Post on 30-Nov-2021
Name of the Candidate (in Capitals)ijh{kkFkhZ dk uke (cM+s v{kjksa esa) :Form Number : in figuresQkWeZ uEcj : vadksa esa
: in words: 'kCnksa esa
Centre of Examination (in Capitals) :ijh{kk dsUæ (cM+s v{kjksa esa) :Candidate’s Signature : Invigilator’s Signature :ijh{kkFkhZ ds gLrk{kj : fujh{kd ds gLrk{kj :
Read carefully the Instructions this Test Booklet.bl ijh{kk iq fLrdk ij fn, funsZ'kk s a dks /;ku ls i<+ s aA
Do not open this Test Booklet until you are asked to do so.bl ijh{kk iqfLrdk dks tc rd uk [kk sysa tc rd dgk u tk,A
)0000CJA102120026)(0000CJA102120026) Test Pattern
egRoiw.kZ funs Z'k :1. ijh{kk iqfLrdk ds bl i`"B ij vko';d fooj.k uhys@dkys ckWy ikbaV
isu ls rRdky HkjsaA isfUly dk iz;ksx fcYdqy oftZr gSaA
2. ijh{kkFkhZ viuk QkeZ ua- (fu/kkZfjr txg ds vfrfjä) ijh{kk
iqfLrdk@mÙkj i= ij dgha vkSj u fy[ksaA
3. ijh{kk dh vof/k 3 ?kaVs gSA4. bl ijh{kk iqfLrdk esa 90 iz'u gaSA vf/kdre vad 300 gSaA
5. bl ijh{kk iq fLrdk esa rhu Hkkx 1, 2, 3 gSa] ftlds izR;sd Hkkx esa HkkSfrdfoKku] jlk;u foKku ,oa xf.kr ds 30 iz'u gSa vkSj izR;sd fo"k;esa 2 [k.M gSA(i) [k.M-I esa 20 cgqfodYih; iz'u gSA ftuds dsoy ,d fodYi
lgh gSaAvad ;k stuk : +4 lgh mÙkj ds fy,] 0 iz;kl ugha djus ijrFkk –1 vU; lHkh voLFkkvksa esaA
(ii) [k.M-II esa 10 la[;kRed eku izdkj ds iz'u gSA fdUgh 5 iz'uksadk mÙkj nhft,A fd;s x;s iz'uksa esa ls dsoy izFke ik¡p iz'uksadks gh vad fn;s tk;saxsaAvad ;kstuk : +4 lgh mÙkj ds fy, rFkk 0 vU; lHkh voLFkkvksa esaA
6. mÙkj i= ds i`"B–1 ,oa i`"B–2 ij okafNr fooj.k ,oa mÙkj vafdr djusgsrq dsoy uhys@dkys ckWy ikbaV isu dk gh iz;ksx djsaA isfUly dk iz;ksxloZFkk oftZr gSA
7. ijh{kkFkhZ }kjk ijh{kk d{k @ gkWy esa ifjp; i= ds vykok fdlh Hkh izdkjdh ikB~; lkexzh eqfær ;k gLrfyf[kr dkxt dh ifpZ;ksa] eksckbyQksu ;k fdlh Hkh izdkj ds bysDVªkfud midj.kksa ;k fdlh vU; izdkjdh lkexzh dks ys tkus ;k mi;ksx djus dh vuqefr ugha gSaA
8. jQ dk;Z ijh{kk iqfLrdk esa dsoy fu/kkZfjr txg ij gh dhft;sA
9. ijh{kk lekIr gksus ij] ijh{kkFkhZ d{k@gkWy NksM+us ls iwoZ mÙkj i= d{kfujh{kd dks vo'; lkSai nsaA ijh{kkFkhZ vius lkFk bl ijh{kk iq fLrdkdks ys tk ldrs gSaA
10. mÙkj i= dks u eksM+ s a ,oa u gh ml ij vU; fu'kku yxk,s aA
Important Instructions :1. Immediately fill in the form number on this page of the
Test Booklet with Blue/Black Ball Point Pen. Use of pencilis strictly prohibited.
2. The candidates should not write their Form Numberanywhere else (except in the specified space) on the TestBooklet/Answer Sheet.
3. The test is of 3 hours duration.4. The Test Booklet consists of 90 questions. The maximum
marks are 300.5. There are three parts in the question paper 1,2,3
consisting of Physics, Chemistry and Mathematicshaving 30 questions in each subject and each subjecthaving Two sections.(i) Section-I contains 20 multiple choice questions
with only one correct option.Marking scheme : +4 for correct answer, 0 if notattempted and –1 in all other cases.
(ii) Section-II contains 10 Numerical Value Typequestions. Attempt any 5 questions.First 5 attempted questions will be considered formarking.Marking scheme : +4 for correct answer and 0 inall other cases.
6. Use Blue/Black Ball Point Pen only for writtingparticulars/marking responses on Side–1 and Side–2 of theAnswer Sheet. Use of pencil is strictly prohibited.
7. No candidate is allowed to carry any textual material,printed or written, bits of papers, mobile phone anyelectronic device etc, except the Identity Card inside theexamination hall/room.
8. Rough work is to be done on the space provided for thispurpose in the Test Booklet only.
9. On completion of the test, the candidate must hand overthe Answer Sheet to the invigilator on duty in the Room/Hall. However, the candidate are allowed to take awaythis Test Booklet with them.
10. Do not fold or make any stray marks on the Answer Sheet.
CLASSROOM CONTACT PROGRAMME(Academic Session : 2020 - 2021)
JEE(Main+Advanced) : ENTHUSIAST & LEADER COURSE
Your Target is to secure Good Rank in JEE(Main) 2021
JEE(Main)AIOT
07-03-2021
Paper : Physics, Chemistry & Mathematicsç'u iq fLrdk : HkkSfrd foKku] jlk;u foKku rFkk xf.kr
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan) INDIA-324005
+91-744-2757575 info@allen.ac.in www.allen.ac.in
Hin
di
ALLEN
Page 2/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026
SECTION–I : (Maximum Marks : 80)
� This section contains TWENTY questions.
� Each question has FOUR options (A), (B), (C)
and (D). ONLY ONE of these four options iscorrect.
� For each question, darken the bubble
corresponding to the correct option in the ORS.
� For each question, marks will be awarded in
one of the following categories :Full Marks : +4 If only the bubblecorresponding to the correct option is darkened.Zero Marks : 0 If none of the bubbles isdarkened.Negative Marks : –1 In all other cases
1. Two blocks A and B of masses m and 2mrespectively are connected by a spring offorce constant k. The masses are moving tothe right with uniform velocity v = 10m/seach, the heavier mass leading the lighterone. The spring in between them is of naturallength during the motion. Block B collideswith a third block C of mass m, at rest. Thecollision being completely inelastic. Findminimum speed (in m/s) of block A in thesubsequent motion.(A) 5 (B) 6 (C) 8 (D) 10
[k.M–I : (vf/kdre vad : 80)
� bl [k.M esa chl iz'u gSa
� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj
(D) gSaA ftuesa dsoy ,d gh lgh gSaA
� izR;sd iz'u ds fy, vks-vkj-,l ij lgh mÙkj fodYi
ds vuq:i cqycqys dks dkyk djsaA
� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa
ls fdlh ,d ds vuqlkj fn;s tk,axs %
iw.kZ vad : +4 ;fn flQZ lgh fodYi ds vuq:i cqycqys
dks dkyk fd;k gSA
'kwU; vad : 0 ;fn fdlh Hkh cqycqys dks dkyk ugha fd;k gSA
½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA
1. æO;eku m rFkk 2m okys nks CykWd Øe'k% A rFkk B dks
cy fu;rkad k okyh ,d fLizax ls tksM+k x;k gSA izR;sd
æO;eku nk¡;h vksj ,dleku osx v = 10m/s ls xfr'khy
gS rFkk Hkkjh æO;eku] gYds æO;eku ls vkxs gSA xfr ds
nkSjku muds e/; yxh fLizax ewy yEckbZ esa gksrh gSA CykWd
B fojkekoLFkk esa fLFkr æO;eku m okys ,d rhljs CykWd
C ls Vdjkrk gSA VDdj iw.kZr;k vizR;kLFk gksrh gSA rnksijkUr
xfr esa CykWd A dh U;wure pky (m/s esa) gS%&
(A) 5 (B) 6 (C) 8 (D) 10
PART 1 - PHYSICS
ALLEN
07032021 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg Page 3/37All India Open Test/Enthusiast & Leader Course 0000CJA102120026
2. In the given circuit diagram, switch wasconnected to position 1 for long time. Att = 0, switch is shifted from position 1 toposition 2. Find the final charge on capacitor2C.
C 2C
C
V 2
1
(A) 6CV
(B) 3CV
(C) 2
3CV
(D) 4
3CV
3. The region between two concentric spheresof radii R1 and R2(R1 < R2) has volume charge
density r =br , where b is constant and r is
the radial distance. A point charge q = 16 µCis placed at the origin, r = 0. Find the valueof b (in SI units) for which the electric fieldin the region between spheres is constant.
(Take : R2 = 2R1 = 4p
mm ).
R1R2 q
(A) 1 (B) 2 (C) 4 (D) 8
2. iznf'kZr ifjiFk esa fLop fLFkfr 1 ij yEcs le; ds fy,tqM+k gqvk FkkA t = 0 ij fLop dks fLFkfr 1 ls fLFkfr 2 ijfoLFkkfir fd;k tkrk gSA la/kkfj= 2C ij vfUre vkos'kKkr dhft;sA
C 2C
C
V 2
1
(A) 6CV
(B) 3CV
(C) 2
3CV
(D) 4
3CV
3. f=T;k R1 rFkk R2(R1 < R2) okys nks ladsUæh; xksyksa ds
eè; izHkkx dk vk;ru vkos'k ?kuRo r =br gS] tgk¡ b
,d vpj rFkk r f=T;h; nwjh gSA ,d fcUnq vkos'kq = 16 µC dks ewy fcUnq (r = 0 ij) j[kk tkrk gS A bdk og eku (SI bdkbZ esa ) Kkr dhft;s] ftlds fy;sxksyksa ds e/; izHkkx esa fo|qr {ks= fu;r gSA
(R2 = 2R1 = 4p
mm ysa)
R1R2 q
(A) 1 (B) 2 (C) 4 (D) 8
ALLEN
Page 4/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026
4. In a football game,a player wants to hit afootball from the ground to one of histeammates, who is running on the field. Takehitter position as origin & receiver's initial
position as ˆ ˆ2i 3j+ , where i & j are in theplane of field. Football's initial velocity
vector is ˆ ˆ ˆ2i 5 j 25k+ + & in the subsequent
run receiver displacement is ˆ5i & ˆ8j , then
ˆ ˆ2 i 4 j+ & then ˆ6 j . How far is the receiverfrom the football when football lands on
ground ? (assume ˆg 10k= -r )
(A) 10 (B) 17 (C) 26 (D) 135. A hollow hemi-sphere of mass m and radius
r is released from rest in the position shown.Knowing that the hemi-sphere rolls withoutsliding. Determine the reaction at thehorizontal surface at the instant when it hasrolled through 90°.
(A) 7mg
4 (B) 5mg
2
(C) 8mg
3 (D) 5mg
3
4. QqVckWy ds [ksy esa ,d f[kykM+h QqVckWy dks /kjkry lsmlds fdlh ,d lkFkh f[kykM+h dh vksj fdd ekjukpkgrk gS] tks fd eSnku ij nkSM+ jgk gSA xsan ij izgkj dhfLFkfr ewyfcUnq ij rFkk xsan idM+us okys f[kykM+h dh
izkjfEHkd fLFkfr ˆ ˆ2i 3j+ ij ekfu;s] tgk¡ i rFkk j eSnkuds ry es a g SA QqVck Wy dk i z kjfEHkd osx lfn'k
ˆ ˆ ˆ2i 5 j 25k+ + gS rFkk rnksijkUr nkSM+us esa xsan idM+us
okys f[kykM+h dk foLFkkiu ˆ5i rFkk ˆ8j , fQj ˆ ˆ2i 4 j+ ;
rFkk fQj ˆ6 j gSA tc QqVckWy /kjkry ij fxjrh gS rksxsan idM+us okys f[kykM+h dh QqVckWy ls nwjh gS%&
(ekuk ˆg 10k= -r )
(A) 10 (B) 17 (C) 26 (D) 135. æO;eku m rFkk f=T;k r okys ,d [kks[kys v¼Z xksys dks
iznf'kZr fLFkfr ls fojkekoLFkk ls NksM+k tkrk gSA ekuk fd
v¼Z xksyk fcuk fQlys yq<+drk gSA tc ;g 90° rd
yq<+d pqdk gksrk gS rks ml {k.k ij {kSfrt lrg ij izfrfØ;k
Kkr dhft;sA
(A) 7mg
4 (B) 5mg
2
(C) 8mg
3 (D) 5mg
3
ALLEN
07032021 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg Page 5/37All India Open Test/Enthusiast & Leader Course 0000CJA102120026
6. Observer standing at the sea coast observes30 waves reaching the coast per minute. Thewavelength of each wave is 10m. Whenobserver drives a boat in sea against waves,with velocity 5 m/sec then n waves strikesthe boat per minute. Velocity of wave andvalue of n is respectively (in SI units)(A) 5, 30 (B) 10, 60(C) 10, 30 (D) 5, 60
7. In a meter Bridge experiment the resistanceof resistance box is 16 W , which is insertedin right gap. The null point is obtained at25cm from the left end. The leastcount formeter scale is 1mm. The percentage errorin calculating unknown resistance isapproximately:(A) 0.23 (B) 0.44 (C) 0.35 (D) 0.53
8. A source contains two phosphorus
radionuclides ( )3215 1/2 1P T T= & ( )33
15 1/2 2P T T= .
Initially 10% of decays comes from 3315P . How
long one must wait until 90% to do so?
(A)
1 2
4 n3t1 1n2T T
=æ ö
-ç ÷è ø
l
l
(B)
1 2
4 n3t1 1n2T T
=æ ö
+ç ÷è ø
l
l
(C)
1 2
2 n3t1 1n2T T
=æ ö
-ç ÷è ø
l
l
(D) None
6. leqæh rV ij [kM+k ,d izs{kd izfr feuV rV ij igq¡pus
okyh 30 rjaxksa dks izsf{kr djrk gSA izR;sd rjax dh rjaxnSè;Z
10m gSA tc izs{kd ,d uko dks leqæ esa rjaxksa ds fo:¼
5m/sec osx ls pykrk gS rks uko ls izfr feuV n rjaxs
Vdjkrh gSA rjax dk osx rFkk n dk eku (SI bdkbZ esa)Øe'k% gS%&(A) 5, 30 (B) 10, 60(C) 10, 30 (D) 5, 60
7. ,d ehVj lsrq iz;ksx esa izfrjks/k ckWDl dk izfrjks/k 16 W gS
tks fd nkfgus fjDr LFkku esa izfo"B djk;k x;k gSA 'kwU;
fcUnq cka;s fljs ls 25 cm nwjh ij izkIr gksrk gSA ehVj
iSekus dk vYirekad 1mm gSA vKkr izfrjks/k dh x.kuk
djus esa izfr'kr = qfV yxHkx gS%&
(A) 0.23 (B) 0.44 (C) 0.35 (D) 0.53
8. ,d L=ksr esa nks QkWLQksjl jsfM;ksU;wDykbM ( )3215 1/2 1P T T=
rFkk ( )3315 1/2 2P T T= gSA izkjEHk esa fo?kVu dk 10%,
3315P ls izkIr gksrk gSA 90% fo?kfVr gksus esa fdruk le;
yxksxk\
(A)
1 2
4 n3t1 1n2T T
=æ ö
-ç ÷è ø
l
l
(B)
1 2
4 n3t1 1n2T T
=æ ö
+ç ÷è ø
l
l
(C)
1 2
2 n3t1 1n2T T
=æ ö
-ç ÷è ø
l
l
(D) dksbZ ugha
ALLEN
Page 6/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026
9. Two blocks A and B of same mass M areconnected with each other with an idealstring of length 2l passing over an idealpulley. The block A is connected to a lightpan C with an ideal string as shown in figure.
A particle of mass 2M
is dropped on pan from
height 2l
as shown. If collision between
particle and pan is perfectly inelastic,acceleration of B just after the collision, is :-
MMA B
l
C
M2
l
2
(A) g (B) 9g
(C) 2g (D) g/18
9. leku æO;eku M okys nks CykWdksa A rFkk B dks ,d vkn'kZ
f?kjuh ij ls xqtjus okyh 2l yEckbZ dh fdlh vkn'kZ jLlh
ls fp=kuqlkj ,d&nwljs ls tksM+k tkrk gSA CykWd A dks ,d
gYds ik= C ls ,d vkn'kZ jLlh ls fp=kuqlkj tksM+k tkrk
gSA æO;eku 2M
okys ,d d.k dks 2l
Å¡pkbZ ls bl ik= ij
fxjk;k tkrk gSA ;fn d.k rFkk ik= ds e/; VDdj iw.kZr;k
vizR;kLFk gS rks VDdj ds Bhd i'pkr~ B dk Roj.k gS%&
MMA B
l
C
M2
l
2
(A) g (B) 9g
(C) 2g (D) g/18
ALLEN
07032021 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg Page 7/37All India Open Test/Enthusiast & Leader Course 0000CJA102120026
10. Figure shows a large closed cylindrical tankcontaining water. Initially, the air trappedabove the water surface has a height h0 andpressure 2P0 where P0 is the atmosphericpressure. There is a hole in the wall of thetank at a depth h1 below the top from whichwater comes out. A long vertical tube isconnected as shown. Find the distance ofwater surface in the long tube from the topof the tank, when the water stops comingout of the hole.
2P0h0 h1
h2
(A) 2h0
(B) h0
(C) h2
(D) h1
10. fp=kuqlkj ikuh ls Hkjk ,d cM+k can csyukdkj Vsad n'kkZ;k
x;k gSA izkjEHk esa ty lrg ds Åij h0 Å¡pkbZ rd ok;q
fo|eku gS rFkk nkc 2P0 gS] tgk¡ P0 ok;qe.Myh; nkc gSA
Vsad dh nhokj esa 'kh"kZ ls uhps h1 xgjkbZ ij ,d fNæ gS
ftlls ty ckgj fudyrk gSA ,d yEch ÅèokZèkj uyh
fp=kuqlkj tqM+h gq;h gSA tc ty fNæ ls ckgj fudyuk
can gks tkrk gS rks yEch uyh esa ty lrg dh Vsad ds 'kh"kZ
ls nwjh Kkr dhft;sA
2P0h0 h1
h2
(A) 2h0
(B) h0
(C) h2
(D) h1
ALLEN
Page 8/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026
11. The K(alpha) X-ray of molybdenum haswavelength 71 pm. If the energy of amolybdenum atom with a K electronknocked out is 23.5 keV, what will be theapprox energy (in keV) of an anothermolybdenum atom when an L electron isknocked out?(A) 6 (B) 4(C) 2 (D) 8
12. A disc of radius 20 cm & mass 1kg is rollingwith slipping on a flat horizontal surface. Ata certain instant, the velocity of its center is4 m/s and its angular velocity is 10 rad/s. Thelowest contact point is O. Find angularmomentum about any point on the groundat shown instant.
O
4m/sP
10rad/sec
(A) 0.60 (B) 0.70(C) 1.00 (D) 1.20
11. eksfyCMsue dh K (,sYQk) X-fdj.k dh rjaxnS/;Z 71pm
gSA ;fn ,d K bysDVªkWu ckgj fudyus ds ckn eksfyCMsue
ijek.kq dh ÅtkZ 23.5 keV gS rks tc ,d L bysDVªkWu
fdlh vU; eksfyCMsue ijek.kq ls ckgj fudyrk gS rks bl
ijek.kq dh yxHkx ÅtkZ (keV esa) Kkr dhft;sA(A) 6 (B) 4(C) 2 (D) 8
12. f=T;k 20 cm rFkk æO;eku 1kg okyh ,d pdrh ,d
lery {kSfrt lrg ij fQlyrs gq, yq<+drh gSA fdlh
fo'ks"k {k.k ij blds dsUæ dk osx 4 m/s gS RkFkk bldk
dks.kh; osx 10 rad/s gSA fuEure laidZ fcUnq O gSA
iznf'kZr {k.k ij èkjkry ij fdlh fcUnq ds lkis{k dks.kh;
laosx Kkr dhft;sA
O
4m/sP
10rad/sec
(A) 0.60 (B) 0.70(C) 1.00 (D) 1.20
ALLEN
07032021 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg Page 9/37All India Open Test/Enthusiast & Leader Course 0000CJA102120026
13. An airplane is in supersonic flight at analtitude h. At what smallest distance a (alongthe horizontal) from the observer on theground is there a point from which the soundemitted by the airplane motors reaches tothe observer earlier than sound emitted byplane when it is at point A that is directlyabove the observer. (vp = velocity of airplane,vs = velocity of sound.)
h
O
AS
a
(A) h
vv
vv
a
s
p
s
p
1
2
2
-÷÷ø
öççè
æ
÷÷ø
öççè
æ
<
(B) h
vv
vv
a
s
p
s
p
12
-÷÷ø
öççè
æ
÷÷ø
öççè
æ
<
(C) not possible for any real value of a
(D) h
vv
vv
a
s
p
s
p
1
2
2
-÷÷ø
öççè
æ
÷÷ø
öççè
æ
>
13. ,d ok;q;ku h Å¡pkbZ ij ijk/ofud mM+ku (supersonic)esa mM+ jgk gSA /kjkry ij [kM+s ,d izs{kd ls {kSfrt dsvuqfn'k fdl U;wure nwjh a ij ,d fcUnq fo|eku gksxk]tgk¡ ls ok;q;ku dh eksVj }kjk mRlftZr /ofu] izs{kd ds BhdÅij fLFkr fcUnq A ij ok;q;ku ds gksus ij blls mRlftZr/ofu dh rqyuk esa izs{kd rd igys igq ¡psxh\ (vp = ok;q;kudk osx] vs = /ofu dk osx)
h
O
AS
a
(A) h
vv
vv
a
s
p
s
p
1
2
2
-÷÷ø
öççè
æ
÷÷ø
öççè
æ
<
(B) h
vv
vv
a
s
p
s
p
12
-÷÷ø
öççè
æ
÷÷ø
öççè
æ
<
(C) a ds fdlh okLrfod eku ds fy, laHko ugh gSA
(D) h
vv
vv
a
s
p
s
p
1
2
2
-÷÷ø
öççè
æ
÷÷ø
öççè
æ
>
ALLEN
Page 10/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026
14. A potential difference is applied between a
conducting sphere and a conducting plate
(“plus” on the sphere and “minus” on the
plate). The dimension of the plate are much
larger than the distance between sphere and
plate. A point positive charge is moved from
point 1 to point 2 parallel to the plate. Using
the above information choose the correct
statement.
1 2
+
(A) Work done by external agent in the
process is zero
(B) Net positive work will be done in moving
charge from 1 to 2 by external agent
(C) Net negative work will be done in
moving charge from 1 to 2 by external
agent
(D) Information is insufficient to give any
assertion regarding work.
14. ,d pkyd xksys rFkk ,d pkyd IysV ds e/; ,d foHkokUrj
vkjksfir fd;k tkrk gSA (xksys ij + rFkk IysV ij –) IysV
dh foek xksys rFkk IysV ds e/; nwjh dh rqyuk esa cgqr
vf/kd gSA ,d fcUnq /kukRed vkos'k dks IysV ds lekUrj
fcUnq 1 ls fcUnq 2 rd xfr djk;h tkrh gSA mijksDr tkudkjh
dk mi;ksx djrs gq;s lgh dFku pqfu;sA
1 2
+
(A) bl izfØ;k esa ckg~; dkjd }kjk fd;k x;k dk;Z 'kwU;
gSA
(B) vkos'k dks 1 ls 2 rd xfr djkus esa ckg~; dkjd }kjk
dqy /kukRed dk;Z fd;k tkrk gSA
(C) vkos'k dks 1 ls 2 rd xfr djkus esa ckg~; dkjd }kjk
dqy ½.kkRed dk;Z fd;k tkrk gSA
(D) dk;Z ds laca/k esa dksbZ Hkh tkudkjh nsus ds fy;s
tkudkjh vi;kZIr gSA
ALLEN
07032021 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg Page 11/37All India Open Test/Enthusiast & Leader Course 0000CJA102120026
15. A long solenoid of cross-sectional radius Rhas a thin insulated copper wire ring tightlyput on its winding. One half of the ring hasthe resistance 10 times that of the other half.The magnetic induction produced by thesolenoid varies with time as B = bt, where bis a constant. Find the magnitude of theelectric field strength in the ring.
(A) Rb119
(B) Rb229
(C) 9 Rb(D) Rb
16. A particle of mass 1kg is moving wherepotential energy varies with displacementU = 10 (1 – cos 2x), then the time period ofsmall oscillation about origin will be
(A) 1220
p (B) 12
10p
(C) 140
p (D) 1
10p
15. vuqizLFk dkV f=T;k R okyh ,d yEch ifjukfydk esa
,d iryh dqpkyd rk¡cs ds rkj dh oy; gS ftls bldh
okbafMax ij dldj yisVk x;k gSA oy; ds vkèks Hkkx dk
izfrjksèk nwljs vk/ks Hkkx ds izfrjks/k ls 10 xquk gSA ifjukfydk
}kjk mRiUu pqEcdh; izsj.k] le; ds lkFk B = bt ds
vuqlkj ifjofrZr gksrk gS] tgk¡ b ,d fu;rkad gSA oy; esa
fo|qr {ks= lkeF;Z dk ifjek.k Kkr dhft;sA
(A) Rb119
(B) Rb229
(C) 9 Rb(D) Rb
16. æO;eku 1kg okyk ,d d. k , s l s L Fk k u ij
xfr'khy gS] tgk¡ fLFkfrt ÅtkZ] foLFkkiu ds lkFk
U = 10 (1 – cos 2x) ds vuqlkj ifjofrZr gksrh gS rks
ewyfcUnq ds lkis{k vYi nksyu dk vkorZdky gS%&
(A) 1220
p (B) 12
10p
(C) 140
p (D) 1
10p
ALLEN
Page 12/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026
17. Two blocks are connected by a string passingover a pulley as shown. Wedge is fixed. M issliding down with speed 12 m/s at an instantand suddenly the string breaks. Thecoefficient of friction between the blocks and
wedge is 12
. Find the velocity (in m/s) of
block M with respect to m after t = 2 secondsfrom the moment the string breaks.
[Take g = 10 ms–2 and 3 1.22
= . Assume that
the length of inclined is sufficiently long.]
30° 30°
Mm
(A) 10 (B) 15 (C) 2.5 (D) 2018. In an experiment, a boy draws graph
between 2v (y-axis) and 2a (x-axis) (wherev = velocity and a = tangential acceleration)for a simple pendulum. The graph is foundto be a straight line of negative slope makingan angle of 30° (with x-axis) whenexperiment was done on the ground and 60°(with x-axis) when experiment was done atheight h above the ground. Then h must be(R = radius of earth)(A) 0.5R (B) 0.24 R
(C) 0.73 R (D) R
17. nks CykWdksa dks ,d f?kjuh ij ls xqtjus okyh ,d jLlh }kjk
fp=kuqlkj tksM+k tkrk gSA ost fLFkj gSA M fdlh {k.k ij
12 m/s pky ls uhps dh vksj xfr'khy gS rFkk vpkud
jLlh VwV tkrh gSA CykWdksa rFkk ost ds e/; ?k"kZ.k xq.kkad
12 gSA jLlh ds VwVus ds {k.k ls t = 2 sec i'pkr~ CykWd
M dk m ds lkis{k osx (m/s esa) Kkr dhft;sA ekuk vkur
ry dh yEckbZ i;kZIr :i ls yEch gSA g = 10 ms–2
rFkk 3 1.22
= ysaA
30° 30°
Mm
(A) 10 (B) 15 (C) 2.5 (D) 2018. fdlh iz;ksx esa ,d yM+dk ,d ljy yksyd ds fy;s
2v (y-v{k) rFkk 2a (x-v{k) ds e/; vkjs[k cukrk gS]
tgk¡ v = osx rFkk a = Li'kZ js[kh; Roj.k gSA tc iz;ksx
èkjkry ij rFkk /kjkry ls h Å¡pkbZ ij fd;k tkrk gS rks
;g vkjs[k x v{k ls Øe'k% 30° o 60° dks.k cukus okyh
½.kkRed <ky dh lh/kh js[kk ds :i esa izkIr gksrk gSA hdk eku gksuk pkfg;s%& (R = i`Foh dh f=T;k)
(A) 0.5R (B) 0.24 R
(C) 0.73 R (D) R
ALLEN
07032021 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg Page 13/37All India Open Test/Enthusiast & Leader Course 0000CJA102120026
19. In the given circuit, initial charge on thecapacitor is 1.5 C with polarity as shown infigure and current in the inductor is zero.Now at t = 0 if the key k is closed, then theminimum current through the key is
0.25F3WD
3VAk
2WC
6.8V 5H 3.8W 2W 2V
B
+ –
(A) 33/25 A (B) 34/25 A(C) 36/25 A (D) none
20. A stationary sound source ‘S’ of frequency334 Hz and a stationary observer ‘O’ areplaced near a reflecting surface moving awayfrom the source with velocity 2 m/sec shownin the figure. If the velocity of the soundwaves in air is V = 332 m/sec, the apparentfrequency of the echo is
SO
2m/s
(A) 332 Hz (B) 326 Hz(C) 334 Hz (D) 330 Hz
19. iznf'kZr ifjiFk esa la/kkfj= ij izkjfEHkd vkos'k dk eku 1.5
C gS rFkk /kzqork fp= esa n'kkZ;h x;h gS] tcfd izsjd dq.Myh
eas /kkjk dk eku 'kwU; gSA vc t = 0 ij dqath k dks can dj
fn;k tkrk gS rks dqath ls izokfgr U;wure /kkjk gksxh%&
0.25F3WD
3VAk
2WC
6.8V 5H 3.8W 2W 2V
B
+ –
(A) 33/25 A (B) 34/25 A(C) 36/25 A (D) dksbZ ugha
20. vkofr 334 Hz okyk ,d fLFkj /ofu L=ksr ‘S’ rFkk ,d
fLFkj izs{kd ‘O’ fp=kuqlkj L=ksr ls nwj 2 m/sec osx ls
xfr'khy ijkorZd lrg ds utnhd j[ks gq;s gSA ;fn ok;q
esa /ofu rjaxks dk osx V = 332 m/sec gS rks izfrèofu dh
vkHkklh vkofr gS%&
SO
2m/s
(A) 332 Hz (B) 326 Hz(C) 334 Hz (D) 330 Hz
ALLEN
Page 14/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026
SECTION-II : (Maximum Marks: 20)� This section contains TEN questions.
Attempt any 5 questions. First 5 attemptedquestions will be considered for marking.
� The answer to each question is aNUMERICAL VALUE.
� For each question, enter the correctnumerical value (If the numerical value hasmore than two decimal places, truncate/round-off the value to TWO decimal places;e.g. 6.25, 7.00, –0.33, –.30, 30.27, –127.30, ifanswer is 11.36777..... then both 11.36 and11.37 will be correct) by darken thecorresponding bubbles in the ORS.For Example : If answer is –77.25, 5.2 thenfill the bubbles as follows.
0 0 0
+
0
–
0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9
•
•
•
•
•
•
•
•
••
0 0 0
+
0
–
0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9
•
•
•
•
•
•
•
•
••
� Answer to each question will be evaluatedaccording to the following marking scheme:Full Marks : +4 If ONLY the correctnumerical value is entered as answer.Zero Marks : 0 In all other cases
[kaM-II : (vf/kdre vad : 20)� bl [kaM esa nl iz'u gSaA fdUgh 5 iz'uksa dk mÙkj nhft,A
fd;s x;s iz'uksa esa ls dsoy izFke ik¡p iz'uksa dks gh vadfn;s tk;saxsaA
� iz R; sd iz'u dk mÙkj ,d l a[; k Red e ku(NUMERICAL VALUE) gSA
� iz R; sd i z'u ds mÙkj ds lgh la[; kRed eku(;fn la[;kRed eku esa nks ls vf/kd n'keyo LFkkugS] rks la[;kRed eku dks n'keyo ds nks LFkkuks a rdVª ad sV@jkm aM vk WQ (truncate/round-off) djs a_ mnkgj.k 6.25, 7.00, –0.33, –.30, 30.27, –127.30,;fn mÙkj 11.36777..... gS] rks 11.36 vkSj 11.37 nksuksalgh gksxsa) dks izfo"B djus ds fy, vks-vkj-,l- esa vuq:icqycqys dks dkyk djsaAmnkgj.k ds fy, : ;fn mÙkj –77.25, 5.2 gS] rks cqycqyksadks fuEu izdkj ls dkyk djsaA
0 0 0
+
0
–
0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9
•
•
•
•
•
•
•
•
••
0 0 0
+
0
–
0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9
•
•
•
•
•
•
•
•
••
� izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstukds vuqlkj gksxk %&iw.kZ vad : +4 ;fn flQZ lgh la[;kRed eku (Numericalvalue) gh mÙkj Lo:i ntZ fd;k x;k gSA'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA
ALLEN
07032021 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg Page 15/37All India Open Test/Enthusiast & Leader Course 0000CJA102120026
1. A biconvex thin lens is prepared from glass
of refractive index µ2 = 32 . The two
converging surface have equal radii of 20cmeach. One of the surface is silvered fromoutside to make it reflecting. It is placed in
a medium of refractive index µ1 = 53 . This
system will behave as concave mirror of focal
length f, find value of f in cm.
2. Find the maximum kinetic energy (in eV) ofthe photoelectron liberated from the surface
of lithium (work function 2.15eVf = ) by
electromagnetic radiation whose electriccomponent varies with time as
E = a (1 + cos wt)cosw0t,where a is a constant, w = 12 × 1014 rads–1 andw0 = 3.6 × 1015 rads–1 (h = 6.6 × 10–34 inSI units)
3. A photosensitive surface is irradiated withlight of wavelength l, the stopping potentialis V. When the same surface is irradiatedwith the light of wavelength 2l, stopping
potential is 3V . Then the ratio of threshold
wavelength (lmax) and the l is
1. ,d f}mÙky irys ysal dks viorZukad µ2 = 32 okys dk¡p
ls cuk;k x;k gSA nksuksa vfHklkjh lrgksa esa izR;sd dh leku
f=T;k 20cm gSA buesa ls fdlh ,d lrg dks ijkorZd
cukus ds fy, ckgj dh vksj ls jtfrr fd;k tkrk gSA bls
viorZukad µ1 = 53 okys ek/;e esa j[kk tkrk gSA ;g
fudk; Qksdl nwjh f okys ,d vory niZ.k dh Hkk¡fr
O;ogkj djrk gSA f dk eku cm esa Kkr dhft;sA
2. fyFkh;e (dk;Z Qyu 2.15eVf = ) dh lrg ls ,sls fo|qr
pqEcdh ; fofdj.k }kjk mRlftZr izdk'k bysDVªkWu dh
vf/kdre xfrt ÅtkZ (eV esa) Kkr dhft;s ftldk fo|qr
?kVd] le; ds lkFk E = a (1 + cos wt)cosw0t,
ds vuqlkj ifjofr Zr gk srk gS _ tgk¡ a fu;rkad gS]
w = 12 × 1014 rads–1 rFkk w0 = 3.6 × 1015 rads–1
o SI bdkbZ esa h = 6.6 × 10–34 gSA
3. ,d izdk'klaosnh lrg ij rjaxnS/; Z l okyk izdk'k vkifrr
fd;k tkrk gS rFkk fujks/kh foHko V gSA ;fn blh lrg ij
rjaxnS/; Z 2l ds izdk'k dks vkifrr fd;k tkrk gS rks
fujksèkh foHko dk eku 3V gSA nSgyh rjaxnSè;Z ( )maxl
rFkk l dk vuqikr Kkr dhft;sA
ALLEN
Page 16/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026
4. One of the circuits for the measurement ofresistance by potentiometer is shown. Thegalvanometer is connected at point A andzero deflection is observed at lengthPJ = 30 cm. In second case the secondarycell is changed. Take ES = 10 V and r = 1W in1st reading and ES = 5V and r = 2W in2nd reading. In second case, the zerodeflection is observed at length PJ = 10 cm.What is the resistance R (in ohm) ?
EP( )
P Q
GJ
AR
rES
5. A uniform triangular plate of triangular area1m2, base length 60 cm and thickness 10 mm(prism like shape) is lying vertically on asmooth ground as shown in figure. Findmaximum value of cotq for which it doesnottopple.
qg
60cm
6. On a particular day, the maximumfrequency reflected from the ionosphere is8 MHz. On next day it was found to increaseto 9MHz. If ratio of maximum electrondensities of first day to maximum electrondensities of next day the ionosphere is n,find value of 81 × n.
4. fp=kuqlkj foHkoekih }kjk izfrjks/k ds ekiu ds fy, ,difjiFk n'kkZ;k x;k gSA xsYosuksehVj dks fcUnq A ij tksM+ktkrk gS rFkk yEckbZ PJ = 30 cm ij 'kwU; fo{ksi izsf{krfd;k tkrk gSA nwljs izdj.k esa f}rh;d lsy dks cnyfn;k tkrk gSA izFke ikB~;kad esa ES = 10 V ,oa r = 1WrFkk f}rh; ikB~;kad eas ES = 5V ,oa r = 2W yhft;sAnwljs izdj.k esa yEckbZ PJ = 10 cm ij 'kwU; fo{ksi izkIrgksrk gSA izfrjks/k R dk eku (vkse esa) Kkr dhft;sA
EP( )
P Q
GJ
AR
rES
5. f=Hkqtkdkj {ks=Qy 1m2, vk/kkj yEckbZ 60 cm rFkk eksVkbZ10 mm (fizTe ln`'; vkd`fr) okyh ,d le:if=Hkqtkdkj IysV ,d fpdus /kjkry ij ÅèokZèkj :i lsfp=kuqlkj j[kh gq;h gSA ;g IysV iyVs ugh blds fy,cotq dk vf/kdre eku Kkr dhft;sA
qg
60cm
6. fdlh fo'ks"k fnu vk;ue.My ls ijkofrZr vf/kdrevkofÙk dk eku 8 MHz gSA vxys fnu ;g c<+dj 9MHzgks tkrh gSA ;fn igys fnu rFkk blds vxys fnuvk;ue.My ds vfèkdre bysDVªkWu ?kuRoksa dk vuqikr ngks rks 81 × n dk eku Kkr dhft;sA
ALLEN
07032021 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg Page 17/37All India Open Test/Enthusiast & Leader Course 0000CJA102120026
7. Find value of base resistance RB (in kW) inthe circuit as shown in figure, if bd,c = 90,VBE = 0.7V; VCE = 4V ?
9VRC 2kW
RB
3V
8. An electric field of 300 V/m is confined to acircular area 10 cm in diameter. If the fieldis increasing at the rate of 20 V/m-s and ifthe magnitude of magnetic field at a point15 cm from the centre of the circle isB × 10–18 T, find value of B.
9. In a given common Emitter transistor,Emitter current is changed by 2.1 mA. Thisresults in a change of 2 mA in the collectorcurrent and a change of 0.05 V in theEmitter-base voltage. The input resistance(in ohm) is :-
10. A rod of ferromagnetic material withdimensions 10cm × 0.5 cm × 0.2 cm isplaced in a magnetic field of strength0.5 × 104 amp/m as a result of which amagnetic moment of 5 amp-m2 is producedin the rod. Find the value of magneticinduction (in SI unit).
7. iznf'kZr ifjiFk esa vk/kkj izfrjks/k RB (kW esa) dk eku Kkr
dhft;s tcfd bd,c = 90, VBE = 0.7V; VCE = 4V gSA
9VRC 2kW
RB
3V
8. ,d 300 V/m dk fo|qr {ks= 10 cm O;kl ds o`Ùkkdkj
{ks=Qy esa ifjc¼ gSA ;fn {ks= dk eku 20 V/m-s dh
nj ls c<+ jgk gS rFkk ;fn o`Ùk ds dsUæ ls 15 cm nwj fLFkr
fcUnq ij pqEcdh; {ks= dk ifjek.k B × 10–18 T gks rks Bdk eku Kkr dhft;sA
9. ,d fn;s x;s mHk;fu"B mRltZd Vªk¡ftLVj esa mRltZd
èkkjk esa 2.1 mA dk ifjorZu fd;k tkrk gSA blds
ifj.kkeLo:i laxzkgd /kkjk esa 2 mA dk ifjorZu rFkk
mRltZd vk/kkj oksYVrk esa 0.05 V dk ifjorZu izsf{kr
gksrk gSA fuos'kh izfrjks/k (vkse esa) dk eku gS%&10. yk SgpqEcdh ; inkFk Z ls cuh ,d NM + dk vkdkj
10cm × 0.5 cm × 0.2 cm g S rFk k bl s0.5 × 104 amp/m lkeF;Z okys pqEcdh; {ks= eas j[kktkrk gS ftlds ifj.kkeLo:i NM+ esa 5 amp-m2 dkpqEcdh; vk?kw.kZ mRiUu gks tkrk gSA pqEcdh; izsj.k dkeku (SI bdkbZ esa) Kkr dhft;sA
ALLEN
Page 18/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026
SECTION–I : (Maximum Marks : 80)� This section contains TWENTY questions.� Each question has FOUR options (A), (B),
(C) and (D). ONLY ONE of these fouroptions is correct.
� For each question, darken the bubblecorresponding to the correct option in theORS.
� For each question, marks will be awardedin one of the following categories :Full Marks : +4 If only the bubblecorresponding to the correct option isdarkened.Zero Marks : 0 If none of the bubbles isdarkened.Negative Marks : –1 In all other cases
1. Ratio of solubilities of gases N2 & O2 in water(as mole of gas in large volume of water) fromair at 25º & 1 atm will be if air is 20 % byvolume of O2 & 80% by volume of N2
Given : KH (N2) = 2 × 104 atmKH (O2) = 104 atm
(A) 8 : 1 (B) 1 : 8(C) 2 : 1 (D) 1 : 2
2. The resistance of 0.05 M solution of oxalic acidis 200 ohm and cell constant is 2.0 cm–1, theequivalent conductance (in S cm2 eq–1) of 0.05 Moxalic acid is :-(A) 100 (B) 0.2(C) 200 (D) 400
PART 2 - CHEMISTRY[k.M–I : (vf/kdre vad : 80)
� bl [k.M esa chl iz'u gSa
� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj
(D) gSaA ftuesa dsoy ,d gh lgh gSaA
� izR;sd iz'u ds fy, vks-vkj-,l ij lgh mÙkj fodYi ds
vuq:i cqycqys dks dkyk djsaA
� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls
fdlh ,d ds vuqlkj fn;s tk,axs %
iw.kZ vad : +4 ;fn flQZ lgh fodYi ds vuq:i cqycqys
dks dkyk fd;k gSA
'kwU; vad : 0 ;fn fdlh Hkh cqycqys dks dkyk ugha fd;k gSA
½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA
1. ;fn ok;q esa vk;ru }kjk 20% O2 rFkk vk;ru }kjk 80%N2 gS] rk s 25º rFkk 1 atm ij ok; q ls ty es a(ty ds vf/kd vk;ru esa xSl ds eksy ds :i esa)N2
rFkk O2 xSlksa dh foys;rkvksa dk vuqikr gksxk &
fn;k gS : KH (N2) = 2 × 104 atmKH (O2) = 104 atm
(A) 8 : 1 (B) 1 : 8(C) 2 : 1 (D) 1 : 2
2. vkWDlsfyd vEy ds 0.05 M foy;u dk izfrjk s/k200 vkse rFkk lsy fu;rkad 2.0 cm–1 gS rks 0.05 MvkWDlsfyd vEy dh rqY;kad pkydrk (S cm2 eq–1 esa)gS :-(A) 100 (B) 0.2(C) 200 (D) 400
ALLEN
07032021 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg Page 19/37All India Open Test/Enthusiast & Leader Course 0000CJA102120026
3. Lyophilic sols are more stable than lyophobicsols because :(A) The colloidal particles have positive
charge(B) The colloidal particles have no charge(C) The colloidal particles are solvated(D) There are strong electrostatic
repulsions between the negativelycharged colloidal particles
4. The decomposition of a gaseous substance(A) to yield gaseous products (B), (C) followsfirst order kinetics. If initially only (A) ispresent and l0 minutes after the start of thereaction, the pressure of (A) is 200 mm Hgand the total pressure is 300mm Hg, thenthe rate constant for reaction A ® B + Cis(A) (1/600) ln 1.25 sec–1
(B) (2.303/10) log 1.5 min–1
(C) (1/10) ln 1.25 sec–1
(D) 2.303log 1.5
600sec–1
5. Which one is non-aromatic?
(A)
1
(B)
(C) (D)
Å
3. nzofojks/kh lkWy dh rqyuk esa nzoLusgh lkWy vf/kd LFkkbZ gksrs
gSa D;ksafd&
(A) dksykbMh d.kksa ij /ku vkos'k gksrk gSA
(B) dksykbMh d.kks a ij dkbZ vkos'k ugha gksrk gSA
(C) dksykbMh d.k foyk;dhÏr gksrs gSaA
(D) ½.kkosf'kr dksykbMh d.kks a ds e/; izcy fo|qr
fLFkSfrd izfrd"kZ.k gksrs gSaA
4. ,d xSlh; inkFkZ (A) ds fo?kVu ls izkIr xSlh; mRikn
(B) rFkk (C) dh vfHkfØ;k izFke dksfV cyxfrdh dk
vuqlj.k djrh gS ;fn izkjEHk esa dsoy (A) mifLFkr gS
rFkk vfHkfØ;k izkjaHk gksus ds l0 feuV ds ckn (A) dk
nkc 200mm Hg rFkk dqy nkc 300mm Hg gS rks
vfHkfØ;k] A ® B + C ds fy, nj fu;arkd gS&
(A) (1/600) ln 1.25 sec–1
(B) (2.303/10) log 1.5 min–1
(C) (1/10) ln 1.25 sec–1
(D) 2.303log 1.5
600sec–1
5. fuEu es ls dkSulk uksu ,sjksesfVd gS\
(A)
1
(B)
(C) (D)
Å
ALLEN
Page 20/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026
6.
OLDA CH =CH–CH –Cl2 2
HOH/H+A (1) LiAlH4(2) conc.H SO2 4
CB
The compound C is -
(A) CH–CH=CH2
(B) CH –CH=CH2 2
(C) CH–CH –CH2 3
(D) CH–CH –CH2 3
7. Which of the following pair(s) is/areEpimers?(A) D-glucose & D-mannose(B) D-mannose & D-galactose(C) D-glucose & D-fructose(D) D-fructose & L-galactose
8. Following conversion can be carried out byusing :-
CH–CH3
Br COOH
(A) Ag2O (B) (i) NaOBr (ii) H+
(C) Aq. KOH (D) Alc. KOH
6.O
LDA CH =CH–CH –Cl2 2HOH/H+A (1) LiAlH4
(2) conc.H SO2 4CB
;kSfxd C gS -
(A) CH–CH=CH2
(B) CH –CH=CH2 2
(C) CH–CH –CH2 3
(D) CH–CH –CH2 3
7. fuEu fdl ; qXe ds ;kSfxd ^,ihej* gS:-
(A) D-XyqdkWl rFkk D-esuksl
(B) D-esuksl rFkk D-xsysDVksl
(C) D-XyqdkWl rFkk D-ÝDVksl
(D) D-ÝDVksl rFkk L-xsysDVksl
8. fuEu :ikarj.k dks] fdlds iz;ksx }kjk djk;k tk ldrk
gS :-
CH–CH3
Br COOH
(A) Ag2O (B) (i) NaOBr (ii) H+
(C) Aq. KOH (D) Alc. KOH
ALLEN
07032021 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg Page 21/37All India Open Test/Enthusiast & Leader Course 0000CJA102120026
9. Select the CORRECT order of size for Ln(III)ions :-
(A) Ce3+
Pr3+
Nd3+
Pm3+
r
(pm)
®
z ®
(size)
(B)
r ®
z ®
Ce3+ Pr3+ Nd3+ Pm3+
(size)(pm)
(C)
r ®
z ®
Ce3+
Pr3+
Nd3+
Pm3+(size)(pm)
(D)
r(size)(pm)
®
z ®
Pm3+
Nd3+
Pr3+
Ce3+
9. Ln(III) vk;uksa ds fy, vkdkj ds lgh Øe dks pqfu, :-
(A) Ce3+
Pr3+
Nd3+
Pm3+
r
(pm)
®
z ®
(size)
(B)
r ®
z ®
Ce3+ Pr3+ Nd3+ Pm3+
(size)(pm)
(C)
r ®
z ®
Ce3+
Pr3+
Nd3+
Pm3+(size)(pm)
(D)
r(size)(pm)
®
z ®
Pm3+
Nd3+
Pr3+
Ce3+
ALLEN
Page 22/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026
10. Which of the following reaction isINCORRECT?
(A) Na + (x+y)NH3(l) ¾® [Na(NH3)x]+ + [e(NH3)y]–
(B) 2NH3 + H2O + CO2 ¾® (NH4)2CO3
(C) Na2CO3.10H2O D¾¾® Na2O + CO2 + 10H2O
(D) Be(OH)2 + 2OH– ¾® [Be(OH)4]2–
11. When H2S gas is passed in acidic mediumwhich set of cations get precipitated?(A) Mn2+, Co2+, Zn2+
(B) Pb2+, Cu2+, Bi3+
(C) Ba2+, Sr2+, Ca2+
(D) Fe3+, Cr3+, Al3+
12. On heating compound X a gas Y is obtained,which is also manufactured by reaction ofBa(NO3)2 + Zn + KOH(aq.).(A) Compound X is (NH4)2Cr2O7
(B) Gas Y is used in Ostwald process(C) Gas Y is colourless and odourless(D) Gas Y produce blue solution in AgNO3(aq.)
13. Calculate % volume occupied by atoms inCsCl type structure assuming ideal crystalhaving anion-anion contact.
[Given +
-
rr = 0.7; p = 3 ; 3 = 1.7]
(A) 67.15 (B) 90.09(C) 78.5 (D) 84
10. fuEu eas ls dkSulh vfHkfØ;k xyr gS ?
(A) Na + (x+y)NH3(l) ¾® [Na(NH3)x]+ + [e(NH3)y]–
(B) 2NH3 + H2O + CO2 ¾® (NH4)2CO3
(C) Na2CO3.10H2O D¾¾® Na2O + CO2 + 10H2O
(D) Be(OH)2 + 2OH– ¾® [Be(OH)4]2–
11. fuEu eas ls dkSuls leqPp; ds /kuk;u] vEyh; ek/;eesa H2S xSl izokfgr fd;s tkus ij vo{ksfir gks tkrs gSa
(A) Mn2+, Co2+, Zn2+
(B) Pb2+, Cu2+, Bi3+
(C) Ba2+, Sr2+, Ca2+
(D) Fe3+, Cr3+, Al3+
12. ;kSfxd X dks xeZ fd;s tkus ij ,d xSl Y izkIr gksrhgSA ftls Ba(NO3)2 + Zn + KOH(aq.) dh vfHkfØ;k
}kjk Hkh cuk;k tk ldrk gS
(A) ;kSfxd X, (NH4)2Cr2O7 gS
(B) xSl Y, vksLVokYV izØe eas iz;ksx esa yh tkrh gS
(C) xSl Y, jaxghu rFkk xa/kghu gS
(D) xSl Y, AgNO3(aq.) esa uhyk foy;u cukrh gS13. ½.kk;u ½.kk;u lEidZ j[kus okys vkn'kZ fØLVy ekurs
gq, CsCl izdkj dh lajpuk esa ijek.kqvksa }kjk ?ksjs x;s
vk;ru dh izfr'krrk dh x.kuk dhft;s -
[Given +
-
rr = 0.7; p = 3 ; 3 = 1.7]
(A) 67.15 (B) 90.09
(C) 78.5 (D) 84
ALLEN
07032021 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg Page 23/37All India Open Test/Enthusiast & Leader Course 0000CJA102120026
14. The DH° for the mutarotation of glucose inaqueous solution,a – D – glucose (aq) ¾® b – D – glucose (aq)has been measured in a microcalorimeterand found to be –1.16 kJ.mol–1. Theenthalpies of solution of the two forms ofglucose have been determined to bea – D – glucose (s) ¾® a – D – glucose (aq)
DH° = 10.72 kJ mol–1
b – D – glucose (s) ¾® b – D – glucose (aq)DH° = 4.68 kJ.mol–1
Calculate DH° (in kJ/mol) for themutarotation of solid a – D – glucose to solidb - D – glucose.
(A) +4.88 kJ/mol (B) –4.88 kJ/mol(C) –2.44 kJ/mol (D) +2.44 kJ/mol
15.
O(i) Cl + anh. AlCl(ii) Na(Hg) + HCl(iii) HNO + H SO
2 3
3 2 4Z (Major),
Z is :
(A)
O
Cl
O2N (B)
NO2
Cl
(C)
Cl
O2N (D)
O
ClON
14. tyh; foy;u esa Xywdkst ds ifjorhZ ?kw.kZu ds fy,]
a – D – Xywdkst (aq) ¾® b – D – Xywdkst (aq),d ekbØks dsykSjhehVj esa DH° ekik x;k gS rFkk
–1.16 kJ.mol–1 ik;k x;k gSA Xywdkst ds nks :iksa ds
foy;u dh Kkr dh xbZ ,UFkSYih fuEu gSa &
a – D – Xywdkst (s) ¾® a – D – Xywdkst (aq)DH° = 10.72 kJ mol–1
b – D – Xywdkst (s) ¾® b – D – Xywdkst (aq)DH° = 4.68 kJ.mol–1
Bksl a – D – Xywdkst ls Bksl b - D – Xywdkst ds
ifjorhZ ?kw.kZu ds fy, DH° (kJ/mol esa) dh x.kuk dhft,
(A) +4.88 kJ/mol (B) –4.88 kJ/mol(C) –2.44 kJ/mol (D) +2.44 kJ/mol
15.
O(i) Cl + anh. AlCl(ii) Na(Hg) + HCl(iii) HNO + H SO
2 3
3 2 4Z (eq[;),
Z gS :
(A)
O
Cl
O2N (B)
NO2
Cl
(C)
Cl
O2N (D)
O
ClON
ALLEN
Page 24/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026
16. Racemic mixture of 3-hydroxy-3-methylcyclohexanone is obtained as a product inreaction -
(A) (i) CH MgCl(1eq.)3
(ii) NH Cl4
O
(B)
O
O
(i) CH MgCl(1eq.)3
(ii) NH Cl4
(C) O O
dil.OH–
low temperature
(D)
O
O (i) CH MgCl(excess)3
(ii) NH Cl4
17. How many of the following will give whiteppt of silver salt when treated withammonical AgNO3 solution?(a) CH3–CH2–CºCH (b) CH3–CºC–CH3
(c) CH2=CH–CH2–Cl (d) CH =C–CH2 3
Cl
(e) Cl (f) Ph–CH2–Cl
(g) I
(A) 3 (B) 4 (C) 5 (D) 6
16. fuEu es ls dkSulh vfHkfØ;k esa 3-gkbMªksDlh-3-esfFkylkbDyksgsDlsuksu dk jslsfed feJ.k mRikn ds :i es izkIrgksrk gS -
(A) (i) CH MgCl(1eq.)3
(ii) NH Cl4
O
(B)
O
O
(i) CH MgCl(1eq.)3
(ii) NH Cl4
(C) O O
dil.OH–
low temperature
(D)
O
O (i) CH MgCl(excess)3
(ii) NH Cl4
17. fuEu ;kSfxdks dks tc veksfudy AgNO3 foy;u ds
lkFk mipkfjr fd;k tkrk gS rks fdrus ;kSfxd flYoj yo.kdk 'osr vo{ksi nsaxs\(a) CH3–CH2–CºCH (b) CH3–CºC–CH3
(c) CH2=CH–CH2–Cl (d) CH =C–CH2 3
Cl
(e) Cl (f) Ph–CH2–Cl
(g) I
(A) 3 (B) 4 (C) 5 (D) 6
ALLEN
07032021 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg Page 25/37All India Open Test/Enthusiast & Leader Course 0000CJA102120026
18. Select INCORRECT statement regardingKMnO4 & K2Cr2O7.(A) Both act as oxidising agent in acidic
medium(B) Both are paramagnetic and coloured(C) O2 is used in the preparation of both(D) The central atom of the anion in both has
d3s hybridisation19. Select CORRECT option regarding
formation of molecular orbital from thecombination of 1s-1s orbitals.
E
®
0(energy)
(1)(2)
y(u)
y(g)
(r) Distance between atoms®
(A) Curve-(1) represents formation of BMO(B) Curve-(2) represents formation of ABMO(C) y(g) = yA + yB
(D) y(u) = yA + yB
20. For a d6 metal ion in an octahedral field, thecorrect electronic configuration is :-
(A) 6 02g gt e when (D0 < P)
(B) 4 22g gt e when (D0 > P)
(C) 6 02g gt e when (D0 > P)
(D) 3 32g gt e when (D0 < P)
18. KMnO4 rFkk K2Cr2O7 ds lUnHkZ esa xyr dFku dk p;udhft,
(A) nksuksa vEyh; ek/;e esa vkWDlhdkjd ds :i esa dk;Z
djrs gSa
(B) nksuksa vuqpqEcdh; rFkk jaxhu gS
(C) O2 dk iz;ksx nksuksa ds fuekZ.k esa fd;k tkrk gS
(D) nksuksa esa] /kuk;u ds dsUæh; ijek.kq dk laØe.k d3s gS
19. 1s-1s d{kdksa ds la;kstu ls vkf.od d{kd fuekZ.k ds
lUnHkZ esa lgh fodYi pqfu,
E
®
0(energy)
(1)(2)
y(u)
y(g)
(r) Distance between atoms®
(A) oØ-(1), BMO ds fuekZ.k dks iznf'kZr djrk gS
(B) oØ-(2), ABMO ds fuekZ.k dks iznf'kZr djrk gS
(C) y(g) = yA + yB
(D) y(u) = yA + yB
20. v"VQydh; {k s= esa d6 /kkrq vk;u ds fy, lghbysDVªkWfu; foU;kl gS :-
(A) 6 02g gt e tc (D0 < P)
(B) 4 22g gt e tc (D0 > P)
(C) 6 02g gt e tc (D0 > P)
(D) 3 32g gt e tc (D0 < P)
ALLEN
Page 26/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026
SECTION-II : (Maximum Marks: 20)� This section contains TEN questions.
Attempt any 5 questions. First 5 attemptedquestions will be considered for marking.
� The answer to each question is aNUMERICAL VALUE.
� For each question, enter the correctnumerical value (If the numerical value hasmore than two decimal places, truncate/round-off the value to TWO decimal places;e.g. 6.25, 7.00, –0.33, –.30, 30.27, –127.30, ifanswer is 11.36777..... then both 11.36 and11.37 will be correct) by darken thecorresponding bubbles in the ORS.For Example : If answer is –77.25, 5.2 thenfill the bubbles as follows.
0 0 0
+
0
–
0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9
•
•
•
•
•
•
•
•
••
0 0 0
+
0
–
0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9
•
•
•
•
•
•
•
•
••
� Answer to each question will be evaluatedaccording to the following marking scheme:Full Marks : +4 If ONLY the correctnumerical value is entered as answer.Zero Marks : 0 In all other cases.
[kaM-II : (vf/kdre vad : 20)� bl [kaM esa nl iz'u gSaA fdUgh 5 iz'uksa dk mÙkj nhft,A
fd;s x;s iz'uksa esa ls dsoy izFke ik¡p iz'uksa dks gh vadfn;s tk;saxsaA
� iz R; sd iz'u dk mÙkj ,d l a[; k Red e ku(NUMERICAL VALUE) gSA
� iz R; sd i z'u ds mÙkj ds lgh la[; kRed eku(;fn la[;kRed eku esa nks ls vf/kd n'keyo LFkkugS] rks la[;kRed eku dks n'keyo ds nks LFkkuks a rdVª ad sV@jkm aM vk WQ (truncate/round-off) djs a_ mnkgj.k 6.25, 7.00, –0.33, –.30, 30.27, –127.30,;fn mÙkj 11.36777..... gS] rks 11.36 vkSj 11.37 nksuksalgh gksxsa) dks izfo"B djus ds fy, vks-vkj-,l- esa vuq:icqycqys dks dkyk djsaAmnkgj.k ds fy, : ;fn mÙkj –77.25, 5.2 gS] rks cqycqyksadks fuEu izdkj ls dkyk djsaA
0 0 0
+
0
–
0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9
•
•
•
•
•
•
•
•
••
0 0 0
+
0
–
0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9
•
•
•
•
•
•
•
•
••
� izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstukds vuqlkj gksxk%&iw.kZ vad : +4 ;fn flQZ lgh la[;kRed eku (Numericalvalue) gh mÙkj Lo:i ntZ fd;k x;k gSA'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA
ALLEN
07032021 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg Page 27/37All India Open Test/Enthusiast & Leader Course 0000CJA102120026
1. 1 mole of monoatomic ideal gas subjected toirreversible adiabatic expansion againstconstant external pressure of 1 atm startingfrom initial pressure of 5 atm and initialtemperature of 300 K till the final pressureis 2 atm. What is the final temperature(in K) in the process? (take R = 2 cal/mol K).
2. The solubility of Pb(OH)2 in water is 5 × 10-6 M.Calculate the pH of buffer solution in whichsolubility of Pb(OH)2 is 0.5 × 10–3 mol litre–1.
3. Total number of lone pair(s) present insulfanilamide
4. For the complex [Pt(H2O)(NH3)(Cl)(Br)].If oxidation number of Pt = xCoordination number = yand total possible geometrical isomers = Zthen find sum of x + y + z = ?
5. By a sample of ground state atomic hydrogen,
UV light of energy 13.6 48 eV49 quanta
´ is
absorbed. How many different wavelengthswill be observed in Balmer region ofhydrogen spectrum ?
6. Root mean square speed of an unknown gasat 727ºC is 105 cm/second. Calculate molarmass of unknown gas (in gram/mole)
[Take R = 253
J/mole-K].
1. 5 atm ds izkjfEHkd nkc rFkk 300 K ds izkjfEHkd rki ls
izkjEHk dj vafre nkc 2 atm rd] 1 eksy ,dy ijek.kq
vkn'kZ xSl dk 1 atm ds fu;r ckg~; nkc ds fo:¼
vuqmRØe.kh; :¼ks"eh; izlkj fd;k x;kA izØe esa vafre
rki (K esa) D;k gSA (fn;k gS R = 2 cal/mol K)
2. ty esa Pb(OH)2 dh ?kqyu'khyrk 5 × 10-6 M gSA cQj
foy;u] ftles a Pb(OH)2 dh ?k qyu'khyrk
0.5 × 10–3 mol litre–1 gS] dh pH Kkr dhft;sA
3. lYQsfuysekbM es mifLFkr ,dkadh bysDVªkWu ; qXeks dh
dqy la[;k crkbZ;s\4. ladqy [Pt(H2O)(NH3)(Cl)(Br)] ds fy,
;fn Pt dh vkWDlhdj.k la[;k = xmilgla;kstu la[;k = yrFkk lEHkkfor dqy T;kfefr; leko;oh = Zrks x + y + z dk ;ksx Kkr dhft,
5. vk| voLFkk ds ijekf.o; gkbMªkstu ds uewus }kjk]
13.6 48 eV49 quanta
´ ÅtkZ dk UV izdk'k vo'kksf"kr fd;k
tkrk gS rks gkbMªkstu LiSDVªe ds ckej {ks= es fdruh
fHkUu&fHkUu rjaxnS/; Z izsf{kr gksxhA
6. 727ºC ij ,d vKkr xSl dk oxZ ek/; ewy osx (rms)105 cm/second gSA vKkr xSl ds eksyj nzO;eku dh
x.kuk (gram/mole esa) dhft,A
[fn;k gS R = 253
J/mole-K].
ALLEN
Page 28/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026
7. The number of condensation polymersamong below mentioned polymers is:-(i) Nylon-6 (ii) Glyptal(iii) Buna-N (iv) PVC(v) Poly vinyl acetate (vi) Terylene(vii) Bakelite (viii) Nylon-6,6(ix) orlone
8. Total number of mono-chloro product(s)obtained (including stereo isomers), whenisopentane react with chlorine in thepresence of sun light?
9. Sum of % p-character used in hybridisationof central atom of given molecules.
XeO3F2 , SO3 & 4NF +
10. XeF6 + 2H2O ¾® X + 4HFNi + 4CO ¾® [Y] (Mond's process)Sum of maximum number of atoms presentin one plane of X & Y.
7. fuEufyf[kr esa ls la?kuu cgqydksa dh la[;k D;k gS \
(i) uk;yksu-6 (ii) fXyIVy
(iii) C;quk-N (iv) PVC
(v) iksyh fouk;y , slhVsV (vi) Vsjhyhu
(vii) csdsykbV (viii) uk;yksu-6,6
(ix) vkjyksu
8. tc vkblksisUVsu lw;Z ds izdk'k dh mifLFkfr esa Dyksfju
ds lkFk fØ;k djrk gS rks izkIr gksus okys eksuksDyksjks mRiknks
(f=foe leko;ofo;ksa dks lfEefyr djrs gq;s) dh dqy
la[;k crkbZ;s\
9. fn;s x;s v.kqvks ds dsUæh; ijek.kq ds ladj.k esa iz;ksx
eas fy;s x;s % p-y{k.k dk ;ksx crkbZ;s
XeO3F2 , SO3 rFkk 4NF +
10. XeF6 + 2H2O ¾® X + 4HFNi + 4CO ¾® [Y] (ekWM izde)X rFkk Y ds ,d ry esa mifLFkr ijek.kqvksa dh vf/kdre
la[;k dk ;ksx gS
ALLEN
07032021 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg Page 29/37All India Open Test/Enthusiast & Leader Course 0000CJA102120026
SECTION–I : (Maximum Marks : 80)� This section contains TWENTY questions.� Each question has FOUR options (A), (B),
(C) and (D). ONLY ONE of these fouroptions is correct.
� For each question, darken the bubblecorresponding to the correct option in theORS.
� For each question, marks will be awardedin one of the following categories :Full Marks : +4 If only the bubblecorresponding to the correct option isdarkened.Zero Marks : 0 If none of the bubbles isdarkened.Negative Marks : –1 In all other cases
1. Let A(1,3), B(a,b), C(4,7) are collinear points
and ( ) ( ) ( ) ( )- + - + - + - =2 2 2 2a 1 b 3 4 a 7 b k ,
then k CANNOT be-(A) 4 (B) 5 (C) 6 (D) 9
2. If a circle passes through point (1, 1) and cutsthe circle x2 + y2 = 8 orthogonally, then thelocus of its centre is -(A) x + y + 5 = 0 (B) x + y = 5(C) x + y = 10 (D) x + y + 10 = 0
3. If the two parabolas y2 = 4x and y2 = (x – k)have a common normal other than x-axis,then value of K cannot be -(A) 1 (B) 2 (C) 3 (D) 4
[k.M–I : (vf/kdre vad : 80)
� bl [k.M esa chl iz'u gSa
� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj(D) gSaA ftuesa dsoy ,d gh lgh gSaA
� izR;sd iz'u ds fy, vks-vkj-,l ij lgh mÙkj fodYids vuq:i cqycqys dks dkyk djsaA
� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esals fdlh ,d ds vuqlkj fn;s tk,axs %
iw.kZ vad : +4 ;fn flQZ lgh fodYi ds vuq:i cqycqysdks dkyk fd;k gSA
'kwU; vad : 0 ;fn fdlh Hkh cqycqys dks dkyk ugha fd;k gSA
½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA
1. ekuk A(1,3), B(a,b), C(4,7) lejs[kh; fcUnq gSa rFkk
( ) ( ) ( ) ( )- + - + - + - =2 2 2 2a 1 b 3 4 a 7 b k
gS] rks k ugha gks ldrk gS-(A) 4 (B) 5 (C) 6 (D) 9
2. ;fn ,d o`Ùk fcUnq (1, 1) ls xqtjrk gS rFkk o`Ùkx2 + y2 = 8 dks yEcdks.kh; dkVrk gS] rks blds dsUæ dkfcUnqiFk gksxk-(A) x + y + 5 = 0 (B) x + y = 5(C) x + y = 10 (D) x + y + 10 = 0
3. ;fn nks ijoy; y2 = 4x rFkk y2 = (x – k) dk x-v{k dsvykok ,d vU; mHk;fu"B vfHkyEc gks] rks K dk ekuugha gks ldrk gS-(A) 1 (B) 2 (C) 3 (D) 4
PART 3 - MATHEMATICS
ALLEN
Page 30/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026
4. Let S = 0 be an ellipse whose vertices arethe extremities of minor axis of the ellipse
E : 2 2
2 2x y 1, a ba b
+ = > . If S = 0 passes through
the foci of E, then its eccentricity is(considering the eccentricity of E as e)
(A) 2
21 2e1 e--
(B) 2
11 e+
(C) 2
21 2e1 e--
(D) 2
2e
1 e+
5.2x 3yx 1
+=
+ is a rectangular hyperbola.
Length of its latus rectum is -
(A) 1 (B) 2 (C) 2 (D) 2 26. Value of
cos20 cos50 cos110sin50 sin110 sin110 .sin20 sin20 sin50
° ° °+ +
° ° ° ° ° °
is -(A) 2 (B) 4 (C) 1 (D) 8
7. The smallest positive root of the equation
sin(1 x) cosx- = is equal to -
(A) 1 32 4
p+ (B)
1 72 4
p+
(C) 1 112 4
p+ (D)
12
4. ekuk nh?k Zo `Ùk S = 0, ftlds 'kh"k Z nh?k Zo `Ùk
E : 2 2
2 2x y 1, a ba b
+ = > ds y?kq v{k ds vfUre fljs
gSA ;fn S = 0, E dh ukfHk ls xqtjrh gks] rks bldhmRdsUærk gksxh (E dh mRdsUærk dks e ekurs gq,)
(A) 2
21 2e1 e--
(B) 2
11 e+
(C) 2
21 2e1 e--
(D) 2
2e
1 e+
5. ,d ledks.kh; vfrijoy; 2x 3yx 1
+=
+ gSA blds
ukfHkyEc dh yEckbZ gksxh -
(A) 1 (B) 2 (C) 2 (D) 2 2
6. cos20 cos50 cos110sin50 sin110 sin110 .sin20 sin20 sin50
° ° °+ +
° ° ° ° ° °
dk eku gksxk &(A) 2 (B) 4(C) 1 (D) 8
7. lehdj.k sin(1 x) cosx- = dk U; wure
èkukRed ewy gksxk-
(A) 1 32 4
p+ (B)
1 72 4
p+
(C) 1 112 4
p+ (D)
12
ALLEN
07032021 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg Page 31/37All India Open Test/Enthusiast & Leader Course 0000CJA102120026
8. Let Ai, i = 1,2,3,......10 be vertices of regular
decagon P. If area of polygon P is 5sin5pæ ö
ç ÷è ø
square units then the value of(A1A2)(A1A3)(A1A4)......(A1A10) is-(A) 10 (B) 9
(C) 45sin5pæ ö
ç ÷è ø
(D) 45cos5pæ ö
ç ÷è ø
9. If A = {1, 3, 5, 7, 9, 11}, B = {2, 4, 6, 8, 10, 12},U is universal set, then A' È ((A È B) Ç B') is-(A) U (B) A'(C) A Ç B (D) A
10. Let I be the set of positve integers. R is arelation on the set I given by
R = {(a,b) Î I × I | 2alogb
æ öç ÷è ø
is a non-negative
integer}, then R is(A) neither symmetric nor transitive but
reflexive.(B) reflexive, transitive but not symmetric(C) neither reflexive nor transitive but
symmetric(D) equivalence relation.
11. If = - +r ˆˆ ˆa i j k and = + +
r ˆˆ ˆb 2i 4j 3k are one of
the sides and medians respectively, of atriangle through the same vertex, then areaof triangle is-
(A) 86 (B) 1 852 (C)
1 842 (D) 80
8. ekuk Ai, i = 1,2,3,......10 lenlHkqt P ds 'kh"kZ gSA ;fn
cgqHk qt P dk {k s=Qy 5sin5pæ ö
ç ÷è ø
oxZ bdkb Z g S]
rk s (A1A2)(A1A3)(A1A4)......(A1A10) dk ekugksxk -(A) 10 (B) 9
(C) 45sin5pæ ö
ç ÷è ø
(D) 45cos5pæ ö
ç ÷è ø
9. ;fn A = {1, 3, 5, 7, 9, 11}, B = {2, 4, 6, 8, 10, 12},, U lkoZf=d leqPp; gS]rks A' È ((A È B) Ç B') gksxk&(A) U (B) A'(C) A Ç B (D) A
10. ekuk I /kukRed iw.kk±dksa dk leqPp; gSA R, leqPp; I ij
,d lEcU/k gS] tks R = {(a,b) Î I × I | 2alogb
æ öç ÷è ø
,d
v½.kkRed iw.kk±d gS} }kjk fn;k x;k gS] rc R gksxk
(A) uk lefer uk gh laØked ijUrq LorqY; gksxkA
(B) LorqY;] laØked ijUrq lefer ugha gksxkA
(C) uk LorqY; uk gh laØked ijUrq lefer gksxkA
(D) rqY;rk lEcU/k gksxkA
11. ;fn = - +r ˆˆ ˆa i j k rFkk = + +
r ˆˆ ˆb 2i 4j 3k leku 'kh"kZ ls
xqtjus okys ,d f=Hkqt dh Øe'k% dksbZ ,d Hkqtk rFkk
ekf/;dk gks] rks f=Hkqt dk {ks=Qy gksxk -
(A) 86 (B) 1 852 (C)
1 842 (D) 80
ALLEN
Page 32/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026
12. Let A(a)r and B(b)r
be points on two skew
lines = + ar a pr r r and = + br b qrr r and the
shortest distance between them is 3 . Ifangle between AB and the line of shortestdistance is 30°, then AB =
(A) 12
(B) 2 (C) 3 (D) 4
13. The range of values of ‘p’ so that both the rootsof the equation (p – 5)x2 – 2px + (p – 4) = 0are positive, one is less than 2 and otheris lying between 2 & 3, is
(A) ÷ø
öçè
æ 24,4
49(B) (5, ¥)
(C) (–¥, 4) U ÷ø
öçè
æ ¥,4
49(D) None of these
14. If square root of 1 1 1 1a 2a 4a 8aa .(2a) .(4a) .(8a) ........ ¥
is 827
, then the value of 'a' is -
(A) 12
(B) 13
(C) 14
(D) 15
15. The negation of ( ) ( )( )® Úp ~ p ~ q is-
(A) Ùp q (B) ®p q
(C) ( )® Ùp p q (D) ( )Ù Úp p q
12. ekuk fcUnq A(a)r rFkk B(b)r
nks fo"ke js[kkvksa = + ar a pr r r
rFkk = + br b qrr r ij fLFkr gS rFkk fo"ke js[kkvksa ds e/;
U;wure nwjh 3 A ;fn AB rFkk y?kqÙke nwjh dh js[kk ds
e/; dks.k 30° gks] rks AB gksxk&
(A) 12
(B) 2 (C) 3 (D) 4
13. ‘p’ ds ekuksa dk ifjlj] rkfd lehdj.k(p – 5)x2 – 2px + (p – 4) = 0 ds nksuksa ewy èkukRedgS] ftlesa ls ,d ewy 2 ls de rFkk vU; ewy 2 rFkk 3ds e/; fLFkr gS] gksxk&
(A) ÷ø
öçè
æ 24,4
49(B) (5, ¥)
(C) (–¥, 4) U ÷ø
öçè
æ ¥,4
49(D) buesa ls dksbZ ugha
14. ;fn 1 1 1 1a 2a 4a 8aa .(2a) .(4a) .(8a) ........ ¥ dk oxZewy
827
gks] rks a dk eku gksxk -
(A) 12
(B) 13
(C) 14
(D) 15
15. ( ) ( )( )® Úp ~ p ~ q dk fu"ks/k gksxk -
(A) Ùp q (B) ®p q
(C) ( )® Ùp p q (D) ( )Ù Úp p q
ALLEN
07032021 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg Page 33/37All India Open Test/Enthusiast & Leader Course 0000CJA102120026
16. An urn contains 3 red, 4 green and certainnumber of white balls. Two balls are drawnsimultaneously and found to be of differentcolour. If the chance that none of then was awhite ball is 2/9, then number of white ballsis equal to -(A) 3 (B) 4 (C) 6 (D) 8
17. The numerically greatest term in theexpansion of (2x + 5y)34, when x = 3 &y = 2, is(A) T21 (B) T22 (C) T23 (D) T24
18. Let ƒ : (3, 6) ® (4, 7) be a function defined by
ƒ(x) = 4x x3 3
ì ü- í ýî þ
(where {.} denotes the
fractional part function), then ƒ–1(x) is equalto -(A) x + 1 (B) x – 1
(C) x
x3
ì ü+ í ýî þ
(D) 3x x4 3
ì ü+ í ýî þ
19. ( )2
x
x2 x 1x
1 1 1lim x x 1 x x ...... x
2 2 2-
-®¥
ì üæ öæ ö æ ö+ + + +í ýç ÷ç ÷ ç ÷è øè ø è øî þ
is
equal to, (x Î N) -(A) 2 (B) ln2 (C) e2 (D) 1
20. If 2
xx 4x 3ƒ(x) tan
5sin x e 7æ öé ù- +
= pç ÷ê úç ÷+ +ë ûè ø,
then number of points, where ƒ(x) isdiscontinuous is ([.] is greatest integerfunction)(A) 0 (B) 1 (C) 2 (D) infinite
16. ,d ik= esa 3 yky, 4 gjh rFkk dqN la[;k esa lQsn
xsans gaSA nks xsankas dks ,d lkFk fudkyk tkrk gS rFkk og
fHkUu jaxks dh ikbZ tkrh gSA muds lQsn xsan ds u gksus
dh izkf;drk 2/9 gS rks lQsn xsanks dh la[;k gksxh&
(A) 3 (B) 4 (C) 6 (D) 817. (2x + 5y)34 ds çlkj esa la[;kRed egÙke in gksxk tc
x = 3 rFkk y = 2 gks(A) T21 (B) T22 (C) T23 (D) T24
18. ekuk ƒ : (3, 6) ® (4, 7) ,d Qyu ƒ(x) = 4x x3 3
ì ü- í ýî þ
ls ifjHkkf"kr gksrk gS (tgk¡ {.} fHkUukRed Hkkx Qyu dks
n'kkZrk gS), rks ƒ–1(x) cjkcj gksxk -(A) x + 1 (B) x – 1
(C) x
x3
ì ü+ í ýî þ
(D) 3x x4 3
ì ü+ í ýî þ
19. ( )2
x
x2 x 1x
1 1 1lim x x 1 x x ...... x
2 2 2-
-®¥
ì üæ öæ ö æ ö+ + + +í ýç ÷ç ÷ ç ÷è øè ø è øî þ
dk
eku gksxk (x Î N) -(A) 2 (B) ln2 (C) e2 (D) 1
20. ;fn 2
xx 4x 3ƒ(x) tan
5sin x e 7æ öé ù- +
= pç ÷ê úç ÷+ +ë ûè ø gks] rks fcUnqvksa
dh la[;k] tgk¡ ƒ(x) vlarr~ gks] gksxh
(tgk¡ [.] egÙke iw.kk±d Qyu gSA)
(A) 0 (B) 1 (C) 2 (D) vuUr
ALLEN
Page 34/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026
SECTION-II : (Maximum Marks: 20)� This section contains TEN questions.
Attempt any 5 questions. First 5 attemptedquestions will be considered for marking.
� The answer to each question is aNUMERICAL VALUE.
� For each question, enter the correctnumerical value (If the numerical value hasmore than two decimal places, truncate/round-off the value to TWO decimal places;e.g. 6.25, 7.00, –0.33, –.30, 30.27, –127.30, ifanswer is 11.36777..... then both 11.36 and11.37 will be correct) by darken thecorresponding bubbles in the ORS.For Example : If answer is –77.25, 5.2 thenfill the bubbles as follows.
0 0 0
+
0
–
0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9
••••••••••
0 0 0
+
0
–
0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9
••••••••••
� Answer to each question will be evaluatedaccording to the following marking scheme:Full Marks : +4 If ONLY the correctnumerical value is entered as answer.Zero Marks : 0 In all other cases
[kaM-II : (vf/kdre vad : 20)� bl [kaM esa nl iz'u gSaA fdUgh 5 iz'uksa dk mÙkj nhft,A
fd;s x;s iz'uksa esa ls dsoy izFke ik¡p iz'uksa dks gh vadfn;s tk;saxsaA
� izR; sd i z'u dk mÙkj ,d l a[; k Red e ku(NUMERICAL VALUE) gSA
� izR; sd i z'u ds mÙkj ds lgh la[; kRed eku(;fn la[;kRed eku esa nks ls vf/kd n'keyo LFkkugS] rks la[;kRed eku dks n'keyo ds nks LFkkuksa rdVª ad sV@jkm aM vk WQ (truncate/round-off) djsa_ mnkgj.k 6.25, 7.00, –0.33, –.30, 30.27, –127.30,;fn mÙkj 11.36777..... gS] rks 11.36 vkSj 11.37 nksuksalgh gksxsa) dks izfo"B djus ds fy, vks-vkj-,l- esa vuq:icqycqys dks dkyk djsaAmnkgj.k ds fy, : ;fn mÙkj –77.25, 5.2 gS] rks cqycqyksadks fuEu izdkj ls dkyk djsaA
0 0 0
+
0
–
0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9
••••••••••
0 0 0
+
0
–
0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9
••••••••••
� izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstukds vuqlkj gksxk%&iw.kZ vad : +4 ;fn flQZ lgh la[;kRed eku (Numericalvalue) gh mÙkj Lo:i ntZ fd;k x;k gSA'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA
ALLEN
07032021 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg Page 35/37All India Open Test/Enthusiast & Leader Course 0000CJA102120026
1. If function ƒ(x) = x3 + ax2 + bx + c ismonotonically increasing " x Î R, where a& b are prime numbers less than 10, thennumber of possible ordered pairs (a,b) is
2. Suppose that ƒ is differentiable for all x suchthat ƒ'(x) £ 2 for all x. If ƒ(1) = 2 and ƒ(4) = 8then ƒ(2) has the value equal to -
3. Consider 2
31
0
I x 1 dx= +ò & 33 2
21
I x 1 dx= -ò ,
then I1 + I2 is equal to
4. If m =2
1
x sin(x(1 x))dx-
-ò and n =2
1
sin(x(1 x))dx-
-ò ,
then mn
is equal to :-
5. Let y = g(x) be the inverse of a bijectivemapping f : R ® R, f(x) = 3x3 + 2x. The areabounded by graph of g(x), the x-axis and theordinate at x = 5 is -
6. A curve satisfying the differentialequation (x2y2 – 1) dy + 2xy3dx = 0, ispassing through the points (1, 1) and(k, 2) then the value of (8k2) is
7. The number of 3 × 3 skew symmetricmatrices A whose entries are chosen from
{–1, 0, 1} and for which the system x 0
A y 0
z 0
é ù é ùê ú ê ú=ê ú ê úê ú ê úë û ë û
has infinite solution is
1. ;fn lHkh x Î R ds fy, Qyu ƒ(x) = x3 + ax2 + bx + c
,dfn"V o/kZeku gS] tgk¡ a rFkk b, 10 ls NksVh vHkkT;
la[;k,¡ gS] rks laHko Øfer ;qXeks (a,b) dh la[;k gksxh
2. ekuk ƒ lHkh x ds fy, vodyuh; gS rFkk ƒ'(x) £ 2
gSA ;fn ƒ(1) = 2 rFkk ƒ(4) = 8 gks] rks ƒ(2) dk eku
gksxk &
3. ekuk 2
31
0
I x 1 dx= +ò rFkk 33 2
21
I x 1 dx= -ò gks] rks
I1 + I2 dk eku gksxk
4. ;fn m =2
1
x sin(x(1 x))dx-
-ò rFkk n =2
1
sin(x(1 x))dx-
-ò ,
rks mn
dk eku gksxk :-
5. ekuk y = g(x), ,dSdh vkPNknd izfrfp=.k f : R ® R,
f(x) = 3x3 + 2x dk izfrykse gSA g(x) ds vkjs[k]
x-v{k rFkk x = 5 ij dksfV }kjk ifjc¼ {ks=Qy gS -
6. ,d oØ] vody lehdj.k (x2y2 – 1) dy + 2xy3dx = 0
dks larq"V djrk gS tks fcUnq (1, 1) rFkk (k, 2) ls xqtjrk
gS] rks (8k2) dk eku gksxk
7. 3 × 3 fo"ke lefer vkO;wg A dh la[;k] ftlds çfof"V;k
{–1, 0, 1} esa ls pquh tkrh gS rFkk ftlds fy;s fudk;
x 0
A y 0
z 0
é ù é ùê ú ê ú=ê ú ê úê ú ê úë û ë û
ds vuUr gy gks] gksxh
ALLEN
Page 36/37 SPACE FOR ROUGH WORK / jQ dk;Z ds fy, txg 07032021All India Open Test/Enthusiast & Leader Course 0000CJA102120026
8. If [.] represents greatest integer function, thesystem of equations 2x – 3y = 4;7x – 2y = 2; 9x – 5y = [4a]
(a is some constant) is consistent, then rangeof values of 'a' is [a, b), where a + b is
9. A jar contains 7 white marbles and 3 bluemarbles. Given that the 4-marbles are chosenfrom the jar at the same time, then thestandard deviation of the number of the blue
marbles choosen is ab
where a and b are
co-prime numbers and a is square free thena + b is
10. If number of arrangements of letters of theword "DHARAMSHALA" taken all at a timeso that no two alike letters appear togetheris (4a.5b.6c.7d), (where a, b, c, d Î N), thena + b + c + d is equal to
8. lehdj.k fudk; 2x – 3y = 4; 7x – 2y = 2;9x – 5y = [4a], tgk¡ [.] egÙke iw.kk±d Qyu dks n'kkZrkgS
(a dksbZ vpj gS )] laxr gS] rks 'a' ds ekuksa dk ifjlj[a, b) gS] tgk¡ a + b dk eku gksxk
9. ,d dy'k esa 7 lQsn ekcZy 3 uhys ekcZy gSA fn;k x;k
gS fd ,d gh ckj esa dy'k ls 4 ekcZy dk p;u fd;k
tkrk gS] rks uhys ekcZy ds pquus dh la [;k dk ekud
fopyu a
b gS] tgk¡ a rFkk b lg vHkkT; la[;k;sa rFkk a
iw.kZ oxZ ugha gks] rks a + b gksxk
10. 'kCn "DHARAMSHALA" ds lHkh v{kjksa dks ,d lkFk
ysdj O;ofLFkr djus dh la[;k rkfd dksbZ Hkh nks le:i
;fn v{kj lkFk&lkFk u vk; s (4a.5b.6c.7d) gSA
(tgk¡ a, b, c, d Î N), rks a + b + c + d dk eku gksxk