paper 1 solutions

7/28/2019 Paper 1 Solutions

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i;k bu funsZ'kksa dks /;ku ls i


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PART - I PHYSICS

SECTION-1 : (Only One option correct type)

[k.M1 : (dsoy ,d lgh fodYi dkj)

This section contains 10 multiple choice qustions. Each question has four choices (A), (B), (C) and (D) out

of which ONLY ONE is correct.bl [k.M esa 10cgqfodYi 'u gSA R;sd 'u esa pkj fodYi (A), (B), (C)vkSj(D)gS] ftuesa ls dsoy ,d lgh gSA

1. A particle of mass m is projected from the ground with an initial speed u0at an angle with the horizontal. At

the highest point of its trajectory, it makes a completely inelastic collision with another identical particle,

which was thrown vertically upward from the ground with the same initial speed u0. The angle that the composite

system makes with the horizontal immediately after the collision is :

,d m nzO;eku ds d.k dks kjfEHkd xfr u0ls {kSfrt ls dks.k ij {ksfir fd;k tkrk gSA ;g d.k {ksI; iFk ds mPpre

fcUnq ij ,d leku nzO;eku ds d.k ds lkFk iw.kZr% vR;kLFk la?k djrk gS] tksfd Hkwry ls /okZ/kj fn'kk esa leku kjfEHkd

xfr u0ls Qsadk x;k FkkA la;qDr fudk; la?k ds rRdky ckn {kSfrt ls fuEu dks.k cuk,xk %

(A)4

(B) a

4

(C) a

4

(D)

4

Ans. (A)

Sol. At the highest point

v1

=2

cosu0 (by applying momentum conservation in horizontal direction)

v2

=2

cosu0 (by applying momentum conservation in vertical direction)

(H =g2

sinu 220 )

= 45Hindi mPpre fcUnq ij

v1

=2

cosu0 ({kSfrt fn'kk esa laosx laj{k.k yxkus ij)

v2

=2

cosu0 (/okZ/kj fn'kk esa laosx laj{k.k yxkus ij)

(H =

g2

sinu 220 )

= 45

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2. The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is

one-third the size of the object. The wavelength of light inside the lens is3

2times the wavelength in free

space. The radius of the curved surface of the lens is :

,d lery mky ySUl ,d okLrfod frfcEc ySUl ds 8m ihNs cukrk gS tks fd oLrq ds vkdkj dk ,d&frgkbZ gSA ySUl

ds vUnj dk'k dh rjaxnS/;Z fuokZr dh rjaxnS/;Z ls32 xquk gSA ySUl ds xksyh; ofr i"B dh ork f=kT;k gS %

(A) 1 m (B) 2 m (C) 3 m (D) 6 m

Ans. (C)

Sol. v = 8 m (magnification =3

1 =

u

v)

u = 24m

R

111

2

3

f

1

R = 3m

Hindi v = 8 m (vko/kZu =3

1 =

u

v)

u = 24m

R

111

2

3

f

1

R = 3m

3. The diameter of a cylinder is measured using a vernier callipers with no zero error. It is found that the zero of

the vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The vernier scale has 50 division

equivalent to 2.45 cm. The 24th division of the vernier scale exactly coincides with one of the main scale

divisions. The diameter of the cylinder is :

,d csyu dk O;kl ekius ds fy, 'kwU; =kqfV jfgr ,d ofuZ;j dSfyilZ dk mi;ksx gksrk gSA ekius ds nkSjku iSekus dk 'kwU;]

eq[;k iSekus ds 5.10 cm vkSj 5.15 cm ds chp esa ik;k tkrk gSA ofuZ;j iSekus 50 Hkkx 2.45 cm ds rqY; gSA bl ofuZ;j iSekus

dk pkSchlok (24th) Hkkx eq[; iSekus ds ,d Hkkx ls lVhd lEikrh gksrk gSA csyu dk O;kl gS %(A) 5.112 cm (B) 5.124 cm (C) 5.136 cm (D) 5.148 cm

Ans. (B)

Sol. 50 VSD = 2.45 cm

1 VSD =50

45.2cm = 0.049 cm

Least count of vernier = 1MSD 1 VSD

= 0.05 cm 0.049 cm

= 0.001 cm

Thickness of the object = Main scale reading + vernier scale reading least count

= 5.10 + (24) (0.001)

= 5.124 cm.

Hindi. 50 VSD = 2.45 cm

1 VSD =50

45.2cm = 0.049 cm

ouhZ;j dk vYirekad = 1MSD 1 VSD= 0.05 cm 0.049 cm

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= 0.001 cm

oLrq dh eksVkbZ = eq[; iSekus dk ikB~;kad + ofuZ;j iSekus dk ikB~;kad vYirekad= 5.10 + (24) (0.001)

= 5.124 cm.

4. The work done on a particle of mass m by a force, K

j

yx

y

i

yx

x

2/3222/322 (K being a constant of

appropriate dimensions), when the particle is taken from the point (a, 0) to the point (0, a) along a circular path

of radius a about the origin in the x-y plane is :

,d cy

j

)yx(

yi

)yx(

xK

2/3222/322 (K ,d mfpr foek dk fLFkjkad gS), ,d m nzO;eku ds d.k dks(a,0) fcUnq ls

(0,a) fcUnq rd ,d a f=kT;k ds okh; iFk ij ys tkrk gS] ftldk dsUnz xy ry dk ewy fcUnq gSA bl cy }kjk fd;k x;k

dk;Z fuEu gS :

(A)aK2 (B)

aK (C)

a2K (D) 0

Ans. (D)

Sol. suppose x = r cosy = r sin

force on particle is jsinricosrr

K3

force is in radial direction so work done by this force along given path (circle) is zero.

Hindi. ekuk fd x = r cosy = r sin

d.k ij cy jsinricosrr

K3

gS

cy f=kT; fn'kk esa gS vr% fn;s x;s okh; iFk ds vuqfn'k bl cy }kjk fd;k x;k dk;Z 'kwU; gS

5. One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal

thin copper wire of length L and radius R. When the arrangement is stretched by a applying forces at two ends,

the ratio of the elongation in the thin wire to that in the thick wire is :

,d 2L yEckbZ o 2R f=kT;k ds eksVs {kSfrt rkj ds ,d fljs dksL yEckbZ o R f=kT;k okys ,d irys {kSfrt rkj ls osfYMax

ds }kjk tksM+k x;k gSA bl O;oLFkk ds nksauks fljksa ij cy yxkdj rkuk tkrk gSA irys o eksVs rkjksa esa rjaxnS/;Z of) dk vuqikrfuEu gS %(A) 0.25 (B) 0.50 (C) 2.00 (D) 4.00

Ans. (C)

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Sol. Y =

L

A

F

1

...(i)

Y =

L2

A4F

2

...(ii)

2

1

= 2

6. A ray of light travelling in the direction

j3i

2

1 is incident on a plane mirror. After reflection, it travels along

the direction j3i2

1 . The angle of incidence is :

,d lery niZ.k ij vkifrr dk'k fdj.k dh xkeh fn'kk j3i2

1 gSA ijkorZu ds ckn xkeh fn'kk j3i

2

1 gks

tkrh gSA fdj.k dk vkiru dks.k gS :(A) 30 (B) 45 (C) 60 (D) 75

Ans. (A)

Sol. Angle between given rays is 120

so angle of incidence is 30

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Hindi. foU;kl 1 esa rqY; rkih; izfrjks/k2

R3gS

foU;kl 2 esa rqY; rkih; izfrjks/k3

RgS

rkih; izfrjks/k mPp rki ls fuEu rki dh vksj "ek izokg esa fy;k x;k le;

8. A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse

is 30mV and the speed of light is 3 10 ms1. The final momentum of the object is :

,d NksVh oLrq] tks kjEHk eas fojke voLFkk eas gS] dk'k dh 100 ns dh ,d LiUn dks iw.kZr;k vo'kksf"kr djrh gSA LiUn dh

'kfDr 30 mW gS o dk'k dh xfr 3 108 ms1gSA oLrq dk vfUre laosx gS :(A) 0.3 1017 kg ms1 (B) 1.0 1017 kg ms1

(C) 3.0 1017 kg ms1 (D) 9.0 1017 kg ms1

Ans. (B)

Sol. Change in momentum = lightofspeed

timetotalpower = c

tP

1.0 1017 kg m/s

Hindi. laosx esa ifjorZu =dh pkykizdk'

le;dqykfDr' =

c

tP

1.0 1017 kg m/s

9. In the Young's double slit experiment using a monochromatic light of wavelength , the path difference (interms of an integer n) corresponding to any point having half the peak intensity is :

,d ;ax f}&fLyV ;ksx esa rjaxnS/;Z ds ,do.khZ dk'k dk ;ksx fd;k tkrk gSA ,sls fcUnq dk ftl ij dk'k dh rhozrk

f'k[kj rhozrk dh vk/kh gS] iFkkUrj gS (iw.kkZad n ds inksa eas) :

(A) 2

1n2

(B) 4

1n2

(C) 8

1n2

(D) 16

1n2

Ans. (B)

Sol. For half of maximum intensity

20=

0+

0+ 2

0cos

(Phase difference) = .........2

5,

2

3,

2

Path difference is

41n2.........

45,

43,

4

Hindi. vf/kdre rhozrk dh vk/kh ds fy,2

0=

0+

0+ 2

0cos

(dykarj) = .........2

5,

2

3,

2

iFkkarj

4

1n2.........

4

5,

4

3,

4

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10. Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partial

pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their densities is:

nks vufHkf;k'khy ,d&ijek.kqd vkn'kZ xSlksa dk ijek.kq nzO;eku 2 : 3 ds vuqikr esa gSA tc budks ,d fLFkjrkih; crZu eas

ifjc) fd;k tkrk gS] rc buds vakf'kd nkcksa dk vuqikr 4 : 3 gSA buds ?kuRo dk vuqikr gS:(A) 1 : 4 (B) 1 : 2 (C) 6 : 9 (D) 8 : 9

Ans. (D)

Sol. P1

=1

1

M

RT...(i)

P2

=2

2

M

RT...(ii)

by (i) and (ii)

9

8

2

1

Hindi. P1

=1

1

M

RT...(i)

P2

=2

2

M

RT...(ii)

(i) o (ii) ls

9

8

2

1

SECTION-2 : (One or more option correct type)

[k.M2 : (,d ;k vf/kd lgh fodYi dkj)

This section contains 5 multiple choice qustions. Each question has four choices (A), (B), (C) and (D) out

of which ONE or MORE are correct.

bl [k.M esa 5cgqfodYi 'u gSA R;sd 'u esa pkj fodYi (A), (B), (C)vkSj(D)gS] ftuesa ls dsoy ,d ;k vf/kdlgh gSA

11. Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities 1and

2respectively,

touch each other. The net electric field at a distance 2R from the centre of the smaller sphere, along the line

joining the centres of the spheres, is zero. The ratio2

1

can be ;

nksR o 2R f=kT;k okys vpkyd Bksl xksydks dks ftu ij e'k%1rFkk

2,dleku vk;ru vkos'k ?kuRo gS] ,d nwljs ls

Li'kZ djrs gq, j[kk x;k gSA nksauks xksydksa ds dsUnzksa ls xqtjrh gqbZ js[kk [khaph tkrh gSA bl js[kk ij NksVs xksyd ds dsUnz ls

2R nwjh ij usV fo|qr {ks=k 'kwU; gSA rc vuqikr2

1

dk eku gks ldrk gS :

(A) 4 (B) 25

32

(C) 2532

(D) 4

Ans. (B,D)

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Sol.

At point P

If resultant electric field is zero

then

RR8

KQ

R4

KQ3

2

2

1

2

1

= 4

At point Q

If resultant electric field is zero

then

0R25

KQ

R4

KQ2

2

2

1

2

1

= 25

32(

1must be negative)

Sol.

fcUnq P ij

;fn ifj.kkeh oS|qr {ks=k 'kwU; gS rc

RR8

KQR4

KQ 3221

2

1

= 4

fcUnq Q ij

;fn ifj.kkeh oS|qr {ks=k 'kwU; gS rc

0R25

KQ

R4

KQ2

2

2

1

2

1

= 25

32(

1_.kkRed gksuk vko';d gS )

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12. A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y(x,

t) = (0.01 m) sin [(62.8 m1) x] cos [(628 s1)t]. Assuming = 3.14, the correct statement(s) is (are) :

nksauks fljksa ij ifjc) {kSfrt rfur Mksjh ikpoh xq.kofk lehdj.k, y(x, t) = (0.01 m) sin [(62.8 m1) x] cos [(628 s1)t]

}kjk dfEir gks jgh gSA ;fn = 3.14 eku tk;s rc fuEu dFku lgh gS@gSa &(A) The number of nodes is 5.

(B) The length of the string is 0.25 m.

(C) The maximum displacement of the midpoint of the string its equilibrium position is 0.01 m.(D) The fundamental frequency is 100 Hz.

(A) fuLianks dh la[;k 5 gSA

(B) Mksjh dh yEckbZ 0.25 m gSA

(C) lkE;koLFkk ls Mksjh ds e/; fcUnq dk vf/kdre foLFkkiu 0.01 m gSA

(D) ewy vkofk 100 Hz gSA

Ans. (B,C)

Sol. (A) There are 5 complete loops.

Total number of nodes = 6

(B) = 628 sec1

k = 62.8 m1 =2

10

1

vw

=k

=

8.62

628= 10 ms1

L = 25.02

5

(C) 2A = 0.01 = maximum amplitude of antinode

(D) f =25.02

10

2

v

= 20 Hz.

Sol. (A) ;gk ij 5 iw.kZ ywi gSAa

fuLiUnksa dh dqy la[;k = 6(B) = 628 sec1

k = 62.8 m1 =2

10

1

vw

=

k

=

8.62

628= 10 ms1

L = 25.02

5

(C) 2A = 0.01 = izLiUnksa dk vf/kdre vk;ke

(D) f =25.02

10

2

v

= 20 Hz.

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13. In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch S1

is pressed first to fully charge the capacitor C1and then released. The switch S

2is then pressed to charge the

capacitor C2. After some time, S

2is released and then S

3is pressed. After some time.

fp=k eas n'kkZ;s ifjiFk eas] nks lekukUrj IysVksa okys la/kkfj=kksa esa R;sd dh /kkfjrk C gSA kjEHk eas fLop S1dks nck;k tkrk gS

rkfd la/kkfj=k C1iw.kZ :i ls vkosf'kr gks tk,A blds ckn S

2dks NksM+ fn;k tkrk gSA blds i'pkr~ la/kkfj=k C

2dks vkosf'kr

djus ds fy;s fLop S2dks nck;k tkrk gSA dqN le; ds ckn S

2dks NksM+k fn;k tkrk gS rFkk S

3dks nck;k tkrk gSA dqN

le; ckn &

(A) the charge on the upper plate of C1is 2CV

0(B) the charge on the upper plate of C

1is CV

0

(C) the charge on the upper plate of C2is 0 (D) the charge on the upper plate of C

2is CV

0

(A) C1dh ijh IysV ij 2CV0vkos'k gSA (B) C1dh ijh IysV ij CV0gSA(C) C

2dh ijh IysV ij 'kwU; vkos'k gSA (D) C

2dh ijh IysV ij CV

0vkos'k gSA

Ans. (B,D)

Sol. When switch S1

is released charge on C1is 2CV

0(on upper plate )

When switch S2is released charge on C

1is CV

0(on upper plate ) and charge on C

2is CV

0(on upper plate)

When switch S3is released charge on C

1is CV

0(on upper plate ) and charge on C

2is CV

0(on upper plate)

Hindi. tc fLop S1nck;k tkrk gS] C

1ij vkos'k 2CV

0gS (ijh IysV ij )

tc fLop S2nck;k tkrk gS] C

1ij vkos'k CV

0gS (ijh IysV ij ) rFkk C

2ij vkos'k CV

0gS (ijh IysV ij)

tc fLop S3nck;k tkrk gS] C

1ij vkos'k CV

0gS (ijh IysV ij ) rFkk C

2ij vkos'k CV

0gS (ijh IysV ij)

14. A particle of mass M and positive charge Q, moving with a constant velocity 11 msi4u

, enters a region of

uniform static magnetic field normal to the x-y plane. The region of the magnetic field extends from x = 0 to x

= L for all values of y. After passing through this region, the particle emerges on the other side after 10

milliseconds with a velocity ji32u2

ms1. The correct statement(s) is (are) :

,d M nzO;eku rFkk Q /ku vkos'k dk d.k] tks 11 msi4u

ds ,dleku osx ls xfr'khy gS] ,dleku fLFkj pqEcdh; {ks=k

eas x-y ry ds vfHkyEcor~ gS rFkk bldk foLrkj {ks=k x = 0ls x = Lrd R;sd y ds eku ds fy, gSA bl pqEcdh; {ks=k dks

;g d.k 10 feyh lSd.M esa ikj dj nwljh vksj ji32u2

ms1osx ls dV gksrk gSA lgh dFku gS@gSa &

(A) The direction of the magnetic field is z direction.(B) The direction of the magnetic field is +z direction

(C) The magnitude of the magnetic fieldQ3

M50units.

(D) The magnitude of the magnetic field isQ3

M100units.

(A) pqEcdh; {ks=k z fn'kk esa gSA (B) pqEcdh; {ks=k +z fn'kk esa gSA

(C) pqEcdh; {ks=k dk ifjek.kQ3

M50bdkbZ gSA (D) pqEcdh; {ks=k dk ifjek.k

Q3

M100bdkbZ gSA

Ans. (A,C)

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Sol. Component of final velocity of particle is in positive y direction.

Centre of circle is present on positive y axis. so magnetic field is present in negative z-direction

Angle of deviation is 30 because

tan =x

y

v

v=

3

1

= 6

t =

= tM

QB

B =Qt

M

B =

Q3

M50

Hindi. d.k ds vfUre osx dk ?kVd /kukRed y fn'kk esa gSok dk dsU /kukRed y v{k ij mifLFkr gSA vr% pqEcdh; {ks=k _.kkRed z-fn'kk esa mifLFkr gS

fopyu dks.k 30 gS D;ksafd

tan =x

y

v

v=

3

1

=6

t =

= tMQB

B =Qt

M

B =

Q3

M50

15. A solid sphere of radius R and density is attached to one end of a mass-less spring of force constant k. Theother end of the spring is connected to another solid sphere of radius R and density 3. The complete arrangementis placed in a liquid of density 2 and is allowed to reach equilibrium. The correct statement(s) is (are)

(A) the net elongation of the spring isk3

gR4 3(B) the net elongation of the spring is

k3

gR8 3

(C) the light sphere is partially submerged. (D) the light sphere is completely submerged.

,d f=kT;k R f=kT;k ?kuRo okys Bksl xksyd dks ,d nzO;eku jfgr fLax ds ,d fljs ls tksM+k x;k gSA bl fLax dk cy

fu;rkad k gSA fLax ds nwljs fljs dks nwljs Bksl xksyd ls tksM+k x;k gS ftldh f=kT;k R o ?kuRo 3gSA iw.kZ foU;kl dks

2?kuRo ds nzo esa j[kk tkrk gS vkSj bldks lkE;koLFkk esa igqpus fn;k tkrk gSA lgh dFku gS@gSa &

(A) fLax dh usV nS/;Zof)k3

gR4 3gSA (B) fLax dh usV nS/;Zof)

k3

gR8 3gSA

(C) gYdk xksyd vakf'kd :i ls Mwck gqvk gSA (D) gYdk xksyd iw.kZ :i ls Mwck gqvk gSAAns. (A,D)

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Page # 13

Sol.

On small sphere

g)2(R34kxg)(R

34 33 ..(i)

on second sphere (large)

kxg)2(R3

4g)3(R

3

4 33 ...(ii)

by equation (i) and (ii)

x =k3

gR4 3

Hindi.

NksVs xksys ij

g)2(R3

4kxg)(R

3

4 33 ..(i)

nwljs cM+s xksys ij

kxg)2(R3

4g)3(R

3

4 33 ...(ii)

lehdj.k (i) o (ii) ls

x =k3

gR4 3

PHYSICS


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SECTION-3 : (Integer value correct Type)

[k.M3 : (iw.kkZad eku lgh dkj)

This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to

9 (both inclusive)

bl [k.M esa 5'u gSA R;sd 'u dk mkj 0ls9rd nksauks 'kfey ds chp dk ,dy vadh; iw.kkZad gSA

16. A uniform circular disc of mass 50kg and radius 0.4 m is rotating with an angular velocity of 10 rad/s1 about its

own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are gently

placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc

and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc

and the system rotates about the original axis. The new angular velocity (in rad s1) of the system is :

,d 50 kg o 0.4 m f=kT;k dh ,dleku fMLd viuh /okZ/kj v{k ds fxnZ10 rad/s1ds dks.kh; osx ls ?kwe jgh gSA nks ,dleku

okkdkj NYys /khj ls fMLd ij lefer rjhds ls ,d nwljs dks Nwrs gq, bl dkj fMLd ry ij j[ks tkrs gS fd os fMLd ds

v{k dks Hkh Li'kZ djsaA R;sd NYys dk nzO;eku 6.25 kg o f=kT;k 0.2 m gSA bl fudk; dk u;k dks.kh; osx (rad s

1esa) fuEugksxk (eku yhft, fd fMLd ,oe~ NYys ds chp ?k"kZ.k bruk gS fd fMLd o NYys ds chp lkis{k xfr 'kwU; gS vkSj fudk; ewy

v{k ij ?kw.kZu dj jgk gS)

Ans. 8

Sol.

1

1=

2

2

2= 1

2

1

=5

4 10 rad/s = 8 rad/s

17. The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stopping

potential versus frequency plot for Silver to that of Sodium is :

pkWnh ,oa lksfM;e ds dk;Z Qyu e'k% 4.6 o 2.3 eV gSA pkWnh o lksfM;e ds fujks/kh foHko ,oa vkofk ds chp xzkQk sa ds


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Page # 16

Sol.

To complete the vertical circle

1g = 2g5

2

1

= 5

Hindi.

/okZ/kj ok iw.kZ djus ds fy,

1g = 2g5

2

1

= 5

19. A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W to the

particle. If the initial speed (in ms1) of the particle is zero, the speed (in ms1) after 5s is :

,d 0.2 kg nzO;eku dk d.k ,d cy ds vUrxZr] tks fd ,d fu;r 'kfDr 0.5 W d.k dks nsrk gS] ,d fn'kk esa xfr'khy gSA

;fn d.k dh izkjafHkd xfr 'kwU; gS rc 5 s ckn bldh xfr (ms1esa) gksxk :

Ans. 5

Sol. E = P.t = 0.5W 5s = 2.5 J =

2

1mv2 v = 5 m/s

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20. A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second. Given

that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will

decay in the first 80s after preparation of the sample is :

,d rqajr rS;kj fd;k gqvk jsfM;ksa vkblksVksi izfrn'kZ] ftldh v)Z&vk;q 1386 s gS] dh lf;rk 103fo?kVu izfr lSd.M gSA

;fn ln 2 = 0.693 gS] rc izFke80 s esa fo?kfVr ukfHkdksa o izkjafHkd dh la[;kvksa dk vuqikr izfr'kr fudVe iw.kkZad esa gSA

Ans. 4

Sol.1386

693.0 = 5 104

Number decayed = N0 N (t)

% age Decayed =0

0

N

)t(NN 100

= (1et) 100

t 100= 5 104 80 100

= 4

Hindi.1386

693.0 = 5 104

fo?kfVr ukfHkdksa dh la[;k = N0 N (t)

izfr'kr fo?kfVr ukfHkdksa dh la[;k =0

0

N

)t(NN 100

= (1et) 100

t 100= 5 104 80 100

= 4

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Page # 18

PART - II : CHEMISTRY

SECTION 1 : (Only One option correct Type)

[k.M 1 : (dsoy ,d lgh fodYi dkj)

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 cgqfodYi 'u gSaA R;sd 'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa] ftuesa ls dsoy ,dlgh gSA

21. The standard enthalpies of formation of CO2(g), H

2O(l) and glucose(s) at 25C are 400 kJ/mol, 300 kJ/mol

and 1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25C is

CO2(g), H

2O(l) rFkk Xywdksl Bksl dh fojpu ekud ,UFkSYiht 25C ij e'k% 400 kJ/eksy], 300 kJ/eksy vkSj 1300

kJ/eksy gSA izfr xzke Xywdksl dh 25C ij ngu ekud ,UFkSYih gSA(A) +2900 kJ (B) 2900 kJ (C) 16.11 kJ (D) + 16.11 kJ

Ans. (C)

Sol. C6H

12O

6(s) + 6O

2(g) 6CO

2(g) + 6H

2O( )

C

H = 6 f

H (CO2

) + 6 f

H (H2

O) f

H (C6

H12

O6

) 6f

(O2

,g)

= 6 (400 300) (1300) 0

= 4200 + 1300

= 2900 KJ/ mol

For one gram of glucose, enthalpy of combustion = 180

2900= 16.11 KJ/g.

Xywdkst ds 1 xzke ds fy, ngu dh ,UFkSYih = 180

2900= 16.11 KJ/g.

22. KI in acetone, undergoes SN2 reaction with each P, Q, R and S. The rates of the reaction vary as

,flVksu esa KI ds foy;u dh izR;sd P, Q, R vkSj S ds lkFk vyx&vyx SN

2 vfHkf;k gksrh gSAbu vfHkf;k dh njksa ds ifjorZu dk lgh e gSA

H3CCl

P Q R S

(A) P > Q > R > S (B) S > P > R > Q

(C) P > R > Q > S (D) R > P > S > Q

Ans. (B)

Sol. Reaction mechanism (SN2)

I effect at carbon (carbon attached to leaving group) will be exerted by hence (s) will be most

reactive.

Sol. vfHkf;k f;kfofk (SN2)

ds izcy I izHkko ds dkj.k dh f;k'khyrk dh nj SN2 vfHkf;k ds fy, vfkdre

gks tkrh gSA

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23. The compound that does NOT liberate CO2, on treatment with aqueous sodium bicarbonate solution, is

(A) Benzoic acid (B) Benzenesulphonic acid

(C) Salicylic acid (D) Carbolic acid (Phenol)

;kSfxd tks tyh; lksfM;e ckbdkcksZusV foy;u }kjk vfHkf;k dj CO2ugha nsrk gS] og gS

(A) csUtksbd vEy (B) csUthulYQksfud vEy(C) lsfyflfyd vEy (D) dkjcksfyd vEy (QhukWy)

Ans. (D)

Sol. Phenol is less acidic than H2CO

3and does not liberate CO

2.

Sol. fQukWy] H2CO

3ls de vEyh; gSA vr% ;s CO

2xSl dk fu"dklu ugha djsxkA

24. Consider the following complex ions, P, Q and R.

P = [FeF6]3, Q = [V(H

2O)

6]2+ and R = [Fe(H

2O)

6]2+.

The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is

fuEufyf[kr ladqy vk;uksa P, Q ,oa R ij fopkj dhft, %P = [FeF

6]3, Q = [V(H

2O)

6]2+ and R = [Fe(H

2O)

6]2+.

ladqy vk;uksa dk lgh e muds izp.k ek=k pqaEcdh; vk?kw.kZ eku (B.M. esa) ds vuqlkj gS(A) R < Q < P (B) Q < R < P

(C) R < P < Q (D) Q < P < R

Ans. (B)Sol. P = [FeF

6]3 ox. no. of Fe = +3 , configuration : - 3d5 4s0

As F is weak ligand, pairing does not take place.

so it has 5 unpaired electron

Q = [V(H2O)

6]2+ ox. no. of V = + 2, configuration 3d3 4s0

It has 3 unpaired electrons.

R = [Fe(H2O)

6]2+, ox. no. of Fe = +2, configuration 3d6, 4s0

As H2O is weak ligand, pairing does not take place, so it has 4 unpaired electron

order of spin only magnetic moment Q < R < Pso, answer is (B).

Sol. P = [FeF6]3 Fe dk vkWDlhdj.k vad = +3 , foU;kl : - 3d5 4s0

F

nqcZy fyxS.M gS vr% ;qXeu ugh gksxkblfy, ;g 5 v;qfXer bysDVkWu j[krk gSQ = [V(H

2O)

6]2+ V dk vkWDlhdj.k vad = + 2, foU;kl : 3d3 4s0

;g 3 v; q f Xer by sDV k Wu j[krk g SR = [Fe(H

2O)

6]2+, Fe dk vkWDlhdj.k vad = +2, foU;kl : 3d6, 4s0

H2O ,d nqcZy fyxS.M gS] vr% ;qXeu ugh gksxk] blfy, ;g 4 v;qfXer bysDVkWu j[krk gSA

izp.k ek=k pqEcdh; vk?kw.kZ dk e Q < R < P

vr% mkj (B) gSA

25. In the reaction,

P + Q R + Sthe time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies

with reaction time as shown in the figure. The overall order of the reaction is

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

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Page # 20

fuEu vfHkf;k]

P + Q R + S

esaPdh 75%vfHkf;k dk le; Pdh 50%vfHkf;k esa fy, x, le; dh rqyuk esa nksxquk gSA Qdh fofHkUu lkUnzrk] vfHkf;kle; vuqlkj fp=k esa n'kkZbZ xbZ gSA bl vfHkf;k dh leLr dksfV gSA

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

Ans. (D)

Sol. For P, if t50%

= x

then t75%

= 2x

This happens only in first order reaction.

so, order with respect to P is 1.

For Q, the graph shows that concentration of Q decreases linearly with time. So rate, with respect to Q,remains constant. Hence, it is zero order wrt Q.

So, overall order is 0 + 1 = 1 Ans. is DSol. P ds fy, ;fn t

50%= x

rc t75%

= 2x

;g izFke dksfV vfHkf;k es gh gksrk gSAvr% P ds lkis{k dksfV 1 ,d gSAQ ds fy,] vkjs[k n'kkZrk gS fd le; ds lkFk Q dh lkUnzrk js[kh; :i ls de gks jgh gSA vr% Q ds lkis{k nj fu;r gS blfy,Q ds lkis{k dksfV 'kwU; gSAvr% vfHkf;k dh dksfV 0 + 1 = 1 gksxh mkj D gSA

26. Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of

lkUnz ukbfVd vEy dk dkQh le; ckn ihys&Hkwjs jax esa ifjofrZr gksuk fdlds cuus ls gksrk gS \(A) NO (B) NO

2(C) N

2O (D) N

2O

4

Ans. (B)

Sol. HNO3decomposes by giving NO

2, O

2, H

2O

4HNO3 2H2O + 2NO2 + 3O2

So, Ans is (B).

Sol. HNO3fo?kfVr gksdj NO

2, O

2, H

2O nsrk gS

4HNO3 2H2O + 2NO2 + 3O2

vr% mkj (B) gSA

27. The arrangement of X ions around A+ ion in solid AX is given in the figure (not drawn to scale). If the radius of

X is 250 pm, the radius of A+ is

,d Bksl AX esa A+vk;u ij Xvk;uksa dh O;oLFkk lgh ekilwpd esa ugha fp=k esa nh xbZ gSA ;fn Xdk v)ZO;kl250 pm gS] rc A+dk v)ZO;kl gksxk

(A) 104 pm (B) 125 pm (C) 183 pm (D) 57 pm

CHEMISTRY


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Page # 21

Ans. (A)

Sol. The given arrangement is octahedral void arrangement.

fn xbZ O;oLFkk v"VQydh; fjfDr dh O;oLFkk gSA

414.0r

r

x

A

250414.0rA

5.103rA

pm.

&

A

A

r

r< 0.732 Ar < 183 pm

So, we have to choose from 104 pm and 125 pm. As no other information is given, we consider exact fit, and

hence 104 pm is considered as answer.

vr% ge104 pm rFkk 125 pm es ls pquko djsxsa ysfdu dksbZ vU; lwpuk ugh nh xbZ] vr% ge104 pm dks mkj ekusaxs D;ksfd;g lfVd lekfgr gksus okyk eki gSA

28. Upon treatment with ammoniacal H2S, the metal ion that precipitates as a sulfide is

,eksfudy H2S ds lkFk vfHkf;k djus ij ftl /kkrq vk;u dk vo{ksi.k lYQkbM ds :i esa gksrk gS] og gS

(A) Fe(III) (B) Al (III) (C) Mg(II) (D) Zn(II)Ans. (D)

Sol. Ammoniacal H2S is group reagent of fourth group cationic radicals. Fe3+ & Al3+ will precipitate Fe(OH)

3and

Al(OH)3respectively. Only Zinc will form white precipitate of ZnS.

veksfudy H2S, {kkjh; ewydks ds prqFkZ lewg ds fy, lewg vfHkdeZd gSA Fe3+rFkkAl3+e'k% Fe(OH)

3 rFkk Al(OH)

3ds

:i esa vo{ksfir gksaxsA dsoy ftad ZnS dk 'osr vo{ksi cuk;sxkA

29. Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at 25 C. For this process, the

correct statement is

(A) The adsorption requires activation at 25C.

(B) The adsorption is accompanied by a decreases in enthalpy.

(C) The adsorption increases with increase of temperature.

(D) The adsorption is irreversible.

25C rkieku ij ,d tyh; foy;u ls esfFkfyu Cyw dk lff;r pkjdksy ij vfk'kks"k.k fd;k x;kA bl ize ds fy, lghdFku gSA(A) vfk'kks"k.k dks25C ij lf;.k dh vko';drk gksrh gSA(B) vfk'kks"k.k ize esa ,UFkSYih ?kVrh gSA(C) vfk'kks"k.k rkieku c


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Page # 22

Sol. Common ore of Ag Ag2S, Cu CuFeS

2,

Pb PbS, Sn SnO2, Mg KCl. MgCl

2. 6H

2O, Al Al

2O

3. xH

2O

So answer is (A)

Sol. Ag ds lkekU; v;LdAg2S, Cu dk lkekU; v;Ld CuFeS

2,

Pb dk lkekU; v;Ld PbS, Sn dk lkekU; v;Ld SnO2, Mg dk lkekU; v;Ld KCl. MgCl

2. 6H

2O

Al dk lkekU; v;Ld Al2O

3. xH

2O

vr% mkj (A) gSA

SECTION 2 : (Only or more options correct Type)

[k.M 2 : (,d ;k vf/kd lgh fodYi dkj)

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) outof which ONE or MORE are correct.

bl [k.M esa 5 cgqfodYi 'u gSaA R;sd 'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa] ftuesa ls ,d;k vf/kdlgh gaSA

31. Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s)is (are)(A) G is positive (B) S

systemis positive

(C) Ssurroundings = 0 (D) H = 0csUthu vkSj usFkyhu lk/kkj.k rkieku ij ,d vkn'kZ foy;u cukrs gSaA bl ize ds fy;s lgh dFku gS gSa(A) G /kukRed gSA (B) S

fudk;/kukRed gSA

(C) Sifjos'k

= 0 (D) H = 0Ans. (B,C,D)Sol. G = ve, S

system= + ve Always for solution formation.

Ssurr.

= 0 No heat exchange between solution and surrounding.

H = 0 For ideal solution.Sol. G = ve, S

system= + ve foy;u fuekZ.k ds fy, lnSo lR; gSA

Ssurr.

= 0 foy;u ,oa ifjos'k ds e/; dksbZ "ek fofue; ughAH = 0 vkn'kZ foy;u ds fy,

32. The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are)(A) [Cr(NH

3)5Cl]Cl

2and [Cr(NH

3)4Cl

2)Cl (B) [Co(NH

3)

4Cl

2]+ and [Pt(NH

3)2(H

2O)Cl]+

(C) [CoBr2Cl

2]2- and [PtBr

2Cl

2]2- (D) [Pt(NH

3)

3(NO

3)Cl and [Pt(NH

3)3Cl]Br

mi&lgla;kstd ;kSfxdksa@vk;Ul ds ;qXe lewg esa tks ,d gh izdkj dh leko;ork n'kkZrs gSa] og gSa(A) [Cr(NH

3)5Cl]Cl

2vkSj [Cr(NH

3)4Cl

2)Cl (B) [Co(NH

3)

4Cl

2]+vkSj [Pt(NH

3)2(H

2O)Cl]+

(C) [CoBr2Cl

2]2-vkSj [PtBr

2Cl

2]2- (D) [Pt(NH

3)

3(NO

3)ClvkSj [Pt(NH

3)

3Cl]Br

Ans. (B,D)Sol. (A) [Cr(NH

3)

5Cl]Cl

2and [Cr(NH

3)4Cl

2)Cl, both compounds will not show either structural or stereoisomerism.

(B) [Co(NH3)

4Cl

2]+ and [Pt(NH

3)

2(H

2O)Cl]+, Ma

4b

2type (octahedral), Ma

2bc type (square planar) and both will

show geometrical isomerism.(C) [CoBr

2Cl

2]2- and [PtBr

2Cl

2]2-, Ma

2b

2type (tetrahedral), Ma

2b

2(square planar).

(D) [Pt(NH3

)3

(NO3

)]Cl and [Pt(NH3

)3

Cl]Br both will show ionisation isomerism.

Sol. (A) nksuks ;kSfxd [Cr(NH3)

5Cl]Cl

2rFkk [Cr(NH

3)

4Cl

2)Cl u rks lajpukRed u gh f=kfoe leko;ork iznf'kZr djsxsaA

(B) [Co(NH3)

4Cl

2]+rFkk [Pt(NH

3)

2(H

2O)Cl]+, Ma

4b

2izdkj (v"VQydh;), Ma

2bc izdkj (oxZ leryh;) rFkk nksuks T;kfefr

leko;ork iznf'kZr djsxsaA(C) [CoBr

2Cl

2]2-rFkk [PtBr

2Cl

2]2-, Ma

2b

2izdkj (prq"Qydh;), Ma

2b

2(oxZleryh;)

(D) [Pt(NH3)

3(NO

3)]Cl rFkk [Pt(NH

3)

3Cl]Br vk;uu leko;ork iznf'kZr djsxsa

33. The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is 1/100th of that of a strong acid(HX, 1M), at 25C. The K

aof HA is

esfFky ,lhVsV (1M) dh nqcZy vEy (HA, 1M) }kjk ty vi?kVu dh izkjafHkd nj 25C ij izcy vEy (HX, 1M) dh rqyukesa1/100 gSA HA ds K

a dk ewY;kadu gS

(A) 1 104 (B) 1 105 (C) 1 106 (D) 1 103

Ans. (A)

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Page # 24

35. Among P, Q, R and S, the aromatic compound(s) is/are

(A) P (B) Q (C) R (D) S

AlCl3

Cl

P

NaH

Q

(NH ) CO4 2 3R

100-115 Co

O O

HClS

O

P, Q, RvkSjSesa ,sjkseSfVd ;kSfxd gS@gSa(A) P (B) Q (C) R (D) S

Ans. (A,B,C,D)

Sol. ;

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Page # 25

SECTION 3 : (Integer value correct Type)

[k.M 3 : (iw.kkd eku lgh dkj)

This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to

9 (both inclusive).

bl [k.M esa 5 'u gSaA R;sd 'u dk mkj 0 ls 9 rd nksuksa 'kkfey ds chp dk ,dy vadh; iw.kkdgSA

36. The total number of lone-pairs of electrons in melamine is

eSySehu ij miyC/k bysDVkWuksa ds ,dkdh ;qXeksa dh dqy la[;k gSAns. 6

Sol. Structure of melamine is as follows

Total no. of lone pairs of electron is '6'.

Sol. eSySfeu dh lajpuk fuEu gS %

dqy ,dkdh bySDVkWu ;qXeksa dh la[;k 6 gSA

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

mRikn P esa dkcksZfDlfyd vEy lewgksa dh dqy la[;k gS

1. H O ,3+

P2. O3

3. H O2 2

O O

O O

O

Ans. 2

Sol.

No. of COOH group is '2'.

COOH lewg dh la[;k 2 gSA

38. The atomic masses of He and Ne are 4 and 20 a.m.u., respectively. The value of the de Broglie wavelength of

He gas at 73 C is "M" times that of the de Broglie wavelength of Ne at 727 C. M isHe vkSj Ne ds ijek.kq nzO;eku e'k% 4 vkSj 20 a.m.u. gSA, He xSl dh73 C ij ns czkWXyh rjax yEckbZNe dh 727 C ijns czkWXyh rjax yEckbZ ls "M" xquk gSA M dk eku gS %

Ans. 5

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Page # 26

Sol. =)KE(m2

hKE T

Ne

He

=

HeHe

NeNe

KEm

KEm=

2004

100020

= 5.

39. EDTA4 is ethylenediaminetetraacetate ion. The total number of NCoObond angles in [Co(EDTA)]1 complex

ion is :

EDTA4,fFkyhu Mkb,sehu VsVk,lhVsV vk;u gSA ladqy vk;u [Co(EDTA)]1 esa NCoOvkca/k dks.kksa dh dqy la[;k gSAns. 8

Sol.

40. A tetrapeptide has COOHgroup on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe)

and alanine (Ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primary

structures) with NH2

group attached to a chiral center is :

,d VsVksisIVkbM esa ,ykuhu ij COOHxzqi fo|eku gSA blds laiw.kZ ty vi?kVu }kjk Xykbflu (Gly), oSyhu (Val), Qsfuy,sykfuu (Phe) rFkk ,sykfuu (Ala) kIr gksrs gSaA bl VsVkisIVkbM dh laHkkfor Ja[kykvksa kFkfed lajpukvksa dh la[;k crk,ftuesa NH

2xqzi fdjsy dsUnz ds lkFk vkcaf/kr gSA

Ans. 4

Sol. Following combinations are possible for tetrapeptideVal Phe Gly Ala

Val Gly Phe Ala

Phe Gly Val Ala

Phe Val Gly Ala

1. In all above sequences C-terminal is alanine

2. Glycine is optically inactive amino acid, hence It should not be N-terminal so, only above combination are

possible.

Sol. VsVkisIVkbM ds fy, fuEu pkj la;kstu laHko gSAVal Phe Gly Ala

Val Gly Phe Ala

Phe Gly Val Ala

Phe Val Gly Ala1. ij fn;s x;s lHkh eksa esa ,ykfuu C vUrLFk gSA2. Xykbflu izdkf'kd fuf"; ,ehuksa vEy gS] vr% ;s N vUrLFk ugha gks ldrk gSA blfy, ij fn;s x;s pkj la;kstu laHkogSA

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PART - III MATHEMATICS

SECTION - I

Straight Objective Type

This section contains 10 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONLY ONE is correct.

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

bl [k.M esa 10 iz'u gSaA izR;sd iz'u ds 4 fodYi (A), (B), (C) rFkk (D) gSa] ftuesa ls flQZ ,d lgh gSA

41. The value of cot

23

1n

n

1k

1 k21cot is

cot

23

1n

n

1k

1 k21cot

dk eku gS&

(A)25

23(B)

23

25(C)

24

23(D)

23

24

Sol. (B)

cot

23

1n

1 n2......6421cot

cot cot1(1 + n(n + 1))

cot tan1

)1n(n1

n1n

cot

23

1n

11 ntan)1n(tan

cot(tan124 tan11)

cot

241

124tan 1

cot23

25

23

25cot 1

42. Let k2ji3PR and k4j3iSQ determine diagonals of a parallelogram PQRS and k3j2iPT

be another vector. Then the volume of the parallelepiped determined by the vectors PQ,PT and PS is

ekukfd k2ji3PR rFkk k4j3iSQ ,d lekUrj prqHkqZt PQRSds fod.kZ fu/kkZfjr djrs gS vkSj k3j2iPT

,d vU; lfn'k gS] rc lfn'kksa PQ,PT rFkk PS }kjk fu/kkZfjr lekUrj "kV~Qyd dk vk;ru gS&

(A) 5 (B) 20 (C) 10 (D) 30

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Page # 28

Sol. (C)

PSPQPR

PSPQSQ

2

SQPRPS

2

SQPRPQ

V = PTPSPQ

V = PT,SQPR,SQPR4

1

V = PT,SQ,PR2

1

321431

213

2

1

2

1( 3 7 10) = 10

43. Let complex numbers and1

lies on circles (x x0)2 + (y y

0)2 = r2 and (x x

0)2 + (y y

0)2 = 4r2,

respectively. If z0

= x0

+ iy0

satisfies the equation 2|z0|2 = r2 + 2, then || =

ekukfd lfEeJ la[;k,a rFkk

1e'k% ok (x x

0)2 + (y y

0)2 = r2rFkk (x x

0)2 + (y y

0)2 = 4r2 ij fLFkr gSA ;fn

z0

= x0

+ iy0 lehdj.k 2|z

0|2 = r2 + 2 dks larq"V djrk gS] rc || =

(A)2

1(B)

2

1(C)

7

1(D)

3

1

Sol. (C)|z z

0| = r

|z z0| = 2r

| z0| = r

0z

1

= 2r 2||

02z

||

= 2r

( z0) 0z = r2 | |2 z0 0z + |z0|

2 = r2

0202

z||

z||

= 4r2 20

2

0

4

2

||

z

||

z

||

||

+ |z

0|2 = 4r2

1 0z 0z + |z0|2 ||2 = 4r2||2

(|2| 1) + |z0|2 (1 ||2) = r2 (1 42)

MATHEMATICS


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Page # 29

(||2 1)

2

2r1

2

= r2(1 4||2)

(||2 1)

2

r 2

= r2(1 4||2)

||2 1 = 2 + 8 ||2

1 = 7||2 || =7

1

44. For a > b > c > 0, the distance between (1, 1) and the point of intersection of the lines ax + by + c = 0 and bx

+ ay + c = 0 is less than 22 . Then

a > b > c > 0 ds fy, (1, 1) rFkk js[kkvksa ax + by + c = 0 o bx + ay + c = 0 ds izfrPNsn fcUnq ds chp dh nwjh 22

ls de gS] rc

(A) a + b c > 0 (B) a b + c < 0 (C) a b + c > 0 (D) a + b c < 0

Sol. (A)(a b)x + (b a)y = 0

x = y

Point of intersection

ba

c,

ba

c

Now 22ba

c1

ba

c1

22

22ba

cba2

a + b c > 0

45. Perpendicular are drawn from points on the line3

z

1

1y

2

2x

to the plane x + y + z = 3. The feet of

perpendiculars lie on the line

ry x + y + z = 3 ij js[kk3

z

1

1y

2

2x

ij fLFkr fcUnqvksa ls yEc Mkys tkrs gSA yEc-ikn fuEu js[kk ij fLFkr gS&

(A)13

2

z8

1

y5x (B)

52

z

31

y

2x (C)

72

z

31

y

4x (D)

52

z

71

y

2x

Sol. (D)

Any point on line2

2x =

1

1y

=3

z=

Let any two points on this line areA( 2, 1, 0), B (0, 2, 3) Put ( = 0, 1)

Let foot of perpendicular from A( 2, 1, 0) on plane is (, , )

1

2=

1

1=

1

0= (say)

Also, + + = 3 2 + 1 + = 3 = 2 M(0, 1, 2)

MATHEMATICS


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Page # 30

Similarly foot of perpendicular from B(0, 2, 3) on plane is N

3

11,

3

4,

3

2

So, equation of MN is

3

2

0x =

3

7

1y

=

3

5

2z .

46. Four persons independently solve a certain problem correctly with probabilities81,

41,

43,

21 . Then the

probability that the problem is solved correctly by at least one of them is

pkj O;fDr Lora=kr;k fdlh ,d leL;k dks izkf;drkvksa8

1,

4

1,

4

3,

2

1ds lkFk Bhd gy djrs gS] rc leL;k ds muesa ls

de ls de ,d O;fDr }kjk Bhd gy fd;s tkus dh izkf;drk gS&

(A)256

235(B)

256

21(C)

256

3(D)

256

253

Sol. (A)P (problem solved by at least one) = 1 P(problem is not solved by by all)

= 1 AP BP CP DP

= 1

8

7

4

3

4

1

2

1= 1

256

21=

256

235

47. The area enclosed by the curves y = sinx + cosx and y = |cosx sinx| over the interval

2

,0 is

vUrjky

2,0 ij oksay = sinx + cosx rFkk y = |cosx sinx| }kjk ifjc) {ks=kQy gS&

(A) 1

24 (B) 1

222 (C) 122 (D) 1222 Sol. (B)Given y = sin x + cos x x [0, /2]

dx

dy= cos x sin x

y = |cos x sin x| =

]2/,4/[xxcosxsin

]4/,0[xxsinxcos

required area =

2/

4/

4/

0

dx|xcos2|dxxsinxcosxcosxsin

=

2/

4/

4/

0

dx|xcos2|dx|xsin2|

= 2/4/4/

0 xsin2xcos2

= 2

2

111

2

1

= 2

2

22

= 222 = 4 22 = 1222

MATHEMATICS


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Page # 31

48. A curve passes through the point

6

,1 . Let the slope of the curve at each point (x, y) be

x

ysec

x

y,

x > 0. Then the equation of the curve is

,d o fcUnq

6

,1 ls xqtjrk gSA ekuk fd izR;sd fcUnq (x, y) ij o dh izo.krk

x

ysec

x

y, x > 0 gS] rc o

dk lehdj.k gS&

(A) sin

x

y= logx +

2

1(B) cosec

x

y= logx + 2

(C) 2xlogx

y2sec

(D)

2

1xlog

x

y2cos

Sol. (A)Given slope at (x, y) is

xy

dxdy + sec(y/x)

letx

y= t y = xt

dx

dy= t + x

dx

dt

t + xdx

dt= t + sec(t)

dxx

1dttcot

sin t = n x + csin(y/x) = n x + c

This curve passes through (1, /6)sin(/6) = n(1) + c c = 1/2sin(y/x) = n x + 1/2

49. Let f :

1,

2

1 R (the set of all real numbers) be a positive, non-constant and differentiable function such that

f(x) < 2 f(x) and f

2

1= 1. Then the value of

1

2/1

dx)x(f lies in the interval

ekuk fd f :

1,

2

1 R (lHkh okLrfod la[;kvksa dk leqPp;) ,d /kukRed] vpjsrj rFkk vodyuh; Qyu gS ftlds fy,

f(x) < 2 f(x) rFkk f

2

1= 1 gS] rc

1

2/1

dx)x(f dk eku fuEu vUrjky esa gS&

(A) (2e 1, 2e) (B) (e 1, 2e 1) (C)

1e,

2

1e(D)

2

1e,0

MATHEMATICS


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Page # 32

Sol. (D)f(x) 2f(x) < 0

dx

d(e2x f(x)) < 0

e2x f(x) is decreasing x > 1/2

e2x

f(x) < 1/e f(x) < e2n1

0 <

1

2/1

1x2

1

2/1

dxedx)x(f

0 < 2

1edx)x(f

1

2/1

50. The number of points in (,), for which x2 x sinx cosx = 0, is

(,) esa fcUnqvksa dh la[;k] ftuds fy, x2 x sinx cosx = 0, gS&

(A) 6 (B) 4 (C) 2 (D) 0

Sol. (C)x2 = x sinx + cos xf(x) = x2

g(x) = x sin x + cos xg(x) = sin x + x cos x sin x

g(x) = x cos x

Only two solution.

SECTION - II

Multiple Correct Answer Type

This section contains 5 questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of

which ONE OR MORE is/are correct.

[k.M- IIcgqy lgh fodYi izdkj

bl [k.M esa 5iz'u gSaA izR;sd iz'u ds mkj ds fy, 4fodYi (A), (B), (C)rFkk (D)gSa] ftuesa ls,d ;k ,d ls vf/kd lghgSA

51. A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is converted into an

open rectangular box by folding after removing squares of equal area from all four corners. If the total area of

removed squares is 100, the resulting box has maximum volume. The lengths of the sides of the rectangular

sheet are

(A) 24 (B) 32 (C) 45 (D) 60

,d fuf'pr ifjeki dh vk;rkdkj pknj dks] ftldh Hkqtkvksa dh yEckb;k8 : 15 ds vuqikr esa gSa] lHkh pkjksa fdukjksa ls leku

{ks=kQy ds oxZ fudky dj ,d [kqyh vk;rkdkj isVh esa ifjofrZr fd;k tkrk gSA ;fn fudkys x;s oxksZ dk dqy {ks=kQy 100

gS] rc ifj.kkeh isVh dk vk;ru egke gSA rc vk;rkdkj pknj dh Hkqtkvksa dh yEckb;k fuEu gS&(A) 24 (B) 32 (C) 45 (D) 60

MATHEMATICS


33/40

Page # 33

Sol. (A,C)

a

a

aa

aa

aa a

a a

Let = 8x, b = 15 x

Volume = (8x 2a) (15x 2a) (a) = 4a3 46a2x + 120 ax2

da

dV= 6a2 46ax + 60 x2

5xatda

dV

= 0

x = 3 and6

5

2

2

da

Vd= 6a 23x

3x&5aat

2

2

da

Vd

< 0,

So, at x = 3 gives maxima

6

5x&5aat

2

2

da

Vd

> 0

So, at x =6

5gives minima.

da

dV= 0 when a = 5 given ( 4a2 = 100 given for maximum volume)

at a = 5

byda

dV= 0 6x2 23x + 15 = 0

x = 3 or 5/6

So by x = 3 (for max volume)8x = 24, 15x = 45 Ans. (A, C)

52. Let Sn

= n4

2

2

)1k(k

)1( k2 . Then Sn

can take value(s)

(A) 1056 (B) 1088 (C) 1120 (D) 1332

ekuk fd Sn

= n4

2

2

)1k(k

)1( k2 , rc SnfuEu eku ys ldrk gS%&

(A) 1056 (B) 1088 (C) 1120 (D) 1332

MATHEMATICS


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Page # 34

Sol. (A D)

Sn

=2

n4

1k

2

)1k(k

k)1(

= 12 22 + 32 + 42 52 62 + 72 + 82 +........= (32 1) + (42 22) + (72 52) + (82 62)......

= 2 ]......2220141264[

termsn2

= 2[(4 + 12 + 20 ......) + (6 + 14 + 22 ........)]n terms n terms

=

)8)1n(62(

2

n)8)1n(24(

2

n2

= 2[n(4 + 4n 4) + n(6 + 4n 4)]= 2(4n2 + (4n + 2)n)= 2(8n2 + 2n)= 4n(4n + 1)(A) 1056 = 32 33 n = 8(B) 1088 = 32 34 n = 8

(C) 1120 = 32 35 n = 8(D) 1332 = 36 37 n = 8

53. A line l passing through the origin is perpendicular to the lines

l1

: (3 + t) i + ( 1 + 2t)j + (4 + 2t) k , < t <

l2

: (3 + 2t) i + (3 + 2t)j + (2 + s) k , < s <

Then, the coordinate(s) of the point(s) on l2at a distance of 17 from the point of intersection of land l1 is(are)

(A)

3

5

,3

7

,3

7

(B) (

1, ,

1, 0) (C) (1, 1, 1) (D)

9

8

,9

7

,9

7

,d js[kk l] tks ewyfcUnq ls xqtjrh gS] js[kkvksa

l1

: (3 + t) i + ( 1 + 2t)j + (4 + 2t) k , < t <

l2

: (3 + 2t) i + (3 + 2t)j + (2 + s) k , < s <

ij yEcor gSA rc, l2ij fLFkr fcUnq fcUnqvksa ds funsZ'kkad] tks js[kkvksa l

rFkk l

1ds frPNsn fcUnq ls 17 dh nwjh ij gSa gSa]

fuEu gS gSa %

(A) 35,

37,

37 (B) (1, ,1, 0) (C) (1, 1, 1) (D) 9

8,97,

97

Sol. (B,D)Let equation of line is

:a

0x =

b

0y =

c

0z = k

This line is perpendicular to given line 1and

2.

Hence a + 2b + 2c = 02a + 2b + c = 0

2

a

=3

b=

2

c

A

B

1

2

MATHEMATICS


35/40

Page # 35

Hence equation of is2

x

=

3

y=

2

z

= k

1, k

2

for

1for

2

Now A(2k1, 3k

1, 2k

1) B(2k

2, 3k

2, 2k

2)

Point A satisfied 1

2k1 i + 3k

1j 2k

1

k = (3 + t)i + (1 + 2t)

j + (4 + 2t) k

3 + t = 2k1

.......(1)1 + 2t = 3k

1.......(2)

4 + 2t = 2k1

.......(3)(2) & (3) 5 = 5k

1 k

1= 1 A (2, 3, 2)

Let any point on 2

(3 + 2S, 3 + 2S, 2 + S)

Given 2 2 2(1 2S) (6 2S) (S) = 17

9S2 + 28S + 37 = 179S2 + 28S + 20 = 09S2 + 18S + 10S + 20 = 09S(S + 2) + 10 (S + 2) = 0S = 2, 10/9

Hence (1, 1, 0) , (7/9, 7/9, 8/9)Ans. (B) & (D)

54. Let f(x) = x sin x, x > 0. Then for all natural numbers n, f (x) vanishes at

(A) a unique point in the interval

2

1n,n (B) a unique point in the interval

1n,

2

1n

(C) a unique point in the interval (n, n +1) (D) two points in the interval (n, n +1)

ekuk fd f(x) = x sin x, x > 0, rc lHkh ?ku&iw.kkdksa n ds fy, f(x) fuEu ij 'kwU; gksrk gS %

(A) varjky

2

1n,n esa ,dek=k ,d fcUnq ij (B) varjky

1n,

2

1n esa ,dek=k ,d fcUnq ij

(C) varjky (n, n +1) esa ,dek=k ,d fcUnq ij (D) varjky (n, n +1) esa nks fcUnqvksa ij

Sol. (B,C)f(x) = x sin x , x > 0f ' (x) = sin x + x cos x = 0

tan x = x

MATHEMATICS


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Page # 36

55. For 33 matrices M and N, which of the following statement(s) is (are) NOT correct ?

(A) NT M N is symmetric or skew symmetric, according as M is symmetric or skew symmetric

(B) M N N M is symmetric for all symmetric matrices M and N

(C) M N is symetric for all symmetric matrices M and N

(D) (adj M) (adj N) = adj(MN) for all invertible matrices M and N

33 vkO;wgksa M rFkk N ds fy, fuEu esa ls dkSu dFku lR; ugha gSa gSa \(A) M ds lefer ;k fo"ke&lefer gksus ds vuqlkj NT M N lefer ;k fo"ke&lefer gSA

(B) lHkh lefer vkO;wgksa M rFkk N ds fy, MN NM fo"ke &lefer gSSA

(C) lHkh lefer vkO;wgksa M rFkk N ds fy, MN lefer gSA

(D) lHkh O;qRe.kh; vkO;wgksa M rFkk N ds fy, (adj M) (adj N) = adj(MN)

Sol. (C,D)(A) (NTMN)T = NT MT N is symmetric if M is symmetric and skew-symmetric if M is skew-symmetric.(B) (MN NM)T = (MN)T (NM)T

= NM MN= (MN NM)

skew symmetric(C) (MN)T = NT MT

= NM

MN hence NOT correct(D) standard result is

adj(MN) = (adjN) (adj M)]

(adjM) (adjN)

SECTION - III : (Integer value correct Type)

[k.M III : (iw.kkd eku lgh dkj)

This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to9 (both inclusive).

bl [k.M esa 5 'u gSaA R;sd 'u dk mkj 0 ls 9 rd nksuksa 'kkfey ds chp dk ,dy vadh; iw.kkdgSA

56. A vertical line passing through the point (h, 0) intersects the ellipse3

y

4

x 22 = 1 at the points P and Q. Let the

tangents to the ellipse at P and Q meet at the point R. If (h) = area of the triangle PQR, 1= 1h2/1

max (h) and

2

= 1h2/1min

(h), then 58

1 8

2=

fcUnq (h, 0) ls xqtjus okyh /okZ/kj js[kk nh?kZok3

y

4

x 22 = 1 dks fcUnqvksaP rFkk Q ij dkVrh gSA ekuk fd fcUnqvksaPrFkk

Q ij nh?kZok dh Li'kZ js[kk, fcUnqR ij feyrh gSA ;fn (h) = f=kHkqt PQRdk {ks=kQy1= 1h2/1

min (h)vkSj 2 = 1h2/1

max

(h) gS] rc5

8

1 8

2=

MATHEMATICS


37/40

Page # 37

Sol. Point of intersection of tangents at P and Q is R(2 sec, 0)

Area of PQR =2

1 32 sin (2 sec 2 cos )

= 32

cos

sin3; where cos

2

1,

4

1

Now

d

d

2

32

cos

)sin(sincossin3.cos32> 0

As i , i cos i

Bmin

= A2

= 32 .

2/1

4/112/3

= 34 .8

33=

8

36

Occues at cos = 1/2

min

= 1

= 32 .

4/1

16/112/3

= 38 .4.4.4

15.15=

16

53.15.32

Occues cos = 1/4

1

=8

455

57. The coefficients of three consecutive terms of (1 + x)n+5 are in the ratio 5 : 10 : 14. Then n =

(1 + x)n+5 ds rhu ekxr inksa ds xq.kkad 5 : 10 : 14 ds vuqikr esa gS] rc n =

Sol. n+5Cr1

: n+5Cr: n+5C

r+1= 5 : 10 : 14

510

C

C

1r5n

r5n

&10

14

C

C

r5n

1r5n

r

1r)5n( = 2 &

5

7

1r

1)1r()5n(

3r

6n

&

5

12

1r

6n

3r =5

12(r + 1)

r = 4 n + 6 = 12 n = 6

MATHEMATICS


38/40

Page # 38

58. Consider the set of eight vectors V = 1,1c,b,a:kcjbia . Three non-coplanar vectors can be chosenfrom V in 2p ways. Then p is

vkB lfn'kksa dk leqPp; V = 1,1c,b,a:kcjbia . yhft, A V ls rhu vleryh; lfn'k 2pdkj ls pqus tk ldrs

gSA rc p dk eku gS%

58. Among set of eight vectors four vectors form body diagonals of a cube, remaining four will be parallel(unlike) vectors.

Numbers of ways of selecting three vectors willbe 4C

3 2 2 2 = 25

Hence p = 5Alternative

Eight vectors

x

kji

y

kji

z k

j

i

kji

kji'x

kji'y

kji'z

kji'

If we take 'x,x

and any one of remaining sin x, vectors will always be coplaner

No. of coplaner vectors = 6

similarly on taking 'y,y

= 6

'z,z

= 6

',

= 6 No. of set of coplaner vectors = 24

Ans. 8C3 24 = 32

AlternativeA(0, 0, 0)B(1, 0, 0)C(1, 0, 1)D(0, 0, 1)

E(0, 1, 1)F(0, 1, 0)G(1, 1, 0)H(1, 1, 1)

kjiAH

kjiBE

kjiCF

kjiDG

Non-coplaner

MATHEMATICS


39/40

Page # 39

59. Of the three independent events E1, E

2and E

3, the probability that only E

1occurs is ,only E

2occurs is and

only E3occurs is . Let the probability p that none of events E

1, E

2or E

3occurs satisfy the equations ( 2)

p = and ( 3) p = 2. All the given probabilities are assumed to lie in the interval (0, 1).

Then3

1

EofoccurrenceofobabilityPr

EofoccurrenceofobabilityPr=

rhu Lora=k ?kVukvksa E1, E2rFkk E3esa ls dsoy E1ds ?kVus dh kf;drk gS] dsoy E2ds ?kVus dh kf;drk gS rFkk dsoyE

3ds ?kVus dh kf;drk gSA ekuk fd ?kVukvksa E

1, E

2;k E

3esa ls fdlh ds Hkh u ?kVus dh kf;drk p, lehdj.kksa ( 2)

p = rFkk ( 3) p = 2dks lUrq"V djrh gSA lHkh kf;drk, vUrjky (0, 1) esa fLFkr ekuk tkrh gSA] rc

drkh kf;ds ?kVus ddrkh kf;ds ?kVus d

3

1

E

E=

Sol. Let x, y, z be probability of E1, E

2, E

3respectively

x(1 y)(1 z) = y(1 x)(1 z) = z(1 x)(1 y) = (1 x)(1 y) (1 z) = PPutting in the given relation we get x = 2y and y = 3z

x = 6z z

x= 6

60. A pack contains n card numbered from 1 to n. Two consecutive numbered card are removed from the pack and

the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is

k, then k 20 =

,d xM~Mh esa n dkMZ gSa tks la[;kvksa 1 ls n }kjk fpfUgr gSA nks ekxr la[;kvksa okys dkMZ xM~Mh ls fudky fn;s tkrs gS vkSj

vof'k"V dkMksZ dh la[;kvksa dk ;ksx 1224 gSA ;fn fudkys x, dkMksZ dh fpfUgr la[;kvksa esa ls y?kqkj la[;k k gS] rc k

20 =

Sol. Numbers removed are k and k + 1

now2

)1n(n k (k + 1) = 1224

n2 + n 4k = 2450 n2 + n 2450 = 4k (n + 50)(n 49) = 4k n > 49

Alternative To satisfy this equation n should be of the form of (4p + 1) or (4p + 2) taking n = 50 4k = 100

k = 25 k 20 = 5Now if we take n = 53k = 103n < kso not possibleHence n 53 will not be possible.

MATHEMATICS


40/40

Write your Name, Roll No. and the name of centre and

sign with pen in the boxes provided on the right part of

ORS. Do not write any of this anywhere else. Darkenthe appropriate bubble UNDER each digit of your Roll

No. in such a way that the impression is created on the

bottom sheet.

C. Questioni Paper Format

The question paper consists of three parts (Physics,

Chemistry and Mathematics). Each part consists of threesections.

Section 1 contains 10 multiple choice questions.

Each question has four choices (A), (B), (C) and (D) outof which ONLY ONE is correct.

Section 2 contains 5 multiple choice questions.

Each questions has four choices (A), (B), (C) and (D)

out of which ONE or MORE are correct.

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

(both inclusive).

D. Marking Scheme

For each question in Section 1, you will be awarded 2

marks if you darken the bubble corresponding to thecorrect answer each zero mark if no bubbles are

darkened. No negative marks will be awarded forincorrect answers in this section.


marks if you darken all the bubble(s) corresponding toonly the correct answer(s) and zero mark if no bubbles

are darkened. In all other cases, minus one (1) markwill be awarded.


marks if you darken the bubble corresponding to onlythe correct answer and zero mark if no bubbles are

darkened. In all other cases, minus one (

1) mark willbe awarded.

13. viuk uke] jksy u-a vkSj ijh{kk dsaUnz dk uke vks- vkj- ,l-ds nk, Hkkx esa fn, x, [kkuksa esa dye ls Hkjs vkSj vius

gLrk{kj djsaA buesa ld dksbZ Hkh tkudkjh dgha vkSj ufy[ksaA jksy uEcj ds gj vad ds uhps vuq:i cqycqys(BUBBLE) dks bl rjg ls dkyk djsa fd fupys i "B ijHkh fu'kku cu tk,A

C. iz'u&i=k dk izk:ibl iz'u&i=k ds rhu Hkkx HkkSfrd foKku] jlk;u foKku]vkSj xf.kr gSaA gj Hkkx ds rhu [k.M gSaA

14. [kaM1 esa 10cgqfodYi iz'u gSaA gj iz'u esa pkj fodYi(A), (B), (C) vkSj (D) gSa ftuesa ls dsoy ,d gh lgh gSA

15. [kaM2 esa5cgqfodYi iz'u gSaA gj iz'u esa pkj fodYi (A),(B), (C) vkSj (D) gSa ftuesa ls ,d ;k vf/kd lgh gSaA

16. [kaM3 esa5iz'u gSaA izR;sd iz'u dk mkj 0 ls9 rd nksuksa'kkfey ds chp ,d ,dy vadh; iw.kkZad gSA

D. vadu ;kstuk17. [kaM 1 es a gj iz'u es a dsoy lgh mkj okys

cqycqys(BUBBLES) dks dkyk djus ij 2 vadvkSj dksbZHkh cqycwyk dkyk ugha djus ij (0)vad iznku fd, tk;saxsA

bl [kaM ds iz'uksa esa xyr mkj nsus ij dksbZ_.kkRed vadugha fn;s tk;saxsA

18. [kaM 2 ds gj iz'u esa dsoy lgh mkjksa mkj okys lHkhcqycqyksacqycqys dks dkyk djus ij 4 vad iznku fd,tk;saxs vkSj dksbZ Hkh cqycqyk dkyk ugha djus ij 'kwU; (0)vad iznku fd, tk;saxsA vU; lHkh fLFkfr;ksa esa _.kkRed,d(1) vad iznku fd;k tk;sxkA

19. [kaM 3 ds gj iz'u esa dsoy lgh mkj okys cqycqys dksdkyk djus ij 4 vad iznku fd, tk;saxs vkSj dksbZ Hkh

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

CODE 3

ijh{kkFkhZ dk ukeName of the Candidate

eSaus lHkh funsZ'kksa dk i

paper 1 solutions

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