allen clas amme · 2015. 12. 21. · 8. bl iqflrdk dh eqgj rksm+us ds i'pkr d`i;k tk¡p ysa fd...

Ïi;k bu fun sZ'kks a dks /;ku ls i

Kota/01CT2140792/40

Target : JEE (Main + Advanced) 2015/18-01-2015/Paper-2

fo"k; [k.M i`"B la[;kSubject Section Page No.

Hkkx-1 HkkSfrd foKku I(i) dsoy ,d lgh fodYi izdkj 03 - 08Part-1 Physics Only One Option Correct Type

I(ii) vuqPNsn izdkj 09 - 12Paragraph Type

I(iii) lqesyu lwpha izdkj 13 - 16Matching List Type

Hkkx-2 jlk;u foKku I(i) dsoy ,d lgh fodYi izdkj 17 - 22Part-2 Chemistry Only One Option Correct Type



Hkkx-3 xf.kr I(i) dsoy ,d lgh fodYi izdkj 30 - 32Part-3 Mathematics Only One Option Correct Type



SOME USEFUL CONSTANTSAtomic No. H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16, Cl = 17,

Br = 35, Xe = 54, Ce = 58,

Atomic masse s : H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24,

Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127,

Xe = 131, Ba=137, Ce = 140,

· Boltzmann constant k = 1.38 × 10–23 JK–1

· Coulomb's law constant pe9

0

1 = 9×104

· Universal gravitational constant G = 6.67259 × 10–11 N–m2 kg–2· Speed of light in vacuum c = 3 × 108 ms–1· Stefan–Boltzmann constant s = 5.67 × 10–8 Wm–2–K–4· Wien's displacement law constant b = 2.89 × 10–3 m–K· Permeability of vacuum µ0 = 4p × 10–7 NA–2

· Permittivity of vacuum Î0 = 20

1

cm· Planck constant h = 6.63 × 10–34 J–s

Enthusiast Course/Score-I/18-01-2015/Paper-2

PHYS

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PART-1 : PHYSICS Hkkx-1 : HkkSfrd foKku

SECTION–I : (i) Only One option correct Type

[k.M-I : (i) dsoy ,d lgh fodYi izdkjThis section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and(D) out of which ONLY ONE is correct.

bl [k.M esa 10 cgq fodYi iz'u gSA izR;sd iz'u es a pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls

dsoy ,d lgh gSA

1. A particle is projected horizontally with velocity 10 m/s from an inclined plane. Incline plane startsmoving with acceleration 10 m/s2 vertically upward as shown. The time after which particle will land on

the plane is :- ( g = 10 m/s2)

,d d.k dks fdlh vkur ry ls 10 m/s osx ls {kSfrt :i ls iz{ksfir fd;k tkrk gSA vkur ry fp=kuqlkj Å/okZ/kj Åij

dh vksj 10 m/s2 ds Roj.k ls xfr djuk izkjEHk djrk gSA fdrus le; i'pkr~ d.k ry ij fxj tk,xk\ ( g = 10 m/s2)

30°

10 m/s

10 m/s2

(A) 3 sec (B) 13 sec (C)

12 3 sec (D) 2 sec

BEWARE OF NEGATIVE MARKINGHAVE CONTROL ¾® HAVE PATIENCE ¾® HAVE CONFIDENCE Þ 100% SUCCESS

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

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2. Figure shows a horizontal circular plate with vertical walls placed on a horizontal surface. Two objectsP and Q of mass m and 2m are kept touching the walls symmetrically about diameter AB. Friction is

absent between P and Q and horizontal surface of plate and coefficient of friction between P and Q and

wall is 1

3. The plate is given some acceleration 'g' as shown.

(A) The angle between friction forces on P and Q is (p – 2q) if q is less than 90°.

(B) The angle between friction forces on P and Q is (p – 2q) only for q ³ 30°

(C) Acceleration of P or Q is ( )g 3 1

2

- for q = 60° w.r.t. plate..

(D) None of these

D

B g(horizontally)

P

Q

A qq

Top view

fp= esa ,d {kSfrt o`Ùkkdkj IysV n'kkZ;h xbZ gS] ftlesa Å/okZ/kj nhokjsa {kSfrt lrg ij fLFkr gSA æO;eku m o 2m okys nks

fi.M Øe'k% P o Q dks O;kl AB ds lkis{k nhokjksa dks Li'kZ djrs gq, lefer :i ls j[kk x;k gSA ;gk¡ P o Q rFkk IysV

dh {kSfrt lrg ds e/; ?k"kZ.k vuqifLFkr gS tcfd P o Q rFkk nhokj ds e/; ?k"kZ.k xq.kkad 1

3 gSA bl IysV dks

fp=kuqlkj dqN Roj.k 'g' fn;k tkrk gSA

(A) P o Q ij yxus okys ?k"kZ.k cyksa ds e/; dks.k (p – 2q) gksxk ;fn q dk eku 90° ls de gksA

(B) dsoy q ³ 30° ds fy, gh P o Q ij yxus okys ?k"kZ.k cyksa ds e/; dks.k (p – 2q) gksxk

(C) IysV ds lkis{k q = 60° ds fy, P ;k Q dk Roj.k ( )g 3 1

2

- gksxkA

(D) buesa ls dksbZ ugha


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3. Block A in the figure is released from the rest when the extension in the spring is x0. The maximum

downward displacement of the block is :-

tc fp= esa iznf'kZr fLizax esa foLrkj x0 gS rks CykWd A dks fojkeoLFkk ls NksM+k tkrk gSA CykWd dk uhps dh vksj vf/kdre

foLFkkiu gksxk%&

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

\\\\\\

\\\\\\

\

MA

K

(A) 0Mg x2k

- (B) 0Mg x2k

+ (C) 02Mg x

k- (D) 0

2Mg xk

+

4. A boy throws a ball with speed u in a well of depth 14 m as shown. On bounce with bottom of the wellthe speed of the ball gets halved. What should be the minimum value of u such the ball may be able toreach his hand again? Assume that his hands are at 1 m height from top of the well while throwing andcatching.

,d yM+dk u pky ls ,d xsan dks 14 m xgjs dq,¡ esa fp=kuqlkj QSadrk gSA dq,¡ ds ry ls Vdjkus ij xsan dh pky vkèkh

gks tkrh gSA u dk U;wure eku D;k gksuk pkfg, rkfd xsan iqu% mlds gkFkksa rd igq¡p lds\ ekuk xsan dks QSadrs rFkk idM+rs

le; yM+ds ds gkFk dq,¡ ds 'kh"kZ ls 1m Å¡pkbZ ij gksrs gSaA

u1m

14m

(A) 10m/s (B) 30m/s (C) 20m/s (D) 40m/s

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5. An undeformed spring of spring constant k is connected to a bead of mass m as shown in figure. Beadcan move along long rigid rod without friction. A particle of mass m moving with velocity ‘v’ in thevertical plane containing spring and rod strikes bead at an angle 45° with horizontal and sticks to bead.Choose correct alternative representing maximum elongation of spring (all particles are in samehorizontal plane).

fLizax fu;rkad k okyh ,d vfo:fir fLizax dks fp=kuqlkj m æO;eku ds eksrh ds lkFk tksM+k tkrk gSA ;g eksrh yEch n`


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7. An ideal diatomic gas goes through a cycle consisting of two isochoric and two isobaric lines. Theaboslute temperature of the gas rises 5 times both in the isochoric heating and in the isobaric expansion.

Efficiency of such a cycle is

,d vkn'kZ f}&ijek.kqd xSl dks nks levk;rfud rFkk nks lenkch; js[kkvksa okys pØ ls xqtkjk tkrk gSA xSl dk ije rki

levk;rfud :i ls xje djus esa rFkk lenkch; :i ls izlkfjr gksus esa nksuksa esa ik¡p xquk c

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10. An unstable radio nuclei X can decay into two stable nuclei Y and Z. A sample containing only X istaken at t = 0, three graphs log

e (N

x) Vs t, N

y Vs t and N

zVs t are drawn as shown below, here N

x, N

y and

Nz represents number of nuclei of X, Y and Z respectively any instant t. Choose the incorrect statement

from the following :

(A) Decay constant for decay of X into Y is ab tan

e

q

(B) Decay constant for decay of X into Z is ac tan

e

q

(C) Number of nuclei of X at t = 0 is ea

(D) Half life of nuclei X is 1

.tan q

a

t q

logeNx

b

t

Ny

c

t

Nz

,d vLFkk;h jsfM;ksukfHkd X nks LFkk;h ukfHkdksa Y rFkk Z esa fo?kfVr gks ldrk gSA ,d izfrn'kZ ftlesa dsoy X fo|eku

gS] t = 0 ij fy;k tkrk gS rFkk blds fy, loge (N

x) o t, N

y o t rFkk N

z o t ds e/; rhu vkjs[k fp=kuqlkj cuk;s tkrs gSaA

;gk¡ Nx, N

y o N

z Øe'k% fdlh {k.k t ij X, Y o Z ds ukfHkdksa dh la[;kvksa dks n'kkZrs gSaA fuEu esa ls xyr dFku pqfu,A

(A) X ds Y esa fo?kVu ds fy, fo?kVu fu;rkad ab tan

e

q gksxkA

(B) X ds Z esa fo?kVu ds fy, fo?kVu fu;rkad ac tan

e

q gksxkA

(C) t = 0 ij X ds ukfHkdksa dh la[;k ea gksxhA

(D) ukfHkd X dh v/kZvk;q 1

tan q gksxhA



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(ii) Paragraph Type

(ii) vuqPNsn izdkjThis section contains 3 paragraphs each describing theory, experiment, data etc. Six questions relateto three paragraphs with two questions on each paragraph. Each question of a paragraph has onlyone correct answer among the four choices (A), (B), (C) and (D).bl [k.M esa fl¼kUrksa] iz;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okys 3 vuqPNsn gSA rhuksa vuqPNsn ls lacaf/kr N% iz'ugSa] ftuesa ls gj vuqPNsn ij nks iz'u gSaA vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesals dsoy ,d lgh gSA

Paragraph for Questions 11 and 12 iz'u 11 ,oa 12 ds fy;s vuqPNsn

There is an interesting way to think of the centripetal acceleration of a particle in circular motion thatstems from our study of collisions and the impulse-momentum theorem. Imaging a particle of mass mmoving at speed v on a frictionless surface inside a fixed horizontal circular hoop of radius r, as shownin figure.

VDdjksa rFkk vkosx&laosx izes; ds v/;;u }kjk ge orqZy xfr esa d.k ds vfHkdsUnzh; Roj.k ds ckjs esa ,d u;s rjhds ls

lksp ldrs gSaA f=T;k r okyh fLFkj {kSfrt oÙ̀kkdkj fNnz ; qDr pdrh ds vUnj ?k"kZ.kjfgr lrg ij v pky ls fp=kuqlkj

xfr'khy ,d m nzO;eku ds d.k ij fopkj dhft;sA

xA

y

ji

r q

initialpoint

Assume the collision of the particle with the circular hoop are elastic. Since the kinetic energy is conservedin elastic collisions and the hoop is fixed, the particle maintains a constant speed before and after eachsuch collision. The magnitude of the momentum of the particle is therefore constant but, clearly, itsdirection is changed with each collision with the hoop. Thus the momentum of the particle is not conservedin its collision with the hoop. Viewed from the center of the circle, successive collisions are separated byan angle q, as indicated in figure.d.k rFkk o`Ùkh; pdrh ds e/; VDdj izR;kLFk ekfu;sA pwafd izR;kLFk VDdj esa xfrt ÅtkZ lajf{kr jgrh gS rFkk pdrhfLFkj gSA vr% bl izdkj dh izR;sd VDdj ls igys rFkk ckn esa d.k dh pky fu;r cuh jgrh gSA d.k ds laosx dk ifjek.kfu;r jgrk gS ijUrq pdrh ls izR;sd VDdj ds dkj.k bldh fn'kk ifjofrZr gks tkrh gSA bl izdkj pdrh ls Vdjkus esa d.kdk laosx lajf{kr ugha jgrk gSA o`Ùk ds dsUnz ls ns[kus ij Øekxr VDdjsa fp=kuqlkj q dks.k ij gksrh gSA


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11. What is average force on the ring between two successive collisions?

nks Øekxr VDdjksa ds e/; oy; ij vkSlr cy D;k gksxk\

(A)

2

sin2

m v

r

qæ öç ÷è ø (B)

2mvr

(C)

2

2 sin2

m v

r

qæ öç ÷è ø (D)

2

4 sin2

m v

r

qæ öç ÷è ø

12. Mark the CORRECT statement.

(A) Angular momentum of particle about initial point is conserved.

(B) Angular momentum about centre of circle is conserved.

(C) Angular momentum just before and just after collision about centre is reversed in sign.

(D) Angular momentum about any point within the circular hoop is conserved.

lR; dFku pqfu;sA

(A) izkjfEHkd fcUnq ds lkis{k d.k dk dks.kh; laosx lajf{kr jgrk gSA

(B) o`Ùk ds dsUnz ds lkis{k dks.kh; laosx lajf{kr jgrk gSA

(C) dsUnz ds lkis{k VDdj ls Bhd igys rFkk ckn esa dks.kh; laosx dk fpUg O;qRØfer gks tkrk gSA

(D) o`Ùkh; pdrh ds vUnj fdlh fcUnq ds lkis{k dks.kh; laosx lajf{kr jgrk gSA



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When an object floats in a liquid, weight of object is balanced by buoyant force. Buoyant force is equal

to weight of displaced liquid by the object. Consider a situation where a ice cube is floating on water and

a small point source of mass "m" is kept on upper surface of ice cube. In the base of vessel a concave

mirror is fixed. Density of ice is equal to 0.7 g/cc. Density of water is 1 g/cc. Side length of cube =

10cm, mass of object m = 100g and mice

= 32

, mwater

= 43

tc ,d fi.M fdlh æo esa rSjrk gS rks fi.M dk Hkkj mRIykou cy }kjk larqfyr gksrk gSA mRIykou cy fi.M }kjkfoLFkkfir æo ds Hkkj ds cjkcj gksrk gSA ,d fLFkfr ij fopkj dhft, tgk¡ ,d cQZ dk ?ku ikuh ij rSj jgk gS rFkk æO;eku"m" okyk ,d NksVk fcUnq L=ksr cQZ ds ?ku dh Åijh lrg ij j[kk gSA ik= ds iSans esa ,d vory niZ.k yxk gqvk gSA cQZdk ?kuRo 0.7 g/cc ds cjkcj gSA ty dk ?kuRo 1 g/cc gSA ?ku dh Hkqtk dh yEckbZ 10 cm gS, fi.M dk æO;eku

m = 100g gS rFkk mice

= 32

, mwater

= 43

gSA

20cm

13. Find out radius of curvature of mirror such that final image is formed on object :-niZ.k dh oØrk f=T;k Kkr dhft, rkfd vfUre izfrfcEc fcEc ij cus %&

(A) 168

9cm (B)

1889

cm (C) 178

9 cm (D)

1888

cm

14. Find out the base area of ice from which light ray enter to the watercQZ dk vk/kkj {ks=Qy Kkr dhft, ftlls izdk'k fdj.k ikuh esa izos'k djrh gS %&(A) 100 cm2 (B) 400» p cm2 (C) 81p cm2 (D) None of these


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A cylinder fitted with a piston which can slide without friction contains one mole of an ideal gas. The

walls of the cylinder and piston are adiabatic. The cylinder contains a resistor of resistance R = 2 kWwhich is connected to a capacitor of capacity C = 75 mF. Initially potential difference across capacitor is

640V

3æ öç ÷è ø

and switch is open. When switch is closed for (2.5 ln 4) min, the gas expands isobarically and

its temperature increases by 72 K. Heat loss through the wires is negligible. (R = 8.3 J mole-1 k-1)

,d csyu esa ,d eksy vkn'kZ xSl Hkjh gq;h gSA bl csyu esa ,d fiLVu yxk gS tks fcuk ?k"kZ.k xfr dj ldrk gSA csyu dh

nhokjas rFkk fiLVu Å"eh; dqpkyd gSA bl csyu esa R = 2 kW dk ,d izfrjks/kd yxk gqvk gS tks C = 75 mF /kkfjrk

okys ,d la/kkfj= ls tqM+k gSA la/kkfj= ij izkjfEHkd foHkokUrj 640

V3

æ öç ÷è ø

gS ,oa fLop [kqyk gqvk gSA tc fLop dks

(2.5 ln 4) min ds fy;s can fd;k tkrk gS rks xSl lenkch; :i ls izlkfjr gksrh gS rFkk bldk rkieku 72 K c


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(iii) Matching List Type (iii) lqesyu lwpha izdkj

This Section contains 4 multiple choice questions. Each question has matching lists. The codes for thelists. have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.bl [k.M esa 4 cgqfodYi iz'u gSaA izR;sd iz'u esa lqesyu lwpha gSA lwfp;ksa ds fy, dksM ds fodYi (A), (B), (C) vkSj (D)gSa ftuesa ls dsoy ,d lgh gSA

17. Consider the figure below. A tank of water (height of water column b) is kept on a electronic weighingscale. A metal cube (side a and density r) is hung from a spring balance and the spring balance is slowlylowered into the tank till the cube reaches the bottom of the tank. The distance between the bottom of thetank and bottom of the cube is denoted by h with initial value h

0.

fp= esa ikuh ls Hkjs ,d Vsad (ty LrEHk dh ÅapkbZ b) dks bysDVªkWfud Hkkjekih e'khu ij j[kk x;k gSA Hkqtk a rFkk ?kuRo r okys,d /kkfRod ?ku dks fLizax rqyk ls yVdkdj fLizax rqyk dks /khjs&/khjs Vsad esa uhps dh vksj ykrs gSa tc rd fd ?ku] Vsad ds iSansrd ugha igqap tkrkA Vsad ds iSans rFkk ?ku ds iSans ds e/; nwjh dks h (izkjfEHkd eku h

0) ls O;Dr fd;k x;k gSA

electronic scale

b hwatertank

cube

springbalance

List-I/lwph&I List-II/lwph&II

(P) Reading of spring balance vs h (1)h h rFkk fLizax rqyk ds ikB~;kad ds e/; vkjs[k

(Q) Reading of electronic scale vs h (2)

hh rFkk bysDVªkWfud Ldsy ds e/; vkjs[k

(R) Force of buoyancy vs h (3)

hh rFkk mRIykou cy ds e/; vkjs[k

(S) Sum of the reading of spring balance (4)hand the reading of electronic scale vs h

fLizax rqyk o bysDVªkWfud Ldsyds ikB~;kadksa ds ;ksx o h ds e/; vkjs[k

Code :P Q R S

(A) 2 3 1 4(B) 4 1 3 2(C) 4 3 2 1(D) 1 2 3 4

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18. An equi-convex lens of refractive index µ2 and focal length f(in air) is kept in medium of refractive index µ

1.

viorZukad µ2 rFkk Qksdl nwjh f (ok;q esa) okys ,d leksÙky ysal dks µ

1 viorZukad okys ,d ek/;e esa j[kk tkrk gSA

y

x

µ1 µ1

µ2

y'

x'

List-I/lwph&I List-II/lwph&II

(P) If lens is cut in two equal parts (1) Lens will act as a glass slabby a plane yy'. (µ

1 = 1)

;fn ysal dks lery yy' }kjk nks ysal dk¡p dh ,d ifêdk dh Hkk¡fr O;ogkj djsxkA

cjkcj Hkkxksa esa dkVk tk;sA (µ1 = 1)

(Q) If lens is cut in two equal parts (2) Lens will be converging and focal length willby a plane xx' . (µ

1 = 1) change.

;fn ysal dks lery xx' }kjk nks ysal vfHklkjh gksxk rFkk bldh Qksdl nwjh ifjofrZr gks

cjkcj Hkkxksa esa dkVk tk;sA (µ1 = 1) tk;sxhA

(R) If µ1 = µ

2. (3) Lens will be converging and focal length will remain same

;fn µ1 = µ

2 gks ysal vfHklkjh gksxk rFkk bldh Qksdl nwjh ifjofrZr ugh gksxhA

(S) If µ1 > µ

2. (4) Lens will be diverging and focal length will change.

;fn µ1 > µ

2 gks ysal vilkjh gksxk rFkk bldh Qksdl nwjh ifjofrZr gks tk;sxhA

Codes :P Q R S

(A) 2 4 1 3(B) 2 3 1 4(C) 3 4 2 1(D) 1 2 3 4



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19. Two charged di-electric spheres A (solid) and B (hollow) of same radius R and masses 2m and mrespectively are kept on a horizontal rough surface. The two spheres A and B have charges 2Q and –Qrespectively and uniformly distributed. They are released from rest when their centres are separated byvery large distance. Assuming there is no sliding.

2m

2Q

ARSolid hollow

B

–Q

R

m

List-I List-II

(P) Kinetic energy of system just before A and B meet (1)2838kQ

2345R

(Q) Kinetic energy of A just before A and B meet (2)2kQ

R

(R) Kinetic energy of B just before A and B meet (3)225kQ

67R

(S) Rotational kinetic energy of system just before A and B meet (4)242kQ

67R

leku f=T;k R rFkk Øe'k% 2m o m nzO;eku ds nks vkosf'kr ijkoS|qr xksyksa A(Bksl) rFkk B ([kks[kys) dks {kSfrt

[kqjnjh lrg ij j[kk x;k gSA nksuksa xksyksa A rFkk B ds vkos'k Øe'k% 2Q o –Q gS rFkk ,dleku :i ls forfjr gSA

tc muds dsUnz cgqr vf/kd nwjh ij gS rks mUgsa fojkekoLFkk ls NksM+k tkrk gSA eku yhft, fd ;gk¡ fQlyu fo|eku ugha gSA

2m

2Q

ARSolid hollow

B

–Q

R

m

lwph&I lwph&II

(P) A rFkk B ds feyus ls Bhd igys fudk; dh xfrt ÅtkZ (1)2838kQ

2345R

(Q) A rFkk B ds feyus ls Bhd igys A dh xfrt ÅtkZ (2)2kQ

R

(R) A rFkk B ds feyus ls Bhd igys B dh xfrt ÅtkZ (3)225kQ

67R

(S) A rFkk B ds feyus ls Bhd igys fudk; dh ?kw.kZu xfrt ÅtkZ (4)242kQ

67R

Codes :P Q R S

(A) 2 3 4 1(B) 1 4 3 2(C) 3 2 1 4(D) 4 1 2 3

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20. A very small current carrying square loop (current I) of side 'L' is placed in y-z plane with centre atorigin of the coordinate system (shown in figure). In List–I the coordinate of the points are given & inList–II magnitude of strength of magnetic field is given. Then :-fp=kuqlkj L Hkqtk okyk ,d cgqr NksVk /kkjkokgh oxkZdkj ywi ftlesa /kkjk I izokfgr gks jgh gS] y-z ry esa bl izdkj j[kkgqvk gS fd bldk dsUnz funsZ'kkad fudk; ds ewy fcUnq ij gSA lwph–I esa fcUnqvksa ds funsZ'kkad fn;s x;s gSa rFkk lwph–II esapqEcdh; {ks= lkeF;Z dk ifjek.k fn;k x;k gSA

y

x

z

P (a, a, 0)3..

P (a, 0, 0)10

I

.P (0, a, 0)2

List–I/lwph&I List II/lwph&II

(P) At point O (0, 0, 0) (1)2

03

I L2 am

pfcUnq O (0, 0, 0) ij

(Q) At point P1 (a, 0, 0) (here a > > L) (2) 0

2 2 ILm

pfcUnq P

1 (a, 0, 0) ij (;gka a > > L)

(R) At point P2 (0, a, 0) (here (a > > L) (3)

20

3

5 I L16 a

mp

fcUnq P2 (0, a, 0) ij (;gka (a > > L)

(S) At point P3 (a, a, 0) (here a > > L) (4)

20

3

I L4 am

pfcUnq P

3 (a, a, 0) ij (;gka a > > L)

Codes :P Q R S

(A) 4 2 3 1(B) 1 1 2 3(C) 4 1 2 3(D) 2 1 4 3



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PART-2 : CHEMISTRY Hkkx-2 : jlk;u foKku

SECTION–I : (i) Only One option correct Type [k.M-I : (i) dsoy ,d lgh fodYi izdkj

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and(D) out of which ONLY ONE is correct.bl [k.M esa 10 cgq fodYi iz'u gSA izR;sd iz'u es a pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa lsdsoy ,d lgh gSA

1.

A(g) KMnO 4 / H+

Zn + dil. HSO2 4CdCl2

B(aq.) Colourless

D(g)

Eyellow

KIO + Starch 3 C (Coloured solution)NaOHColourless

solution

Gas ‘A’ is

(A) O3

(B) H2S (C) SO

2(D) NH

3

A(g) KMnO 4 / H+

Zn + HSOruq 2 4CdCl2

B(aq.) jaxghu

D(g)

Eihyk

KIO + 3 LVkpZ C ( )jaxhu foy;uNaOH

jaxghufoy;u

‘A’ xSl gS

(A) O3

(B) H2S (C) SO

2(D) NH

3


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2. Which of the following Werner’s complex has least electrical conductivity ?fuEu esa ls dkSuls cuZj ladqy dh oS|qr pkydrk U;wure gS ?

(A) Co3+ Cl

NH3NH3

Cl

NH3NH3 Cl

(B) Co3+

NH3NH3

H N3

ClH N3 NH3

Cl

Cl

(C) Co3+

ClNH3

Cl

ClH N3 NH3

(D) Co3+

NH3NH3

Cl

NH3Cl NH3

Cl

3. Which of the following reaction does not take place during the extraction of Ag from Ag2S by cyanide

process ?

(A) Ag2S + CN– ® [Ag(CN)

2]–+S2–

(B) Zn + 2[Ag(CN)2]– ® [Zn(CN)

4]2– + 2Ag¯

(C) S2– + O2 ® SO

42–

(D) None of these

lk;ukbM izØe }kjk Ag2S ls Ag ds fu"d"kZ.k ds nkSjku] fuEu esa ls dkSulh vfHkfØ;k ugha gksrh gS ?

(A) Ag2S + CN– ® [Ag(CN)

2]–+S2–

(B) Zn + 2[Ag(CN)2]– ® [Zn(CN)

4]2– + 2Ag¯

(C) S2– + O2 ® SO

42–

(D) buesa ls dksbZ ugha


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4. Select the correct order of H – M – H bond angle

H – M – H ca/k dks.k dk lgh Øe pqfu,sa

(A) PH3 > PH

4+ (B) P

2H

4 > PH

4+ (C) PH

3 > NH

4+ (D) PH

4+ > NH

3

5. Ethylmethanoate ¾¾¾¾¾¾¾¾¾® ++ E555Fuctorganic prod

W X(i)MeMgBr(2eq)

(ii)H O3

2CaOCl3W CHCl Y¾¾¾® +

2CaOCl3X CHCl Z¾¾¾® +

DryDistillationY Z+ ¾¾¾¾® Organic product(s) + CaCO3

Which of the following organic product can be formed on dry distillation reaction

(I) CH3 – COOH (II) CH

3– CO – CH

3(III) CH

3 – CH = CH – CH

3(IV) HCHO

(A) I and II only (B) II and III only (C) I, II and III only (D) II and IV only

,sfFkyesFksuks,V +

3

(i) MeMgBr (2eq)(ii) H O

¾¾¾¾¾¾¾¾® +E5555FW X

dkcZfud mRikn

2CaOCl3W CHCl Y¾¾¾® +

2CaOCl3X CHCl Z¾¾¾® +

¾¾¾¾®Y + Z'kq"d

vklou dkcZfud mRikn + CaCO3

fuEu esa ls dkSuls dkcZfud mRikn 'kq"d vklou vfHkfØ;k djkus ij cuk;s tk ldrs gaS&

(I) CH3 – COOH (II) CH

3– CO – CH

3(III) CH

3 – CH = CH – CH

3(IV) HCHO

(A) dsoy I rFkk II (B) dsoy II rFkk III (C) dsoy I, II rFkk III (D) dsoy II rFkk IV

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6. Correct IUPAC name of the following compound is :

OH

BrCN

(A) 4-Bromo-3-cyanophenol (B) 2-Bromo-5-hydroxybenzonitrile.

(C) 2-Cyano-4- hydroxy bromobenzene. (D) 6- Bromo - hydroxy benzonitrile.

fuEu ;kSfxd dk lgh IUPAC uke gS&

OH

BrCN

(A) 4-czkseks-3-lk;uksfQukWy (B) 2-czkseks-5-gkbMªkWDlhcsUtksukbVªkby

(C) 2-lk;uks-4- gkbMªkWDlh czksekscsUthu (D) 6- czkseks&gkbMªkWDlh csUtksukbVªkby


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7. Consider the following Aldoses/ketoses. Which of them can give same osazone on reaction withPhNHNH

2

(I) CHO

H–C OH–HO–C–H

H–C OH–

H–C OH–CH OH2

(II) CHO

HO–C H–HO–C H–

H–C OH–H–C OH–

CH OH2

(III) CHO

HO–C H–

H–C H–O

H–C OH–

CH OH2

(IV) CH OH2

HO–C H–H–C H–O

H–C OH–

CH OH2

C= 0

(A) I, II only (B) I, II and III only (C) I, II and IV only (D) I, II, III and IV

fuEu , sYMkst@dhVkst ij fopkj dhft;sA buesa ls dkSu PhNHNH2 ds lkFk vfHkfØ;k djkus ij leku vkslktksu ns

ldrs gSa&

(I) CHO

H–C OH–HO–C–H

H–C OH–

H–C OH–CH OH2

(II) CHO

HO–C H–HO–C H–

H–C OH–

H–C OH–

CH OH2

(III) CHO

HO–C H–

H–C H–O

H–C OH–

CH OH2

(IV) CH OH2

HO–C H–H–C H–O

H–C OH–

CH OH2

C= 0

(A) dsoy I, II (B) dsoy I, II rFkk III (C) dsoy I, II rFkk IV (D) I, II, III rFkk IV

8. Number of nearest neighbours in which of the following is not correct -

(A) simple cubic unit cell = 6 (B) Body centered cubic unit cell = 6

(C) Face centered cubic unit cell = 12 (D) hexagonal close packed unit cell = 12

fuEu esa ls fdlesa fudVre iMksfl;ksa dh la[;k lgh ugha gS-

(A) ljy ?kuh; bdkbZ lSy = 6 (B) dk; dsfUnzr ?kuh; bdkbZ lSy = 6

(C) Qyd dsfUnzr ?kuh; bdkbZ lSy = 12 (D) "kVdks.kh; cUn ladqfyr bdkbZ lSy = 12


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9. Which of the following are correct statement(A) Spontaneous adsorption of gases on solid surface is an endothermic process as entropy

decreases during adsorption

(B) Formation of micelles takes place when temperature is below Kraft Temperature (Tk) and concen-

tration is above critical micelle concentration (CMC)

(C) A colloid of Fe(OH)3 is prepared by adding a little excess (required to completely precipitate Fe3+

ions as Fe(OH)3)) of NaOH in FeCl

3 solution the particles of this sol will move towards anode

during electrophoresis.

(D) According to Hardy-Schulze rule for the coagulation of As2S

3 sol flocculation value of Fe3+ ion will

be more than Ba2+ or Na+.

fuEu esa ls dkSulk lgh dFku gS

(A) Bksl lrg ij xSlksa dk lrr~ vf/k'kks"k.k Å"ek'kks"kh izØe gksrk gS D;kasfd vf/k'kks"k.k ds nkSjku ,.VªkWih ?kVrh gS

(B) felsYl (micelles) dk fuekZ.k rc gksrk gS tc rki Øk¶V rki (Tk) ls de rFkk lkUnzrk] ØkfUrd felsy lkUnzrk

(CMC) ls vf/kd gksrh gS

(C) Fe(OH)3 ds ,d dksykbM dk fuekZ.k FeCl

3 foy;u esa NaOH ds cgqr de vkf/kD; (Fe3+ vk;uks a dks

Fe(OH)3) esa iw.kZ vo{ksfir djus ds fy, vko';d) dks feykus ls cuk;k tkrk gS bl lkWy ds d.k] oS|qr d.k

lapyu ds nkSjku ,uksM dh vksj xfr djrs gS

(D) As2S

3 lkWy ds LdUnu ds fy, gkMsZ&Lyqt (Hardy-Schulze) fu;e ds vuqlkj Fe3+ vk;u dk Å.kZu eku Ba2+ ;k

Na+ dh rqyuk esa vf/kd gksxkA

10 For a solution of 0.849 g of mercurous chloride in 50 g of HgCl2(l) the freezing point depression is1.24ºC. Kf for HgCl2 is 34.3. What is the state of mercurous chloride in HgCl2 ? (Hg – 200, Cl – 35.5)

(A) as Hg2Cl2 molecules (B) as HgCl molecules

(C) as Hg+ and Cl– ions (D) as Hg22+ and Cl– ions

50 g HgCl2(l) esa 0.849 g eD;qZjl DyksjkbM ds ,d foy;u ds fy, fgekad fcUnq esa voueu 1.24ºC gksrk gS

HgCl2 ds fy, Kf = 34.3 gksrk gSA HgCl2 esa eD;qZjl DyksjkbM dh voLFkk D;k gksrh gS ? (Hg – 200, Cl – 35.5)

(A) Hg2Cl2 v.kqvksa ds leku (B) HgCl v.kqvksa ds leku

(C) Hg+ rFkk Cl– vk;uksa ds leku (D) Hg22+ rFkk Cl– vk;uksa ds leku



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(ii) Paragraph Type (ii) vuqPNsn izdkj

This section contains 3 paragraphs each describing theory, experiment, data etc. Six questions relateto three paragraphs with two questions on each paragraph. Each question of a paragraph has onlyone correct answer among the four choices (A), (B), (C) and (D).bl [k.M esa fl¼kUrksa] iz;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okys 3 vuqPNsn gSA rhuksa vuqPNsn ls lacaf/kr N% iz'ugSa] ftuesa ls gj vuqPNsn ij nks iz'u gSaA vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesals dsoy ,d lgh gSA


A colorless substance 'A' (C6H

7N) is sparingly soluble in H

2O but when treated with mineral acid gives

'B' which is water soluble. 'A' when reacts with CHCl3/KOH produces a foul smelling compound 'C'.

'A' also reacts with PhSO2Cl and produces compound 'D' which is soluble in alkali. 'A' reacts with

HNO2 to produce 'E' which on reaction with phenol/OH– gives an orange dye 'F'

,d jaxghu ;kSfxd 'A' (C6H

7N), H

2O esa vYi&foys; gksrk gSA ijUrq tc [kfut vEy ds lkFk mipkfjr fd;k

tkrk gS rks 'B' nsrk gS tks ty esa foys; gksrk gS] 'A' tc CHCl3/KOH ds lkFk fØ;k djrk gS rks nqxZa/k;qDr ;kSfxd

'C' cukrk gSA 'A', PhSO2Cl ds lkFk Hkh fØ;k djrk gS vkSj ;kSfxd 'D' cukrk gS tks {kkj esa foys; gksrk gSA

'A' , HNO2 ds lkFk fØ;k djds 'E' cukrk gS tks QhukWy@OH– ds lkFk vfHkfØ;k djus ij ,d xqykch jatd

'F' nsrk gSA11. Which of the following reaction can be used to prepare 'A' :

fuEu vfHkfØ;k esa ls fdldk mi;ksx 'A' dks cukus esa fd;k tk ldrk gS&

(A)

C

CNK

O

O

(i) Ph–Br

(ii) H O3+ (B)

(i) Sn/HCl

(ii) OH–

NO2

(C) LiAlH4

CONH2

(D) Na/EtOH

CN

12. Which one of the following cannot react easily with 'E'

(A)

OH

in basic medium (B) OH

in basic medium

(C)

OCH3

(D)

NH2

fuEu esa ls dkSulh ,d] 'E' ds lkFk vklkuh ls fØ;k ugha dj ldrh gS&

(A) {kkjh; ek/;e esa

OH

(B) {kkjh; ek/;e esa OH

(C)

OCH3

(D)

NH2

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Paragraph for Questions 13 and 14Consider the following sequence

ore (A)Consisting

of two metals

Na O solution2 2 Boiled and filtered

(B) + (C)reddishbrown ppt.

E brick

red ppt.

¯

dil.H SO (excess)2 4

(D)orange

solution

H O H2 2 / + FBlue solution

G + O (g)Greensolution

2

AgNO3

On standing

13. Pick the INCORRECT statement(s)(A) ‘B’ is soluble in dil. HCl but insoluble in excess of NH3 .

(B) ‘G’ is soluble in excess of NaOH.

(C) Both ‘C’ and ‘D’ involves three d-orbitals in hybridisation.

(D) The oxidation state of central atom in ‘F’ is + 10

14. 'B' HCl¾¾® 'X'

Pick the correct statement for 'X'

(A) It is a yellow color solution

(B) It gives yellow precipitate with CH3COONH

4

(C) It gives a yellow color solution with excess of NaF

(D) All are correct



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iz'u 13 ,oa 14 ds fy;s vuqPNsn

fuEu vfHkfØ;k Øe ij fopkj dhft,

v;Ld (A)ftlesa nks /kkrq,sa gSa

Na O2 2 foy;umckysa vkSj Nkusa

(B) + (C) ykyHkwjk vo{ksi

E ¯ bZaV tSlk

yky vo{ksi

ruq vkf/kD;H SO ( )2 4

(D)ukjaxh foy;u

H O H2 2 / + Fuhyk foy;u

G + O (g)2

AgNO3

j[kk jgus ij

gjk foy;u

13. xyr dFku pqfu,sa

(A) ‘B’ xeZ lkUæ HCl esa foys; ysfdu NH3 ds vkf/kD; esa vfoys;

(B) ‘G’ NaOH ds vkf/kD; esa foys; gS

(C) ‘C’ rFkk ‘D’ nksuks a ladj.k esa rhu d-d{kdksa dks lfEefyr djrs gS

(D) ‘F’ esa dsUnzh; ijek.kq dh vkWDlhdj.k voLFkk + 10 gS

14. 'B' HCl¾¾® 'X'

'X' ds fy, lgh dFku pqfu,sa

(A) ;g ihys jax dk foy;u gS

(B) ;g] CH3COONH

4 ds lkFk ihyk vo{ksi nsrk gS

(C) ;g NaF ds vkf/kD; ds lkFk ihys jax dk foy;u cukrk gS

(D) lHkh lgh gS


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Paragraph for Questions 15 and 16

iz'u 15 ,oa 16 ds fy;s vuqPNsn

Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential

(E°) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example

is a Daniel cell in which zinc goes into solution and copper gets deposited.

jsMkWDl vfHkfØ;kvksa dh jlk;u rFkk tho foKku esa ,d egRoiw.kZ Hkwfedk gksrh gSA nks v/kZ lsy vfHkfØ;kvksa ds ekud

jsMkWDl foHko (E°) ds eku fu/kkZfjr djrs gS fd vfHkfØ;k fdl fn'kk esa c


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17. Match the given reaction in List I with the reagent used in List IIList–I List–IIReaction Reagent

(P) 1-(3-Hydroxycyclopentyl) ethanone ® 3-Ethylcyclopentanol (1) NH2 – NH

2 /OH–/glycol

(Q) Cyclohexanone ® cyclohexanol (2) DIBAL-H

(R) Cyclohexane carbonyl chloride ®Cyclohexane carbaldehyde (3) LiAlH4

(S) Phenylbenzoate ® Benzaldehyde (4) H2/Pd/BaSO

4/Quinoline

lwph I es nh x;h vfHkfØ;k dks lwph II esa mi;ksx esa vkus okys vfHkdeZd ds lkFk lqesfyr dhft;s&lwph–I lwph–IIvfHkfØ;k vfHkdeZd

(P) 1-(3-gkbMªksDlh lkbDyksisfUVy) ,sFksukWu ® 3-,sfFkylkbDyksisUVsukWy (1) NH2 – NH

2 /OH–/XykbdkWy

(Q) lkbDyksgsDlsukWu ® lkbDyks gsDlsukWy (2) DIBAL-H

(R) lkbDyks gsDlsu dkcksZfuy DyksjkbM ®lkbDyksgsDlsu dkcsZfYMgkbM (3) LiAlH4

(S) Qsfuy csUtks,V ® csUtsfYMgkbM (4) H2/Pd/BaSO

4/ Dohuksyhu

Codes :

P Q R S

(A) 1 2 4 3

(B) 1 3,2 4,2 2

(C) 1 2 4,2 3

(D) 4 2 1 3


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18. Match the reaction given in List I with the information about product given in list II.

List–I List–II(Reaction) (Product Information)

(P) CH – CH – COOH2 2CH – CH – COOH2 2

D (1) Final product formation involves dehydration

only

(Q)CH –COOH2

CH –COOH2(2) Final product formation involves

decarboxylation only

(R)

O

CH – CH – C – CH COOH3 2 2 D (3) Final product is an anhydride

(S)

O

CH – C– NH3 2 DP O2 5 (4) Final product invovles dehydration as well as

decarboxylation both

lwph I es nh x;h vfHkfØ;k dks lwph II esa mRikn ds ckjs esa nh x;h lwpuk ds lkFk lqesfyr dhft;s&lwph–I lwph–IIvfHkfØ;k (mRikn lwpuk)

(P) CH – CH – COOH2 2CH – CH – COOH2 2

D (1) vfUre mRikn ds fuekZ.k esa dsoy futZyhdj.k

lfEefyr gksrk gSA

(Q)CH –COOH2

CH –COOH2(2) vfUre mRikn ds fuekZ.k esa dsoy fodkcksZfDlyhdj.k

lfEefyr gksrk gSA

(R)

O

CH – CH – C – CH COOH3 2 2 D (3) vfUre mRikn ,sugkbMªkbM gSA

(S)

O

CH – C– NH3 2 DP O2 5 (4) vfUre mRikn es a futZyhdj.k d s lkFk&lkF k

fodkcksZfDlfydj.k nksuksa lfEefyr gksrs gSACodes :

P Q R S(A) 1 4 2 3(B) 4 3 2 1(C) 2 3 1 4(D) 1 3 4 2



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19. List–I List–II(P) CaH

2(1) Paramagnetic anion

(Q) K2O

2(2) Homodiatomic, diamagnetic anion

(R) KO2

(3) Neutral aqueous solution

(S) NaCl (4) Gives hydrogen on hydrolysis

lwph–I lwph–II

(P) CaH2

(1) vuqpqEcdh; ½.kk;u

(Q) K2O

2(2) lef}ijekf.o;, izfrpqEcdh; ½.kk;u

(R) KO2

(3) mnklhu tyh; foy;u

(S) NaCl (4) ty vi?kVu ij gkbMªkstu nsrk gSCodes :

P Q R S(A) 3 2 1 4(B) 4 2 3 1(C) 4 3 2 1(D) 4 2 1 3

20. List–I List–II

(P) Constant boiling mixture (1) 2nd order reaction

(Q) Cryoscopic constant (2) Azeotrope

(R) t3/4 = 2t1/2 (A – B, single step) (3) 1st order reaction

(S) 1st t1/2 : 2nd t1/2 : 3

rd t1/2 : : 1 : 2 : 4 (4) Solvent

lwph–I lwph–II

(P) fLFkj DoFkukad feJ.k (1) 2nd dksfV vfHkfØ;k

(Q) Økb;ksLdksfid fu;rkad (2) fLFkjDokFkh

(R) t3/4 = 2t1/2 (A – B, ,dy in) (3) 1st dksfV vfHkfØ;k

(S) 1st t1/2 : 2nd t1/2 : 3

rd t1/2 : : 1 : 2 : 4 (4) foyk;d

Code :P Q R S

(A) 3 2 1 4

(B) 2 3 1 4(C) 2 4 3 1(D) 3 4 2 1


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PART-3 : MATHEMATICS

Hkkx-3 : xf.krSECTION–I : (i) Only One option correct Type

[k.M-I : (i) dsoy ,d lgh fodYi izdkjThis section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and(D) out of which ONLY ONE is correct.

bl [k.M esa 10 cgq fodYi iz'u gSA izR;sd iz'u es a pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa ls

dsoy ,d lgh gSA

1. The minimum distance of origin from the locus of point satisfying the vector equation r a b´ =

(b 0)¹ is equal to

lfn'k lehdj.k r a b´ = (b 0)¹ dks lUrq"V djus okys fcUnq dk fcUnqiFk ls ewyfcUnq dh U;wure nwjh gksxh -

(A) 2

a.b

a(B) 2

a a b

a

+ ´(C) 2

a a b

a

- ´(D)

2

a b

a

´

2. Number of values of 'a' for which Rolle's theorem is valid for 2 ax 3 , x 1

f(x)2x | a |, x 1

ì + ¹= í

+ =î on [–2, 2] is

equal to

a ds ekuksa dh la[;k ftlds fy, jksys izes;] vUrjky [–2, 2] ij 2 ax 3 , x 1

f(x)2x | a |, x 1

ì + ¹= í

+ =î ds fy, lR; gks] gksxh

(A) 0 (B) 1 (C) 2 (D) 3

3. Pair of tangents are drawn from P(–1, 2) to y2 = 4x touching it at A, B, then area of triangle PAB

(in sq. units) is equal to

fcUnq P(–1, 2) ls ijoy; y2 = 4x ij Li'kZjs[kk ; qXe [khaph tkrh gS] tks ijoy; dks fcUnq A rFkk B ij Li'kZ djrh

gS] rks f=Hkqt PAB dk {ks=Qy (oxZ bdkbZ esa) gksxk

(A) 8 2 (B) 4 2 (C) 2 2 (D) 2



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4. The range of the function 2 2f(x) x x 1 x x 1= + + - - + is (a, b), then (a + b) is

(A) positive (B) negative (C) 0 (D) an irrational number

Qyu 2 2f(x) x x 1 x x 1= + + - - + dk ifjlj (a, b) gks] rks (a + b) dk eku gksxk

(A) /kukRed (B) ½.kkRed (C) 0 (D) ,d vifjes; la[;k

5. The value of the integral

1

21

( 2)4

sin (sin(2x 1)) dx

p-

-

- p-

é ù+ë ûò (where [.] denotes greatest integer function) is

equal to

lekdy

1

21

( 2)4

sin (sin(2x 1)) dx

p-

-

- p-

é ù+ë ûò dk eku gksxk (tgk¡ [.] egÙke iw.kk±d Qyu dks n'kk Zrk gS) -

(A) p – 2 (B) 2 (C) 12

p- (D) 1

6. Let f : R ® R & g : R ® R be two differentiable functions satisfying f(x) – f(y) = (x3 – y3) g(x – y)"x, y Î R (where f, g are not identically zero) then

(A) f is many one onto and g is one-one onto

(B) f is one-one onto and g is many one into

(C) f is one-one into and g is many one into

(D) f is many one into and g is many one into

ekuk f : R ® R rFkk g : R ® R nks vodyuh; Qyu gS] tks f(x) – f(y) = (x3 – y3) g(x – y) "x, y Î R(tgk¡ f, g leku :i ls 'kwU; ugha gS) dks lUrq"V djrs gS] rc

(A) f, cgq,dSdh vkPNknd rFkk g ,dSdh vkPNknd gSA

(B) f, ,dSdh vkPNknd rFkk g cgq,dSdh vUr%{ksih gSA

(C) f, ,dSdh vUr%{ksih rFkk g cgq,dSdh vUr%{ksih gSA

(D) f cgq,dSdh vUr%{ksih rFkk g cgq,dSdh vUr%{ksih gSA

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7. The value of 6 6 6

12 12 12 12 12n

n 32n 243n 32lim .....

1 n 2 n 3 n 4097n®¥æ ö

+ + + +ç ÷+ + +è ø is equal to

6 6 6

12 12 12 12 12n

n 32n 243n 32lim .....

1 n 2 n 3 n 4097n®¥æ ö

+ + + +ç ÷+ + +è ø dk eku gksxk

(A) 11 tan (64)

6-

(B) 11 tan (32)

6-

(C) 11 tan (16)

6-

(D) 11 tan (48)

6-

8. Centres of two circles having radius 2 and 1 units are 5 units apart. The area of the quadrilateralformed by joining the points of contact of external tangents drawn to two circles is equal to (in sq.

units)

2 rFkk 1 bdkbZ f=T;k ds nks o`Ùkk sa ds dsUæ 5 bdkbZ nwjh ij gSA nksuks a o`Ùkksa ij [khaph xbZ cká Li'kZjs[kkvksa ds Li'kZ fcUnqvks adks tksM+us ls fufeZr prqHkq Zt dk {ks=Qy gksxk (oxZ bdkbZ esa)

(A) 172 6

25(B)

96 6

25(C)

144 6

25(D)

72 6

25

9. Let xe

f(x)x

-

= , then identify the correct statement

(A) f(x) = 1 has no solution (B) f(x) = –2 has exactly one solution

(C) f(x) = –3 has no solution (D) f(x) = – 4 has exactly two solutions

ekuk xe

f(x)x

-

= gks] rks lgh dFku dk p;u dhft,

(A) f(x) = 1 dk dksbZ gy ugha gSA (B) f(x) = –2 dk Bhd ,d gy gSA

(C) f(x) = –3 dk dksbZ gy ugha gSA (D) f(x) = – 4 ds Bhd nks gy gSaA

10. Area bounded by the curve xy2 = xe2x is equal to (in sq. units)

oØ xy2 = xe2x }kjk ifjc¼ {ks=Qy gksxk (oxZ bdkbZ esa)

(A) 2 (B) 4 (C) 1 (D) e2


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(ii) Paragraph Type

(ii) vuqPNsn izdkjThis section contains 3 paragraphs each describing theory, experiment, data etc. Six questions relateto three paragraphs with two questions on each paragraph. Each question of a paragraph has onlyone correct answer among the four choices (A), (B), (C) and (D).

bl [k.M esa fl¼kUrksa] iz;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okys 3 vuqPNsn gSA rhuksa vuqPNsn ls lacaf/kr N% iz'u

gSa] ftuesa ls gj vuqPNsn ij nks iz'u gSaA vuqPNsn esa gj iz'u ds pkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa

ls dsoy ,d lgh gSA


iz'u 11 ,oa 12 ds fy;s vuqPNsnNine persons Pi, i = 1, 2, ...., 9 of equal strength are playing a tournament such that they are first

grouped into three groups A, B, C each containing three persons at random and a winner from each

group is selected and a new group is formed and finally from this group the winner of the tournament

is decided.

leku {kerk ds 9 O;fDr Pi, i = 1, 2, ...., 9, ,d izfr;ksfxrk bl izdkj [ksyrs gSa fd loZizFke os rhu lewgks a A, B, Cesa cVrs gks] ftuesa ;kn`PN;k rhu O;fDr gS rFkk izR;sd lewg ls ,d fotsrk dk p;u fd;k tkrk gS rFkk ,d u;k lewgcuk;k tkrk gS vUrr% bl lewg ls izfr;ksfxrk ds fotsrk dk fu.kZ; fd;k tkrk gSA

11. Probability that P2 and P4 were in different groups given that P4 is the winner of the tournament isequal to

P2 rFkk P4 ds vyx lewgksa esa gksus fd izkf;drk tcfd fn;k x;k gS fd P4 izfr;ksfxrk dk fotsrk gS] gksxh -

(A) 2

3(B)

1

2(C)

1

4(D)

3

4

12. Probability that P1 is not part of the new group and P2 is the winner of the tournament is equal toP1 ds u;s lewg dk lnL; uk gksus rFkk P2 ds izfr;ksfxrk ds fotsrk gksus dh izkf;drk gksxh -

(A) 1

12(B)

1

4(C)

1

8(D)

1

9

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Let z be a complex number satisfying 3

3

z 1arg

z 1 2

æ ö- p=ç ÷+è ø

on the argand plane

ekuk vkxZ.M lery ij ,d lfEeJ la[;k z, 3

3

z 1arg

z 1 2

æ ö- p=ç ÷+è ø

dks lUrq"V djrh gSA

13. Length of the arc of the locus of z for which Re(z) > 0 and Im(z) < 0 is equal to

z ds fcUnqiFk ds pki dh yEckbZ ftlds fy, Re(z) > 0 rFkk Im(z) < 0 gks] gksxh

(A) 6

p(B)

3

p(C)

9

p(D)

4

p

14. Length of the arc traced by z for which Re(z) > 0 is equal to

z }kjk fufeZr pki dh yEckbZ] ftlds fy, Re(z) > 0 gks] gksxh

(A) 3

p(B)

2

p(C)

2

3

p(D)

5

6

p


iz'u 15 ,oa 16 ds fy;s vuqPNsnConsider a plane P : x + y + z = 3 and a circle x2 + y2 = 1, z = 0

ekuk ,d lery P : x + y + z = 3 rFkk ,d o`Ùk x2 + y2 = 1, z = 0 gSA15. If A(x1, y1, 0) and B(x2, y2, 0) are the coordinates of the points on the circle which are at maximum

and minimum distance from the plane P, then 1 1 2 2(x y ) (x y )+ - + is equal to

;fn A(x1, y1, 0) rFkk B(x2, y2, 0) o`Ùk ij fLFkr fcUnqvksa ds funsZ'kkad gS] tks lery P ls vf/kdre rFkk U;wure

nwjh ij gS] rks 1 1 2 2(x y ) (x y )+ - + dk eku gksxk

(A) 0 (B) 2 (C) 2 2 (D) 2

16. Image of the points on the circle in the plane P lies onlery P esa o`Ùk ij fLFkr fcUnqvksa dk izfrfcEc fuEu ij fLFkr gksxk -(A) x + y – 2z = 1 (B) x + y + z = 5 (C) 2x + y + 2z = 10 (D) 2x + 2y – z = 6


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17. Let m1, m2, m3 (m1< m2 < m3) be the slopes of the distinct normals to 2 2x y

116 12

+ = , passing through

1 2,

2 2 2 3

æ öç ÷ç ÷è ø

, then

List-I List-II

(P) m1 + m2 + m3 is equal to (1) 2 3

(Q) m1m2m3 is equal to (2)8

3 3-

(R) m1 + m3 – m2 is equal to (3)2

43

-

(S) m3 – m1 – m2 is equal to (4)2

43

+

ekuk m1, m2, m3 (m1< m2 < m3), 2 2x y

116 12

+ = ds fofHkUu vfHkyEcksa dh izo.krk;sa gS] tks 1 2,2 2 2 3

æ öç ÷ç ÷è ø

ls xqtjrh

gS] rc

lwpha-I lwpha-II

(P) m1 + m2 + m3 dk eku gksxk (1) 2 3

(Q) m1m2m3 dk eku gksxk (2)8

3 3-

(R) m1 + m3 – m2 dk eku gksxk (3)2

43

-

(S) m3 – m1 – m2 dk eku gksxk (4)2

43

+

Codes :P Q R S

(A) 1 2 3 4(B) 3 2 1 4(C) 4 2 1 3(D) 2 2 1 3

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18. Consider a twice differentiable function f(x) satisfying f(x) + f ''(x) = 2f '(x) where f(0) = 0, f(1) = eList-I List-II

(P) f '(–1) is equal to (1) 0

(Q) f ''(1) is equal to (2) 1

(R)1

0f(x)dxò is equal to (3) 2e2

(S) Number of roots of f(x) = 0 is equal to (4) 3e

ekuk nks ckj vodyuh; Qyu f(x), f(x) + f ''(x) = 2f '(x) dks lUrq"V djrk gS] tgk¡ f(0) = 0, f(1) = e gSA

lwpha-I lwpha-II(P) f '(–1) dk eku gksxk (1) 0(Q) f ''(1) dk eku gksxk (2) 1

(R)1

0f(x)dxò dk eku gksxk (3) 2e2

(S) f(x) = 0 ds ewyks a dh la[;k gksxh (4) 3e

Codes :

P Q R S

(A) 2 3 4 1

(B) 2 3 1 4

(C) 1 3 2 4

(D) 1 4 2 2


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19. Let (1 + x2 + x3)40 = a0 + a1x + a2x2 + ..... + a120x

120

List-I List-II

(P) Number of values of r for which ar = 0, is equal to (1) 0

(Q) a0 – a2 + a4 – ..... + a120 is equal to (2) 1

(R) a1 – a3 + a5 – ..... – a119 is equal to (3) 2

(S) 2a

20 is equal to (4) –1

ekuk (1 + x2 + x3)40 = a0 + a1x + a2x2 + ..... + a120x

120 gSA

lwpha-I lwpha-II

(P) r ds ekuksa dh la[;k ftlds fy, ar = 0 gks] gksxh (1) 0

(Q) a0 – a2 + a4 – ..... + a120 dk eku gksxk (2) 1

(R) a1 – a3 + a5 – ..... – a119 dk eku gksxk (3) 2

(S) 2a

20 dk eku gksxk (4) –1

Codes :

P Q R S

(A) 3 2 1 3

(B) 2 2 1 3

(C) 4 1 2 3

(D) 2 1 2 3

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20. Straight lines are formed passing through (4, 6) and forming triangles of given area with the coordinateaxes (each line forms a triangle of fixed area with the axes).

List-I List-II(P) Number of lines which form an area of (1) 1

36 sq. units is equal to

(Q) Number of lines which form an area of (2) 2


(R) Number of lines which form an area of (3) 3


(S) Number of lines which form an area of (4) 4


fcUnq (4, 6) ls xqtjus okyh ljy js[kk;sa [khaph tkrh gS rFkk funsZ'kh v{kks a ds lkFk fn, x, {ks=Qy ds f=Hkqt cuk;s tkrsgS (izR;sd js[kk funsZ'kh v{kksa ds lkFk fuf'pr {ks=Qy dk ,d f=Hkqt cukrh gS).

lwpha-I lwpha-II(P) js[kkvksa dh la[;k] tks 36 oxZ bdkbZ {ks=Qy cukrh gks] gksxh (1) 1(Q) js[kkvksa dh la[;k] tks 48 oxZ bdkbZ {ks=Qy cukrh gks] gksxh (2) 2(R) js[kkvksa dh la[;k] tks 60 oxZ bdkbZ {ks=Qy cukrh gks] gksxh (3) 3(S) js[kkvksa dh la[;k] tks 40 oxZ bdkbZ {ks=Qy cukrh gks] gksxh (4) 4

Codes :P Q R S

(A) 3 1 4 2

(B) 2 3 1 4

(C) 2 3 4 2

(D) 3 1 1 4


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D. vadu ;kstuk / Marking scheme :

12. [kaM-I (i, ii & iii) ds gj iz'u esa dsoy lgh mÙkjksa (mÙkj) okys lHkh cqycqyksa (cqycqys) dks dkyk djus ij 3 vad vkSj dksbZ Hkh cqycqyk

dkyk ugha djus ij 'kwU; (0) vad iznku fd;k tk;sxkA vU; lHkh fLFkfr;ksa esa ½.kkRed ,d (–1) vad iznku fd;k tk;sxkA

For each question in Section-I (i, ii & iii), you will be awarded 3 marks if you darken all the bubble(s) corresponding to

only the correct answer(s) and zero mark if no bubbles are darkened. In all other cases minus one (–1) mark will be awarded

13. g = 10 m/s2 iz;qDr djsa ] tc rd fd vU; dksbZ eku ugha fn;k x;k gksA / Take g = 10 m/s2 unless otherwise stated.

Name of the Candidate / ijh{kkFkhZ dk uke

I have read all the instructions and shall abide by them.eSusa lHkh vuqns'kksa dks i

allen clas amme · 2015. 12. 21. · 8. bl iqflrdk dh eqgj rksm+us ds i'pkr d`i;k tk¡p ysa fd...

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