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MAJOR TEST (MT)(JEE MAIN PATTERN)

TARGET : JEE (MAIN+ADVANCED) 2016COURSE : VIKAAS (JA) & VIPUL (JB)

Date¼fnukad½:27-01-2015 Time¼le;½: 3 Hours ¼?k.Vs½ Max. Marks¼egÙke vad½: 279Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

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

INSTRUCTIONS / funsZ'k :A. General : A. lkekU; %1. This booklet is your Question Paper. Do not break the

seals of this booklet before being instructed to do soby the invigilators.

1. ;g iqfLrdk vkidk iz'ui=k gSA bldh eqgjsa rc rd u rksMsatc rd fujh{kdksa ds }kjk bldk funsZ'k u fn;k tk;sA

2. Blank spaces and blank pages are provided in thequestion paper for your rough work. No additionalsheets will be provided for rough work.

2. dPps dke ds fy;s [kkyh i`"B vkSj [kkyh txg bl iqfLrdkesa gh gSA dPps dke ds fy, dksbZ vfrfjDr dkxt ugha fn;ktk;sxkA

3. Blank papers, clipboards, log tables, slide rules,calculators, cameras, cellular phones, pagers andelectronic gadgets are NOT allowed inside theexamination hall.

3. dksjs dkxt] fDyi cksMZ (CLIP BOARD)] ykWx rkfydk]LykbM:y] dSYdqysVj] dSejk] lsyQksu] istj vkSj fdlhizdkj ds bysDVªkfud midj.k ijh{kk de esa vuqefr ugha gSA

4. Write you name and roll number in the space providedon the back cover of this booklet.

4. bl iqfLrdk ds fiNys i`"B ij fn, x, LFkku esa viuk ukevkSj jksy uEcj fyf[k,A

5. Using a black ball point pen, darken the bubbleson the upper original sheet.

5. Åijh ewy i`"B ds cqycqyksa (BUBBLES) dks dkys ckWyIokbaV dye ls dkyk djsaA

6. DO NOT TAMPER WITH/MUTILATE THE ORS ORTHE BOOKLET.

6. vks-vkj-,l- (ORS) ;k bl iqfLrdk esa gsj&Qsj@foÑfr udjsaA

7. On breaking the seals of the booklet check that itcontains all the 84 questions and correspondinganswer choices are legible. Read carefully theInstructions printed at the beginning of each section.

7. bl iqfLrdk dh eqgjsa rksM+us ds i'pkr~ Ñi;k tk¡p ysa fd blesalHkh 84 iz'u vkSj muds mÙkj fodYi Bhd ls i<+s tk ldrsgSaA lHkh [kaMksa ds izkjaHk esa fn;s gq, funsZ'kksa dks /;ku ls i<+saA

B. Filling the ORSUse only Black ball point pen only for filling the ORS.

8. Write your Roll no. in the boxes given at the top leftcorner of your ORS with black ball point pen. Also,darken the corresponding bubbles with Black ball pointpen only. Also fill your roll no on the back side of your ORSin the space provided (if the ORS is both side printed).

B. vks-vkj-,l (ORS) HkjukORS dks Hkjus ds fy, dsoy dkys ck¡y iSu dk mi;ksx dhft,A

8. ORS ds lcls Åij cka;s dksus esa fn, x, ck¡Dl esa viukjksy uEcj dkys ck¡y ikbUV ls fyf[k, rFkk laxr xksys Hkhdsoy dkys isu ls Hkfj;sA ORS ds ihNs dh rjQ Hkh viukjksy uEcj fyf[k, (;fn ORS nksuksa rjQ Nih gqbZ gSA)

9. Fill your Paper Code as mentioned on the Test Paperand darken the corresponding bubble with Black ballpoint pen.

9. ORS ij viuk isij dksM fyf[k, rFkk laxr xksyksa dks dkysck¡y isu ls dkys dhft,A

10. If student does not fill his/her roll no. and paper codecorrectly and properly, then his/her marks will not bedisplayed and 5 marks will be deducted (paper wise)from the total.

10. ;fn fo|kFkhZ viuk jksy uEcj rFkk isij dksM lgh vkSjmfpr rjhds ugha Hkjrk gS rc mldk ifj.kke jksd fy;ktkosxk rFkk iz'u&i=k esa izkIrkad ls 5 vad dkV fy, tkosaxsaA

11. Since it is not possible to erase and correct pen filledbubble, you are advised to be extremely careful whiledarken the bubble corresponding to your answer.

11. pawfd isu ls Hkjs x, xksys feVkuk vkSj lq/kkjuk laHko ugha gSblfy, vki lko/kkuh iwoZd vius mÙkj ds xksyksa dks HkjsaA

Please read the last page of this booklet for rest of the instructionsÑi;k 'ks"k funsZ'kksa ds fy;s bl iqfLrdk ds vfUre i`"B dks i<+sA

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

Ph.No. : +91-744-3012222, 6635555 | Toll Free : 1800 200 2244 | 1800 102 6262 | 1800 258 5555Reg. Office : J-2, Jawahar Nagar, Main Road, Kota (Raj.)-324005 | Ph. No.: +91-744-3192222 | FAX No. : +91-022-39167222

Website : www.resonance.ac.in | E-mail : [email protected] | CIN: U80302RJ2007PTC024029

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

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

Website : www.resonance.ac.in | E-mail : [email protected] Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PTC024029

JAMTMAIN270115C0-1


PART - I : PHYSICS

SECTION - IStraight Objective Type

This section contains 19 multiple choice questions. Each question has 4 choices (1), (2), (3) and(4) for its answer, out of which ONLY ONE is correct.

[k.M- Ilh/ks oLrqfu"B izdkj

bl [k.M esa 19 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,dlgh gSA

1. The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In

another 10s it will decrease to times its original magnitude, where equals.

5 lSd.M esa voefUnr nksfy=k dk vk;ke ?kVdj vius izkjfEHkd eku dk 0.9 xquk gks tkrk gSA vU; 10 lSd.M

esa ;g ?kVdj blds izkjfEHkd eku dk xquk gks tkrk gS] tgkW fdlds cjkcj gksxkA(1) 0.7 (2) 0.81 (3) 0.729 (4) 0.6

2. A uniform elastic rod of mass m in vertical position is pulled by a constant vertical upward force of

magnitude F applied at its top end as shown. The natural length of rod is L, Young's modulus of

rod is Y and cross section area of rod is A. Then the extension in rod is : (acceleration due to

gravity is g)

(1) FL2YA

(2) (F mg)L2YA (3) (F mg)L

2YA (4) None of these

Å/okZ/kj fLFkr m nzO;eku dh ,d ,dleku izR;kLFk NM+ dks Åijh fljs ij fu;r cy F yxkdj Å/okZ/kj

Åij dh vksj fp=kkuqlkj [khapk tkrk gSA NM+ dh LokHkkfod yEckbZ L gSA NM+ dk ;ax izR;kLFkrk xq.kkad Y rFkk

vuqizLFk dkV {ks=kQy A gS rks NM+ esa foLrkj gksxk & (g xq:Roh; Roj.k gS)

(1) FL2YA

(2) (F mg)L2YA (3) (F mg)L

2YA (4) buesa ls dksbZ ugh

F

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JAMTMAIN270115C0-2

3. Which of the following will have maximum average total kinetic energy at temperature 300 K.(1) 1 kg of H2 (2) 1 kg of He

(3) 12

kg of H2 +12

kg of He (4) 14

kg of H2 +34

kg of He

fuEu fn;s x;s esa fdldh dqy vkSlr xfrt ÅtkZ 300K ij egRre gksxhA(1) H2 dk 1 kg (2) He dk 1 kg

(3) H2 dk 12

kg + He dk 12

kg (4) H2 dk 14

kg + He dk 34

kg

4. Figure shows a particle P moving with constant velocity V along the positive x axis and four pointsa, b, c and d with their x and y coordinates. If L1, L2, L3 and L4 are the magnitudes of angularmomentum of the particle about the points a, b, c and d respectively then which of the following isincorrect :fp=kkuqlkj ,d d.k P fu;r osx V ls /kukRed x v{k ds vuqfn'k xfr dj jgk gS rFkk pkj fcUnw a, b, c rFkk d

vius x rFkk y funsZ'kkad ds lkFk iznf'kZr gSA ;fn L1, L2, L3 rFkk L4 Øe'k% fcUnqvksa a, b, c rFkk d ds lkis{k d.kds dks.kh; laosx ds ifjek.k gS rks fuEu esa ls dkSu ls vlR; gSA

(1) L1 > L3 (2) L2 = L4 (3) L1 = L3 (4) L1 > L2

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JAMTMAIN270115C0-3

5. Some liquid is filled in a cylindrical vessel of radius R. Let F1 be the magnitude of force applied by

the liquid on the bottom of the cylinder. Now the same liquid is poured in to a vessel of uniform

square cross-section of side R. Let F2 be the magnitude of force applied by the liquid on the

bottom of this new vessel, both vessels are placed on horizontal surface. Then : (Neglect

atmospheric pressure)

R f=kT;k ds ,d csyukdkj crZu es dksbZ nzo Hkjk x;k gS] tks fd F1 ifjek.k dk cy mijksDr csyukdkj crZu ds

iSans (bottom) ij vkjksfir djrk gSA vc mlh nzo dks R Hkqtk ds ,dleku oxkZdkj vuqizLFk dkV ds ik=k esa

Hkjk x;k gSA vc ;fn nwljs ik=k ds iSans ij ;g F2 ifjek.k dk cy vkjksfir djrk gS rks (;g ekfu;s fd nksuksa

ik=k {kSfrt lrg ij j[ks x;s gSA½ ¼ok;qe.Mfy; nkc ux.; gSA½

(1) F1 = F2 (2) F1 = 2F

(3) F1 = F2 (4) F1 = F2

6. A Carnot engine, having an efficiency of = 1/10 as heat engine, is used as a refrigerator. If the

work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower

temperature is

= 1/10 n{krk okyk ,d dkuksZa Å"ek batu 'khryd ds :i esa iz;qDr fd;k tkrk gS] ;fn fudk; ij fd;k

x;k dk;Z 10 J gS, rks fuEu rki ij Hk.Mkj ls vo'kksf"kr ÅtkZ dh ek=kk gksxhA

(1) 99 J (2) 90 J (3) 1 J (4) 100 J

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JAMTMAIN270115C0-4


7. A boy of mass 30 kg starts running from rest along a circular path of radius 6 m with constant

tangential acceleration of magnitude 2 m/s2. After 2 sec from start he feels that his shoes started

slipping on ground. The friction coefficient between his shoes and ground is : (Take g = 10 m/s2)

30 kg nzO;eku dk ,d yM+dk 6 m f=kT;k ds o`Ùkkdkj iFk ds vuqfn'k 2 m/s2 ds fu;r Li'kZjs[kh; Roj.k ls

fojkekoLFkk ls nkSM+uk 'kq: djrk gSA mlds nkSM+us ds 2 lS- i'pkr~ og eglwl djrk gS fd mlds twrs tehu

ij fQlyuk izkjEHk djrs gSA mlds twrs rFkk tehu ds e/; ?k"kZ.k xq.kkad gksxkA (g = 10 m/s2 yhft;s)

(1) 12

(2) 13

(3) 14

(4) 15

8. A small bob of mass 10kg is pulled along the circular arc by means of an ideal string which passes

over a small pulley Q. A horizontal constant force F = 200 N is acting on the string towards right as

shown in figure. The bob starts from rest at P. Assuming that string remains taut :

(1) The speed of bob at Q would be 20 m/s (2) The speed of bob at Q would be 10 3 m/s

(3) The speed of bob at Q would be 15 m/s (4) The speed of bob at Q would be 18 m/s

10kg nzO;eku dk ,d NksVk xksyd vkn'kZ Mksjh tks NksVh f?kjuh Q ls xqtj jgh gS] dh lgk;rk ls o`Ùkkdkj

pki ds vuqfn'k [khapk tkrk gSA F = 200 N dk ,d fu;r {kSfrt cy Mksjh ij fp=kkuqlkj nka;h vksj dk;Zjr gSA

xksyd fcUnq P ij fojke ls çkjEHk gksrk gSA ;g ekfu, fd Mksjh ruh gqbZ jgrh gSA

(1) fcUnq Q ij xksyd dh pky 20 m/s gksxkA (2) fcUnq Q ij xksyd dh pky 10 3 m/s gksxhA

(3) fcUnq Q ij xksyd dh pky 15 m/s gksxhA (4) fcUnq Q ij xksyd dh pky 18 m/s gksxhA

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JAMTMAIN270115C0-5

9. A block of mass m1 lies on top of fixed wedge as shown in figure-1 and another block of mass m2lies on top of wedge which is free to move as shown in figure-2. At time t = 0, both the blocks arereleased from rest from a vertical height h above the respective horizontal surface on which thewedge is placed as shown. There is no friction between block and wedge in both the figures. LetT1 and T2 be the time taken by block in figure-1 and block in figure-2 respectively to just reach thehorizontal surface, then :(1) T1 > T2(2) T1 < T2(3) Magnitude of normal reaction on block of mass m2 is less than that on the block of mass m1.(4) After falling through same height (less than h) from top of wedge, the speed of block of mass

m2 is less than that of block of mass m1.

h

horizontal surfacefigure-1

m1

h

smooth horizontal surfacefigure-2

m2

fixedwedge

h

{kSfrt lrgfp=k-1

m1

h

fpduh {k Sfrt lrgfp=k-2

m2

fLFkj ost

fp=k-1 esa m1 nzO;eku dk CykWd fLFkj ost ds Åijh fljs ij j[kk gqvk gS rFkk fp=k-2 esa m2 nzO;eku dk nwljkCykWd nwljs ost ds Åijh fljs ij j[kk gqvk gS tks xfr djus ds fy, LorU=k gSA t = 0 le; ij nksuksa CykWdksadks fojkekoLFkk ls mudh {kSfrt lrg tgkW osx j[kk gS] ls Å/okZ/kj h Å¡pkbZ ls NksM+k tkrk gS gSA nksuksafLFkfr;ksa esa CykWd o ost ds e/; ?k"kZ.k vuqifLFkr gSA ekuk {kSfrt lrg rd igq¡pus esa nksuksa CykWdksa ¼fp=k (1)rFkk ¼2½ esa ½ }kjk fy;k x;k le; Øe'k% T1 rFkk T2 gS rks &(1) T1 > T2(2) T1 < T2

(3) m2 nzO;eku ds CykWd ij vfHkyEc izfrfØ;k dk ifjek.k m1 nzO;eku ds CykWd ij vfHkyEc izfrfØ;k dsifjek.k ls de gSA

(4) ost ds 'kh"kZ ls (h ls de) leku Å¡pkbZ ls fxjk;s x;s m2 nzO;eku ds CykWd dh pky m1 nzO;eku ds CykWddh pky ls de gSA

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JAMTMAIN270115C0-6

10. A source of frequency ‘ f ’ is stationary and an observer starts moving towards it at t = 0 with

constant small acceleration. Then the variation of observed frequency f registered by the

observer with time is best represented as :

‘ f ’ vko`fr dk L=kksr fLFkj gS rFkk t = 0 ij ,d isz{kd mldh vksj vYi vpj Roj.k ls xfr izkjEHk djrk gSA rc

izs"kd }kjk le; ds lkFk ekih xbZ izsf{kr vkofÙk f dks lcls lgh iznf'kZr djrk gS

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

11. Kinetic energy versus time graph of a particle of mass m executing SHM under the effect of a net

force F is shown. Then maximum value of F2/2m (in SI unit) is :

dqy cy F ds izHkko esa ljy vkorZ xfr djrs gq, m nzO;eku ds d.k dk xfrt ÅtkZ o le; ds e/; xzkQ

fp=kkuqlkj iznf'kZr gS rc F2/2m (SI bdkbZ esa) dk vf/kdre eku gksxkA

(1) 50 2 (2) 100 2 (3) 200 2 (4) 400 2

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12. A uniform solid sphere of radius 'r' is placed on a horizontal surface. A horizontal impulse isapplied on it at a height 'h' above the center as shown in the figure. If soon after the impact spherestarts rolling (without slipping) the ratio of h/r is:r f=kT;k dk ,d ,dleku Bksl xksyk ,d {kSfrt lrg ij j[kk gSA ,d {kSfrt vkosx bl ij dsUnz ls h ÅWpkbZij fp=kkuqlkj vkjksfir fd;k tkrk gSA ;fn VDdj ds Bhd ckn xksyk yq<+duk ¼fcuk fQlys½ izkjEHk djrk gS rksvuqikr h/r gS %

h

r

(1) 1/2 (2) 1/5 (3) 2/5 (4) 1/4

13. A uniform square plate of mass m and length 2 L is hinged at point A. The square plate canrotate in the vertical plane about the horizontal, smooth and fixed axis passing through A. Thesquare plate is released from a position where it's diagonal AC is horizontal as shown. The angularvelocity of the square when the diagonal AC becomes vertical is :,d nzO;eku m rFkk 2 L yEckbZ dh ,dleku oxkZdkj IysV dks fcUnq A ij dhyfdr fd;k x;k gSA oxkZdkjIysV A ls xqtjus okyh {kSfrt] ?k"kZ.k jfgr ,oa fLFkj v{k ds ifjr% Å/okZ/kj ry esa ?kw.kZu dj ldrh gSA oxkZdkjIysV dks fp=kkuqlkj ml fLFkfr ls tgk¡ ij bldk fod.kZ AC {kSfrt gS] NksM+k tkrk gSA tc fod.kZ AC Å/okZ/kjgks tkrk gS rc oxkZdkj IysV ds dks.kh; osx dk eku Kkr djks &

(1) 3g2L

(2) 2gL

(3) 4gL

(4) 3g5L

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JAMTMAIN270115C0-8


14. Two moles of an ideal monoatomic gas at an initial temperature T0 = 300K is cooled isochorically

such that its pressure becomes half the initial value. After this the gas is isobarically allowed toexpand till its temperature again comes back to initial value, that is, T0 = 300K. Then the

magnitude of total heat absorbed by gas in this complete process is (Take R = 8.3 mol–1 J K–1),d vkn'kZ ,dijek.kqd xSl ds nks esky izkjfEHkd rki T0 = 300K ij gS rFkk bldks levk;rfud :i ls blrjg ls B.Mk fd;k tkrk gS fd bldk nkc izkjfEHkd nkc dk vk/kk gks tkrk gS blds ckn bl xSl dkslenkch; :i ls rc rd QSyrh gS tc rd bldk iqu% ogh izkjfEHkd rki vk;s vFkkZr~ T0 = 300K ij okilvk;s] rks bl iwjs izØe ds nkSjku dqy vo'kksf"kr Å"ek dk ifjek.k gS & (R = 8.3 mol–1 J K–1)

(1) 1245 J (2) 2290 J (3) 2490 J (4) 3735 J

15. A sphere of mass m and radius r is projected in a gravity free space with speed v. If coefficient of

viscosity of the medium in which it moves is 16

, the distance travelled by the body before it stops is :

m nzO;eku ,oa r f=kT;k dk ,d xksyk ,d xq:Rojfgr vkdk'k esa v osx ls ç{ksfir fd;k tkrk gSA ;fn ek/;e

ftlesa ;g xfr'khy gS dk ';kurk xq.kakd 16

gS ] rks oLrq }kjk :dus ls iwoZ r; dh x;h nwjh gksxh :

(1) mv2r

(2) 2mvr

(3) mvr

(4) mv4r

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JAMTMAIN270115C0-9

16. One mole of an ideal gas at pressure P0, volume V0 and temperature T0 is expanded isothermally

to twice its volume and then compressed at constant pressure to (V0/2) and the gas is brought

back to original state by a process in which P V (Pressure is directly proportional to volume).

The correct representation of process is

,d eksy vkn'kZ xSl nkc P0, vk;ru V0 rFkk rkieku T0 ls nqxqus vk;ru rd lerkih; :i ls izlkfjr dh

tkrh gS rFkk fQj lenkch izØe ls (V0/2) rd laihfMr dh tkrh gS rFkk xSl dks okil izkjfEHkd voLFkk esa

,sls izØe ls ykrs gS ftlesa P V (nkc] vk;ru ds lekuqikrh gksrk gS) gS rks mijksDr izØe dks iznf'kZr fd;k

tk ldrk gS

(1) (2)

(3) (4)

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17. The given system is displaced by distance ‘A’ and released from rest. Both the blocks movetogether without relative slipping in the whole process. The magnitude of frictional force between

them at time ‘t’ is : (where = K2m

)

fn;s x;s fudk; dks ‘A’ nwjh ls foLFkkfir fd;k tkrk gS rFkk fojke voLFkk ls NksM+k tkrk gSA lEiw.kZ çfØ;k esanksauks CykWd fcuk lkisf{kd fQlyu ls lkFk&lkFk xfr djrs gSA fdlh le; ‘t’ ij buds e/; yxus okys ?k"kZ.k

cy dk ifjek.k gksxk : (tgk¡ = K2m

)

m

m

very rough ( )cgqr [kqjnjk

smooth ( )?k"kZ.kjfgrx=0

k

(1) KA2

|cos t| (2) KA2

cos t (3) KA2

|sin t| (4) KA |cos t|

18. A uniform rod AB of mass m and length at rest on a smooth horizontal surface. An impulse P isapplied to the end B as shown in figure. The time taken by the rod to turn through a right angle is :m nzO;eku o yEckbZ dh ,d ,dleku NM+ AB ,d {kSfrt fpdus ry ij fLFkj j[kh gq;h gSA blds fljs Bij ,d vkosx P vkjksfir fd;k tkrk gS tSlk fd fp=k esa n'kkZ;k x;k gSA NM+ }kjk ledks.k ls ?kwe tkus esayxk le; gS &

(1) 2 mP (2) m

3P (3) m

12P (4) 2 m

3P

19. Assuming the xylem tissues through which water rises from root to the branches in a tree to be ofuniform cross-section find the maximum radius of xylem tube in a 10 m high coconut tree so thatwater can rise to the top. (surface tension of water = 0.1N/m, Density of water = 1000 kg/m3,Angleof contact of water with xylem tube= 60° and g=10m/s2)ekuk tkbZfye mÙkdksa }kjk isM+ks esa ikuh tM+ks ls mldh Vgfu;ksa rd le:i vuqizLFk dkV {ks=k }kjk Åij p<+rkgSA ukfj;y ds 10 eh- Å¡ps isM+ esa ikuh p<+us ds fy, tkbfye uyh dh vf/kdre f=kT;k Kkr djksA (ikuh dki`"B ruko = 0.1N/m, ikuh dk ?kuRo = 1000 kg/m3,ikuh o tkbZfye uyh dk Li'kZ dks.k = 60° rFkkg=10m/s2)(1) 1 cm (2) 1 mm (3) 10–5 m (4) 10–6 m

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JAMTMAIN270115C0-11

SECTION - IIReasoning Type

This section contains 4 reasoning type questions. Each question has 4 choices (1), (2), (3) and (4),out of which ONLY ONE is correct.

[k.M - IIdkj.k&izdkj

bl [k.M esa 4 dkj.k ds ç'u gSA çR;sd ç'u ds 4 fodYi (1), (2), (3) rFkk (4) gS] ftlesa ls flQZ ,d lgh gSA

20. STATEMENT–1: The mechanical energy between consecutive node and antinode will remain

conserved in standing wave.

STATEMENT–2: The mechanical energy does not flow between node and antinode in standing

wave.

(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 True; Statement-2 is NOT a correct explanation for Statement-1

(3) Statement-1 is True, Statement-2 is False

(4) Statement-1 is False, Statement-2 is True

oDrO;-1: vizxkeh rjax esa Øekxr fuLian rFkk izlianks ds chp ;kaf=kd ÅtkZ lajf{kr jgsxhA

oDrO;-2: vizxkeh rjax esa fuLian rFkk izlianks ds e/; ;kaf=kd ÅtkZ izokg ugh gksrk gSA

(1) oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO; -1 dk lgh Li"Vhdj.k gSA

(2) oDrO;-1 lR; gS, oDrO;-2 lR; gS ; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k ugha gSA

(3) oDrO; -1 lR; gS, oDrO;-2 vlR; gSA

(4) oDrO; -1 vlR; gS , oDrO;-2 lR; gSA

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JAMTMAIN270115C0-12


21. STATEMENT-1 : A SHM may be assumed as composition of many SHM's.STATEMENT-2 : Superposition of many SHM's (along same line) of same frequency will be aSHM.(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 True; Statement-2 is NOT a correct explanation for Statement-1(3) Statement-1 is True, Statement-2 is False(4) Statement-1 is False, Statement-2 is TrueoDrO;-1 : ,d ljy vkorZ xfr dks dbZ lkjh ljy vkorZ xfr;ksa ds la;qXeu ls cuk eku ldrs gSAoDrO;-2 : ,d gh leku js[kk ij leku vko`fÙk okyh dbZ ljy vkorZ xfr;ksa ds v/;kjksi.k ls ,d ljy vkorZxfr gksxhA(1) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA(2) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k ugha gSA(3) oDrO;-1 lR; gS] oDrO;-2 vlR; gS ;(4) oDrO;-1 vlR; gS] oDrO;-2 lR; gS

22. STATEMENT-1 : A man standing in a lift which is moving upward, will feel his weight to be greaterthan when the lift was at rest.STATEMENT-2 : If the acceleration of the lift is ‘a’ upward, then the man of mass m shall feel hisweight to be equal to normal reaction (N) exerted by the lift given by N = m(g+a) (where g isacceleration due to gravity)(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 True; Statement-2 is NOT a correct explanation for Statement-1(3) Statement-1 is True, Statement-2 is False(4) Statement-1 is False, Statement-2 is TrueoDrO;-1 : ,d vkneh fy¶V esa [kM+k gS rFkk fy¶V Åij dh rjQ py jgh gS rks og O;fDr bl ckj fy¶V dsfojke esa gksus dh rqyuk esa ges'kk vf/kd Hkkj eglwl djsxkAoDrO;-2 : ;fn fy¶V dk Åij dh rjQ Roj.k 'a' gS rks m nzO;eku dk O;fDr }kjk eglwl fd;k x;k Hkkjfy¶V }kjk vkneh ij yxk;s x;s vfHkyEc cy N ds cjkcj gksxk tgka N = m(g+a) ( ;gka g xq:Rokd"kZ.k dsdkj.k Roj.k gS)(1) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA(2) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k ugha gSA(3) oDrO;-1 lR; gS] oDrO;-2 vlR; gS ;(4) oDrO;-1 vlR; gS] oDrO;-2 lR; gS

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JAMTMAIN270115C0-13

23. STATEMENT-1: During beat formation intensity of sound at a particular position of space keep on

changing with time.

STATEMENT-2: Beats are formed due to super position of sound waves of unequal frequencies.

(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 True; Statement-2 is NOT a correct explanation for Statement-1


(4) Statement-1 is False, Statement-2 is True.

oDrO;-1 : foLian cuus ds nkSjku fdlh fo'ks"k fLFkfr ij /ofu dh rhozrk le; ds lkFk ifjofrZr gksrh gSA

oDrO;-1 : foLian vleku vko`fÙk;ksa dh /ofu rjaxksa ds v/;kjksi.k ds dkj.k curs gSaA

(1) oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO; -1 dk lgh Li"Vhdj.k gSA


(3) oDrO; -1 lR; gS, oDrO;-2 vlR; gSA

(4) oDrO; -1 vlR; gS , oDrO;-2 lR; gSA

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JAMTMAIN270115C0-14

SECTION - IIIInteger value correct Type

This section contains 5 questions. The answer to each question is a single digit integer, rangingfrom 0 to 9 (both inclusive).

[k.M - IIIiw.kkZad eku lgh izdkj

bl [k.M esa 5 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd ¼nksuksa 'kkfey½ ds chp dk ,dy vadh; iw.kkZad gSA

24. A transverse sinusoidal wave moves along a string in the positive x-direction at a speed of10 cm/s. The wavelength of the wave is 0.5 m and its amplitude is 10 cm. At a particular time t, the

snap-shot of the wave is shown in figure. The speed of point P is x 2 cm/s when its

displacement is 102

cm, then the value of x is :

,d vuqçLFk T;koØh; (transverse sinusoidal) rjax ,d Mksjh esa 10 cm/s dh pky ls /kukRed x-v{k dhvksj pyrh gSA bldh rjaxnS/;Z 0.5 m rFkk vk;ke 10 cm gSA ,d fo'ks"k le; t ij, rjax dk vk'kqfp=k (snap-

shot) fp=k esa fn[kk;k x;k gSA tc fcUnq P dk foLFkkiu 102

cm gS rc fcUnq P dh pky x 2 cm/s gS rks x

dk eku Kkr djksA

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JAMTMAIN270115C0-15

25. A particle moves along x-axis in positive direction. Its acceleration 'a' is given as a = cx + d, where

x denotes the x-coordinate of particle, c and d are positive constants. For velocity-position graph of

particle to be of type as shown in figure, find the value of speed (in m/s) of particle at x = 0.

Take c = 1 s–2 and d = 3 ms–2

x-v{k ds vuqfn'k /kukRed fn'kk esa xfr djrs gq;s ,d d.k ds fy, Roj.k a = cx + d }kjk fn;k tkrk gS tgk¡

x d.k dk x-funsZ'kkad iznf'kZr djrk gS] c rFkk d /kukRed fu;rkad gSA osx&fLFkfr xzkQ fn[kk;s x;s fp=kkuqlkj

gksus ds fy, x = 0 ij d.k dh pky ¼m/s esa½ dk eku Kkr djksA yhft;s c = 1 s–2 rFkk d = 3 ms–2

26. 4th harmonic of an open organ pipe has frequency 33 Hz lesser than 5th harmonic of same open

organ pipe. Find the length of the pipe (in meters). (Assume velocity of sound in air = 330 m/s.

Neglect end correction.)

,d [kqys vkWxZu ikbZi dh 4th lauknh dh vko`fÙk blh ikbZi dh 5th lauknh dh vko`fÙk ls 33 Hz U;wu gSA ikbi

dh yEckbZ ehVj esa Kkr djksA (ok;q esa /ofu osx = 330 m/s ekfu,) vUR; la'kks/ku ux.; gSA

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JAMTMAIN270115C0-16


27. An ideal gas undergoes a process A B C shown in P-V graph. Then total heat given to thegas during the whole process (A B C) is n × 104 J, then n is:,d vkn'kZ xSl dks iznf'kZr P–V xzkQ esa izØe A B C ls xqtkjk tkrk gS rc lEiw.kZ izØe(A B C) ds nkSjku xSl dks nh xbZ dqy Å"ek n × 104 J gS] rc n gksxkA

28. A ball collides elastically with a massive wall moving towards it with a velocity of v as shown. Thecollision occurs at a height of h above ground level and the velocity of the ball just before collisionis 2v in horizontal direction. Then the distance between the foot of the wall and the point on the

ground where the ball lands (at the instant the ball lands) is 2hn vg

, where n is :

fp=k esa n'kkZ;s vuqlkj ,d xsan] ,d Hkkjh nhokj tks xsan dh rjQ v osx ls xfr'khy gS] ls çR;kLFk VDdj djrhgSA VDdj i`Foh ry esa h Å¡pkbZ ij gksrh gS rFkk VDdj ls Bhd iwoZ xsan dk {kSfrt fn'kk esa osx 2v gSA rcnhokj ds vk/kkj (foot) ,oa /kjkry ij fLFkr ij ml fcUnq ds e/; dh nwjh ¼/kjkry ij xsan ds fxjus ds {k.k

ij½ tgk¡ xsan fxjrh gS] 2hn vg

gS] tgkW n gksxk &

h

v 2v

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Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005Website : www.resonance.ac.in | E-mail : [email protected] Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PTC024029

JAMTMAIN270115C0-17

PART : II CHEMISTRYAtomic 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]





29. For the redox reaction MnO4– + C2O4

2– + H+ Mn2+ + CO2 + H2O,

the correct whole number stoichiometric coefficients of MnO4–, C2O4

2– and H+ are respectively :

mikip;h vfHkfØ;k MnO4– + C2O4

2– + H+ Mn2+ + CO2 + H2O esa

MnO4–, C2O4

2– rFkk H+ ds fy, iw.kk±d jllehdj.kfefr xq.kkad Øe'k% fuEu gSa %

(1) 2, 5, 16 (2) 5, 2, 8 (3) 2, 5, 8 (4) 5, 2, 16

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30. For a liquid, enthalpy of fusion is 1.435 × 103 cal / mol. K and molar entropy change is

5.26 cal/ mol. Melting point of the liquid (in °C)

,d nzo ds fy, laxyu dh ,UFkSYih 1.435 × 103 cal / mol K o eksyj ,UVªksih ifjorZu 5.26 cal/ mol gS rks

nzo dk DoFkaukd (°C esa) gksxkA(1) 273 K (2) 300 K (3) 473 K (4) 0 K

31. Which of the following graph is/are correct regarding ideal gas ?

(1)

P

density

(T=Constant) (2)n

T

(P,V=Constant)

(3)

n

P

(P,V=Constant) (4)

P(T=Constant)

1V

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JAMTMAIN270115C0-19

fuEu esa lss dkSulk xzkQ vkn'kZ xSl ds fy, lgh gS\

(1)P

?kuRo

(T=fu;r)(2)

n

T

(P,V=fu;r)

(3)n

P

(P,V=fu;r) (4)

P(T=fu;r)

1V

32. The mass of Glucose (C6H12O6) required to prepare 50 mL of its M9

aqueous solution is :

Xywdksl (C6H12O6) dk og nzO;eku] tks blds M9

tyh; foy;u ds 50 mL dks cukus ds fy, vko';d gS]

fuEu gS %(1) 10 g (2) 1 mg (3) 1 g (4) 100 mg

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33. An insulated container is divided into two equal portions. One portion contains an ideal gas at

pressure P and temperature T. The other portion is a perfect vaccum, if a hole is opened

between two portion then

(1) temperature remain constant

(2) temperature increases

(3) temperature decreases

(4) temperature initially constant and then increases

,d vojks/kh (insulated) ik=k dks nks cjkcj Hkkxksa esa foHkkftr djrs gSA nkc P rFkk rki T ij ik=k ds ,d Hkkx

esa vkn'kZ xSl mifLFkr gSA nwljk Hkkx iw.kZr% fuokZfrr gSA ;fn nksuksa Hkkxksa ds e/; ,d fNnz cuk fn;k tk, rc %

(1) rkieku fu;r jgrk gSA

(2) rkieku c<+ tkrk gSA

(3) rkieku ?kV tkrk gSA

(4) rkieku izkjEHk esa fu;r rFkk fQj c<+rk gSA

34. The ratio of radius difference between 4th & 3rd orbit of of He+ ion & radius difference between 4th

& 3rd orbit of Li2+ ion is :

He+ vk;u dh 4th rFkk 3rd d{kkvksa dh f=kT;kvksa ds e/; vUrj o Li2+ vk;u dh 4th rFkk 3rd d{kkvksa dh

f=kT;kvksa ds e/; vUrj dk vuqikr gS %(1) 3:2 (2) 2:3 (3) 1:3 (4) 3:1

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35. AB(s) A (g) + B(g) Kp = 4, H = + ve

In a container, A (g) and B (g) are filled to partial pressure of 1 atm each. Now AB (s) is added

(in excess quantity). Which of the following is INCORRECT ? (No other gas is present in

container):

(1) The pressure of B(g) increases with time and become constant.

(2) On increasing temperature, only rate of forward reaction increases.

(3) At equilibrium, the total pressure in the container is 4 atm.

(4) At equilibrium, the total pressure in the container is more than 4 atm, if temperature is

increased.

AB(s) A (g) + B(g) Kp = 4, H = + ve

,d ik=k esa A (g) o B (g) izR;sd 1 atm ds vkaf'kd nkc ij Hkjs tkrs gSaA vc AB (s) dks feyk;k (vkf/kD; ek=kk

esa) tkrk gSA fuEu esa ls dkSulk dFku xyr gSA (ik=k esa vU; dksbZ xSl mifLFkr ugha gS) :

(1) le; ds lkFk B(g) ds nkc esa o`f) gksdj ;g fu;r gks tkrk gSA

(2) rkieku esa of) djus ij dsoy vxz vfHkfØ;k dh nj esa o`f) gksrh gSA

(3) lkE; ij] ik=k esa dqy nkc 4 atm gSA

(4) lkE; ij] ik=k esa dqy nkc 4 atm ls vf/kd gS, ;fn rkieku esa o`f) gksrh gksA

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36. For the reaction N2(g) + O2(g) 2NO(g), the equilibrium constant is K1. The equilibrium

constant is K2 for the reaction 2NO(g) + O2(g) 2NO2(g). What is K for the reaction

NO2(g) ½N2(g) + O2(g) ?

vfHkfØ;k] N2(g) + O2(g) 2NO(g) ds fy, lkE; fLFkjkad K1 gS rFkk vfHkfØ;k]

2NO(g) + O2(g) 2NO2(g) ds fy, K2 gSA vfHkfØ;k NO2(g) ½N2(g) + O2(g) ds fy, lkE;

fLFkjkad] K dk eku D;k gksxk \

(1) 1 / (2K1K2) (2) 1 / (4K1K2) (3) 1 / [K1K2]½ (4) 1 / (K1K2)

37. Densities of two gases are in the ratio 1 : 2 and their temperatures are in the ratio 2 : 1, then the

ratio of their respective pressure is :

nks xSlksa ds ?kuRo dk vuqikr 1 : 2 gS o muds rkieku dk vuqikr 2 :1 gS rks muds laxr nkc dk vuqikr fuEu

gS %

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

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38. An ideal gas cannot be liquefied because :

(1) its critical temperature is very high.

(2) its vander waal's constants 'a' and 'b' are very high

(3) the intermolecular forces among gaseous molecules are zero.

(4) None of these

,d vkn'kZ xSl dks nzfor ugha fd;k tk ldrk gS D;ksafd %

(1) bldk ØkfUrd rki cgqr T;knk gksrk gSA

(2) blds okUMjokYl fu;rkad 'a' rFkk 'b' ds eku cgqr vf/kd gksrs gSA

(3) xSlh; v.kqvksa ds e/; vUrjkf.od cy 'kwU; gksrk gSA

(4) buesa ls dksbZ ugha

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39. The IUPAC name of the compound shown below is

Br

Cl|

(1) 2 - Bromo-6-chlorocyclohex-1-ene (2) 6-Bromo-2-chlorocyclohexene

(3) 3-Bromo-1-chlorocyclohex-1-ene (4) 1-Bromo-3-chlorocyclohexene

fuEu ;kSfxd dk IUPAC uke gSA

Br

Cl|

(1) 2-czkseks-6-DyksjkslkbDyksgSDl-1-bZu (2) 6-czkseks-2-DyksjkslkbDyksgSDlhu

(3) 3-czkseks-1-DyksjkslkbDyksgSDl-1-bZu (4) 1-czkseks-3-DyksjkslkbDyksgSDlhu

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40. Which of the following order is incorrect for given property ?

fuEu esa ls dkSulk Øe nh xbZ lkeF;Zrk ds vuqlkj lgh ugha gS \

(1) ClCH2CH2CH2COOH < CH3–CH–CH2–COOH

Cl

< CH3–CH2–CH–COOH

Cl

(Acidic strength) ¼vEyh; lkeF;Zrk½

(2)

NH2

<N

<

NH2

(Basic strength) ¼{kkjh; lkeF;Zrk½

(3)

COOHOH

<

COOH

CH3

<

COOH

(Acidic strength) ¼vEyh; lkeF;Zrk½

(4) CH3–C–C–CH3

O O <

OO

(% of enol) ¼bZuksy dk %½

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JAMTMAIN270115C0-26

41.

CHO

OWhat is not true about the given compound

(1) It is the isomer of 5356 HOCCHC||O

(2) It gives precipitate with ammonical silver nitrate(3) It gives iodoform test(4) It librates hydrogen gas with Na metal

CHO

OmijksDr ;kSfxd ds fo"k; esa fuEu esa ls dkSulk dFku lgh ugha gS \

(1) ;g 5356 HOCCHC||O

dk leko;oh gSA(2) ;g veksfu;ke; flYoj ukbVªsV ds lkFk vo{ksi nsrk gSA(3) ;g vk;ksMksQkWeZ ijh{k.k nsrk gSA(4) ;g Na /kkrq ds lkFk gkbMªkstu xSl eqDr djrk gSA

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42. Compound (X) with molecular formula C5H10 on catalytic hydrogenation gives compound Y, which

undergoes photochemical monochlorination to give 4 structural products. Compound (X), on

ozonolysis gives an equimolar mixture of acetone & acetaldehyde. Identify compound (X).(1) 2–methylbut–2–ene (2) 3–methyl–1–butene

(3) 1–Pentene (4) 2–methylbut–1–ene

v.kqlw=k C5H10 okyk ;kSfxd (X), mRizsjdh; gkbMªkst+uhdj.k ij ;kSfxd Y nsrk gS tks izdk'kjklk;fud

eksuksDyksjhuhdj.k ds i'pkr~ 4 lajpukRed mRikn nsrk gSA ;kSfxd (X) ds vkstksuhvi?kVu ls ,slhVksu rFkk

,slhVSfYMgkbM dk leeksyj feJ.k izkIr gksrk gS] rks ;kSfxd (X) dks igpkuksa \

(1) 2–esfFkyC;wV–2–bZu (2) 3–esfFky–1–C;wVhu

(3) 1–isUVhu (4) 2–esfFkyC;wV–1–bZu

43. Which of the following statements is incorrect ?

(1) Generally the radius trend and the ionization energy trend across a period are opposites.

(2) Electron gain enthalpy values of elements may be exothermic (negative) or endothermic

(positive).

(3) Amongst Li–, Be– , B– and C– , Li– is least stable ion .

(4) Te2– > I– > Cs+ > Ba2+ represents the correct decreasing order of ionic radii.

fuEu esa ls dkSulk dFku xyr gS \

(1) vkorZ esa ck;sa ls nk;sa tkus ij lkekU;r% f=kT;k esa ifjorZu rFkk vk;uu ÅtkZ esa ifjorZu foijhr gksrs gSaA

(2) rRoksa dh bysDVªkWu xzg.k ,UFkSYih dk eku Å"ek{ksih ¼_.kkRed½ ;k Å"ek'kks"kh ¼/kukRed½ gks ldrk gSA

(3) Li–, Be– , B– rFkk C– esa Li– lcls de LFkk;h vk;u gSA

(4) Te2– > I– > Cs+ > Ba2+ vk;fud f=kT;k ds lgh Øe dks iznf'kZr djrk gSA

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JAMTMAIN270115C0-28

44. Identify correct statements about NO molecule :

(i) When NO is ionized to NO+, the electron is removed from the *2p orbital

(ii) Bond order of NO is 2.5 and bond order of NO+ is 3.0.

(iii) Bond length of NO+ is greater than that of NO.

(iv) It is similar to N2– in all respect.

(1) (i) and (ii) (2) (i) and (iii)

(3) (ii) and (iv) (4) All are correct

NO v.kq ds lanHkZ esa lgh dFku dh igpku dhft;s :

(i) tc NO] NO+ esa vk;fur gksrk gS] rc bySDVªkWu *2p d{kd ls eqDr gksrk gSA

(ii) NO dk ca/k Øe 2.5 rFkk NO+ dk ca/k Øe 3.0 gSSA

(iii) NO+ dh ca/k yEckbZ NO dh rqyuk esa vf/kd gksrh gSA

(iv) ;g lHkh xq.kksa ds lanHkZ esa N2– ds leku gSA

(1) (i) rFkk (ii) (2) (i) rFkk (iii)

(3) (ii) rFkk (iv) (4) lHkh lgh gSA

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45. Which of the following is not correctly matched according to stability ?

fuEu esa ls dkSu LFkkf;Ro ds vuqlkj lgh lqesfyr ugh gS \

(1) (III > II > I)

(2) (II > I > III)

(3) (II > I > III)

(4)

I II III

(III > I > II)

46. The principal products obtained on heating iodine with cold and concentrated caustic sodasolution:vk;ksMhu dks B.Ms ,oa lkUnz dkfLVd lksMk foy;u ds lkFk xeZ djus ij eq[; mRikn fuEu izkIr gksxk &(1) NaIO + NaI (2) NaIO + NaIO3 (3) NaIO3 + NaI (4) NaIO4 + NaI

47. When orthoboric acid is heated to red heat the residue is :(1) Boron (2) Boron oxide (3) Pyroboric acid (4) Metaboric acidtc vkFkksZcksfjd vEy dks yky m"ek esa xeZ fd;k tkrk gSA rks izkIr vo'ks"k gS &(1) cksjksu (2) cksjksu vkWDlkbM (3) ikbjkscksfjd vEy (4) esVkcksfjd vEy

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48. Statement-1 : When a gas at high pressure expands against vacuum, the magnitude of work

done is maximum.

Statement-2 : Work done in expansion depends upon the pressure inside the gas and increase in

volume in reversible process.

(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 True; Statement-2 is NOT a correct explanation for

Statement-1.

(3) Statement-1 is True, Statement-2 is False.

(4) Statement-1 is False, Statement-2 is True.

oDrO;-1 % tc ,d xSl dk mPp nkc ij] fuokZr ds fo:) izlkj fd;k tkrk gS rks fd;s x, dk;Z dk ifjek.k

vf/kdre gksrk gSA

oDrO;-2 % mRØe.kh; izØe esa izlkj esa fd;k x;k dk;Z vk;ru esa o`f) rFkk xSl ds vkUrfjd nkc ij fuHkZj

djrk gSA

(1) oDÙkO;&1 lR; gS] oDÙkO;&2 lR; gS ; oDÙkO;&2, oDÙkO;&1 dk lgh Li"Vhdj.k gSA

(2) oDÙkO;&1 lR; gS] oDÙkO;&2 lR; gS ; oDÙkO;&2, oDÙkO;&1 dk lgh Li"Vhdj.k ugha gSA

(3) oDÙkO;&1 lR; gS] oDÙkO;&2 vlR; gSaA

(4) oDÙkO;&1 vlR; gS] oDÙkO;&2 lR; gSaA

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49. Statement-1 : During the reaction

A(g) + 2 B(g) C()

volume contraction takes place.

Statement-2 : Volume contraction always takes place when one of the product is in liquid state



Statement-1



oDrO;–1 : A(g) + 2 B(g) C()

vfHkfØ;k ds nkSjku vk;ru dk ladqpu gksrk gSA

oDrO;–2 : vk;ru ladqpu ges'kk rc gksrk gS tc ,d mRikn nzo voLFkk esa gksrk gSA

(1) oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k gSA


(3) oDrO;-1 lR; gS, oDrO;-2 vlR; gSA

(4) oDrO;-1 vlR; gS, oDrO;-2 lR; gSA

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50. Statement-1 : InO

CH –3 CO–

all C–O bond lengths are equal.

Statement-2 : It has two resonating structures of equal energies.



Statement-1



oDrO;–1 :O

CH –3 CO– esa lHkh C–O cU/k yEckbZ;kW leku gksrh gSA

oDrO;–2 : ;g leku Åtkvksa dh nks vuquknh lajpuk;sa j[krk gSA





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51. Statement-1 : Benzoate ion is less stable than phenoxide ion.

Statement-2 : In benzoate ion negative charge is poor or weakly delocalised while in phenoxide

ion negative charge is better or more delocalised.



Statement-1



oDrO;–1 : csUtks,V vk;u] fQukWDlkbM vk;u dh rqyuk esa de LFkk;h gksrk gSA

oDrO;–2 : csUtks,V vk;u esa _.kk;u dk foLFkkuhdj.k nqcZy ,oa de gksrk gS tcfd fQukWDlkbM vk;u eas

_.kk;u dk foLFkkuhdj.k izcy ,oa vPNk gksrk gSA





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52. How many of the following solution can be classified as buffer systems.

fUkEu es ls fdrus foy;uksa dks cQj ra=k ds :i esa oxhZd`r fd;k tk ldrk gSa %

(a) KH2PO4 / H3 PO4 (b) NaClO4 / HCIO4 (c) C5H5N / C5H5NHCl

(d) KF / HF (e) KBr / HBr (f) Na2CO3 / NaHCO3

(g) NH3 / NH4 NO3 (h) Na2SO3/ NaHSO3 (j) KCl / HCl

(i) NaI / H

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53. Find the solubility of As2S3 in a 10–2M Na2S solution assuming no hydrolysis or complexation of

cationic or anionic part.

Given : Ksp for As2S3 = 1625

× 10–24. Report your answer as Y where : solubility (in mol/L)

= Y × 10–11.

,d 10–2M Na2S foy;u esa As2S3 dh foys;rk Kkr dhft;s \ ¼ekuk fd] /kuk;fud vFkok _.kk;fud Hkkx

dk fdlh Hkh izdkj dk ty vi?kVu vFkok ladqyu ugha gksrk gS½ ¼fn;k gS : As2S3 ds fy,

Ksp = 1625

× 10–24) viuk mÙkj Y ds :i esa nhft;s tgk¡ foys;rk (mol/L esa) = Y × 10–11.

54. CH3COOH (60 ml, 0.1M) is titrated against 0.1M NaOH solution. Calculate the pH at the addition

of 10 ml of NaOH. Ka of CH3COOH is 2 × 10–5. [log 2 = 0.3]

CH3COOH (60 ml, 0.1M) foy;u dks 0.1 M NaOH foy;u ds lkFk vuqekfir fd;k x;kA NaOH ds 10 ml

feykus ij pH dh x.kuk dhft;sA CH3COOH dk Ka = 2 × 10–5 gSA [log 2 = 0.3]

55. Number of sp2 hybrid boron atoms in the anion of borax, Na2B4O7.10H2O ; is ______

cksjsDl (Na2B4O7.10H2O) ds _.kk;u esa sp2 ladfjr cksjksu ijek.kqvksa dh la[;k gS %

56. Total number of position isomers of trimethyl cyclohexane are :

VªkbZesfFky lkbDyksgsDlsu ds lHkh fLFkfr leko;oh dh dqy la[;k D;k gksxh \

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

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PART-III MATHEMATICS





57. Set of all real values of p such that both the roots of the equation (p – 5)x2 – 2px + (p – 4) = 0 arepositive, one is less than 2 and other is lying between 2 and 3, isp ds lHkh okLrfod ekuksa dk leqPp; ftlds fy, lehdj.k (p – 5)x2 – 2px + (p – 4) = 0 ds nksuksa ewy/kukRed rFkk ,d ewy 2 ls NksVk vkSj nwljk ewy 2 vkSj 3 ds e/; gS] gS&

(1) 49 , 244

(2) (4, 24) (3) (–, 4) 49 ,4

(4) 494,4

58. If , , are the roots of the quadratic equation ax2 + bx + c = 0 then the roots of the quadraticequation (a + b + c) x2 – (b + 2c) x + c = 0 are;fn f}?kkr lehdj.k ax2 + bx + c = 0 ds ewy , gS] rc f}?kkr lehdj.k(a + b + c) x2 – (b + 2c) x + c = 0 ds ewy gS&

(1) 1

, 1

(2) 1

, 1

(3)1

,1

(4)– 1

,– 1

59. If a circle touches the y - axis and cuts off a fixed length on the x -axis, then the centre of the circlelies on a(1) Parabola. (2) Hyperbola. (3) Ellipse (4) Circle;fn ,d oÙk y-v{k dks Li'kZ djrk gS rFkk x-v{k ij ,d fuf'pr vUr% [k.M dkVrk gS] rc oÙk dk dsUnz fLFkr gksxk-(1) ijoy; ij (2) vfrijoy; ij (3) nh?kZo`r ij (4) o`Ùk ij


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60. If statements p and q take truth values as TT, TF, FT, FF in order, then the respective truth values ofstatement (pq)(~ p ~ q) are;fn dFku p rFkk q ds lR; eku Øe'k% TT, TF, FT, FF gks rks dFku (p q) (~ p ~ q) ds lR; ekuØe'k% gksxsa&(1) T,F,F,T (2) T,F,F,F (3) F,F,F,F (4) T,T,T,T

61. Let x1, x2, x3 , y be the elements of the set A = {1, 2, 3, 5, 6, 10, 15, 30}, then the number ofpositive integral solution of x1x2x3 = y is

ekuk x1, x2, x3 , y leqPp; A = {1, 2, 3, 5, 6, 10, 15, 30} ds vo;o gS] rc x1 x2 x3 = y ds /kukRed iw.kkZadgyks dh la[;k gS–(1) 64 (2) 27 (3) 81 (4) 80

62. A (–2, 4), B(–1, 2), C(1, 2), D(2, 4) are the vertices of a quadrilateral ABCD. E is a point on theside AD such that area (quad. BCDE) = area (ABE), then the abscissa of the point E isA (–2, 4), B(–1, 2), C(1, 2), D(2, 4) ,d prqHkqZt ABCD ds 'kh"kZ gSA Hkqtk AD ij fcUnq E bl izdkj gS fd{ks=kQy (prqHkqZt BCDE) = {ks=kQy (ABE) gS] rks fcUnq E dk Hkqt gS &

(1) 0 (2) 12

(3) 1 (4) 2

63. Equation of line which is parallel to the line common to the pair of lines given by 6x2 – xy – 12y2 = 0and 15x2 + 14xy – 8y2 = 0 and 5 unit away from origin and having positive x and y intercept, is6x2 – xy – 12y2 = 0 vkSj 15x2 + 14xy – 8y2 = 0 ds ;qXe dh mHk;fu"B js[kk ds lekUrj rFkk ewy fcUnq ls 5

bdkbZ nwjh ij fLFkr js[kk dk lehdj.k gksxk tks /kukRed x o y vUr%[k.M j[krh gS] gS&(1) 4x + 3y = 12 (2) 3x + 4y = 25 (3) 12x + 5y = 65 (4) 15x + 8y = 85


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64. If x satisfies the equation x2 – 2x cos +1 = 0, then the value of nn

1xx

for (n N – {1} is

;fn x lehdj.k x2 – 2x cos +1 = 0 dks lUrq"V djrk gS] rc nn

1xx

(n N – {1} ds fy,) dk eku

gksxk -

(1) 2n cosn (2) 2n cosn (3) 2 cosn (4) 2 cosn

65. A man throws a fair coin number of times and gets 2 points for each head and 1 point for each tailthe probability that he gets exactly 6 points is,d O;fDr ,d fu"i{k flDds dks dbZ ckj Qasdrk gS rFkk fpÙk vkus ij 2 vad rFkk iV vkus ij 1 vad çkIrdjrk gS mlds }kjk Bhd 6 vad çkIr djus dh çkf;drk gS&

(1) 2132

(2) 1364

(3) 4364

(4) 2332

66. Iftan – i sin cos

2 2

1 2isin2

is purely imaginary, then the number of values of [0, 2] is

;fntan – i sin cos

2 2

1 2isin2

fo'kq) dkfYiud gS] rc [0, 2] ds ekuks dh la[;k gS -

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


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67. If and are angles in the first quadrant such that tan = 1/7 and sin = 1/ 10 then ( + 2)may be equal to;fn dks.k rFkk izFke prqFkkZa'k esa bl izdkj gS fd tan = 1/7 rFkk sin = 1/ 10 rc ( + 2) dk eku gksldrk gS -(1) 90° (2) 75° (3) 45° (4) 135°

68. If y = 2x is a chord of the circle x2 + y2 – 10x = 0, then the equation of a circle with this chord as

diameter is;fn y = 2x o`Ùk x2 + y2 – 10x = 0 dh ,d thok gS] rc bl thok dks O;kl ekudj [khaps x;s o`Ùk dk lehdj.kgS &(1) x2 + y2 – 2x – 4y = 0 (2) x2 + y2 + 2x – 4y = 0(3) x2 + y2 – 2x + 4y = 0 (4) x2 + y2 + 2x + 4y = 0

69. Twenty coupons are numbered 1, 2, 3......, 20 respectively. Four coupons are selected at randomwithout replacement. If maximum and minimum numbers on two selected coupons are 12 and 3respectively. Then the probability that the maximum number on the remaining selected coupons is9 is -20 dwiuks ij Øe'k% 1, 2, 3......, 20 vafdr fd;k x;k gSA pkj dwiuks dks fcuk izfrLFkkfir fd;s pquk tkrk gSA;fn pwus x;s nks dwiuks ij vf/kdre rFkk U;wure la[;k Øe'k% 12 rFkk 3 gSA rc 'ks"k pqus x;s dwiuksa ijvf/kdre la[;k 9 gksus dh izkf;drk gksxh -

(1) 556

(2) 528

(3) 1528

(4) 856


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70. The remainder when 3(13)51 is divided by 15, istc 3(13)51 dks 15 ls foHkkftr fd;k tkrk gS rks 'ks"kQy gS&(1) 3 (2) 2 (3) 1 (4) 6

71. Set A contains 4 elements & set B contains 5 element and C is a subset of A × B. If representsnumber of elements in set C then which of the following is FALSE ?(1) If = 3, then number of set C equals 20C3

(2) If = 10, then number of set C equals 20C10

(3) If 2, then number of set C equals 220–21(4) If = 18, then number of set C equals 220–21leqPp; A esa 4 vo;o gS rFkk leqPp; B esa 5 vo;o gS rFkk C; A × B dk mileqPp; gSA ;fn leqPp; C esavo;oks dh la[;k dks iznf'kZr djrk gS rc fuEu esa ls vlR; dFku gS ?

(1) ;fn = 3 rc leqPp; C dh la[;k 20C3 gS

(2) ;fn = 10 rc leqPp; C dh la[;k 20C10 gS

(3) ;fn 2 rc leqPp; C dh la[;k 220–21 gS

(4) ;fn = 18 rc leqPp; C dh la[;k 220–21 gS

72. Let a, b, c be the lengths of the sides of a ABC and A,B,C be angles opposite to sides a,b,crespectively such that b + c 1, c – b 1 and a 1. If b c c b c b c blog a log a 2log a.log a then ekuk a, b, c, ABC dh Hkqtkvksa dh yEckb;k¡ gSA rFkk A,B,C Øe'k% Hkqtkvksa a, b, c ds lEeq[k dks.k bl izdkjgS fd b + c 1, c – b 1 rFkk a 1 gSA ;fn b c c b c b c blog a log a 2log a.log a rc -(1) 2 2 2sin A sin B 2sin C (2) tanA tanB 1 (3) A + B = C (4) 2cos2A + cos2B = 1


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73. If the ellipse2x

4+

2y1

= 1 meets the ellipse2x

1 +

2

2ya

= 1 in four distinct points and

a = b2 – 10b + 25, then the number of integral values of b is

(1) 3 (2) 4 (3) 6 (4) infinite

;fn nh?kZo`Ùk2x

4+

2y1

= 1 nh?kZo`Ùk2x

1 +

2

2ya

= 1 dks pkj fHkUu fHkUu fcUnqvksa ij feyrk gS rFkk

a = b2 – 10b + 25 rc b ds iw.kkZad ekuks dh la[;k gksxh -(1) 3 (2) 4 (3) 6 (4) vuUr

74. Let R be a relation on the set of integers given by aRb a = 2k.b for some integer k. Then R is

(1) reflexive but not symmetric, not transitive (2) Symmetric and reflexive but not transitive

(3) reflexive and transitive but not symmetric (4) an equivalence relation

ekuk R iw.kkZadks ds leqPp; esa ,d lac/k gSA tks fd bl izdkj fn;k tkrk gS aRb a = 2k.b dqN iw.kkZd k ds

fy, rc R gS -

(1) LorqY; fdUrq lefer ugha] LorqY; ugh (2) lefer rFkk LorqY; fdUrq laØked ugh

(3) LorqY; o laØked fdUrq lefer ugh (4) ,d rqY;rk lac/k

75. How many polynomial P(x) (with leading coefficient = 1) are there such that P(x) Q(x) = x4 – 1 forsome other polynomial Q(x), where the coefficients of P(x) and Q(x) can be complex numbers.P(x) ,d cgqin gS] ftldk vxzt xq.kkad (leading coefficient) 1 gS rFkk P(x) Q(x) = x4 – 1 gS] fdlh vU;cgqin Q(x) ds fy,, tgk¡ P(x) rFkk Q(x) ds xq.kkad lfEeJ la[;k gks ldrs gSA lHkh lEHkkfor cgqinksa P(x)

dh la[;k Kkr dhft,A(1) 4 (2) 16 (3) 15 (4) 64


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76. Statement-1 : The length of the latus-rectum of the ellipse whose parametric equation isx = t2 + t + 1 and y = t2 – t + 1, t R is 2.

Statement-2: Length of latus-rectum of the ellipse2

2xa

+2

2yb

= 1, a < b is22a

b(1) Statement -1 is True, Statement -2 is True ; Statement -2 is a correct explanation forStatement -1(2) Statement-1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation forStatement-1(3) Statement -1 is True, Statement -2 is False(4) Statement -1 is False, Statement -2 is TrueoDrO;-1 : ml nh?kZo`Ùk ds ukfHkyEc dh yEckbZ ftldk izkpfyd lehdj.k x = t2 + t + 1 vkSj y = t2 – t + 1,t R gS] 2 gSA

oDrO;-2 : nh?kZo`Ùk2

2xa

+2

2yb

= 1 ds ukfHkyEc dh yEckbZ a < b gksus ij22a

b gksrh gSA

(1) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA(2) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k ugha gSA(3) oDrO;-1 lR; gS] oDrO;-2 vlR; gSA(4) oDrO;-1 vlR; gS] oDrO;-2 lR; gSA


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77. Let a, b, c are non zero real numbers such that ax2 + bx + 7c = 0 and 2x2 + 2x + 7 = 0 have a

common root.

Statement -1 : a = b = 2c.

Statement -2 : Two quadractic equations connot have only one imaginary root common.

(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 True ; Statement-2 is NOT a correct explanation for

Statement-1

(3) Statement -1 is True, Statement -2 is False

(4) Statement -1 is False, Statement -2 is True

ekuk a, b, c v'kwU; okLrfod la[;k,a bl izdkj gS fd ax2 + bx + 7c = 0 rFkk 2x2 + 2x + 7 = 0 dk ,d ewy

mHk;fu"B gSA

oDrO;-1 : a = b = 2c.

oDrO;-2 : nks f}?kkr lehdj.kksa dk ,d mHk;fu"B dkYifud ewy ugha gks ldrk gSA

(1) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA

(2) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k ugha gSA

(3) oDrO;-1 lR; gS] oDrO;-2 vlR; gSA

(4) oDrO;-1 vlR; gS] oDrO;-2 lR; gSA


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78. Some particulars of the distribution of weights of boys and girls are given belowBoys Girls

Number 100 50Mean Weight 60 kg 45 kgvariance 25 9Statement -1 : Distribution of the boys is more variable than girls.Statement -2 : distribution with more coefficient of variance is more variable.(1) Statement -1 is True, Statement -2 is True ; Statement -2 is a correct explanation forStatement -1(2) Statement-1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation forStatement-1(3) Statement -1 is True, Statement -2 is False(4) Statement -1 is False, Statement -2 is True

dqN yM+ds rFkk yM+fd;ksa ds Hkkjksa dk forj.k fuEu çdkj gS&yM+ds yM+fd;k¡

la[;k 100 50ek/; Hkkj 60 kg 45 kgçlj.k 25 9oDrO;-1 : yM+dksa dk forj.k yM+fd;ksa ds forj.k dh rqyuk esa vf/kd ifjofrZr gksrk gSAoDrO;-2 : vf/kd çlj.k xq.kkad okys forj.k vf/kd ifjofrZr gksrs gSA(1) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA(2) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k ugha gSA(3) oDrO;-1 lR; gS] oDrO;-2 vlR; gSA(4) oDrO;-1 vlR; gS] oDrO;-2 lR; gSA


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79. Statement-1 : If A = {a, b, c} and B = {1, 2} then number of relations from A to B are 26 and

number of functions from A to B are 23

Statement-2 : Number of functions number of relations from A to B.



Statement-1



oDrO;–1 : ;fn A = {a, b, c} rFkk B = {1, 2} rc A ls B esa ifjHkkf"kr lEcU/kksa dh la[;k 26 gS rFkk Qyuksa

dh la[;k 23 gSA

oDrO;–2 : A ls B esa Qyuksa dh la[;k A ls B esa lEcU/kksa dh la[;kA

(1) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA

(2) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k ugha gSA

(3) oDrO;-1 lR; gS] oDrO;-2 vlR; gSA

(4) oDrO;-1 vlR; gS] oDrO;-2 lR; gSA


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80. Find the term independent of x in the expansion of8

1/ 3 –1/ 51 x x2

.

O;tad8

1/ 3 –1/ 51 x x2

esa x ls Lora=k in Kkr dhft,A

81. Number of integral values of x satisfying the inequation sinx {x} in [–, 2], is (where {.} denotes the fractional part function)vlfedk sinx {x} , x [–, 2] dks larq"V djus okys x ds iw.kk±d ekuksa dh la[;k gS& (tgk¡ {.} fHkUukRed Hkkx Qyu dks O;Dr djrk gS)

82. Let C1, C2, C3 ....... be circles such that the radius of Cn equals to the length of diameter Cn+1, where

n 1. If the radius of C1 is10

cm then the least value of n for which area of Cn is less than 1 sq.

cm. is -ekuk C1, C2, C3 ....... o`Ùk bl izdkj gS fd Cn dh f=kT;k Cn+1 ds O;kl ds cjkcj gS tgk¡ n 1 ;fn C1 dh

f=kT;k 10

cm gS] rc n dk og U;wure eku ftlds fy, Cn dk {ks=kQy 1 oxZ lseh ls de gksA


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83. Find the number of integers between 1 and 200, each of which has 10 divisors.

1 rFkk 200 ds e/; mu iw.kk±dksa dh la[;k Kkr dhft, ftuds Hkktdksa dh la[;k 10 gksA

84. Let ABC be right angle at vertex C and side a < b and a, b, c, N. If perimeter of the triangle is

equal to the area of the triangle, then find the number of ordered pair (a, b)

ekuk ABC 'kh"kZ C ij ledks.k cukrk gS a < b rFkk a, b, c, N. ;fn f=kHkqt dk {ks=kQy mlds ifjeki ds

cjkcj gks rc Øfer ;qXe (a, b) dh la[;k Kkr dhft,A

Code0MAJOR TEST (MT)

JEE MAIN PATTERNDate: 27-01-2015 COURSE : VIKAAS (JA) &

VIPUL (JB)

12. Neither try to erase / rub / scratch the option normake the Cross (X) mark on the option once filled. Donot scribble, smudge, cut, tear, or wrinkle the ORS.Do not put any stray marks or whitener anywhere onthe ORS.

12. fodYi dks u feVk,a@u Ldzsp djsa vkSj u gh xyr (X) fpUgdks HkjsaA ORS dks dkVs u gh QkMs u gh xUnk ugha djsa rFkkdksbZ Hkh fu'kku ;k lQsnh ORS ij ugha yxk;sA

13. If there is any discrepancy between the written dataand the bubbled data in your ORS, the bubbled datawill be taken as final.

13. ;fn ORS esa fdlh izdkj dh fy[ks x, vkadMksa rFkk xksysfd, vkadMksa esa fojks/kkHkkl gS] rks xksys fd, vkadMksa dks ghlgh ekuk tkosxkA

C. Question Paper FormatThe question paper consists of Three parts (PhysicsChemistry and Mathematics). Each part consists ofthree sections.

14. Section 1 contains 19 multiple choice questions.Each question has Four choices (A), (B), (C) and (D)out of which only ONE is correct.

C. iz'u&i=k dk izk:ibl iz'u&i=k rhu Hkkx ¼HkkSfrd foKku ] jl;ku foKku vkSjxf.kr½ gSA gj Hkkx ds rhu [k.M gSaA

14. [kaM 1 esa 19 cgqfodYi iz'u gSaA gj iz'u esa pkj fodYi(A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,d lgh gSA

15. Section 2 contains 4 multiple choice questions.Each question has Four choices (A), (B), (C) and (D)out of which only ONE is correct.

15. [kaM 2 esa 4 cgqfodYi iz'u gSaA gj iz'u esa pkj fodYi (A),(B), (C) vkSj (D) gSa ftuesa ls dsoy ,d lgh gSA

16. Section 3 contains 5 questions. The answer to eachquestion is a single-digit integer, ranging from 0 to 9(both inclusive).

16. [kaM 3 esa 5 iz'u gSaA izR;sd iz'u dk mÙkj 0 ls 9 rd¼nksuksa 'kkfey½ ds chp ,d ,dy vadh; iw.kkZad gSA

D. Marking Scheme17. For each question in Section 1, you will be awarded

3 marks if you darken the bubble corresponding toonly the correct answer and zero mark if no bubblesare darkened. In all other cases, minus one (–1)mark will be awarded.

D. vadu ;kstuk17. [kaM 1 esa gj iz'u esa dsoy lgh mÙkj okys cqycqys(BUBBLES)

dks dkyk djus ij 3 vad vkSj dksbZ Hkh cqycwyk dkyk ughadjus ij (0) vad iznku fd, tk;saxsA vU; lHkh fLFkfr;ksa esa_.kkRed ,d(–1) vad iznku fd;k tk;sxkA

18. For each question in Section 2, you will be awarded4 marks if you darken all the bubble(s) correspondingto only the correct answer(s) and zero mark if nobubbles are darkened. In all other cases, minus one(–1) mark will be awarded.

18. [kaM 2 esa gj iz'u esa dsoy lgh mÙkjksa ¼mÙkj½ okys lHkhcqycqys(BUBBLES) dkyk djus ij 4 vad vkSj dksbZ Hkhcqycwyk dkyk ugha djus ij (0) vad iznku fd, tk;saxsA vU;lHkh fLFkfr;ksa esa _.kkRed ,d (–1) vad iznku fd;k tk;sxkA

19. For each question in Section 3, you will be awarded4 marks if you darken the bubble corresponding toonly the correct answer and zero mark if no bubblesare darkened. No negative marks will be awarded forincorrect answers in this section.

19. [kaM 3 esa gj iz'u esa dsoy lgh mÙkj okyscqycqys(BUBBLES) dks dkyk djus ij 4 vad vkSj dksbZ Hkhcqycwyk dkyk ugha djus ij (0) vad iznku fd, tk;saxsAbl [kaM ds iz'uksa esa xyr mÙkj nsus ij dksbZ _.kkRed vadugha fn;s tk;saxsA

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

I have read all the instructions and shallabide by themeSaus lHkh funsZ'kksa dk i<+ fy;k gS vkSj eSa mudkvo'; ikyu d:¡xk@d:¡xhA

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

ijh{kkFkhZ ds gLrk{kj

Roll Number (jksy uEcj) :

I have verified all the information filled bythe candidate.ijh{kkFkhZ }kjk Hkjh xbZ lkjh tkudkjh dks eSusatk¡p fy;k gSA

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

ijh{kd ds gLrk{kj

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