COURSE : VIJETA (JP) | BATCH : JPB*,JPAB CODE 1 le; TARGET ...

Resonance Eduventures Limited REGISTERED & CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005

Ph.No. : 0744-2777777, 0744-2777700 | Toll Free : 1800 258 5555 | FAX No. : +91-022-39167222 | 73400 10333 Website : www.resonance.ac.in | E-mail : [email protected] | CIN: U80302RJ2007PLC024029

®

MAIN PATTERN CUMULATIVE TEST-1 (MCT-1)

TARGET : JEE (MAIN + ADVANCED) 2021

COURSE : VIJETA (JP) | BATCH : JPB*,JPAB

Ñi;k bu funsZ'kksa dks /;ku ls i<+saA vkidks 5 feuV fo'ks"k :i ls bl dke ds fy, fn;s x;s gSaA

(Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose).

INSTRUCTIONS TO CANDIDATES

DATE

31-05-2020

CODE

1

Time (le;) : 3 Hours (?k.Vs) Maximum Marks (vf/kdre vad) : 300

Question paper has three (03) parts: Physics, Chemistry and Mathematics.

ç'u&i=k eas rhu (03) Hkkx gS % (HkkSfrdh ] jlk;u foKku ,oa

xf.kr)

Each part has a total twenty five (25) questions divided

into two (02) sections (Section-1 and Section-2).

çR;sd Hkkx esa dqy iPphl (25) iz'u gS tks nks (02) [kaMks esa

foHkkftr gS ¼[kaM 1 vkSj [kaM 2)

Total number of questions in Question paper are Seventy Five (75) and Maximum Marks are Three Hundred only (300).

iz'u&i=k esa ç'uksa dh dqy la[;k % fipgÙkj (75) ,oa vf/kdre

vad % rhu lkS (300) gSaA

ç'uksa ds çdkj vkSj ewY;kadu ;kstuk,¡ (Type of Questions and Marking Schemes)

[kaM–1 ¼vf/kdre vad : 80½ | SECTION-1 (Maximum Marks : 80)

This section contains TWENTY (20) questions bl [kaM esa chl (20) iz'u gSaA

Each question has FOUR options (1), (2), (3) and (4) ONLY ONE of these four option is correct)

izR;sd iz'u esa pkj fodYi (1), (2), (3) rFkk (4) gSaA bu pkj

fodYiksa esa ls dsoy ,d fodYi lgh gSaA

Marking scheme: Full Marks : +4 If the corresponding to the

answer is darkened Zero Marks : 0 If none of the options is chosen

(i.e. the question is unanswered). Negative Marks : –1 In all other cases.

vadu ;kstuk

iw.kZ vad % +4 ;fn flQZ lgh fodYi gh pquk x;k gSA

'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS

¼vFkkZr~ iz'u vuqÙkfjr gS½A

_.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA

[kaM–2 ¼vf/kdre vad : 20½ | SECTION – 2 : (Maximum Marks : 20)

This section contains 5 questions. bl [k.M esa 5 ç'u gSaA

The answer to each question is a Single Digit Integer, ranging from 0 to 9 (both inclusive).

çR;sd ç'u dk mÙkj 0 ls 9 rd ¼nksuksa 'kkfey½ ds chp

dk ,dy vadh; iw.kk±d gSA

Marking scheme Full Marks : +4 If the corresponding to the

answer is darkened Zero Marks : 0 If none of the options is

chosen (i.e. the question is unanswered). Negative Marks : –1 In all other cases.

vadu ;kstuk

iw.kZ vad % +4 ;fn flQZ lgh fodYi gh pquk x;k gSA

'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS

¼vFkkZr~ iz'u vuqÙkfjr gS½A


ijh{kkFkhZ dk uke : ………………………………………………………………jksy uEcj : ………………………..…………………

eSaus lHkh funsZ'kksa dks i<+ fy;k gS vkSj eSa mudk eSusa ijh{kkFkhZ dk ifjp;] uke vkSj jksy uEcj

vo'; ikyu d:¡xk@d:¡xhA dks iwjh rjg tk¡p fy;k gSA

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

ijh{kkFkhZ ds gLrk{kj ijh{kd ds gLrk{kj

Resonance Eduventures Limited REGISTERED & CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005

Ph.No. : 0744-2777777, 0744-2777700 | Toll Free : 1800 258 5555 | FAX No. : +91-022-39167222 | 73400 10333 Website : www.resonance.ac.in | E-mail : [email protected] | CIN: U80302RJ2007PLC024029

vkWfIVdy fjLikal 'khV (ORS) Hkjus ds funsZ’'k (INSTRUCTIONS FOR OPTICAL RESPONSE SHEET (ORS)

Darken the appropriate bubbles on the original by applying

sufficient pressure.

Åijh ewy i`"B ds vuq:i cqycqyksa (BUBBLES) dks i;kZIr ncko

Mkydj dkyk djsaA

The original is machine-gradable and will be collected by the

invigilator at the end of the examination.

ewy i`"B e'khu&tk¡p gS rFkk ;g ijh{kk ds lekiu ij fujh{kd ds

}kjk ,d=k dj fy;k tk;sxkA

Do not tamper with or mutilate the ORS. vks-vkj-,l- dks gsj&Qsj@fod`fr u djsaA

Write your name, roll number and the name of the

examination centre and sign with pen in the space provided

for this purpose on the original. Do not write any of these

details anywhere else. Darken the appropriate bubble

under each digit of your roll number.

viuk uke] jksy ua- vkSj ijh{kk dsanz dk uke ewy i`"B esa fn, x,

[kkuksa esa dye ls Hkjsa vkSj vius gLRkk{kj djsaA buesa ls dksbZ Hkh

tkudkjh dgha vkSj u fy[ksaA jksy uEcj ds gj vad ds uhps vuq:i

cqycqys dks dkyk djsaA

ORS ij cqycqyksa dks dkyk djus dh fof/k % (DARKENING THE BUBBLES ON THE ORS) :

Use a BLACK BALL POINT to darken the bubbles in the

upper sheet.

Åijh ewy i`"B ds cqycqyksa dks dkys ckWy ikbUV dye ls dkyk

djsaA

Darken the bubble COMPLETELY. cqycqys dks iw.kZ :i ls dkyk djsaA

Darken the bubble ONLY if you are sure of the answer. cqycqyksa dks rHkh dkyk djsa tc vkidk mÙkj fuf'pr gksA

The correct way of darkening a bubble is as shown here :

cqycqyksa dks dkyk djus dk mi;qDr rjhdk ;gk¡ n'kkZ;k x;k gS %

There is NO way to erase or "un-darkened bubble. dkys fd;s gq;s cqycqys dks feVkus dk dksbZ rjhdk ugha gSA

The marking scheme given at the beginning of each section

gives details of how darkened and not darkened bubbles

are evaluated.

gj [k.M ds izkjEHk esa nh x;h vadu ;kstuk esa dkys fd;s x;s rFkk dkys

u fd;s x;s cqycqyksa dks ewY;kafdr djus dk rjhdk fn;k x;k gSA.

For example, if answer ‘SINGLE DIGIT’ integer type below :

0 1 2 3 4 5 6 7 8 9

mnkgj.k ds fy, , ;fn mÙkj ‘,dy vadh;’ iw.kkZad gS rc :

0 1 2 3 4 5 6 7 8 9

For example, if answer ‘DOUBLEDIGIT’ integer type below :

0 0

1 1

2 2

3 3

4 4

5 5

6 6

7 7

8 8

9 9

mnkgj.k ds fy, , ;fn mÙkj ‘f)&vadh;’ iw.kkZad gS rc :

0 0

1 1

2 2

3 3

4 4

5 5

6 6

7 7

8 8

9 9

FOR DECIMAL TYPE QUESTIONS OMR LOOKS LIKE :

1 2 . 3 4

0 0 0 0

1 1 1 1

2 2 2 2

3 3 3 3

4 4 4 4

5 5 5 5

6 6 6 6

7 7 7 7

8 8 8 8

9 9 9 9

COLUMN

n'keyo iw.kkZad@la[;kRed vadksa ds fy, ORS fuEu çdkj gS :

1 2 . 3 4

0 0 0 0

1 1 1 1

2 2 2 2

3 3 3 3

4 4 4 4

5 5 5 5

6 6 6 6

7 7 7 7

8 8 8 8

9 9 9 9

COLUMN

If answer is 3.7, then fill 3 in either 1st or 2nd column and 7 in

3rd or 4th column.

;fn mÙkj 3.7 gS rc 3 dks 1st ;k 2nd dkWye esa Hkjsa rFkk 7 dks 3rd

;k 4th dkWye esa HkjsaA

If answer is 3.07 then fill 3 in 1st or 2nd column ‘0’ in 3rd

column and 7 in 4th column.

;fn mÙkj 3.07 gS rks 3 dks 1st dkWye ;k 2nd dkWye esa Hkjsa rFkk ‘0’

dks 3rd dkWye esa rFkk 7 dks 4th dkWye esa HkjsaA

If answer is, 23 then fill 2 & 3 in 1st and 2nd column

respectively, while you can either leave column 3 & 4 or fill

‘0’ in either of them.

;fn mÙkj 23 gS rc 2 dks 1st dkWye esa ] 3 dks 2nd dkWye esa tcfd

3rd vkSj 4th dkWye dks [kkyh NksM+ nsa ;k ‘'kwU; Hkj nsaA

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

®

Reg. / Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005

Website : www.resonance.ac.in | E-mail : [email protected] P01JPAMCT1310520C1-1

Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029

PART-A

SECTION – 1 : (Maximum Marks : 80)

This section contains TWENTY (20) questions.

Each question has FOUR options (1), (2), (3)

and (4) ONLY ONE of these four option is

correct

Marking scheme :

Full Marks : +4 If ONLY the correct option is

chosen.

Zero Marks : 0 If none of the options is chosen

(i.e. the question is unanswered).

Negative Marks : –1 In all other cases

1. A thin prism of glass is placed in air and water

respectively. If ng = 3/2 and n

w = 4/3, then the

ration of deviation produced by the prism for a

small angle of incidence when placed in air and

water separately is :

(1) 9 : 8 (2) 4 : 3

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

2. Distance between two images formed by upper

and lower part for the point object placed at 30

cm from given lens.

(1) 66 cm (2) 36 cm

(3) 72 cm (4) 42 cm

Hkkx– A

[kaM 1 : (vf/kdre vad : 80)

bl [kaM esa chl (20) iz'u gSaA

izR;sd iz'u esa pkj fodYi (1), (2), (3) rFkk (4) gSaA

bu pkjksa fodYiksa esa ls dsoy ,d fodYi lgh gSaA

vadu ;kstuk :

iw.kZ vad % +4 ;fn flQZ lgh fodYi gh pquk x;k

gSA

'kwU; vad % 0 ;fn dksbZ Hkh fodYi ugha pquk x;k

gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A


1. ,d dk¡p dk iryk fizTe Øe'k% ok;q o ty esa j[kk

x;k gSA ;fn ng = 3/2 rFkk n

w = 4/3, gS] rks Øe'k%

ok;q rFkk ty esa j[kus ij fizTe }kjk vYi vkiru

dks.kksa ds fy, mRiUu fopyu dk vuqikr gksxk &

(1) 9 : 8 (2) 4 : 3

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

2. fn;s x;s ySUl ls 30 cm nwjh ij fLFkr oLrq ds nks

çfrfcEc Åijh Hkkx rFkk fupys Hkkx ls curs gS]

çfrfcEcksa ds e/; nwjh gksxh&

(1) 66 cm (2) 36 cm

(3) 72 cm (4) 42 cm


®




3. A thin beam of light, incident on a transparent

sphere from air, gets focussed on opposite

surface of the sphere as shown in figure then

refractive index of the material of the sphere is :

(1) 1.5 (2) 2.0

(3) 2.5 (4) 3.0

4. The distance between an object and the screen

is 100 cm. A lens produces an image on the

screen when the lens is placed at either of the

positions 40 cm apart. The power of the lens is

nearly :

(1) 3 diopters

(2) 5 diopters

(3) 2 diopters

(4) 9 diopters

5. A person AB of height 170 cm is standing infront

of a plane mirror. His eyes are at height 164 cm.

At what distance from P should a hole be made

in the mirror so that he cannot see the top of his

head.

(1) 167 cm

(2) 161 cm

(3) 163 cm

(4) none of these

3. çdk'k dk ,d iryk iqat gok ls ,d ikjn'khZ xksys ij

vkifrr gksrk gS rFkk xksys dh foijhr lrg ij

fp=kkuqlkj Qksdl gksrk gSA rc xksys ds inkFkZ dk

viorZukad gksxkA

(1) 1.5 (2) 2.0

(3) 2.5 (4) 3.0

4. oLrq vkSj ijns ds chp nwjh 100 cm gSA yasl dks 40

cm nwj nks fLFkfr;ksa ij j[kus ij oLrq dk izfrfcEc

ijns ij curk gSA ysal dh yxHkx {kerk gksxhA

(1) 3 Mk;vkWIVlZ

(2) 5 Mk;vkWIVlZ

(3) 2 Mk;vkWIVlZ

(4) 9 Mk;vkWIVlZ

5. ,d O;fDr ftldh Å¡pkbZ 170 cm gS ,d lery niZ.k

ds lkeus [kM+k gSA mldh vk¡[ks 164 cm Å¡pkbZ ij gSA

P ls fdl nwjh ij Nsn cuk;k tk;s ftlls og viuk

flj dk Åijh fljk ugh ns[k ldrk A

(1) 167 cm

(2) 161 cm

(3) 163 cm

(4) buesa ls dksbZ ugh


®




6. In the figure two concentric conducting shells of

radius R & 2 R are shown. The inner shell is

charged with Q and the outer shell is uncharged.

The amount of energy dissipated when the

shells are connected by a conducting wire is:

(1)2k Q

4 R

(2) 2k Q

2 R

(3) 2k Q

8 R

(4) 23kQ

4

7. A ring of radius R having a linear charge density

moves towards a solid imaginary sphere of

radius R

2, so that the centre of ring passes

through the centre of sphere. The axis of the ring

is perpendicular to the line joining the centres of

the ring and the sphere. The maximum flux

through the sphere in this process is :

(1) 0

R

(2) 0

R

2

(3) 0

R

4

(4)0

R

3

6. fp=k esa R o 2R f=kT;k ds nks ladsUnzh xksyh; dks'k

fn[kk, x, gSaA vkUrfjd dks'k Q vkos'k ls vkosf'kr gS

rFkk cká dks'k vukosf'kr gSA nksuksa dks'kksa dks rkj ls

tksM+us ij O;; ÅtkZ gksxhA

(1)2k Q

4 R

(2) 2k Q

2 R

(3) 2k Q

8 R

(4) 23kQ

4

7. ,d oy; ftldh f=kT;k R rFkk js[kh; vkos'k ?kuRo

gS] ,d dkYifud Bksl xksys ¼f=kT;k R/2½ dh vksj

bl izdkj xfr djrh gS fd oy; dk dsUnz xksys ds

dsUnz ls gksdj xqtjrk gSA oy; dk v{k oy; rFkk

xksys ds dsUnzksa dks tksM+us okyh js[kk ds yEcor~ gSA bl

izfØ;k esa xksys ls xqtjus okys vf/kdre ¶yDl dk

eku gksxk &

(1) 0

R

(2) 0

R

2

(3) 0

R

4

(4)0

R

3


®




8. Dipole is placed parallel to the electric field. If W

is the work done in rotating the dipole by 60°,

then work done in rotating it by 180° is :

(1) 2W (2) 3W

(3) 4W (4) W/2

9. A graph of the x component of the electric field

as a function of x in a region of space is shown.

The Y and Z components of the electric field are

zero in this region. If the electric potential is 10 V

at the origin, then potential at x = 2.0 m is :

(1) 10 V

(2) 40 V

(3) – 10 V

(4) 30 V

10. Two smooth spherical non conducting shells

each of radius R having uniformly distributed

charge Q & Q on their surfaces are released

on a smooth non-conducting surface when the

distance between their centres is 5 R. The mass

of A is m and that of B is 2 m. The speed of A

just before A and B collide is: [Neglect

gravitational interaction] (take K =0

1

4)

(1)22kQ

5mR (2)

24k Q

5mR

(3) 28k Q

5mR (4)

216k Q

5mR

8. ,d f}/kzqo fo|qr {kS=k ds lekukUrj j[kk gSA bls 60°

?kqekus esa vko';d dk;Z W gks rks bls 180° ?kqekus esa

vko';d dk;Z gksxkA

(1) 2W (2) 3W

(3) 4W (4) W/2

9. fo|qr {ks=k ds x ?kVd dk x ds Qyu ds :i esa xzkQ

fp=k esa çnf'kZr gSA bl {ks=k esa fo|qr {ks=k ds Y rFkk Z

?kVd 'kwU; gSA ;fn ewy fcUnq ij fo|qr foHko 10 V gks

rks x = 2.0 m ij foHko gksxk :

(1) 10 V

(2) 40 V

(3) – 10 V

(4) 30 V

10. nks fpdus xksyh; dqpkyd xksyh; dks'kksa (çR;sd dh

f=kT;k R) ds i`"B ij vkos'k Q o Q ,dleku :i ls

forfjr gSA tc muds dsUnzksa ds chp dh nwjh 5 R gS rks

os ,d fpdus dqpkyd i`"B ij NksM+s tkrs gsA A dk

nzO;eku m rFkk B dk nzO;eku 2 m gSA A o B ds

VDdj ds Bhd igys A dh pky gksxh & [xq:Roh;

vUrZfØ;k izHkko dks ux.; ekfu,A] (K =0

1

4)

(1)22kQ

5mR (2)

24k Q

5mR

(3) 28k Q

5mR (4)

216k Q

5mR


®




11. Two concentric uniformly charged spheres of

radius 10 cm & 20 cm are arranged as shown in

the figure. Potential difference between the

spheres is :

(1) 4.5 1011 V

(2) 2.7 1011 V

(3) 0

(4) none of these

12. The resultant electric field at centre of a ring due

to ring is zero. Which of the following is

incorrect:

(1) The total charge of the ring may be zero,

although every part of the ring has charge.

(2) The charge on the ring must be uniformly

distributed.

(3) The charge on the ring may be distributed

nonuniformly.

(4) Total charge on the ring may be positive.

13. Two spherical conductors B and C having equal

radii and carrying equal charges in them repel

each other with a force F when kept apart at

some distance. A third spherical conductor

having same radius as that of B but uncharged,

is brought in contact with B, then brought in

contact with C and finally removed away from

both. The new force of repulsion between B and

C is :

(1) F/4 (2) F/2

(3) F/8 (4) 3F/8

11. nks ladsUnzh ,d leku vkosf'kr xksyksa dh f=kT;k,sa

10 cm o 20 cm gSA budks fp=kkuqlkj O;ofLFkr fd;k

x;k gSA xksyksa ds chp foHkokUrj gSA

(1) 4.5 1011 V

(2) 2.7 1011 V

(3) 0

(4) buesa ls dksbZ ugha

12. oy; ds dkj.k oy; ds dsUnz ij ifj.kkeh fo|qr {ks=k

'kwU; gSA fuEu esa ls dkSulk vlR; gSA

(1) oy; dk dqy vkos'k 'kwU; gks ldrk gS ;|fi oy;

dk izR;sd Hkkx vkosf'kr gSA

(2) oy; ij vkos'k ,d leku :i ls forfjr gksxkA

(3) oy; ij vkos'k vleku :i ls forfjr gks ldrk gSA

(4) oy; ij dqy vkos'k /kukRed gks ldrk gSA

13. nks leku xksyh; pkyd B vkSj C ij cjkcj vkos'k

nsdj dqN nwjh ij j[kus ij] nksuksa ,d nwljs dks F cy

ls izfrdf"kZr djrs gSaA ,d rhljk leku vukosf'kr

xksyh; pkyd dks igys B ds lkFk vkSj fQj C ds lkFk

lEidZ esa ykdj gVk fn;k tkrk gS rks B vkSj C ds chp

u;k izfrdf"kZr cy D;k gksxk \

(1) F/4 (2) F/2

(3) F/8 (4) 3F/8


®




14. Two equal charges are separated by a distance

d. A third charge placed on a perpendicular

bisector at x distance from centre will experience

maximum coulomb force, when :

(1) x = d/ 2

(2) x = d/2

(3) x = d/2 2

(4) x = d/2 3

15. The following data are given for a crown glass

prism ;

refractive index for blue light nb = 1.521

refractive index for red light nr = 1.510

refractive index for yellow light ny = 1.550

Dispersive power of a parallel glass slab made

of the same material is :

(1) 0.01

(2) 0.02

(3) 0.03

(4) 0

16. An object moves in front of a fixed plane mirror.

The velocity of the image of the object is

(1) Equal in the magnitude and in the direction to

that of the object.

(2) Equal in the magnitude and opposite in

direction to that of the object.

(3) Equal in the magnitude and the direction will

be either same or opposite to that of the object.

(4) Equal in magnitude and makes any angle

with that of the object depending on direction of

motion of the object.

14. nks leku vkos'kksa dks d nwjh ij j[kk x;k rFkk ,d

rhljs vkos'k dks d ds yEc v)Zd ij x nwjh ij j[kk

tkrk gS tgk¡ ml ij vf/kdre cy yxs] rks:

(1) x = d/ 2

(2) x = d/2

(3) x = d/2 2

(4) x = d/2 3

15. Økmu dk¡p ds fizTe ds fuEu vkadsMs fn;s x;s gS &

uhys izdk'k dk viorZukad nb = 1.521

yky izdk'k dk viorZukad nr = 1.510

ihys izdk'k dk viorZukad ny = 1.550

bl inkFkZ ls cuh lekUrj dk¡p ifV~Vdk dh fo{ksi.k

{kerk gS &

(1) 0.01

(2) 0.02

(3) 0.03

(4) 0

16. ,d fcEc ,d fLFkj lery niZ.k ds lkeus xfr djrk

gSA fcEc ds izfrfcEc dk osx &

(1) fcEc ds osx ds ifjek.k vkSj fn'kk ds leku gSA

(2) fcEc ds osx ds ifjek.k ds leku gS vkSj fn'kk ds

foijhr gSA

(3) fcEc ds osx ds ifjek.k ds leku gS vkSj fn'kk ;k

rks mlds ¼fcEc ds½ leku gksxh ;k mlds ¼fcEc ds½

foijhr gksxhA

(4) fcEc ds osx ds ifjek.k ds leku gS vkSj bldh

fn'kk fcEc ds lkFk ,slk dksbZ Hkh dks.k cukrh gS tks

fcEc dh xfr dh fn'kk ij fuHkZj djrh gSA


®




17. Find the displacement of the ray after it imerges

from CD

(1) 2.5 cm

(2) 5 cm

(3) 1 cm

(4) 13

3 cm

18. Electric flux through a surface of area 100 m2

lying in the xy plane is (in V-m) if

ˆ ˆ ˆE i 2 j 3 k

(1) 100

(2) 141.4

(3) 173.2

(4) 200

19. Two similar very small conducting spheres

having charges 40 C and –20 C are some

distance apart. Now they are touched and kept

at same distance. The ratio of the initial to the

final force between them is :

(1) 8 : 1

(2) 4 : 1

(3) 1 : 8

(4) 1 : 1

17. fdj.k dk CD ls fudyus ds i'pkr~ foLFkkiu gS &

(1) 2.5 cm

(2) 5 cm

(3) 1 cm

(4) 13

3 cm

18. 100 m2 {ks=kQy ds i`"B ls tks xy ry esa j[kk gS]

fuxZr ¶yDl (V-m esa) D;k gksxk ;fn

ˆ ˆ ˆE i 2 j 3 k gSA

(1) 100

(2) 141.4

(3) 173.2

(4) 200

19. cgqr NksVs nks leku pkyd xksys ftu ij vkos'k

40 C vkSj –20 C gS ] dqN nwjh ij j[ks gSaA mu

nksuksa dks Li'kZ djokrs gSa rFkk leku nwjh ij j[krs gSaA

muds chp çkjfEHkd o vfUre cy dk vuqikr gS :

(1) 8 : 1

(2) 4 : 1

(3) 1 : 8

(4) 1 : 1


®




20. A non-conducting disc of mass 2kg, total

charge = +1C uniformly distributed, is placed on

a rough horizontal nonconducting surface with its

cross-section in vertical plane as shown. A

uniform horizontal electric field E is now switched

on. Find maximum magnitude of electric field E

in (N/C) so that the disc rolls purely. [g = 10 ms–2]

(1) 10

(2) 14

(3) 12

(4) 18


This section contains 5 questions.

The answer to each question is a Single Digit

Integer, ranging from 0 to 9 (both inclusive).


chosen.




21. A positive charge +q1 is located to the left of a

negative charge –q2. On a line passing through

the two charges, there are two places where the

total potential is zero. The reference is assumed

to be at infinity. The first place is between the

charges and is 4.00 cm to the left of the negative

charge. The second place is 7.00 cm to the right

of the negative charge. If q2 = –12/11 C, what is

the value of charge q1 in C.

20. fdlh vpkyd pdrh dk nzO;eku 2kg gS rFkk bl ij

+1C vkos'k ,d leku :i ls forfjr gSA bls [kqjnjh

{kSfrt vpkyd lrg ij fp=kkuqlkj bl izdkj j[krs gS

fd bldk vuqizLFk dkV Å/okZ/kj ry esa gksA vc ,d

le:i {kSfrt fo|qr {ks=k E pkyw fd;k tkrk gSA ;g

pdrh 'kq) ykSVuh xfr djs] blds fy;s fo|qr {ks=k E dk

vf/kdre ifjek.k (N/C esa) D;k gksxk? [g = 10 ms–2]

(1) 10 (2) 14 (3) 12

(4) 18

[kaM 2 ¼vf/kdre vad% 20)

bl [k.M esa 5 ç'u gSaA

çR;sd ç'u dk mÙkj 0 ls 9 rd ¼nksuksa 'kkfey½ ds

chp dk ,dy vadh; iw.kk±d gSA

vadu ;kstuk :


gSA




21. ,d /kukos'k +q1 ,d _.kkos'k –q

2 ds cka;h vksj j[kk x;k

gSA nksuksa vkos'kksa ls ,d js[kk xqtjrh gS] bl ij nks

LFkkuksa ij dqy foHko 'kwU; gSA funsZ'k fcUnq dks vuUr ij

ekuk tk;sA izFke LFkku vkos'kksa ds e/; rFkk _.kkos'k ls

4.00 cm ij cka;h vksj gSA nwljk LFkku _.kkos'k ls

7.00 cm ij nka;h vksj gSA ;fn q2 = –12/11 C ]

vkos'k q1 C esa½ dk eku D;k gksxk \


®




22. A ray is incident on a sphere of refractive index

2 as shown in figure. Angle of refraction of the

ray inside sphere is 30º. If total deviation

suffered by the ray is 30xº, then find x.

23. If direct rays only strikes on plane mirror,

distance of first image formed by concave mirror

is 15 cm. from concave mirror, then find .

40 cm 40 cm

O

R = 80 cm

24. A dipole ˆ ˆ(i 2 j)

P3

10–9 C m is placed

at origin. Calculate potential in volt at a point

having coordinate (1m, 2 m).

22. ,d çdk'k dh fdj.k fp=kkuqlkj 2 viorZukad ds

xksys ij vkifrr gks jgh gSA xksys ds vUnj fdj.k dk

viorZu dks.k 30º gSA ;fn fdj.k }kjk cuk;k x;k dqy

fopyu 30xº gS] rks x Kkr djksA

23. ;fn izdk'k fdj.ksa lh/ks dsoy lery niZ.k ls Vdjkrh

gS] rc vory niZ.k }kjk cus izfrfcEc dh vory niZ.k

ls nwjh 15 cm gSA rc Kkr dhft,A

40 cm 40 cm

O

R = 80 cm

24. ,d fo|qr f}/kzqo ftldk f}/kzqo vk?kw.kZ

ˆ ˆ(i 2 j)P

3

10–9 C m dks ewy fcUnq ij

j[kk x;k gS rks (1m , 2 m) fcUnq ij fo|qr foHko dk

eku D;k gksxkA


®




25. In the figure shown a converging lens and a

diverging lens of focal lengths 20 cm and –5 cm

respectively are separated by a distance such

that a parallel beam of light of intensity

100 watt/m2 incident on the converging lens

comes out as parallel beam from the diverging

lens. The intensity of the outgoing beam in

watt/m2 is 200x, find x.

25. n'kkZ;s x;s fp=k esa Øe'k% 20 cm o –5 cm Qksdl

nwjh ds ,d vfHklkjh ySUl o ,d vilkjh ySUl dqN

nwjh ij bl çdkj j[ks gS fd 100 watt/m 2 rhozrk

dk lekUrj çdk'k iqat vfHklkjh ySUl ij vkifrr gksrk

gS vkSj vilkjh ySUl ls lekUrj çdk'k iqUt ds :i esa

ckgj fudyrk gSA fuxZr iqUt dh rhozrk watt/m 2 esa

200x gS rks x Kkr djksA


®

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) – 324005

Website : www.resonance.ac.in | E-mail : [email protected] CJPB*MCT131052020C1-9


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]



Each question has FOUR options (1), (2), (3) and

(4) ONLY ONE of these four option is correct

Marking scheme :

Full Marks : +4 If ONLY the correct option

is chosen.

Zero Marks : 0 If none of the options is

chosen (i.e. the question is

unanswered).


26. Calculate the value of Ecell for the given cell:

Pt(s) |)bar1(

2 )g(H |12

3 3 2(0.1M) (0.1M) (Ksp 4 10 )

CH COOH CH COONa || Ag S(aq) |Ag(s)

Given 2.303RT

F= 0.06 V,Ka(CH3COOH)= 10–5,

0

Ag / AgE = 0.8 V, log 5 = 0.70

(1) 0.459 V (2) 0.623 V (3) 0.878 V (4) None of these

27. Select incorrect statements:

(I) Boric acid is an acid because its

molecule contains replaceable H+ ion.

(II) Boric acid is an acid because its

molecule accepts OH– from water

releasing proton.

(III) Silicon has a strong tendency to form

polymers like silicones. The chain length

of silicone polymer can be controlled by

adding MeSiCl3.

(IV) Silicon has a strong tendency to form

polymers like silicones. The chain length

of silicone polymer can be controlled by

adding Me3SiCl.

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

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

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]



izR;sd iz'u esa pkj fodYi (1), (2), (3) rFkk (4) gSaA bu

pkjksa fodYiksa esa ls dsoy ,d fodYi lgh gSaA

vadu ;kstuk :


gSA




26. fn;s x;s lSy ds fy, ElSy ds eku dh x.kuk djks&

Pt(s) |)bar1(

2 )g(H |12

3 3 2(0.1M) (0.1M) (Ksp 4 10 )

CH COOH CH COONa || Ag S(aq) |Ag(s)

fn;k x;k gS 2.303RT

F= 0.06 V,Ka (CH3COOH) =10–5,

0

Ag / AgE = 0.8 V, log 5 = 0.70

(1) 0.459 V (2) 0.623 V

(3) 0.878 V (4) buesa ls dksbZ ugha

27. xyr dFkuksa dks pqfu;s &

(I) cksfjd vEy ,d vEy gS D;ksafd blds v.kq esa

izfrLFkkiuh; H+ vk;u gksrk gSA

(II) cksfjd vEy ,d vEy gS D;ksafd bldk v.kq ty

ls OH– xzg.k dj izksVkWu fu"dkflr djrk gSA

(III) flfydu dh flfydksuksa tSls cgqyd cukus dh

izcy izof̀Ùk gksrh gSA flfydksu cgqyd dh

Ük`a[kyk dh yEckbZ dks MeSiCl3 feykdj

fu;af=kr fd;k tk ldrk gSA

(IV) flfydu dh flfydksuksa tSls cgqyd cukus dh

izcy izof̀Ùk gksrh gSA flfydksu cgqyd dh

Ük`a[kyk dh yEckbZ dks Me3SiCl feykdj

fu;af=kr fd;k tk ldrk gSA

(1) (I), (III) (2) (II), (IV) (3) (II), (III) (4) (I), (IV)


®




28. The following electrochemical cell is taken.

Cu | Cu2+(aq) || Ag+ (aq) | Ag, and has emf E1 > 0.

By which of the following actions, E1

increases?

(1) Adding NH3 to the cathodic chamber.

(2) Adding HCl to the cathodic chamber.

(3) Adding AgNO3 to the anodic chamber.

(4) Adding NH3 to the anodic chamber.

29. On electrolysis of aqueous CuSO4 solution

using silver electrodes, which of the following

statement is correct ?

(1) The pH of the solution after electrolysis

changes

(2) [Cu2+] decreases

(3) Weight of anode rod decreases

(4) Weight of cathode rod dereases

30. Two liquids A and B form ideal solution.

At 25ºC, vapour pressure of the solution

containing 2 mol of X and 3 mol of Y is

320 torr. If on adding 1 mol of A, the vapour

pressure decreases by 20 torr , then the

vapour pressure of pure A and pure B at

25ºC is :

(1) 400 torr, 500 torr (2) 200 torr, 400 torr


31. The solubility of a solid in H2O at different

temperatures is indicated in the

accompanying diagram. What mass of the

solid will crystallize when 40 mL of a solution

that is saturated at 80º C is cooled to 20º C ?

Solu

bili

ty (

g-s

alt/1

00 m

l H

2O

)

100

75

50

25

20 40 60 80 100

Temperature (ºC)

(1) 12 g (2) 24 g

(3) 30 g (4) 36 g

28. fuEu fo|qrjklk;fud lsy

Cu | Cu2+(aq) || Ag+ (aq) | Ag fy;k x;k gS rFkk

bldk emf E1 > 0 gSA fuEu esa ls fdl izØe }kjk

E1 c<+rk gS \

(1) dSFkksM Hkkx esa NH3 feykus ij

(2) dSFkksM Hkkx esa HCl feykus ij

(3) ,uksM Hkkx esa AgNO3 feykus ij

(4) ,uksM Hkkx esa NH3 feykus ij

29. flYoj bySDVªksM+ dk mi;ksx djds tyh; CuSO4

foy;u dk oS|qr vi?kVu fd;k x;k rc fuEu es ls

dkSulk dFku lgh gSA

(1) oS|qr vi?kVu ds i'Pkkr~ foy;u dh pH

ifjofrZr gksxh

(2) [Cu2+] deh gksxh

(3) ,uksM NM+ ds ekj esa deh gksxhA

(4) dSFkksM NM+ ds ekj esa deh gksxhA

30. nks nzO; A rFkk B vkn'kZ foy;u cukrs gSaA

25ºC ij, 2 eksy X o 3 eksy Y j[kus okyk

foy;u dk ok"i&nkc 320 torr gSA ;fn 1 eksy A

feykus ij] ok"i nkc esa 20 torr rd deh gksrh gS]

rks 25°C ij 'kq) A o 'kq) B dk ok"i&nkc

fuEu gS %



31. fuEu fn;s x;s fp=k esa fofHkUu rki ij H2O esa ,d

Bksl dh foys;rk n'kkZ;h x;h gS tc bl Bksl ds

40 mL lar`Ir foy;u dks 80º C ls 20º C rd

B.Mk fd;k tkrk gS rc Bksl dk fdruk nzO;eku

fØLVyhd`r gksxk \

Solu

bili

ty (

g-s

alt/1

00 m

l H

2O

)

100

75

50

25

20 40 60 80 100

Temperature (ºC)

(1) 12 g (2) 24 g (3) 30 g (4) 36 g


®




32. Which of the following combination has

maximum tendency to form adduct?

(1) BF3 + NH3 (2) BCl3 + NH3

(3) BBr3 + NH3 (4) 3B + NH3

33. The reduction potential of hydrogen half-cell

will be positive, if :

(1) p(H2) = 1 bar and [H+] = 2.0 M

(2) p(H2) = 1 bar and [H+] = 1.0 M

(3) p(H2) = 2 bar and [H+] = 1.0 M

(4) None of these

34. At 300 K, the vapour pressure of an ideal

solution containing 3 mole of A and 2 mole of

B is 600 torr. At the same temperature, if

1.5 mole of A & 0.5 mole of C(non-volatile)

are added to this solution the vapour

pressure of solution increases by 30 torr.

What is the value of 0BP in torr?

(1) 940 (2) 405

(3) 90 (4) 900

35. Which of the following statement about the

zeolite is false?

(1) They are used as cation exchangers

(2) They have open structure which enables

them to take up small molecules

(3) Zeolites are aluminosilicates having three

dimensional network

(4) Some of the 44SiO units in silicates are

replaced by 54AlO and 9

6AlO ions to

form zeolites

32. fuEu esa ls dkSulk la;kstu ;ksxt (adduct) cukus

ds fy, vf/kdre izo`fr j[krk gS \

(1) BF3 + NH3 (2) BCl3 + NH3

(3) BBr3 + NH3 (4) 3B + NH3

33. gkbMªkstu v)Z lSy dk vip;u foHko /kukRed

gksxk] ;fn %

(1) p(H2) = 1 bar rFkk [H+] = 2.0 M

(2) p(H2) = 1 bar rFkk [H+] = 1.0 M

(3) p(H2) = 2 bar rFkk [H+] = 1.0 M

(4) bues ls dksbZ ugh

34. 300 K ij] 3 eksy A rFkk 2 eksy B ;qDr ,d

vkn'kZ foy;u dk ok"inkc 600 Vksj gSA leku rki

ij ;fn 1.5 eksy A rFkk 0.5 eksy C(vok"i'khy)

dks bl foy;u esa feyk;k tkrk gS rks foy;u dk

ok"inkc 30 Vksj c<+rk gSA 0BP dk eku Vksj esa

D;k gS \

(1) 940 (2) 405

(3) 90 (4) 900

35. fuEu esa lsa dkSulk dFku ftvksykbV ds ckjs esa

vlR; gS \

(1) ;g /kuk;u fofue;d (cation exchangers)

ds :i esa iz;qDr gksrk gSA

(2) ;s [kqyh lajpuk j[krs gS tks budks NksVs v.kqvksa

dks xzg.k djus ds fy;s l{ke cukrh gSA

(3) ftvksykbV f=kfofe; tkyd j[kus okys

,Y;qfeuksflfydsV gSA

(4) flyhdsV es dqN 44SiO bdkbZ;ka] 5

4AlO rFkk

96AlO vk;uksa }kjk izfrLFkkfir gksdj

ftvksykbV cukrh gSA


®




36. Which of the following will produce a buffer

solution when mixed in equal volume ?

(1) 0.1 mol dm–3 NH4OH and 0.1 mol dm–3 HCl



(4) 0.1 mol dm–3 CH3COONa and 0.1 mol dm–3 NaOH

37. 250 mL sample of a 0.20 M Cr3+ is

electrolysed with a current of 96.5A. If the

remaining [Cr3+] is 0.1M then the duration of

process is:

(Assume volume remain constant during

process)

(1) 25 sec (2) 225 sec

(3) 150 sec (4) 75 sec

38.

O

Cl HO

HO 2

(1) PhMgBr

(2)H /H O Product

(3º alcohol)

Number of equivalent of PhMgBr consumed

in above reaction is/are:

(1) 1 (2) 2

(3) 3 (4) 4

39. Which of the following compound undergoes

ELECTROPHILIC SUBSTITUTION on

aromatic ring at fastest rate.

(1)

(2)

N

H

(3)

O

(4)

N–H

36. fuEufyf[kr esa ls fdUgsa leku vk;ru esa feyk,

tkus ij cQ+j foy;u izkIr gksxk\

(1) 0.1 mol dm–3 NH4OH rFkk 0.1 mol dm–3 HCl



(4) 0.1 mol dm–3 CH3COONa rFkk 0.1 mol dm–3 NaOH

37. 0.20 M Cr3+ ds 250 mL izkn'kZ dks 96.5 ,Eih;j

/kkjk ds lkFk oS|qr vi?kfVr fd;k tkrk gSA ;fn

[Cr3+] 0.1M 'ks"k jgrk gS] rks izØe dh vof/k gS&

(ekuk izØe ds nkSjku vk;ru fu;r jgrk gS)

(1) 25 sec (2) 225 sec

(3) 150 sec (4) 75 sec

38.

O

Cl HO

HO 2

(1) PhMgBr

(2)H /H O mRikn

(3º ,YdksgkWy)

mijksDRk vfHkfØ;k esa fdrus lerqY;kad PhMgBr

mi;ksfxd gksrs gS \

(1) 1 (2) 2

(3) 3 (4) 4

39. fuEu esa ls dkSulk ;kSfxd rhozre nj ls ,sjkseSfVd

oy; ij bysDVªkWu Lusgh izfrLFkkiu nsrk gSA

(1)

(2)

N

H

(3)

O

(4)

N–H


®




40. Which of the following will not react with

acetyl chloride ?

(1) H2O (2)

(3) (4)

41. Esterification of the acid (P) with the alcohol

(Q) will gives.

(P) ; (Q)

(R– configuration) (±)

(1) only one enantiomer

(2) a mixture of diastereomer

(3) a mixture of enantiomer

(4) only one fraction on fractional distillation

42. Which of the following compound gives

fastest nucleophilic addition reaction :

(1) (2)

(3) (4)

40. buesa esa ls dkSulk ;kSfxd ds lkFk

vfHkfØ;k ugha djsxk ?

(1) H2O (2)

(3) (4)

41. vEy (P) dk ,YdksgkWy (Q) ds lkFk ,sLVjhdj.k

djus ij izkIr gksxkA

(P) ; (Q)

(R– foU;kl) (±)

(1) flQZ ,d izfrfcEc :i leko;oh

(2) foofje :i leko;fo;ksa dk ,d feJ.k

(3) izfrfcEc :i leko;fo;ksa dk ,d feJ.k

(4) izHkkth vklou djus ij izkIr ,d izHkkt

42. fuEu esa ls dkSulk ;kSfxd ukfHkdLusgh ;ksxkRed

vfHkfØ;k rhozre xfr ls nsrk gS%

(1) (2)

(3) (4)


®




43.

3CH Mg Br 3CH Mg Br

3CH Mg Br 2H O

Product.

(1) +

(2) +

(3)

(4) +

44.

O||

HCCH3 + KCN/H

CN

CH3

H OH

In above reaction NC

attackes on

acetaldehyde in its

(1) HOMO (2) LUMO *

(3) LUMO (4) HOMO *

43 3CH Mg Br 3CH Mg Br

3CH Mg Br

2H O mRikn

(1) +

(2) +

(3)

(4) +

44.

O||

HCCH3 + KCN/H

CN

CH3

H OH

mijksDr vfHkfØ;k] esa NC

dk vkØe.k

,sflVSfYMgkbM ds fuEu d{kd esa gksrk gSA

(1) HOMO (2) LUMO *

(3) LUMO (4) HOMO *


®




45. The major product of the reaction is :

C—OH

O 18

+

OH 42SOH.w arm/.Conc

(I)

C—O

O

18

(II)

C—O

O 18

(III)

C—O

O

18

18

(IV)

C—O

O

(1) I & II (2) II & IV

(3) I, II & IV (4) Only II






chosen.




46. A tube of uniform cross sectional area 1 cm2

is closed at one end with semi permeable

membrane. A solution having 6 g glucose per

100 ml is placed inside the tube and dipped

in pure water at 27C. When the equilibrium

is established, what will be the osmotic

pressure, height developed in vertical column

(in meters). Density of glucose solution is

1 gm/ml. (Take R = 0.08 L atm mol–1 k–1,

density of Hg = 13.8 g/cc) (Give your answer

in nearest single digit integer after diving by

12.

47. To get the silicone R3Si–(OSiR

2)

n–SiR

3 having

six Si–O–Si linkage x unit of R2SiCl

2 and y

unit of R3SiCl is taken. Value of (x + y) is:

45. mijksDr vfHkfØ;k dk eq[; mRikn gS&

C—OH

O 18

+

OH 42SOH./. xeZlkUnz

(I)

C—O

O

18

(II)

C—O

O 18

(III)

C—O

O

18

18

(IV)

C—O

O

(1) I rFkk II (2) II rFkk IV

(3) I, II rFkk IV (4) dsoy II





vadu ;kstuk :


gSA




46. 1 cm2 vuqizLr dkV {ks=kQy ;qDr ,d le:i

ufydk dks ,d fljk v)ZikjxE; f>Yyh ls can gSaA

ufydk esa] izfr 100 ml, 6gm Xywdksl ;qDr foy;u

Hkjk tkrk gS rFkk ;g ufydk 27C ij 'kq) ty esa

j[kh tkrh gSA tc lkE; LFkkfor gksrk gS rc]

ijklj.k nkc (ehVj esa) D;k gS (m)Zokgj dkWye dh

izkIr ÅpkbZ ds :i esa) Xywdksl foy;u dk ?kuRo

1 gm/ml gSA

(fyft;s R = 0.08 L atm mol–1 k–1, Hg dk

?kuRo = 13.8 g/cc)

(vius mÙkj dks 12 ls Hkkx ns single digit integer

esa nhft,½

47. N% (six) Si–O–Si cU/ku ;qDr flfydkWu

R3Si–(OSiR

2)

n–SiR

3 izkIr djus ds fy, R

2SiCl

2

dh x bdkbZ o R3SiCl dh y bdkbZ yh tkrh gSA

(x + y) dk eku gS&


®




48. What will be the pH of 0.1 M CH3COONH4 at

25ºC?

49. Total number of carboxylic acid group in

product is :

O

O

O

Br

O

O

O

O

42SOH.w arm/.dil

50. In how many reactions correct major product

is given :

(1) CH3–CH=O

Ph-MgBr

Ether

H2O CH3–CH–Ph

OH

(2) CH3–CH=O

LiAlH4

1 Eq.

D2O CH3–CH2–OD

(3)

OH

COOC2H5

CH3MgBr

1 Eq.

H2O CH4

(4)

O

–CN

HCN

CN

OH

(5)

O

O–C2H5 CH3

H3O + C2H5OH (2 moles)

(6)

CH3–CH2–C–Cl

O

Ph MgBr

THF

H2O

CH3–CH2–C–Ph

OH

Ph

(7) CH3–NH2

CH3–C–O–C–CH3

O O

CH3–NH–C–CH3 + CH3–COOH

O

48. 25°C ij 0.1 M CH3COONH4 dh pH D;k gksxh ?

Ka = 1.8 × 10–5 ; Kb = 1.8 × 10–5

49. mRikn esa dkcksZfDtfyd ,flM lewg dh dqy la[;k

gS \

O

O

O

Br

O

O

O

O

42SOH/xeZruq

50. fuEu esa ls fdruh vfHkfØ;kvksa es lgh eq[; mRikn

fn;k x;k gS &

(1) CH3–CH=O

Ph-MgBr

bZFkj

H2O CH3–CH–Ph

OH

(2) CH3–CH=O

LiAlH4

1 Eq.

D2O CH3–CH2–OD

(3)

OH

COOC2H5

CH3MgBr

1 Eq.

H2O CH4

(4)

O

–CN

HCN

CN

OH

(5)

O

O–C2H5 CH3

H3O + C2H5OH (2 eksy)

(6)

CH3–CH2–C–Cl

O

Ph MgBr

THF

H2O

CH3–CH2–C–Ph

OH

Ph

(7) CH3–NH2

CH3–C–O–C–CH3

O O

CH3–NH–C–CH3 + CH3–COOH

O

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005

Website : www.resonance.ac.in | E-mail : [email protected] MJPBMCT1310520C1-23

Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029

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

PART – C



Each question has FOUR options (1), (2), (3) and

(4) ONLY ONE of these four option is correct

Marking scheme :

Full Marks : +4 If ONLY the correct option

is chosen.


chosen (i.e. the question is

unanswered).


51. The probabilities of different faces of biased

dice to appear are as follows

Face number

1 2 3 4 5 6

Probability

0.1 0.32 0.21 0.15 0.05 0.17

The dice is thrown and it is known that either

the face number 1 or 2 will appear. Then, the

probability of the face number 1 to appear is

(1) 5/21

(2) 5/12

(3) 7/23

(4) 3/10

52. The mean and variance of a binomial variate

X are 4 and 3 respectively. Then P(X 1) =

(1) 1 –

161

4

(2) 1 –

163

4

(3) 1 –

162

3

(4) 1–

161

3

Hkkx – C



izR;sd iz'u esa pkj fodYi (1), (2), (3) rFkk (4) gSaA bu

pkjksa fodYiksa esa ls dsoy ,d fodYi lgh gSaA

vadu ;kstuk :


gSA




51. ,d fu"i{kikrh ikls ds fofHkUu lrgksa dh izkf;drk,a

fuEu gS %

lrg la[;k

1 2 3 4 5 6

izfk;drk

0.1 0.32 0.21 0.15 0.05 0.17

,d iklk Qsadk tkrk gS vkSj bldh lrg 1 ;k 2

vkrh gS rc lrg ds 1 vkus dh izkf;drk gS&

(1) 5/21

(2) 5/12

(3) 7/23

(4) 3/10

52. f}in pj X dk ek/; vkSj izlj.k Øe'k% 4 vkSj 3 gS

rc P(X 1) =

(1) 1 –

161

4

(2) 1 –

163

4

(3) 1 –

162

3

(4) 1–

161

3





53. If P(A) = PA

B

= 1

4 and P

B

A

= 1

2, then

P (A B) =

(1) 1

8

(2) 1

4

(3) 3

4

(4) 7

8

54. Given the relation R = {(1,2), (2,3)} on the set

A = {1,2,3} the minimum number of ordered

pairs which when added to R make it is an

equivalence relation is

(1) 5

(2) 6

(3) 7

(4) 4

55. Let R = {(x, y): x2 + y2 = 1; x, y R} be a

relation in R, then the relation R is:

(1) reflexive

(2) symmetric

(3) transitive

(4) equivalence

53. ;fn P(A) = PA

B

= 1

4 vkSj P

B

A

=1

2 gks]

rks P (A B) =

(1) 1

8

(2) 1

4

(3) 3

4

(4) 7

8

54. fn;k x;k gS fd lEcU/k R = {(1,2), (2,3)} leqPp;

A = {1,2,3} ij ifjHkkf"kr gSA R dks rqY;rk lEcU/k

cukus ds fy, vko';d vfrfjDr U;wure Øfer

;qXeksa dh la[;k gS&

(1) 5

(2) 6

(3) 7

(4) 4

55. ekuk R = {(x, y): x2 + y2 = 1; x, y R}, R esa ,d

lEcU/k gS] rc lEcU/k R gS&

(1) LorqY;

(2) lefer

(3) laØked

(4) rqY;rk





56. If 2 2

2 2y yf 2x , 2x

8 8

= xy, then find

value of 1

28{f(60, 48) + f(80, 48) + f(13, 5)}.

(where x,y > 0)

(1) 2

(2) 4

(3) 6

(4) 8

57. Let f be a real valued function such that

x3x

2002f2)x(f

for all x > 0. The value of

f(2) is

(1) 1000

(2) 2000

(3) 3000

(4) 4000

58. If domain of f(x) is [1, 3], then the domain of

))2x3x((logf 22 will be

(1) [–5, –4] [1, 2]

(2) [–13, –2] [3/5, 5]

(3) [4, 1] [2, 7]

(4) [–3, 2]

56. ;fn 2 2

2 2y yf 2x , 2x

8 8

= xy gks rks 1

28{f(60,

48) + f(80, 48) + f(13, 5)} dk eku gS –

(tgk¡ x,y > 0)

(1) 2

(2) 4

(3) 6

(4) 8

57. lHkh x > 0 ds fy,] ekuk f ,d okLrfod ekuh; Qyu

bl izdkj gS fd x3x

2002f2)x(f

gS] rc f(2)

dk eku gS -

(1) 1000

(2) 2000

(3) 3000

(4) 4000

58. ;fn f(x) dk izkUr [1, 3] gS rc

))2x3x((logf 22 dk izkUr gksxk -

(1) [–5, –4] [1, 2]

(2) [–13, –2] [3/5, 5]

(3) [4, 1] [2, 7]

(4) [–3, 2]





59. If 2

2

9f x 2 x 1 x 2x

x 2x

then

number of integers in domain of f x

(1) 3

(2) 4

(3) 5

(4) 6

60. Given f (x) = 8 8

1 x 1 x

and

g (x) = 4 4

f(sin x) f(cos x) then g(x) is

(1) periodic with fundamental period /2

(2) periodic with fundamental period

(3) periodic with fundamental period 2

(4) aperiodic

61. f : R R, f(x) = 2

2

3x mx n

x 1

, if the range of

this function is [–4, 3) then

(1) m = 0, n = 2

(2) m = 0, n = 4

(3) m = 24 , n = 2

(4) m = 5, n =12

23

59. ;fn 2

2

9f x 2 x 1 x 2x

x 2x

gks] rks

f x ds izkUr esa iw.kk±dksa dh la[;k gS&

(1) 3

(2) 4

(3) 5

(4) 6

60. fn;k x;k gS f (x) = 8 8

1 x 1 x

vkSj

g (x) = 4 4

f(sin x) f(cos x) rc g(x) gS -

(1) /2 ds ewyHkwr vkorZ

(2) ds ewyHkwr vkorZ

(3) 2ds ewyHkwr vkorZ

(4) vukorhZ

61. f : R R, f(x) = 2

2

3x mx n

x 1

;fn bl Qyu

dk ifjlj [–4, 3) gks rc&

(1) m = 0, n = 2

(2) m = 0, n = 4

(3) m = 24 , n = 2

(4) m = 5, n =12

23





62. If , satisfy the equation

tan–1x + tan–1(1 – x) = tan–1 9

7

then the

value of 9(2 + 2) is

(1) 1

(2) –3

(3) 5

(4) 7

63. If two angles of a triangle are 1 1sin

5

and

1 1sin

10

then third angle is

(1) 6

(2) 4

(3) 3

(4) 4

3

64. cot –1 1

–2

+ cot –1 1

–3

is equal to

(1) 3

4

(2) 5

4

(3) 4

(4) –3

4

62. ;fn , lehdj.k tan–1x + tan–1(1 – x) = tan–1

9

7

dks larq"V djrs gS] rc 9(2 + 2) dk eku

gS&

(1) 1

(2) –3

(3) 5

(4) 7

63. ;fn f=kHkqt ds nks dks.k 1 1sin

5

vkSj 1 1sin

10

gS rc rhljk dks.k gS -

(1) 6

(2) 4

(3) 3

(4) 4

3

64. cot –1 1

–2

+ cot –1 1

–3

cjkcj gS&

(1) 3

4

(2) 5

4

(3) 4

(4) –3

4





65. Let 1 1 1cos cos 2 cos 3 x x x = . If x

satisfies the equation 3 2 1 0ax bx cx .

Then the value of 2

a

is

(1) 4

(2) 5

(3) 6

(4) 7

66. A fair coin is tossed 100 times. The

probability of getting tails an odd number of

times is

(1) 1

2

(2) 1

8

(3) 3

8

(4) 1

16

67. A man throws a fair coin number of times and

gets 2 points for each head and 1 point for

each tail the probability that he gets exactly 6

points is

(1) 21

32

(2) 13

64

(3) 43

64

(4) 23

32

65. ekuk 1 1 1cos cos 2 cos 3 x x x = ;fn

x lehdj.k 3 2 1 0ax bx cx dks larq"V

djrk gS rks 2

a dk eku gksxk -

(1) 4

(2) 5

(3) 6

(4) 7

66. ,d fu"i{kikrh ikls dks 100 ckj Qsadk x;k gS]

fo"ke ckj esa iV vkus dh izkf;drk gS&

(1) 1

2

(2) 1

8

(3) 3

8

(4) 1

16

67. ,d O;fDr ,d fu"i{k flDds dks dbZ ckj Qasdrk gS

rFkk fpÙk vkus ij 2 vad rFkk iV vkus ij 1 vad

çkIr djrk gS mlds }kjk Bhd 6 vad çkIr djus

dh çkf;drk gS&

(1) 21

32

(2) 13

64

(3) 43

64

(4) 23

32





68. A = {1, 2, 3, …., 9}. If three numbers are

selected from set A and arrange them, find

the probability that three numbers are either

in increasing order or in decreasing order.

(1) 1

6

(2) 2

3

(3) 1

3

(4) 1

2

69. The probability that atleast one of the events

A, B happens is 0.6. If probability of their

simultaneously happening is 0.2, then

P (A) + P (B) =

(1) 0.4

(2) 0.8

(3) 1.2

(4) 1.4

70. Consider the following statements :

S1 : For an odd function f(x), graph of y = f(x)

always passes through origin.

S2 : If f and g are two bijective function then

f(g(x)) is also bijective.

S3 : All points of intersection of y = f (x) and

y = f – 1 (x) lies on y = x only.

State, in order, whether S1, S2, S3, S4 are

true or false

(1) T F T

(2) T T F

(3) F T T

(4) F F F

68. ekuk A = {1, 2, 3, …., 9} leqPp A ls rhu la[;k,a

pquh tkrh gS rFkk mudksa O;ofLFkr fd;k tkrk gS

izkf;drk gksxh fd rhu la[;k,a ;k rks o/kZeku Øe

gS ;k Ðkleku Øe gS&

(1) 1

6

(2) 2

3

(3) 1

3

(4) 1

2

69. ;fn ?kVukvksa A vkSj B esa ls de ls de ,d ?kVuk

ds ?kfVr gksus dh izkf;drk 0.6 gks rFkk ;fn muds

,d lkFk ?kfVr gksus dh izkf;drk 0.2 gks] rks

P (A) + P (B) =

(1) 0.4

(2) 0.8

(3) 1.2

(4) 1.4

70. fuEu dFkuksa ij /;ku nhft,&

S1 : f(x) fo"ke Qyu ds fy, y = f(x) dk vkjs[k

lnSo ewy fcUnq ls xqtjrk gSA

S2 : ;fn f vkSj g nks ,dSdh vkPNknd Qyu gks]

rc f(g(x)) Hkh ,dSdh vkPNknd Qyu gSA

S3 : y = f (x) vkSj y = f – 1 (x) ds lHkh izfrPNsn

fcUnq dsoy y = x ij fLFkr gksxsaA

Øe esa crk,¡ fd S1, S2, S3 lR; ;k vlR; gSA

(1) T F T

(2) T T F

(3) F T T

(4) F F F










chosen.


chosen (i.e. the question is unanswered).


71. A three digit number, which is multiple of 11,

is chosen at random. Probability that the

number so chosen is also a multiple of 9 is

k

27. Then find the value of k.

Ans: 3

72. The maximum number of equivalence

relations on the set A = {, {} , {{}}} are

Ans. (5)

73. Let f be a function such that f(xy) = y

)x(f for

all positive real number x and y. If f(30) = 20,

then find the value of 3

1f(40).

Ans. 5

74. Let are the roots of the equation

x2 + 9x – 1 + k – k2 = 0; Rk then value of

1

tan1

tantantan 1111 is

Ans. 0

75. If f : R R, defined by f(x) = x5 + e3x and

g(x) is inverse function of f(x), then find the

value of 1

g (1)

Ans. 3





vadu ;kstuk :


gSA




71. rhu vadks dh ,d ,slh la[;k tks 11 ds xq.kt gS

;kn`fPNd :i ls pquh tkrh gSA bl izdkj pquh xbZ

la[;k ds 9 dk Hkh xq.kt gksusa dh izkf;drk k

27 gS

rc k dk eku Kkr dhft,A

72. leqPp; A = {, {} , {{}}} ij rqY;rk lEcU/kks dh

vf/kdre la[;k gS -

73. ekuk fd f ,d Qyu gS tks lHkh /kukRed okLrfod

la[;k x rFkk y ds fy, f(xy) = y

)x(f ;fn

f(30) = 20 rc 3

1f(40) dk eku cjkcj gS&

74. ekuk lehdj.k x2 + 9x – 1 + k – k2 = 0;

Rk ds ewy gS rc

1

tan1

tantantan 1111 dk eku

cjkcj gS -

75. f(x) dk izfrykse Qyu g(x) gS] rks 1

g (1) dk eku

Kkr dhft,A

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005

Website : www.resonance.ac.in | E-mail : [email protected] SOLJPB*MCT131052020C1-1


MAIN PATTERN CUMULATIVE

TEST-1 (MCT-1)

TARGET : JEE (MAIN+ADVANCED) 2021

DATE : 31-05-2020 SET/CODE-1 COURSE : VIJETA (JP) | BATCH : (JPB*)

HINTS & SOLUTIONS

PART : A PHYSICS

1. A thin prism of …………..

Sol. a = 3

12

× A =

A

2

W = 3 / 2

1 A4 / 3

=

A

8

air

water

=

4

1

2. Distance between

Sol.

L

1

f =

31

2

1 1

20 20

FL = 20 cm

1

f =

1

10 –

2

20 = –

1

5

1

v +

1

30 =

1

5

1

v

= 1

30 –

1

5 v = –6 cm

for lens 1

v–

1

30 =

1

20 ySUl ds fy, v = –60 cm,

Hence vr% d = 60 + 6 = 66 cm

3. A thin beam of light

Sol. 2R

–

1

=

1

R

= 2.

4. The distance between Sol. At first position of lens, let the distance of lens from object and

screen be x and y respectively.

x + y = 100 ...........(1)

At second position of lens the distance of lens from object and screen shall be y and x respectively.

y – x = 40 ...........(2) solving equation (1) and (2) we get

y = 70 cm = 70

100 m and x = 30 cm =

30

100m

The power of lens is,

1

f =

1

y +

1

x =

100

70 +

100

30

= 100

21 5 diopters

5. A person AB of height ………….. Sol.

Figure in self explanatory 6. In the figure two

Sol. Ui =

2KQ

2R Uf =

2KQ

2 2R (whole charge will get transfered to

outer sphere)

so heat =

2KQ

2 2R =

2KQ

4R

7. A ring of radius ………….. Sol. Flux will be maximum when maximum length of ring is inside the

sphere.

This will occur when the chord AB is maximum. Now maximum length of chord AB = diameter of sphere. In this case the arc of

ring inside the sphere subtends an angle of 3

at the centre of

ring.

Charge on this arc =R

3

.

=

0

R

3

=

0

R

3

8. Dipole is placed …………..

Sol. Work done in moving a dipole by angle in a given electric field E

W = pE (1 – cos)

= PE (1 – cos 60º) = PE

2

For = 180º W’ = PE (1 – cos 180º) = 2pE = 4W. 9. A graph of the x …………..

Sol. VB – VA = – xE dx = – [Area under Ex – x curve]

VB – 10 = – 1

2.2. (–20) = 20

VB = 30 V.

10. Two smooth spherical …………..




Sol. From conservation of momentum, if speed of sphere A is v,

then speed of sphere B is v

2.

From conservation of energy

221 1 v

mv (2m)2 2 2

=

2 2kQ kQ

5R 2R

or 23

mv4

=

23 kQ

10 R or v =

22 kQ

5 mR

11. Two concentric uniformly ………….. Sol. potential difference due to inner 10C charge

= K 10

2.0

1

1.0

1 = 9 × 1010 (5) = 45 × 1010 = 4.5 ×

1011V

potential difference due to outer charge = 0

p.d. = 4.5 × 1011V 12. The resultant electric ………….. Sol. Even for non uniform charge distribution field may be zero. E.g. when charges on diametrically opposite points are equal. 13. Two spherical ………….. Sol. Let charges on B and C = q Let distance between B and C = d

F =

2

2

Kq

d

When third conductor A is brought in contact with B both will have equal potential (charge will transfer from B to A).

1K(q q )

R

=

1Kq

R

q1 = q/2

When A is brought in contact with C. Let q2 charge in

transfered from C to A.

1 2K(q q )

R

=

2K(q q )

R

q1 – q = 2q2

–q

2 = 2q2 q2 = –q/4

Final charge on B = q/2

Final charge on C = q – q/4 = 3q

4

Force between B & C = 2

K.q.3q

2.4.d

=

2

2

3Kq

8d =

3

8 F

14. Two equal charges …………..

Sol. F = 2

KQq

r

d Q

r

F F

Q

x

q

Fsin Fsin

2Fcos

r = 2 2x d / 4 , cos =

x

r =

2 2

x

x d / 4

Net force on q,

P = 2F cos= 2KQq. 2 2 3 / 2

x

(d / 4 x )

For maximum value of P,

dP

dx= 0 x =

d

2 2 .

15. The fol lowing …………..

Sol. Dispersive power () = b r

y

n n

n 1

=

1.521 1.510

1.550 1

= 0.02

16. An object moves ………….. Sol. When object moves normal to the mirror, image velocity will be

opposite to it. When object moves parallel to the mirror, image velocity will be in the same direction.

17. Find the displacement …………..

Sol. by snell's law 2sin30° = 13 sinr

1 = 13 sinr

sinr = 1

13, tanr =

1

12

So, lateral displacement = t sin(i r)

cosr

= t[sin(i)cosr cos(i)sinr]

cosr

= t[sin i – cos i tanr]

= 10[sin 30 – cos 30 × 1

12]

= 10

32

1

2

3

2

1 = 2.5 cm

18. Electric flux through …………..

Sol. = E . ds = (i + 2 j + 3 k). (100 k) = 173.2 Ans.

19. Two similar very …………..

Sol. Finitial = 2

k(40 C)(20 C)

r

= F1

Final charge on each spheres is 40 20

2

= 10 C

Ffinal =2

k(10 C)(10 C)

r

= F2

1

2

F 8

F 1

20. A non-conducting disc …………..

Sol. aCM = r




qE mg

m

=

2

mgrr

mr / 2

qE – mg = 2mg

qE

m= 3g E =

3 gm

q

=

3 0.2 10 2

1

= 12

21. A positive charge …………..

Sol.

– q 2 q 1

x

4 7

1 2q q

x 4 4

= 0

1 2q q

x 7 7

= 0

1

2

q

q=

x 4

4

1

2

q

q =

x 7

7

x 4

4

=

x 7

7

7x – 28 = 4x + 28 3x = 56

x = 56

3

1

2

q

q

=

567

3

7

=11

3

| q2 | = +12

11 c q1 =

12

11 ×

11

3 = 4 c

22. A ray is incident on …………..

Sol. sini

2sinr

i = 45º

Total deviation = (45º – 30º) + 180º –2(30º) + (45º

– 30º) = 30º + 120º = 150º = 30x x = 5. 23. If direct rays only ………….. Sol.

40 cm 40 cm

O

40 cm

1

1 1 2

V 120 80

1 1 1 3

V 120 40 3

=

2

120

V = –60 cm

24. A dipole

ˆ ˆ(i 2 j)P

3

10–9 C …………..

Sol.

9 –99 10 (1 2) 10

( 3 ) (3 3 )

= 3 V.

25. In the f igure shown ………….. Sol.

()out =

2

0

2

0

100 ( r )

r

4

= 1600 W/m2

PART : B CHEMISTRY

26. Calculate …………

Sol. Anode : 1

2H2 H+ + e–

Cathode : Ag+ + e– Ag(s)

1

2H2(g) + Ag+ (aq) H+(aq) + Ag(s)

Ecell = 0

cell

0.06 [H ]E log

1 [Ag ]

Ag2S 2Ag+ + S–2

Ksp = 4 × 10–12 = [Ag+]2 [S2–]

[Ag+] = 2 ×10–4 M

pH = pKa + log

salt

acid

= 5 + log0.1

0.1

= 5

Ecell = 0.8 – 0.06

1 log

5

4

10

2 10

= 0.878 V

28. The following ……………




Sol. Cu | Cu2+(aq) || Ag+ (aq) | Ag. E1 > 0.

(1) Due to NH3 addition, concentration of Ag+

decreases due to complex formation. Hence,

E decrease.

(2) Due to addition of HCl, Ag+ is consumed

to form AgCl solid. Concentration of Ag+

decreases. So, E decreases.

(3) Adding AgNO3 to anodic chamber result

in direct reaction between Cu and Ag+

Cu + 2Ag+ Cu2+ + 2Ag

More Cu2+ produced, hence E decreases.

(4) Adding NH3 in anodic chamber results in

complex formation, [Cu(NH3)

4 ]2+. Hence Cu2+

concentration decreases. So E increases.

29. On electrolysis ……………

Sol. At anode : Ag(s) Ag+(aq.) + e–

At cathode : Ag+(aq.) + e– Ag(s)

pH = no change

[Cu2+] = no change

30. Two liquids ……………

Sol. Moles of A = 2 XA = 0.4

Moles of B = 3 XB = 0.6

PT = PA0 XA + PB

0 XB

320 = PA0 × 0.4 + PB

0 × 0.6

3200 = 4PA0 + 6PB

0 (1)

On adding 1 mole of A, XA = 0.5, XB = 0.5

(320 –20) = PA0 × 0.5 + PB

0 × 0.

3000 = 5PA0 + 5PB

0

600 = PA0 + PB

0 (2)

From (1) and (2)

PA0 = 200 torr

PB0 = 400 torr

31. The solubility ……………

Sol. 80–100

0–100

x–100

50–100x = 90 at 80º C

20–100

0–100

y–100

50–100 y = 60 at 20º C

Precipitation = 90 – 60 = 30g for 100 mL

For 40 ml g12400

4030

32. Which of ……………

Sol. 3B is best Lewis acid, due to least back bonding.

Hence it has greater tendency to accept lone pair.

33. The reduction ……………

Sol. 2H+ (aq) + 2e– H2 (g)

Ered = Eºred – 0.0591

nlog

H22

P

(H )

Ered is found to be positive for (1) 34. At 300 K ……………

Sol. 600 =

5

3P0

A +

5

2P0

B ….(1)

630 =

7

5.4P0

A +

7

2P0

B ….(2)

by solving 0BP = 90 Torr

36. Which of …………… Sol. Basic buffer NH4OH + NH4Cl will be formed by

reaction. 37. 250 mL …………… Sol. Initial moles of Cr3+ = 0.25 × 0.2 = 0.05 mol Final moles of Cr3+ = 0.25 × 0.1 = 0.025 mol Therefore moles of Cr3+ reduced is : 0.05 – 0.025 = 0.025 mol

or eq. of Cr3+ reduced 0.025×3 = t 96.5

96500

t = 75 sec 39. Which of …………… Sol. (2) Most e– rich ring will show fastest rate towards

electrophile.

41. Esterification ……………

Sol. +




+

46. A tube of ……………

Sol. V = nRT

30008.0180

61.0

= 8.0 atm

= h × d × g 8.0 × 76 × 13.8 × 981 = h × 1 × 981 h = 83.9 m 47. To get the …………… Sol. x = 5, y = 2

48. What will ……………

Sol. bKaKWKH p–pp2

1p

bKaKpp

7)14(2

1p

2

1p WKH

50. In how many ……………

Sol. In (3) acid base reaction takes place.

PART : C MATHEMATICS

51. The probabilities.................. Sol. The probability of the face number 1 to

appear = P(1)

P(1) P(2)

= 0.1

0.1 0.32 =

5

21

52. The mean and ...................... Sol. Mean = np = 4 Variance = npq = 3

n = 16, 3 1

q , p4 4

So P(X 1) = 1 – P(X = 0) =

163

1–4

53. If P(A) = PA

B

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

Sol. Since P(A) = PA

B

, therefore, A and B are independent events.

Thus P(B) = PB

A

=1

2

P (A B) = 1 – P (A B) = 1 – P (A B)

= 1 – [P(A) – P(A B)]

= 1 – P(A) + P(A) . P(B) = 7

8

54. Given the relation................. Sol. R is reflexive if it contains (1, 1), (2, 2), (3, 3)

(1,2) R, (2,3) R.

R is symmetric if (2,1), (3,2) R. Now R = {(1,1), (2,2), (3,3), (2,1), (3,2),(2,3),(1,2)}

R will be transitive if (3,1), (1,3) R. Thus R becomes an equivalence relation by adding (1,1), (2,2), (3,3) (2,1),(3,2), (1,3), (3,1).

Hence the total no. of ordered pairs is 7.

55. Let R = {(x, y): x2 ......................

Sol. R = {(x, y) : x2 + y2 = 1; x, y R}

Let 1 R and 12 + 12 = 2 1

(1, 1) R R is not reflexive

Let (x, y) R

x2 + y2 = 1 y2 + x2 = 1 (y, x) R

R is symmetric

(0, 1) and (1, 0) R but (0, 0) R as 02 + 02 1

R is not transitive

56. If

2 22 2y y

f 2x , 2x8 8

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

Sol.

2 22 2y y

f 2x , 2x8 8

= xy =

2 22 2

2 2y y2x 2x

8 8

1

28{f(60, 48) + f(80, 48) + f(13, 5)}

= 1

28

2 2 2 2 2 260 48 80 48 13 5

= 1

28[36 + 64 + 12] = 4

57. Let f be a real ...................... Sol. Putting x = 2, we get f(2) + 2f (1001) = 6 Putting x = 1001, we get f(1001) + 2f(2) = 3003 Solving for f(2), we get f(2) = 2000 58. If domain of f(x) ......................

Sol. 3)2x3x(log1 22

8)2x3x(2 2

2x1or4x5




59. If 2

2

9f x 2 x 1 x 2x

x 2x

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

Sol.

22

2

x 2 x 92 x 1 0

x 2 x

x 1 x 3

0x x 2

0 x 1 or 2 x 3

60. Given f (x) ......................

Sol. f (x) = 2

4

1 x; f (sinx) =

4

| cos x | and

f (cosx) = 4

| sin x | ;

hence g (x) = | sin x | + | cos x |

(1) ]

61. f : R R, f(x) ......................

Sol. y =

2

2

3x mx n

x 1

x2(y – 3) – mx + (y + n) = 0

x R, D 0

m2 – 4(y – 3)(y + n) 0

m2 – 4(y2 + ny – 3y – 3n) 0

4y2 – 4y(–n + 3) – 12n – m2 0 ......(1)

(y + 4)(y – 3) 0 comparing (1) & (2) , (1) & (2) m = 0, n = – 4 (m – n) = 4

62. If , satisfy ......................

Sol. tan–1

)x–1(x–1

x–1x = tan–1

7

9

tan–1

1x–x

12

= tan–1

7

9

x2 – x + 1 = 9

7

9x2 – 9x + 2 = 0

9x2 – 6x – 3x + 2 = 0

(3x – 1)(3x – 2) = 0

= 3

1, =

3

2

so 9(2 + 2) = 5 63. If two angles...................... Sol. Required angle

1 11 1sin sin

10 5

1 1 1 11 1 1 6 1 3cos cos cos cos

410 5 5 2 5 2 2

64. cot –1 1

–2

+ cot ......................

Sol. – cot–1 1

2

+ – cot –1 1

3

2– [ cot –1 1

2

+ cot –1 1

3

]

= + tan –1 1

2

+ tan –1 1

3

= + tan –1 1/ 2 1/3

1– 1/ 6

= + 4

=

5

4

65. Let 1 1 1cos cos 2 cos 3 x x x ............

Sol. 1 1cos 2 cos 3 x x = – cos–1x

cos–1 (6x2 – 21 4x .

21 9x ) = cos–1 (–x)

(6x2 – 21 4x .

21 9x ) = (–x)

12x3 + 14x2 –1 = 0. Hence a = 12 66. A fair coin is...................... Sol. The total number of cases is 2100. The number of favourable

ways is 100C1 + 100C3 + ............ + 100C99 = 2100–1 = 299.

Therefore the probability of the required event is

99

100

2

2 =

1

2

67. A man throws ...................... Sol. To get 6 points he needs atleast 3 throws and at most 6 throws (2,2,2), (2,2,1,1), (2,1,1,1,1), (1,1,1,1,1,1) so required probability

= 1

8 +

4

2 2.

1

16+

5

4.

1

32 +

1

64

68. A = {1, 2, 3, …., 9}....................... Sol. Favourable cases = 2.9C3 Total arrangement = 9 × 8 × 7

Required probability =

932 C

9 8 7

=

1

3

69. The probability ......................

Sol. P(A B) = 0.6

P(A B) = 0.2

P(A) + P(B) = P(A B) + P(A B) = 0.6 + 0.2 = 0.8.

P (A) + P (B) = 0.4 +0.8 = 1.2

70. Consider the ...................... Sol. S1 & S2 :Obviously S3 : y = f(x) and y = f–1(x) can intersect at points other than y = x

e.g. y = – x + c or y = 21 x

71. A three digit ...................... Sol: Three digit no. divisible by 11 will be {10×11, 11×11,........,90×11} = 81 Numbers the no.that is also divisible by 9 {18×11, 27×11,........,90×11} = 9 Numbers

Hence 9 1 k

p81 9 27

= = =

k = 3 72. The maximum ...................... Sol. Possible equivalence relations on A are as follows

R1 = {(, ), ({},{}),({{}},{{}})}




R2 = R1 {(, {}), ({}, )}

R3 = R1 {(, {{}}), ({{}}, )}

R4 = R1 {({}, {{}}), ({{}}, {})} R5 = A × A (universal relation)

73. Let f be a function......................

Sol. Given f(xy) = y

)x(f

put x = 1, f(y) = y

)1(f

or f(30) = 30

)1(f f(1) = 30 × f(30) = 30 × 20 = 600

f(40) = 40

)1(f =

40

600 = 15

74. Let are ......................

Sol. 01kk2

so negative

022

1tantan

1tantan 1111

75. If f : R R, .................... Sol. f(0) = 1 g(1) = 0

Now g(f(x)) = x g(f(x)) . f(x) = 1

g(f(x)) = 1

f (x) g(f(0)) =

1

f (0)

g(1) = 1

f (0) =

1

3

1

g (1) = 3.





TEST-1 (MCT-1)

TARGET : JEE (MAIN+ADVANCED) 2021DATE : 31-05-2020 SET/CODE-1 COURSE : VIJETA (JP) | BATCH : (JPB*)

ANSWER KEY

SET/CODE-1

PART : A PHYSICS

1. (4) 2. (1) 3. (2) 4. (2) 5. (1) 6. (1) 7. (4)

8. (3) 9. (4) 10. (1) 11. (1) 12. (2) 13. (4) 14. (3)

15. (2) 16. (4) 17. (1) 18. (3) 19. (1) 20. (3) 21. (4)

22. (5) 23. (4) 24. (3) 25. (8)

PART : B CHEMISTRY

26. (3) 27. (1) 28. (4) 29. (3) 30. (2) 31. (1) 32. (4)

33. (1) 34. (3) 35. (4) 36. (3) 37. (4) 38. (4) 39. (2)

40. (4) 41. (2) 42. (3) 43. (3) 44. (2) 45. (2) 46. (7)

47. (7) 48. (7) 49. (4) 50. (6)


51. (1) 52. (2) 53. (4) 54. (1) 55. (2) 56. (2) 57. (2)

58. (1) 59. (2) 60. (1) 61. (2) 62. (3) 63. (4) 64. (2)

65. (3) 66. (1) 67. (3) 68. (3) 69. (3) 70. (4) 71. (3)

72. (5) 73. (5) 74. (0) 75. (3)





TEST-1 (MCT-1)

TARGET : JEE (MAIN+ADVANCED) 2021DATE : 31-05-2020 SET/CODE-1 COURSE : VIJETA (JP) | BATCH : (JPB*)

ladsr ,oa gy

PART : A PHYSICS

1. ,d dk¡p dk ……………

Sol. a = 3

12

× A =

A

2

W = 3 / 2

1 A4 / 3

=

A

8

air

water

=

4

1

2. fn;s x;s ySUl ……………

Sol.

L

1

f =

31

2

1 1

20 20

FL = 20 cm

1

f =

1

10 –

2

20 = –

1

5

1

v +

1

30 =

1

5

1

v =

1

30 –

1

5 v = –6 cm

1

v–

1

30 =

1

20 ySUl ds fy, v = –60 cm,

vr% d = 60 + 6 = 66 cm

3. çdk'k dk ,d ……………

Sol. 2R

–

1

=

1

R

= 2.

4. oLrq vkSj ijns ……………

Sol. ySal dh igyh fLFkfr esa] ;fn oLrq dh ySal vkSj insZ ls nwjh

Øe'k% x vkSj y gS

x + y = 100 ...........(1)

ySal dh nwljh fLFkfr esa ySal dh oLrq vkSj inkZ ls nwjh Øe'k%

y vkSj x gksxh

y – x = 40 ...........(2)

(1) vkSj (2) dks gy djus ij] gesa feyrk gS

y = 70 cm = 70

100 m vkSj x = 30 cm =

30

100m

ySal dh 'kfDr gksxh

1

f =

1

y +

1

x =

100

70 +

100

30

= 100

21 5

5. ,d O;fDr ftldh ……………

Sol.

fp=k Lo;a gh le> esa vkuk gSA

6. fp=k esa R o 2R ……………

Sol. Ui =

2KQ

2R Uf =

2KQ

2 2R(dqy vkos'k ckg~; xksys ij LFkkukarfjr)

vr% Å"ek =

2KQ

2 2R =

2KQ

4R

7. ,d oy; ftldh ……………

Sol. tc oy; dh vf/kdre yEckbZ xksys ds vUnj gksxh] rc ¶yDl Hkh

vf/kdre gksxh

;g rHkh gksxk tc thok AB vf/kdre gksxkA thok AB dh vf/kdre

yEckbZ = xksys dk O;klA bl fLFkfr esa xksys ds vUnj oy; dk pki]

oy; ds dsUæ ij 3

dks.k cukrk gSA

bl pki ij vkos'k =R

3

.

=

0

R

3

=

0

R

3

8. ,d f}/kzqo fo|qr ……………

Sol. fn;s x;s fo|qr {ks=k E esa f}/kzqo dks dks.k ls ?kqekus esa fd;k x;k

dk;Z

W = pE (1 – cos)

= PE (1 – cos 60º) = PE

2

For = 180º

W’ = PE (1 – cos 180º) = 2pE = 4W.

9. fo|qr {ks=k ds ……………

Sol. VB – VA = – xE dx = –[Ex – x oØ ls f?kjk gqvk {ks=kQy ]

VB – 10 = – 1

2.2. (–20) = 20

VB = 30 V.




10. nks fpdus xksyh; ……………

Sol. laosx ds laj{k.k ls] ;fn A xksys dh pky v gS] rks B xksys dh pky

v

2 gksxhA

ÅtkZ ds laj{k.k ls]

221 1 v

mv (2m)2 2 2

=

2 2kQ kQ

5R 2R

or 23

mv4

=

23 kQ

10 R or v =

22 kQ

5 mR

11. nks ladsUnzh ,d ……………

Sol. vkUrfjd 10C vkos'k ds dkj.k foHkokUrj

= K 10

2.0

1

1.0

1 = 9 × 1010 (5) = 45 × 1010 = 4.5 ×

1011V

ckº; vkos'k ds dkj.k foHkokUrj = 0

p.d. = 4.5 × 1011V

12. oy; ds dkj.k ……………

Sol. vleku vkos'k forj.k ds fy, {ks=k 'kwU; gks ldrk gSA

tc vkos'k O;klr% forjhr fcUnqvksa ij leku gks

13. nks leku xksyh; ……………

Sol. ekuk fd B vkSj C dk vkos'k = q

ekuk fd B vkSj C ds chp dh nwjh = d

F =

2

2

Kq

d

tc rhljs pkyd A dks B ds lEidZ esa yk;k x;k] rks nksuksa dk

foHko cjkcj gksxk (vkos'k B ls A esa tk;sxk).

1K(q q )

R

=

1Kq

R

q1 = q/2

Tc A dks C ds lEidZ esa yk;k x;k] ekuk fd q2 vkos'k C ls A esa

x;kA

1 2K(q q )

R

=

2K(q q )

R

q1 – q = 2q2

–q

2 = 2q2 q2 = –q/4

B ij vafre vkos'k = q/2

C ij vafre vkos'k = q – q/4 = 3q

4

B vkSj C ds chp u;k cy = 2

K.q.3q

2.4.d

=

2

2

3Kq

8d =

3

8 F

14. nks leku vkos'kksa ……………

Sol. F = 2

KQq

r

d Q

r

F F

Q

x

q

Fsin Fsin

2Fcos

r = 2 2x d / 4 , cos =

x

r =

2 2

x

x d / 4

q ij dqy cy,

P = 2F cos= 2KQq. 2 2 3 / 2

x

(d / 4 x )

P ds vf/kdre eku ds fy;s

dP

dx= 0 x =

d

2 2 .

15. Økmu dk¡p ds ……………

Sol. fo{ksi.k {kerk () = b r

y

n n

n 1

=

1.521 1.510

1.550 1

= 0.02

16. ,d fcEc ,d ……………

Sol. tc oLrq niZ.k ds yEcor~ xfr djrh gS] izfrfcEc dk osx blds

yEcor~ gksxk tc oLrq niZ.k ds lekUrj xfr djrh gS rks izfrfcEc dk

osx mlh leku fn'kk esa gksxkA

17. fdj.k dk CD ls

Sol. Lusy ds fu;e ls 2sin30° = 13 sinr

1 = 13 sinr

sinr = 1

13, tanr =

1

12

ik'oZ foLFkkiu = t sin(i r)

cosr

= t[sin(i)cosr cos(i)sinr]

cosr

= t[sin i – cos i tanr]

= 10[sin 30 – cos 30 × 1

12]

= 10

32

1

2

3

2

1 = 2.5 cm

18. 100 m2 {ks=kQy ……………

Sol. = E . ds = (i + 2 j + 3 k). (100 k) = 173.2 Ans.

19. cgqr NksVs nks ……………

Sol. FizkjfEHkd = 2

k(40 C)(20 C)

r

= F1

izR;sd xksys ij vafre vkos'k 40 20

2

= 10 C




Fvafre = 2

k(10 C)(10 C)

r

= F2

1

2

F 8

F 1

20. fdlh vpkyd ……………

Sol. aCM = r

qE mg

m

=

2

mgrr

mr / 2

qE – mg = 2mg

qE

m= 3g E =

3 gm

q

=

3 0.2 10 2

1

= 12

21. ,d /kukos'k +q1 ……………

Sol.

– q 2 q 1

x

4 7

1 2q q

x 4 4

= 0

1 2q q

x 7 7

= 0

1

2

q

q=

x 4

4

1

2

q

q =

x 7

7

x 4

4

=

x 7

7

7x – 28 = 4x + 28 3x = 56

x = 56

3 1

2

q

q =

567

3

7

=11

3

| q2 | = +12

11 c q1 =

12

11 ×

11

3 = 4 c

22. ,d çdk'k dh fdj.k ……………

Sol. sini

2sinr

i= 45º

dqy fopyu = (45º – 30º) + 180º –2(30º) + (45º – 30º) = 30º

+ 120º = 150º = 30x x = 5.

23. ;fn izdk'k fdj.ksa ……………

Sol.

40 cm 40 cm

O

40 cm

1

1 1 2

V 120 80

1 1 1 3

V 120 40 3

=

2

120

V = –60 cm

24. ,d fo|qr f}/kzqo ……………

Sol.

9 –99 10 (1 2) 10

( 3 ) (3 3 )

= 3 V.

25. n'kkZ;s x;s ……………

Sol.

()out =

2

0

2

0

100 ( r )

r

4

= 1600 W/m2

PART : B CHEMISTRY

26. fn;s x;s ………….

Sol. (,uksM+) : 1

2H2 H+ + e–

(dSFkksM+) : Ag+ + e– Ag(s)

1

2H2(g) + Ag+ (aq) H+(aq) + Ag(s)

Ecell = 0

cell

0.06 [H ]E log

1 [Ag ]

Ag2S 2Ag+ + S–2

Ksp = 4 × 10–12 = [Ag+]2 [S2–]

[Ag+] = 2 ×10–4 M

pH = pKa + log

salt

acid

= 5 + log0.1

0.1

= 5




Ecell = 0.8 – 0.06

1 log

5

4

10

2 10

= 0.878 V

28. fuEu ………….

Sol. Cu | Cu2+(aq) || Ag+ (aq) | Ag. E1 > 0.

(1) NH3 feykus ij ladqy fuekZ.k ds dkj.k Ag+ dh

lkUnzrk ?kVrh gSA vr% E ?kVrk gSA

(2) HCl feykus ij Ag+ Bksl AgCl fuekZ.k esa

iz;qDr gksrk gSA ftlls Ag+ dh lkUnzrk ?kV tkrh

gSA blfy, E ?kVrk gSA

(3) ,uksM Hkkx esa AgNO3 feykus ds ifj.kkeLo:i

Cu o Ag+ ds e/; lh/kh vfHkfØ;k (direct

reaction) gksrh gSA

Cu + 2Ag+ Cu2+ + 2Ag

vf/kd Cu2+ mRiUu gksrs gSa vr% E ?kVsxkA

(4) ,uksM Hkkx esa NH3 feykus ij ladqy

[Cu(NH3)

4 ]2+ fuekZ.k gksrk gS vr% Cu2+ dh

lkUnzrk ?kVrh gS blfy, E c<+rk gSA

29. flYoj bySDVªksM+ ………….

Sol. ,uksM ij : Ag(s) Ag+(aq.) + e–

dSFkksM ij : Ag+(aq.) + e– Ag(s)

pH = dksbZ ifjorZu ugha

[Cu2+] = dksbZ ifjorZu ugha

30. nks nzO; ………….

Sol. A ds eksy = 2 XA = 0.4

B ds eksy = 3 XB = 0.6

PT = PA0 XA + PB

0 XB

320 = PA0 × 0.4 + PB

0 × 0.6

3200 = 4PA0 + 6PB

0 (1)

1 eksy A dks feykus ij, XA = 0.5, XB = 0.5

(320 –20) = PA0 × 0.5 + PB

0 × 0.5

3000 = 5PA0 + 5PB

0

600 = PA0 + PB

0 (2)

(1) o (2) ls

PA0 = 200 torr

PB0 = 400 torr

31. fuEu fn;s………….

Sol. 80–100

0–100

x–100

50–100x = 90 at 80º C

20–100

0–100

y–100

50–100 y = 60 at 20º C

vo{ksi.k = 90 – 60 = 30g for 100 mL

For 40 ml g12400

4030

32. fuEu esa ls ………….

Sol. 3B U;wure i'p ca/ku ds dkj.k Js"B yqbZl vEy

gSA blfy, ;g ,dkdh ;qXe xzg.k djus dh vf/kd

izo`fr j[krk gSA

33. gkbMªkstu ………….

Sol. 2H+ (aq) + 2e– H2 (g)

Evi- = Eºvi- – 0.0591

nlog

H22

P

(H )

fodYi (1) esa Evipf;r dk eku /kukRed ik;k tkrk

gSA

34. 300 K ij ………….

Sol. 600 =

5

3P0

A +

5

2P0

B ….(1)

630 =

7

5.4P0

A +

7

2P0

B ….(2)

gy djus ij 0BP = 90 Torr

36. fuEufyf[kr ………….

Sol. {kkjh; cQj foy;u NH4OH + NH4Cl vfHkfØ;k

}kjk cusxkA

37. 0.20 M Cr3+ ………….

Sol. Cr3+ ds izkjfEHkd eksy = 0.25×0.2 = 0.05 mol

Cr3+ ds vfUre eksy = 0.25 × 0.1 = 0.025 mol

blfy, Cr3+ ds vipf;r eksy gS&

0.05 – 0.025 = 0.025 mol

;k vipf;r Cr3+ ds rqY;kad 0.025×3 = t 96.5

96500

t = 75 sec





Sol. lokZf/kd bysDVªkWu /kuh oy; bysDVªkWuLusgh ds izfr

rhozre nj n'kkZ;sxhA

41. vEy (P) ………….

Sol. +

(foofje

leko;oh)

+

46. 1 cm2 vuqizLr ………….

Sol. V = nRT

30008.0180

61.0

= 8.0 atm

= h × d × g 8.0 × 76 × 13.8 × 981 = h × 1 × 981 h = 83.9 m

47. N% (six) ………….

Sol. x = 5, y = 2

48. 25°C ij ………….

Sol. bKaKWKH p–pp2

1p

bKaKpp

7)14(2

1p

2

1p WKH


Sol. (3) esa vEy {kkj vfHkfØ;k gksrh gSA


51. ,d fu"i{kikrh ......................

Sol. lrg 1 vkus dh izkf;drk

= P(1)

P(1) P(2)=

0.1

0.1 0.32 =

5

21

52. f}in pj X dk ......................

Sol. ek/; = np = 4

izlj.k = npq = 3

n = 16, 3 1

q , p4 4

blhfy, P(X 1) = 1 – P(X = 0) =

163

1–4

53. ;fn P(A) = PA

B

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

Sol. pwafd P(A) = PA

B

, vr% A vkSj B Lora=k ?kVuk,¡ gSaA

vr% P(B) = PB

A

=1

2

P (A B) = 1 – P (A B) = 1 – P (A B)

= 1 – [P(A) – P(A B)]

= 1 – P(A) + P(A) . P(B) = 7

8

54. fn;k x;k gS fd ......................

Sol. R LorqY; gksxk ;fn blesa (1, 1), (2, 2), (3, 3) fo|eku gksA

(1,2) R, (2,3) R.

R lefer gksxk ;fn (2,1), (3,2) R.

vc, R = {(1,1), (2,2), (3,3), (2,1), (3,2),(2,3),(1,2)}

R laØked gksxk ;fn (3,1), (1,3) R vr% R dks rqY;rk lEcU/k

cukus ds fy, vfrfjDr vko';d Øfer ;qXe (1,1), (2,2), (3,3)

(2,1),(3,2), (1,3), (3,1) gksxsaA

vr% dqy Øfer ;qXeksa dh la[;k 7 gksxhA

55. ekuk R = {(x, y): ......................

Sol. R = {(x, y) : x2 + y2 = 1; x, y R}

ekuk 1 R vkSj 12 + 12 = 2 1

(1, 1) R R LorqY; ugha gSA

ekuk (x, y) R

x2 + y2 = 1 y2 + x2 = 1 (y, x) R

R lefer gSA

(0, 1) vkSj (1, 0) R ijUrq (0, 0) R D;kasfd 02 + 02 1

R laØked ugha gSA

56. ;fn

2 22 2y y

f 2x , 2x8 8

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

Sol.

2 22 2y y

f 2x , 2x8 8




= xy =

2 22 2

2 2y y2x 2x

8 8

1

28{f(60, 48) + f(80, 48) + f(13, 5)}

= 1

28

2 2 2 2 2 260 48 80 48 13 5

= 1

28[36 + 64 + 12] = 4

57. lHkh x > 0 ds fy, ...................

Sol. x = 2 j[kus ij

f(2) + 2f (1001) = 6

x = 1001 j[kus ij

f(1001) + 2f(2) = 3003

(2) dks gy djus ij f(2) = 2000

58. ;fn f(x) dk izkUr........................

Sol. 3)2x3x(log1 22

8)2x3x(2 2

2x1or4x5

59. ;fn 2

2

9f x 2 x 1 x 2x

x 2x

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

Sol.

22

2

x 2 x 92 x 1 0

x 2 x

x 1 x 3

0x x 2

0 x 1 ;k 2 x 3

60. fn;k x;k gS ......................

Sol. f (x) = 2

4

1 x; f (sinx) =

4

| cos x | and

f (cosx) = 4

| sin x | ;

g (x) = | sin x | + | cos x |

(1) ]

61. f : R R, f(x) ......................

Sol. y =

2

2

3x mx n

x 1

x2(y – 3) – mx + (y + n) = 0

x R, D 0

m2 – 4(y – 3)(y + n) 0

m2 – 4(y2 + ny – 3y – 3n) 0

4y2 – 4y(–n + 3) – 12n – m2 0 ......(1)

(y + 4)(y – 3) 0

(1) & (2) dh rqyuk ls

m = 0, n = – 4 (m – n) = 4

62. ;fn , lehdj.k................

Sol. tan–1

)x–1(x–1

x–1x = tan–1

7

9

tan–1

1x–x

12

= tan–1

7

9

x2 – x + 1 = 9

7

9x2 – 9x + 2 = 0

9x2 – 6x – 3x + 2 = 0

(3x – 1)(3x – 2) = 0

= 3

1, =

3

2

blfy, 9(2 + 2) = 5

63. ;fn f=kHkqt ds .....................

Sol. vHkh"V dks.k 1 11 1

sin sin10 5

1 1 1 11 1 1 6 1 3cos cos cos cos

410 5 5 2 5 2 2

64. cot –1 1

–2

+ cot ......................

Sol. – cot–1 1

2

+ – cot –1 1

3

2– [ cot –1 1

2

+ cot –1 1

3

]

= + tan –1 1

2

+ tan –1 1

3

= + tan –1 1/ 2 1/3

1– 1/ 6

= + 4

=

5

4

65. ekuk 1 1 1cos cos 2 cos 3 x x x ..........

Sol. 1 1cos 2 cos 3 x x = – cos–1x

cos–1 (6x2 – 21 4x .

21 9x ) = cos–1 (–x)

(6x2 – 21 4x .

21 9x ) = (–x)

12x3 + 14x2 –1 = 0. vr% a = 12

66. ,d fu"i{kikrh......................

Sol. dqy fLFkfr;ksa dh la[;k 2100 gSA vuqdqy fLFkfr;ka 100C1 + 100C3 + ............ + 100C99 = 2100–1 = 299.

vHkh"V ?kVuk dh izkf;drk

99

100

2

2 =

1

2 gSA

67. ,d O;fDr ,d ......................

Sol. 6 vad çkIr djus ds fy, flDds dks de ls de 3 ckj rFkk vf/kdre

6 ckj Qsaduk iMsxkA

(2,2,2), (2,2,1,1), (2,1,1,1,1), (1,1,1,1,1,1)

vHkh"B çkf;drk = 1

8 +

4

2 2.

1

16+

5

4.

1

32 +

1

64




68. ekuk A = {1, 2, 3, …., 9}...............

Sol. vuqdwy fLFkfr;ka = 2.9C3

dqy Øep; = 9 × 8 × 7

vHkh"V izkf;drk =

932 C

9 8 7

=

1

3

69. ;fn ?kVukvksa A vkSj .................

Sol. P(A B) = 0.6

P(A B) = 0.2

P(A) + P(B) = P(A B) + P(A B) = 0.6 + 0.2 = 0.8.

P (A) + P (B) = 0.4 +0.8 = 1.2

70. fuEu dFkuksa ij ....................

Sol. S1 rFkk S2 : Li"Vr;k

S3 : y = f(x) vkSj y = f–1(x), y = x ds vykok vU; fcUnqvksa ij Hkh

izfrPNsn dj ldrs gSA

mnkgj.kkFkZ y = – x + c

;k y = 21 x

71. rhu vadks dh ,d.................

Sol: rhu vadks dh 11 ls HkkT; la[;kvksa dh la[;k

{10×11, 11×11,........,90×11} = 81 la[;k,¡

,slh la[;k,¡ 9 ls Hkh HkkT; gS

{18×11, 27×11,........,90×11} = 9 la[;k,¡

vr% 9 1 k

p81 9 27

= = =

k = 3

72. leqPp; A = {,......................

Sol. A ij rqY;rk lEcU/k gS

R1 = {( , ), ({},{}),({{}},{{}})}

R2 = R1 {(, {}), ({}, )}

R3 = R1 {(, {{}}), ({{}}, )}

R4 = R1 {({}, {{}}), ({{}}, {})} R5 = A × A (universal relation)

73. ekuk fd f ,d Qyu......................

Sol. fn;k x;k gS f(xy) = y

)x(f

x = 1 j[kus ij f(y) = y

)1(f

;k f(30) = 30

)1(f f(1) = 30 × f(30) = 30 × 20 = 600

f(40) = 40

)1(f =

40

600 = 15

74. ekuk lehdj.k.............

Sol. 01kk2

blfy, = _.kkRed gS

022

1tantan

1tantan 1111

75. ;fn f(x) = x5 + e3x .........

Sol. f(0) = 1 g(1) = 0

vc g(f(x)) = x g(f(x)) . f(x) = 1

g(f(x)) = 1

f (x) g(f(0)) =

1

f (0)

g(1) = 1

f (0) =

1

3

1

g (1) = 3.





TEST-1 (MCT-1)

TARGET : JEE (MAIN+ADVANCED) 2021

DATE : 31-05-2020 SET/CODE-1 COURSE : VIJETA (JP) | BATCH : (JPB*)

ANSWER KEY

SET/CODE-1

PART : A PHYSICS

1. (4) 2. (1) 3. (2) 4. (2) 5. (1) 6. (1) 7. (4)

8. (3) 9. (4) 10. (1) 11. (1) 12. (2) 13. (4) 14. (3)

15. (2) 16. (4) 17. (1) 18. (3) 19. (1) 20. (3) 21. (4)

22. (5) 23. (4) 24. (3) 25. (8)

PART : B CHEMISTRY

26. (3) 27. (1) 28. (4) 29. (3) 30. (2) 31. (1) 32. (4)

33. (1) 34. (3) 35. (4) 36. (3) 37. (4) 38. (4) 39. (2)

40. (4) 41. (2) 42. (3) 43. (3) 44. (2) 45. (2) 46. (7)

47. (7) 48. (7) 49. (4) 50. (6)


51. (1) 52. (2) 53. (4) 54. (1) 55. (2) 56. (2) 57. (2)

58. (1) 59. (2) 60. (1) 61. (2) 62. (3) 63. (4) 64. (2)

65. (3) 66. (1) 67. (3) 68. (3) 69. (3) 70. (4) 71. (3)

72. (5) 73. (5) 74. (0) 75. (3)

COURSE : VIJETA (JP) | BATCH : JPB*,JPAB CODE 1 le; TARGET ...

Documents

Transcript of COURSE : VIJETA (JP) | BATCH : JPB*,JPAB CODE 1 le; TARGET ...