course : vijeta (jp) | batch : jpb*,jpab code 1 le; target
TRANSCRIPT
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
_.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA
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)
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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
_.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA
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
Space for Rough Work / (dPps dk;Z ds fy, LFkku)
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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] P01JPAMCT1310520C1-2
Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029
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
Space for Rough Work / (dPps dk;Z ds fy, LFkku)
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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] P01JPAMCT1310520C1-3
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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
Space for Rough Work / (dPps dk;Z ds fy, LFkku)
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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] P01JPAMCT1310520C1-4
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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
Space for Rough Work / (dPps dk;Z ds fy, LFkku)
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Reg. / Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
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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
Space for Rough Work / (dPps dk;Z ds fy, LFkku)
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Reg. / Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
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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
Space for Rough Work / (dPps dk;Z ds fy, LFkku)
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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
Space for Rough Work / (dPps dk;Z ds fy, LFkku)
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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
SECTION – 2 : (Maximum Marks : 20)
This section contains 5 questions.
The answer to each question is a Single Digit
Integer, ranging from 0 to 9 (both inclusive).
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
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 :
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
_.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA
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 \
Space for Rough Work / (dPps dk;Z ds fy, LFkku)
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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
Space for Rough Work / (dPps dk;Z ds fy, LFkku)
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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
Space for Rough Work / (dPps dk;Z ds fy, LFkku)
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PART – B
Atomic masses : [H = 1, D = 2, Li = 7, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27, Si = 28, P = 31, S = 32, Cl = 35.5, K = 39, Ca = 40, Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65, As = 75, Br = 80, Ag = 108, I = 127, Ba = 137, Hg = 200, Pb = 207]
SECTION – 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
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]
[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
_.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA
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)
Space for Rough Work / (dPps dk;Z ds fy, LFkku)
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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
(3) 500 torr, 400 torr (4) 400 torr, 200 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 %
(1) 400 torr, 500 torr (2) 200 torr, 400 torr
(3) 500 torr, 400 torr (4) 400 torr, 200 torr
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
Space for Rough Work / (dPps dk;Z ds fy, LFkku)
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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
Space for Rough Work / (dPps dk;Z ds fy, LFkku)
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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
(2) 0.05 mol dm–3 NH4OH and 0.1 mol dm–3 HCl
(3) 0.1 mol dm–3 NH4OH and 0.05 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
(2) 0.05 mol dm–3 NH4OH rFkk 0.1 mol dm–3 HCl
(3) 0.1 mol dm–3 NH4OH rFkk 0.05 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
Space for Rough Work / (dPps dk;Z ds fy, LFkku)
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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)
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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 *
Space for Rough Work / (dPps dk;Z ds fy, LFkku)
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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
SECTION – 2 : (Maximum Marks : 20)
This section contains 5 questions.
The answer to each question is a Single Digit
Integer, ranging from 0 to 9 (both inclusive).
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
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
[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 :
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
_.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA
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&
Space for Rough Work / (dPps dk;Z ds fy, LFkku)
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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
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Space for Rough Work / (dPps dk;Z ds fy, LFkku )
PART – C
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
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
[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
_.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA
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
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Space for Rough Work / (dPps dk;Z ds fy, LFkku )
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
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Space for Rough Work / (dPps dk;Z ds fy, LFkku )
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]
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Space for Rough Work / (dPps dk;Z ds fy, LFkku )
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
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Space for Rough Work / (dPps dk;Z ds fy, LFkku )
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
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Space for Rough Work / (dPps dk;Z ds fy, LFkku )
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
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Space for Rough Work / (dPps dk;Z ds fy, LFkku )
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
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Space for Rough Work / (dPps dk;Z ds fy, LFkku )
SECTION – 2 : (Maximum Marks : 20)
This section contains 5 questions.
The answer to each question is a Single Digit
Integer, ranging from 0 to 9 (both inclusive).
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
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
[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 :
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
_.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA
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
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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 …………..
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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
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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 ……………
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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. +
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+
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
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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 = {(, ), ({},{}),({{}},{{}})}
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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.
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MAIN PATTERN CUMULATIVE
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)
PART : C MATHEMATICS
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)
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MAIN PATTERN CUMULATIVE
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.
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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
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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
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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
39. fuEu esa ls ………….
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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
50. fuEu esa ls ………….
Sol. (3) esa vEy {kkj vfHkfØ;k gksrh gSA
PART : C MATHEMATICS
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
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= 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
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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.
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MAIN PATTERN CUMULATIVE
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)
PART : C MATHEMATICS
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)