ct-2 (paper-1)_01_09_2013

ELPD CT-02 Date: 01-09-2013

PAPER�1

Q.No. Subject Nature of Questions No. of Questions Marks Negative Total

1 to 18 SCQ 18 3 �1 54

19 to 24 Comprehension (3 Com. × 2 Q.) 6 3 0 18

25 to 42 SCQ 18 3 �1 54


49 to 66 SCQ 18 3 �1 54


72 216

Paper-1

Total Total

Physics

Chemistry

Maths

SECTION - IStraight Objective Type

This section contains 18 multiple choice questions. Each question has choices (A), (B), (C) and (D),out of which ONLY ONE is correct.

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

bl [k.M esa 18 cgq&fodYih iz'u gSA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa, ftuesa ls flQZ ,d lgh gSA

1. dx)2x3( 5 =

(A) 6

)2x3( 6 + C (B*)

18)2x3( 6

+ C (C) 5 (3x + 2)4 + C (D) 15 (3x + 2)4 + C

Sol. dx)2x3( 5

3x + 2 = t3dx = dt

dx = 3dt

dx)2x3( 5 =

31

3dt

t5

6t6

+ C

= 18

)2x3( 6 + C

2. y = x3 + x2 + 2 then the value of dxdy

at x = 2 is :

y = x3 + x2 + 2 rks x = 2 ij dxdy

dk eku gksxk :

(A) 14 (B*) 16 (C) 18 (D) None of these

Sol.dxdy

= 3x2 + 2x

= 3 × 22 + 2 × 2

= 16

3. The resultant of the given three vectors is :uhps fn;s x;s rhu lfn'kksa dk ifj.kkeh gksxk &

(A) 9 m (B) 13 m (C) 29 m (D*) 45 m

Sol.

R = 22 63 = 45 m

4 . k�4j�3i�2A

and k�8j�6i�4B

then. A

B

is equal to :

k�4j�3i�2A

rFkk k�8j�6i�4B

gS rks A

B

dk eku gksxk &

(A) �2 (B) +2 (C) 21

(D*) A

B

is meaning less (vFkZghu gS)

Sol. A quantity can't be divided by vector.,d lfn'k jkf'k nwljs ls foHkkftr ugha gksrh gSA

5. Mark the correct statement : lgh dFku pqfu, &(A) Scalar product of two non-zero colinear vectors is zero.(B*) Vector product of two non-zero colinear vectors is zero.(C) If vector product of two non-zero vectors is zero than they must be parallel.(D) none of the above(A) nks v'kwU; lajs[kh; lfn'kksak dk vfn'k xq.kuQy 'kwU; gksrk gSA(B*) nks v'kwU; lajs[kh; lfn'kksak dk lfn'k xq.kuQy 'kwU; gksrk gSA(C) ;fn nks v'kwU; lfn'kks dk lfn'k xq.kuQy 'kwU; gS rks os lekUrj gksaxsA(D) buesa ls dksbZ ugha

Sol. BA

= |B||A|

sin = 0 for = 0º or 180º.

6. The minimum possible resultant of vectors of magnitude 10 N, 5 N and 3 N is :lfn'kksa ftuds ifjek.k 10 N, 5 N rFkk 3 N ds ifj.kkeh dk U;wure eku fuEu gks ldrk gS :(A*) 2 N (B) 3 N (C) 0 N (D) none of these bues ls dksbZugha

Sol.

R = 10 � 5 � 3 = 2

7. A balloon of Mass M is rising vertically upwards with constant velocity under the action of constant force andits weight. If some mass m is removed from balloon then acceleration becomes (Assume that constant forceremains same and acceleration due to gravity is g) :M nzO;eku dk ,d xqCckjk fu;r cy rFkk blds Hkkj ds çHkko esa fu;r osx ls Å ij mB jgk gSA ;fn xqCckjs ls dqN nzO;ekum gVk fn;k tk, rks xqCckjs dk Roj.k gksxkA (fu;r cy dks ,d leku ekusaA xq:Roh; Roj.k g gS :

(A*) )m�M(mg

(B) )m�M(Mg

(C) )m�M()mM(

g (D) g

Sol. Initialy to balance Mg, F = MgçkjEHk eas lkE;koLFkk esa Mg, F = Mg

F

Mg

afterwards blds cknm�M

)m�M(�Fg = a

8. Velocity of rain w.r.t. ground is (� 3 i� � 3 j� ) m/s. If rain appears to be falling vertically to a man, then velocity

of man w.r.t. ground in m/s is

tehu ds lkis{k ckfj'k dk osx (� 3 i� � 3 j� ) m/s gSA ,d vkneh dks ckfj'k Å /okZ/kj fxjrh gqbZ çrhr gksrh gS rks vkneh dk

osx tehu ds lkis{k m/s esa gksxk �

j

i

(A) 3 i� (B*) � 3 i� (C) � 3 j� (D) 3 j�

9. If tension in string A and string B are TA and T

B then find out

B

A

TT

(assume that strings are light and

g = 10 m/s2) :

;fn jLlh A rFkk jLlh B esa ruko Øe'k% TA rFkk T

B gks rks

B

A

TT

Kkr djksA (ekuk jfLl;k¡ gYdh rFkk g = 10 m/s2) :

(A) 2 (B*) 1 (C) 2

1(D) 2

Sol.2

TC = TTA = T

B.

TC

2

TC

2

TA

TB

10. Position�time graph for a particle moving along x-direction is as shown in the figure. Average speed of the

particle from t = 0 to t = 4 is :x-v{k ds vuqfn'k xfr'khy d.k ds fy, fLFkfr≤ oØ fp=k esa çnf'kZr gSAt = 0 ls t = 4 rd d.k dh vkSlr pky gS :

(A*) 4

15m/s (B)

310

m/s (C) 25

m/s (D) 45

m/s

Sol. <speed> <pky> = 4105

= 4

15m/s.

11. Two particle A and B are projected as shown in figure. Maximum height is same for both the particles. uA and

uB are initial speeds of A and B respectively then :

nks d.kksa A rFkk B dks fp=kkuqlkj ç{ksfir fd;k tkrk gSA nksauks d.kksa dh vf/kdre Å ¡pkbZ leku gSA A rFkk B dh çkjfEHkd pkyu

A rFkk u

B gks rks :

(A*) uA < u

B(B) u

A > u

B(C) u

A = u

B(D) T

B > T

A

Sol. TA = T

B

uAsin

A = u

B sin

B

A >

B

uA < u

B .

12. If a force F acts on a body of mass m it produces an acceleration of magnitude a. Then acceleration of thesame body in the situation shown in figure :m nzO;eku dh oLrq ij vkjksfir cy F, a ifjek.k dk Roj.k mRiUu djrk gS rks bl oLrq ds fy, uhps çnf'kZr fLFkfr esaRoj.k Kkr djks

120º 120º

120º

2F

FF

(A) 2a (B) 0 (C) a (D*) 3a

Sol. F3 Fnet

FF

FSo, vr%a = 3a

13. Find out velocity of block B in the given figure : (all the pulleys are ideal and string are inextensible massless)fn;s x;s fp=k esa B dk osx Kkr djks : (lHkh f?kjfu;ka vkn'kZ gS rFkk lHkh jfLl;ka nzO;ekughu rFkk vfoLrkfjr gS)

80 m/s

A

B

20 m/s

j

i

(A) 80 m/s j� (B) � 40 m/s j� (C*) 0 m/s (D) � 80 m/s j�

Sol. 1 +

2 = 0

20 + 20 � x = 0

x = 40

3 +

4 = 0

40 � y + 40 � 80 = 0

80 m/s

A

B

20 m/s

l3

l 1l2

l 4

x

y

y = 0

14. A particle is moving along straight line with initial velocity +7 m/sec and uniform acceleration �2 m/sec2. Thedistance travelled by the particle in 4th second of its motion is :,d d.k ljy js[kk ds vuqfn'k izkjfEHkd osx +7 m/sec rFkk le:i Roj.k �2 m/sec2 ls xfr'khy gSA bldh xfr ds nkSjku4th lSd.M esa d.k }kjk r; nwjh gS :(A) 0 (B) 0.25 m (C*) 0.5 m (D) 7 m

Sol. Distance travelled r; nwjh = |S3�3.5

| + |S3.5

� 4

|

= 5.02

)1(05.0

2

01

= 0.25 + 0.25 = 0.5 m.

15. A particle is moving along straight line whose position x at time t is described by x = t3 � t2 where x is inmeters and t is in seconds. Then the average acceleration from t = 2 sec. to t = 4 sec. is :ljy js[kk ds vuqfn'k xfr'khy d.k dh fLFkfr x le; t ds lkFk x = t3 � t2 }kjk nh tkrh gSA ;gkW x-ehVj esa rFkk le;&lSd.Mesa gSA t = 2 sec. ls t = 4 sec. ds e/; vkSlr Roj.k D;k gksxk :(A*) 16 m/s2 (B) 18 m/s2 (C) 22 m/s (D) 10 m/s2

Sol. v = dtdx

= 3t2 � 2t

v4 = 3 × 42 � 2 × 4 = 40

v2 = 3 × 22 � 2 × 2 = 8

<a> = 24vv 24

= 24840

= 16 m/s2.

16. Two men A and B, A standing on the extended floor nearby a building and B is standing on the roof of thebuilding. Both throw a stone towards each other. Then which of the following will be correct.

nks O;fDr A o B, A Å ¡ps Q'kZ ij ,d ehukj ds utnhd [kM+k gS rFkk B ehukj dh Nr ij [kM+k gSA nksuksa ,d nwljs dh vksjiRFkj QSadrs gSa dkSulk dFku lR; gSA

(A) stone will hit A, but not B iRFkj A ls Vdjkrk gS] ijUrqB ls ugha(B) stone will hit B, but not A iRFkj B ls Vdjkrk gS] ijUrq A ls ugha(C) stone will not hit either of them, but will collide with each other iRFkj fdlh ls ugha Vdjkrk] ijUrq vkil esa Vdjkrs(D*) none of these. mijksDr esa ls dksbZ ugha

Sol. (D)Path of stone thrown by one person wirht respect to other person is not straight line but parabolic henceneither stone will hit any person. Condition of collision will depend upon direction as well as velocities ofprojection which are not given.

,d O;fDr ds lkis{k nwljs O;fDr dk iFk ljy js[kk ugha gksxk cfYd ijoy;kdkj gksxkA vr% dksbZ Hkh iRFkj vkneh ls ughaVdjkrkA VDdj dh 'krZ fn'kk ds lkFk&lkFk iz{ksi.k osx ij Hkh fuHkZj djrh gSA tks ugha fn;k x;k gSA

17. In the given figure, if velocity of block C at a particular instant is �20 m/s j� , then velocity of rod (A) at that

instant will be : (string is attached to wedge)

fn;s x;s fp=k esa fdlh {k.k CykWd C dk osx �20 m/s j� gks rks NM+ (A) dk bl {k.k ij osx D;k gksxk : (l string is attached

to wedge)

C

= 37º

B

Aj

i

(A*) 15 m/s j� (B) � 15 m/s j� (C) 20 m/s j� (D) � 20 m/s j�

Sol.

C

37º

B

A

l 1

l 2 u

1 +

2 = 0

20 � u = 0

u = 20 m/s

By wedge constraint urry ca/ku }kjk

u sin 37º = cos 37º

= u × 43

= 15 m/s

= 15 j�

18. A block of mass 10 kg is placed on a horizontal ground surface as shown in the figure and a force F = 100 Nis applied on the block as shown in the figure. The block is at rest with respect to ground. The contact forcebetween block and ground is : (Take g = 10 m/s2)10 kg nzO;eku dk CykWd {kSfrt lrg ij fp=kkuqlkj j[kk gS rFkk fp=kkuqlkj F = 100 N cy CYkWkd ij vkjksfir gSA çkjEHkesa CykWd tehu ds lkis{k fLFkj gS rks tehu rFkk CykWd ds e/; lEidZ cy gksxk : (g = 10 m/s2)

(A) 50 N (B*) 100 N (C) 50 5 N (D) 50 2 NSol. F sin30º

= 50 N

So normal is 50 Nblfy, vfHkyEc cy 50 N gSAand rFkkf

max = N = 100 N > Fcos 30º

So block is in equlibriumCykWd lkE;koLFkk esa gS vr%

0 Mg F R

120º

100

100R

)Mg F( � R

R = 100 N

SECTION - IIComprehension Type

This section contains 3 paragraphs. Based upon each paragraph, 2 multiple choice questions have tobe answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

[k.M - IIc) cks/ku çdkj

bl [k.M esa 3 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 2 cgq&fodYih ç'u ds mÙkj nsus gSA çR;sd ç'uds 4 fodYi (A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA

Paragraph for Question Nos. 19 to 20iz'u 19 ls 20 ds fy, vuqPNsn

Water is flowing in a river from left to right with velocity 4 m/s as shown in figure. The width of river is 60m. Theswimmer 'A' can swim with speed 5 m/s with respect to water :unh esa ikuh fp=kkuqlkj ck,W ls nkW, rjQ 4 m/s osx ls izokfgr gSA unh dh pkSM+kbZ 60m gSA rSjkd A ikuh ds lkis{k5 m/s ls rSj ldrk gS :

19. If the swimmer wants to cross the river in minimum time, then this minimum time taken by swimmer to crossthe river is :;fn rSjkd U;wure~ le; esa unh ikj djuk pkgs rks unh dks ikj djus esa fy;k x;k U;wure~ le; gS :(A) 20 sec. (B) 15 sec. (C*) 12 sec. (D) None of these buesa ls dksbZ ughaA

20. If the swimmer swims in such a way that his displacement in the direction of river flow is zero then the anglebetween his velocity with respect to river and direction of river flow is :;fn rSjkd bl izdkj rSjrk gS fd unh izokg ds vuqfn'k bldk foLFkkiu 'kwU; gS] rks unh ds lkis{k blds osx rFkk unh izokgdh fn'kk ds e/; dks.k gksxk :

(A) 127º (B*) 143º (C) 37º (D) 53º

Sol.(19 to 20)For crossing the river in minimum time, he must swim perpendicular to river flow

So, t = river w.r.t.swimmer of velocitywidth

= 5

60 = 12 sec. Ans.

U;wure~ le; esa unh ikj djus ds fy,] rSjkd dks unh izokg ds yEcor~ rSjuk gksxkA

vr%, t = x lkis{k osnh dsrSjkd dk u

MkbZunh dh pkS =

560

= 12 sec. Ans.

So vr%, S = m 4112 Ans.

5m/s

4m/s

5 sin = 4

sin = 54

= 53º

So, required angle is 90º + 53º = 143º Ans.vr%, vko';d dks.k 90º + 53º = 143º Ans.


Comprehension�2 : (21 to 22)A particle is projected with velocity 50 m/s such that its initial velocity makes an angle 37º with east

direction and is in vertical plane. The particle has constant acceleration 10 m/s2 towards north due towind and 10 m/s2 downwards due to gravity.,d d.k 50 m/s ds osx ls bl izdkj iz{ksfir fd;k tkrk gS fd bldk izkjfEHkd osx Å /okZ/kj ry esa iwoZ fn'kk ls 37º dks.kcukrk gSA gok ds dkj.k d.k dks fu;r Roj.k 10 m/s2 mÙkj dh rjQ rFkk xq:Ro ds dkj.k 10 m/s2 uhps dh rjQ izkIrgksrk gSA

21. The time of flight of particle is :d.k dk mM~M;u dky gksxkA(A) 10 sec. (B) 8 sec. (C*) 6 sec. (D) None of these buesa ls dksbZ ughaA

22. The trajectory of particle as seen from above is :Å ij ls ns[kus ij d.k dk iFk gksxkA

(A*)

N

E

S

W(B)

N

E

S

W (C)

N

E

S

W (D)

N

E

S

W

Sol.(21 to 22)

Å ij 30m/s

E 40m/s

uhps 10m/s2

W

50m/s

37º

,D rjQ ls ns[kus ij

T = 10

302 = 6 sec. Ans.

N

E

S

W

uhps

Å ij

10m/s2

40m/s

D.k dk iFk

10m/s2 30m/s

Å ij ls ns[kus ij

Paragraph for Question Nos. 23 to 24

iz'u 23 ls 24 ds fy, vuqPNsn

Two bodies A and B of masses 10 kg and 5 kg are placed very slightly separated as shown in figure. Thecoefficient of friction between the floor and the blocks is = 0.4. Block A is pushed by an external forceF. The value of F can be changed. When the friction between block A and ground reaches the limitingvalue, block A will start pressing block B and when friction of B also reaches the limiting value, block Bwill start pressing the vertical wall (take g = 10 m/s2) �

nks fi.M A o B ft ud s nzO;eku 10 kg o 5 kg gS cgqr gh d e nwjh ij j[ks x;s gS t Slk fp=k esa çnf'kZr gSA fi.M vkSjry d s chp ?k"kZ.k xq.kkad = 0.4. gSA fi.M A d ks ckº; cy F ls /kd syk t krk gSA F d k eku ifjorZu'khy gSA t cfi.M A vkSj t ehu d s chp osfYMax VwV t krh gSA rc fi.M A, fi.M B d ks nckuk izkjEHk d jrk gS A t c fi.M B d hosfYMax VwV t krh gS rks fi.M B Å /okZ/kj nhokj d ks nckuk çkjEHk d jrh gSA

23. If F = 20 N, with how much force does block A presses the block B;fn F = 20 N, rks fi.M A, fi.M B d ks fd rus cy ls nck;sxk &(A) 10 N (B) 20 N (C) 30 N (D*) Zero 'kwU;

Sol. If F = 20 N, 10 kg block will not move and it would not press 5 kg block So N = 0.;fn F = 20 N, 10 kg d k CykWd xfr ugha d jsxk rFkk ;g 5 kg d s CykWd d ks ugha nck;sxk vr% N = 0

24. What should be the minimum value of F, so that block B can press the vertical wallF dk U;wure eku D;k gksxk , ftlds dkj.k CykWd B Å /okZ/kj nhokj dks nck ldsaA(A) 20 N (B) 40 N (C*) 60 N (D) 80 N

Sol. If F = 50 N, force on 5 kg block = 10 N;fn F = 50 N, rks 5 kg d s CykWd ij cy = 10 N yxsxkA

So friction force = 10 Nvr% ?k"kZ.k cy = 10 N

Page No. # 1

TEST PATTERN

CT-02 Date : 0109.2013

Physical Syllabus

Mole Concept, Atomic Structure & Gaseous State

Paper-1SCQ (8)


1 to 18 SCQ 18 3 �1 54


25 to 42 SCQ 18 3 �1 54


49 to 66 SCQ 18 3 �1 54


72 216

Chemistry

Maths

Paper-1

Total

Paper-2

Total

Physics

Paper-1

Physical SCQ (10)


This section contains 18 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONLY ONE is correct.

[k.M- I lh/ks oLrqfu"B izdkj

bl [k.M esa 18 iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls flQZ ,d lgh gSA

25. Which of the following sets of quantum numbers can be correct for an electron in 4f-orbitals :4f-d{kd esa ,d bysDVªkWu ds fy, pkjksa DokaVe la[;kvksa dk dkSulk lewg lR; gS %

(A) n = 3, = 2, m = �2, s = +21

(B) n = 4, = 4, m = �4, s = �21

(C*) n = 4, = 3, m = +1, s = +21

(D) n = 4, = 3, m = +4, s = +21

Sol. For n = 4, 4, for = 3, m 4gy . n = 4 ds fy, 4, = 3 ds fy, m 4

26. A solution of density 'd' has mole fraction of solute 'a' and the mole fraction of solvent is 'b'. Molar mass ofsolute and solvent is m and M respectively. Molarity of solution can be calculated by formula :

(A) 1000bMd

ma

(B) bMam1000db

(C*) bMam

1000da

(D) Cannot be calculated

?kuRo 'd' dk ,d foy;u] foys; dh eksy&fHkUu 'a' o foyk;d dh eksy fHkUu 'b' j[krk gSA foys; o foyk;d dk eksyj nzO;ekuØe'k% m rFkk M gSaA fuEu lw=k }kjk foy;u dh eksyjrk dks ifjdfyr fd;k tk ldrk gS %

(A) 1000bMd

ma

(B) bMam

1000db

(C*)

bMam1000da

(D) x.kuk ugha dh tk ldrh

27. An ion of a hypothetical element, Xn+ having mass number equal to 19 is iso-electronic with oxygen atom.What will be the value of 'n', if it contains (Z + 1) neutrons (where z is atomic number of X) :(A*) 1 (B) 2 (C) 3 (D) 4,d dkYifud rRo dk vk;u Xn+ gSA bldh nzO;eku la[;k 19 gS rFkk ;g vkWDlhtu ijek.kqds lkFk lebySDVªkWfud gSA'n' dk eku D;k gksxk] ;fn ;g (Z + 1) U;wVªkWu j[krk gS % (tgka z, X dk ijek.kq Øekad gS) :(A*) 1 (B) 2 (C) 3 (D) 4

Page No. # 2

Sol. Z + Z + 1 = 19 or Z = 9n = 9 � 8 = 1

28. A sample of SO3(g) contains 6.02 10y molecules and has mass equal to the mass of 5.6 litres O2(g) at STP.y is (NA = 6.02 × 1023) :SO3 xSl dk ,d izkn'kZ 6.02 10y v.kq j[krk gS rFkk bldk nzO;eku STP ij 5.6 yhVj O2 xSl ds cjkcj gksrk gSA yfuEu gS (NA = 6.02 × 1023) %(A) 21 (B*) 22 (C) 23 (D) 24

Sol. Mass of O2 gas = 324.22

6.5 = 8 gm

Moles of SO3 gas = 808

=101

Molecules are 101 6.02 1023

= 6.02 1022

gy% O2 xSl dk nzO;eku = 324.22

6.5 = 8 gm

SO3 xSl ds eksy = 808

=101

v.kq 101 6.02 1023 gSa

= 6.02 1022

29. Consider three electron jumps described below for the hydrogen atomX : n = 3 to n = 1Y : n = 4 to n = 2Z : n = 5 to n = 3

For which transition will the electron experience the longest change in orbit radius ?(A) X (B) Y (C*) Z (D) Same for each transition

gkbMªkstu ijek.kq ds fy, rhu bysDVªkWu ds Nykax dk o.kZu fuEu gSX : n = 3 to n = 1Y : n = 4 to n = 2Z : n = 5 to n = 3

dkSuls laØe.k ds fy, bySDVªkWu] d{k dh f=kT;k esas vf/kdre ifjorZu vuqHko djsxk \(A) X (B) Y (C*) Z (D) izR;sd laØe.k ds fy, leku

30. In a photoelectric experiment, kinetic energy of photoelectrons was plotted against the frequency of incidentradiation (), as shown in the figure. Which of the following statement is correct?

(A) The threshold frequency is 1.

(B*) The slope of this straight line is equal to Planck�s constant.

(C) As the frequency of incident photon increases beyond threshold frequency, kinetic energy of photoelectronsdecreases.(D) It is impossible to obtain such a graph.izdk'kfo|qr iz;ksx esa] fp=k esa n'kkZ;s vuqlkj QksVks bysDVªkWuksa dh xfrt Å tkZ o vkifrr fofdj.k () dh vkofr ds e/; xzkQcuk;k x;k gSA fuEu esa ls dkSulk dFku lgh gS \

Page No. # 3

(A) nsgyh vkofÙk 1 gSA

(B*) bl lh/kh js[kk dk <ky Iykad fu;rkad ds cjkcj gSA(C) tSls vkifrr QksVksu dh vkofÙk nsgyh vkofÙk ls c<+rh gS] QksVksbysDVªkWuks dh xfrt Å tkZ ?kV tkrh gSA(D) bl izdkj dk xzkQ izkIr djuk vlEHako gSA

31. Spin only magnetic moments of V (Z = 23), Cr (Z = 24) and Mn (Z = 25) are x, y, z in their ground state.Hence :vk| voLFkk esa] V (Z = 23), Cr (Z = 24) o Mn (Z = 25) dk dsoy pØ.k pqEcdh; vk?kw.kZ Øe'k% x, y, z gS vr%(A) x = y = z (B) x < y < z (C*) x < z < y (D) z < y < x

32. The moles of O2 required for reacting with 6.8 g of ammonia is :(......NH3 + ....... O2 ...... NO + ..... H2O)

6.8 g veksfu;k ds lkFk fØ;k djus ds fy, O2 ds vko';d eksyksa dh la[;k gS %(......NH3 + ....... O2 ...... NO + ..... H2O)

(A) 5 (B) 2.5 (C) 1 (D*) 0.5

Sol. 2NH3 +

25

O2 2NO + 3H2O

from mole-mole analysis (eksy&eksy fo'ys"k.k ls)

2

n3NH

= 2/5

n2O

217

8.6

= 25

n2O

2On = 0.5 mole.

33. What will be molality of 1% w/w sucrose solution in water (molar mass of sucrose = 342 gram)ty esa 1% Hkkj/Hkkj (w/w) lqØksl foy;u dh eksyyrk D;k gksxhA (lqØksl dk eksyj nzO;eku = 342 xzke)(A) 0.06 m (B*) 0.03 m (C) 0.09 m (D) 0.12 m

Sol. m = 99342

10001

34. A man consumes 100 gram glucose (C6H

12O

6) to get energy and converts all carbon atoms present in

glucose into CO2 gas. What will be volume of CO

2 gas thus released at STP.

,d O;fDr Xywdksl (C6H

12O

6) ds 100 xzke dk mi;ksx Å tkZ izkIr djus ds fy, djrk gS rFkk Xyqdksl esa mifLFkr lHkh dkcZu

ijek.kqvksa dks CO2 xSl esa ifjofrZr djrk gS bl izdkj STP ij eqDr CO

2 xSl dk vk;ru D;k gksxk \

(A) 6.22 Litre (B) 12.44 litre (C*) 74.67 litre (D) None of these buesa ls dksbZugha

Sol. nglucose

6 = 2COn (applying POAC)

180100

6 = 2COn = 3

10

Page No. # 4

volume of CO2 =

310

22.4 = 74.61 litre

Sol. nXyqdksl

6 = 2COn (POAC iz;qDr djus ij)

180100

6 = 2COn = 3

10

CO2 dk vk;ru =

310

22.4 = 74.61 litre

Organic SCQ (8)35. Which is the correct order of 2nd .E ?

2nd vk;uu Å tkZ dk lgh Øe dkSulk gSa ?(A) Si < S < P < Cl (B) Si < P < S < Cl (C) P < Si < Cl < S (D*) Si < P < Cl < S

Sol. Valence shell configurationonSi+ � 2s1

P+ � 2p2

S+ � sp3

Cl+ � sp4

Clearly S+ has half filled orbital.la;ksth dks'k foU;klvk;uSi+ � 2s1

P+ � 2p2

S+ � sp3

Cl+ � sp4

Li"V gS fd S+ esa v)Z iwfjr d{kd mifLFkr gSaA

36. Following compound are :

(A) Chain Isomers (B) Positional Isomers (C*) Functional Isomers (D) Metamers

fn;s x;s ;kSfxd gSa %

(A) Ja[kyk leko;oh (B) fLFkfr leko;oh (C*) fØ;kRed leko;oh (D) e/;ko;oh

37. How many of the following are tertiary amines ?fuEu esa ls fdrus dkcZfud ;kSfxd] rrh;d ,sehu gSa\

, , , , ,

, , ,

(A*) 3 (B) 4 (C) 5 (D) 6

Page No. # 5

Sol. , ,

38. Which of the following can not have any positional isomer ?fuEu esa ls fdl ;kSfxd esa fLFkfr leko;oh lEHko ugha gS ?

(A) (B) (C*) (D)

39. Member of which of the following pair of isomers is metamers ?

fuEu esa ls dkSulk leko;oh ;qXe e/;ko;oh gS \

(A*) & (rFkk) (B) & (rFkk)

(C) & (rFkk) (D) & (rFkk)

Sol. & are metamers

rFkk e/;ko;oh gSA

40. Among F, F�, O and O�2, the species with smallest radius is :

(A*) F (B) F� (C) O (D) O2�

F, F�, O o O�2 esa ls fuEure~ f=kT;k j[kus okyh Lih'kht dkSulh gS %(A) F (B) F� (C) O (D) O2�

Sol. Correct order of radius : F < O < F� < O�2. So, largest species is O2�.

f=kT;k dk lgh Øe : F < O < F� < O�2. vr% lcls cM+h Lih'kht O2� gSA

41. Which of the following is correct order of increasing Z-effective ?

fuEu esa ls Z- izHkko ds c<+us dk lgh Øe gS ?(A*) O2� < F� < Na+ < Mg2+ (B) O2� > F� > Na+ > Mg2+

(C) O2� > F� < Na+ < Mg2+ (D) O2� < F� < Na+ > Mg2+

42. From the ground state electronic configurations of the elements given below, pick up to one with highestvalue of second ionization enthalpy :

uhps rRoksa ds vk| voLFkk bysDVªkWfud vfHkfoU;kl fn;s x;s gSaA buesa ls ml rRo ds vfHkfoU;kl dk p;u fdft;s ftlds

fy, f}rh; ,UFkSYih dk eku vf/kdre gS %(A) 1s2 2s2 2p6 3s2 (B*) 1s2 2s2 2p6 3s1 (C) 1s2 2s2 2p6 (D) 1s2 2s2 2p5.

Sol. Here, IE2 involves the removal of an electron from the inert gas configuration (1s2 2s2 p6).

gy ;gk¡ IE2 ls rkRi;Z vfØ; xSl vfHkfoU;kl (1s2 2s2 p6) ls ,d bysDVªkWu fudkyus ds fy, vko';d Å tkZ ls gSA

Page No. # 6


This section contains 3 paragraphs. Based upon each paragraph, there are 2 questions. Each question has4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

[k.M - IIcks/ku çdkj

bl [k.M esa 3 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 2 ç'u gSA çR;sd ç'u ds 4 fodYi(A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA

Physical Comprehension : (2)Paragraph for Question Nos. 43 to 44

iz'u 43 ls 44 ds fy, vuqPNsnH

2SO

4 is one of the important chemical reagent.

On large scale it is prepared by contact process.Reactions involved are.

(i) S8 + O

2 SO

2

(ii) Oxidation of SO2

2SO2 + O

2

catalyst 2 SO3

(iii) SO3 + H

2O H

2SO

4

Assume 100% yield of all the reactions.H

2SO

4 ,d egRoiw.kZ jklk;fud vfHkdeZd gSA

ogn~ ek=kk esa bldk fuekZ.k lEidZ fof/k }kjk fd;k tkrk gSAblesa fughr vfHkfØ;k;sa fuEu gSa &

(i) S8 + O

2 SO

2

(ii) SO2 dk vkWDlhdj.k

2SO2 + O

2

catalyst 2 SO3

(iii) SO3 + H

2O H

2SO

4

ekuk dh lHkh vfHkfØ;kvksa ds fy;s yfC/k 100% gSA

43. How many gram of S8 is required to obtain 24.5 Kg of H

2SO

4 :

H2SO

4 ds 24.5 Kg izkIr djus ds fy;s fdrus xzke S

8 dh vko';drk gS %

(A*) 8 Kg (B) 500 gm (C) 800 gm (D) 8000 Kg

Sol. Mole of H2SO

4 =

98105.24 3

= 250

mole of S8 =

81

250

weight of S8 =

81

250 32 8 = 8000 gm

Sol. H2SO

4 ds eksy =

98105.24 3

= 250

S8 ds eksy =

81

250

S8 dk Hkkj =

81

250 32 8 = 8000 gm

Page No. # 7

44. Calculate volume of air at STP containing 20% O2 by mole to oxidise 5 mole of SO

2 in step (ii)

in (ii) esa] SO2 ds 5 eksy dks vkWDlhd r djus ds fy;s] eksy ls 20% O

2 j[kus okyh ok;q ds vk;ru dh x.kuk dhft;sA

(STP ij)

(A) 2000 litre (B) 2

25litre (C*) 280 litre (D) None of these buesa ls dksbZ

ugh

Sol. mole of O2 =

25

mole of air = 2

25

volume of air = 2

25 22.4 = 280 litre

Sol. O2 ds eksy =

25

ok;q ds eksy = 2

25

ok;q dk vk;ru = 2

25 22.4 = 280 litre


If we want to find, which energy level of one H-like species (A) has same energy as energy level of another H-like species (B), then we can simply equate their energies.;fn ge Kkr djuk pkgrs gS fd ,d H-leku Lih'khT+k (A) ds dkSuls Å tkZ Lrj dh Å tkZ] ,d vU; H-leku Lih'khT+k (B)

ds Å tkZ Lrj ds leku gS] rc ge budh Å tkZvksa dks cjkcj dj nsrs gSA

i.e. (vFkkZr) �E0 × 2

2

A

A

n

Z = �E0

2

2

B

B

n

Z

A

A

nZ

= B

B

nZ

nA = B

A

ZZ

nB

Hence the energy level in He+ having same energy as 2nd energy level of H-atom is :vr% He+ esa Å tkZ Lrj ; H-ijek.kq ds 2nd Å tkZ Lrj ds leku Å tkZ j[krk gS :

n = 12

× 1 = 2.

On this basis, answer the following. (bl vk/kkj ij fuEu iz'uksa ds mÙkj nhft,A)

45. The transition of an electron in the Li2+ ion from n2 to n1 would have the same wavelength as the transitionof electron from n = 4 to n = 2 of the He+ ion. What is the value of (n1 + n2) ?

Li2+ vk;u esa n2 ls n1 esa ,d bysDVªkWu ds laØe.k ds fy, rjax}S/;Z dk eku He+ vk;u esa n = 4 ls n = 2 rd bysDVªkWu

ds laØe.k ds leku gksrk gS] rks (n1 + n2) dk eku Kkr dhft, \(A) 5 (B) 6 (C) 8 (D*) 9

Sol. The transition of an electron in the Li2+ ion from n2 = 6 to n1 = 3 would have the same wavelength as thetransition of electron from n = 4 to n = 2 of the He+ ion.

Page No. # 8

n1 + n2 = 6 + 3 = 9

Sol. Li2+ vk;u esa n2 = 6 ls n1 = 3 esa ,d bysDVªkWu ds laØe.k ds fy, rjax}S/;Z dk eku He+ vk;u esa n = 4 ls n = 2 rd

bysDVªkWu ds laØe.k ds leku gksrk gSAn1 + n2 = 6 + 3 = 9

46. Which electronic in 2He+ will produce same energy photon as is produced by 4Be3+ from 4 to 2 ?

(A) 8 4 (B*) 2 1 (C) 16 8 (D) None of these

2He+ esa dkSuls bysDVªkWfud laØe.k ls izkIRk QksVksu dh Å tkZ 4Be3+ esa 4 ls 2 laØe.k }kjk izkIr QksVksu dh Å tkZ ds leku

gksrh gS ?

(A) 8 4 (B*) 2 1 (C) 16 8 (D) buesa ls dksbZ ugha

Sol. 4 3

4Be2,

42

× 4 He2

42

× 2

= 2 = 1So, 2 1.

Organic Comprehension : (1) (1 × 2)Paragraph for Question Nos. 47 to 48

iz'u 47 ls 48 ds fy, vuqPNsnThe amount of energy required to remove the most loosely bound electron from an isolated gaseous atom iscalled as first ionization energy (E1). Similarly the amount of energies required to knock out second, thirdetc. electrons from the isolated gaseous cations are called successive ionization energies and E3 > E2 >E1.(i) Nuclear charge (ii) Atomic size (iii) penetration effect of the electrons (iv) shielding effect of the innerelectrons and (v) electronic configurations (exactly half filled & completely filled configurations are extrastable) are the important factors which affect the ionisation energies.Similarly the amount of energy released when a neutral isolated gaseous atom accepts an extra electron toform gaseous anion is called electron affinity.

X (g) + e� (g) X� (g) + energy

A positive electron affinity indicates that the ion X� has a lower more negative energy than the neutral atom X.The second electron affinity for the addition of a second electron to an initially neutral atom is negativebecause the electron repulsion out weighs the nuclear attraction, e.g.,

O(g) + e� Exothermic O�(g) ; Ea = + 141 kJ mol�1 .............. (i)

O�(g) + e� cEndothermi O2�(g) ; Ea = � 780 kJ mol�1 .............. (ii)

The electron affinity of an element depends upon (i) atomic size (ii) Nuclear charge & (iii) electronic configuration.In general, ionisation energy and electron affinity increases as the atomic radii decrease and nuclear chargeincreases across a period. In general, in a group, ionisation energy and electron affinity decrease as theatomic size increases.The members of third period have some higher (e.g. S and Cl) electron affinity values than the members ofsecond period (e.g. O and F) because second period elements have very small atomic size. Hence there isa tendency of electron-electron repulsion, which results in less evolution of energy in the formation ofcorresponding anion.

Page No. # 9

vuqPNsn # 2 ,d foyfxr xSlh; ijek.kq ds <hys ca/ks bysDVªkWu dks iFkd~ djus ds fy, vko';d Å tkZ dh ek=kk dks izFke vk;uuÅ tkZ (E1) dgrs gSA blh izdkj foyfxr xSlh; /kuk;u ls f}rh;] rrh; vkfn bysDVªkWu dks iFkd~ djus ds fy, vko';dÅ tkZ dh ek=kk dks Øekxr vk;uu Å tkZ,sa dgrs gS vkSj E3 > E2 > E1A(i) ukfHkdh; vkos'k (ii) ijek.oh; vkdkj (iii) bysDVªkWu dk Hksnu izHkko (iv) vkrafjd bysDVªkWu dk ifjj{k.k izHkko vkSj (v)

bysDVªksfud foU;kl ¼v)Ziwfjr vkSj iw.kZiwfjr foU;kl dk vfrfjDr LFkk;hRo½ vk;uu foHko dks izHkkfor djrs gSAblh izdkj tc ,d mnklhu foyfxr xSlh; ijek.kq ,d vfrfjDr bysDVªkWu xzg.k dj xSlh; _ .kk;u cukrk gS vkSj tks Å tkZfudyrh gS] mls bysDVªkWu ca/kqrk dgrs gSA

X (g) + e� (g) X� (g) + Å tkZ

/kukRed bysDVªkWu cU/kqrk ;g n'kkZrh gS fd X� vk;u mnklhu ijek.kq X dh rqyuk esa vf/kd _ .kkRed Å tkZ j[krk gSA mnklhuijek.kq esa nwljk bysDVªkWu tksM+us ij f}rh; bysDVªkWu cU/kqrk dk eku _ .kkRed gksrk gS D;ksafd bysDVªkWu izfrd"kZ.k ukfHkdh;vkd"kZ.k dh rqyuk esa vf/kd gks tkrk gSA

O(g) + e� ek{kssihÅ " O�(g) ; Ea = + 141 kJ mol�1 .............. (i)

O�(g) + e� khkks"ek'Å " O2�(g) ; Ea = � 780 kJ mol�1 .............. (ii)

,d rRo dh bysDVªkWu ca/kqrk (i) ijek.oh; vkdkj (ii) ukfHkdh; vkos'k vkSj (iii) bysDVªkWfud foU;kl ij fuHkZj djrh gSAlkekU;r% ,d vkorZ esa vk;uu Å tkZ vkSj bysDVªkWu ca/kqrk ijek.oh; f=kT;k ?kVus ij vkSj ukfHkdh; vkos'k c<+us ij c<+rhgSA ,d oxZ esa ijek.oh; vkdkj c<+us ij vk;uu Å tkZ vkSj bysDVªkWu ca/kqrk ?kVrh gSA rrh; vkorZ ds rRoksa (mnk- S vkSjCl) dh bysDVªkWu ca/kqrk dk eku f}rh; vkorZ ds rRoksa (mnk- O vkSj F) ls T;knk gksrk gS D;ksafd f}rh; vkorZ ds rRoksa dkijek.oh; vkdkj vR;f/kd NksVk gksrk gS vr% buesa bysDVªkWu&bysDVªkWu ds e/; izfrd"kZ.k gksrk gSA ftlds ifj.kkeLo:ilacafèkr _ .kk;u ds fuekZ.k esa de Å tkZ mRlftZr gksrh gSA

47. Which order is incorrect for first Ionisation enthalpy ?izFke v;uu Å tkZ ds fy, dkSulk Øe xyr gS \(A) F > Cl > Br > I(B*) S > O > Se > Te(C) N > C > Be > B(D) Si > Mg > Al > Na

Sol. O > S > Se > Te

48. Which is energy releasing process ?Å tkZ fu"dkflr izØe dkSulk gS \(A) Cl�

(g)Cl

(g) + e� (B) O�

(g) O

(g) + e�

(C) O(g)

O+(g)

+ e� (D*) O(g)

+ e� O�

(g)

Sol. O(g)

+ e� O�

(g is energy releasing process. (;g Å tkZ fu"dkflr izØe gSA)

Page # 1

TEST PATTERN

CT-02 Date : 1/9/2013

Syllabus : FOM-1, Quadratic Equation, Set-Relation & Function, sequence & series


This section contains 18 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONLY ONE is correct.

[k.M- I

lh/ks oLrqfu"B izdkj

bl [k.M esa 18 iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls flQZ ,d lgh gSA

49. A quadratic equation whose roots are

a

a a b is

f}?kkr lehdj.k ftlds ewy

a

a a b gS&

(A) 2 2ax b bx b 0 (B*) 2 2bx 2a ax a 0

(C) ax2 � abx + b = 0 (D) 2 2 2a x abx b 0

Sol. On rationalizing, the roots areifjes; djds djus ij ewy

baab

a

S = b

aa2, P =

b

a2

50. If a + b + c = 0 then ab

c

ca

b

bc

a

xxx

222

.. is equal to

;fn a + b + c = 0 gks rks ab

c

ca

b

bc

a

xxx

222

.. cjkcj gS&(A) 1 (B) x (C) x2 (D*) x3

Sol. L.H.S. is ck;ka Hkkx 33333

xxx abc

abc

abc

cba

a3 + b3 + c3 = 3abc

when tc a + b + c = 0

51. If x3 + 3x2 - 9x + c is of the form xx 2 then c can be equal to


1 to 18 SCQ 18 3 �1 54


25 to 42 SCQ 18 3 �1 54


49 to 66 SCQ 18 3 �1 54


72 216

Paper-1

Total

Paper-2

Total

Physics

Chemistry

Maths

Page # 2

;fn x3 + 3x2 - 9x + c dk :i xx 2 gksrks c dk eku cjkcj gks ldrk gS&

(A*) �27 (B) 27 (C) �5 (D) 15Sol. If f(x) = 0, has a root of order n then f�(x) = 0 has the same root of order n � 1.

;fn f(x) = 0, n de dk ,d ewy gS rc f�(x) = 0, n � 1 Øe dk leku ewy j[krk gS

f(x) = x3 + 3x2 � 9x + c is of the form xx 2 , showing that is a double root so that f�(x) = 0

f(x) = x3 + 3x2 � 9x + c dk :i xx 2 , tks ds iqujkofÙk ewy dks f�(x) = 0

i.e., 3x2 + 6x � 9 = 0

or x2 + 2x � 3 = 0

or (x + 3) (x � 1) = 0

has the root once which can be either �3, or 1.

dk ,d ewy gS rc �3, ;k 1 gks ldrk gSAIf = 1 then f(x) = 0 gives c � 5 = 0 or c = 5

;fn = 1 rc f(x) = 0 fn;k gS c � 5 = 0 ;k c = 5If = -3 then f(x) = 0 gives

;fn = -3 rc f(x) = 0 fn;k gS�27 + 27 + 27 + c = 0

c = � 27

52. If n(A) = 3, n(B) = 6 and BA , then the number of elements in BA is equal to

;fn n(A) = 3 rFkk n(B) = 6 vkSj BA , rc BA esa vo;oksa dh la[;k cjkcj gS&(A) 2 (B*) 3 (C) 4 (D) 6

Sol. Since BA

pwafd BA

BA = AA

BAn = n(A) = 3

53. Range of y = 2

2

x4�x31

4�x3x

, x R, is

y = 2

2

x4�x31

4�x3x

, x R dk ifjlj gS&

(A) R (B)

4

1�,1� (C*) R �

4

1�,1� (D) none of these buesa ls dksbZ ughaA

54. If ,cbacb

bcx

ca

acx

ba

abx

then x =

;fn ,cbacb

bcx

ca

acx

ba

abx

rc x =

(A) a (B*) ba (C) abc (D) none of these buesa ls dksbZ ughaA

Sol.

0c

ba

abx

or ;k 0

ba

cabcabx

or abx [constant] = 0

;k abx [vFkkZr] = 0

Page # 3

55. The sum of n terms of a G.P. is n

n

2

1

4

33

then the common ratio is equal to

xq.kksÙkj Js.kh ds n inksa dk ;ksxQy n

n

2

1

4

33

gS rc lkoZvuqikr cjkcj gS&

(A*) 16

3(B)

256

3(C)

256

39(D) none of these buesa ls dksbZ

ughaA

Sol. Sn = n

n

2

1

4

33

. Putting j[kus ij n = 1, 2

T1 = S

1 =

16

93 =

16

39

S2 =

256

273 = TT

1 + T

2

T2 = S

2 � T

1

r = 1

2

T

T =

1172563916

= 117 16256 39

=

16

3

56. The sum of n terms of the series

........2

1

2

11

2 is less than 1.999. The greatest value of n is

Js.kh ........2

1

2

11

2 ds inksa dk ;ksxQy 1.999. ls de gS rc n dk vf/kdre eku gS&

(A*) 10 (B) 11 (C) 9 (D) 5

Sol.

n1

1. 12

1.9991

12

1 �

n

21

<

20001999

n

21

>

20001

2n < 2000 n 10

57. The A.M. of two positive numbers exceeds their G.M. by 15 and H.M. by 27. The numbers are

nks /kukRed la[;kvksa dk lekUrj ek/;] muds xq.kksÙkj ek/; ls 15 vf/kd gS rFkk gjkRed ek/; ls 27 vf/kd gS rcla[;k,sa gS&

(A) 100, 50 (B*) 120, 30 (C) 90, 60 (D) none of these buesa ls dksbZughaA

Sol. Let the numbers be a, b and their A.M., G.M. and H.M. be denoted by A , G and H respectively.

Page # 4

also we know that A, G, H are in G.P.

or G2 = AH .....................(1);k G2 = AH .....................(1)Since A - G = 15 and A - H = 27pwafd A - G = 15 vkSj A - H = 27

(A - 15)2 = G2 = AH = A (A - 27)

or ;k A = 75 = 2

ba

a + b = 150 ....................(2)Since A - G = 15pwafd A - G = 15 75 - G = 15

or G = 60 = ab

;k G = 60 = abab = 3600 .....................(3)

Hence from (2) and (3) we conclude that a and b are the roots oft2 � 150t + 3600 = 0

vr% (2) o (3) ls fu"d"kZ fudyrk gS fd a vkSj b lehdj.kt2 � 150t + 3600 = 0 ds ewy gS&

or (t - 120) (t - 30) = 0;k (t - 120) (t - 30) = 0 t = 120, 30

58. If a, b, c are distinct and are in H.P., then a2 (b - c)2, 2222

,4

bacacb

are in

;fn a, b, c fofHkUu gS rFkk gjkRed Js.kh esa gS rc a2 (b - c)2, 2222

,4

bacacb

gS&

(A) H.P. (B) G.P. (C) A.P. (D*) All of these mijksDr lHkh

Sol. b = ca

ac

2

or ;k b (a + c) = 2ac ...................(1)T

1 + T

3= b2 (a2 + c2) - 4a2c2 + 2a2c2 by (1)= b2 (a2 + c2) - 2a2c2

= b2 (a2 + c2) - 2.b2

4

2ca by (1)

= 2222

22

cacab

= 22

2

.22

Tacb

T1, T

2, T

3 are in A.P.

59. The solution set of the inequality ||x| � 1| < 1 � x is equal to

vlfedk ||x| � 1| < 1 � x dk gy leqPp; cjkcj gS&

(A) ,0 (B) ,1

(C) 1,1 (D*) 0,Sol. Here because of |x| which is equal to either x or �x we shall consider the cases x > 0 or x < 0

Page # 5

;gk¡ D;kasfd |x| tks x ;k �x ds cjkcj gSA blfy, nks fLFkfr;k¡ x > 0 ;k x < 0Then we will have |x - 1| = |x + 1| when x < 0rc |x - 1| = |x + 1| tc x < 0

CasefLFkfr I : x > 0thenrks |x - 1| < (1 - x)

or;k |1 - x| < (1 - x) i.e. |t| < tAbove is not true for any value of t.t ds fdlh eku ds fy, lR; ugha gSACasefLFkfr II : x < 0,|-x - 1| < 1 - xor;k |x + 1| < (1 - x) ................(1)

If ;fn x + 1 > 0 i.e. x > -1 and vkSj x < 0

i.e. �1 < x < 0 i.e. x [-1, 0) thenrks lehdj.k (1) reduces to lsx + 1 = 1 - x

or;k 2x < 0 or ;k x < 0If x + 1 < 0 i.e. x < -1 then (1) reduce to;fn x + 1 < 0 i.e. x < -1 rks lehdj.k (1) ls- (x + 1) < (1 - x) or -2 < 0 is true for all x < - 1 ...........(2)- (x + 1) < (1 - x) ;k -2 < 0 lHkh x < - 1 ds fy, lR; gSA ...........(2)

Hence from (1) and (2) 0,x

vr% (1) o (2) ls 0,x

60. If Rx , the solution set of the in-equation 042.74 5.0 xx is equal to

;fn Rx vlfedk 042.74 5.0 xx dk gy leqPp; cjkcj gS&

(A)

2

7,2 (B*) ,2 (C) ,2 (D) ,

Sol. Put 2-x = t and 40.5 = 41/2 = 2j[kus ij 2-x = t vkSj 40.5 = 41/2 = 2 2t2 - 7t - 4 < 0 (t - 4) (2t + 1) < 0

42

1 t

But t = 2-x is +ive being exponential function henceijUrq t = 2-x /kukRed pj?kkrkadh Qyu gS&

422

1

x

2220 x

or ;k

2

2

1

2

1

2

1

x

Since pwafd 12

1

-2 < x <

,2x

61. Let A and B be two sets such that n A B = 6. If three elements of A x B are (3, 2), (7, 5), (8, 5), then

ekuk A vkSj B nks leqPp; bl izdkj gS fd n A B = 6. ;fn A x B ds rhu vo;o (3, 2), (7, 5), (8, 5) gS, rc

Page # 6

(A*) A = {3, 7, 8} (B) A = {2, 5, 7} (C) B = {2, 1} (D) none of these buesa ls dksbZ ughaASol. Since (3, 2), (7, 5), (8, 5) A x B, we have 3, 7, 8 A and 2, 5 B.

pwafd (3, 2), (7, 5), (8, 5) A x B ;gk¡ 3, 7, 8 A A vkSj 2, 5 B.Also n(A x B) = 6 = 3 x 2rFkk n(A x B) = 6 = 3 x 2 A = {3, 7, 8} and B = {2, 5}

A = {3, 7, 8} vkSj B = {2, 5}

62. Let A and B be two non-empty subsets of a set X such that A is not a subset of B, then

(A) A is always a subset of the complement of B(B) B is always a subset of A.(C) A and B are always disjoint.(D*) A and the complement of B are always non-disjoint.leqPp; X ds nks vfjDr mileqPp; A vkSj B bl izdkj gS fd A, B dk mileqPp; ugha gS] rc&(A) A lnSo B ds iwjd dk mileqPp; gSA(B) B lnSo A dk mileqPp; gSA(C) A lnSo B dk folaf?kr (disjoint) leqPp; gSA(D*) A lnSo B dk iwjd leqPp; folaf?kr leqPp; ugha gSA

Sol. A and the complement of B are always non-disjoint.A, B dk iwjd leqPp; lnSo folaf?kr leqPp; ugha gSA[ A is not a subset of B

A, B dk ,d mileqPp; ugha gSA some pt. of A will not be a point of B

A ds dqN fcUnqvksa ds fy, B dk dksbZ fcUnq ugha gksxkA that point will belong to Bc ]

tks Bc esa gSA]

63. Consider the equation x2 + x � n = 0, where n is an integer lying between 1 to 100. Total number of different

values of �n� so that the equation has integral roots is

ekuk lehdj.k x2 + x � n = 0 gS tgk¡ n, 1 ls 100 ds e/; ,d iw.kk±d gSA n ds fofHkUu ekuksa dh la[;k tcfd lehdj.kds iw.kk±d ewy gSa] gS&(A) 6 (B) 4 (C*) 9 (D) 8

Sol. x = 2

n411

As x is integer pawfd x iw.kk±d gSA 1 + 4n must be a perfect square of an integer1 + 4n ,d iw.kk±d dk oxZ gksxkA n = 2,6,12,20,30,42,56,,72,90 nine values of n. n ds 9 eku gSA

64. The equation 1x1x = x41 has

(A*) no solution (B) one solution (C) two solutions (D) more than two solutions

lehdj.k 1x1x = x41 ds gS&

(A*) dksbZ gy ugha (B) ,d gy (C) nks gy (D) nks ls vf/kd gy

Sol. Form domain of definition. izkUr dh ifjHkk"kk ls

x �1 and vkSj x 1 and vkSj x 41

Domain of definition izkUr dh ifjHkk"kk x

Equation has no solution. lehdj.k dk dksbZ gy ugha

65. If A = {1, 2, 3, 4} then which of the following is function from A to itself ?

Page # 7

;fn A = {1, 2, 3, 4} gks rc A ls A esa fuEu esa lsa dkSulk Qyu gS&

(A) 1:,1 xyyxf (B) 4:,2 yxyxf

(C) xyyxf :,3 (D*) 5:,4 yxyxf

Sol. 4,3,3,2,2,11 f

2f 1,4 , 2,3 , 2,4 ,(3,2),(3,3), 3,4 , 4,4 , 4,1 , 4,2 , 4,3

3,4,2,4,1,4,2,3,1,3,1,23 f

1,4,2,3,3,2,4,14 f f

4 is a function from A to itself.

f4, A ls A esa ,d Qyu gSA

and f1, f

2, f

3 are not functions from A to itself.

rFkk A ls A esa f1, f

2, f

3 Qyu ugha gSA

66. If a,b,c Q and a < c < b, then the roots of 012222 xcbaxba are

(A*) imaginary (B) rational (C) equal (D) irrational

;fn a,b,c Q vkSj a < c < b, rc 012222 xcbaxba ds ewy gS&

(A*) dkYifud (B) ifjes; (C) cjkcj (D) vifjes;

Sol. or ;k = 4 222 bacba

or ;k = 4 ,2222 cbca by L2 � M2

or ;k = 16 ivebcac

as c lies between a and bpwafd a o b ds e/; c fLFkr gSA

Roots are complex ewy lfEeJ gksxsaA


This section contains 3 paragraphs. Based upon each paragraph, there are 2 questions. Each questionhas 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

[k.M - IIcks/ku çdkj

bl [k.M esa 3 vuqPNsn (paragraphs) gSA çR;sd vuqPNsn ij vk/kkfjr 2 ç'u gSA çR;sd ç'u ds 4 fodYi(A), (B), (C) rFkk (D) gS , ftuesa ls flQZ ,d lgh gSA


If y = f(x) and y = g(x) are two functions, then number of solutions of the equation f(x) = g(x) is equal to thenumber of points of intersection of two curves y = f(x) and y = g(x).;fn y = f(x) vkSj y = g(x) nks Qyu gS rc lehdj.k f(x) = g(x) ds gyksa dh la[;k dks nks oØks y = f(x) vkSj y = g(x) dsizfrPNsn fcUnqvksa dh la[;k ls izkIr dh tkrh gSA

67. Number of solutions of the equation |2 � x| = x2 + x � 2 are

lehdj.k |2 � x| = x2 + x � 2 ds gyksa dh la[;k gS&(A) 0 (B) 1 (C*) 2 (D) 3

68. Number of solutions of the equation 2||x| � 1| = 1 are

lehdj.k 2||x| � 1| = 1 ds gyksa dh la[;k gS&(A) 1 (B) 2 (C) 3 (D*) 4

Page # 8

Sol. 67 68.


A survey shows that 63% of the Americans like cheese where as 76% like apples. If % of the Americanslike both cheese and apples then,d losZ esa 63% vesfjdu iuhj dks tcfd 76% lso dks ilUn djrs gS ;fn vesfjdu dk % nksuks iuhj o lso dks ilUndjrs gS rc

69. Which of the following is CORRECT ?fuEu esa ls dkSulk lgh gS \(A) < 39 (B) < 63 (C) > 63 (D*) 39 63

70. Number of Integral values are ds iw.kkZad ekuks dh la[;k gS&(A) 38 (B*) 25 (C) 37 (D) none of these buesa ls dksbZughaA

Sol. Let total number of americans be 100, let x, c, A be the sets of all amercians, American liking cheese,americans liking apples respectively.ekuk vesfjdu dh dqy la[;k 100 gS ekuk x, c, A Øe'k% lHkh vesfjdu dk leqPp;ksa gS] iuhj dks ilUn djus okysvesfjduksa dk leqPp; rFkk lsc dks ilUn djus okys vesfjduksa dk leqPp; gSA

n(x) = 100, n(c) = 63, n(A) = 76, n C A

n(A C) = n(A) + n(C) � n (A C)n(A C) = 76 + 63 �

n(A C) 100139 � 100 39

Also rc n(C A) n(A) 63Number of integral values of are 25.ds iw.kk±d ekuksa dh la[;k 25 gSA


If roots of ax3 + bx2 + cx + d = 0 are real and given by ,, then b

a ,

ac

andad

.

ekuk lehdj.k ax3 + bx2 + cx + d = 0 ds ewy okLrfod rFkk ,, gS rc b

a ,

ac

vkSjad

71. If ,, are in A.P. then;fn ,, lekUrj Js.kh esa gSA(A*) 2b3 - 9abc + 27a2d = 0 (B) 2b3 + 9abc - 27a2d = 0(C) 2b3 + 9abc + 27a2d = 0 (D) none of these

Page # 9

Sol.71. Let roots areekuk ewyA - D, A , A + DSum of rootsewyksa dk ;ksx

3A = �ab

A = �a3

b

putting the value of A in the given equation and simplifyingwe getnh xbZ lehdj.k esa A dk eku j[kus ij rFkk ljy djus ij2b3 - 9abc + 27a2d = 0

72. If ,, are in H.P. then;fn ,, gjkRed Js.kh esa gS rc(A) 27ad3 = 9bcd2 - 2c3d (B) 27ad3 = abcd2 - 2c3d(C) ad3 = bcd2 - c3d (D*) 27ad2 � 9bcd + 2c3 = 0

Sol..72.Let roots are DA

1

, A1

, DA

1

then

ekuk ewy DA

1

, A1

, DA

1

gS] rc

ac

DADA1

DA1

A1

A1

.DA

1

putting the value of d

cA

3 in the given equation and simplifying

we get 27ad2 � 9bcd + 2c3 = 0

gy djus ij d

cA

3 bl eku dks lehdj.k esa gy djus ij 27ad2 � 9bcd + 2c3 = 0

ct-2 (paper-1)_01_09_2013

Documents