e ek=kk ds lkFk fefJr fd;k tkrk

FeCl3 a colloidal solution is obtained, this is

gS] rks dksykWbMh foy;u izkIr gksrk gS] ;g fuEu dk

an example of : (1) Protective action

mnkgj.k gS&

(2) Dissolution

(1) j{kh

fØ;k

(2) ?kqyu'khyrk

(3) Peptization

(3) isIVhdj.k

(4) Dialysis

(4) viksgu

39.

Which

of

the

following

statement

is

39.

INCORRECT ?

fuEu esa ls dkSulk dFku xyr gS \ (1) C60 esa

(1) C60 contains twelve-six membered rings and twenty -five membered rings

ckjg&N% lnL;h; oy; rFkk chl&ik¡p

lnL;h; oy; gksrh gSA

(2) Fullerenes are cage-like molecules

(2) Qqyfju

fiatjs ds leku lajpuk ;qDr v.kq gSaA

(3) Graphite is thermodynamically most

(3) xzsQkbV

Å"ekxfrd :i ls dkcZu dk lokZf/kd

stable allotrope of carbon (4) Graphite is slippery and therefore is used as dry lubricant in machines

LFkk;h vij:i gSA (4) xzsQkbV

fpduk gksrk gS rFkk blfy, e'khuksa es

'kq"d Lusgd ds :i es iz;qDr gksrk gSA Space for Rough Work / (dPps

dk;Z ds fy, LFkku )

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CSET101JRMMT2301218-16

40.

41.

40.

How many of the following are planar ?

fuEu esa ls fdrus leryh; gSa\

XeF2, ClF3, H2O, [XeF5]–,  3– , BCl3, XeF4, SF4,

XeF2, ClF3, H2O, [XeF5]–,  3– , BCl3, XeF4, SF4,

PCl5, SF6, IF7.

PCl5, SF6, IF7.

(1) 7

(1) 7

(2) 5

(2) 5

(3) 6

(3) 6

(4) 10

(4) 10

Amongst the following ions which one has the

41.

highest magnetic moment value ? (1)

[Cr(H2O)6]3+

(2)

[Fe(H2O)6]2+

fuEu vk;uksa esa ls dkSulk ,d vk;u mPpre pqEcdh; vk?kw.kZ eku j[krk gS \ (1) [Cr(H2O)6]3+ (2) [Fe(H2O)6]2+

(3) [Zn(H2O)6]2+

(3) [Zn(H2O)6]2+

(4) [Cu(NH3)4]2+

42.

(4) [Cu(NH3)4]2+

Find the solubility of As2S3 in a 10–2 M Na2S solution

(assuming

no

hydrolysis

42.

or

complexation of cationic or anionic part).

1 Given : Ksp for As2S3 = × 10–24. 625 (1) 1 × 10–11 mol/L (2) 2 ×

10–11

mol/L

(3) 2 ×

10–10

mol/L

(4) 2 ×

10–13

,d 10–2 M Na2S foy;u esa As2S3 dh foys;rk Kkr dhft;s ¼ekuk fd] /kuk;fud vFkok _.kk;fud Hkkx dk fdlh Hkh izdkj dk ty vi?kVu vFkok ladqyu ugha gksrk gS½A ¼fn;k x;k gS : Ksp As2S3 =

1 × 10–24) 625

(1) 1 × 10–11 mol/L (2) 2 × 10–11 mol/L (3) 2 × 10–10 mol/L

mol/L

(4) 2 × 10–13 mol/L

Space for Rough Work / (dPps

dk;Z ds fy, LFkku )

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CSET101JRMMT2301218-17

43.

5    Cv,m  2 R   

A gas

behaving ideally is

43.

and

:)ks"eh; :i ls 1 yhVj ls 32 yhVj rd izlkfjr

adiabatically from 1 litre to 32 litre. Its initial

dh tkrh gSA bldk izkjfEHkd rki 327°C gSA izØe

temperature is 327°C. The molar enthalpy

ds fy, eksyj ,UFkSYih ifjorZu gS&

change for the process is :

(1) –1125 R

allowed

to

expand

reversibly

(1) –1125 R

(2) –625 R

(2) –625 R

(3) –1575 R

(3) –1575 R

(4) buesa

(4) None of these 44.

5    Cv,m  2 R  mRØe.kh; rFkk  

,d vkn'kZ xSl

For 1st order reaction graph is :

44.

1st

ls dksbZ ugha

dksfV vfHkfØ;k ds fy, xzkQ gS ¼tgk¡ ladsr

muds lkekU; vFkZ j[krs gS½& n Ct

45º n Ct

45º

Time (min)

le; (min)

then t1/2 of reaction is (where symbols have

rc vfHkfØ;k dk t1/2 gS&

their usual meaning):

(1) 1 min (1) 1 min (2) 0.693 min (2) 0.693 min (3) 0.231 min

(3) 0.231 min

(4) buesa

(4) None of these 45.

XeF6 + H2O  A + B Compound A & B are

45.

ls dksbZ ugha

XeF6 + H2O  A + B

respectively ;

gSa&

(1) XeO4, HF

(1) XeO4, HF

(2) XeO3, F2

(2) XeO3, F2

(3) XeF2, Xe

(3) XeF2, Xe

(4) XeO3, HF

(4) XeO3, HF Space for Rough Work / (dPps

;kSfxd A rFkk B Øe'k%

dk;Z ds fy, LFkku )

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CSET101JRMMT2301218-18

46.

In the given process A and B are

46.

respectively: CaC2 + A

fn;s x;s izØe esa A rFkk B Øe'k% gSa& CaC2 + A

  

B

H2O  

H2O   CaCO3 + NH3   B 

CaCO3 + NH3

(moZjd)

(Fertilizer)

(1) NH3, CaCN2+C

(1) NH3, CaCN2+C

(2) N2, Ca3N2+C

(2) N2, Ca3N2+C

(3) N2O, CaCN2+C

(3) N2O, CaCN2+C

(4) N2, CaCN2+C

(4) N2, CaCN2+C

47.

What is the IUPAC name of Hg[Co(SCN)4] ?

47.

Hg[Co(SCN)4] dk IUPAC uke

D;k gS ?

(1) Mercury(II) tetrathiocyanato-S-cobaltate(II)

(1) edZjh(II) VsVªkFkk;kslk;usVks-S-dksckYVsV (II)

(2) Mercury(I) tetrathiocyanato-S-cobaltate(II)

(2) edZjh(I) VsVªkFkk;kslk;usVks-S-dksckYVsV (II)

(3) Mercury(II) tetrathiocyanato-N-cobaltate(II)

(3) edZjh(II) VsVªkFkk;kslk;usVks-N-dksckYVsV (II)

(4) Mercury(II) tetrathiocyanato-S-cobalt(II) (4) edZjh(II) VsVªkFkk;kslk;usVks-S-dksckYV (II)

48.

In the electrolysis of AgNO3 (aq) solution pH

48.

of solution :

AgNO3 (aq)

foy;u ds oS|qrvi?kVu esa foy;u

dh pH&

(1) Increases (1) c<+rh

gSA

(2) ?kVrh

gSA

(3) leku

jgrh gSA

(4) buesa

ls dksbZ ugha

(2) Decreases (3) remain same (4) None of these

Space for Rough Work / (dPps

dk;Z ds fy, LFkku )

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CSET101JRMMT2301218-19

49.

Which combination will be best to prepare

49.

anisole ?

,uhlkWy cukus ds fy, dkSulk la;kstu lcls mi;qDr gksxk\ ONa

ONa

+ CH3 – F

(1)

+ CH3 – F

(1)

Br

Br

(2)

(2)

+ CH3 ONa

+ CH3 ONa ONa

ONa

(3)

(3)

+ CH3O SO2 OCH3

+ CH3O SO2 OCH3 I

I

50.

Condensation

+ CH3 – ONa

(4)

+ CH3 – ONa

(4)

polymerisation

among

the

50.

fuEu cgqydksa esa ls dkSulk la?kuu cgqyd gS\

following polymer is :

51.

(1) Teflon

(2) Polystyrene

(1) Vs¶yku

(2) ikWyhLVhfju

(3) PVC

(4) Dacron

(3) PVC

(4) MsØksu

Which of the following species/compounds

51.

are stable and polar – CH2

(I)

Fulvene O

Triafulvene

(III)

CH2

(II)

(IV)

fuEu esa ls dkSulh Lih'kht@;kSfxd LFkk;h rFkk /kqozh; gSa& CH2

(I)

CH2

(II)

Qqyohu

VªkbZQqyohu

O (III)

(IV) ,tqyhu

Azulene

(V) (V)

(1) I,IV (3) II,V

(2) III,IV (4) I,II,III Space for Rough Work / (dPps

(1) I,IV

(2) III,IV

(3) II,V

(4) I,II,III

dk;Z ds fy, LFkku )

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CSET101JRMMT2301218-20

52.

Select the incorrect order.

xyr Øe dk p;u dhft,A

52.

COOH

COOH

COOH

COOH

CH3 (1)

>

(Acidic strength)

(1)

CH3 (vEyh;

>

lkeF;Z)

OH

(2)

> H3C

OH

(2)

Cl

>

H3C

Cl

(Acidic strength) (3)

>

(3)

(vEyh;

lkeF;Z)

({kkjh;

lkeF;Z)

>

(Basic strength)

(4)

53.

>

What

is

(Basic strength)

true

about

the

following

(4)

CH2OH

CH2OH H HO

H

H OH

H

H

OH

O

H H O

CH2OH

HO

H OH

H OH

O

H H O

H H

H

OH

H

H

OH

(1) ;g

eksukslsdsjkbM gSA



(2) ;g

,d vipk;d 'kdZjk gSA



(3)



(2) It is a reducing sugar.



(3) It shows the phenomenon of inversion of

CH2OH

OH

H H

(1) It is monosaccharide .

;g vEyh; ek/;e esa 'kdZjk dk izfriu iznf'kZr djrk gSA

sugar in acidic medium. (4) It shows mutarotation.

O

OH





lkeF;Z)

fuEu dkcksgZ kbMªsV ds fy;s D;k lR; gS \

53.

carbohydrate:

O

({kkjh;

>



Space for Rough Work / (dPps

(4) ;g

E;wVkjksVs'ku iznf'kZr djrk gSA

dk;Z ds fy, LFkku )

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CSET101JRMMT2301218-21

54.

Which of the following is narcotic analgesic?

54.

fuEu esa ls dkSulk eknd ihMkgkjh (narcotic analgesic) gS

(1) Aspirin

\

(1) ,Lizhu

(2) Paracetamol

(2) isjklsVkeksy (3) Morphine

(3) ekWfQZu

(4) Analgin

(4) ,ufYtu O

O 55.

CH3—C—OH

(1) PCl5 (2) H2/Pd/BaSO4

O CH3—C—OH

(1) PCl5

dil.NaOH P  (Major)

(2) H2/Pd/BaSO4

(1) PCl5 ruq NaOH CH P Q 3—C—OH (2) H2/Pd/BaSO4  (eq[;) (Major) (1) PCl5 ruq NaOH CH3—C—OH P Q (2) H2/Pd/BaSO4 Q  (eq[;) (eq[;) (Major) eq[; mRikn Q gS &

dil.NaOH 55. P O (Major)

Major product Q is :

(1) CH3–CH=CH–CH2–OH

(1) CH3–CH=CH–CH2–OH (2) CH3–CH=CH–CH=O

(2) CH3–CH=CH–CH=O

(3) CH3–COOH

(3) CH3–COOH

(4) CH2=CH–CH=O

(4) CH2=CH–CH=O

56.

(i) CO + NaOH, 

   2     (P > Q) % yield 

(i) CO + NaOH, 

   2     (P > Q) % 

56.

(ii) H

(ii) H

Select the correct option :

lgh fodYi dk p;u dhft;sA

(1) Boiling point , (P > Q)

(1)

(2) Melting point , (Q > P)

(2) xyukad , (Q > P)

(3) Water solubility , (P > Q)

(3) ty

(4) Acid Strength , (Q > P)

(4) vEyh;

Space for Rough Work / (dPps

yfC/k gS

DoFkukad , (P > Q)

esa foys;rk , (P > Q) lkeF;Z , (Q > P)

dk;Z ds fy, LFkku )

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CSET101JRMMT2301218-22

(eq[

57.

The three compounds x, y and z can be

57.

distinguished (from each other) by using, CH3– C  C – H

ls½ foHksfnr fd;k tk ldrk gSA

CH3 – CH = O

(x)

C6HCH NO 5 –– C 2 C – H

CH3 – CH = O

3

(y)

(x)

(y)

C6H5 – NO2

C6H5 – NO2

(z)

(z)

58.

rhu ;kSfxd x, y rFkk z dks fdlds }kjk ¼,d nwljs

(1) Tollen's reagent (AgNO3 + NH4OH)

(1) VkWysu

(2) Fehling solution (Cu+2 / OH–)

(2) Qsgfyax

(3) Zn / NH4Cl

(3) Zn / NH4Cl

(4) Cu2Cl2 / NH4OH

(4) Cu2Cl2 / NH4OH

In

the

Newman

projection

for

2,

2-

58.

vfHkdeZd (AgNO3 + NH4OH) foy;u (Cu+2 / OH–)

2, 2-MkbDyksjksC;wVsu

ds U;weku iz{ksi.k ds fy,

Dichlorobutane

dkSulk dFku lgh gS\

Which one statement is correct?

(1) ;fn ‘X’, H gS

rc ‘Y’, CH3 gksxk

(2) ;fn ‘X’, H gS

rc ‘Y’, Cl gksxk

(3) ;fn ‘X’, H gS

rc ‘Y’, C2H5 gksxk

(4) ;fn ‘X’, Cl gS

rc ‘Y’, CH3 gksxk

(1) If ‘X’ is H then ‘Y’ will be CH3 (2) If ‘X’ is H then ‘Y’ will be Cl (3) If ‘X’ is H then ‘Y’ will be C2H5 (4) If ‘X’ is Cl then ‘Y’ will be CH3

Space for Rough Work / (dPps

dk;Z ds fy, LFkku )

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CSET101JRMMT2301218-23

C

59.

60.

After the reaction (if any) pH of solution will

59.

vfHkfØ;k ds ckn ¼;fn laHko gks½ foy;u dh pH esa

become :

ifjorZu gksxk \

(i)

(i)

(ii)

(ii)

(1) Acidic in I and no change in II

(1) I esa

vEyh; vkSj II esa vifjorZuh;

(2) Acidic in II and no change in I

(2) II esa

vEyh; vkSj I esa vifjorZuh;

(3) Basic in I and acidic in II

(3) I esa

{kkjh; vkSj II esa vEyh;

(4) Basic in II and acidic in I

(4) II esa

{kkjh; vkSj I esa vEyh;

In which of the following option, second

60.

fuEu esa ls fdl fodYi esa] f}rh; ;kSfxd izFke

compound gives a precipitate more rapidly

;kSfxd dh rqyuk esa vf/kd 'kh?kzrk ls vo{ksi nsrk gS

then the first compound when reacted with

tc budh vfHkfØ;k ,FksukWy dh mifLFkfr esa

AgNO3 in ethanol. (1)

and

(2)

(3)

and

AgNO3 ds

lkFk djrs gSa %

(1)

rFkk rFkk

(2)

and

(3)

(4) Ph–CH2–Br and CH2=CH–CH2–Br

Space for Rough Work / (dPps

rFkk

(4) Ph–CH2–Br rFkk CH2=CH–CH2–Br

dk;Z ds fy, LFkku )

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CSET101JRMMT2301218-24

PART – C

Hkkx – 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.

lh/ks oLrqfu"B izdkj bl [k.M esa 30 cgq&fodYih iz'u gSaA izR;sd iz'u ds

61.

61.

Number

of

solutions

of

equation

4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA [0, 4] esa lehdj.k 6sinx + sin2x – 4sin3x = 5sec2x

6sin x + sin2 x – 4sin3 x = 5sec2 x, in [0, 4] is

ds gyksa dh la[;k gS -

(1) 8

(1) 8

(2) 10

(2) 10

(3) 12

(3) 12

(4) 0

(4) 0 

62.

The number of positive integral values of ‘x’ 

satisfying the equation

 o

63.

62.

lehdj.k

o

 n tdt   n 2 is.  x2  t 2 4

okys x ds /kukRed iw.kkZad ekuksa dh la[;k gS -

(1) 1

(1) 1

(2) 2

(2) 2

(3) 0

(3) 0

(4) 3

(4) 3

If H1,H2,H3…H100 are 100 harmonic means

 n tdt   n 2 dks larq"B djus  2  t2 4

x

63.

;fn H1, H2, H3…H100 'a' vkSj'b' ds e/; 100 gjkRed ek/; gS (a > 0, b > 0) rc

between ‘a’ and ‘b’ (a > 0, b > 0) then value of

H1  a H100  b  H1  a H100  b

is

H1  a H100  b  H1  a H100  b dk eku gS

(1) 100

(1) 100

(2) 50

(2) 50

(3) 200

(3) 200

(4) 25

(4) 25

-

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

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MSET101JRMM2301218-25

64.

A man in a boat rowed away from a cliff

64.

150m high takes 2 minutes to change the

pksVh ds 'kh"kZ dk mUu;u dks.k] igkM+h ls nwj tkus

angle of elevation of the top of the cliff from

ij 2 feuV esa 60o ls 45o rd cnyrk gSA rc uko

60o to 45o. The speed of boat is (in Km/h).

dh pky (fdeh / ?k.Vk esa gSA)

(1)

9 3 3 2

(1)

93 3 2

(2)

93 3 2

(2)

93 3 2

(3)

9 3 2

(3)

9 3 2

(4) 9

(4) 9

65.

Value of lim x 0

66.

,d uko esa cSBs O;fDr ds 150 ehVj igkM+h dh

x 600   sin x  x 2  sin x 

600

600

is

65.

lim x 0

x 600   sin x  x 2  sin x 

(1) 600

(1) 600

(2) 100

(2) 100

(3) 0

(3) 0

(4) does not exists

(4) fo|eku gS A

Let P is the midpoint of side AD of rectangle ABCD. If and AC is: (1) 60o

AD  AB

2 then angle between BP

66.

600

600

dk eku gS -

ekuk P vk;r ABCD dh Hkqtk AD dk e/; fcUnq gS ;fn

AD  AB

2 rc BP rFkk AC ds e/; dks .k

gS (1) 60o

(2) 45o

(2) 45o

(3) 30o

(3) 30o

(4) 90o

(4) 90o

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

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MSET101JRMM2301218-26

67.

Let orthocenter of the  formed by the lines

3 x + y = 3, y = a (a, b) then = b (1)

(2) (3)

(4)

68.

67.

ekuk js[kkvksa

3 4 3 3 4 4

(2)

(3)

(4)

If p(x) is a polynomial of degree 4 having

68.

3 3 4 4

3 3 2 3

;fn x = 1, 2 ij pje ekuksa dk pkj ?kkr dk cgqin p(x) gS rFkk lim  1  x 0

then p(2) =

(2) 1

(2) 1

(3) 2

(3) 2

(4) –1

(4) –1



 Let a  2iˆ  ˆj  2kˆ and b  iˆ  ˆj if vector      c is such that a.c | c |,| c  a | 2 2 and    angle between ( a  b ) and c is the 30o then    | (a  b )  c | is equal to

(2)

3 2

(3) 2 (4) 3



p( x )    2 rc p(2) gS x2 

(1) 0

(1) 0

2 3

a = b

3 4

(1)

3 3 2 3

(1)

3 x – 3 vkSj

y = 2 ls cus f=kHkqt dk yEc dsUnz (a, b) ij gS rc

p( x) extrema’s at x = 1, 2 and lim 1  2   2 x 0 x  

69.

3 x + y = 3, y =

3 x – 3 and y = 2 is at

69.

  ekuk a  2iˆ  ˆj  2kˆ vkSj b  iˆ  ˆj ;fn lfn'k      c bl izdkj gS fd a.c | c |,| c  a | 2 2 rFkk    ( a  b ) vkSj c ds e/; dks .k 30o gS rc    | (a  b )  c | dk eku cjkcj gS -

2 3 3 (2) 2 (1)

(3) 2 (4) 3

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

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MSET101JRMM2301218-27

70.

Let

   sin   x  if  x  is odd   4  f ( x)   cos   1  x   x   if  x  is even    4 

70.

tgk¡ [x] egÙke iw.kkZad Qyu gS rFkk {x} fHkUukRed Hkkx Qyu gS -

where [x] is G.I.F and {x} is FPF then

71.

(1) f(x) is continuous at x = 1, 2, 3

(1) f(x), x = 1, 2, 3 ij lrr~ gS A

(2) f(x) = is continuous at x = 4

(2) f(x), x = 4 ij lrr~ gS A

(3) f(x) is differentiable,  x  R

(3) f(x), lHkh x R ds fy, vodyuh; gS A

(4) f’(x) does not exists for all integer points

(4) f ’(x), lHkh iw.kkZad fcUnqvksa ds fy, fo|eku ugh gSA

Let A, B, C are angles of an acute angled triangle such that tanA, tanB, tanC are in HP

71.

then minimum. Value of cotB can be (1) 1

ekuk A, B, C U;wudks.k f=kHkqt ds dks.k gS tcfd tanA, tanB, tanC gjkRed Js .kh esa gS rc cotB dk U;wure eku gks ldrk gS (1) 1

(2)

3

(2)

3

(3)

1 3

(3)

1 3

(4) 2 72.

ekuk

   sin  x ;fn  x  fo"ke gSA   4  f ( x)   cos   1  x   x   ;fn  x  le gSA    4 

(4) 2

If standard deviation of ‘n’ observations

x1, x2 , x3..........xn is 4 and another set of ‘n’ observations y1, y2, y3 ……….. yn is 3 then standard deviation of x1 – y1, x2 – y2,

72.

;fn x1, x2 , x3..........xn n izs{k.kksa dk ekud fopyu 4 gS rFkk vU; ‘n’ isz{k.kksa y1, y2, y3 …… yn dk

ekud fopyu 3 gS rc x1 – y1, x2 – y2,

x3 – y3……….xn – yn is

x3 – y3……….xn – yn dk ekud fopyu gS -

(1) 1

(1) 1

(2) 2

(2) 2

(3)

3

(4) Data insufficient

(3)

3

(4) vkdMsa vi;kZIr gS A

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

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MSET101JRMM2301218-28

73.

Let P, Q, R be invertible matrices of 3rd order such that A=2PQ -1 , B=3QR -1 , C=

74.

75.

73.

1 RP -1 2

ekuk P, Q, R rhu Øe ds izfrykseh; vkO;wg bl izdkj gS fd A=2PQ -1 , B=3QR -1 , C= 1 RP -1 2

then value of det (ABC + BCA + CAB) =

rc det(ABC + BCA + CAB) dk eku gS -

(1) 9

(1) 9

(2) 27

(2) 27

(3) 81

(3) 81

(4) 729

(4) 729

Let

  x  13  2 2  x  1  3 f  x    x  1 1  x  1 then  2 1 x  2    x  1 

74.

ekuk

  x  1 3  2 2  x  1  3 f  x    x  1 1  x  1  2 1 x  2   x  1 

rc

number of points where function f(x) attains a

fcUnqvksa dh la[;k] tgk¡ Qyu f(x) LFkkuh; mfPPk"B

local maximum or a local minimum is

;k LFkkuh; fufEu"B j[krk gS -

(1) 2

(1) 2

(2) 3

(2) 3

(3) 4

(3) 4

(4) 5

(4) 5

Let f(x) is a cubic polynomial which has local

75.

ekuk f(x) ,d ?kuh; cgqin gS tks x = – 1 ij

maximum at x = – 1 and f ’(x) has a local

LFkkuh; mfPp"B j[krk gSA rFkk f (x), x = 1 ij

minimum at x = 1. If f(– 1) = 15 and

LFkkuh; fufEu"B j[krk gSA ;fn f(– 1) = 15 vkSj

f(3) = – 22 then the distance between its two

f(3) = – 22 rc blds nks {kSfrt Li'kZ js[kkvksa ds

horizontal tangents is

e/; nwjh gS -

(1) 37

(1) 37

(2) 32

(2) 32

(3) 47

(3) 47

(4) 42

(4) 42

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

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MSET101JRMM2301218-29

76.

 ,  

Let

and

 ,  

are two points on

76.

the parabola y2 =8x such that the normal’s at them meet in

18,12 

fcUnq gS bl izdkj gS fd mu ij [khps x, vfHkyEc]

then value of

18,12  ij feyrs gS rc

 

(2) 256

(2) 256 (3) 64

(3) 64

(4) 16

(4) 16

,  ,  are

If

dk eku gS

(1) 512

(1) 512

77.

ekuk  ,   vkSj  ,   ijoy; y2 = 8x ij nks

roots

of

equation

77.

x3 + 3x2 + 4x – 11 = 0 and   ,   , 

;fn ,  ,  lehdj.k x3 + 3x2 + 4x – 11 = 0 ds ewy gSA rFkk    ,    ,   lehdj.k

are roots of equation x3 + ax2 + bx + C = 0 x3 + ax2 + bx + C = 0 ds ewy gS rc C dk eku gS

then value of C = (1) 21

(1) 21

(2) 22

(2) 22

(3) 23

(3) 23

(4) 25

(4) 25



78.

If

 

 | sin x | dx  8 and



4

then



4

x cos 3 xdx = 1  sin 4 x





| cos x | dx  9

78.

0

 4

rc

sin 2 xdx 0 sin 4 x  cos 4 x ecos x dx 0 ecos x  e cos x

x 2 cos3 x dx 4  1 1  x

(4) 0

| cos x | dx  9

0

x cos 3 xdx = 1  sin 4 x

sin 2 xdx 0 sin 4 x  cos 4 x



(2)

1

(3)





(1)



(2)



4 



(1)

 

 | sin x | dx  8 rFkk

;fn

ecos x dx 0 ecos x  e cos x 1

(3)

x 2 cos 3 x dx 4  1 1  x

(4) 0

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

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MSET101JRMM2301218-30

79.

Two squares are chosen at random from the small squares on a chessboard.

79.

The

'krajt cksMZ ij NksVs oxksZ ls ;kn`fPNd nks oxZ pqus tkrs gS rc bu pqus x, oxksZ esa Bhd ,d dksuk

probability that the two squares chosen

mHk;fu"B gksus dh izkf;drk gS&

have exactly one corner in common is:

80.

(1)

11 144

(1)

11 144

(2)

1 12

(2)

1 12

(3)

7 144

(3)

7 144

(4)

5 144

(4)

5 144

Let |z| = 2 and w = z +

1 then area of the z

80.

region bounded by the locus formed by

ekuk |z| = 2 vkSj w = z +

1 rc lfEeJ la[;k w z

ls cus fcUnqiFk ls ifjc) {ks=k dk {ks=kQy gS -

complex number w is: (1)

(1)

15 2

(2) 25

(2) 25

(3) 9

(3) 9

(4)

81.

15 2

15 4

(4)

Area of  formed by common tangents of circles

x2

+

y2

– 6x = 0 and

x2

+

y2

+ 2x = 0

81.

15 4

o`Ùkksa x2 + y2 – 6x = 0 vkSj x2 + y2 + 2x = 0 dh Li'kZ js[kkvksa ls cus f=kHkqt dk {ks=kQy gS -

is (1) 2 3

(1)

2 3

(2)

4 3

(2) 4 3

(3)

3 3

(3)

3 3

(4)

3

(4)

3

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

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MSET101JRMM2301218-31

82.

Perpendicular is drawn from origin to the line

82.

x + 2y + 3z + 4 = 0 = 2x + 3y + 4z + 5 then

ewy fcUnq ls js[kk x+2y+3z+4=0=2x+3y+4z+5 ij yEc Mkyk tkrk gS rc yEc ikn ds funsZ'kkad

co-ordinates of foot of perpendicular is

 ,  ,   (1)  

(2)  

(3)   (4)

83.

84.



 ,  ,   gS rc fuEu esa ls dkSulk lgh gS

then which of the following is true.

(1)  

2 3

(2)  

5 3

(3)  

4 3

(4)

=1

The converse of p  (q r) is

83.



2 3 5 3

4 3

=1

p  (q r) dk izfryks e gS -

(1)  q  ~ r   p

(1)  q  ~ r   p

(2)  ~ q  r   p

(2)  ~ q  r   p

(3)  q  ~ r   ~ p

(3)  q  ~ r   ~ p

(4)  q  ~ r   p

(4)  q  ~ r   p

Let [aij] is a square matrix of order 2 where aij  {0,1,2,3,4,6} then number of matrices A -1

such that A exists is (given that all elements of matrix A are distinct)

84.

?

ekuk [aij] 2 Øe dk oxZ vkO;wg bl izdkj gS fd tgk¡ aij {0,1,2,3,4,6} rc vkO;wgks A dh la[;k gksxh tcfd A-1 fo|eku gS -(fn;k x;k gS fd vkO;wg A ds lHkh vo;o fHkUu fHkUu gSA)

(1) 300 (2) 240

(1) 300 (2) 240

(3) 360 (4) 344

(3) 360 (4) 344

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

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MSET101JRMM2301218-32

85.

Let y = f(x) satisfies the differential equation

85.

  dy  2 y tan x  sin x, given that f    0 dx 3

ekuk y = f(x) vodyuh; lehdj.k

  dy  2 y tan x  sin x, f    0 dks la rq"B dx 3

20 

20

 f  x  dx 

then

djrk gS tcfd

0

86.

(1) –20

(2) 20

(2) 20

(3) 15

(3) 15

(4) –15

(4) –15

The coefficient of x50 in the expansion of

86.

(1+x2)25 (1+x25) (1+x40)(1+x45)(1+x50) is:

87.

(1+x2)25 (1+x25) (1+x40)(1+x45)(1+x50) ds foLrkj

esa x50 dk xq.kkad gS -

25

C5

(1) 25C5

(2) 2 + 25C5

(2) 2 + 25C5

(3) 3 + 25C5

(3) 3 + 25C5

(4) 1 + 25C5

(4) 1 + 25C5

Number

x2 4x 2 9 x2 (1) 6 (2) 1

;g fn;k x;k gS

0

(1) –20

(1)

 f  x  dx

of

real 2

roots

of

equation

lehdj.k

x2

2

 x  1  x  2  2 2  2 x  1  2 x  2  2 2  3x  1  3x  2 

87.

 2 x  6 is

4x 2 9 x2

2

2

 x  1  x  2  2 2  2 x  1  2 x  2  2 2  3x  1  3x  2 

 2x  6

ds

okLrfod ewyksa dh la[;k gS (1) 6 (2) 1

(3) 3 (4) 4

(3) 3 (4) 4

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

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MSET101JRMM2301218-33

88.

Let PQRS is a parallelogram such that

 PQ =



ˆi

+ ˆj + kˆ



88.



, PQ  iˆ  ˆj  kˆ then

volume



, PQ 

of

parallelopiped whose adjacent edges are



+ ˆj + kˆ , QS = 2 ˆi + 3 ˆj + 4 kˆ

iˆ  ˆj  kˆ  rc

lekUrj

   PR, QR and PQ  iˆ  ˆj  kˆ is (1) 4

(1) 4

(2) 5

(2) 5

(3) 6

(3) 6

(4) 7

(4) 7

If

y  x  



is a variable tangent



to

89.

2 parabola x  8y then locus of point w is a



ijoy; x2  8y dh pj Li'kZ

(2) 4

(2) 4

(3) 2

(3) 2

1 2

(4)

The sum of greatest and least values of is

90.

1 2

sin8 x  cos8 x dk vf/kdre o U;wure ekuksa dk ;ksxQy gS -

(1)

9 8

(1)

9 8

(2)

5 4

(2)

5 4

(3)

3 2

(3)

3 2

(4) 2

  PR, QR

(1) 1

(1) 1

sin8 x  cos8 x

y  x  

fdukjs

dk

'kkado C ds ukfHkyEc dh yEckbZ gS -

is

(4)

;fn

iV~Qyd

js[kk gS rc fcUnq w dk fcUnqiFk ,d 'kkado C gS rc

conic C then length of latus rectum of conic C

90.

ˆi

vk;ru gksxk ftlds  rFkk PQ  iˆ  ˆj  kˆ gS -



89.

ekuk PQRS ,d lekUrj prqHkqZt bl izdkj gS fd  PQ =

 , QS = 2 ˆi + 3 ˆj + 4 kˆ

(4) 2

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

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MSET101JRMM2301218-34

P02-18

MAIN M AJ OR T EST (MMT ) (JEE MAIN PATT E RN) COURSE : ALL INDIA TEST SERIES (VIKALP) | CLASS - XII/XIII DAT E : 30-12-2018

SET - 1

IMPORT ANT INSTRUCTI ONS / egÙoi. w kZ funZs'k A. General % 1.

2.

3. 4.

A.

Immediately fill the particulars on this page of the Test 1. Booklet with Blue / Black Ball Point Pen. Use of pencil is strictly prohibited. The Answer Sheet is kept inside this Test Booklet. When you 2. are directed to open the Test Booklet, take out the Answer Sheet and fill in the particulars carefully. The Test Booklet consists of 90 questions. The maximum 3. marks are 360. There are three parts in the question paper A, B, C 4. 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.

5.

Candidates will be awarded marks as stated above in 5. 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.

6.

There is only one correct response for each question. Filling 6. up m ore 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. Filling the Top-half of the ORS : B. Use only Black ball point pen only for filling the ORS. Write your Roll no. in the boxes given at the top left corner of your ORS with black ball point pen. Also, darken the 7. 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. 7.

8.

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

9.

If student does not fill his/her roll no. and paper code 9. correctly and properly, then his/her marks will not be displayed and 5 marks will be deducted from the total.

lkekU; : 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 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 bl ijh{kk iqfLrdk esa 90 iz'u gSA vf/kdre vad 360 gSA bl ijh{kk iqfLrdk es rhu Hkkx A, B, C gSA ftlds izR;sd Hkkx esa HkkSf rd 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 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 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 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 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) ORS ij viuk isij dksM fyf[k, rFkk laxr xksyksa dks dkys ck¡y isu ls dkys dhft,A ;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 pawfd isu ls Hkjs x, xksys feVkuk vkSj lq/kkjuk laHko ugha gS blfy, vki lko/kkuh iwoZd vius mÙkj ds xksyksa dks HkjsaA

10. Since it is not possible to erase and correct pen filled bubble, 10. you are advised to be extremely careful while darken the bubble corresponding to your answer. 11. Neither try to erase / rub / scratch the option nor make the 11. fodYi dks u feVk,a@u Ldzsp djsa vkSj u gh xyr (X) fpUg dks HkjsaA Cross (X) mark on the option once filled. Do not scribble, ORS dks u dkVs u gh QkMs u gh xUnk ugha djsa rFkk dksbZ Hkh smudge, cut, tear, or wrinkle the ORS. Do not put any stray fu'kku ;k lQsnh ORS ij ugha yxk,aA marks or whitener anywhere on the ORS. 12. If there is any discrepancy between the written data and the 12. ;fn ORS esa fdlh izdkj dh fy[ks x, vkadMksa rFkk xksys fd, vkadMksa esa bubbled data in your ORS, the bubbled data will be taken as final. fojks/kkHkkl gS] rks xksys fd, vkadMksa dks gh lgh ekuk tkosxkA

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

Roll Number (jksy uEcj) :

I have read all the instructions and shall abide by them

I have verified all the information filled by the candidate.

eSaus lHkh funsZ'kksa dks i<+ fy;k gS vkSj eSa mudk vo'; ikyu d:¡xk@d:¡xhA

ijh{kkFkhZ }kjk Hkjh xbZ lkjh tkudkjh dks eSusa tk¡p fy;k gSA

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

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

ijh{kkFkhZ ds gLrk{kj

ijh{kd ds gLrk{kj

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80034 44888

MAIN MAJOR TEST (MMT) (JEE MAIN PATTERN)

TARGET : JEE (MAIN + ADVANCED) 2019

DATE : 30-12-2018

SET-1

|C CLASS XII/XIII | COURSE : ALL INDIA TEST SERIES (VIKALP)

HINTS & SOLUTIONS PART-A : PHYSICS 1. For the given ……………………… Sol. The output D for the given combination D=

(A  B).C  (A  B)  C

If A = B = C = 0 then D = (0  0)  0

5. In the system ……………………… Sol. 4a = 4g – T 4a = T T = 20 N a=5

 00 50 =

=1+1=1 If A = B = 1, C = 0 then D =

n 2  0.6

20 1 20

(1  1)  0  1  0 n=3

=0+1=1 2.

A particle ………………………

Sol.

A V   A2    3

0.2 =

2

1 2 5t 2 0.08

t= t1 = 0

8A2 9

V = 

Vnew = 2V =

=

4 2 3

2 2 3

t2 = 0.08

t3 = 0.16

A.

0.16 –

0.08

t1  t 2

(A)

0.08

1 2 –1

So the new amplitude is given by 6.

 (A new )2 – x 2

Vnew =

4 2 A A   (A new )2 –   3 3

2

Initially two ………………………

Sol. f =

1 2

T 

nf = –n2 +

2

32 2 A A  (A new )2 – 9 9 Anew2 =

3.

33A 2 9

33A 3

Anew =



nT –

1 T 2 T 6 1 2 = f 2 100

1 2

n

=

 f = 600 Hz.

A disc of mass ………………………

Sol. Torque = I about the point of contact. Kx(R) = (

4.

f f

1 2

MR 2 2

+ MR2) , x = R

A uniform rope ………………………

Sol. v =

7. Two sources ……………………… Sol. Apparent frequency of S1 and S2 heard by observer is f1 =

v f v u

 Beat = f1 – f2 –

and f2 =

v f vu

2uv 2

v  u2

f

xg

dx  dt

10

xg

 t = 2sec.



 0

dx x

t



 g dt

8. Two light waves ……………………… Sol. I E2   

0



I1 I2

=

22 32

=4/ 9

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SOLSET1JRMMT2301218-1

9.

In the circuit ………………………

1  2500 2 fc

8 – 0.5

7.5 Sol. I = = mA = 3.4 mA 3 2.2 2.2  10

10 6 F 2  50  2500

C 10. If the magnetic ……………………… Sol. Let us compute the magnetic field due to any one segment :

40 4   1.27  F 10 

Vmax  i0 Z  2 2  103 10002  25002  2 2  1  6.25  7.62V 13. B=

=

0 i (cos00  cos(180  )) 4 (dsin )

0 i 1  cos   4 (dsin )

=

In an atom ………………………

2R V

Sol. T =

But R  n2

0 i  tan 4 d 2

&V

Resultant field will be : Bnet = 2B =

 0i  tan 2d 2

k =

0 i 2d

1 n

So V 

1 R

So T  R3/2

11. Find current ……………………… Sol.

T1  R    T2  4R  14.

3/2

1 8

=

Two radioactive ………………………

Sol. NA = N0e–5t NB = N0 e–t 15.

A large open ………………………

Sol.

A

dy dt

=

a 2gy

2A  H  H n a 2g  

12. A 50 Hz ac source ……………………… Sol. 1000 W

 2A  H  n  0 a 2g  

= T1

= T2

T1 = T2 n = 4.

VC  5V

16.

Surface tension ………………………

VR  2V IRms

V 2  R  Amp R 1000

Sol. P0 –



= 2mA

IRms  X C  VC 2mA  XC  5 XC 

5000  2500W 2

17.

2S R1

+gh –

2S R2

=P0

2S 2S   gh R1 R2

In determination ………………………

Sol. (A)  =

F    AY (F / A) Y

= slope of curve

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SOLSET1JRMMT2301218-2

 Y

=

21. A plank P is ……………………… Sol. Let velocity of c.m. of sphere be v. The velocity of the plank = 2v.

(4  2)  10 3 4000  103

Given l = 1m  Kinetic energy of plank =

4000  103 2  10 3

Y=

= 2 × 109 N/m 2

1 × m × (2v)2 2

Kinetic energy of cylinder = 18.

Two moles ………………………

Sol.

Vrms 

3RT Mm i x .

=

RT Mmix.

Vsound 

3RT Mm i x.

2

=

1 2 1 1 2 2 mv + ×  mR   2 2 2 

1 1 3 1  mv 2 1   = . mv 2 2 2   2 2



2 Vsound

vrms =

=2mv2

K.E. of plank K.E. of sphere

=

2mv 2 3 mv 2 4

=

8 . 3

22. A particle of ……………………… Sol. Befor collision

RT Mmix

After collision

3 r 2 rmix 

n1CP1  n2 CP2 [By momentum conservation in both direction]

n1Cv1  n2 Cv 2

mv0 = 2mv cos +

7R 5R  n 3 2 2  5R 3R 2 2  n 2 2 2

19.

 30 + 9n = 28 + 10n  n = 2

Two uniform ………………………

T1 2  2 T2 1

mv 0 2 23.

•

Q = 4r2eT4

2

•

=

–ms

Sol. P =

4

 r2     r1 

2

v0

.

2 2

1 mv 2 2

t

= 4000 watt

24.

A block of ………………………

Sol. Resultant force = N

1  2

= mg cos 

1  2

 32

A semi circular ………………………

Sol.

4R  1  3  2 2 Mg  R     MR   3  2  2  

4g  4  1   3R  3 

v =

3

20.



1

= 5.36 HP

d 4 d  – r 3s dt 3 dt

 d   •   dt   1  Q1   •    d   Q2     dt   2

= 2mv.

mgh 

r  T   1   1  4 • Q 2  r2   T2  Q

..........(ii)

A pump is ………………………

•

Q1

– 2mv sin

tan = 45º Now again momentum conservation in y–direction

Sol.  1T1 = 2T2

Rate of heat loss

mv 0 2

..........(i)

By (i) & (ii),

3 14  5n  2 10  3n



0=

mv 0 2

25.

A perfect smooth ………………………

Sol. N cos 30° = g m A sin 30° + m B g sin 30° N = 20

3N

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SOLSET1JRMMT2301218-3

26. At an instant ……………………… Sol.

2 2 10   R 40 75 =

1 2  20 15



2 R

=

=



15  40 20  15

255 3  15

R = 24 cm.

PART-B : CHEMISTRY 31. The correct statement ............ Sol. Facts

vBA = vB – vA =

[4iˆ  3ˆj]  [3iˆ  4ˆj]

=

ˆi  7ˆj

vapp = 4 cos45º + 3 cos45º + 3 cos45º – 4 cos45º = 6 cos45º =

32. If urea is added in ............ Sol. Non volatile solute is added in the solution therefore. V.P., RLVP, Tf , Tf , Tb , Tb,  

3 2 m/s 33.

27.

In a compound XY............

A point charge ………………………

Sol. Flux due to +q is 1 =

q 240

 = 1 + 2 =

34.

5q 240

2

  7  x   

tºC > 982.8 – 273 = 709.8ºC

2

× 5 = 5

1 F

=

1 2  fm f

1 2 2    7.5 R 40

When borax is ............ 2B(OH)3 + 2[B(OH)4]

Therefore molality of resulting solution =

Power will be max when 7 + x is minimum i.e., for x = 0 30. A beam of ……………………… Sol. fm = –R/2 f = 40 m

= 982.8 K

—

36. A 1.5 m solution ............ Sol. Let mass of solvent in 1.5 m solution = m 1 Kg and let mass of solvent in 3 m solution = m 2 Kg.

Power generated in 5W

  7  x  

131 .3 0.1336

Sol. [B4O5(OH)4] 2– + 5H2O

 7x



T>

35.

29. In the given ……………………… Sol. Current in the circuit is given by

=

Calculate the temperature............

Sol. G = H – TS 0 = +131.3 – T(+0.1336) (for spontaneity)

28. Three identical ……………………… Sol. The net charge on middle plate is zero and is placed in uniform electric field. Hence net force on middle plate is zero.

i=

= 0.45

Radius ratio lies between 0.414 and 0.732 and represent octahedral void. Y– is larger than X+ so Y– will form FCC and X+ will occupy voids . So co-ordination no of X+ = 6

Flux due to –q is also in same direction because it is kept below the square. 2 =

90 200

Sol. Radius ratio =

q 60

1.5m1  3m2 = 2. m1  m2

 m1 = 2 m2 1 Therefore required ratio = 109 . 59 37.

The value of Kp for ............

Sol. t =0 0 t = equi m

CO2 (g) + C(s) 0.48

0.48 - x

—

2 CO(g)

— 2x

2

KP =

(2x) (0.48  x)

=3

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SOLSET1JRMMT2301218-4

38. When freshly ............ Sol. Theory based

Hm  nCP T 

39. Which of the following ............ Sol. Refer theory. 40.

44. For 1st order reaction ............ Sol. Ct = C0 e–Kt ln Ct = ln C0 –Kt y = C + mx

How many of the ............

Sol. (i) Planar molecules : XeF2, ClF3, H2O, [XeF5] –,

Slope = –1  K =1 min–1 Half life = 0.693 min

 3– , BCl3, XeF4. (ii) SF4 – See - Saw shape PCl5 – Trigonal bipyramidal SF6 – Square bipyramidal IF7 – Pentagonal bipyramidal

+3

[Ar]

3d

3

[Ar]

4s

3d6

4s

XeF6 + H2 O  A + B............

Sol.

XeF6 + 3H2O  XeO3 + 6HF

46.

In the given process ............ 

H O

47. What is the IUPAC ............ Sol . Mercury(II) tetrathiocyanato-S-cobaltate(II)

4p

48. In the electrolysis of ............ Sol. At anode the reaction is H2O  O2 + 4H+ + 4e

unpaired electrons = 3 (2) [Fe(H2 O)6 ]2+ Fe+2

45.

2 Sol. CaC2+N  CaCN2+C  CaCO3+ NH3 2

41. Amongst the ............ Sol. (1) [Cr(H2 O)6]3+ Cr

7 R 150  600   1575R 2

4p

4d

unpaired electrons = 4

49.

Which combination ............

(3) [Zn(H2 O)6 ]2+

O

OCH3

O

Zn+2(Ar] 3d10

4s

Sol.

4p

unpaired electron = 0

O

SN2 + CH 3–O–S–O–CH3   O

+ CH3–O–S–O O

Anisole

50. Condensation ............ Sol. Dacron is condensation polymer of Glycol and Terphthalic acid.

(4) [Cu(NH3)4 ]2+ 3d9

[Ar]

in presence of  NH3

3d8 xx

4s

4p

xx

xx xx

51. Which of the following ............ Sol. A

unpaired electron = 1 42.

Find the solubility ............ Sol. As2S3 2As3+ + 3S2– 2s 3s + 10–2

B

 s= 2 × 10–11 mol/L.

Sol. Basic-strength :

1

Sol.

T2  V1    T1  V2  or ;k T2 = 150 K



CH2

Aromatic



52.

5    Cv,m  2 R   



CH2 

 (2s)2 (10–2)3 =  1  10 –24   625   

A gas

CH2

C

  10–2 3+ 2 2– 3 Now, [As ] [S ] = Ksp

43.

 

CH2

Select the incorrect ............ NH2

behaving ............

53.

NH2

>

CH3

(due to ortho effect)

What is true about ............

Sol. It shows the phenomenon of inversion of sugar in acidic medium. C12H22O11

C6H12O6 + C6H12O6

Sucrose

Glucose fructose

 = + 66.5º

+52.5º

–92.7º

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SOLSET1JRMMT2301218-5

O

55.

CH3—C—OH

PART-C : MATHEMATICS

(1) PCl5

............

61.

(2) H2/Pd/BaSO4

Sol.

Sol. P is CH3–CH=O

Number of solutions…………..

3sinx  sin 3 x  sin 2 x  5 sec 2 x [-5,5]

[5, ]

Q is CH3 –CH=CH–CH=O 62.

The number of…………..

Sol. t = x tan  /2

56.



(i) CO2 + NaOH, 

        ............ 

o

(ii) H

1 x

OH OH CHO

Sol.

(i) CHCl3 NaOH,  (ii) H

CHO (Q)

(P)

57.

The three compounds ............ 63. Sol. T.R

CH3–CH=O   CH3COO + Ag (silver mirror)

58.

T.R  

   n x   n tan   d  o

If H1,H2,H3…H100 …………..

1 1 1 1 1 , , ,..... , a H1 H 2 Hn b 1 1  d H1 a

No reaction

are in AP

1 1  d Hn b

In the Newm an projection ............

H1  Sol.

n2 4

 Two values = x = 2, 4

T.R Sol. CH3 –CHC–H   CH3–CC–Ag (white ppt.)

C6H5 – NO2

 /2

n x  n 2  x 2 4 n x n 2  x 2

+

 

 n x   n tan  n 2  x sec 2  d   x 2 sec 2  4



,



a 1  ad

Hn 

b 1  bd

H1  a Hn  b  H1  a Hn  b 59.

After the reaction (if any) ............

Sol.

+HBr

(strong acid) +

=

a b a b 1  ad  1  bd a b a b 1  ad 1  bd

=

2  ad 2  bd   ad bd

=

21 1 2      2   n  1 2 d b a

(SN1 product)

No reaction.

Because aryl halide have resonance stabilized C – X bond,

= 2n

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64.

A man in a boat…………..

66.

Let P is the midpoint…………..

Sol.

x =2 

Sol. Slope of AC =

a b

2=

C

(0,a) D (0,a/2) P 150m

x tan 60o =

y

m BP =

150 y

67.

150  150 3

x = 150 - 50 x = 50 (3 -

Sol. A:

 1  ,2   3 

B:

 5  ,2   3 

C:



H:

4   3,  3 

3

3)





25 3  3 60 Velocity =

km/h

1000 =

3 3 3 2





=

a  0 a 2  b 2b

=

1 2

Let orthocenter…………..

x + y = 150 x+

B(b,0)

A

60o

45o

93 3 2

3, 0



(0,3)

A y=2 B

65.

Sol.

Value of …………..

lim x 0

x 600   sin x   sin x  x2    x 

C

600

600

(

x 600

3 ,0)

600

x lim x0

600

 x3    x  ....  6   602 x

 x2    x 600  x 600  600 C0  600 C1    ...    6    lim 602 x 0 x = 100

68. If p(x) is a polynomial………….. Sol. p(x) = 2x2 + ax3 + bx4 p’(x) = 4x + 3ax2 + 4bx3 4 + 3a + 4b = 0 8 + 12a + 32b = 0 2 + 3a + 8b = 0 –2 + 4b = 0

69. Sol.

 a  2iˆ  ˆj  2kˆ …………..       | c  a |2  (c  a ).( c  a )  (2 2) 2

Let

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SOLSET1JRMMT2301218-7

   c 2  a 2  2c.a  8   c 2  2 | c | .  a2  8   c 2  2 | c | 1  0   (c  1)2  0  |c | 1       | (a  b )  c || a  b || c | sin 30o 1    (a  b ) 2

3cot 2B  1 cot B  72.

Sol.

iˆ ˆj kˆ   a  b  2 1 2  2iˆ  2 ˆj  kˆ 1 1

Sol.

Let

 

2

n



 xi  yi  x  y







2





 2  xi  x

 y  y   i





 xi  x yi  y



n

73.

Let P, Q, R be…………..

Sol.

ABC=3I

…………..

BCA=3I CAB=3I

f  2 n  1  0

f

2



n

 9  16  2

f

2

n



 det  ABC  BCA  CAB 

   2n  1  x  2n  sin  4 x    f  x   cos   1  x  2n  x  2n  1    4 

 



 xi  x   yi  y

   sin  x   4  f ( x)   cos   1  x   x      4 

1 2 1 f 2n  2 1 f 2n  2



 xi  x

S .D 

0

f  2n  

71.

If standard deviation…………..



  | (a  b ) | 3    3 | (a  b )  c | 2

70.

1 3

  2 n  1   0

 det  9I   729

74.



Let

  x  1 3  2 2  x  1  3 f  x    x  1 1  x  1 …………..  2 1 x  2   x  1 

Sol.

  2 n  1   1 

Y

Let A, B, C are angles…………..

Sol. cotA cotB + cotB cotC + cotA cotC = 1

-2 2 cotB = cotA + cotC

-1

0

1

X

cotB (2 cotB) + cotA cotC = 1

-1

cotA+cotC  cotAcotC 2 cotB  1-2cot 2 B

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75. Let f(x) is a cubic………….. Sol. As x = 1 is point of inflection of curve so x = –1 and x = 3 and point of local maximum and local minimum hence

(-1,15)

79. Two squares are………….. Sol. n(S) = 64C2 = 63 x 32 n(E) = 7 x 7 x 2 ={number of ways of choosing 2 x 2 square}x 2 p(E) =

80.

Let |z| = 2 and…………..

Sol.

Z  2ei w=

(3,-22) Distance = 37

= 76.

Let

 ,  

and………….. =

Sol. Equation of normal y = mx – 4m- 2m 3 12 = 18 m – 4 m – 2m 3 m3 – 7m + 6 = 0 m = 1, – 3, 2 Point of parabola (2m 2, – 4m)

If









 

       

2





 

 





=

 2    2   2   2   2   2  2

=

                    

 –3  = –3 (4) -  = –3

+  (–3 –

1 2e i

1  i e 2 5 3 cos   i sin  2 2

 

2ei 

 .a .b

5 3 . . 2 2 15 = 4 =

, ,  are roots…………..

Sol. –C =

2e i 

w represent an ellipse then area of region =

 ,   ,  ,     2, 4  & 8, 8  77.

7 7 2 7  63  32 144

81.

Area of  form ed…………..

Sol.

 ,0  

1.3  3.1  ,0   3 1 

)

= –12 – 11 C = 23 

78.

If

 | sin x | dx  8 …………..



Sol.

     4     4 9  17     2   4 17 

I 

x cos 3 xdx I  1  sin 4 x





17 

2I 





    3, 0  CAB = 30o tan 30o = 17 





3

16  x  cos 3 xdx 82.

Perpendicular is…………..

i

3

cos xdx 0 1  sin 4 x 0

 144 2 

1 2 3 3  3 3 2

2

16 . cos xdx cos 3 xdx  16  .9 0 1  sin 4 x 1  sin 4 x 

area =

1  sin 4 x

CB  CB  3 3

Sol. drs of line =

j

k

1 2 3

= –i + 2j – k

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SOLSET1JRMMT2301218-9

Point on the line (1, –1, –1) Equation of line

y . sec2 x =

x 1 y 1 z 1    1 2 1

2

 sin x.sec

x dx

y sec2 x = sec x + C

   1,  2   1,   1

C = –2 sec2 x.y = sec x -2



- 1 + 4  +2+  -1 = 0 20 

6  = -2



20 

  cos x  2 cos x  dx  2  cos 2

0



1 3

=

2

xdx

0



= -2. 20

 cos

2

xdx

0

83.

The converse…………..

 /2

 cos

= -80

Sol. Converse of p  q is q  p

2

xdx   20

0

(q r)  p

q r 

p 86.

=

~ q r  

=

q





~r



p

The coefficient………….. (1+x2 )25 (1+x25 ) …………..

Sol. G.T. of (1+x2 )25 = 25Cr x2r



p Coefficient of x50 in (1+x2 )25 (1+x25 + x40 + x45 + x50) Coefficient of x50 = 25C25 + 25C5 + 1

84.

Let [aij] is a…………..

Sol. for

A-1

exists |A|



= 1 + 25 C5 + 1

0

Case – I : Exactly one element is zero 5C 3

C3  C3 – C1

Case – II : No elem ent is zero

A=

a c

Number of real…………..

Sol. C2  C2 – C1

4 = 240

x

87.

x2

b d

|A| = ad – bc = 0

ad = bc = 6

2x  1

4x

2

9x

2

4x  1

C3  C3 – 2C2

4x  4

x2

8x  4  4x

6 x  1 12 x  4

9x

2x  1 2

2

4x 1 2

2

6x  1 2

8 R1  R1 – R2

ad = bc = 12



5C 4

x

8

4 - 16 = 120 – 16

= 104 Total = 344

R2  R2 – R3

3x2

2 x

0

5x2

2 x

0  2 6 x3 10 x3

9x

2





6x  1 2

= –8x3 = 2x + 6 85.

Let y=f(x) satisfies………….. = 4x3 + x + 3 = 0 one real root

Sol. If =

2 tan xd x e  e 2  n (sec x )

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SOLSET1JRMMT2301218-10

88.

Let PQRS is a…………..

Sol.

  QR  a  b  3i  4 j  5 k  PS  PR  2 a  b  4i  5 j  6 k

S

R b

a b

P

Q

a

i j  PQ   i  j  k   1 1 1 1 Volume =

k 1

=

2i  2 j

1

3

4 5

4

5 6  5 18   6  6  8   90  84  6

2 2 0

89.

Sol.

If y   x   is…………..

y  mx  2m2

y  x   2 m   ,  2m  

  2 2  x2  90.

y 2

The sum of…………..

sin8 x  cos8 x ………….. Sol.

sin8 x  cos8 x  cos2 x  sin2 x  Max value = 1

sin8 x  cos8 x  sin8 x  cos8 x 2 sin 8 x  cos 8 x  2 sin 2 x cos 4 x 

1 sin 2 2 x 8

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SOLSET1JRMMT2301218-11

MAIN MAJOR TEST (MMT) (JEE MAIN PATTERN)

TARGET : JEE (MAIN + ADVANCED) 2019

DATE : 30-12-2018

|C CLASS XII/XIII | COURSE : ALL INDIA TEST SERIES (VIKALP)

ANSWER KEY SET-1 PART-A (PHYSICS) 1.

(4)

2.

(3)

3.

(3)

4.

(4)

5.

(3)

6.

(1)

7.

(3)

8.

(2)

9.

(1)

10.

(2)

11.

(1)

12.

(3)

13.

(3)

14.

(2)

15.

(3)

16.

(4)

17.

(1)

18.

(2)

19.

(2)

20.

(1)

21.

(3)

22.

(2)

23.

(3)

24.

(2)

25.

(4)

26.

(4)

27.

(4)

28.

(4)

29.

(4)

30.

(2)

PART-B (CHEMISTRY) 31.

(4)

32.

(4)

33.

(2)

34.

(2)

35.

(3)

36.

(2)

37.

(2)

38.

(3)

39.

(1)

40.

(1)

41.

(2)

42.

(2)

43.

(3)

44.

(2)

45.

(4)

46.

(4)

47.

(1)

48.

(2)

49.

(3)

50.

(4)

51.

(4)

52.

(4)

53.

(3)

54.

(3)

55.

(2)

56.

(2)

57.

(1)

58.

(3)

59.

(1)

60.

(3)

PART-C (MATHEMATICS) 61.

(4)

62.

(2)

63.

(3)

64.

(1)

65.

(2)

66.

(4)

67.

(2)

68.

(1)

69.

(2)

70.

(2)

71.

(3)

72.

(4)

73.

(4)

74.

(2)

75.

(1)

76.

(1)

77.

(3)

78.

(4)

79.

(3)

80.

(4)

81.

(3)

82.

(3)

83.

(1)

84.

(4)

85.

(1)

86.

(2)

87.

(2)

88.

(3)

89.

(4)

90.

(1)

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SOLSET1JRMMT2301218-12

MAIN MAJOR TEST (MMT) (JEE MAIN PATTERN)

TARGET : JEE (MAIN + ADVANCED) 2019

DATE : 30-12-2018

SET-1

|C CLASS XII/XIII | COURSE : ALL INDIA TEST SERIES (VIKALP)

la d s r ,oa gy PART-A : PHYSICS 1.

fn;s x;s rkfdZd

Sol.

nh xbZ fLFkfr ds fy, fuxZr D=

……………………… D

5.

n'kkZ;s x;s fudk;

Sol.

gksxkA

4a = T T = 20 N

(A  B).C  (A  B)  C

a=5

;fn A = B = C = 0 rc D = (0  0)  0  0  0

50 =

=1+1=1

;fn A = B = 1, C = 0 rc D = (1  1)  0  1  0

n 2  0.6

0.2 =

,d d.k

Sol.

A V   A2    3

A ……………………… t=

8A2 9

V = 

4 2 3 vr% u;k vk;ke gSA Vnew = 2V =

t1 = 0

t1  t 2 6.

33A 2 9

Anew =

Sol.

cy vk?kw.kZ = I lEidZ fcUnq ds ifjr%

4.

MR 2 Kx(R) = ( + MR2) , x = R 2 M=0.1kg vkSj ………………………

………………………

xg

dx  dt

xg

 t = 2sec.



 0

dx x

t



T 

 g dt

1 2

nf = –n2 +



=

1 T 2 T

6 f

=

1 2 2 100

S1

1 2

nT –

n

 f = 600 Hz.

7.

nks L=kksr

Sol.

izs{kd }kjk S1 o S2 dh lquh xbZ vkHkklh vko`fÙk f1 =



8. 10

………………………

2

33A 3

‘m’ nzO ;eku

Sol. v =

0.08

2 –1

1 2

f f

3.

vkSj

0.16 –

1

çkjEHk esa nks

Sol. f =

32 2 A2 A  (A new )2 – 9 9 Anew =

t3 = 0.16

(A)

4 2 A A   (A new )2 –   3 3

2

t2 = 0.08

0.08

A.

 (A new )2 – x 2

Vnew =

1 2 5t 2 0.08

2

2 2 3

=

20 1 20

n=3

=0+1=1 2.

………………………

4a = 4g – T

foLiUn

rFkk ………………………

v f v u

and f2 =

= f1 – f2 –

nks izdk'k rjaxs

v f vu

2uv 2

v  u2

f

E1 = 2 ………………………

Sol. I E2 

0



I1 I2





2

=

2 32

=4/ 9

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SOLSET1JRMMT2301218-1

9.

ifjiFk esa]

Sol. I =

………………………

8 – 0.5 2.2  103 'P' ij

=

7.5 mA = 3.4 mA 2.2

1  2500 2 fc

10.

;fn

Sol.

fdlh ,d [k.M ds dkj.k pq- {ks- dh x.kuk djrs gS&

pqEcdh;

5000  2500W 2

XC 

………………………

10 6 F 2  50  2500

C

40 4   1.27  F 10 

Vmax  i0 Z

 2 2  103 10002  25002  2 2  1  6.25  7.62V  0i (cos00  cos(180   )) 4 (dsin  )

B= =

 i 0i  1  cos   = 40d tan 2 4(dsin  )

13.

11.

dqath

S

cUn

………………………

2R V

Sol. T =

ifj.kkeh {ks=k gksxk Bnet = 2B =

,d ijek.kq esa

But R  n2

 0i  tan 2d 2

 i k = 0 2d

&V

1 n

vr% V 

………………………

1 R

Sol.

vr% T  R3/2 T1  R    T2  4R 

3/2

1 8

=

 14.

nks jsfM;ks ,fDVo

………………………

Sol. NA = N0e–5t

12.

50 Hz vko`fr

dk

15.

,d foLr`r [kqyk

Sol.

A

………………………

dy dt

=

NB = N0 e–t ………………………

a 2gy

2A  H  H n a 2g  

Sol. 1000 W

 2A  H  n  0 a 2g  

= T1

= T2

T1 = T2

VC  5V

n = 4.

VR  2V IRms 

VR 2  Amp R 1000

16.

,d nzo dk i`"B

Sol. P0 –

= 2mA

IRms  X C  VC 2mA  XC  5



2S R1

………………………

+gh –

2S R2

=P0

2S 2S   gh R1 R2

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SOLSET1JRMMT3301218-2

17.

,d rkj ds ;ax

………………………

F    AY (F / A) Y

Sol. (A)  =

 Y

Sol.

gkbMªkstu ds

dk
Sol.

4R  1  3   Mg  R  MR2  2  3  2  2  

4000  103 2  10 3

= 2 × 109 N/m 2 21.

CykWd

Sol.

ekuk xksys dk nzO;eku dsUnz dk osx

RT Mmix.

dks {kSfrt

………………………

RT Mmix

2

=

csyu dh xfrt ÅtkZ

=

1 2 1 1 2 2 mv + ×  mR   2 2 2 

= 2v.

=2mv2

1 1 3 1  mv 2 1   = . mv 2 2 2 2 2 

CykWad dh xfrt ÅtkZ csyu dh xfrt ÅtkZ

2mv 2 3 mv 2 4

=

22.

m 0 nzO ;eku ………………………

Sol.

VDdj ds igys

n1Cv1  n2 Cv 2

CykWad dk osx

1 × m × (2v)2 2



n1CP1  n2 CP2

v gSA

=

=

3 2

rmix 

P

CykWad dh xfrt ÅtkZ

2 Vsound

3RT Mm i x.

………………………

………………………

Vsound  vrms =

rFkk

4g  4  1   3R  3 



3RT Mm i x .

Vrms 

r

‘M’ nzO;eku

(4  2)  10 3 4000  103

=

fn;k gS l = 1m Y =

18.

= oØ

20.

=

8 . 3

VDdj ds ckn

7R 5R 2  n 3 2 2  5R 3R 2 2  n 2 2

3 14  5n  2 10  3n



19.

[ nksauks

mv0 = 2mv cos +

mv 0 2

..........(i)

mv 0 – 2mv sin 2 (i) rFkk (ii), ls

………………………

nks ,d leku

fn'kkvksa esa laosx laj{k.k ls]

 30 + 9n = 28 + 10n  n = 2

0=

Sol.  1T1 = 2T2

T1 2  2 T2 1

..........(ii)

tan = 45º

nqckjk

•

y–fn'kk

esa laosx laj{k.k ls

Å"ek ál dh nj Q = 4r2eT4 •

2

mv 0 2

4

r  T   1   1  4 • Q 2  r2   T2  Q1

•

Q

=

–ms

d 4 d  – r 3s dt 3 dt

23.

= 2mv.

Sol. P =

 r2     r1 

 32

v0

.

2 2

………………………

,d iEi }kjk

3

v =

2

mgh   d   •   dt   1  Q1   •    d  Q   2  dt   2

1

t

1 mv 2 2

= 4000 watt

= 5.36 HP

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SOLSET1JRMMT3301218-3

24.

5 kg

nzO;eku dk

………………………

Sol. Resultant force = N

1  2

   7 x  

= mg cos 

2

=

1  2

   7 x  

2

× 5= 5

7 + x U;wure

'kfDr egRre gksxh tc

gksxk

i.e., for x = 0

vFkkZr

25.

2kg nzO;eku

dk

30.

………………………

Sol. N cos 30° = g m A sin 30° + m B g sin 30° N = 20 26.

3

N

fdlh {k.k ij

x = 0 ds

fy,A ………………………

,d leryksÙky

Sol. fm = –R/2 f = 40 m

1 F

………………………

Sol.



=

1 2  fm f

1 2 2   7.5 R 40

2 2 10   R 40 75 =



2 R

1 2  20 15 =

=



15  40 20  15

255 3  15

R = 24 cm. vBA = vB – vA =

[4iˆ  3ˆj]  [3iˆ  4ˆj]

=

PART-B : CHEMISTRY

ˆi  7ˆj

vapp = 4 cos45º + 3 cos45º + 3 cos45º – 4 cos45º = 6 cos45º = 27.

–q

Bksl esa =kqfV;ksa ds

Sol.

rF;

a

ds

dkj.k ¶yDl

……………………… 1 =

q 60

32.

;fn ;wfj;k dks 'kdZjk

Sol.

vok"i'khy foys; dks foy;u esa feyk;k tkrk gS vr%

............

V.P., RLVP, Tf , Tf , Tb , Tb,  

ds dkj.k ¶yDl Hkh leku fn'kk es gh gksxk] D;ksafd bls oxZ

XY

q 240

 = 1 + 2 =

5q 240

28.

rhu ,d leku

………………………

Sol.

e/; IysV ij dqy vkos'k 'kwU; gS vkSj ,d&leku fo|qr {ks=k esa

X+ ............

33.

,d ;kSfxd

Sol.

f=kT;k vuqikr =

90 200

= 0.45

f=kT;k vuqikr

0.414

rFkk

esa]

ds uhps j[kk x;k gS 2 =

............

3 2 m/s

Hkqtk yEckbZ

Sol. +q ds

31.

0.732

ds e/; fLFkr gksrk gS rFkk

v"VQydh; fjfDrdk dks iznf'kZr djrk gSA Y– , X+

ls cM+k gS vr%

Y– FCC

cuk;sxk rFkk

X+

fjfDrdkvksa dks xzg.k

djsxkA

gSA vr% bl ij ifj.kkeh cy 'kwU; gksxkA

vr% 34.

X+

dh leUo; la[;k = 6

ml rkieku dh x.kuk

............

Sol. G = H – TS 29.

fn;s x;s ifjiFk

Sol.

ifjiFk esa /kkjk O;Dr dh tkrh gS  7x esa mRiUu 'kfDr gSA

i= 5W

………………………

0 = +131.3 – T(+0.1336)

¼Lor% vfHkfØ;k ds fy,½ T>

131 .3 0.1336

= 982.8 K

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SOLSET1JRMMT3301218-4

35.

tc cksjsDl dks ty............

Sol. [B4O5(OH)4]2– + 5H2O

2B(OH)3 + 2[B(OH)4]

—

2m

foy;u

(m

............

Sol.

ekuk

1.5 m

foy;u esa foyk;d dk nzO;eku = m 1 Kg rFkk ekuk

foy;u

3

4p

=0

(4) [Cu(NH3)4]2+

foy;u esa foyk;d dk nzO;eku = m 2 Kg gSA

bl izdkj ifj.kkeh foy;u dh eksyyrk = 1.5m1  3m2 m1  m2

42.

= 109 . 59

ij vfHkfØ;k............

t =0 0 t = equim

dh mifLFkfr esa

—

—

2x

xx

4s

4p

xx

xx xx

=1

foy;u............

2As3+ + 3S2– 3s + 10–2

2s

  10–2 vc, [As3+]2 [S2–] 3 = Ksp  (2s)2 (10–2)3 =  1  10 –24   625   

2

(2x) (0.48  x)

10–2 M Na2S

2 CO(g)

0.48 0.48 - x

,d

Sol. As2S3

CO2 (g) + C(s)

Sol.

KP =

3d8 NH

3  

v;qfXer bysDVªk Wu

bl izdkj vko';d vuqikr 1000 K

3d9

[Ar]

= 2.

 m1 = 2 m2 1

37.

4s

v;qfXer bysDVªk Wu

36.

m

Zn+2(Ar] 3d10

=3

 s= 2 × 10–11 mol/L.

x = 0.33 ckj 38.

tc rktk vo{ksfir

43.

............

Sol. Theory based

5   ,d vkn'kZ xSl  Cv,m  R  ............ 2   1

39.

fuEu esa ls dkSulk

Sol.

F;ksjh ns[ksaA

40.

fuEu esa ls fdrus

Sol. (i)

leryh; v.kq

............

Hm  nCP T 

............ : XeF2, ClF3, H2O, [XeF 5]–,

44.

41.

<+k y = –1  K =1 min–1 v)Z vk;q

oxZ f}fijkfeMh;

IF7 –

iapdks.kh; f}fijkfeMh; ............

Sol. (1) [Cr(H2 O)6]3+ +3

Cr

[Ar]

3d

3

4s

v;qfXer bysDVªkWu (2) [Fe(H2 O)6 Fe+2

v;qfXer bysDVªkWu (3) [Zn(H2 O)6 ]2+

4p

=4

45.

XeF6 + H2 O  A + B............

Sol.

XeF6 + 3H2O  XeO3 + 6HF

46.

fn;s x;s izØe esa

A ............



=3

4s

= 0.693 min

H O

2 Sol. CaC2+N  CaCN2+C  CaCO3+ NH3 2

4p

]2+ [Ar] 3d6

............

y = C + mx

f}fijkfeMh;

SF6 –

fuEu vk;uksa esa ls

dksfV vfHkfØ;k ds fy,

ln Ct = ln C0 –Kt

<s+dqyh vkÑfr

PCl5 – f=kdks.kh;

1st

7 R 150  600   1575R 2

Sol. Ct = C0 e–Kt

 3– , BCl3, XeF4. (ii) SF4 –

T2  V1    T1  V2  or ;k T2 = 150 K

Sol.

47.

Hg[Co(SCN)4]

Sol .

edZjh(II) VsVªkFkk;kslk;usVks-S-dksckYVsV (II)

48.

AgNO3 (aq)

Sol.

,uksM+ ij vfHkfØ;k gSA H2O  O2 + 4H+ + 4e

dk

............

4d

foy;u ds............

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SOLSET1JRMMT3301218-5

49.

,uhlkWy cukus ds fy,

............

T.R

CH3–CH=O   CH3COO + Ag (jtr

O

OCH3 O

O SN2

O

, ful kW y

50.

fuEu cgqydksa esa ls

Sol.

MsØksu] Xykbdksy o VjFkSfyd vEy dk la?kuu cgqyd gSA

51.

fuEu esa ls dkSulh

58.

Sol.

ds

............

,





............ 59. 



CH2

B

2, 2-MkbDyksjksC;wVsu

............

Sol. A

T.R   dksbZ vfHkfØ;k ughaA

+ CH3–O–S–O

+ CH3–O–S–O–CH3  O

Sol.

C6H5 – NO2

niZ.k)

............

gy.

CH2



CH2

vfHkfØ;k ds ckn

+HBr

(izcy



CH2

(SN1mRikn)

vEy) +

,sjksesfVd



dksbZ vfHkfØ;k ugha

C 

52.

xyr Øe dk p;u

Sol.

{kkjh; lkeF;Z

D;ksafd ,fjygSykbM esa vr% ;s

............

NH2

:

S N vfHkfØ;k

NH2 CH3

>

(vkFkksZ

61.

fuEu dkcksZgkbMªsV ds

Sol.

;g vEyh; ek/;e esa 'kdZjk dk izfriu iznf'kZr djrk gSA

............

C12H22O11

Sol.

[0, 4]

3sinx  sin 3 x  sin 2 x  5 sec 2 x [-5,5] 

62.

Xywdksl

lehdj.k

+52.5º

–

 /2

 o

O CH3—C—OH

(1) PCl5

1 x

............

(2) H2/Pd/BaSO4

Sol. P , CH3–CH=O

56.

 n x   n tan  n 2  x sec 2  d   2 2 x sec  4

 /2

   n x   n tan   d  o

gSA

Q, CH3 –CH=CH–CH=O

 n tdt   n 2 …………..  2  t2 4

Sol. t = x tan

92.7º

55.

[5, ]

x o

ÝDVksl  = + 66.5º

ugha nsrsA

esa lehdj.k…………..

C6H12O6 + C6H12O6

lqØksl

ca/k] vuqukn }kjk LFkk;hd`r gksrk gS]

PART-C : MATHEMATICS

izHkko ds dkj.k)

53.

C – X

n2 4

n x  n 2  x 2 4 n x n 2  x 2

gSA

(i) CO + NaOH, 

   2     ............ 



(ii) H

nks eku

= x = 2, 4

OH OH

63.

CHO

Sol.

(i) CHCl3 NaOH, 

+

  (ii) H

(P)

57.

rhu ;kSf xd

CHO (Q)

x, y rFkk z ............

T.R Sol. CH3 –CHC–H   CH3–CC–Ag (lQsn

vo{ksi)

Sol.

;fn

H1, H2, H3 …H100…………..

1 1 1 1 1 , , ,..... , a H1 H 2 Hn b 1 1  d H1 a

lekUrj Js.kh gSA

1 1  d Hn b

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SOLSET1JRMMT3301218-6

H1 

a 1  ad

Hn 

b 1  bd

H1  a Hn  b  H1  a Hn  b =

=

=

65.

x =2 

 sin x  x2    x  600

lim x0

600

x 600

 x3    x  ....  6   602 x

= 2n

= 100

ekuk

P

vk;r

ABCD…………..

izo.krk =

2= a b

C

(0,a) D (0,a/2) P

150m

tan 60o =

60o y m BP =

150 y 67.

150  150 3

x = 150 - 50

x = 50 (3 -

=

 1  ,2   3 

B:

 5  ,2   3 

C:



H:

4   3,  3 



km/h

3, 0

=

1 2

…………..

Sol. A:

3)



a 0 a 2  b 2b

ekuk js[kkvksa

3

25 3  3 60 osax =

B(b,0)

A

x + y = 150

x+

600

 x2   x 600  x 600  600 C0  600 C1    ...   6   lim 602 x 0 x

Sol. AC dh

x

…………..

600

x 600   sin x 

lim

x

66.

45o

600

600

21 1 2      2   n  1 2 d b a

Sol.

x 2  sin x 

x 0

a b a b 1  ad  1  bd a b a b 1  ad 1  bd 2  ad 2  bd   ad bd

,d uko esa cSBs…………..

lim x 0

Sol.

64.

x 600   sin x 



1000

3 3 3 2





=

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SOLSET1JRMMT3301218-7

(0,3)

70.

A y=2 B C (

68.

Sol.

3 ,0)

   sin  x   4  ekuk f ( x )   …………..    cos 1  x   x       4    2n  1  x  2n  sin  4 x    f  x    cos   1  x  2n  x  2 n  1    4 

1 2

f  2n 

;fn x = 1, 2 ij pje…………..

f  2 n  1  0

1 2 1 f 2n  2

Sol. p(x) = 2x2 + ax3 + bx4 p’(x) = 4x + 3ax2 + 4bx3 4 + 3a + 4b = 0 8 + 12a + 32b = 0

f 2n 

 

f

  2 n  1   0

 

f

  2 n  1   1





2 + 3a + 8b = 0 71. –2 + 4b = 0

2 cotB = cotA + cotC

69.

 ekuk a  2iˆ  ˆj  2kˆ …………..

Sol.

      | c  a |2  (c  a ).(c  a )  (2 2)2

cotB (2 cotB) + cotA cotC = 1

cotA+cotC  cotAcotC 2 cotB  1-2cot 2 B

   c 2  a 2  2c .a  8   c 2  2 | c |.  a 2  8

3cot 2B  1 cot B 

  c 2  2 | c | 1  0   ( c  1) 2  0  |c | 1       | (a  b )  c || a  b || c | sin 30o

1    (a  b ) 2

  | ( a  b ) | 3

72.

Sol.

;fn

0



1 3

x1, x2 , x3 …………..



 xi  x

S .D 



iˆ ˆj kˆ   a  b  2 1 2  2iˆ  2 ˆj  kˆ 1 1

ekuk A, B, C U;wudks.k…………..

Sol. cotA cotB + cotB cotC + cotA cotC = 1



2

n



2

 xi  x   yi  y





 xi  yi  x  y n





2

 9  16  2



2



 2  xi  x

 y  y  i

n





 xi  x yi  y



n

   3 | (a  b )  c | 2 Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029

SOLSET1JRMMT3301218-8

73.

ekuk P, Q, R rhu…………..

ijoy; ij fcUnq (2m 2, – 4m)

Sol.

ABC=3I BCA=3I CAB=3I

 ,   ,  ,     2,  4  & 8,  8  77.

 det  ABC  BCA  CAB 







 

-1

0

2

2





2

                     +  (–3 –

)

= –12 – 11 C = 23

1

X

 | sin x | dx  8 …………..

78.

;fn

Sol.

     4     4 9  17     2   4 17 

I



x cos 3 xdx I 1  sin 4 x 17 





pwfd

2

=

2I 

Sol.

 

2

  

               2



ekuk f(x) ,d ?kuh; …………..



=

 –3  = –3 (4) - 

-1

75.

2

2



Y

x = 3 ij



= –3

Sol.

x=1

lehdj.k…………..

      

  x  1 3  2 2  x  1  3 ekuk f  x    x  1 1  x  1 …………..  2 1 x  2   x  1 

-2

,  , 

Sol. –C =

 det  9I   729

74.

;fn

17 



16  x  cos 3 xdx 1  sin 4 x



2

16 . cos 3 xdx cos 3 xdx  16  .9 0 1  sin 4 x 1  sin 4 x 

ij oØ dk ufrijfjorZu fcUnq gS blfy,

x = –1

cos 3 xdx 0 1  sin 4 x 0

 144 2 

rFkk

LFkkuh; mfPp"B rFkk LFkkuh; fufEu"B j[krk gSA 79.

(-1,15)

'krajt cksMZ ij NksVs…………..

Sol. n(S) = 64C2 = 63 x 32 n(E) = 7 x 7 x 2 ={number of ways of choosing 2 x 2 square}x 2 n(E) = 7 x 7 x 2 ={ 2 x 2 ds p(E) =

Distance = 37

ekuk |z| = 2 vkSj…………..

Sol.

Z  2ei



w=

76.

ekuk  , 

Sol.

vfHkyEc dk lehdj.k y = mx – 4m- 2m 3

7 7 2 7  63  32 144

80.

(3,-22)

2e i 

vkSj………….. =

12 = 18 m – 4 m – 2m 3 m3 – 7m + 6 = 0 m = 1, – 3, 2

=

oxZ pquus ds rjhd} x 2

1 2e i

1  i e 2 5 3 cos   i sin  2 2

2ei 

 

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SOLSET1JRMMT3301218-9

w =

=

81.

nh?kZo`Ùk dks O;Dr djrk gS rc {ks=k dk {ks=kQy gS

=

 .a .b

q r 

5 3 . . 2 2 15 4

o`Ùkksa

84.

x2 + y2 – 6x = 0…………..

=

~ q r  

=

q

Sol.



-1



~r



p



p

[aij] 2 Øe…………..

ekuk

Sol. A

1.3  3.1  ,0   ,0    3 1 

p

ds fy, fo|eku gS

|A|



0

fLFkfr-I : Bhd,d vo;o 'kwU; gSA 5C 3

4 = 240

x

fLFkfr-II : dksbZ vo;o 'kwU; ughA

a c

A=

b d

|A| = ad – bc = 0

8

ad = bc = 12



5C 4

x

8

ad = bc = 6

4 - 16 = 120 – 16

= 104

   3, 0 

85.

CAB = 30o tan 30o =

{ks=kQy

82.

=

js[kk ds fnd~ vuqikr

=

= 344

ekuk

y = f(x) …………..

2 tan xd x ;fn = e   e 2  n (sec x ) 2 ;fn = sec x

y . sec2 x =

j

20 

k

1 2 3

 = –i + 2j – k

20 



=

cos 2 xdx

0

 cos

2

xdx

0

x 1 y 1 z 1    1 2 1

 /2

= -80

 cos

2

xdx   20

0

+2+  -1 = 0

1 3

86.

(1+x2 )25 (1+x25 ) …………..

Sol. (1+x2 )25 = 25Cr x2r (1+x2 )25 (1+x25 + x40 + x45 + x50 )

83.





= -2. 20

   1,  2   1,   1





cos x  2 cos 2 x dx  2

0

js[kk ij fcUnq (1, –1, –1) gSA

 - 1+ 4 6  = -2

x dx

sec2 x.y = sec x -2

2 3 4

js[kk dk lehdj.k

2

 sin x.sec

y sec2 x = sec x + C C = –2

ewy fcUnq ls js[kk…………..

i Sol.

Sol.

CB  CB  3 3 1 2 3 3  3 3 2

dqy

esa

x50 dk

xq.kkad gSA

p  (q r) dk …………..

Sol. p  q dk

izfrykse q  p gSA

(q r)  p

x50 dk

xq.kkad

= 25 C25 + 25C5 + 1

= 1 + 25 C5 + 1 Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029

SOLSET1JRMMT3301218-10

2

x2 87.

lehdj.k

  2 2

2

 x  1  x  2  2 2  2 x  1  2 x  2  ………….. 2 2  3x  1  3x  2 

4x2 9x2

 x2 

Sol. C2  C2 – C1 C3  C3 – C1

x

2

2x 1

4x  4

x

2x  1 2

4x 2

4 x  1 8x  4  4 x2

4x  1 2

9 x2

6 x  1 12 x  4

9x 2

6x 1 2

sin8 x  cos8 x  cos2 x  sin2 x

sin 8 x  cos 8 x  2 sin 2 x cos 4 x 

3x2

2 x

0

2

2 x

0  2 6 x 3 10 x3

9x

Sol.

sin8 x  cos8 x  sin8 x  cos8 x 2

R2  R2 – R3

2

sin8 x  cos8 x …………..  vf/kdre = 1

R1  R1 – R2

5x

90. C3  C3 – 2C2 2

y 2



1 sin 2 2 x 8



6x  1 2

= –8x3 = 2x + 6 = 4x3 + x + 3 = 0

88. Sol.

dsoy ,d okLrfod ewy gSA

ekuk PQRS ,d…………..

  QR  a  b  3i  4 j  5 k  PS  PR  2 a  b  4i  5 j  6 k S

R b

a b

P

Q

a

i j  PQ   i  j  k   1 1 1 1

k 1

=

2i  2 j

1 =

vk;ur

3

4 5

4

5 6  5 18   6  6  8   90  84  6

2 2 0 89. Sol.

;fn y   x  

…………..

y  mx  2m2

y  x   2

m   ,  2m   Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029

SOLSET1JRMMT3301218-11

MAIN MAJOR TEST (MMT) (JEE MAIN PATTERN)

TARGET : JEE (MAIN + ADVANCED) 2019

DATE : 30-12-2018

|C CLASS XII/XIII | COURSE : ALL INDIA TEST SERIES (VIKALP)

ANSWER KEY SET-1 PART-A (PHYSICS) 1.

(4)

2.

(3)

3.

(3)

4.

(4)

5.

(3)

6.

(1)

7.

(3)

8.

(2)

9.

(1)

10.

(2)

11.

(1)

12.

(3)

13.

(3)

14.

(2)

15.

(3)

16.

(4)

17.

(1)

18.

(2)

19.

(2)

20.

(1)

21.

(3)

22.

(2)

23.

(3)

24.

(2)

25.

(4)

26.

(4)

27.

(4)

28.

(4)

29.

(4)

30.

(2)

PART-B (CHEMISTRY) 31.

(4)

32.

(4)

33.

(2)

34.

(2)

35.

(3)

36.

(2)

37.

(2)

38.

(3)

39.

(1)

40.

(1)

41.

(2)

42.

(2)

43.

(3)

44.

(2)

45.

(4)

46.

(4)

47.

(1)

48.

(2)

49.

(3)

50.

(4)

51.

(4)

52.

(4)

53.

(3)

54.

(3)

55.

(2)

56.

(2)

57.

(1)

58.

(3)

59.

(1)

60.

(3)

PART-C (MATHEMATICS) 61.

(4)

62.

(2)

63.

(3)

64.

(1)

65.

(2)

66.

(4)

67.

(2)

68.

(1)

69.

(2)

70.

(2)

71.

(3)

72.

(4)

73.

(4)

74.

(2)

75.

(1)

76.

(1)

77.

(3)

78.

(4)

79.

(3)

80.

(4)

81.

(3)

82.

(3)

83.

(1)

84.

(4)

85.

(1)

86.

(2)

87.

(2)

88.

(3)

89.

(4)

90.

(1)

Corp. / Reg. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029

SOLSET1JRMMT3301218-12