jee advanced 2016 mock test paper

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[Type text] Page 1 JPT-1 (JEE ADVANCE) TEST DATE : | BATCH : JP,JF,JR FACULTY NAME: TARGET DATE : 16-06-2014 Syllabus : XI and XII S.No. Subject Nature of Questions No. of Questions Marks Negative Total 1 to 15 MCQ 15 4 0 60 16 to 19 Match matrix listing 4 3 1 12 20 to 34 MCQ 15 4 0 60 35 to 38 Match matrix listing 4 3 1 12 39 to 53 MCQ 15 4 0 60 54 to 57 Match matrix listing 4 3 1 12 57 216 Paper-2 Total Total 1. Let A = 0 0 0 and (A + ) 50 50A = a b c d then [MT-AL] [301] ekuk A = 0 0 0 vkSj (A + ) 50 50A = a b c d rc (A*) a + d = 2 (B*) a + b = 1 (C*) b + c = 0 (D) a + d = 1 Sol. A = 0 0 0 A 2 = 0 0 0 0 = 0 A 3 = 0 (A + ) 3 = A 3 + 3A 2 + 3A + Similarly blh izdkj (A + ) 50 = 50A + ......(1) (A + ) 50 50A = 1 0 0 1 ans. A,B,C a =1, b = 0, c = 0, d = 1 2. From a pack of 52 playing cards, all face cards are removed. From the remaining pack, cards are dealt one by one until an ace appear. Let P 1 denote the probability that exactly 10 cards are dealt before the first ace

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Page 1: JEE Advanced 2016 Mock test paper

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JPT-1 (JEE ADVANCE) TEST DATE : | BATCH : JP,JF,JR

FACULTY NAME:

TARGET DATE : 16-06-2014

Syllabus : XI and XII

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

1 to 15 MCQ 15 4 0 60

16 to 19 Match matrix listing 4 3 �1 12

20 to 34 MCQ 15 4 0 60

35 to 38 Match matrix listing 4 3 �1 12

39 to 53 MCQ 15 4 0 60

54 to 57 Match matrix listing 4 3 �1 12

57 216

Paper-2

TotalTotal

1. Let A =0

0 0

and (A + )50 � 50A = a b

c d

then [MT-AL] [301]

ekuk A =0

0 0

vkSj (A + )50 � 50A = a b

c d

rc

(A*) a + d = 2 (B*) a + b = 1 (C*) b + c = 0 (D) a + d = 1

Sol. A = 0

0 0

A2 = 0 0

0 0

= 0

A3 = 0

(A + )3 = A3 + 3A2 + 3A +

Similarly blh izdkj

(A + )50 = 50A + ......(1)

(A + )50 � 50A = 1 0

0 1

ans. A,B,C

a =1, b = 0, c = 0, d = 1

2. From a pack of 52 playing cards, all face cards are removed. From the remaining pack, cards are dealt one by one until an ace appear. Let P1 denote the probability that exactly 10 cards are dealt before the first ace

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appear. If cards continue to dealt until a second ace appears. Let P2 denote the probability that exactly 20

cards are dealt before the second ace. If 130 P2 = mn

(m, n are co-prime natural numbers) then-

rk'k d h 52 iÙkksa dh ,d xM~Mh ls lHkh psgjs okys iÙks ¼face cards½ gVk;s tkrs gSA 'ks"k cph gqbZ xM~Mh ls ,d ,d

djds iÙks dks [khapk tkrk gS tc rd dh bDdk u vk tk;sA ekuk P1 ] bDdk vkus ls igys Bhd 10 iÙksa fudkyus dh

izkf;drk dks O;Dr d jrk gSA ;fn iÙkksa dks yxkrkj fudkyk tkrk gS tc rd fd nwljk bDdk u vk tk;sA ekuk P2 ]

nwljk bDdk vkus ls igys Bhd 20 iÙksa fudkyus dh izkf;drk dks O;Dr djrk gSA ;fn 130 P2 = mn

(m, n lgvHkkT;

izkd r̀ la[;k gS) rc [PR-CD] [302]

(A*) m + n = 46 (B) m � n = 28 (C*) m × n = 333 (D*) n �1

m= 4

Sol. 10 card are drawn before Ist ace

izFke bDds ls igys 10 iÙks [khps tkrs gSA

first 10 cards are non ace and 11th card is an ace

izFke 10 iÙks] bDds ugh gS rFkk 11 ok¡ iÙkk bDdk gSA

P1 = 36

1040

10

C 4.

C 30

P1 = 27.28.29

10.37.38.39

Position by pack now vc xM~Mh ls fLFkfr

293aces

26non ace

P2 = 26

9129

9

C 3 9. .P

C 20 10.13.37

130 P2 = 9

37

3. If m and n are positive integers more than or equal to 2, m > n, then (mn)! is divisible by �

;fn m vkSj n, 2 ls vf/kd ;k cjkcj /kukRed iw.kkZad gS] m > n rc (mn)! fdlls foHkkftr gS - [PC-GP] [304]

(A*) (m!)n (B*) (n!)m (C*) (m + n)! (D*) (m � n)!

Sol. n

(mn)!

(m!)is the number of ways of distributing mn distinct objects in n persons equally.

mn fofHkUu oLrqvksa dks n O;fDr;ksa esa leku :i ls ckaVus ds rjhds n

(mn)!

(m!)

Hence n

(mn)!

(m!) is an integer (m!)n| (mn)!. Similarly (n!) m | (mn)!

vr% n

(mn)!

(m!) ,d iw.kk±d gSA (m!)n| (mn)!. blh izdkj (n!) m | (mn)!

10

11th 21th

2nd ace1st ace non-ace 9

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Further iqu% m + n < 2 m mn (m + n)! | (mn) and vkSj m � n < m < mn (m � n)! | (mn) 4. Let be an operation defined on points in Cartesian plane such that if 'P' and 'Q' are 2 points, then

PQ= R, where PQR is an equilateral triangle with P, Q, R being in anticlockwise sense, then

[MI-STAR] [303]

(A*) If PQ = R RP = Q

(B) If PQ = R PR = Q

(C) (PQ)S = P(QS)

(D*) If PQ = R, RQ = S and QS = T, then ratio of areas of PQR and QST is 1

ekuk dkfrZ; lery esa ifjHkkf"kr fcUnqvksa ij ,d lafØ;k bl izdkj gS fd P vkSj Q nks fcUnq ds fy, PQ= R, tgk¡

PQR leckgq f=kHkqt gS vkSj P, Q, R dks okekorZ fn'kk esa fy;k x;k gS] rc

(A*) ;fn PQ = R RP = Q

(B) ;fn PQ = R PR = Q

(C) (PQ)S = P(QS)

(D*) ;fn PQ = R, RQ = S vkSj QS = T, rc PQR vkSj QST ds {ks=kQyksa dk vuqikr 1 gSA

Sol. Clearly if PQ = R, then as long as cyclic order is maintained by operating on any two we will get the third point, i.e. PQ = R RP = Q QR = P. Now for three points the operation need not be associative for example if PQ= S, then (C) will not hold true. In all the cases in (D) triangles have same side lengths hence their area are equal hence (A) and (D).

Li"Vr;k% ;fn PQ = R, rc pØh; Øe esa dksbZ nks fcUnq ds fy, lafØ;k dks bl izdkj fy;k x;k gS fd rhljs fcUnq

dks izkIr djrk gSA vFkkZr~ PQ = R RP = Q QR = P. vc rhu fcUnqvksa ds fy, lafØ;k dks lkgp;Z gksus dh

t:jh ugha gSA mnkgj.k ds fy, ;fn PQ= S, rc (C) lR; ugha gksxhA bu lHkh fLFkfr;ksa esa (D) f=kHkqtksa dh leku

yEckbZ;k¡ gS vr% muds {ks=kQy cjkcj gSA vr% (A) vkSj (D).

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5. Let C1 and C2 be centres of two circles whose radii are 2 and 4 respectively. Also C1C2 = 10 and direct common tangents of these circles touch them at P,Q,R,S. Another circle of radius '' is drawn passing through P, Q, R, S, then [CR-CT] [301]

(A*) Mid-point of C1C2 is centre of the circle passing through P,Q,R,S.

(B) Centre of the circle passing through P,Q,R,S divides C1C2 in the ratio 1 : 2.

(C*) 2 = 33

(D) 2 = 35

ekuk C1 vkSj C2 nks oÙ̀kksa ds dsUnz gS ftldh f=kT;k,a Øe'k% 2 vkSj 4 gS rFkk C1C2 = 10 vkSj bu oÙ̀kksa dh vuqØe

mHk;fu"B Li'kZ js[kk,a gS P,Q,R,S ij mudksa Li'kZ djrh gSA ;fn '' f=kT;k dk vU; oÙ̀k P, Q, R, S ls xqtjrk gS] rc

(A*) P,Q,R,S ls xqtjus okys oÙ̀k dk dsUnz] C1C2 dk e/; fcUnq gSA

(B) P,Q,R,S ls xqtjus okys oÙ̀k dk dsUnz] C1C2 dks 1 : 2 esa foHkkftr djrk gSA

(C*) 2 = 33

(D) 2 = 35

Sol.

C1 C2

QS

RP T

Let T be the mid point of PR, perpendicular from T to PR meets C1C2 at M which is mid point of C1C2. Also MP2 = MR2 = MQ2 = MS2 = MT2 + TR2

Hence M is centre of the required circle

PR = LDCT = 2 2

1 2 1 2C C � r � r = 100 � 4 = 96

Now 2 = MT2 + TR2

= 2

2 42

+ TR2 = 9 + 964

= 9 + 24 = 33

Hindi

C 1 C 2

Q S

R P T

ekuk T, PR dk e/; fcUnq gSA T ls PR ij C1C2 dks M ij yEc gS tks C1C2 dk e/; fcUnq gSA

rFkk MP2 = MR2 = MQ2 = MS2 = MT2 + TR2

vr% M vHkh"V oÙ̀k dk dsUnz gS

PR = LDCT = 2 2

1 2 1 2C C � r � r = 100 � 4 = 96

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vc 2 = MT2 + TR2

= 2

2 42

+ TR2 = 9 + 964

= 9 + 24 = 33

6. If complex number z and w satisfy the relation z + w = zw

and 1w

+ z = wz then which of the following

are correct. [CR-CT] [301]

(A*) |z| + |w| = 2 if arg(z) = (B*) If arg(z) = then arg(w) = ± 3

(C*) If w = � 1 then Re(z) = 12

(D) If w = 1 then Re(z) = 12

;fn lfEeJ la[;k z vkSj w lEcU/k z + w = zw

vkSj 1w

+ z = wz dks larq"B djrk gS rc fuEu esa ls dkSulk lgh gS -

(A*) |z| + |w| = 2 ;fn dks.kkad (z) = (B*) ;fn dks.kkad (z) = rc dks.kkad (w) = ± 3

(C*) ;fn w = � 1 rc Re(z) = 12

(D) ;fn w = 1 rc Re(z) = 12

Sol. z + w = zw

...........(1) and rFkk 1w

+ z = wz .........(2)

(1) � (2) w � 1w

= z1� w

w

1� w

w

(1 + z ) = 0 z = � 1 or w = 1w

If z = � 1 then (1) becomes � 1 + w = � 1w

w2 � w + 1 = 0

;fn z = � 1 rc (1) � 1 + w = � 1w

w2 � w + 1 = 0

w = 1 3i

2

arg(w) = ± 3

If w = 1w

then (1) becomes z + w = zw

;fn w = 1w

rc (1) z + w = zw

w = �1 z � 1 = � z z + z = 1 2Re(z) = 1

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7. If f(k) = k

r 1

1r

and 2013

r 1

f(r)

= a f(b) + c, then [SS-SS] [302]

;fn f(k) = k

r 1

1r

vkSj 2013

r 1

f(r)

= a f(b) + c rc -

(A) b = c (B*) a � b = 1 (C*) b + c = 0 (D*) a + c = 1

Sol. f(k) = k

r 1

1r

= 1 +12

+ 13

+..........+ 1k

f(1) = 1

f(2) = 1 + 12

f(3) = 1 + 12

+ 13

:

f(2013) = 1 +12

+ 13

+ 14

..........+ 1

2013

Adding tksMus ij 2013

r 1

f(r)

= 1 × 2013 + 12

× 2012 + 13

× 2011 + ....... + 1

2013× 1

= 1 × 2013 + 12

× (2013 � 1) + 13

× (2013 � 2) + ....... + 1

2013 × (2013 � 2012)

= 20131 1 1 1 2 3 2012

1 ...... � ..........2 3 2013 2 3 4 2013

= 2013 f(2013) � 1 1 1 1

1� 1� 1� ...... 1�2 3 4 2013

= 2013 f(2013) � 1 × 2012 + 1 1 1

1 ........2 3 2013

= 2013 f(2013) � 2013 + f(2013) = 2014 f(2013) � 2013

a = 2014, b = 2013 , c = � 2013

8. A function f is defined by f(x) = 0

cos tcos(x t)dt

; 0 x 2 then which of the following hold(s) good?

[DI-SP] [301] (A) f (x) is continuous but not differentiable in (0, 2)

(B) Maximum value of f is

(C*) There exists atleast one c (0, 2) such that f '(c) = 0.

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(D*) Minimum value of f is � 2

,d Qyu f, f(x) =0

cos tcos(x t)dt

; 0 x 2 }kjk ifjHkkf"kr gS] rc fuEu esa ls lR; gS ?

(A) f(x), (0, 2) esa lrr~ gS ijUrq vodyuh; ugh gSA

(B) f dk vf/kdre eku gSA

(C*) de ls de ,d c (0, 2) bl izdkj gS fd f '(c) = 0

(D*) f dk U;wure eku �2

gSA

Sol. f (x) = 0

cos tcos(x t)dt

....(1)

= 0

cos t·cos(x t)dt

f (x) = 0

cos t·cos(x t)dt

....(2)

(1) + (2) gives

2 f (x) = 0

cos t(2cos x·cos t)dt

f (x) = cos x 2

0

cos t dt

= 2 cos x2

2

0

cos t dt

f (x) = cos x2

Now verify vc lR;kfir.

9. Let Pn = n(3n)!(2n)!

(n = 1, 2, 3,.......) then n

n

P2lim

3 n is [LT-LG] [302]

ekuk Pn = n(3n)!(2n)!

(n = 1, 2, 3,.......) rc n

n

P2lim

3 n gS -

(A) 274e

(B*) 9e

1�cos x

2x 0

e �1lim

x

(C) e3

23 (D*)

92e

Sol. nPn

= 1/n

n

((2n 1)(2n 2).......(2n n))lim

n

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nPn

= 1/n

2n 1 2n 2 2n 3 2n n.........

n n n n

logenP

n=

n

er 1

1 2n rlog

n n

logenP

n=

n

enr 1

1 rlim log 2

n n

dx)x2(lognP

loglim1

0

en

en

nen

Plim log

n=

27n

4e

nlim nP

n=

274e

10. If yx � xy = 1, then the value of dydx

at x = 1 is - JPT-1 [MD-GN] [303]

;fn yx � xy = 1 rc x = 1 ij dydx

dk eku gS -

(A) n(4e2) (B*) n2e

4

(C*) 2(1 � n 2) (D) 2(1 + n 2)

Sol. yx � xy = 1 [if x = 1 then y = 2] [;fn x = 1 rc y = 2]

Let ekuk P = yx

n P = x n y

1P

dPdx

= n y + xy

dydx

dPdx

= Px dy

nyy dx

dPdx

= yx x dyny

y dx

....... (1)

Let ekuk Q = xy

n Q = y n x

1Q

dQdx

=yx

+ n x . dydx

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dQdx

= xy y dy

nx.x dx

........ (2)

Now vc yx � xy = 1 P � Q = 1

differentiate both sides w.r.t. x

x ds lkis{k vodyu djus ij

dPdx�

dQdx

= 0 yx x dyny

y dx

� xy

y dynx.

x dx

= 0 {from (1) and (2)} {(1) vkSj (2) ls}

when tc x = 1, y = 2 dydx

= 2(1 � n 2)

11. The curve, with the property that the projection of the ordinate on the normal is constant and has a length equal to 'a', may be (Where c is are abitrary constant) [TN-NR] [302]

oØ ftldh] dksfV dk vfHkyEc ij iz{ksi vpj gS rFkk a ds cjkcj yEckbZ gS] gks ldrk gS&

(tgk¡ c LoSPN vpj gS)

(A*) 2 2aln y a y x c (B) 2 2x a y c

(C) (y � a)2 = cx (D*) | y + 2 2y � a | = cex/a

Sol. Ordinate = PM. Let P (x, y) [T/S]

dksfV = PM. ekuk P (x, y)

Projection of ordinate on normal = PN

dksfV dk vfHkyEc ij iz{ksi = PN

PN = PM cos = a (given fn;k x;k gS)

2

ya

1 tan

y = 21a 1 (y )

2 2y ady

dx a

2 2

adydx

y a

2 2aln|y y a | x c

12. If z � axis be vertical, then the equation of the line of greatest slope through the point (2, �1, 0) on the plane 2x + 3y � 4z = 1 is [TD-PP] [304]

;fn z- v{k m/okZ/kj gS] rc lery 2x + 3y � 4z = 1 ij fLFkr fcUnq (2, �1, 0) ls xqtjus okyh vf/kdre izo.krk dh

ljy js[kk dk lehdj.k gksxk&

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(A*) x 6 y 13 z 13

8 12 13

(B) x � 2 y 1 z � 0

3 4 5

(C) x � 2 y 1 z � 0

3 4 12

(D*) x � 2 y 1 z

8 12 13

Sol. Let equation of the line with greatest slope is x � 2 y 1 z

a b c

ekuk vf/kdre izo.krk dh ljy js[kk dk lehdj.k x � 2 y 1 z

a b c

gSA

Where tgk¡ 2a + 3b � 4c = 0 ���.(1)

We know that equation of horizontal plane is z = 0

ge tkurs gS fd {kSfrt lery dk lehdj.k z = 0 gksrk gSA

i.e. vFkkZr~ , 0.x + 0.y + 1.z = 0

Now, a vector along the line of intersection of given plane and horizontal plane is

� � �i j k� �V 0 0 1 3i 2 j

2 3 �4

vr% fn;s x;s lery ,oa {ksfrt lery dh izfrPNsnu js[kk ds vuqfn'k ,d lfn'k

� � �i j k� �V 0 0 1 3i 2 j

2 3 �4

Since the line of greatest slope is also perpendicular to the vector V

.

pwafd vf/kdre izo.krk dh js[kk Hkh lfn'k V

ds yEcor~ gSA

Hence vr% , �3a + 2b = 0 ���.(2)

From (1) & o (2) ls a b c8 12 13

Equation of the line of greatest slope = x � 2 y 1 z

8 12 13

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vf/kdre izo.krk dh js[kk dk lehd j.k = x � 2 y 1 z

8 12 13

13. If , , are roots of cubic equation x3 + 3x + 2 = 0. If the equation whose roots are [QE-TE] [302]

(� ) (� ), (�) (� ), ( � ) (� ) is x3 + ax2 + bx + c = 0 then which of the following is/are correct.

;fn ?kuh; lehdj.k x3 + 3x + 2 = 0 ds ewy , , gSA ;fn lehdj.k x3 + ax2 + bx + c = 0 ds ewy

(� ) (� ), (�) (� ), ( � ) (� ) gS] rc fuEu esa ls lgh gS&

(A*) 24a + b + c = 0 (B*) b = 0 (C) a + b + c = 9 (D*) abc = 0

Sol. + + = 0

+ + = 3

= �2

Let ekuk y = (� ) (� )

y = 2 � (+ ) +

= 2 � (+ + ) + 2

y = 2 � 3 + 2

y = 3 � 3� 4

we know that 3 = � 3� 2

ge tkurs gS 3 = � 3� 2

y = � 6(+ 1) = �6

y 6

so equation is y3 + 9y2 � 216 = 0

lehdj.k y3 + 9y2 � 216 = 0 gSA

14. Sum of first n terms of the series 1 + 3 + 5 + 7 + ..... is 100. If 3 consecutive terms be removed from these n terms then sum of the remaining terms and sum of original terms are in the ratio 17 : 20. Which of the following is/are correct. [SS-AP] [303]

(A*) sum of the square of removed terms is 83

(B*) all 3 removed terms are prime

(C*) n = 10

(D) If , , are removed terms then coefficient of x in (x + )(x + )(x + ) is 81

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Js.kh 1 + 3 + 5 + 7 + ..... ds izFke n inksa dk ;ksxQy 100 gSA ;fn 3 Øekxr inksa dks bu n la[;kvksa ls gVk;k tk,

rc 'ks"k inksa ds ;ksxQy vkSj ewy inksa ds ;ksxQy dk vuqikr 17 : 20 gS fuEu esa ls lgh gS -

(A*) 83 gVk, x, inksa ds oxksZ dk ;ksxQy gSA

(B*) lHkh 3 gVk, x, in vHkkT; gSA

(C*) n = 10

(D) ;fn , , gVk, x, in gS rc (x + )(x + )(x + ) esa x dk xq.kkad 81 gSA

Sol. n2

[2 + (n � 1)2] = 100

n2 = 100 n = 10

sumof remainingterms

sumof original terms =

1720

= 85

100

so sum of 3 removed terms = 15

let these are a � d, a, a + d

a = 5

d = 2 terms are 3, 5, 7 hence A, B, C are correct

Hindi n2

[2 + (n � 1)2] = 100

n2 = 100 n = 10

'ks"k ink sadk ;ksxQy ewy ink sadk ;ksxQy

= 1720

= 85

100

blfy, gVk;h xbZ 3 la[;kvksa dk ;ksxQy gS = 15

ekuk ;s la[;k,¡ a � d, a, a + d

a = 5

d = 2 in 3, 5, 7 gSA vr% A, B, C lgh gSA

15. The range of values of m for which the line y = mx and the curve y =2

x

x 1enclose a region, lies in set

m ds ekuksa dk ifjlj ftlds fy, js[kk y = mx vkSj y =2

x

x 1 ,d {ks=k dks ifjc) djrs gS] ftl leqPp; esa fLFkr gS]

og gS& [SL-MS] [304]

(A*) (�1, 1) (B*) (0, 1) (C*) [0, 1] (D) (1, )

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Sol. Solving gy djus ij

mx =x

x 1 x2 + 1 =

1m

or x = 0

x2 = 1m

� 1 > 0 for a region {ks=k ds fy,

m 1

m

< 0 m (0, 1)

Note: for m = 0 or m = 1 the line does not enclose a region. uksV: m = 0 ;k m = 1 ds fy, js[kk {ks=k dks ifjc) ugha djrh gSA

SECTION � 3 : Matching List Type (Only One Option Correct) [k.M � 3 : lqesyu lwph izd kj (dsoy ,d fodYi lgh)

This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List-I and List-II are given as options (A),(B),(C) and (D), out of which ONE is correct.

bl [k.M esa 4 cgqfodYi iz'u gSA izR;sd iz'u esa lqesyu lwfp;k¡ gSA lwfp;ksa ds fy, dwV ds fodYi (A), (B), (C) rFkk (D) gS , ftuesa ls dsoy ,d lgh gSA

16. Match List I with List II and select the correct answer using the code given below the lists : lwph I dks lwph II ls lqesfyr dhft, rFkk lwfp;ksa ds uhps fn, x, dksM dk ç;ksx djds lgh mÙkj pqfu;s % [MTC-MS] [302] List-I List-II lwph- I lwph - II (P) If quadratic eqution (2 cos )x2 � (2 cosec)x + (cosec2

) = 0 1. 2 has only one solution for x, then number of values of

(0 2)are

(Q) The value of 2

2 2

0

x(sin (sinx) cos (cos x))dx

is equal to 2. 3

(R) If e is eccentricity of ellipse whose area is half the area of its auxiliary 3. 4

circle then 4e2 =

(S) If t0 is term independent of x in expansion of 9

2 1x �

x

and 1mt ,

2mt 4. 7

are coefficient of two middle terms then 1 20 m mt t t

12

feyku dhft, -

Page 14: JEE Advanced 2016 Mock test paper

[Type text] Page 14

lwph- I lwph - II (P) ;fn f}?kkr lehdj.k (2 cos )x2 � (2 cosec)x + (cosec2

) = 0 1. 2

x ds fy, dsoy ,d gy j[krk gS rc (0 2) ds ekuksa dh la[;k gS

(Q) 2

2 2

0

x(sin (sinx) cos (cos x))dx

dk eku cjkcj gS& 2. 3

(R) ;fn nh?kZoÙ̀k d h mRdsUnzrk e gS ftldk {ks=kQy] blds lgk;d oÙ̀k 3. 4

ds {ks=kQy dk vk/kk gS] rc 4e2 =

(S) ;fn 9

2 1x �

x

ds foLrkj esa x ls Lora=k in t0 gS rFkk 1mt ,2mt 4. 7

nks e/; inksa dk xq.kkad gS] rc 1 20 m mt t t

12

gS&

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

Sol. (P) If ;fn cos = 0 = 2

,32

If ;fn cos0 D = 0

4 cosec �4.2.cos .cosec= 0

1 = 2 cos sin 0 24

sin2= 1 2= 2

, 52

Four values of

ds pkj ekuksa ds fy,

(Q) Let ekuk I = 2 2

0

x(sin (sinx) cos (cos x))dx

= / 2

2 2 2 2

0

x(sin (sinx) cos (cos x)) � x sin (sinx) cos cos x dx

= / 2

2 2

0

(sin (sinx) cos (cos x))dx

Page 15: JEE Advanced 2016 Mock test paper

[Type text] Page 15

= / 2 / 2 / 2

2 2

0 0 0

(sin (sinx)dx dx � (sin (cos x)dx

= . 2

= 2

2

Ans. = 2

(R) a2 = 2ab

a = 2b

b2 = a2(1 � e2)

1 = 4(1�e2)

4e2 = 3

(S) t r+1 = 9c r (x2)q� r r

�1

x

= 9cr.x18�3r(�1)r

put 18 � 3r = 0 j[kus ij

r = 6

t0 = 9C6

9

2 1x �

x

index in odd lwpukad fo"ke esa gSA

two middle term nks e/; in 1mt = 9c4 x

4, 2mt =9c4(�x)5

1 29

0 m m 6t t t c 0

12 12

=

8412

17. List-I [MTC-MS] [301] List-II lwph- I lwph - II

(P) If a,b,c are non zero real numbers, system of equations y + z = a + 2x, 1. � 20

x + z = b + 2y, x + y = c + 2z are consistent and b = 4a + c4

, then

sum of roots of quadratic equation at2 + bt + c = 0 is

(Q) Let A be a matrix of order 3, such that A2 = 2A � , where is an 2. �8

identity matrix of order 3. If A5 = A + then =

(R) Let an ant start from origin (O) travels 3 units on negative x-axis, 3. �3

travels 1 unit parallel to positive y-axis, travels 2 units parallel to negative z-axis to

reach point A : � � �p i � 2 j k

and q

be such that resultant of

p

and q

is � � �3i � 3 j 4k , then [p OA q]

=

Page 16: JEE Advanced 2016 Mock test paper

[Type text] Page 16

(S) If z1,z2 are two complex number numbers satisfying 1 2

1 2

z � 3z

3 � z z = 1, 4 1

|z1| 3, then |z2| is equal to

lwph- I lwph - II

(P) ;fn a, b, c v'kwU; okLrfod la[;k,¡ gS] lehd j.k fudk; y + z = a + 2x, 1. � 20

x + z = b + 2y, x + y = c + 2z laxr gS rFkk b = 4a + c4

, rc

f}?kkr lehdj.k at2 + bt + c = 0 ds ewyksa dk ;ksxQy gS&

(Q) ekuk A, 3 Øe dh vkO;wg bl izdkj gS fd A2 = 2A � tgk¡ 3 dksfV 2. �8

dh rRled vkO;wg gS ;fn A5 = A + rc =

(R) ekuk ,d phVh ewy fcUnq (O) ls 3 bdkbZ _ .kkRed x-v{k ij pyrh gS] 3. �3

/kukRed y v{k ds lekUrj 1 bdkbZ pyrh gS] _ .kkRed z-v{k ds lekUrj

2 bdkbZ py dj A = � � �p i � 2 j k

fcUnq ij igqaprh gS rFkk lfn'k

q

bl izdkj gS fd lfn'k p

vkSj q

dk ifj.kkeh � � �3i � 3 j 4k gS

rc [p OA q]

=

(S) ;fn z1, z2 nks lfEeJ la[;k,¡ bl izdkj gS fd 1 2

1 2

z � 3z

3 � z z = 1, |z1| 3 4 1

gS] rc |z2| cjkcj gS&

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

Sol. (P) Adding tksMus ij

2(x+ y +z) = a+ b + c + 2x + 2y +2z

0 = a + b + c ��..(1)

b = 4a + c4

16a � 4b + c = 0 ��..(2)

comparing at2 + bt + c = 0

Page 17: JEE Advanced 2016 Mock test paper

[Type text] Page 17

at2 + bt + c = 0 rqyuk djus ij

by (1) t = 1

(1) ls t = 1

by (2) t = �4

(2) ls t = �4

sum of roots = � 3

ewyks dk ;ksxQy = � 3

t = �4

(Q) A3 = AA2

=A(2A �)

=2A2 � A

= 2(2A � ) � A = 3A � 2

= 3A � 2

A4 = 4A � 3

A5 = 5A � 4

=5, = �4 = �20

(R) A =(�3, 1,�2)

p q = (3, �3, 4)

q =(3 �1, �3 + 2, 4 � 1)

=(2, �1, 3) p OA q

=

1 �2 1

�3 1 �2

2 �1 3

= 1(3, �2)+ 2 (�9 + 4) + 1(3 �2)

= + 1 + 2 (�5) + 1

= � 8

(S) |z1|2 � (z13 2z + 1z 3z2) +9|z2|

2 = 9 �(3z1 2z + 3 1z z2) + |z1|2|z2|

2

|z1|2 � |z1|

2|z2|2 + 9|z2|

2 �9 = 0

|z1|2 (1 � |z2|

2 ) + 9(|z2|2 � 1) = 0

(|z1|2 � 9) (1 � |z2|

2) = 0

Page 18: JEE Advanced 2016 Mock test paper

[Type text] Page 18

|z1| 3 |z2| = 1

18. List-I [MTC-MS] [302] List-II lwph- I lwph - II

(P) Given that the side length of a rhombus is the geometric mean 1. 11

of the lengths of its diagonals. The acute angle between the sides

of rhombus is k

then k equals

(Q) If y = mx + 1 is tangent to y2 = 4x, then m is 2. 1

(R) The absolute value of the difference of two distinct value of x 3. e

which minimizes the value of expression (x2 �x � 21) (x2

� x � 39) is

(S) If x1, x2, x3 are the three real solutions of the equation 4. 6

1(2 nx�1) 9x e

=

191 e

ln x�0.5x , where x, 1, x2 1, x3 1,

then value of x1x2x3 is

lwph- I lwph - II

(P) fn;k x;k gS fd leprqHkZqt dh Hkqtk dh yEckbZ blds fod.kksZ dh 1. 11

yEckbZ dk xq.kksÙkj ek/; gS ;fn leprqHkqZt dh Hkqtkvksa ds e/; U;wudks.k

k

gS] rc k cjkcj gS&

(Q) ;fn y = mx + 1 ijoy; y2 = 4x dh Li'kZ js[kk gS rc m gS 2. 1

(R) x ds nks fofHkUu ekuksa ds vUrj dk fujis{k eku ftuds fy, O;atd 3. e

(x2 �x � 21) (x2

� x � 39) dk eku U;wure gks] gksxk&

(S) ;fn lehd j.k1

(2 nx�1) 9x e =

191 e

ln x�0.5x ds rhu 4. 6

okLrfod gy x1, x2, x3 gS tgk¡ x, 1, x2 1, x3 1,

rc x1x2x3 dk eku gS&

Codes : P Q R S (A*) 4 2 1 3

Page 19: JEE Advanced 2016 Mock test paper

[Type text] Page 19

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

Sol. (P) tana

2 b

, x = a b

x2 = 2 2a b

4 4

Area {ks=kQy = x2.sin = 12

a.b

sin = 12

= 6

a/2

x

/2 b/2

(Q) c = am

1 = 1m

m = 1

(R) Let ekuk y = x2 � x � 21

General equation = (y) . (y � 18)

O;kid lehdj.k = (y) . (y � 18)

= y2 � 2.y.9 + 81 � 81

= (y � 9)2 � 81

minimum at y = 9

y = 9 ij U;wure

x2 � x � 21 = 9

x2 � x � 30 = 06

�5 difference vUrj |6 � (�5)| = 11

(S) Let ekuk t = x 2ln x�11

ln x� 22x x

t2 + e1/9 = (1 + e1/9)t

t2 � (1 + e1/9) t + e1/9 = 0

Page 20: JEE Advanced 2016 Mock test paper

[Type text] Page 20

t = 1 or t = e1/9

2ln x�1

2x = 1 or

2ln x�1

2x = e1/9

2lnx �1

2 = 0 take loge

lnx =12

2ln x�1

2 1lnx

9

x = e1/2 2y �1 1

y2 9

, y = lnx

2y2 � y = 29

y2 + y3 = 12

lnx2 + lnx3 = 12

x2x3 = e1/2

x1x2x3 = e1/2.e1/2 = e

19. List-I [MTC-MS] [302] List-II lwph- I lwph - II

(P) The number of solutions of equation x 2

0

xx ln t dt

3 , x R+ are 1. 1

(Q) Points (t�1, 2t + 2) and (2t + 1, t +1) are image of each other with 2. 2

respect to line L. If line L passes through (�1, 0) then value of �2t is

(R) If f(x) = 3 2x 2x ax 6 ; x 1

2x b ; x 1

is differentiable for x R, then 3. 4

value of (b � a) is

(S) If z0 is the only point of contact of curves z + z = 2|z � 1|, arg (z + 1 + i) = , 4. 7

(0, ), then |z0|2 may be

Page 21: JEE Advanced 2016 Mock test paper

[Type text] Page 21

lwph- I lwph - II

(P) lehdj.k x 2

0

xx ln t dt

3 , x R+ ds gyksa dh la[;k gS& 1. 1

(Q) (t�1, 2t + 2) vkSj (2t + 1, t +1) js[kk L ds lkis{k ,d nqljs ds izfrfcEc gS ;fn 2. 2

L, (�1, 0) ls xqtjrh gS rc�2t dk eku gS&

(R) ;fn f(x) = 3 2x 2x ax 6 ; x 1

2x b ; x 1

, x R ds fy, vodyuh; gS rc 3. 4

(b � a) dk eku gS&

(S) ;fn z0 oØ z + z = 2|z � 1|, d ks.kkad (z + 1 + i) = , (0, ) dk dsoy 4. 7

,d Li'kZ fcUnq gS] rc |z0|2 dk eku gks ldrk gS&

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

Sol. (P) x + x(lnx � 1) = 3x2

xlnx = 3x2

lnx = 3x

two solutions nks gy

(Q) mid-point 3t 3t 3

,2 2

and given point (�1, 0)

e/; fcUnq 3t 3t 3

,2 2

vkSj fn;k x;k fcUnq (�1, 0)

slope izo.krk =

32

t

1t

12t3

0�)1t(23

slope of given points nks fcUnqvksa dh izo.krk = 2�t�

1t1�t2�1�t

1t

Page 22: JEE Advanced 2016 Mock test paper

[Type text] Page 22

product of slopes izo.krkvksa dk xq.kuQy =

)2t(�1t

32

t

1t = �1

(t + 1)2 =

32

t (t + 2) 1 � 4 2t3 3

t2 + 2t+1 = t2 + 2t + 3t2

+ 34

� 1 = 2t

y=x

y=x/3

(R) 1 + 2 + a + 6 = 2 + b

7 = b � a

(S) 2x = 2 22 y)1�x(

arg(Z � (�1 � i)) =

x2 = x2 � 2x + 1 + y2

y2 = 2x � 1

y2 = 2

21

�x

y = m

21

�x + m

2/1

�1 = m

23

+ m21

m2

1m22m3

3m2 = 2m + 1

m =1 tangent is Li'kZ js[kk y = x

solving with parabola ijoy; ds lkFk gy djus ij

y = 2x � 1

x = 1 y = 1 P(1, 1) z0 1 + i

Page 23: JEE Advanced 2016 Mock test paper

[Type text] Page 23

|z0|2 = 2

Z0

Page 24: JEE Advanced 2016 Mock test paper

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P2-1

BATCH : P, F, R_(JPT-1) DATE : 10-05-2015

Factulty :- SYLLABUS : FULL SYLLABUS

TIME : 3 Hrs + 3 Hrs (Morning + Evening)

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

1 to 15 MCQ 15 4 0 60

16 to 19 Match matrix listing 4 3 �1 12

20 to 34 MCQ 15 4 0 60

35 to 38 Match matrix listing 4 3 �1 12

39 to 53 MCQ 15 4 0 60

54 to 57 Match matrix listing 4 3 �1 12

57 216

Paper-2

TotalTotal

PAPER-2

SECTION-1 : (One or more option correct type) [k.M�1 : (,d ;k vf/kd lgh fodYi çdkj)

This section contains 15 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.

bl [k.M esa 15 cgqfodYi ç'u gSA çR;sd ç'u esa pkj fodYi (A), (B), (C) vkSj (D) gS] ftuesa ls dsoy ,d ;k vf/kd lgh gSA

MCQ._(15)

20. Two waves are simultaneously passing through a string, their equations are y1 = A1 sin k(x � vt) and y2 = A2 sin k (x � vt + x0) respectively, where k = 3.14 cm�1 and x0 = 3.0 cm, A1 = 6 mm, A2 = 5.2 mm, select the correct statements. [ST-SO](102)

(A*) Their phase difference is 3

(B*) They produce destructive interference

(C*) Their resulting amplitude is 0.8 mm

(D) Their resulting amplitude is 11.2 mm

nks rjax ,d Mksjh ls ,d lkFk xqtj jgh gSA muds lehdj.k Øe'k% y1 = A1 sin k(x � vt) rFkk

y2 = A2 sin k (x � vt + x0) tgk¡ k = 3.14 cm�1 rFkk x0 = 3.0 cm, A1 = 6 mm, A2 = 5.2 mm gSA lR; dFkuksa

dk p;u dhft,A

Page 25: JEE Advanced 2016 Mock test paper

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P2-2

(A*) mudk dykUrj 3 gksxkA

(B*) os fouk'kh O;frdj.k mRiUu djrk gSA

(C*) mudk ifj.kkeh vk;ke 0.8 mm gSA

(D) mudk ifj.kkeh vk;ke 11.2 mm gSA

Sol. (a) phase difference dykUrj = kx0 = 3.14 × 3 = 3

(b) odd multiple of . ; dk fo"ke xq.kt

(c) Amin = |A1 - A2| = |6 - 5.2| = 0.8 mm 21. A black body has a temperature of 'T' (in kelvin) and wavelength �� corresponding to maximum

energy density. When body cools, if the wavelength corresponding to the maximum energy density changes by 900% then choose the correct alternative(s) [HT-FA](104)

,d d f̀".kd k oLrq dk rki 'T' (dSfYou esa) rFkk vf/kdre~ Å tkZ ?kuRo ds laxr rjaxnS/;Z �� gSA tc oLrq BaMh

gksrh gSA ;fn vf/kdre~ Å tkZ ?kuRo ds laxr rjaxnS/;Z 900% ls ifjorfrZr gksrh gS rc lgh fodYiksa dk p;u

dhft,A

(A) Temperature of black body to which it is cooled is T9

(B*) Ratio of rate of emission i.e. final to initial will be 1 : 10000

(C*) The magnitude of difference in two temperature 9T10

(D) Ratio of rate of emission i.e. final to initial will be 1 : 6561

(A) d f̀".kd k oLrq dk rki ftl ij ;g BaMh gksrh gS] T9

gSA

(B*) mRltZu dh nj dk vuqikr vFkkZr~ vfUre rFkk izkjfEHkd dk vuqikr 1 : 10000 gSA

(C*) nksuksa rkiksa esa vUrj dk ifjek.k 9T10

gSA

(D) mRltZu dh nj dk vuqikr vFkkZr~ vfUre rFkk izkjfEHkd dk vuqikr 1 : 6561 gSA

Sol.(bc) = 9 fTf = iTi

f � = 9 10.Tf = T

f = 10 f

TT

10

magnitude change in 'T' esa ifjorZu dk ifjek.k 9T10

4

f f

i i

Rate T

Rate T

1 : 10000

22. In the given circuit potential of the point A is 9V higher than potential of the point B. Choose correct alternative(s) [CE-KL] (104)

fn;s x;s ifjiFk esa fcUnq A dk foHko fcUnq B ds foHko ls 9V mPp gSA lgh fodYiksa dk p;u djksA

Page 26: JEE Advanced 2016 Mock test paper

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P2-3

(A) Value of resistance R is 1

izfrjks/k R dk eku 1 gS

(B*) Value of resistance R is 7

izfrjk/sk R dk eku 7 gS

(C*) magnitude of potential difference between B and D is 30 V.

B rFkk D ds e/; foHkokUrj dk ifjek.k 30V gSA

(D) magnitude of potential difference between B and C is 15 V.

B rFkk C ds e/; foHkokUrj dk ifjek.k 15V gSA

Ans. (bc)

Sol. 24 15 6

IR 1 2 1

A BV V 6 Ir

33

9 6R 4

R = 7

VBC = 15 � 3(2) = 9V

VBD = 30V

23. A block of mass 'm' rests on a fixed incline plane of inclination '' with horizontal. Assume friction is large enough to make the block stationary. Then choose correct alternative(s). [FR-MQ] (104)

m nzO;eku dk ,d CykWd {kSfrt ls dks.k ij >qds fLFkj urry ij fojke ij gSA ;g ekfu, fd ?k"kZ.k CykWd

dks fLFkj j[kus ds fy, i;kZIr gSA rc lgh fodYiksa dk p;u djksA

(A) Maximum value of horizontal component of friction is mg

?k"kZ.k cy ds {kSfrt ?kVd dk vf/kdre eku mg gS

(B*) Angle at which horizontal component of friction is maximum is /4

dks.k , ftl ij ?k"kZ.k cy dk {kSfrt ?kVd vf/kdre gS, /4 gksxk

(C*) Maximum value of horizontal component of friction is mg/2

?k"kZ.k cy ds {kSfrt ?kVd dk vf/kdre eku mg/2 gS

(D) Angle at which horizontal component of friction is maximum is /3

dks.k , ftl ij ?k"kZ.k cy dk {kSfrt ?kVd vf/kdre gS, /3 gksxk

Ans. (BC)

Sol.

Page 27: JEE Advanced 2016 Mock test paper

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P2-4

fx = mg sin

fx = f cos = mg sin cos

xmg

f sin 22

fxmax = mg/2 (4

)

24. A particle travels along the path y = a + bx + cx2 where a, b, c are positive constants. If v0 is the

speed of the particle and it is constant. Then choose the correct options, if at the given instant particle is at x = 0. [CM-VT](104)

,d d.k iFk y = a + bx + cx2 ds vuqfn'k xfr'khy gS tgk¡ a, b, c /kukRed fu;rkad gSaA ;fn v0 d.k dh pky

gS rFkk ;g fu;r gSA rc lgh fodYiksa dk p;u djks] ;fn fn;s x;s {k.k ij d.k x = 0 ij gS

(A*) radius of curvature is 2 3/2(1 b )

2c

oØrk f=kT;k 2 3/2(1 b )

2c

gksxh

(B*) magnitude of net acceleration is

20

3/22

2cv

1 b

dqy Roj.k dk ifjek.k

20

3/22

2cv

1 b gS

(C) radius of curvature is

2 20

3.22

2c v

1 b

oØrk f=kT;k

2 20

3.22

2c v

1 b gksxh

(D) magnitude of centriptal acceleration 2

02

v

1 b

vfHkdsfUnz; Roj.k dk ifjek.k 2

02

v

1 b gksxk

Sol. (ab)

R =

3/22

2 3/2

2

2

dy1

dx (1 b )2cd y

dx

r ta a a

ta

= 0. (speed constant) (pky fu;r gS)

| a

| = | ra

| = 20v

R =

20

3/22

2cv

1 b

25. In the circuit shown below the switch between A & B is closed at t = 0, then choose the correct options. (Consider circuit to be in steady state at t < 0 )

Page 28: JEE Advanced 2016 Mock test paper

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P2-5

[Made AMG Sir_2013-14] [EI-CI](102)

uhps n'kkZ;s x;s ifjiFk esa A o B ds e/; dqath t = 0 ij cUn dh tkrh gS, rc lgh fodYi@fodYiksa dk p;u

dhft,A (ekfu;s fd ifjiFk t < 0 ij LFkk;h voLFkk esa gSA)

B

R1

L

A E

R2

(A*) Current through R1 and R2 will not change just after the switch is closed

dqath cUn djus ds Bhd ckn R1 o R2 ls xqtjus okyh /kkjk ifjofrZr ugha gksrh gSA

(B) Current through R1 and R2 will change just after switch is closed

dqath cUn djus ds Bhd ckn R1 o R2 ls xqtjus okyh /kkjk ifjofrZr gksrh gSA

(C) Current through L will be different at both instants. i.e., just after switch is closed and after long time.

nksuksa {k.kksa ¼vFkkZr~ dqath cUn djus ds Bhd ckn ,oa cgqr yEcs le; i'pkr~½ ij L ls xqtjus okyh /kkjk

fHkUu&fHkUu gksxhA

(D*) Current through R2 will be same at t < 0 and t .

R2 ls xqtjus okyh /kkjk t < 0 ij rFkk t ij leku gksxhA

Sol. Since initially inductor L has current 1R

E flowing through it, so when switch is closed inductor will

give some induced emf, hence (A), (D)

pqafd izkjEHk esa izsjdRo L ls xqtjus okyh /kkjk 1R

E gS rFkk tc dqath dks cUn fd;k tkrk gS izsjdRo dqN izsfjr

fo-ok-cy nsrk gS vr% (A), (D)

26. Two spring of spring constant K and 4K are attached with a block of mass m and other end of springs are free, system is placed between two rigid wall W1 and W2. Springs are in its natural

length at instant shown, at t = 0 a sharp impulse is give to the block towards wall W1 then

[SH-SM](104)

K 4K

Smooth surface

W1W2

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P2-6

nks fLiazx ftuds cy fu;rkad Øe'k% K rFkk 4K gS] fp=kkuqlkj m nzO;eku ds ,d CykWd ls tqM+h gqbZ gS rFkk

fLiazxksa ds nwljs fljs eqDr gSA ;g fudk; nks n<̀+ nhokjksa W1 rFkk W2 ds e/; j[kk gqvk gSA t = 0 ij, nksuksa fLiazx

lkekU; yEckbZ esa gS rFkk n'kkZ;s x;s bl {k.k ij CykWd dks ,d rh{.k vkosx nhokj W1 dh rjQ fn;k tkrk

gS] rks

K 4K

W1 W2

fpduh lrg

(A) Block perform SHM

(B) Time period of motion of block is m

25K

(C*) Block perform oscillatory motion (but not SHM) and its time period is 3 m2 K

(D) Amplitude of SHM of block is 5mK

(A) CykWd ljyvkorZ xfr djsxkA

(B) CykWd dk vkorZdky m

25K

gksxkA

(C*) CykWd nksyu xfr djsxk (ijUrq ljyvkorZ xfr ugha djsxk½ rFkk bldk vkorZdky 3 m2 K

gksxkA

(D) CykWd dh ljyvkorZ xfr dk vk;ke 5mK

gksxkA

Sol. Two springs are not compressed or extended together and compression of both springs are not same. So block not perform SHM, it perform oscillatory motion and time period is given by

nksuksa fLiazx ,d lkFk lEihfMr ;k izlkfjr ugha gksrh gS rFkk nksuksa fLiazx esa lEihM+u leku ugha gSA vr% ljy

vkorZ xfr ugha djsaxk] ;g nkSyu xfr djsxk rFkk vkorZ dky fuEu izdkj gksxkA

T = m4K

+ mK

= 3 m2 K

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P2-7

27. A point object is kept at a distance of .OP u The radius of curvature of spherical surface APB is

.CP R The refractive index of the media are 1n and 2n which are as shown in the diagram, then

(consider only paraxial rays)

[GO-RS](103)

,d fcUnq oLrq nwjh OP = u ij j[kh gqbZ gSA xksykdkj lrg APB dh oØrk f=kT;k CP = R gSA ek/;e dk

viorZukad n1 o n2 gS tks fp=k esa n'kkZ;suqlkj gS] rc (dsoy lek{kh; fdj.ks ekusa)

(A*) If 1 2n n image is virtual for all values of u

(B*) If 2 12n n image is virtual when R u

(C) The image is real for all values of u, 1n and 2n

(D) If 2 1n n then image will be always real

(A*) ;fn 1 2n n gS rc u ds lHkh ekuksa ds fy, vkHkklh izfrfcEc cusxkA

(B*) ;fn 2 12n n gS rc R u ds fy, vkHkklh izfrfcEc cusxkA

(C) u, n1 rFkk n2 ds lHkh ekuksa ds fy, okLrfod izfrfcEc cusxkA

(D) ;fn 2 1n n rc ges'kk okLrfod izfrfcEc cusxkA

Sol. R

nn)u(

nvn 1212

O u

n1 n2

C

ri

A

P

B

un

Rnn

vn 1122

(a) if ;fn n1 > n2 v is � ve ; v _ .kkRed gS (a)

(b) un

Rn

vn2 111

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P2-8

u1

R1

v2

if ;fn R > u v is -ve _ .kkRed gS

28. An infinitely long straight wire carrying a current 1 is partially surrounded by ABCD loop as shown in figure, arc AD and BC have circular shape and the infinite wire passes through their centre C1C2. The loop has a length L, radius R and carries a current 2. The axis of the loop coincides with the wire, ABCD plane and infinite length wire are coplanar. Then

[EM-TL](102)

1 /kkjk dk ,d vuUr yEckbZ dk lh/kk /kkjkokgh rkj fp=k esa n'kkZ;suqlkj ywi ABCD }kjk vkaf'kd ifjc) gS]

pki AD rFkk BC oÙ̀kkdkj vkdkj j[krs gS ,oa vuUr yEckbZ dk rkj muds dsUnz C1C2 ls xqtjrk gSA ywi dh

yEckbZ L, f=kT;k R rFkk izokfgr /kkjk 2 gSA ywi dh v{k rkj ds lEikrh gS] ry ABCD o vuUr yEckbZ dk

rkj leryh; gSA rc

L 1

2

R D

A

B C

C2

C1

(A) Net force exerted on the loop by the wire is zero.

(B*) Net force exerted on the wire by the loop is 0 1 2I I LR

, in the plane ABCD and towards CD.

(C) Net force exerted by the wire on the loops is 0 1 2I I L2 R

, in the plane ABCD and towards CD.

(D*) Net torque acting on the loop, about axes C1C2 is zero.

(A) rkj }kjk ywi ij vkjksfir dqy cy 'kwU; gSA

(B*) ywi }kjk rkj ij vkjksfir dqy cy 0 1 2I I LR

gS, tks ry ABCD esa rFkk CD dh vksj gSA

(C) rkj }kjk ywi ij vkjksfir dqy cy 0 1 2I I L2 R

gS, tks ry ABCD esa rFkk CD dh vksj gSA

(D*) v{k C1C2 ds lkis{k ywi ij dk;Zjr~ dqy cyk?kw.kZ 'kwU; gSA

Sol. Force is exerted only by straight portions of loop

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P2-9

cy dsoy ywi ds lh/ks Hkkx }kjk vkjksfir gSA

Fnet = R2

L210

+

R2L210

option fodYi (B) & o (D)

29. A charged particle (+q) is moving simple harmonically on the x-axis with its mean position at origin. Amplitude of the particle is A and its angular frequency is .

,d vkosf'kr d.k (+q), x-v{k ij ljy vkorZ xfr djrk gS] ftldh ek/; fLFkfr eqy fcUnq ij gSA d.k dk

vk;ke A rFkk bldh dks.kh; vkof̀Ùk gSA [EM-FQ](104)

y

x (A, 0) (�A, 0) (0, 0)

(A) The magnitude of magnetic field at (2A, 0) will change periodically with period 2/.

(B*) The maximum magnitude of the magnetic field at (0, A) is Aq

40

(C*) The magnetic field at (A, A) at the moment the particle passes through

0,

2A

, will be

A55

q3 0

(D*) The magnitudes of magnetic field at (0, A) and (0, �A) will be same at any time.

(A) (2A, 0) ij pqEcdh; {ks=k dk ifjek.k vkorZdky 2/ ls vkorhZ :i ls ifjofrZr gksxkA

(B*) (0, A) ij pqEcdh; {ks=k dk vf/kdre ifjek.k Aq

40

gSA

(C*) d.k tc

0,

2A ls xqtjrk gS ml {k.k (A, A) ij pqEcdh; {ks=k

A55

q3 0

gksxkA

(D*) (0, A) rFkk (0, �A) ij pqEcdh; {ks=k dk ifjek.k fdlh Hkh le; ij leku gksxkA

Sol. (a) For (2A, 0), = 0° or 180°

B = 0 permanent zero LFkk;h :i ls 'kwU;

(b) Magnetic field will be max at (0, A) when the particle passes through (0, 0)

(0, A) ij pqEcdh; {ks=k vf/kdre gksxk tc d.k (0, 0) ls xqtjrk gSA

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P2-10

Bmax =

40 × 2A

90sin)A(q =

Aq

40

(0, A)

Vmax = A

(c) sin = 5

2

2/5A

A

A/2 A/2

A

(A, A)

v

4A

A2

2

r = 2

5A4

AA

22

V = 22 xA = 2

2

2A

A

=

23A

B =

40 ×

2rsinqv

=

40 × 2

25

A

5

223

Aq

=

40

Aq

54

5

223

= A55

q3 0

(d) same r leku, same leku

same B leku.

(0, A) r

(0, A) r

30. The radii of a spherical capacitor are equal to a and b (b > a). The space between them is filled with a dielectric of dielectric constant K and resistivity . At t = 0, the inner electrode is given a charge q0. Choose the correct options : [CP-EQ](103) [Capacitance]

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P2-11

[Made CPG sir 2011-12]

(A*) Charge q on the inner electrode as a function of time is given by 0

t

K0q q e

(B) In a short time, the charge on the inner electrode will become zero

(C*) After a long time, the charge on the outer sphere will become q0

(D*) The total amount of heat generated during the spreading of charge will be given by 2

0

0

q1 1H

a b 8 K

,d xksfy; la/kkfj=k dh f=kT;k a rFkk b (b > a) gSA buds e/; Hkjs inkFkZ dk ijkoS|qrkad K rFkk çfrjks/kdrk

gSA le; t = 0, ij vkUrfjd bysDVªksM+ dks vkos'k q0.fn;k x;k gks rkss

lgh dFku@dFkuksa dk p;u dhft,A

(A*) vkUrfjd bysDVªksM+ ij vkos'k q, le; ds Qyu ds :i esa 0

t

K0q q e

gksxkA

(B) vYi le; vUrjky es vkarfjd bysDVªksM++ ij vkos'k 'kwU; gks tk,xk

(C*) ,d YkEcs le; ds ckn] ckgjh xksys ij vkos'k q0 gks tk,xk

(D*) vkos'k ds forj.k dss nkSjku mRiUu dqy Å "ek dk eku 2

0

0

q1 1H

a b 8 K

gksxk

Sol. C = 04 abK

b a

R = b

2a

dr

4 r

= (b a)

4 ab

= RC = K0

q = q0e�t/

q = 0

tK

0q e

t = , q = 0

Amount of heat generated mRiUu Å "ek

2

0

0

q1 1H

a b 8 K

31. A small object moves counter clockwise along the circular path whose centre is at origin as shown in figure. As it moves along the path, its acceleration vector continuously points towards point S. Then the object [CH-HZ](104)

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P2-12

y

x

B

OSC A

D

(A*) Speed up as it moves from A to C via B. (B) Slows down as it moves from A to C via B.

(C*) Slows down as it moves from C to A via D. (D) Speed up as it moves from C to A via D.

,d NksVh oLrq oÙ̀kkd kj iFk ij okekorZ fn'kk esa xfr'khy gS] rFkk bl oÙ̀kkdkj iFk dk dsUnz fp=kkuqlkj ewy

fcUnq gSA iFk ds vuqfn'k xfr djrs gq, bldk Roj.k lfn'k yxkrkj fcUnq S dh rjQ funsZf'kr gS] rks oLrq

y

x

B

OSC A

D

(A*) B ls gksdj A ls C rd xfr ds nkSjku pky c<+rh gSA

(B) B ls gksdj A ls C rd xfr ds nkSjku pky ?kVrh gSA

(C*) D ls gksdj C ls A rd xfr ds nkSjku pky ?kVrh gSA

(D) D ls gksdj C ls A rd xfr ds nkSjku pky c<+rh gSA

Sol. (Moderate) As the object moves from A to C via B the angle between acceleration vector and velocity vector decreases from 90° and then increases back to 90°. Since the angle between

velocity and acceleration is acute, the object speeds up. As the object moves from C to A via D the angle between acceleration vector and velocity vector

increases from 90° and then decreases back to 90°. Since the angle between velocity and

acceleration is obtuse, the object slows down.

tSlk fd oLrq B ls gksdj A ls C xfr djrh gSA Roj.k lfn'k ,oa osx lfn'k ds e/; dks.k 90° ls ?kV+rk gS ,oa

rc c<+dj okil 90° gks tkrk gSA pwafd osx o Roj.k ds e/; U;wu dks.k gSA vr% oLrq dh pky c<+rh gSA

tSlk fd oLrq D ls gksdj C ls A xfr djrh gSA Roj.k lfn'k ,oa osx lfn'k ds e/; dks.k 90° ls c<+rk gS ,oa

rc ?kVdj okil 90° gks tkrk gSA pwafd osx o Roj.k ds e/; vf/kd dks.k gSA vr% oLrq /kheh gksxhA

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P2-13

y

x

B

OSC A

D

v

a

a v

32. In a radioactive decay reaction :

,d jsfM;ks/kehZ {k; vfHkfØ;k esa % [MADE: CSS 2012-13] [NP-DL](102)

A 2

B 9

C

Select correct alternative/s at the instant the number of the particles of B is maximum :

(A*) Activity of A is equal to activity of B (B*) No of atoms of A is 4.5 times of B

(C) Activity of A is more then activity of B (D) Activity of A is minimum

lgh fodYi@fodYiksa dk p;u ml {k.k dhft, tc B ds d.kksa dh la[;k vf/kdre gS %

(A*) A dh lfØ;rk B dh lfØ;rk ds rqY; gSA (B*) A ds ijek.kqvksa dh la[;k B dh 4.5 xquk gSA

(C) A dh lfØ;rk B dh lfØ;rk ls vf/kd gSA (D) A dh lfØ;rk U;wure gSA

Sol. NA 2 = NB 9

NA = 4.5 NB

33. In series LCR circuit voltage drop across resistance is 8 V and voltage across inductor to is 6 V across capacitor is 12 volt. Then select incorrect alternative/s : [AC-RC](102)

Js.kh LCR ifjiFk esa izfrjks/k ds fljksa ij foHko iru 8 V gS rFkk izsjd dq.Myh ds fljks ij oksYVst 6 V gS rFkk

la/kkfj=k ds fljks ij 12 V gSA rc vlR; fodYi@fodYiksa dk p;u dhft,A

(A*) Voltage of the source will be leading current in the circuit

ifjiFk esa L=kksr oksYVst /kkjk ls vkxs ¼leading½ gksxkA

(B*) Voltage drop across each element will be less than the applied voltage

izR;sd vo;o ds fljks ij foHko iru vkjksfir oksYVst ls de gksxkA

(C*) Power factor of circuit will be 4/3

ifjiFk dk 'kfDr xq.kkad 4/3 gksxkA

(D*) Source voltage is 26 volt

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P2-14

Lkzksr oksYVst 26 oksYV gSA

Sol. Since, cos = Z

R =

IZ

IR =

10

8 =

5

4

pwafd , cos = Z

R =

IZ

IR =

10

8 =

5

4

(cos dHkh Hkh ,d ls vf/kd ugha gks ldrk gSA)

Also ¼rFkk½, IxC > IxL XC > XL

Current will be leading (/kkjk vkxs gksxh)

In a LCR circuit (LCR ifjiFk esa)

2 2L C RV (V � V ) V 2 2(6 � 12) 8 10

V = 10 ; which is less than voltage drop across capacitor.

V = 10 ; tks la/kkfj=k ds fljksa ij foHko iru ls de gksxkA

34. Hydrogen gas and oxygen gas have volume 1cm3 each at N.T.P.

N.T.P. ij gkbMªkstu xSl o vkWDlhtu xSl dk ¼izR;sd dk½ vk;ru 1cm3 gSA [TH-FL] (103)

(A*) Number of molecules is same in both the gases.

nksuksa xSlksa esa v.kqvksa dh la[;k leku gSA

(B) The rms velocity of molecules of both the gases is the same.

nksuksa xSlkas ds v.kqvksa dh oxZ ek/; ewy pky leku gSA

(C*) The internal energy of each gas is the same.

çR;sd xSl dh vkUrfjd Å tkZ leku gSA

(D*) The average velocity of molecules of each gas is the same.

çR;sd xSl ds v.kqvksa dk vkSlr osx leku gSA

Sol. At NTP Same volume means same molecules f1 = f2 , so same tempereature means same

internal energy (U =f2

nRT). Average velocity is zero.

mÙkj NTP ij leku vk;ru dk vFkZ gS leku v.kq ] f1 = f2 vr% leku rki dk vFkZ gS leku vkarfjd Å tkZ (U

=f2

nRT) vkSlr xfr 'kwU; gSA

SECTION � 2 : (Matching List Type)

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P2-15

[k.M � 2 : (lqesyu lwph çdkj)

This section contains 4 multiple choice questions. Each questions has matching lists. The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

bl [k.M esa 4 cgqfod Yi ç'u gSaA çR;sd ç'u esa lqesyu lwph gSA lwfp;ksa ds fy, dksM ds fodYi (A), (B), (C)

vkSj (D) gSa] ftuesa ls dsoy ,d lgh gSA

Match List_(4) 35. Type of ideal gas and process followed by it is given in list-I, and in list-II related known

parameters (P,V,T) or quantities or nature is given. Match list-Iwith list-IIand select the correct answer using the codes given below the lists . (where R is universal gas constant) [TH-AD](102)

lwph-I esa vkn'kZ xSl dk izdkj rFkk blds }kjk vuqlfjr izØe fn;k x;k gS rFkk lwph-II eas lEcfU/kr izkpy

(P,V,T) ;k jkf'k;k¡ ;k izÑfr nh xbZ gSA lwph-Idks lwph-IIls lqesfyr dhft, vkSj lwfp;ksa ds uhps fn;s x;s

dksM dk iz;ksx djds lgh mÙkj pqfu;s % (tgk¡ R lkoZf=kd xSl fu;rkad gS)

List-I List-II

lwph-I lwph-II

(P) PV2/3 = constant (1) molar specific heat of gas is 4R

monoatomic gas

PV2/3 = fu;r] ,d ijek.kqd xSl xSl dh eksyj fof'k"V Å "ek 4R gSA

(Q) T2 V (2) volume expansion coefficient is 3T

diatomic gas f}ijekf.od xSl vk;ru izlkj xq.kkad 3T

gS

(R) 3T

V constant (3) as volume increases temperature

mono atomic gas increases

3T

V fu;r] ,dijek.kqd xSl vk;ru c<+us ds lkFk rkieku c<s+xk

(S) P3 V = constant (4) Volume expansion coefficient is 2T

diatomic gas

P3V = fu;r] f}ijekf.od xSl vk;ru izlkj xq.kkad 2T

gSA

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

Sol. (P) 2, 3 ; (Q) 3, 4 ; (R) 2, 3 ; (S) 1, 3

Sol. (P) constant molar heat capacity fu;r eksyj fof'k"V Å "ek

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P2-16

v

RC C

1 N

3R R 9R

22 213

PV2/3 = constant fu;rkad (PV = n R T)

2/3T.V

V= constant fu;rkad

TV�1/3 = constant fu;rkad

3T

V = constant fu;rkad

So, V increases T must increase

pwafd V c<+rk gS vr% T c<s+xk

Now vc,

3 1T V constant fu;rkad

3 nT nV nK

3dT dv

0T v

3 dvT vdT volume expansion coefficient vk;ru izlkj xq.kkad

So, for (P) 2,3

(Q) 2T V 2 1T V constant fu;rkad

2 2 1P V V constant fu;rkad

2P V constant fu;rkad

1/2PV constant fu;rkad

5R R 9R

C12 212

Now vc,

2 1T V constant fu;rkad = K

2 nT nV nK

2dT dV

0T V

2 dVT VdT Volume expansion coefficient vk;ru izlkj xq.kkad

as pwafd 2T V

So, if V increases T must increase

pwafd V c<+rk gS vr% T c<s+xk

for (Q) ds fy, 3, 4

for (R) same as (P)

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P2-17

(R), (P) ds leku gS

(R) 2, 3

for (S) ds fy, 3P V constant fu;rkad

1/3PV constant fu;rkad

5R R

C12 13

5R 3R

C 4R2 2

Volume expansion coefficient is vk;ru izlkj xq.kkad 3

2T

As 1/3PV K

2/3TV K

2

nT nV nK3

dT 2 dV

0T 3 V

3 dV

2T VdT (Volume expansion coefficient) (vk;ru izlkj xq.kkad )

Also if V increases T increases

pwafd V c<+rk gS vr% T c<s+xk

(S) 1, 3

36. Three metallic bars 1, 2, 3 are arranged as shown is figure, with number of free charge carriers in ratio N1 : N2 : N3 = 1 : 3 : 2; resisitivity ratio 1 2 3: : 2 : 1: 3; lengths in ratio

1 2 3: : 2 : 2 : 3 for 1, 2, and 3 bars respectively (radius of cross-section shown in figure),

carry current i as shown. Match list-with list-and select the correct answer using the codes given below the lists : [CE-DF](103)

rhu /kkfRod NM+s 1, 2, 3 fp=kkuqlkj O;ofLFkr gSaA muds eqDr vkos'k okgdksa dh la[;k dk vuqikr Øe'k% N1 :

N2 : N3 = 1 : 3 : 2; izfrjks/kdrk dk vuqikr Øe'k% 1 2 3: : 2 : 1: 3; yEckbZ dk vuqikr Øe'k%

1 2 3: : 2 : 2 : 3 gSA (vuqizLFk dkV dh f=kT;k fp=k esa iznf'kZr gSA) fp=kkuqlkj /kkjk i izokfgr gSA lwph-dks

lwph-ls lqesfyr dhft, vkSj lwfp;ksa ds uhps fn;s x;s dksM dk iz;ksx djds lgh mÙkj pqfu;s %

List lwph�I List lwph�II

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P2-18

(P) If P1, P2 and P3 are power dissipiated (1) 8132

across AB, BC, and CD then 2 32

1

P P

P

AB, BC rFkk CD ds fljksa ij O;f;r

'kfDr Øe'k% P1, P2 rFkk P3 gS rc 2 32

1

P P

P

(Q) If E1, E2 and E3 are magnitude of (2) 1

32

Electric-fields across AB, BC, and

CD Then 2

3

1 2

E

E .E

AB, BC rFkk CD ds fljksa ij fo|qr

{ks=k ds ifjek.k Øe'k E1, E2 rFkk E3 gS rc 2

3

1 2

E

E .E

(R) If 1 2d d, and

3d are drift speeds (3) 14

in bar AB, BC and CD respectively

Then 2 3

1

d d

2d

equals

NM+ AB, BC rFkk CD esa viogu pky

Øe'k% 1 2d d, rFkk

3d gS rc 2 3

1

d d

2d

(S) If V1, V2, V3 are potential differences (4) < 1

across AB, BC and CD Then 2 32

1

V .V

V

AB, BC rFkk CD ds fljksa ij

foHkokUrj Øe'k% V1, V2, V3 gS rc 2 32

1

V .V

V

Codes : P Q R S

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P2-19

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

Ans. (P) 2, 4 ; (Q) 1 ; (R) 3, 4 ; (S) 2, 4

Sol. 1 2

2R 2 . 4 .

Ar

2 2

2 2R . .

9 A3r

3 2

3 9R 3

4 A(2r)

2 3 2 32 2

1 1

2 9P P R R 19 4

16 32P R

E = J

dI

J nevA

.I

EA

1 2

2 .I 2 IE

Ar

2 2

.I IE

9A3r

3 2

3 .I 3 IE

4A2r

1 223

12E E 9

9E16

=3281

Similarly value of blh izdkj

2 3

1

d d

2d

=

14

2 32

1

V .V

V=

132

37. The figures in list- show some charge and current distribution with a charged particle projected in some specific direction list- gives certain conditions which may exist in the subsequent motion of

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P2-20

the charged particle. Match list with list and select the correct answer using the codes given below the lists :

[EM-EB](104)

lwph-I ds fp=k esa dqN vkos'k rFkk /kkjk forj.k n'kkZ;k gS tc ,d vkosf'kr d.k dks fo'ks"k fn'kk esa iz{ksfir djrs

gSA lwph-II esa dqN fo'ks"k 'krs± nh xbZ gS] tks vkosf'kr d.k ds v/kksfyf[kr xfr es fo|eku gks ldrh lwph-dks

lwph-ls lqesfyr dhft, vkSj lwfp;ksa ds uhps fn;s x;s dksM dk iz;ksx djds lgh mÙkj pqfu;s % ¼vkosf'kr d.k

ij xq:Ro dk izHkko ux.; gS½

List -I List-II

(P) A positively charge projected along the axis of two (1) Magnitude of acceleration of

coaxial carrying wires, carrying currents in opposite the charged particle is

directions as shown constant

1 2

v + q

(Q) Two long line charges having equal charge density (2) Kinetic energy of the charged

parallel to z-axis passing through points A (a,0,0) particle is constant

and B (-a,0,0) and a positive charge projected along

Y-axis from point C(0,2a,0)

Z

+ q

X

A

C B

Y

(R) A positive charge is projected from an inside point (3) Angular momentum of the

on the axis of a long solenoid at some angle charged particle about the

with axis as shown. Charge does not hit the solenoid of projection is constant

+ q

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P2-21

(S) A positive charge projected from a point in between (4) Path of the charged particle

two large parallel oppositely charged horizontal plates is straight

at some angle with horizontal as shown. Charge

does not hit the plate.

+ q

� � � � � � � � � � � �

+ + + + + + + + + + + +

lwph-I lwph-II

(P) ,d /kukRed vkosf'kr d.k dks nks lekf{k; /kkjkokgh (1) vkosf'kr d.k ds Roj.k dk ifjek.k

rkjksa ds ywi ftuesa /kkjk foijhr fn'kk esa fp=kkuqlkj fu;r gSA

izokfgr gS] dh v{k ds vuqfn'k iz{ksfir fd;k tkrk gS

1 2

v + q

(Q) leku vkos'k ?kuRo ds nks yEcs js[kh; vkos'k forj.k (2) vkosf'kr d.k dh xfrt Å tkZ fu;r gSA

z-v{k ds lekUrj fcUnq A (a, 0, 0) o B (-a, 0, 0) ls xqtjrs gS

rFkk ,d /kukRed vkosf'kr d.k dks n'kkZ;suqlkj Y-v{k ds

vuqfn'k fcUnq C(0, 2a, 0) ls iz{ksfir djrs gSA

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P2-22

Z

+ q

X

A

C B

Y

(R) ,d /kukRed vkosf'kr d.k dks ,d yEch ifjukfydk dh (3) vkosf'kr d.k dk dks.kh; laosx iz{ksi.k

v{k ds fdlh fcUnq ls v{k ls n'kkZ;s vuqlkj dqN dks.k fcUnq ds lkis{k fu;r jgrk gSA

cukrs gq, iz{ksfir djrs gSA vkosf'kr d.k ifjukfydk ls

ugha Vdjkrk gSA

+ q

(S) ,d /kukRed vkosf'kr d.k dks foifjr vkos'k ls (4) vkosf'kr d.k dk iFk ljy js[kk gS

vkosf'kr nks foLrfjr {kSfrt IysVksa ds e/; fcUnq ls

n'kkZ;suqlkj {kSfrt js[kk ls dqN dks.k cukrs gq, iz{ksfir

djrs gSA vkosf'kr d.k IysV ls ugha Vdjkrk gSA

+ q

� � � � � � � � � � � �

+ + + + + + + + + + + +

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

(D*) 3 4 2 1

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P2-23

Ans. (P) � 1, 2, 3, 4 ; (Q) � 3, 4 ; (R) � 1, 2 ; (S) � 1

Sol. (P) Direction of magnetic field is parallel or antiparallel to velocity, hence no change in K.E. (1), (2)

pqEcdh; {ks=k dh fn'kk osx ds lekUrj ;k izfrlekUrj gS] vr% xfrt Å tkZ esa dksbZ ifjorZu ugha gS (1), (2)

Angular momentum L = 0 (3), (4)

dks.kh; laosx L = 0 (3), (4)

(Q) The two infinite line charges will repel �q� along (�y)-axis 3, 4

nks vuUr yEckbZ js[kh; vkos'k] vkos'k �q� dks (�y)-v{k ds vuqfn'k izfrd"kZ.k djsxsa 3, 4

(R) The charge will undergo helical path inside solenoid 1, 2

vkos'k ifjukfydk ds vUnj gSyhdy iFk esa xfr djsxk 1, 2

(S) A uniform electric field will provide constant acceleration towards � ve plate only 1

,dleku oS|qr {ks=k dsoy _ .kkRed IysV dh vksj fu;r Roj.k iznku djsxsa 1

K.E., L 0

38. An object O (real) is placed at focus of an equi-convex lens as shown in figure. The refractive

index of material of lens is = 1.5 and the radius of curvature of either surface of lens is R. The lens is surrounded by air. In each statement of list-I some changes are made to situation given above and information regarding final image formed as a result is given in list-II. The distance between lens and object is unchanged in all statements of list-I. Match list-with list-and select the correct answer using the codes given below the lists : [M.Bank_GO_12.14]

[GO-LE](102)

,d fcEc O (okLrfod ) fp=k ds vuqlkj ,d lemÙky (equi-convex) ysUl ds Qksdl ij fLFkr gSA ysUl ds

inkFkZ dk viorZukad = 1.5 gS rFkk ysUl ds fdlh Hkh lrg dh oØrk f=kT;k R gSA ysUl ok;q ls f?kjk gqvk

gSA lwph-I ds izR;sd dFku esa Å ij nh xbZ fLFkfr esa dqN ifjorZu fd;s x;s gS rFkk ifj.kkeLo:i cuus okys

vfUre izfrfcEc ls lEcfU/kr lwpuk;sa lwph-II esa nh xbZ gSaA lwph-I ds lHkh dFkuksa esa ysUl rFkk oLrq ¼fcEc½ ds

chp nwjh vifjofrZr gSA lwph-dks lwph-ls lqesfyr dhft, vkSj lwfp;ksa ds uhps fn;s x;s dksM dk iz;ksx

djds lgh mÙkj pqfu;s %

R R ok;q ok;q O

f

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P2-24

List-I List-II

(P) If the refractive index of the lens is (1) final image is real

doubled (that is, made 2 ) then

(Q) If the radius of curvature is doubled (2) final image is virtual

(that is, made 2R) then

(R) If a glass slab of refractive index = 1.5 (3) final image becomes smaller in size

is introduced between the object in comparison to size of image before

and lens as shown, then the change was made

R R

O

slab

(S) If the left side of lens is filled with a medium (4) final image is inverted.

of refractive index = 1.5 as shown, then

R R

O

air

lwph& I lwph& II

(P) ;fn ysUl dk viorZukad nqxquk fd;k tkrk gS (1) vfUre izfrfcEc okLrfod gksrk gSA

(vFkkZr~ 2 fd;k tkrk gS) rc

(Q) ;fn oØrk f=kT;k nqxquh dh tkrh gS (2) vfUre izfrfcEc vkHkklh gksrk gSA

(vFkkZr~ 2R dh tkrh gS) rc

(R) ;fn viorZukad = 1.5 ds dk¡p dh iêh (3) ifjorZu djus ds igys ds izfrfcEc

dks fcEc rFkk ysUl ds chp ds vkdkj dh rqyuk esa vc vfUre

fp=kkuqlkj j[kk tkrk gS] rks izfrfcEc vkdkj esa NksVk curk gSA

R R

O

iêh

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P2-25

(S) ;fn ysUl ds cka;h vksj viorZukad = 1.5 (4) vfUre izfrfcEc mYVk gksrk gSA

dk ,d ek/;e fp=kkuqlkj Hkjk tkrk gSA rks

R R

O

ok;q

Codes : P Q R S (A) 1 2 3 4 (B*) 4 3 2 3 (C) 4 3 2 1 (D) 3 1 4 2 Ans. (P) � 1,3,4 ; (Q) 2, 3 ; (R) 2, 3 ; (S) 2, 3

Sol. Initially the image is formed at infinity.

(P) As increases the focal length decreases. Hence the object is at a distance larger than focal length. Therefore final image is real. Also final image becomes smaller in size in comparision to size of image before the change was made.

(Q) If the radius of curvature is doubled, the focal length increases. Hence the object is at a distance lesser than focal length. Therefore final image is virtual. Also final image becomes smaller in size in comparision to size of image before the change was made.

(R) Due to insertion of slab the effective object for lens shifts right wards. Hence final image is virtual. Also final image becomes smaller in size in comparision to size of image before the change was made.

(S) The object comes to centre of curvature of right spherical surface as a result. Hence the final image is virtual. Also final image becomes smaller in size in comparision to size of image before the change was made.

izkjEHk esa izfrfcEc vuUr ij curk gSA

(P) tc dks c<+k;k tkrk gS] rks Qksdl nwjh ?kVrh gSA blfy;s fcEc Qksdl nwjh ls vf/kd nwjh ij gksxkA

blfy;s vfUre izfrfcEc okLrfod gSA lkFk gh ifjorZu djus ds igys ds izfrfcEc ds vkdkj dh rqyuk esa

vfUre izfrfcEc vkdkj esa NksVk curk gSA

(Q) ;fn oØrk f=kT;k nqxquh dh tkrh gS] rks Qksdl nwjh ?kV tkrh gSA blfy;s fcEc Qksdl nwjh ls de nwjh

ij gksxkA blfy;s vfUre izfrfcEc vkHkklh gksxkA lkFk gh ifjorZu djus ds igys ds izfrfcEc ds vkdkj

dh rqyuk vfUre izfrfcEc vkdkj esa NksVk curk gSA

(R) iV~Vh ds izos'k ds dkj.k ysUl ds fy, izHkkoh fcEc nak;h vkSj foLFkkfir gks tkrk gSA blfy;s vfUre

izfrfcEc vkHkklh gksxkA lkFk gh ifjorZu djus ds igys ds izfrfcEc ds vkdkj dh rqyuk esa vfUre izfrfcEc

vkdkj esa NksVk curk gSA

(S) ifj.kkeLo:i fcEc nak;s xksyh; lrg ds oØrk dsUnz ij vk tkrk gSA blfy;s izfrfcEc vkHkklh gksxkA lkFk

gh ifjorZu djus ds igys ds izfrfcEc ds vkdkj dh rqyuk esa vfUre izfrfcEc vkdkj esa NksVk curk gSA

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P2-26

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Page # 1

Course : JP, JF, JR JPT-1 (JEE ADVANCE) Test Date : 10-05-2015

Test Type : JEE ADVANCED (ELPD)

SYLLABUS : FULL SYLLABUS

Test Pattern :

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

1 to 12 SCQ 12 3 �1 36

13 to 20 Comprehenstion (4 x 2Ques) 8 3 �1 24

21 to 25 Integer Type Questions (Two Digits Answer) 5 3 0 15

26 to 37 SCQ 12 3 0 36

38 to 45 Comprehenstion (4 x 2Ques) 8 3 �1 24

46 to 50 Integer Type Questions (Two Digits Answer) 5 3 0 15

51 to 62 SCQ 12 3 0 36

63 to 70 Comprehenstion (4 x 2Ques) 8 3 �1 24

71 to 75 Integer Type Questions (Two Digits Answer) 5 3 0 15

75 225

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

1 to 15 MCQ 15 4 0 60

16 to 19 Match matrix listing 4 3 �1 12

20 to 34 MCQ 15 4 0 60

35 to 38 Match matrix listing 4 3 �1 12

39 to 53 MCQ 15 4 0 60

54 to 57 Match matrix listing 4 3 �1 12

57 216

Paper-1

Paper-2

Total Total

TotalTotal

Maths

Physics

Chemistry

Physical Inorganic Chemistry Paper-1 Organic Chemistry Paper-1 SCQ (8) SCQ (4) Comp.(4 x 2Q.) (2) Comp.(4 x 2Q.) (2) Integer (Double digit) (3) Integer (Double digit) (2) Physical Inorganic Chemistry Paper-2 Organic Chemistry Paper-2 MCQ (10) MCQ (5) Match Listing type (2) Match Listing type (2)

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Page # 2

Remarks (if any):

Page 52: JEE Advanced 2016 Mock test paper

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JEE (ADVANCED) CHEMISTRY PAPER SKELETON

Faculty Name : Test Name : JR (JPT-1) Faculty preparing the TEST PAPER should fill it according to paper pattern and submit it with finalisation of paper at SMD.

PAPER-2

S. No. TYPE (P) (I) (O) TOPIC(S) SUBTOPIC(S)

DIFFICULTY LEVEL : Easy (E), Moderate (M), Tough (T)

Question Nature

Is the solution there (1/0)

39 MCQ (P) SCP SCP-EBP M 207 1

40 MCQ (I) PBC PBC-XVII M 206 1

41 MCQ (I) PBC PBC-XIII M 206 1

42 MCQ (P) CEQ CEQ-HEE M 206 1

43 MCQ (P) IEQ IEQ-AS M 206 1

44 MCQ (P) SST SST-RR M 206 1

45 MCQ (I) PBC PBC-XV M 206 1

46 MCQ (I) SBC SBC-CAM M 206 1

47 MCQ (P) CEQ CEQ-SEQ M 207 1

48 MCQ (P) ECH ECH-AOE M 205 1

49 MCQ (O) HYC HYC-RAA M 204 1

50 MCQ (O) AC AC-AA E 203 1

51 MCQ (O) AC AC-AN M 203 1

52 MCQ (O) BP BP-CBH E 203 1

53 MCQ (O) AC AC-EAS M 204 1

54 MTC (I) PBC PBC-XV E 203 1

55 MTC (I) MTL MTL-POIM M 203 1

56 MTC (O) AK AK-HHDR M 204 1

57 MTC (O) CAD HYC

CAD-AEA HYC-RAA M 204 1

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Page # 4

PAPER-2 MCQ (10) 39. NaCl (aq.) solution is taken at 0ºC and 1 atm. Now some ice is added in it keeping temperature and

pressure constant then : (SCP-EBP_207(P)) (A*) Vapour pressure of solution will increase (B) Osmotic pressure of solution will increase (C*) Whole of the added ice will melt (D) Some amount of ice(s) will exist in solution NaCl (tyh;) foy;u dks 0ºC rFkk 1 atm ij fy;k x;k gS vc rki rFkk nkc dks fu;r j[krs gq, blesa dqN cQZ

feykrs gS] rc & (A*) foy;u dk ok"inkc c<+sxkA (B) foy;u dk ijklj.k nkc c<+sxkA (C*) feyk;h xbZ lEiw.kZ cQZ fi?ky tk,xhA (D) cQZ dh dqN ek=kk foy;u esa fo|eku gksxhA Sol. Since freezing point of solution is lower than pure water so whole of the added ice will melt. Due to this

solution will be diluted and vapour pressure will increase. pwafd foy;u dk fgekad 'kq) ty ls de gS blfy, feyk;h xbZ lEiw.kZ cQZ fi?ky tk,xhA ftlds dkj.k foy;u ruq

gksxk rFkk ok"inkc c<+sxkA 40. Cold NaCl aqueous solution is electrolysed with vigorous stirring then correct option(s) is/are : (PBC-XVII_206(I)) (A*) Process will involve disproportionation reaction. (B*) Final product has bleaching action. (C) Same product is formed at high temperature (D*) Gas formed at cathode can not be dried by conc. H2SO4. BaMs NaCl ds tyh; foy;u dks rsth ls foyksfMr (stirring) gq, bls fo|qr vi?kfVr djrs gS tc lgh fodYi@fodYiksa

dk p;u dhft,A (A*) izØe esa fo"kekuqikrhdj.k vfHkfØ;k gksxh (B*) vfUre mRikn fojatd fØ;k djrk gSA (C) mPp rki ij leku mRikn curk gSA (D*) dSFkksM ij fufeZr xSl dks lkUnz H2SO4 }kjk 'kq"d ugha fd;k tk ldrk gSA Sol. At Cathode dSFkksM ij 2H2O + 2e� H2 + 2OH� At Anode ,uksM ij 2Cl� Cl2 + 2e�

Cl2 + OH� Cold ( )

B.Mk Cl� + OCl�

41. Colemanite 2 3Na CO x HCl

yMonoprotic

acid

z Mg

wan element

Identify incorrect statement. (PBC-XIII_206(I)) (A) Oxidation number of boron in colemanite is +3 (B*) z when heated with Co gives blue colour in oxidizing flame and red in reducing flame. (C*) Almost all compounds of w are ionic. (D*) Decahydrated x has 4 O�H bonds.

dksysesukbV 2 3Na CO x HCl

y,dy izksfVd vEy

z Mg

W,d rRo

xyr dFku pqfu;sA (A) dksysesukbV esa cksjksu dh vkWDlhdj.k la[;k +3 gSA

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Page # 5

(B*) z dks tc Co ds lkFk xeZ djrs gS] rks ;g vkWDlhdkjh Tokyk esa uhyk jax rFkk vipk;ddkjh Tokyk esa yky jax nsrk gSA

(C*) w ds lHkh ;kSfxd izk;% vk;fud gksrs gSA (D*) Msdkty;ksftr x 4 O�H ca/k j[krk gSA

Sol. Ca2B6O11 + Na2CO3 2 4 7(X)

Na B O HCl 3 3

(Y)H BO

2 3(Z)

B O Mg

(W)B

Decahydrated Borax is Na2[B4O5(OH)4].8H2O Msdkty;ksftr cksjsDl Na2[B4O5(OH)4].8H2O gSA So it has 20 O�H bonds. vr% ;g 20 O�H ca/k j[krk gSA 42. Which of following represents equilibrium condition? (CEQ-HEE_206(P)) (A*) 10 g ice and 5 g H2O () at 0ºC & 1 atm (B*) 5 g ice and 5 g H2O () at 0ºC & 1 atm (C*) 10 g H2O () and 5 g H2O (v) at 100ºC & 1 atm (D*) 5 g H2O () and 5 g H2O (v) at 100ºC & 1 atm fuEu esa ls dkSu lkE; ifjfLFkfr iznf'kZr djrs gSa\ (A*) 0ºC rFkk 1 atm ij 10 g cQZ rFkk 5 g H2O () (B*) 0ºC rFkk 1 atm ij 5 g cQZ rFkk 5 g H2O ()

(C*) 100ºC rFkk 1 atm ij 10 g H2O () rFkk 5 g H2O (v)

(D*) 100ºC rFkk 1 atm ij 5 g H2O () rFkk 5 g H2O (v) Sol. P and T condition decides equilibrium for such equilibria (not amount). nkc rFkk rki ifjfLFkfr bl izdkj ds lkE;ksa ds fy, lkE; dks fuf'pr djrh gSA ¼ek=kk ugha½ 43. Salt AB undergoes anionic hydrolysis and its 0.1 M solution has pOH as 5 then. Identify correct. (IEQ-AS_206(P)) (A*) Kh is equal to Kb of B� (B*) pKa of HB is 5 (C*) h is 0.01% (D*) pH of 0.1 M HB is 3 yo.k AB _ .kk;fud ty vi?kVu nsrk gS rFkk bldk 0.1 M foy;u 5 pOH j[krk gS] rc lgh dFku igpkfu;sA (A*) Kh , B

� ds Kb ds cjkcj gksrk gSA (B*) HB dk pKa 5 gSA (C*) h 0.01% gSA (D*) 0.1 M HB dh pH 3 gSA Sol. B� + H2O HB + OH� c(1�h) ch ch c = 0.1 M ch = 10�5 h = 10�4 Kh = ch2 = 10�9 Ka(HB) = 10�5 Kb of B� = 10�9

pH = 12

(5 � (�1)) = 3

44. A solid A+B� has r 2 2R 2

then correct statement is : (SST-RR_206(P))

(A*) B� touches total 18 ions (B*) A+ touches 6 ions (C*) Each body diagonal passes through the centre of 2 cation and 1 anion (D*) Each body diagonal passes through the centre of 2 anion and 1 cation

,d Bksl A+B�r 2 2R 2

eku j[krk gS] rc lgh dFku gSa &

(A*) B� dqy 18 vk;uksa dks Li'kZ djrk gSA (B*) A+ 6 vk;uksa dks Li'kZ djrk gSA (C*) izR;sd dk; fod.kZ 2 /kuk;uksa rFkk 1 _ .kk;u ds dsUnz ls xqtjrk gSA (D*) izR;sd dk; fod.kZ 2 _ .kk;uksa rFkk 1 /kuk;u ds dsUnz ls xqtjrk gSA

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Sol. This radius ratio suggests exact fitting in rock salt type structure. ;g f=kT;k vuqikr jkWd lkYV izdkj dh lajpuk esa iw.kZr% O;ofLFkr gksrk gSA 45. Which is/are correct about SO2Cl2 ? (PBC-XV_206(I))

(A*) OSO > OSCl > ClSCl bond angle (B*) Number of p� p bonds < Number of p�d bonds (C*) 1 mol of this reacts with 0.1 mol P4 (D*) Its aqueous solution does not give pink colour with phenolphthaleine SO2Cl2 ds lUnHkZ esa dkSulk@dkSuls dFku lgh gS@gSa\

(A*) OSO > OSCl > ClSCl ca/k dks.k (B*) p� p ca/kksa dh la[;k < p�d ca/kksa dh la[;k (C*) blds 1 eksy 0.1 eksy P4 ds lkFk fØ;k djrs gSA (D*) bldk tyh; foy;u fQukW¶Fksyhu ds lkFk xqykch jax ugha nsrk gSA

Sol. O

S O Cl Cl

2 (p�d) bonds ca/k No (p�p) bonds ca/k P4 + 10SO2Cl2 4PCl5 + 10SO2 vf la;kstdrk xq.kkad = 20 vf la;kstdrk xq.kkad = 2 SO2Cl2 + H2O H2SO4 + HCl

46. Compound of Na Na, A + colourless gas B. (SBC-CAM_206(I))

(x) Identify true statement : (A*) Cl2 disproportionates in aqueous solution of A. (B*) B does not show allotropy (C*) X can give red colouration if treated with NH2CSNH2 + CH3COOH then followed by FeCl3 (D) B is not combustible but combustion supporter.

Na dk ;kSfxd Na, A + jaxghu xSl B.

(x) lR; dFku igpkfu;sA (A*) A ds tyh; foy;u esa Cl2 fo"kekuqikrhd r̀ gksrk gSA (B*) B vij:irk ugha n'kkZrk gSA (C*) X yky jax ns ldrk gS ;fn bls igys NH2CSNH2 + CH3COOH ds lkFk rFkk blds i'pkr~ FeCl3 ds lkFk

mipkfjr fd;k t krk gSA (D) B nguh; ugh gS] ysfdu ;g ngu lgk;d gksrk gSA Ans. x NaNO2 or ;k NaNO3

A Na2O B N2

47. CaCO3(s) CaO(s) + CO2(g) (CEQ-SEQ_207(P))

CO2(g) CO(g) + 12

O2 (g)

For above simultaneous equilibrium if CO2 is added from out side at equilibrium then :

mijksDr ,d lkFk gksus okys lkE;ks ds fy, ;fn CO2 dks lkE; ij vyx ls ¼ckgj ls½ feyk;k tkrk gS rc &

(A) 2COP will increase (B)

2COP will decrease

(C*) No shift in 2nd equilibrium (D*) Backward shift in 1st equilibrium (A)

2COP c<sxkA (B) 2COP ?kVsxkA

(C*) 2nd lkE; esa foLFkkiu ugha gksxkA (D*) 1st lkE; esa i'p fn'kk dh vksj foLFkkiu gksxkA Sol. Second equilibrium will not be affected by CO2 addition only first will shift backward.

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f}rh; lkE; CO2 feykus ls izHkkfor ugha gksxk] dsoy izFke lkE; gh i'p fn'kk d h vksj foLFkkfir gksxkA 48. An aqueous solution of ZnCl2 with conc. H2SO4 is electrolysed using zinc electrodes at anode and cathode

then which of following options are correct ? (ECH-AOE_205(P)) (A) Cl2 gas can evolve at anode due to over voltage conditions (B*) H2 gas evolve at cathode so pH of solution get increases (C) Zinc will oxidise at anode but conc. of Zn+2 of solution remain constant. (D*) conc. of Zn+2 in electrolyte increases but conc. of anion remain same lkUnz H2SO4 ds lkFk ZnCl2 ds tyh; foy;u dks ,uksM rFkk dSFkksM ij ftad bysDVªkWM dk iz;ksx djds fo|qr

vi?kfVr fd;k tkrk gS rc fuEu esa ls d kSuls fodYi lgh gSa \ (A) Cl2 xSl vf/kd oksYVst ifjfLFkfr;ks ds dkj.k fu"dkflr gks ldrh gSA (B*) H2 xSl dSFkksM ij fu"dkflr gksrh gS blfy, foy;u dh pH c<+rh gSA (C) ftad ,uksM ij vkWDlhd r̀ gksxh ysfdu foy;u ds Zn+2 dh lkUnzrk fu;r jgrh gSA (D*) fo|qr vi?kV; esa Zn+2 dh lkUnzrk c<rh gSA ysfdu _ .kk;u dh lkUnzrk leku jgrh gSA Sol. Electrolyte ZnCl2 and conc. H2SO4 electrode Zn rod

Anode reaction Zn Zn+2 + 2e�

Cathode reaction 2H+ + 2e� H2 So conc. of [Zn+2] , conc. of Cl� and SO4

2� remain same and pH

gy% fo|qr vi?kV~; ZnCl2 rFkk lkUnz H2SO4 bysDVªkWM Zn NM+ gSA

,uksM vfHkfØ;k Zn Zn+2 + 2e�

dSFkksM vfHkfØ;k 2H+ + 2e� H2 blfy;s [Zn+2] dh lkUnzrk , Cl� rFkk SO4

2� dh lkUnzrk leku jgrh gS rFkk pH

49.

H2 + Pd + BaSO4

Na / liq. NH3

Lindlar's catalyst

Compound (Q) C8H18 H2/Pt

Compound (P) C8H12

(Optically active)

(Optically inactive)

Compound (R) C8H14 (Optically active)

Compound (S) C8H14 (Optically inactive)

Which amongs the following is/are correct for above sequence of reactions ? (A*) Oxidative ozonolysis product of P, R & S will be achiral (B*) R & S are diastereomers (C*) Product Q & R are formed by syn addition & S is formed by anti addition (D) P must have cis configuration

H2 + Pd + BaSO4

Na / liq. NH3

fy.Mykj dk mRizsjd

;kSfxd (Q) C8H18 H2/Pt

;kSfxd (P) C8H12

(izdkf'kd lfØ;)

(izdkf'kd vfØ;)

;kSfxd (R) C8H14 (izdkf'kd lfØ;)

;kSfxd (S) C8H14 (izdkf'kd vfØ;)

mijksDr vfHkfØ;k vuqØe ds fy, lgh dFku@dFkuksa dk p;u dhft,& (HYC-RAA(O)(M)_203) (A*) P, R o S dk vkWDlhdkjh vkstksuhvi?kVu mRikn vfdjsy gksxkA (B*) R o S foofje leko;oh gSA (C*) mRikn Q o R flu ;ksx ls fufeZr gksrk gS o mRikn S ,UVh ;ksx ls fufeZr gksrk gSA (D) ;kSfxd P dk lei{k foU;kl gksuk pkfg,A

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Sol.

H3C�CH=CH�CH�CC�CH3

| CH3

(P)

H3C�CH2�CH2�CH�CH2�CH2�CH3

| CH3

CH3

(Trans) | (Cis) CH3�CH=CH�CH�CH=CH�CH3

CH3

| CH3�CH=CH�OH�CH=CH�CH3

(Trans) (Trans)

Trans

Sol.

H3C�CH=CH�CH�CC�CH3

| CH3

(P)

H3C�CH2�CH2�CH�CH2�CH2�CH3

| CH3

CH3

(foi{k) | (lei{k) CH3�CH=CH�CH�CH=CH�CH3

CH3

| CH3�CH=CH�OH�CH=CH�CH3

(foi{k) (foi{k)

foi{k

50. Which statements are correct in the following ? (AC-AA(O)(E)_203) (A*) Cyclopentadiene reacts with NaNH2 and releases NH3 gas however cyclopropene does not. (B*) Cyclopropenone has a greater dipole moment than acetone (C*) Cyclobutadiene does not exist at room temperature but cyclobutene exists at room temperature (D*) Cycloheptatrienyl bromide ionizes faster however cyclopentadienyl bromide does not fuEu esa ls dkSuls dFku lgh gS\ (A*) lkbDyksisUVkMkbbZu NaNH2 ds lkFk fØ;k djds NH3 eqDr djrh gS tcfd lkbDyksizksihu ughaA (B*) lkbDyksizksiukWu dk f}/kzqo vk?kw.kZ ,lhVksu dh rqyuk esa vf/kd gksrk gSA (C*) lkbDyksC;wVkMkbbZu dejs ds rki ij ugha ik;k tkrk gS ysfdu lkbDyksC;wfVu dejs ds rki ij ik;k tkrk gSA (D*) lkbDyksgsIVkVªkbZbukbZy czksekbM rsth ls vk;fur gks tkrk gS ;|fi lkbDyksisUVkMkbZukbZy czksekbM ugha gSA

Sol. (A)

2NaNH

+ NH3

Aromatic

,

2NaNH

+ NH3

Antiaromatic

(B)

O

is aromatic & highly polar.

(C) is antiaromatic & is nonaromatic.

(D)

Br

Br

Aromatic,

Br

�Br

Antiaromatic

Sol. (A)

2NaNH

+ NH3

,sjkseSfVd

,

2NaNH

+ NH3

,UVh,sjkseSfVd

(B)

O

,jkseSfVd o mPp /kzqfo; gSA

(C) ,UVh,sjkseSfVd gS o ukWu,sjkseSfVd gS

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(D)

Br

Br

,sjkseSfVd ,

Br

�Br

,UVh,sjkseSfVd

51. Optically active amines having molecular formula C5H13N on reaction with NaNO2 + HCl produces tertiary

optically inactive alcohol. Find out structures of amines. v.kqlw=k C5H13N ;qDr izdkf'kd lfØ; ,ehu dh NaNO2 + HCl ds lkFk vfHkfØ;k djkus ij rr̀h;d izdkf'kd vfØ;

,sYdksgkWy curk gSA ,ehu dh lajpuk D;k gksxhA (AC-AN(O)(M)_203)

(A*)

NH2 (B)

N

(C*) NH2

(D)

NH2

Sol. (1) Needs primary amine for preparation of alcohol with HNO2. (NaNO2 + HCl) (2) Only (A) & (C) are optically active which after reaction with NaNO2 + HCl will give optically inactive

alcohols. Sol. (1) HNO2 (NaNO2 + HCl) ds lkFk ,Ydksgy ds fuekZ.k ds fy, izkFkfed ,ehu dh vko';drk gksrh gSA (2) dsoy (A) o (C) izdkf'kd lfØ; gS tks NaNO2 + HCl ds lkFk vfHkfØ;k ds i'pkr~ izdkf'kd vfØ; ,Ydksgy

nsrh gSA 52. Which of the following are reducing sugar ? (BP-CBH(O)(E)_203) fuEu esa ls dkSulh vip;ukRed 'kdZjk gS\

(A)

OH

O OH

HO

OH

O HO

O

OH OH

HO

(B*)

OH

O OH

HO

OH O

OH O

OH

OH

OH

(C*)

(CHOH3)3 OH

HO O

(D*)

OH

O OH

HO

OH

O OH

O

OH

OH

OH

Sol. -Hydroxy carbonyl groups & sugars having hemiacetal group will be reducing sugar. -gkbMªksDlh dkcksZfuy lewg o 'kdZjk gsfe,sflVsy lewg j[krh gS tks vip;ukRed 'kdZjk gksxhA 53. Which of the following reactions give same product ? fuEu esa ls dkSulh vfHkfØ;k leku mRikn nsrh gSA (AC-EAS(O)(M)_204)

(A*)

NO2

H3C

Sn

HCl 2NaNO

HCl 3 4H PO

(B*)

3

CO HCl

AlCl

4LiAlH

TsCl

4NaBH

(C)

3

3

O||

CH C Cl

AlCl

4KMnO H

4LiAlH

(D*)

Cl

AlCl3

4KMnO H

Red P

H

Sol. (A*)

NO2

H3C

Sn

HCl

NH2

H3C

2NaNO

HCl

N2Cl

H3C

3 4H PO

H3C

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Page # 10

(B*)

3

CO HCl

AlCl

CHO

4LiAlH

CH2�OH TsCl

CH2OTs

4NaBH

CH3

(C)

3

3

CH COCl

AlCl

COCH3 4KMnO

COOH

4LiAlH

CH2OH

(D*)

Cl

AlCl3

4KMnO H

COOH Red P

H

P

H

yky CH3

Match Listing type (MTC)(2) 54. Match the following : Column � I Column � II (PBC(INO)) (P) SiO2 (1) Reacts with HF

(Q) [Co(CO)4]� (2) Pseudo halide

(R) I� (3) Gives compound with Cu2+ via redox reaction (S) N2 (4) Inert towards reaction LrEHk � I LrEHk � II (P) SiO2 (1) HF ds lkFk fØ ;k

(Q) [Co(CO)4]� (2) Nn~e gSykbM

(R) I� (3) ;kSfxd Cu2+ ds lkFk jsMkWDl vfHkfØ ;k nsrk gSA (S) N2 (4) vfHkfØ ;k ds izfr vfØ ; gksrk gSA Code dwV :

(P) (Q) (R) (S)

(A) 4 3 2 1

(B*) 1 2 3 4

(C) 3 2 4 1

(D) 2 4 1 3

Sol. (A) SiO2 + 4HF SiF4 + 2H2O SiF4 + 2HF H2SiF6

(B) A common complex pseudohalide is [Co(CO)4]�

e.g. like HX, HCo(CO)4 is formed.

(C) white

Cu5Cu2 22�2 �

3III

(D) N2 is inert towards reaction.

Sol. (A) SiO2 + 4HF SiF4 + 2H2O, SiF4 + 2HF H2SiF6

(B) [Co(CO)4]�

,d lkekU; tfVy Nn ~e gSykbM gSA mnk- HX dh rjg HCo(CO)4 Hkh curk gSA

(C) white

Cu5Cu2 22�2 �

3III

(D) N2 vfHkfØ ;k ds izfr vfØ ; gksrk gSA 55. Match the gases with solution / substances in which they are absorbed. xSalksa dks foy;uksa vFkok inkFkks± ds lkFk lqesfyr dhft, ftuesa ;s vo'kksf"kr gksrh gSA List-I List-II

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lwph-I lwph-II (P) CO (1) Absorbed by ethanolamine ,sFksukWy,sehu ds }kjk vo'kksf"kr (Q) CO2 (2) Absorbed by FeSO4 solution FeSO4 foy;u ds }kjk vo'kksf"kr (R) NO (3) Absorbed by ammonical Cu2Cl2 veksfudy Cu2Cl2 ds }kjk vo'kksf"kr (S) O3 (4) Absorbed by turpentine oil rkjihu ds rsy }kjk vo'kksf"kr Code : dwV %

(P) (Q) (R) (S)

(A) 1 3 2 4

(B*) 3 1 2 4

(C) 2 1 3 4

(D) 3 2 1 4

Sol. Fact rF;

56. Match the List ; (AK-HHDR(O)(M)_204) List-I List-II

(P)

CH=O OH

(1) Nucleophilic addition

(Q) CH3�C�CH3

O

OH

(2) Nucleophilic acyl

(R) CH3�CH=O HCN (3) Hydride transfer

(S) CH3�C�OC2H5

O

2 5C H O

(4) Carbanion

lwph dk feyku dfj;s % lwph-I lwph-II

(P)

CH=O OH

(1) ukfHkdLusgh ;ksx

(Q) CH3�C�CH3

O

OH

(2) ukfHkdLusgh ,sfly

(R) CH3�CH=O HCN (3) gkbMªkbM LFkkukUrj.k

(S) CH3�C�OC2H5

O

2 5C H O

(4) dkcZ_ .kk;u

Code dwV : (P) (Q) (R) (S) (A) 1,3 1 2,4 1,4 (B) 1 2,4 1,3 1,4 (C*) 1,3 1,4 1 2,4 (D) 1,4 1,3 1 2,4

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Ans. (A) � p, r ; (B) � p, s ; (C) � p ; (D) � q, s Sol. (A) is cannizzaro reaction (B) is Aldol condensation (C) is Nucleophilic addition (D) is cleaisen condensation Sol. (A) dSfutkjks vfHkfØ;k gSA (B) ,YMksy la?kuu gSa (C) ukfHkdLusgh ;ksxkRed vfHkfØ;k gSA (D) dSylu la?kuu gSA 57. Match the reaction in List-I with appropriate option in List-II. (CAD-AEA(O)(M)_204) (HYC-RAA(O)(M)_204) List-I List-II

(P) HBr (1) Substitution reaction

(Q) (2) Addition reaction

(R) (3) Coupling reaction

(S) (4) Racemic mixture

dkWye-I esa nh xbZ vfHkfØ;kvksa ds fy, dkWye-II ls mfpr feyku dhft;A lwph -I lwph-II

(P)

HBr (1) izfrLFkkiu vfHkfØ;k

(Q) (2) ;ksxkRed vfHkfØ;k

(R) (3) ;qXeu vfHkfØ;k

(S) (4) jslsfed feJ.k (Racemic mixture)

Code dwV : (P) (Q) (R) (S)

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(A) 1,3 2,4 1,4 2,4 (B*) 2,4 1,3 1,4 2,4 (C) 2,4 1,3 2,4 1,4 (D) 1,4 2,4 1,3 2,4 Ans. (A) � q, s; (B) � p, r; (C) � p, s; (D) � q, s Sol. (A) The reactant undergoes electrophilic addition having a carbocation intermediate forming a racemic-

mixture. (B) The reactant undergoes an electrophilic substitution which is a coupling reaction without any

carbocation intermediate. (C) The reactant undergoes an electrophilic substitution reaction having a carbocation rearrangement

forming a racemic mixture. (D) The reactant undergoes nucleophilic addition forming a racemic mixture without any carbocation

intermediate. gy % (A) fØ;kdkjd bysDVªkWuLusgh ;ksx djrk gS] tks dkcZ/kuk;u e/;orhZ j[krk gS rFkk jslsfed feJ.k cukrk gSA (B) fØ;kdkjd bysDVªkWuLusgh izfrLFkkiu djrk gS] tks fcuk dkcZ/kuk;u e/;orhZ ds ,d ;qXeu vfHkfØ;k gSA (C) fØ;kdkjd bysDVªkWuLusgh izfrLFkkiu djrk gS] tks dkcZ/kuk;u iqufoZU;kl j[krk gS jslsfed feJ.k cukrk gSA (D) fØ;kdkjd ukfHkdLusgh ;ksx djrk gS tks fcUkk dkcZ/kuk;u e/;orhZ ds jslsfed feJ.k cukrk gSA