JPT 2 Q

JEE (MAIN+ADVANCED)
Academic Session: 2023-24
P01-23
TARGET TEST PATTERN JEE (MAIN)

JEE (MAIN+ADVANCED)
EXAMINATION
JEE PREPRATORY TEST
TEST TYPE
TARGET YEAR 2024 (JPT)
TEST CODE &
PAPER NO. ONE SEQUENCE
JPT 2
PAPER CODE 1 MAX. MARKS 300

TEST
CLASS XII 3 Hrs
DURATION
TEST DATE 07th JANUARY 2024
COURSE NAME VIJETA, ANOOP
TEST DAY SUNDAY
COURSE CODE JP, EP TEST TIME Start: 09:30 AM
(PAPER-1) End : 12:30 PM
PHASE CODE(S) JP, EP
TOTAL NO. OF
BATCH CODE(S) JP, EP PAGES IN PAPER 36
BOOKLET
TEST PAPER DETAILS MARKING SCHEME

Full If No
Section Subject Type of No. of (–)ve Total Subject
Qs. No. Marks Per Option
No. Sequence Qs.* Qs. Marks Marks Total
Qs. Chosen
1 to 20 1 MCQ 20 4 0 –1 80
Physics 100
21 to 30 2 NVQ 10** 4 0 –1 20
31 to 50 1 MCQ 20 4 0 –1 80
Chemistry 100
51 to 60 2 NVQ 10** 4 0 –1 20
61 to 80 1 MCQ 20 4 0 –1 80
Maths 100
81 to 90 2 NVQ 10** 4 0 –1 20
Total 90 MAXIMUM MARKS 300
* Please turn overleaf to understand the meaning of coding for types of Questions.
**5 out of 10 Qs. with answers as a numerical value have to be answered.
A. GENERAL INSTRUCTIONS B. DARKENING THE BUBBLES ON THE ORS :
1. Use a BLACK BALL POINT to darken the
1. Darken the appropriate bubbles on the original
bubbles in the upper sheet.
by applying sufficient pressure. 2. Darken the bubble COMPLETELY.
2. The original is machine-gradable and will be 3. Darken the bubble ONLY if you are sure of the
collected by the invigilator at the end of the answer.
examination. 4. The correct way of darkening a bubble is as
3. Do not tamper with or mutilate the ORS. shown here :
5. There is NO way to erase or "un-darkened
4. Write your name, roll number and the name of
bubble.
the examination centre and sign with pen in the 6. The marking scheme given at the beginning of
space provided for this purpose on the original. each section gives details of how darkened and
Do not write any of these details anywhere not darkened bubbles are evaluated.
else. Darken the appropriate bubble under each 7. Zero marks ‘0’ If none of the options is chosen
digit of your roll number. (i.e. the question is unanswered).
C. FOR INTEGER TYPE QUESTIONS OMR LOOKS D. FOR DECIMAL TYPE QUESTIONS OMR LOOKS
LIKE: LIKE:
1. For example, if answer ‘SINGLE DIGIT’ integer 1.
type below : COLUMN
0 1 2 3 4 5 6 7 8 9 1 2 . 3 4
2. For example, if answer ‘SINGLE DIGIT’ integer 0 0 0 0
with positive / negative type below : 1 1 1 1
0 2 2 2 2
– 1
3 3 3 3
2
4 4 4 4
3
5 5 5 5
4
6 6 6 6
5
7 7 7 7
6
8 8 8 8
7
9 9 9 9
8
9 2. If answer is 3.7, then fill 3 in either 1st or 2nd
3. For example, if answer ‘DOUBLE DIGIT’ integer column and 7 in 3rd or 4th column.
type below :
0 0 0 0 0 0 0 0 0 3. If answer is 3.07 then fill 3 in 1st or 2nd column ‘0’
1 1 1 1 1 1 1 1 1 in 3rd column and 7 in 4th column.
2 2 2 2 2 2 2 2 2 If answer is, 23 then fill 2 & 3 in 1st and 2nd
3 3 3 3 3 3 3 3 3 column respectively, while you can either leave
4 4 4 4 4 4 4 4 4 column 3 & 4 or fill ‘0’ in either of them.
5 5 5 5 5 5 5 5 5
6 6 6 6 6 6 6 6 6
7 7 7 7 7 7 7 7 7
8 8 8 8 8 8 8 8 8
9 9 9 9 9 9 9 9 9
TYPE WISE CODES FOR QUESTIONS

SR# QUESTION TYPE CODE
1 MULTIPLE CHOICE QUESTION (ONLY ONE CORRECT OPTION) MCQ
2 MULTIPLE SELECT (ONE OR MORE THAN ONE CORRECT OPTIONS QUESTION) MSQ
3 NUMERICAL VALUE QUESTION NVQ
4 COMPREHENSION / PARAGRAPH BASED QUESTION CBQ
5 MATCH THE LISTING QUESTION MLQ
6 COLUMN MATCH QUESTION CMQ
7 FILL IN THE BLANKS TYPE QUESTION FBQ
8 TRUE OR FALSE TYPE QUESTION TFQ
9 ASSERTION & REASON / STATEMENT TYPE QUESTION ARQ
10 SUBJECTIVE TYPE QUESTION STQ
PHYSICS
PART-A (Hkkx-A) : PHYSICS (HkkSfrd foKku)
SECTION A (Maximum Marks: 80)
 This section contains Twenty (20) questions for each subject.
 Each question has four options (1), (2), (3) and (4). ONLY ONE of these four options is the most
appropriate or best answer, which will be considered the correct answer.
 The answer to each question will be evaluated according to the following 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
[kaM A (vf/kdre vad : 80)
 bl [kaM esa izR;sd fo"k; ds fy, chl (20) iz'u gSaA
 izR;sd iz'u ds pkj fodYi (1), (2), (3) rFkk (4) gSaA bu pkjksa fodYiksa esa ls dsoy ,d gh lcls mi;qDr ;k loksZÙke mÙkj
gS] ftls lgh ekuk tk,xkA
 izR;sd iz'u ds mÙkj dk ewY;kadu fuEufyf[kr vadu ;kstuk ds vuqlkj fd;k tk,xk%
 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. A conical flask of mass 10 kg and base area 103 cm2 is floating in liquid of specific gravity 1.2 as shown in
the figure. The force that liquid exerts on curved surface of conical flask is (Neglect Patm) : (g = 10 m/s2)
(1) 20 N in downward direction (2) 40 N in downward direction
(3) 20 N in upward direction (4) 40 N in upwards direction
10 cm
10 kg nzO;eku rFkk 103 cm2 vk/kkj {ks=kQy okyk 'kaDokdkj tkj 1.2 fof'k"V xq:Ro okys nzo esa fp=kkuqlkj rSj jgk gSa
tkj dh oØkdkj lrg ij nzo }kjk vkjksfir cy D;k gksxk ( Patm ux.; ekusa) : (g = 10 m/s2)
Space for Rough Work / (dPps dk;Z ds fy, LFkku)
(1) 20 U;wVu uhps dh rjQ (2) 40 U;wVu uhps dh rjQ
(3) 20 U;wVu Åij dh rjQ (4) 40 U;wVu Åij dh rjQ
Reg. & Corp. Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 PAGE
Website : www.resonance.ac.in | E-mail : contact@resonance.ac.in
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029
P-01JP-EP-JPT2-070124-C1 1
PHYSICS
2. In the graph variation of magnetic field with time ' t ' applied perpendicular to the plane of the ring is
shown:
(1) at t = 2 sec and current flowing in ring is equal to zero

(2) current will change its direction two times in time intervals t = 0 to t = 8 sec
(3) current will change its direction only once in the above interval
(4) flux in ring is same at t = 0 & t = 4 sec.
,d oy; ds lery ds yEcor~ vkjksfir pqEcdh; {ks=k dk le; ' t ' ds lkFk oØ fuEu fp=k esa iznf'kZr gSA
(1) t = 2 sec ij oy; esa izokfgr /kkjk 'kwU; gksxhA

(2) t = 0 rFkk t = 8 sec ds varjky esa /kkjk nks ckj viuh fn'kk cnysxh
(3) dsoy ,d gh ckj mijksDr varjky esa /kkjk dh fn'kk cnysxhA
(4) t = 0 rFkk t = 4 sec ij oy; ds ¶yDl 'kwU; gSA
P-01JP-EP-JPT2-070124-C1 2
PHYSICS
3. A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry
to infinity. The time taken by it to reach height h is______s.
fdlh fi.M dks i`Foh ds i`"B ls Å/okZ/kj mifjeq[kh ml osx ls Qssdk x;k gS tks mls vuUr rd ys tkus ds fy,
i;kZIr gSA Å¡pkbZ h rd ig¡pus esa bl fi.M }kjk fy;k x;k le; lsd.Mksa esa gksxkA
 3   3 

2R e h 2  R e 
1 
h 2 
  1
(1) 1  R   1 (2)
2g  R e 
 e 
g
  
   
 3   3 
1 R e  h 2  1 2R e  h 2 
(3) 1    1 (4) 1    1
3 2g  R e  3 g  R e 
   
   
4. Binding Energy per nucleon of a fixed nucleus XA is 6 MeV. It absorbs a neutron moving with KE = 2 MeV,
and converts into Y, emitting a photon of energy 1 MeV. The Binding Energy per nucleon of Y (in MeV)
is
,d fLFkj ukfHkd XA dh izfr ukfHkdh; d.k ca/ku ÅtkZ 6 MeV gSA ;g ,d U;wVªkWu dk vo'kks"k.k djrk gS ftldh
xfrt ÅtkZ KE = 2 MeV gS rFkk ukkfHkd Y esa :ikUrfjr gks tkrk gS ,oa lkFk esa 1 MeV ÅtkZ dk QksVksu mRlftZr
djrk gSA ukkfHkd Y esa izfr ukfHkdh; d.k cU/ku ÅtkZ (MeV esa ) D;k gksxhA
(6A  1) (6A  1) ( 6 A  1)
(1) (2) (3) 7 (4)
(A  1) (A  1) ( A  1)
5. A uniform rope having uniformly distributed mass hanges vertically from a rigid support. A transverse
wave pulse is produced at the lower end. The speed (v) of the wave pulse varies with height (h) from
the lower end as:
,dleku forfjr nzO;eku dh ,d ,dleku Mksjh ,d n`<+ vk/kkj ls Å/okZ/kj yVdh gqbZ gSA ,d vuqizLFk rjax LiUn
blds fupys fljs ij mRiUu fd;k tkrk gSA rjax LiUn dh pky (v) fupys fljs ls ÅWpkbZ (h) ds lkFk fdl izdkj
ifjofrZr gksxhA
(1) (2) (3) (4)

P-01JP-EP-JPT2-070124-C1 3
PHYSICS
6. The voltage drop across a forward biased diode is 0.7 volt. In the following circuit, the voltages across
the 10 ohm resistance in series with the diode and 20 ohm resistance are :
,d vxzfnf'kd vfHkfufr Mk;ksM ij oksYVst MªkWi 0.7 oksYV gSA fuEufyf[kr ifjiFk esa] Js.kh Øe esa lEc) 10ds
izfrjks/k o Mk;ksM ds chp rFkk lekUrj Øe esa yxs gq, 20ds izfrjks/k ds chp foHkokUrj gksxk :
(1) 0.70 V, 4.28 V (2) 3.58 V, 4.28 V (3) 5.35 V, 2.14 V (4) 3.58 V, 9.3 V
7. An electron from various excited states of hydrogen atom emit radiation to come to the ground state.
Let n, g be the de Broglie wavelength of the electron in the n th state and the ground state respectively.
Let n be the wavelength of the emitted photon in transition from the n th state to the ground state. For
large n, (A, B are constants)
,d bysDVªkWu fdlh gkbMªkstu ijek.kq ds fofHkUu mÙksftr voLFkkvksa ls fofdj.k mRlftZr djds fuEure voLFkk esa vk
tkrk gSA ekuk fd n rFkk g noha voLFkk rFkk fuEure voLFkk essa bysDVªkWu dh Mh czksxyh rajxnS/;Z gSA ekuk noha
voLFkk ls fuEure voLFkk esa laØe.k }kjk mRlftZr QksVksu dh rjaxnS?;Z n gSA n ds cM+s eku ds fy, ¼;fn A rFkk
B fLFkjkad gS)
(1) n2  A  Bn2 (2) n2  
B
(3) n  A  (4) n  A  Bn
 n2
P-01JP-EP-JPT2-070124-C1 4
PHYSICS
8. A transparent cylinder has its right half polished so as to act as a mirror. A paraxial light ray is incident
from left, that is parallel to principal axis, exits parallel to the incident ray as shown. The refractive index
n of the material of the cylinder is :
,d ikjn'khZ csyu ds nk;s vk/ks Hkkx dks ikWfy'k fd;k x;k gS] ftlls ;g niZ.k dh Hkkafr O;ogkj djrk gSA v{k ds
utnhd] eq[;v{k ds lekUrj fdj.ksa bl ij ck;ha vksj ls vkifrr gksrh gS rFkk vkifrr fdj.kksa ds lekUrj gh ckgj
fudyrh gSA csyu ds inkFkZ dk viorZukad n gSa &
(1) 1.2 (2) 1.5 (3) 1.8 (4) 2.0
9. A uniform cylinder of mass M and radius R is to be pulled over a step of height a (a < R) by applying a
force F at its centre 'O' perpendicular to the plane through the axes of the cylinder on the edge of the
step (see figure). The minimum value of F required is :
nzO;eku M rFkk f=kT;k R ds ,d csyu (cylinder) dks a (a < R) Å¡pkbZ dh ,d lh<+h ds Åij [khapuk gSA bld fy;s
bld dsUnz O ij ,d cy F, tks fd csyu ds v{k vkSj lh<+h ds fdukjs ls gksdj tkus okys lery ds yEcor~ gS]
yxk;k tkrk gSA (fp=k ns[ksa) F dk U;wure eku gS &
R a
2
a2  R  a
(1) Mg 1    (2) Mg 1  (3) Mg   1 (4) Mg
R a
2
 R  R R
P-01JP-EP-JPT2-070124-C1 5
PHYSICS
10. The electromagnetic radiations are in descending order of wavelength in the following sequence
(1) infra-red waves, radio waves, X-rays, visible light rays
(2) radio-waves, infra-red waves, visible light, X-rays
(3) radio waves, visible light, infra - red waves, X-rays
(4) X-rays, visible light, infra-red wave, radio waves
rjaxnS/;Z ds ?kVrs gq, Øe esa] oS|qr pqEcdh; fofdj.k dh Js.kh fuEufyf[kr gksxhA
(1) vojDr rjaxs] jafM;ks rjaxsa] x-fdj.ksa, n`'; çdk'k fdj.kas
(2) jsfM;k rjaxs] vojDr rjaxsa] n`'; çdk'k] x-fdj.ksa
(3) jsfM;k rjaxs] n`'; çdk'k] vojDr rjaxsa] x-fdj.ksa
(4) x-fdj.ksa] n`'; çdk'k] vojDr rjaxsa] jsfM;ks rjaxs]
11. A ball is thrown upward at an angle 30º to the horizontal and lands on the top edge of a building that is
20 m away and 5m high. How fast was the ball thrown.
(1) 10 m/s (2) 20 m/s (3) 40 m/s (4) 80 m/s
,d xsan dks {kSfrt ls 30º ds dks.k ij Åij Qsadk tkrk gS] rFkk xsan 20 m nwj rFkk 5m Å¡ph bekjr ds dkssus ij
Vdjkrh gSA rks xsan fdl pky ls Qsadh tkrh gSA
(1) 10 m/s (2) 20 m/s (3) 40 m/s (4) 80 m/s
12. Interference pattern is observed at 'P' due to superimpostion of two rays coming out from a source 'S'
as shown in the figure. The value of '  ' for which maxima is obtained at 'P' is :
(R is perfect reflecting surface) :
,d L=kksr 'S' ls fudy jgh nks fdj.ksa ds v/;kjksi.k ls 'P' ij ,d O;frdj.k fp=k ik;k tkrk gS] tSlk fd fp=k esa
n'kkZ;k x;k gSA '  ' dk og eku] ftlds fy, 'P' ij izkIr fp=k esa egÙke rhozrk gS] gS :
(R ,d iw.kZr;k ijkorhZ i``"B gS)
2n
Space for Rough Work / (dPps dk;Z
(2nds– 1fy,
) 3LFkku) ( 2n – 1)
(1)   (2)   (2n – 1) (3)   (4)  
3 1 2( 3  1) 4(2  3 ) 3 1
P-01JP-EP-JPT2-070124-C1 6
PHYSICS
13. A uniformly charged ring of radius R is rotated about its axis with constant linear speed v of each of its
particles. The ratio of electric field to magnetic field at a point P on the axis of the ring distant x = R from
centre of ring is (c is speed of light)
f=kT;k R dh ,dleku vkosf'kr oy; dks bldh v{k ds ifjr% fp=kkuqlkj bl izdkj ?kqek;k tkrk gS fd blds izR;sd
d.k dh fu;r js[kh; pky v gSA oy; dh v{k ij blds dsUnz ls x = R nwjh ij ,d fcUnq P ij fo|qr {ks=k rFkk
pqEcdh; {ks=k dk vuqikr gksxk & (c izdk'k dk osx gS)
c2 v2 c v
(1) (2) (3) (4)
v c v c
14. An organ pipe of length L is open at one end and closed at other end. The wavelengths of the three
lowest resonating frequencies that can be produced by this pipe are
yEckbZ L dk ,d vkWxZu ikbZi ,d fljs ls [kqyk gS o nwljs fljs ls cUn gSA rhu U;wure vko`fÙk;k¡ tks bl ikbZi ls
mRiUu gks ldrh gSa ] dh rjaxnS/;Z gS&
(1) 4L, 2L, L (2) 2L, L, L/2 (3) 2L, L, 2L/3 (4) 4L, 4L/3, 4L/5
15. A particle of mass m is fixed to one end of a light spring of force constant k and unstretched length .
k
The system is rotated about the other end of the spring with an angular speed  (   ) in gravity
m
free space. The increase in length of the spring is :
m2 m2 m2
(1) (2) (3) (4) none of these
k k  m2 k  m2

m nzO;eku ds ,d d.k dks vladqfpr  yEckbZ okyh gYdh fLçax ds ,d fljs ls tksM+k tkrk gSA vc fudk; dks fLçax
ds nwljs fljs ds lkFk  dks.kh; osx ls ux.; xq:Ro {ks=k esa ?kqek;k tkrk gSA fLçax esa foLrkj gksxk &
m2 m2 m2
(1) (2) (3) (4) buesa ls dksbZ ugha
k k  m2 k  m2
P-01JP-EP-JPT2-070124-C1 7
PHYSICS
16. A non – conducting semicircular disc (as shown in figure) has a uniform surface charge density . The
ratio of electric field to electric potential at the centre of the disc will be :
,d vpkyd v/kZ&o`Ùkkdkj pdrh ¼fp=kkuqlkj½ dk ,dleku i`"Bh; vkos'k ?kuRo gSA pdrh ds dsUnz ij fo|qr {ks=k
rFkk fo|qr foHko dk vuqikr gksxk &
1 nb / a 2 1 n(b / a)2 (b  a)

(1) (2) (3) (4)
 (b  a)   (b  a) 2 n(b / a)
17. A Soap bubble of radius R is surrourded by another soap bubble of radius 2R, as shown. Take surface
tension = S. Then, the pressure inside the smaller soap bubble, in excess of the atmospheric pressure
will be :
R f=kT;k dk lkcqu dk cqycqyk fp=kkuqlkj 2R f=kT;k ds vU; lkcqu ds cqycqys esa fLFkr gSA i`"B ruko = S gS rks
NksVs xksys ds vUnj dk nkc] ok;qe.Myh; nkc dh rqyuk esa fdruk T;knk gS –
Atmosphere
(ok;qe.My)
2R
R
Air
4S 6S 8S 3S
(1) (2) (3) (4)
R R R R
18. The time taken by a particle performing SHM on a straight line to pass from point A to B where its
velocities are same is 2 seconds. After another 2 seconds it returns to B. The time period of oscillation
is (in second):
ljy js[kk esa ljy vkorZ xfr dj jgs ,d d.k }kjk A ls B fcUnq tgk¡ blds osx leku gS] rd igqapus esa yxk le;
2 lSd.M gSA vxys 2 lSd.M ckn ;g B ij ykSVrk gSA nksyu dk vkorZdky ¼lSd.M+ esa½ gS&
(1) 2 (2) 4 (3) 6 (4) 8
P-01JP-EP-JPT2-070124-C1 8
PHYSICS
19. When a metallic surface is illuminated with monochromatic light of wavelength , the stopping potential
is 5 V0. When the same surface is illuminated with light of wavelength 3, the stopping potential is V0.
Then the work function of the metallic surface is :
tc ,d /kkfRod lrg dks , rjaxnS/;Z ds ,do.khZ; çdk'k ls çdkf'kr fd;k tkrk gS rks fujks/kh foHko 5 V0 gSA tc
bl lrg dks 3 rjaxnS/;Z ds çdk'k ls çdkf'kr fd;k tkrk gS rks fujks/kh foHko V0 gSA /kkfRod lrg dk dk;ZQyu gS
hc hc hc 2hc
(1) (2) (3) (4)
6 5 4 4
20. A particle of mass m is moving in a circular path of constant radius r such that its centripetal
acceleration ac is varying with time as ac= k2 r t2, where k is a constant. The power delivered to the
particle by the forces acting on it is:
m æO;eku dk ,d d.k fu;r f=kT;k r okys o`Ùkh; iFk ij bl izdkj xfr'khy gS fd bldk vfHkdsUæh; Roj.k le;
ds lkFk ifjofrZ gS ac = k2 r t2, tgk¡ k ,d fu;rkad gSA d.k ij dk;Zjr cy }kjk d.k dks iznku dh x;h 'kfDr gksxh
(1) 2 m k2 r2 t (2) m k2 r2 t (3) (1/3) m k4 r2 t5 (4) 0
SECTION B (Maximum Marks: 20)

 This Section contains Ten (10) questions. Out of these Ten (10) questions, only Five (05) Questions need
to be attempted.
 Candidates are advised to do the calculations with the constants given (if any) in the questions. The
answer should be rounded off to the nearest Integer.
 For each question, enter the correct integer value of the answer using the mouse and the on-screen
virtual numeric keypad in the place designated to enter the answer.
 Zero Marks : 0 If none of the values are entered (i.e. the question is unanswered).
 Negative Marks : –1 In all other cases.
[kaM B ¼vf/kdre vad% 20)
 bl [kaM esa nl (10) iz'u gSA bu nl (10) iz'uksa esa ls dsoy ik¡p (05) iz'uksa dks gy djus dh vko';drk gSA
 mEehnokjksa dks lykg nh tkrh gS fd os iz'uksa esa fn, x, fLFkjkadksa ¼;fn dksbZ gks½ ds lkFk x.kuk djsaA mÙkj dks fudVre
iw.kk±d rd iw.kk±fdr fd;k tkuk pkfg,A
 izR;sd i'u ds fy,] mÙkj ntZ djus ds fy, fufnZ"V LFkku ij ekml vkSj vkWu&LØhu opqZvy U;wesfjd dhiSM dk mi;ksx
djds mÙkj dk lgh iw.kk±d ekuSpace aA Rough Work / (dPps dk;Z ds fy, LFkku)
ntZ djsfor
 'kwU; vad % 0 ;fn dksbZ Hkh eku ugha Hkjk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A
P-01JP-EP-JPT2-070124-C1 9
PHYSICS
21. A series RLC, AC circuit operates at half its resonant frequency and at a power factor of 0.8. Then write
XC
the ratio of :
R
,d Js.kh RLC, AC ifjiFk bldh vuquknh vko`fÙk dh vk/kh vko`fÙk ij rFkk 0.8 ds 'kfDr xq.kkad ij dk;Z djrk gS
XC
rks dk vuqikr gksxk :
R
22. Find the current (in A) through the battery after the switch S is closed if L/R = RC = 1 ms.
fLop S cUn djus ds i'pkr~ cSVjh ls çokfgr /kkjk (A esa) Kkr djksA ;fn L/R = RC = 1 ms
23. A block slides down a 53° incline in twice the time it would take to slide down a frictionless 53° incline,
covering same distances in both. Find the value of 100 where is the coefficient of kinetic friction
between the block and the incline plane.
,d xqVdk ,d ur ry tks fd {kSfrt ds lkFk 53° dk dks.k curk gS ml ij fQlyrk gSA ;fn ur ry [kqjnjk ys
rks mlds }kjk fdlh leku nwjh dks r; djus esa yxk le; ur ry dks fpduk ysus dh rqyuk esa yxs le; dk
nksxquk gS rks 100 dk eku Kkr djksa tgk¡ CykWd ,oa ur ry ds e/; xfrd ?k"kZ.k xq.kkad gSA
P-01JP-EP-JPT2-070124-C1 10
PHYSICS
24. An engine operates by taking a monatomic ideal gas through the cycle shown in the figure. The
percentage efficiency of the engine is close to _____________
,d batu ds izpkyu esa dksbZ ,dijek.kqd vkn'kZ xSl vkjs[k esa n'kkZ;s x;s pØ ls xqtjrh gSA bl batu dh fudVLFk
n{krk (izfr'kr esa) gksxh_____________
25. Find the equivalent emf of the three batteries as shown in the figure.
fp=k esa n'kkZbZ xbZ rhu cSVjh dk rqY; fo-ok-cy Kkr dhft,A
26. A person with a normal near point (25 cm) is using a compound microscope in normal adjustment with
objective focal length f0 = 8.0 mm and an eye-piece of focal length fe = 2.5 cm can bring an object
placed 9.0 mm from the objective in sharp focus. How much is the magnifying power of the
microscope?
lkekU; fudV fcUnq (25 cm) dk ,d O;fDr vfHkn`';d Qksdl nwjh f0 = 8.0 mm rFkk vfHkus=k Qksdl nwjh
fe = 2.5 cm ds ,d la;qDr lw{en'khZ ds mi;ksx ls vfHkn`';d ls 9.0 mm ij j[ks fcEc dks Qksdflr djds Li"V
izfrfcEc fudV fcUnq ij izkIr djrk gSA lw{en'khZ dh vko/kZu {kerk fdruh gksxhA
P-01JP-EP-JPT2-070124-C1 11
PHYSICS
27. A long straight wire is carrying current I1 = 2/5A in +z direction. The x - y plane contains a closed
circular loop carrying current I2 = 5/2A and not encircling the straight wire, then the force (in newton) on
the loop will be ? (radius of the circular loop R = 3/4m).
,d yEcs lh/ks rkj esa +z fn'kk esa I1 = 2/5A /kkjk izokfgr gks jgh gSA x - y ry esa ,d cUn o`Ùkkdkj ywi gS] ftlesa
I2 = 5/2A /kkjkizokfgr gks jgh gS] rFkk ;g o`Ùkkdkj ywi lh/ks rkj ds vUnj ugh gSA rc ywi ij cy (U;wVu esa) gksxkA
(o`Ùkkdkj ywi dh f=kT;k R = 3/4m).
28. A compound rod, 2m long is constructed of a solid steel core, 1cm in diameter surrounded by copper
casing whose outside diameter is 2cm. The outer surface of rod is thermally insulated. One end is
maintained at 100ºC and other at 0ºC. What fraction (in %) of heat current is carried by steel core.
(Ksteel = 12 cal/m-K, Kcopper = 92 cal/m-K).
2m yEch la;qDr NM+, 1cm O;kl ds Bksl LVhy ØksM+ ls cuh gS tks 2cm ckº; O;kl dh rkez vkoj.k ls f?kjh gqbZ gSA
NM+ dk ckº; i`"B Å"eh; :) gSA bldk ,d fljk 100ºC rki ij rFkk nwljk fljk 0ºC rki ij j[kk x;k gSA LVhy
ØksM+ }kjk Å"eh; /kkjk dk fdruk Hkkx (% esa) ogu fd;k x;k gSA (Ksteel = 12 cal/m-K, Kcopper = 92 cal/m-K).
29. A sphere of mass m and radius r is projected in a gravity free space with speed v. If coefficient of
1 xmv
viscosity is , the distance travelled by the body before it stops is . Find x.
6 r
m nzO;eku ,oa r f=kT;k dk xksyk ,d xq:Rojfgr vkdk'k esa v osx ls ç{ksfir fd;k tkrk gSA ;fn ';kurk xq.kakd
1 xmv
gS ] rks oLrq }kjk :dus ls iwoZ r; dh x;h nwjh gSA x Kkr dhft,A
6 r
30. A particle of mass m = 1 kg is made to move with uniform speed v = 3 m/s along the perimeter of a
regular hexagon. Find the magnitude of impulse applied at each corner of the hexagon in kg m-s–1.
m = 1 kg nzO;eku dk ,d d.k le"kVHkqt dh Hkqtk ds vuqfn'k v = 3 m/s dh ,d leku pky ls xfr djrk gSA
"kVHkqt ds izR;sd dksus ij vkjksfir vkosx dk ifjek.k kg m-s–1 esa gksxkA
P-01JP-EP-JPT2-070124-C1 12
CHEMISTRY
PART-B (Hkkx-B) : CHEMISTRY (jlk;u foKku)
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]
 'kwU; vad % 0 ;fn dksbZ Hkh fodYi ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A
31. Given below are two statements:

Statement I: According to Bohr's model of hydrogen atom, the angular momentum of an electron in a
given stationary state is quantised.
Statement II: The concept of electron in Bohr's orbit, violates the Heisenberg uncertainty principle.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Statement I is incorrect but Statement II is correct
(2) Both Statement I and Statement II are incorrect
(3) Statement I is correct but Statement II is incorrect
(4) Both Statement I and Statement II are correct
C-01JP-JPT2-070124-C1 13
CHEMISTRY
uhps nks dFku fn, x, gS%
dFku I : gkbMªkstu ijek.kq ds cksj ekWMy ds vuqlkj] fdlh LFkk;h d{kk esa bysDVªkWu dk dks.kh; laosx DokaVhÑr gSA
dFku II : cksj d{kk esa bysDVªkWu dh ladYiuk] gkbtsucxZ ds vfuf'Pkrrk fl)kar dk ikyu ugha djrk gSA
mi;qZDRk dFkuksa ds lanHkZ esa] uhps fn, x, fodYiksa esa ls lokZf/kd mi;qDRk mÙkj pqusa%
(1) dFku I xyr gS ijUrq dFku II lgh gSA
(2) dFku I ,oa dFku II nksuksa xyr gSA
(3) dFku I lgh gS ijUrq dFku II xyr gSA
(4) dFku I ,oa dFku II nksuksa lgh gSA
32. A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g)
is 0.04 atm. What is Kp for the given equilibrium ?
2HI (g) H2 (g) + 2 (g)
HI(g) dk ,d izkn'kZ 0.2 atm nkc ij ,d ¶ykLd esa fy;k x;k gSA lkE; ij] HI(g) dk vkaf'kd nkc 0.04 atm gSA
fn;s x;s lkE; ds fy, Kp D;k gS\
2HI (g) H2 (g) + 2 (g)
(1) 16 (2) 8 (3) 4 (4) 1/4
33. Given below are two statements:

Statement I: SO2 and H2O both possess V-shaped structure.
Statement II: The bond angle of SO2 is less than that of H2O.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) Statement I is incorrect but Statement II is correct
(2) Both Statement I and Statement II are correct
(3) Statement I is correct but Statement II is incorrect
(4) Both Statement I and Statement II are incorrect
C-01JP-JPT2-070124-C1 14
CHEMISTRY
uhps nks dFku fn;s x;s gSA
dFku I : SO2 vkSj H2O nksuksa dh lajpuk V-vkdkj dh gSA
dFku II : SO2 dk vkca/k dks.k H2O ls de gksrk gSA
mijksDRk dFkuksa ds vk/kkj ij uhps fn;s x;s fodYiksa esa ls lcls mfpr mÙkj dks pqusaAࡏ
(1) dFku I vlR; gS ijUrq dFku II lgh gSA (2) dFku I vkSj II nksuksa lR; gSA
(3) dFku I lgh gS ijUrq dFku II vlR; gSA (4) dFku I vkSj II nksuksa vlR; gSA
34. PtCl4.6H2O can exist as a hydrated complex. 1 molal aq. solution has depression in freezing point of
3.72ºC. Assume 100% ionisation and Kf (H2O) = 1.86ºC mol–1Kg, then complex is:
PtCl4.6H2O ty;ksftr ladqy ds :i es fo|eku gks ldrk gSA 1 eksyy tyh; foy;u 3.72ºC fgekad es voueu
j[krk gSA 100% fo;kstu rFkk Kf (H2O) = 1.86ºC mol–1Kg ekfu,] rc ladqy gS &
(1) [Pt(H2O)6]Cl4 (2) [Pt (H2O)4Cl2]Cl2 . 2H2O
(3) [Pt(H2O)3Cl3]Cl. 3H2O (4) [Pt(H2O)2Cl4].4H2O
35. A(g) → 2B(g) + C(g) is a first order reaction. The initial pressure of the system was found to be 800
mm Hg which increased to 1600 mm Hg after 10 min. The total pressure of the system after 30 min will
be ________mm Hg.
A(g) → 2B(g) + C(g),d izFke dksfV dh vfHkfØ;k gSA fudk; dk izkjfEHkd nkc 800 mm Hg Fkk tksfd 10 min
ckn 1600 mm Hg gks x;kA fudk; dk dqy nkc 30 min ckn ________mm Hg gksxkA
(1) 2200 (2) 2100 (3) 2000 (4) 1900
36. The pair of lanthanides in which both elements have high third-ionization energy is:
ySUFksuk;M+ksa dk og ;qXe ftles nksuksa rRoksa dh r`rh; vk;uu ÅtkZ mPPk gS
(1) Eu, Yb (2) Eu, Gd (3) Dy, Gd (4) Lu, Yb
C-01JP-JPT2-070124-C1 15
CHEMISTRY
37. Given below are two statement one is labelled as Assertion A and the other is labelled as Reason R :
Assertion A : The amphoteric nature of water is explained by using Lewis acid/base concept
Reason R : Water acts as an acid with NH3 and as a base with H2S.
In the light of the above statements choose the correct answer from the options given below :
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true but R is NOT correct explanation of A.
(3) A is true but R is false.
(4) A is false but R is true.
uhps nks dFku fn, x, gSaA ,d vfHkdFku A gS vkSj nwljk dkj.k RA
vfHkdFku A : ty dh mHk;/kehZ izd`fr dh O;k[;k djus ds fy, ywbl vEy@{kkj dh ladYiuk dk mi;ksx djrs gSA
dkj.k R : ty veksfu;k ds lkFk vEy rFkk H2S ds lkFk {kkj dk dk;Z djrk gSA
mi;qZDr dFkuksa ds vk/kkj ij uhps fn, x, fodYiksa esa ls lgh mÙkj pqusa&
(1) A rFkk R nksuksa lR; gSa vkSj R, A dh lgh O;k[;k gSA
(2) A rFkk R nksuksa lR; gSa vkSj R, A dh lgh O;k[;k ugha gSA
(3) A lR; gS] ijUrq R vlR; gSA
(4) A vlR; gS] ijUrq R lR; gSA
38. Match list I with List II.

lwph I dk feyku lwph II ls djsa
List – I Complex List – II CFSE (0)
lwph – I ladqy lwph – II CFSE (0)
A. [Cu(NH3)6]2+ I. –0.6
B. [Ti(H2O)6]3– II. –2.0
C. [Fe(CN)6]3– III. –1.2
D. [NiF66] 4– IV. –0.4
Choose the correct answer from the options given below.
uhps fn, x, fodYiksa esa ls lgh mÙkj pqusa%
(1) A-I, B-IV, C-II, D-III (2) A-III, B-IV, C-I, D-II (3) A-II, B-III, C-I, D-IV (4) A-I, B-II, C-IV, D-III
C-01JP-JPT2-070124-C1 16
CHEMISTRY
39. When a solution of mixture having two inorganic salts was treated with freshly prepared ferrous
sulphate in acidic medium a dark brown ring was formed whereas on treatment with neutral FeCl 3, it
gave deep red colour which disppeared on boiling and a brown red ppt was formed. The mixture
contains
nks vdkcZfud lkYVksa ds feJ.k ds foy;u dks tc vEyh; ek/;e esa rqjar fufeZr Qsjl lYQsV ds foy;u ls
mipkfjr djrs gS rks Hkwjs jax dk oy; curk gS tcfd mnklhu FeCl3 ls mipkfjr djus ij xgjk yky jax izkIRk
gksrk gSA ;g jax foy;u dks mckyus ij mM+ tkrk gS vkSj Hkwjs&yky jax dk vo{ksi feyk gSA feJ.k esa mifLFkr gSA
(1) C2O42– & NO3– (2) SO32– & C2O42– (3) CH3COO– & NO3– (4) SO32– & CH3COO–
40. In given cyclic process, the work included is -

fn;s x;s pØh; izØe esa lfEefyr dk;Z gS–
2V
V
V
P 2P
P
(1) (–/4) PV (2) (/4) PV (3) 2 PV (4) (/2) PV
41. What is the value of pKb(CH3COO-), if m = 390 and m = 7.8 for 0.04 M of a CH3COOH solution at
25ºC?
pKb(CH3COO-) dk eku D;k gS] ;fn 25ºC ij CH3COOH foy;u ds 0.04M ds fy, m = 390 rFkk m = 7.8 gS\
(1) 9.3 (2) 9.2 (3) 4.7 (4) 4.8
C-01JP-JPT2-070124-C1 17
CHEMISTRY
42. The complex that dissolves in water is :
og ladqy tks ty esa ?kqyrk gS] gS&
(1) [Fe3(OH)2(OAc)6]Cl (2) K3[Co(NO2)6]
(3) (NH4)3[As(Mo3O10)4] (4) Fe4[Fe(CN)6]3
O N–OH H
N O
NH OH H O/ 
43. 
[O]
 
2
 
Re arrangement
 
2
 H2N–(CH2)5–COOH
( X)
-amino caproic acid

Polymerisaction
Nylon-6
The rearrangement (X) is called-
(1) Reformatsky reaction (2) Hoffmann's rearrangement
(3) Beckmann's rearrangement (4) Hoffmann bromamide degradation
O N–OH H
N O
NH OH iquZfoU;kl H O/ 

[O]
 
2
 
 
2
 H2N–(CH2)5–COOH
(X)
-,ehuks dsizksbd vEy
cgqyhdj.k

 ukbyksu-6
iquZfoU;kl (X) dgykrk gS &

(1) fjQkWesZV~Ldh vfHkfØ;k (2) gkWQeku iquZfoU;kl
(3) csdeku iquZfoU;kl (4) gkWQeku czksesekbM fuEuuhdj.k
C-01JP-JPT2-070124-C1 18
CHEMISTRY
CH3
(CH3CO)2 O Br2 / Fe H O / H
44.     
2
X
NH2
'X' is :
'X' gS &
CH3 CH3 CH3 CH3
Br COCH3
(1) (2) (3) (4)
Br COCH3
NH2 NH2 NH2 NH2
45. Find 'A' in the given sequence

fn;s x;s vuqØe esa 'A' Kkr dhft, &
O3 / H2O 
[A]  2[B]  2CH3COOH + 2CO2
(1) (2) (3) (4)
46. Which one is correct order acidic nature of nitrophenols.

(1) picric acid > p-nitrophenol > o-nitrophenol > m-nitrophenol
(2) p-nitrophenol > o-nitrophenol > m-nitrophenol > picric acid
(3) p-nitrophenol > m-nitrophenol > o-nitrophenol > picric acid
(4) p-nitrophenol > picric acid > m-nitrophenol > o-nitrophenol
fuEu esa ls dkSulk ,d ukbVªksfQukWy dh vEyh; izd`fr dk lgh Øe gS &
(1) fifØd vEy > p-ukbVªksfQukWy > o-ukbVªksfQukWy > m-ukbVªksfQukWy
(2) p-ukbVªksfQukWy > o-ukbVªksfQukWy > m-ukbVªksfQukWy > fifØd vEy
(3) p-ukbVªksfQukWy > m-ukbVªksfQukWy > o-ukbVªksfQukWy > fifØd vEy
(4) p-ukbVªksfQukWy > fifØd vEy > m-ukbVªksfQukWy > o-ukbVªksfQukWy
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CHEMISTRY
47. Identify 'z' in the reaction given below :
uhps nh x;h vfHkfØ;k esa 'z' dks igpkfu, &
NH2
(1) HNO2 (280K) NaOH CH3I

X Y Z
(2) H2O, boil
 –
NH–CH3 N2Cl OCH3 OCH3
H3C CH3 H3C CH3 HO OH
(1) (2) (3) (4)
CH3 CH3 OH
48. Among the bromides given below, the order of reactivity towards SN1 reaction is :
uhps fn;s x;s czksekbMks esa] SN1 vfHkfØ;k ds izfr fØ;k'khyrk dk Øe gS &
Br
Br Br
O
(I) (II) (III)
(1) III > II > I (2) II > III > I (3) III > I > II (4) II > I > III
49. The increasing order of dehydration of the following compounds is :

fuEu ;kSfxdks ds futZyhdj.k dk vkjksgh Øe gS &
OH OH OH OH
(I) (II) (III) (IV)

(1) I < II < III < IV (2) II < III < IV < I (3) I < III < II < IV (4) I < IV < II  III
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CHEMISTRY
50. Identify 'Z' in the following reaction :
uhps nh x;h vfHkfØ;k esa 'Z' dks igpkfu, &
OH
CHCl3 + NaOH NaOH

X Z
Conc.
OH OH OH OH
CH CH2–OH COONa
(1) (2) +
OH
OH OH OH OH
C C–CH
(3) (4)
O O OH

to be attempted.
C-01JP-JPT2-070124-C1 21
CHEMISTRY
djds mÙkj dk lgh iw.kk±d eku ntZ djsaA
51. See the following chemical reaction:

Cr2O72  XH  6Fe2  YCr 3   6Fe3   ZH2O
The sum of X, Y and Z is _________.

fuEUkfyf[kr vfHkfØ;k dks ns[ksa%
Cr2O72  XH  6Fe2  YCr 3   6Fe3   ZH2O
X, Y vkSj Z dk ;ksx_________ gSA
52. How many of the following complexes may be optically active ?

fuEUk ladqyksa esa ls fdrus ladqy izdkf'kd lfØ; gks ldrs gSa \
(a) [Ni(NH3)Br(Cl)(CN)]– (b) [Pt(NH3)Br(Cl)(CN)] –
(c) [Pt(gly)3]Cl (d) [Cu(gly)2]–
(e) K3[Fe(ox)3] (f) K3[Co(CO3)3]
(g) [Ni(dmg)2] (h) [Fe(acac)(H2O)2(gly)]Br
(i) [Co(gly)(NO2)4]2–
C-01JP-JPT2-070124-C1 22
CHEMISTRY
53. To 500 mL of xM NaOH, when 500 ml of 0.6 M CuSO 4 solution is added, the pH of NaOH solution
decreased by 1 unit. The final solution turns red litmus blue. Find 2x. (Assume no dissociation of
Cu(OH)2(s))
xM NaOH ds 500 mL esa tc 0.6 M CuSO4 dk 500 ml foy;u feykrs gS] rks NaOH foy;u dh pH 1 bdkbZ ls
?kVrh gSA vfUre foy;u yky fyVel dks uhyk djrk gSA 2x Kkr dhft,A ¼ekuk Cu(OH)2(s) dk fo;kstu ugh
gksrk gSA½

54. A  B (zero order reaction)
C  D (first order reaction)
If first order reaction gets 75% completed in 40 min and zero order reaction gets 75% completed in 30
min then calculate the value of Z. Where
Half life period of first order reaction
Z=
Half life period of zero order reaction
A  B ¼'kwU; dksfV vfHkfØ;k½
C  D ¼izFke dksfV vfHkfØ;k½
;fn izFke dksfV vfHkfØ;k 40 feuV esa 75% iw.kZ gksrh gS rFkk 'kwU; dksfV vfHkfØ;k 30 feuV es 75% iw.kZ gksrh gS] rks
Z dk eku ifjdfyr dhft,] tgk¡
izFke dksfV vfHkfØ;k dk v)Z vk;q dky
Z=
'kwU; dksfV vfHkfØ;k dk v)Z vk;q dky
55. Calculate acid dissociation constant for 0.1 M HCOOH if its solution shows a resistance of 50  when
filled in a cell having separation between parallel electrodes 4 cm and cross section area of electrode
10 cm2

Given, m [Ca(HCOO)2 ] = 230 Scm2 mol–1
m [CaCl2 ]  280 Scm2 mol–1
m [HCl] = 425 Scm2 mol–1
(Fill your answer after multiplying the answer with 103)
C-01JP-JPT2-070124-C1 23
CHEMISTRY
0.1 M HCOOH ds fy, vEy fo;kstu fLFkjkad ifjdfyr dhft;s ;fn bldk foy;u 50  dk izfrjks/k n'kkZrk gS
tc lekUrj bysDVªkWMks ds e/; 4 cm nwjh j[kus okys ,d lsy esa bls Hkjk tkrk gS rFkk bysDVªkWM dk vuqizLFk dkV
{ks=kQy 10 cm2 gSA
fn;k gS m [Ca(HCOO)2 ] = 230 Scm2 mol–1
m [CaCl2 ]  280 Scm2 mol–1
m [HCl] = 425 Scm2 mol–1
(viuk mÙkj 103 ls xq.kk djus ds i'pkr~ nhft;s)
56. For a reversible reaction A B, the Hforward reaction = 20 kJ mol–1. The activation energy of the
uncatalysed forward reaction is 300 kJ mol–1. When the reaction is catalysed keeping the reactant
concentration same, the rate of the catalysed forward reaction at 27°C is found to be same as that of
the uncatalysed reaction at 327°C. The activation energy of the catalysed backward reaction
is________ kJ mol–1.
mRØe.kh; vfHkfØ;k A B ds fy,] Hforward reaction = 20 kJ mol–1 gSA fcuk mRizsjd ds vxz vfHkfØ;k ds fy,
lfØ;.k mtkZ 300 kJ mol–1 gSA vfHkdkjdksa dh lkanzrk leku j[krs gq, tc vfHkfØ;k dks mRizsjd ds lkFk lEikfnr
fd;k tkrk gS] rks 27°C ij mRizsfjr vxz vfHkfØ;k dh nj 327°C ij fcuk mRizsjd ds vfHkfØ;k ds leku ik;h tkrh
gSA mRizsfjr i'Pkkr vfHkfØ;k dh lfØ;.k mtkZ gS%________ kJ mol–1.
57. A2 + B2  2AB. H0f = –200 kJ mol–1

AB, A2 and B2 are diatomic molecules. If the bond enthalpies of A2, B2 and AB are in the ratio 1 : 0.5 : 1,
then the bond enthalpy of A2 is ____________kJ mol–1 (Nearest integer)
A2 + B2  2AB. H0f = –200 kJ mol–1
AB, A2 rFkk B2 f}ijekf.Okd v.kq gSA ;fn A2, B2 rFkk AB dh vkcU/k ,UFkSYih;k¡ 1 : 0.5 : 1 vuqikr esa gSa rks A2 dh
vkcU/k ,UFkSYih____________kJ mol–1 gksxh (fudVre iw.kkZad esa)
C-01JP-JPT2-070124-C1 24
CHEMISTRY
58. Sum of total number of optically active and optical inactive isomers of following compound.
fuEufyf[kr ;kSfxd ds izdkf'kd lfØ; rFkk izdkf'kd vlfØ; leko;oh;ksa dh dqy la[;k dk ;ksx gS &
59. The total number of carboxylic acid groups in the product P is :

mRikn P esa dkcksZfDlfyd vEy lewgksa dh dqy la[;k gS &
60. Total number of products formed in the following reaction is :

fuEu vfHkfØ;k esa fufeZr mRiknksa dh dqy la[;k gS &
C-01JP-JPT2-070124-C1 25
MATHEMATICS
PART-C (Hkkx-C) : MATHEMATICS (xf.kr)
 'kwU; vad % 0 ;fn dksbZ Hkh fodYi ugh pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A
1 1
61. The set of complete values of x which satisfy the inequation x–1
< x .
1– 2 2 –1
1 1
vlfedk x–1
< x dks larq"V djus okys x ds ekuksa lEiw.kZ leqPp; gS&
1– 2 2 –1
(1) (1, 2) (2) (0, log2 4 / 3)
(3) x  (0, log2 4/ 3)  (1, ) (4) (log2 8 / 3, )
M-01JP-JPT2-070124-C1 26
MATHEMATICS
62. A spherical balloon is expanding. If the radius is increasing at the rate of 2 inches per minute, the rate
at which the volume increases (in cubic inches per minute) when the radius is 5 inches is
,d xksykdkj xqCckjk QSy jgk gSA ;fn f=kT;k esa o`f) nj 2 bap@feuV gks] rks vk;ru esa o`f) nj ¼?ku@feuV½ tc
f=kT;k 5 bap gks] gS&
(1) 10  (2) 100  (3) 200  (4) 50 

63. The function f(x) = 1 + x(sinx)[cos x],0< x  /2, where [.] denotes greatest integer function.
(1) is discontinuous in (0,/2) (2) is strictly decreasing in (0,/2)
(3) is strictly increasing in (0,/2) (4) has global maximum value 1.
Qyu f(x) = 1 + x(sinx)[cos x],0 < x  /2, tgk¡ [.] egÙke iw.kk±d Qyu dks O;Dr djrk gS&
(1) (0,/2) ij vlrr~ gSA (2) (0,/2) esa
fujUrj áleku gSA
(3) (0,/2) esa fujUrj o/kZeku gSA (4) ije mfPp"B eku 1 Gsa
 51  50   49   48   47   47  n
  +   +   +   +   +   is equal to (where   denotes Cr)
64. n
3 3  3   3   3   4 r 
 51  50   49   48   47   47  n n
  +   +   +   +   +   (tgk¡   , Cr dks iznf'kZr djrk gS) dk eku gS&
3 3  3   3   3   4 r 
 52   52   52   52 
(1)   (2)   (3)   (4)  
 1  2 3  4
65. The equation of the diameter of the circle (x – 2)2 + (y + 1)2 = 16 which bisects the chord cut off by the
circle on the line x – 2y – 3 = 0 is
(1) x + 2y = 0 (2) 2x + y – 3 = 0 (3) 3x + 2y – 4 = 0 (4) none of these
o`Ùk (x – 2) + (y + 1) = 16 ds O;kl dk og lehdj.k tks o`Ùk }kjk ljy js[kk x – 2y – 3 = 0 ij dkVh xbZ thok
2 2
dks lef}Hkkftr djrk gS &

(1) x + 2y = 0 (2) 2x + y – 3 = 0 (3) 3x + 2y – 4 = 0 (4) buesa ls dksbZ ugha
M-01JP-JPT2-070124-C1 27
MATHEMATICS
z – 1  i 
66. If is purely imaginary then z lies on
z  1  i 
(1) a straight line (2) a circle (3) a line segment (4) None of these
z – 1  i 
;fn fo'kq) dkYifud gks] rks z fLFkr gS&
z  1  i 
(1) ljy js[kk ij (2) o`Ùk ij (3) ,d js[kk[k.M ij (4) buesa ls dksbZ ugha
67. STATEMENT-1 : Arg (2 + 3i) + Arg (2 – 3i) = 0 (Arg z stands for principal argument of z)
STATEMENT-2 : Arg z + Arg z = 0, z = x + iy,  x, y  R (Arg z stands for principal argument of z)
(1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(2) Statement-1 is True, Statement-2 is False
(3) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(4) Statement-1 is False, Statement-2 is True
dFku-1 : Arg (2 + 3i) + Arg (2 – 3i) = 0 (Arg z, z ds eq[; dks.kkad dks iznf'kZr djrk gS)
dFku-2 : Arg z + Arg z = 0, z = x + iy,  x, y  R (Arg z, z ds eq[; dks.kkad dks iznf'kZr djrk gS)
(1) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gS
(2) oDrO;-1 lR; gS] oDrO;-2 vlR; gS
(3) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k ugha gS
(4) oDrO;-1 vlR; gS] oDrO;-2 lR; gS
68. If f(x) = |x – a| (x), where (x) is continuous function, then

(1) f (a+) = –(a) (2) f (a –) = –(a) (3) f (a+) = f (a –) (4) none of these
;fn f(x) = |x – a| (x), tgk¡ (x) lrr~ Qyu gS] rc
(1) f (a+) = –(a) (2) f (a –) = –(a) (3) f (a+) = f (a –) (4) buesa ls dksbZ ugha
M-01JP-JPT2-070124-C1 28
MATHEMATICS
69. Consider the following statements :
If f and g be two functions such that f(x)  0, g(x)  f(x)  0 and g(x)  0 for all x, then
 f ' f (fg) f  g (f  g) f  g (f / g) f  g
(i)   = (ii) = + (iii) = + (iv) = –
g g fg f g f g f g f /g f g
which of these statements are correct ?
(1) i and ii (2) ii and iii (3) ii and iv (4) iii and iv
ekukfd dFku :
;fn f vkSj g nks Qyu bl izdkj gS fd f(x)  0, g(x)  f(x)  0 vkSj g(x)  0 lHkh x ds fy,, rc
 f ' f (fg) f  g (f  g) f  g (f / g) f  g
(i)   = (ii) = + (iii) = + (iv) = –
g g fg f g f g f g f /g f g
fuEu esa ls dkSuls dFku lgh gS ?
(1) i vkSj ii (2) ii vkSj iii (3) ii vkSj iv (4) iii vkSj iv
a a 
1  sin2 t
  esin t dt  is equal to
2
70. lim e dt –
x 0 x  
y xy 
a a 
1  sin2 t
  esin t dt  cjkcj gS&
2
lim e dt –
x 0 x  
y xy 
2 2
(1) 1 (2) 0 (3) esin y
(4) sin2y. esin y
dy
71. If c is an arbitrary constant then a solution of the differential equation (tany) = sin(x + y) + sin(x – y), is
dx
dy
;fn c LoSPN vpj gS] rc vody lehdj.k (tany) = sin(x + y) + sin(x – y) dk ,d gy gS&
dx
(1) secy + 2cosx = c (2) secy – 2cosx = c
(3) cosy – 2sinx = c (4) none of these ¼buesa ls dksbZ ugha½
M-01JP-JPT2-070124-C1 29
MATHEMATICS
72. The ellipse x2 + 4y2 = 4 is inscribed in a rectangle alingent with the coordinate axes, which in turn is
inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is :
nh?kZo`Ùk x2 + 4y2 = 4 funsZ'kkad v{kksa ls lajsf[kr ,d vk;r ds vUrxZr gS tks Lo;a fcUnq (4, 0) ls tkus okys nwljs
nh?kZo`Ùk ds vUrxZr gSA rc bl nh?kZo`Ùk dk lehdj.k gS %
(1) x2 + 12y2 = 16 (2) 4x2 + 48y2 = 48 (3) 4x2 + 64y2 = 48 (4) x2 + 16y2 = 16
x 1
73. If f(x) = , then f(f(ax)) is equal to
x 1
x 1
;fn f(x) = gks] rks f(f(ax)) dk eku gSµ
x 1
1 f ( x)  1 f ( x)  1 f ( x)  1
(1) (2) (3) (4)
ax a( f ( x )  1) a( f ( x )  1) a( f ( x )  1)
1
74. A hyperbola having the transverse axis of length unit is confocal with the ellipse 3x2 + 4y2 = 12, then
2
which of the following is INCORRECT ?
x2 y2 1
(1) equation of hyperbola is – 
15 1 16
(2) eccentricity of the hyperbola is 4
1
(3) distance between the directrices of the hyperbola is units
8
15
(4) length of latus rectum of the hyperbola is unit.
2
1
,d vfrijoy; dh bdkbZ yEckbZ dh vuqizLFk v{k gS] ftldh nh?kZo`Ùk 3x2 + 4y2 = 12 ds lkFk ukfHk;ka laikrh gS
2
rc fuEu esa ls dkSulk xyr gS \
x2 y2 1
(1) vfrijoy; dk lehdj.k –  gSA
15 1 16
(2) vfrijoy; dh mRdsUnzrk 4 gSaA
1
(3) vfrijoy; dh fu;rkvksa ds e/; nwjh bdkbZ gSA
8
Space for Rough
15 Work / (dPps dk;Z ds fy, LFkku)
(4) vfrijoy; ds ukfHkyEc dh yEckbZ bdkbZ gSA
2
M-01JP-JPT2-070124-C1 30
MATHEMATICS
x –1
75. x x 1
is equal to
x –1
x x 1
cjkcj gS&
(1) n x  x 2 – 1 – sec –1 x  C (2) n x  x 2  1 – sec –1 x  C
(3) n x  x 2 – 1 – tan–1 x  C (4) n x  x 2  1  tan–1 x  C
1
76. lim ( nx)1– nx
is equal to
xe
1
lim ( nx)1–
xe
nx
dk eku gS&
(1) 0 (2) e (3) e–1 (4) e2
77. If the equation ax + 4y + z = 0, bx + 3y + z = 0, cx + 2y + z = 0 have non-trivial solution, then a + c =

;fn lehdj.k ax + 4y + z = 0, bx + 3y + z = 0, cx + 2y + z = 0 ds v'kwU; gy gS] rc a + c =
(1) b2 (2) 0 (3) 2b (4) b
78. If R is a relation defined on the set of integers as follows :

'aRb  a  2k.b , for some integer k, then R is
(1) reflexive but not symmetric, transitive (2) reflexive and symmetric but not transitive
(3) reflexive and transitive but not symmetric (4) equivalence
;fn iw.kk±dksa ds leqPp; ij ifjHkkf"kr lEcU/k R fuEu gS :
'aRb  a  2k.b tgk¡ k ,d iw.kk±d gS rc R gS&
(1) LorqY; ijUrq lefer] laØked ugha (2) LorqY; vkSj lefer ijUrq laØked ugha
(3) LorqY; vkSj laØked ijUrq lefer ugha (4) rqY;rk
M-01JP-JPT2-070124-C1 31
MATHEMATICS
79. If a, b, c > 0 and ab + bc + ca = A and abc = G then H.M. of a, b, c is
;fn a, b, c > 0 vkSj ab + bc + ca = A rFkk abc = G gS rc a, b, c dk gjkRed ek/; gS&
3G 3G G G
(1) (2) (3) (4)
2A A 3A A
a.(b  c)
80. If b and c are two perpendicular unit vectors and a is any vector the (a.b) b  (a.c) c  (b  c)
| b  c |2
is equal to
a.(b  c)
;fn b vkSj c nks yEcor~ bZdkbZ lfn'k gS vkSj a dksbZ lfn'k gS] rks (a.b) b  (a.c) c  (b  c) cjkcj
| b  c |2
gS&
(1) 0 (2) a (3) b (4) c

to be attempted.
M-01JP-JPT2-070124-C1 32
MATHEMATICS
djds mÙkj dk lgh iw.kk±d eku ntZ djsaA
81. Number of solution of equation 52 log5 x = x + 6 is (are)

lehdj.k 52 log5 x = x + 6 ds gyksa dh la[;k gS -
82. A line L varies such that length of perpendicular on it from origin O is always 4 units. If L cuts x-axis
and y-axis at A and B respectively then minimum value of (OA)2 + (OB)2 is
,d js[kk L bl izdkj ifjofrZr gks jgh gS fd ml ij ewyfcUnq O ls Mkys x, yEc dh yEckbZ lnSo 4 bdkbZ jgrh
gSA ;fn L, x-v{k rFkk y-v{k dks Øe'k% A rFkk B ij dkVrh gS] rks (OA)2 + (OB)2 dk U;wure eku gS &
83. Last three digits of the number N = 7100 – 3100 are

N = 7100 – 3100 ds vafre 3 vad gS&
84. If , ,  are the roots of the equation x2 ( cx + d ) = e ( x + 1 ). Then the value of the determinant
1  1 1
1 1  1 is
1 1 1 
1  1 1
;fn , ,  lehdj.k x2 ( cx + d ) = e ( x + 1 ) ds ewy gS] rks lkjf.kd 1 1  1 dk eku gS&
1 1 1 
M-01JP-JPT2-070124-C1 33
MATHEMATICS
85. Number of divisors of 2 .3 .5 which are divisible by 2 but not divisible by 6 and 8, is –
5 2 3
25. 32.53 ds 2 ls foHkkftr ysfdu 6 vkSj 8 ls foHkkftr ugha gksus okys Hkktdksa dh la[;k gS &
86. We are required to form different words with the help of the letters of the word INTEGER. Let m 1 be the
number of words in which I and N are never together and m 2 be the number of words which begin with I
and end with R, then m1 / m2 is given by –
gesa 'kCn INTEGER ds v{kjksa dh lgk;rk ls fHkUu fHkUu 'kCn cukus gSA ekuk ,sls 'kCn ftuesa I vkSj N dHkh Hkh lkFk
u gks] dh la[;k m1 gS rFkk ,sls 'kCn tks I ls izkjEHk gksdj R ij [kRe gks] dh la[;k m2 gS] rc m1 / m2 gksxk&
87. The marks of some students were listed out of a maximum 75. The SD of marks was found to be 9.
Subsequently the marks were raised to a maximum of 100 and variance of new marks was calculated.
The new variance is
dqN fo|kfFkZ;ksa dks vf/kdre 75 vadksa esa ls vad fn;s tkus ij vadksa dk ekud fopyu 9 ik;k x;kA Øfed :i ls
vadksa dks vf/kdre 100 rd c<k;k tkrk gS vkSj u;s vadksa ds izlj.k dh x.kuk dh tkrh gSA u;k izlj.k eku gksxk&
x–3 y–8 z–3 x3 y7 z–6
88. The square of shortest distance between the lines
= = and = = is
3 –1 1 –3 2 4
x–3 y–8 z–3 x3 y7 z–6
js[kkvksa = = ,oa = = ds e/; U;wure nwjh gS&
3 –1 1 –3 2 4
89. A line makes the same angle  with each of the x and z axes. If the angle , which it makes with y-axis,
is such that sin2  = 2 sin2 , then 2cos2  equals
,d js[kk x-v{k rFkk z-v{k ds lkFk lekudks.k cukrh gSA ;fn dks.k tks ;g js[kk y-v{k ds lkFk cukrh gS bl
izdkj gS fd sin2  = 2 sin2 , rks 2cos2  cjkcj gS &
90. The area of the region(s) enclosed by the curves y = x 2 and y = | x | is A, then 81A is –
oØks y = x2 vkSj y = | x | ls ifjc) {ks=k dk {ks=kQy A gS rc 81A gS-

M-01JP-JPT2-070124-C1 34

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JEE (MAIN+ADVANCED)

Academic Session: 2023-24

TARGET TEST PATTERN JEE (MAIN)

PAPER CODE 1 MAX. MARKS 300

TEST PAPER DETAILS MARKING SCHEME

TYPE WISE CODES FOR QUESTIONS

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

(1) at t = 2 sec and current flowing in ring is equal to zero

(1) t = 2 sec ij oy; esa izokfgr /kkjk 'kwU; gksxhA

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

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

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

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

(1) 1.2 (2) 1.5 (3) 1.8 (4) 2.0

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

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

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

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

1 nb / a 2 1 n(b / a)2 (b  a)

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

SECTION B (Maximum Marks: 20)

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

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

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

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

31. Given below are two statements:

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

33. Given below are two statements:

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

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

38. Match list I with List II.

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

40. In given cyclic process, the work included is -

(1) (–/4) PV (2) (/4) PV (3) 2 PV (4) (/2) PV

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

iquZfoU;kl (X) dgykrk gS &

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

45. Find 'A' in the given sequence

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

46. Which one is correct order acidic nature of nitrophenols.

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

(1) HNO2 (280K) NaOH CH3I

49. The increasing order of dehydration of the following compounds is :

(I) (II) (III) (IV)

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

CHCl3 + NaOH NaOH

SECTION B (Maximum Marks: 20)

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

51. See the following chemical reaction:

The sum of X, Y and Z is _________.

X, Y vkSj Z dk ;ksx_________ gSA

52. How many of the following complexes may be optically active ?

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

57. A2 + B2  2AB. H0f = –200 kJ mol–1

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

59. The total number of carboxylic acid groups in the product P is :

60. Total number of products formed in the following reaction is :

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

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