JEE Advaced 2016 Mock test paper

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P2JPT2(ADV.)170515C0-1

TEST PATTERN

AIOT (JEE ADVANCE) - 2 TEST DATE : 03-05-2015 BATCH :

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

1 to 8 MCQ 8 3 0 24

9 to 14 Comprehension (3 Comp. x 2 Q.) 6 3 �1 18

15 to 20 Integer Type Questions (Double Digits Answer) 6 3 0 18

21 to 28 MCQ 8 3 0 24



41 to 48 MCQ 8 3 0 24



60 180TotalTotal

Maths

Physics

Chemistry

Paper-2 (AIOT-2 JEE ADVANCE)

MCQ � 8 1. Let a = sin�1(sin3) + sin�1(sin4) + sin�1(sin5), [FN-IN] [307]

f(x) = |x|x2

e , domain of f(x) be [a, ), range of f(x) be [b, ),

g(x) = (4cos4x � 2cos2x �21

cos4 x � x7)1/7, domain and range of g(x) is set of real numbers. Which

of the following are CORRECT ? (A*) a = � 2 (B*) b + a = � 1

(C*) f(gog(b)) =e2 (D)both f(x), g(x) are non-invertible functions

ekuk a = sin�1(sin3) + sin�1(sin4) + sin�1(sin5), f(x) = |x|x2

e , f(x) d k izkUr [a, ) gS] f(x)

d k ifjlj [b, ), g(x) = (4cos4x � 2cos2x �21

cos4 x � x7)1/7 , g(x) d k izkUr vkSj ifjlj] okLrfod

la[;kvksa d k leqPp; gS fuEu esa ls d kSulk lgh gS ? (A*) a = � 2 (B*) b + a = � 1

(C*) f(gog(b)) =e2 (D) f(x), g(x) nksuksa izfry kse Q y u ugha gSA

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mailto:[email protected]

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P2JPT2(ADV.)170515C0-2

Sol. a = ( � 3) + ( � 4) + (5 � 2)=�2 f(�2) = f(2) f(x) is many one non invertible f(�2) = f(2) f(x) cgq,Sd h gSA izfry kseh; ugha gSA

3

4 4 5

Let ekuk t = x2+ |x| t )

X

t

f(x) [1, ) b = 1 a + b = �1

g(x) =((1 + cos 2x)2 �2cosx �

21

(2cos22x�1)�x7)1/7

=(1 + 2cos 2x+cos22x �2cos2x�cos22x + 21

� x7)1/7

g(x) =7/1

7x�23

gog(x) =7/1

7))x(g(�23

=7/1

7x�23

�23

= (x7)1/7

= x f(gog(b)) = f(b) = e1+1 = e2 2. Following usual notation, if in a [ST-HA] [303] triangle ABC, r1 + r2 = 3R, r2 + r3 = 2R, then smallest angle is 10nº, n N. The equation (x � 1) (x � 2)(x � 3) = 24 has real root and imaginary root ± iR.

Which of the following are CORRECT ? ABC esa lkekU; l ad srkuqlkj r1 + r2 = 3R, r2 + r3 = 2R, rc lcls NksVk d ks.k 10nº, n N. lehd j.k (x � 1) (x � 2)(x � 3) = 24 d s okLrfod ewy vkSj d kYifud ewy ± iR. fuEu

esa l s d kSulk lgh gS ? (A*) n = 3 (B*) n + = 8 (C*) 2 +2 = 6 (D*) + 2= 2n

Sol. r1 + r2 = 3R b�sa�s

= 3.

4abc

43

)b�s)(a�s()c(2

abc

4s(s � c) = 3ab

2

)c�ba(2

)cba(4 = 3ab a2 + b2 � c2 = ab



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P2JPT2(ADV.)170515C0-3

cos c = 21

r2 + r3 = 2R cos A = 0 angles are d ks.k 60º, 90º, 30º

smallest lcls NksVk 10nº = 10×3º n = 3 cubic ?kuh; x3 � 6x2 + 11x � 30 = 0 (x � 5) (x2 � x + 6) = 0 = 5, (+ i)(� i) = 6

+ 2 = 6

3. The line x � 2

3=

y �1

2 =

z �1

�1 intersects the curve x2 � y2 = a2, z = 0 if a is equal to

js[kk x � 2

3=

y �1

2 =

z �1

�1 oØ x2 � y2 = a2, z = 0 d ks izfrPNsn d jrh gS] ;fn a cjkcj gS&

(A*) 4 (B) 5 (C*) �4 (D) � 5 Sol. For the point where the line intersects the curve, we have z = 0, so Revi. PT-3-29-4-2015 fcUnq d s fy ,] js[kk oØ d s fy , izfrPN sn d jrh gSA z = 0, blfy , [TD-MS] [303]

x � 2

3=

y �1

2=

0 �1

�1 x = 5 and vkSj y = 3

Put these values in x2 � y2 = a2, we get a2 = 16 a = ±4 x2 � y2 = a2 esa bu ekuksa d ks j[kus ij, a2 = 16 a = ±4

4. Consider the graph of the function f(x) =x 3

nx 1e

then which of the following is correct (A) Range of the function is (1, ) [FN-DN] [302]

(B*) f(x) has no zeroes (C*) Graph lies completely above the x-axis (D*) Domain of f is (� , �3) (�1, )

ekukfd Q y u f(x) =x 3

nx 1e

d k vkjs[k d s fy , rc fuEu esa ls d kSulk lgh gS& (A) Q y u d k ifjlj (1, ) gSA

(B*) f(x) 'kwU; ugha gSA (C*) vkjs[k iw.kZr;k x-v{k ls Å ij gSA

(D*) f d k izkUr (� , �3) (�1, ) gSA

Sol. y = 1x3x

> 0 x < � 3 or ;k x > � 1

or ;k x �3 x x �1 x 1 (0, 1) (1, )

5. If

0i2

2

0jji )1a()1�a(

a

a.a

1 where i j and a >1 then possible values of may be

;fn

0i2

2

0jji )1a()1�a(

a

a.a

1 t gk¡ i j vkSj a > 1 rc d s laHko eku gks ld rs gS&

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

Sol. When no restriction on i and j [SS-GP] [303] t c i vkSj j ij d ksbZ izfrcU/k ugha gSA

S =

0i2

22

20j

ji )1�a(

a.......

a

1a1

1a.a

1 i j



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P2JPT2(ADV.)170515C0-4

S = 2

2

a

(a � 1) (a 1)

> 2

6. A manufacturer of airplane parts makes a certain engine that has a probabil ity p of

failing on any given f light. There are two planes fitted with this type of engine. One plane has 3 such engines and other planes has 5. A plane crashes if more than half the engines fit ted in it fail. If the two plane models have the same probabil ity of crashing then the value of p can be [PR-BT] [301]

,d gokbZ t gkt cukus d h d Eiuh] d qN bat u cukrh gS] mld s fd lh nh xbZ mMku d s [kjkc gksus d h izkf;d rk p gSA bl izd kj d s bat uksa ls nks gokbZ t gkt cuk, t krs gSA ,d gokbZ t gkt esa bl izd kj d s 3 bat u gS vkSj nwljs gokbZ t gkt esa 5 bat u gSA ,d gokbZ t gkt d h nq?kZVuk gksrh gSA ;fn y xk, x, vk/ks ls T;knk bat u [kjkc gksrs gSA nks gokbZ t gkt d h nq?kZVuk ?kfVr gksus d h izkf;d rk leku gS] rc p d k eku gks ld rk gS&

(A*) 0 (B*) 1 (C*) 1/2 (D) 3/5 Sol. The two engines have the same probabil ity of fail ing if nks bat u leku izkf;d rk d s gS 5C3p3(1 � p)2 + 5C4p4(1�p) + p5 = 3C2p2(1�p) + p3 10p3(1�p)25 + p4(1 � p) + p5 = 3p2(1 � p) + p3 2p3

� 5p2 + 4p � 1 = 0 p = 0 or ;k p = ½ or ;k p = 1

7. Let [.] denotes the greatest integer function and f(x) =

3

sin[x]a , x 0

x2 , x 0

sinx xb , x 0

x

.

If f(x) is continuous at x = 0 , then b is equal to : [CD-MS] [301]

ekukfd [.] egÙke iw.kk±d Qyu dks iznf'kZr djrk gS rFkk f(x) =

3

sin[x]a , x 0

x2 , x 0

sinx xb , x 0

x

.

;fn f(x), x = 0 ij lr~r gks rks b = (A*) a2 � 1 (B*) a + 1 (C) a + 2 (D) a � 2 Sol.

0xlim f(x) =

0xlim a = a = 2

Since pawfd 3x

xxsin =

3

753

x

x....!7

x!5

x!3

xx

= � !3

1 +

!5x2

� !7

x4

+ .....

0xlim f(x) =

0xlim b +

3x

xxsin = b � 1

b = 3 = a2 � 1 = a + 1 8. The parabola x2 = ay makes an intercept of length 40 on the l ine y � 2x = 1, if a is

equal to [PB-PL] [301] ijoy ; x2 = ay js[kk y � 2x = 1 ij 40 y EckbZ d k vUr[k.M cukrh gS] rc a cjkcj gS&

(A) � 1 (B*) � 2 (C*) 1 (D) 2 Sol. x2 = a(1 + 2x) Revi. PT-3-29-4-2015 x2 � 2ax � a = 0



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P2JPT2(ADV.)170515C0-5

x1 + x2 = 2a x1 x2 = � a

length of chord t hok d h y EckbZ = |x1 � x2 | 21 m

= 24a 4a 1 4 = 40

i.e. ;k a2 + a � 2 = 0 i.e. ;k (a + 2) (a � 1) = 0

i.e. ;k a = � 2, 1.

Paragraph for Question Nos. 9 to 10 (iz'u 9 ls 10 ds fy, vuqPNsn)

Let A,B,C be the three points on the ellipse 2 2

2 2

x y

a b = 1 having eccentric angles

respectively. [EL-EA] [301]

ekuk A,B,C nh?kZoÙk 2 2

2 2

x y

a b = 1 ij rhu fcUnq gSA ft ud s mRd sUnz d ks.k gSA

9. Area of triangle PQR formed by corresponding points on auxiliary circle is JPT-2 Adv. 2015

(A*) ab

(area ABC) (B) ba

(area ABC) (C) 2ab

(area ABC) (D) (area ABC)

lgk;d oÙk ij l axr fcUnqvksa l s cuk f=kHkqt PQR d k {ks=kQ y gS&

(A*) ab

(ABC d k {ks=kQ y ) (B) ba

(ABC d k {ks=kQ y )

(C) 2ab

(ABC d k {ks=kQ y ) (D) (ABC d k {ks=kQ y )

Sol. Let ABC is formed by the points A(a cos, bsin), B(a cos, bsin) and C(a cos, bsin) on 2 2

2 2

x y

a b = 1

ekuk nh?kZoÙk 2 2

2 2

x y

a b = 1 ] A(a cos, bsin), B(a cos, bsin) and C(a cos, bsin) fcUnqvksa l s

,d f=kHkqt cuk;k t krk gSA

Let ekuk P(a cos, asin) Q(a cos, asin) ; R(a cos, asin) be point on auxiliary circle. Q(a cos, asin) ; R(a cos, asin) lgk;d oÙk ij gSA

Area of ABC d k {ks=kQ y = 12

acos bsin 1

acos bsin 1

acos bsin 1

= 12

ab

cos sin 1

cos sin 1

cos sin 1

area of PQR d k {ks=kQ y = 12

a2

cos sin 1

cos sin 1

cos sin 1

Area of ABC d k {ks=kQ y = ba

Area of PQR d k {ks=kQ y .

10. The eccentric angles of the vertices of triangle of maximum area inscribed in an ellipse differ by nh?kZoÙk d s vUrxZr vf/kd re {ks=kQ y d s f=kHkqt d s 'kh"kZ d s mRd sUnz d ks.kksa d k vUrj gS&

(A) 3

(B) 2

(C*) 23

(D)

Sol. QOR = 120º



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P2JPT2(ADV.)170515C0-6

� = 23

Similarlyblh izd kj � = 23

, � = 23

eccentric angle of vertices of Differ by 23

d s 'kh"kksZ d k mRd sUnz d ks.kksa d k vUrj 23

gSA


Let < an> and < bn> be the arithmetic sequences each with common difference 2 such that a1 < b1

and let cn =

n

1kka , dn =

n

1kkb . Suppose that the points An(an, cn), Bn(bn, dn) are all lying on the

parabola C: y = px2 + qx + r where p,q,r are constants. [SS-MS] [303] ekuk < an> vkSj < bn> lekUrj vuqØ e gS ft uesa izR;sd d k lkoZvUrj 2 bl izd kj gS fd

a1 < b1 vkSj ekuk cn =

n

1kka , dn =

n

1kkb . ekukfd fcUnq An(an, cn), Bn(bn, dn) ijoy ;

C: y = px2 + qx + r ij fLFkr gS t gk¡ p,q,r vpj gSA

11. The value of p equals p d k eku cjkcj gS&

(A*) 41

(B) 31

(C) 21

(D) 2

12. If r = 0 then the value of a1 and b1 are

(A) 21

and 1 (B) 1 and 23

(C*) 0 and 2 (D) 21

and 2

;fn r = 0 gks] rks a1 vkSj b1 d s eku gS&

(A) 21

vkSj 1 (B) 1 vkSj 23

(C*) 0 vkSj 2 (D) 21

vkSj 2

Sol. 17. Given fn;k x;k gS cn =a1 + a2 + a3 + �.. + an where a1, a2, ��., an are in A.P. with d = 2 t gk¡ a1, a2, ��., an l ekUrj Js<h esa gS ft ld k lkoZvUrj d = 2 gSA and dn = b1 + b2 + b3 + ��+ bn are in A.P. with d = 2 vkSj dn = b1 + b2 + b3 + ��+ bn lekUrj Js<h esa gSA ft ld k lkoZvUrj d = 2 gSA also (an, cn) lies on y = px2 + qx + r rFkk (an, cn) oØ y = px2 + qx + r ij fLFkr gSA Nowvc cn = pan + qan + r �..(1) cn�1 = pa2

n�1 + qa n�1 + r �.(2) From (1) and (2), we get (1) vkSj (2) ls gesa izkIr gksrk gSA cn � cn�1 = p(a2

n � a2

n�1) + q(an �an�1) an = p(an + an�1) (an �an�1) + q(an �an�1) an = (an � an�1) [p(an + an�1)+q] �.(3) (an � an�1 = d) on putting n = 2 and 3 in equation (3), we get n = 2 vkSj 3 esa j[kus ij (3), we get a2 = d[p(a2 + a1) + q] �.(4) a3 = d[p(a3 + a2) + q] �.(5) Now vc (5) � (4) , we get 3 2 3 1

d 2d

a � a d[p(a � a )]



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P2JPT2(ADV.)170515C0-7

4p = 1 p = 41

ans.

adding (4) andvkSj (5) t ksMus ij q = 12

18. if ;fn r = 0, then t c c1 = pa1

2 + qa1

a1 = 41

a12 +

21

a1 ( c1 = a1)

a21 = 2a1 = 0 a1 = 0 or ;k a1 = 2

AlsorFkk d1 = 41

b12 + qb1

11 bdand

21

q

d1 = 41

b12 +

21

b1

b12 � 2b1 = 0 b1 = 0 or ;k b1 = 2

but ijUrq a1 < b1 a1 = 0 and vkSj b1 = 2 ans.


If f : R R be a differentiable function such that (f(x))7 = x � f(x) then [AR-MS] [307] ;fn f : R R vod y uh; Q y u bl izd kj (f(x))7 = x � f(x) rc

13. The value of 2

0

1� dx)x(f is

2

0

1� dx)x(f d k eku gS&

(A) 3 2 (B) 2 (C*) 3 (D) 1 14. The area bounded by curve y = f(x) between the ordinates x = 0, x = 3 and x-axis is

(A*) 7f( 3 )

8 3 � f 3 � 4f( 3)8

sq. units (B)

73f�38

8)3(f

sq. units

(C)

893

�3f3 sq. units (D) none of these

oØ y = f(x) d ksfV x = 0, x = 3 vkSj x-v{k ls ifjc) {ks=k d k {ks=kQ y gS&

(A*) 7f( 3 )

8 3 � f 3 � 4f( 3)8

oxZ bd kbZ (B)

73f�38

8)3(f

oxZ bd kbZ

(C)

893

�3f3 oxZ bd kbZ (D) buesa l s d ksbZ ugha

Sol. f(x)[f(x)6 + 1] = x f(0)[(f(0))6 + 1] = 0 f(0) = 0 and vkSj 7 (f(x))6.f(x) = 1 � f1(x) f(x)[7(f(x))6 + 1] = 1 f(x) > 0 x R Hence f(x) is increasing function x R vr% f(x) , x R ,d o/kZeku Q y u gSA so there exists an inverse of f(x) such that f�1(0) = 0 blfy , f(x) d k izfry kse fo|eku gS t cfd f�1(0) = 0 x7 + x = f�1(x)



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P2JPT2(ADV.)170515C0-8

2

0

282

0

7

2x

8x

dxxx = 2 + 1 = 3

Now, we know that vc gesa izkIr gksrk gS dx)x(fdx)x(f)a(f

0

1�a

0 = af(a)

Hence vr% )3(f3dx)x(fdx)x(f)3(f

0

1�3

0

f ( 3 )3

7

0 0

f(x)dx 3f( 3) � (x x)dx

)3(f4�3f�38

8)3(f

dx)x(f7

3

0

INTEGER TYPE - 6 15. A game is played with special fair cubic die which has one red side, two blue sides, and three

green sides. The result is the colour of the top side after the dice is been rolled. If the die is rolled repeatedly. The probability that second blue result occurs on or before the tenth roll can be

expressed in the form of p q

r

3 � 2

3 where p,q,r are positive integers then find the value of p + r � q.

,d [ksy ] ,d fof'k"V fu"i{kikrh ?kuh; ikls ls [ksy k t krk gS ft ld h ,d Hkqt k y ky ] nks Hkqt k,a uhy h rFkk rhu Hkqt k,a gjh gSA ikls d ks Q sd us d s ckn Å ijh Hkqt k ij jax ifj.kke gSA ;fn ikls d ks y xkrkj Q sad k t krk gS] rks 10 ckj esa ;k mll s igy s iklk Q sad us ij nwljh ckj

uhy k ifj.kke vkus d h izkf;d rk d ks p q

r

3 � 2

3 d s : i esa O;Dr fd ;k t krk gSA t gk¡ p,q,r /kukRed

iw.kk±d gS] rc p + r � q d k eku gS& [PR-BT] [303] [JPT-2_Adv. 2015] Ans. 07 Sol. Now 2nd blue result occur on or before the 10th roll is equivalent to the occurrence of blue face

atleast twice is 10 rolls 10oha ckj ;k mll s igy s iklk Q sad us ij nwljh ckj uhy k ifj.kke vkus d h izkf;d rk] 10 ckj

esa d e l s d e nks ckj uhy h lrg vkus d s rqY; gSA

n = 10, p =13

q =23

P(r 2) = 1 � (P(0) + p(1))

= 1 � 10 9

2 1 210.

3 3 3

1 � 9 11 9 11

9 9 9

2 .4 2 3 � 21

3 3 3

p = 9 , q = 11, r = 9 p + r � q = 9 + 9 � 11 = 7

16. Let y be an element of the set A = {1, 2, 3, 5, 6, 10, 15, 30} and x1 x2 x3 = y, then the number of

positive integral solution of x1x2x3 = y is m. Find m. [PC-MT] [304] [JPT-2_Adv. 2015]

ekuk l eqPp; A = {1, 2, 3, 5, 6, 10, 15, 30} d k vo;o y gS vkSj x1 x2 x3 = y, rc x1x2x3 = y d s gy ksa

d s /kukRed iw.kk±d gy ksa d h la[;k m gSA m Kkr d hft ,A Ans. 64



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P2JPT2(ADV.)170515C0-9

Sol. The number of solutions of the given equation is the same as the number of solution of the equation x1x2x3 x4 = 30 = 2 × 3 × 5 (here x4 is dummy variable)

fn, x, lehd j.k d s gy ksa d h la[;k] lehd j.k x1x2x3 x4 = 30 = 2 × 3 × 5 d s gy ksa d h la[;k d s

cjkcj gS (;gk¡ x4 xqIr pj gS) Hence number of solutions is 43 = 64 vr% gy ksa d h la[;k 43 = 64 17. If area of triangle formed by the points (2, ),(+ , 2+ ) and (2, 2) is 8 square unit then

area of triangle whose vertex are ( + , � ), (3 � , + 3) and (3 � , 3 � ) is ;fn (2, ),(+ , 2+ ) vkSj (2, 2) ls cus f=kHkqt d k {ks=kQ y 8 oxZ bd kbZ gS rFkk

( + , � ), (3 � , + 3) vkSj (3 � , 3 � ) ls cus f=kHkqt d k {ks=kQ y gS& [JA_CT-2_21-09-2014] [JPT-2_Adv. 2015] [SL-AR] [301]

Ans.32

Sol.

P(2 ,2 )b aR(2 , )a b

C(3 � ,3 � )b b aaQ( + ,2 + )b b aa

D(3 � , +3 )b a b a

Area of ABC = 4(PQR) ABC d k {ks=kQ y = 4(PQR) = 4 × 8 = 32 square unit oxZ bd kbZ 18. If A1, A2, A3, A4 be the area of a triangular faces of a tetrahedron and h1, h2, h3, h4 be the

corresponding altitude of the tetrahedron. If volume of tetrahedron is 5 cubic units then find the

minimum value of 1 2 3 4 1 2 3 4(A A A A )(h h h h )

24

[SS-IG] [304] [JPT-2_Adv.

2015] ;fn A1, A2, A3, A4 ls cus prq"Q y d d s f=kHkqt kd kj lrgksa d k {ks=kQ y gS rFkk h1, h2, h3, h4

prq"Q y d d s l axr 'kh"kZ y Ec gSA ;fn prq"Q y d d k vk;ru 5 ?ku bd kbZ gS rc

1 2 3 4 1 2 3 4(A A A A )(h h h h )

24

d k U;wure eku gS&

Ans. 10

Sol. V = 13

h1A1 h1 =1

3VA

h2 = 2

3VA

h3 = 3

3VA

h4 = 4

3VA

(A1 + A2 + A3 + A4) 1 2 3 4

3V 3V 3V 3VA A A A

3V(A1 + A2 + A3 + A4) 1 2 3 4

1 1 1 1A A A A

1 2 3 4

1 2 3 4

A A A A 41 1 1 14A A A A

(A1 + A2 + A3 + A4) 1 2 3 4

1 1 1 1A A A A

16

Min value U;wure eku : 3V(16) = 48 V = 240

min value of U;wure eku 1 2 3 4 1 2 3 4(A A A A )(h h h h )

24

= 10



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P2JPT2(ADV.)170515C0-10

19. If roots of the cubic equation (z � ab)3 = a3, a 0 represents the vertex of a triangle, then the length of one of the sides of the triangle will be , when a = 3 [JPT-2_Adv. 2015]

;fn ?kuh; l ehd j.k (z � ab)3 = a3, a 0 d s ewy f=kHkqt d s 'kh"kZ gS] rc f=kHkqt d h Hkqt kvksa d s

,d d h y EckbZ gksxh] t c a = 3 [CN-CR] [302] Ans. 03 Sol. (z � ab)3 = a3

z � ab = a(1)1/3, z � ab = a, a, a2 z1 = ab + a, z2 = ab + a, z3 = ab + a2 |z1 � z2| = |a � a| = |a||1�| = 3 | a |

= 3 . 3 = 3

20. If the value of the definite integral = 3

6 3 21

(2x � 1)dx

x 2x 9x 1

can be expressed in the form of

�1A Ccot

B Dwhere A,B,C,D are rationals in their lowest form then the value of 2(A + B + C + D) is

equal to [DI-II] [301]

;fn fuf'pr lekd y = 3

6 3 21

(2x � 1)dx

x 2x 9x 1

d s eku d ks �1A Ccot

B Dd s : i esa O;Dr d jrs gS t gk¡

A,B,C,D mud s ljy re : i esa ifjes; la[;k,a gS] rc 2(A + B + C + D) d k eku cjkcj gS& Ans. 18

Sol. = 2

412

12x � dx

x1

x 2x 9x

put x2 +

1x

= t j[kus ij

= 2

2

dtt 9

= 13

cot�1 23



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

Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

P2JPJFJRAIOT220215C0-1

BATCH : JP, JF, JR AIOT-2 (JEE ADVANCE) DATE : 03-05-2015 SYLLABUS : (XI & XII SYLLABUS)

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

1 to 8 MCQ 8 3 0 24



21 to 28 MCQ 8 3 0 24



41 to 48 MCQ 8 3 0 24



60 180TotalTotal

Maths

Physics

Chemistry

Paper-2 (AIOT-2 JEE ADVANCE)



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PAPER-2

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

This section contains 8 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 8 cgqfodYi ç'u gSaA çR;sd ç'u esa pkj fodYi (A), (B), (C) vkSj (D) gSa] ftuesa ls ,d ;k vf/kd lgh gSA

21. A light wire of length and radius r is welded to another light wire of length 2 are radius 2r. The

free end of 1st wire is fixed and the free end of 2nd wire supports a mass m. The young's modulus of each wire is Y. [EC-GE](104)

yEckbZ rFkk r f=kT;k dk rkj, 2 yEckbZ rFkk 2r f=kT;k ds vU; rkj ds lkFk tqM+k gqvk gSA igys rkj dk eqDr fljk tM+or~ gS rFkk nwljs rkj ds eqDr fljs ls fp=kkuqlkj m nzO;eku yVdk gqvk gSA izR;sd rkj dk ;ax xq.kkad Y gSA

m

2

(A*) Extension in upper half of upper wire is 2

mg

2 r Y

Å ijh rkj ds Å ijh v)Z Hkkx esa izlkj 2

mg

2 r Y

gSaA

(B*) Extension in lower wire is 2

mg

2 r Y

fupys rkj esa izlkj 2

mg

2 r Y

gSA

(C) Total elongation in the composite wire is 2

5 mg2 r Y

la;qDr rkj esa dqy izlkj 2

5 mg2 r Y

gSaA

(D*) Total elongation in the composite wire is 2

3 mg2 r Y

la;qDr rkj esa dqy izlkj 2

3 mg2 r Y

gSA

Sol. Here stretching force is same but stress will be different for wires. ;gk¡ ruu cy leku gS] ijUrq nksuksa rkjksa esa izfrcy fHkUu&fHkUu gSA For upper wire Åijh rkj ds fy,

Stress izfrcy = Y × strain foÑfr

2

Mg

r = Y

1 = 2

mgYr

for lower wire fupys rkj ds fy,



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2= 2 2

mg 2 mgY(2r) 2 r Y

total elongation dqy foLrkj = 2

3 mg2 r Y

22*. A particle of mass 'm' is suspended with the help of an ideal string from the ceiling of a car as

shown in figure. Initially the whole system is at rest. At time t = 0 the car starts accelerating towards right with uniform acceleration 'a' such that a << g. Then choose the correct statement(s).

[CM-VT](104) ,d vkn'kZ jLlh dh lgk;rk ls 'm' nzO;eku dk d.k fp=kkuqlkj dkj dh Nr ls yVdk gqvk gSA izkjEHk esa

lEiw.kZ fudk; fLFkj gSA t = 0 ij dkj nka;h vksj le:i Roj.k a ls Rofjr gksrh gS ,oa a << g gSA rks lgh dFku@dFkuksa dk p;u dhft,A

m

(A*) The particle will execute simple harmonic motion with respect to car. d.k dkj ds lkis{k ljy vkorZ xfr djsxk (B*) The particle will be at equilibrium position w.r.t. car for the 2nd time at instant

t = 2 2

32 a g

dkj ds lkis{k d.k t = 2 2

32 a g

le; i'pkr~ okil ek/; fLFkfr ij nwljh ckj igqprk gSA

(C) The maximum speed of the particle with respect to ground during subsequent motion will be

2 2 1/ 2(g a ) tan�1 ag

v/kksfyf[kr xfr ds nkSjku tehu ds lkis{k d.k dh vf/kdre pky 2 2 1/ 2(g a ) tan�1 ag

gSA

(D*) The speed of the particle with respect to ground at the instant the string becomes vertical

again for the first time is aT where T = 22 2a g

tc jLlh nqckjk izFke ckj Å /okZ/kj fLFkfr esa vkrh gS rc d.k dh tehu ds lkis{k pky aT gSa ;gk¡ T = 2

2 2a g

gSA

Sol. The particle will execute S.H.M with angular amplitude tan�1 ag

with respect to car and of time

period T = 22 2a g

.

d.k dkj ds lkis{k dks.kh; vk;ke tan�1 ag

rFkk vkorZdky T = 22 2a g

ds lkFk ljy vkorZ xfr djrk gSA

It will be at eqb. position for second time at time t = 3

T4



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;g nwljh ckj lkE;koLFkk ij le; t = 3

T4

ij igq¡prk gSA

The max. speed of the particle with respect to car will be A = 2 2g a

. tan�1 ag

.

dkj ds lkis{k d.k dh vf/kdre pky A = 2 2g a

. tan�1 ag

gSA

The particle will come to rest at with respect to car at the instant string becomes vertical & its

speed will be same as speed of car i.e., aT where T = 22 2a g

dkj ds lkis{k d.k fojkekoLFkk esa rc vkrk gS tc jLlh Å/okZ/kj gksrh gS rFkk bldh pky bl le; dkj dh pky ds

cjkcj gksrh gS vFkkZr~, aT ds cjkcj gksrh gS ;gk¡ T = 22 2a g

gSA

23. The figure here shows a water tank provided with a square gate (of side 2m) which can rotated

freely about z axis passing through shown point A. The gate is held in its place due to buoyant

force acting on the light balloon attached to the gate via a light thread. Considering the information

that the gate just opens when level of water reaches the height of 4m above point A and hinge

force is zero, the shown point P is point at which the thread is attached to the gate

[FL-BY](103) iznf'kZr fp=k esa oxkZdkj njokts (Hkqtk 2m) ;qDr ,d ikuh dk VSad iznf'kZr gSA njoktk fp=k esa iznf'kZr fcUnq A

ls xqtjus okyh z-v{k ds ifjr% eqDr :i ls ?kw.kZu dj ldrk gSA njokts dks blls tqM+h gqbZ gYdh jLlh }kjk

ca/ks gq, gYds xqCckjs ij vkjksfir mRiykod cy }kjk nh xbZ fLFkfr esa fLFkj j[kk tkrk gSA ;g ekuk tkrk gS]

tc ikuh dk Lrj fcUnq A ls Å ij 4m vkrk gS] rc njoktk Bhd [kqy tkrk gSA fuyEcu }kjk vkjksfir cy

'kwU; gSA fp=kkuqlkj fcUnq P ij jLlh njokts }kjk cU/kh gqbZ gSA

Gate ¼njoktk½

P Water (ikuh) B

Light balloon (gYdk xqCckjk) A

b = 2m

a = 4m

z

y

x

(A*) Volume of water displaced by the balloon is 20 m3

xqCckjs }kjk foLFkkfir ikuh dk vk;ru 20 m3 gSA

(B) Volume of water displaced by the balloon is 30 m3 xqCckjs }kjk foLFkkfir ikuh dk vk;ru 30 m3 gSA

(C*) distance between A and P is approximately 1.06 m.

A rFkk P ds e/; nwjh yxHkx 1.06 m gSA

(D) Force on gate due to Water is 15kN

njokts ij ikuh }kjk vkjksfir cy 15kN gSA

Sol. At the instant of just opening of gate njokts ds Bhd [kqyrs le;,



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Ffluid Fbuoyant

PMID × Area = Vg

g

2b

a × b2 = Vg

V = 2b2b

a

= 20m3

Also, about the axis passing through A ; A ls xqtjus okyh v{k ds ifjr%

fluid tension

a b

a

gh bdh(h a)

= Vgx

x = m06.1V

dh)ah(hba

a

24. A solid sphere of mass m is released on the plank of mass M which lies on an inclined plane of

inclination as shown in figure. There is sufficient frictional force between sphere and plank and the minimum value of co-efficient of friction between plank and surfaces is to keep the plank at rest. Then [RB-CD](104)

m nzO;eku dk Bksl xksyk M nzO;eku ds r[rs ij NksM+k tkrk gSA r[rk dks.k okys urry ij fp=kkuqlkj j[kk gqvk gSA xksy rFkk r[rs ds e/; mifLFkr ?k"kZ.k cy i;kZIr gS ,oa r[rs o urry ds e/; mifLFkr ?k"kZ.k xq.kkad dk U;wure eku gS, tks fd r[rs ds fLFkj j[kus ds fy, i;kZIr gS] rks

m

M

(A*) Frictional force between sphere and plank is 27

mg sin , when plank is at rest

xksys rFkk r[rs ds e/; ?k"kZ.k cy 27

mg sin gS, tc r[rk fLFkj gSA

(B*) The value of is 7M 2m7(M m)

tan

dk eku 7M 2m7(M m)

tan gSA

(C) If there is no friction between the plank and inclined plane, then acceleration of plank is less than g sin . ;fn r[rs rFkk urry ds e/; dksbZ ?k"kZ.k cy vkjksfir ugha gks rks r[rs dk Roj.k g sin ls de gSA

(D*) If there is no friction between plank and inclined plane, then friction force on the sphere is zero.

;fn r[rs rFkk urry ds e/; dksbZ ?k"kZ.k cy vkjksfir ugha gks rks xksys ij vkjksfir ?k"kZ.k cy 'kwU; gSA Sol. If plank is at rest acceleration of cylinder ;fn r[rk fLFkj gks rks xksys dk Roj.k

a =

2

gsin 5gsin71

mR



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mg sin � f = ma f = 27

mgsin

2f mgsin

7

M

Mg sin

(M + m)g cos

Mgsin + 27

mgsin = (M + m)g cos

= 7M 2m7(M m)

tan

25. Consider an infinite mesh as shown in figure Each side of the mesh has resistance R. Consider

hexagonal part ABCDEF of infinite mesh and equivalent resistance between any two points of hexagon is measured. Choose the correct option(s) : [CE-EQ](104)

fp=k esa ,d vuUr vkdkj dk tky iznf'kZr gSA izR;sd tky dh izR;sd Hkqtk dk izfrjks/k R gSA vuUr tky ds ,d "kV~dks.kh; Hkkx ABCDEF dh dYiuk dhft, rFkk bl Hkkx ds fdUgh nks fcUnqvksa ds e/; rqY; izfrjks/k ekik tkrk gSA rks lgh fodYiksa dk p;u dhft,A

A B

C

D E

F

(A*) equivalent resistance between A and B is 2R3

A rFkk B ds e/; rqY; izfrjks/k 2R3

gSA

(B*) equivalent resistance between A and C is R A rFkk C ds e/; rqY; izfrjks/k R gSA

(C*) equivalent resistance between A and D is 7R6

A rFkk D ds e/; rqY; izfrjks/k 7R6

gSA

(D*) equivalent resistance between A and E is R A rFkk E ds e/; rqY; izfrjks/k R gSA Sol. If we connect positive terminal of a battery to A and negative to infinity then the current through

branches will be as shown. If we connect negative to B and positive to infinity current will be in

opposite direction by supper -position of two situation net current in AB branch will be 23

.



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Potential difference between A and B will be 2

R3

. Effective resistance between A and B will be

2R3

. Similarly we can do for other terminals.

;fn cSVjh ds /kukRed fljs dks A ls rFkk _ .kkRed fljs dks vuUr ls tksM+ ns rks bl ifjiFk dh 'kk[kkvksa esa /kkjk fp=kkuqlkj izokfgr gksxhA ;fn cSVjh ds _ .kkRed fljs dks B ls rFkk /kukRed fljs dks vuUr ls tksM+ ns rks /kkjk

foifjr fn'kk esa izkokfgr gksxh] rFkk bu nksuksa fLFkfr;ksa esa izokfgr /kkjk ds v/;kjksi.k }kjk 'kk[kk AB esa dqy /kkjk 23

izokfgr gksxhA A rFkk B ds e/; foHkokUrj 2

R3

gksxkA blfy, A rFkk B ds e/; rqY; izfrjks/k 2R3

gksxk] vkSj blh

izdkj vU; fljks ds fy, Hkh Kkr dj ldrs gSA

A

/3

/6 /6

/3 /3 /6 /6

26. Consider an imaginary cube of side '' as shown in figure : [ES-GT](103) fp=k esa iznf'kZr '' Hkqtk ds ,d dkYifud ?ku dh dYiuk dhft, :

A

E

C

G

B

F

D

H

Q

P

An infinite large sheet of surface charge density passes through PQGF. P and Q are mid points

of AB and DC respectively. Out of six faces of cube the electric flux linked : i"Bh; vkos'k ?kuRo dh ,d vUur vkdkj dh ifV~Vdk PQGF ls xqtjrh gSA P rFkk Q Øe'k% AB rFkk DC

ds e/; fcUnq gSA blds N% Qydksa ls lEcfU/kr ¶yDl esa ls : (A*) is zero with three faces rhu Qydksa ds fy, 'kwU; gSA (B*) with face ADHE and face BCGF are equal Qyd ADHE rFkk Qyd BCGF ds fy, leku gSA (C*) is minimum (non zero) with FGHE FGHE ds fy, U;wure (v'kwU;) gSA (D) is zero with two faces nks Qydksa ds fy, 'kwU; gSA Sol. Consider the projected area of any face to calculate the flux ¶yDl Kkr djus ds fy, fdlh Qyd ds {kS=kQy dh dYiuk djrs gSA Electric flux with ABCD, ABEF, DCGH is zero ABCD, ABEF, DCGH ls lEcfU/kr fo|qr ¶yDl 'kwU; gSA



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27. Figure shows top view of a horizontal surface. Two blocks each of mass m are placed on the surface and connected with a string. The friction coefficient is for each block. A horizontal force F is applied on one of the block as shown in the figure. F is maximum so that there is no sliding at any contact. [FR-SG](104)

,d {kSfrt lrg dk Å ijh n'; fp=k esa iznf'kZr gSA leku nzO;eku m ds nks CykWd bl lrg ij fLFkr gS rFkk vkil esa ,d jLlh ls cU/ksa gq, gSA izR;sd CykWd ds fy, ?k"kZ.k xq.kkad gSA fp=kkuqlkj ,d CykWd ij ,d {kSfrt cy F vkjksfir fd;k tkrk gSA F vf/kdre gS ftlds dkj.k fdlh Hkh lEidZ ij fQlyu ugha gSA

F

Top view

m

(A*) If ;fn = 30° F = 3

mg2 & T < mg

(B*) If ;fn = 45° F = mg2 & T = mg

(C) If ;fn = 60° F = 2mg & T < mg

(D*) If ;fn = 60° F = mg3 & T = mg Sol.

f T

F

T cos = f sin but T mg T cos = mg sin F = T sin + f cos

= mg

cossin

sin + mg cos

= mg

cos)cos(

For Fmax = < 45° Fmax = mg sec Fmax ds fy,] = < 45° Fmax = mg sec But tension in the string can not be more than mg jLlh esa ruko mg ls T;knk ugha gks ldrk

T = mg sin cos

< mg should be less than 45°

T = mg sin cos

< mg ] 45° de gksuk pkfg,A

Now if (vc ;fn) > 45° T = mg = 90° � F = mg sin + mg cos = 2mg sin If (;fn) < 45° then (rks) Fmax = mg sec

If (;fn) > 45° then (rks) Fmax = 2mg sin



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28. Consider a hemisphere of radius R with centre of curvature at origin O, as shown. Refractive index

of material of the hemisphere varies as = 2R

2R x. Where x is x-coordinate of material point. A

ray travelling in air in xy-plane is grazingly incident at O, as shown R f=kT;k ds ,d v)Zxksys dh dYiuk dhft,] ftldk oØrk dsUnz ewyfcUnq O ij fp=kkuqlkj gSA v)Zxksys ds

inkFkZ dk viorZukad = 2R

2R x ds vuqlkj ifjofrZr gksrk gSA tgk¡ x inkFkZ ds fcUnq dk x-funsZ'kkad gSA ,d

fdj.k ok;q esa xfr djrh gqbZ] fp=kkuqlkj xy-ry esa fcUnq O ij i"BLi'khZ vkifrr gksrh gSA [GO-RP](105) y

O x

(A*) Trajectory followed by ray as it travels inside the hemisphere is circular v)Z xksys eas xfr ds nkSjku fdj.k dk iFk oÙkkdkj gSA (B) y-coordinate of the point of hemisphere where the ray comes out of the hemisphere lies

between 0.5R and 0.75R fdj.k v)Zxksys ds ftl fcUnq ls ckgj fudyrh gS] v)Zxksys ds ml fcUnq dk y-funsZ'kkad 0.5R ls 0.75R ds

e/; fLFkr gSA (C*) Deviation suffered by the ray just before it comes out of the hemispherical surface lies

between 0° and 30° xksyh; lrg ls Bhd ckgj fudyrs le; fdj.k }kjk izkIr fopyu 0° ls 30° ds e/; fLFkr gSA (D) Deviation suffered by the ray just before it comes out of the hemispherical surface lies between

30° and 45° xksyh; lrg ls Bhd ckgj fudyrs le; fdj.k }kjk izkIr fopyu 30° ls 45° ds e/; fLFkr gSA Sol.

O

x 90°

A

The figure shows a strip at a distance x of thickness dx. As of material increases with x, ray will

deviate continuously as shown. By snell's law, between O and A, fp=k esa x nwjh ij dx eksVkbZ dh ifêdk iznf'kZr gSA pqafd nwjh x ds lkFk inkFkZ dk viorZukad c<+rk gS] bl

dkj.k fdj.k fp=kkuqlkj yxkrkj fopfyr gksrh gSA O ls A rd Lusy ds fu;e }kjk



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1 × sin90° = 2R

sin2R x

tan = 2 2

2R x

(2R) (2R x)

2 2

dy 2R xdx (2R) (2R x)

y x

2 20 0

2R xdy dx

(2R) (2R x)

y = 2 2

2R 2R x

y2 + (x-2R)2 = (2R)2 which is equation of circle of radius 2R centered at (2R, 0) tksfd ,d 2R f=kT;k ds oÙk dh lehdj.k gS ftldk dsUnz (2R, 0) ij gSA also, equation of hemispherical surface is blh izdkj v)Zxksys dh lrg dh lehdj.k fuEu gS x2 + y2 = (R)2 y2 = R2 � x2 Putting this value of y2 in equation of trajectory we can find coordinates of point where the ray

comes out of hemisphere, as bl y2 ds eku dks iFk lehdj.k esa j[kus ij mu fcUnqvksa ds funsZ'kkad dj ldrs gS tgk¡ ij fdj.k v)Zxksys ls

ckgj vkrh gS R2 + x2 + (x � 2R)2 = 4R2

x = R4

y = 2

2 2 2 R 15RR x R 0.97R

4 4

Now, to find deviation () consider the diagram here vc, fopyu () Kkr djus ds fy, fuEu fp=k dk iz;ksx djrs gSA

y

y 2R

Clearly Li"V :i ls, sin =

y 15 16 1or

2R 8 8 2

0 < < 30°

SECTION � 2 : (Paragraph Type)

[k.M � 2 : (vuqPNsn çdkj)

This section contains 3 paragraphs each describing theory, experiment, data etc. Six questions relate to three paragraphs with two questions on each paragraph. Each question of a paragraph has only one correct answer among the four choices (A), (B), (C) and (D).



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bl [k.M esa fl)karksa] ç;ksxksa vkSj vk¡dM+ksa vkfn dks n'kkZus okys 3 vuqPNsn gSaA rhuksa vuqPNsnksa ls lacaf/kr N%

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

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

Paragraph for Question Nos. 29 to 30 iz'u 29 ls 30 ds fy, vuqPNsn

Comp.

The rectangular box shown in the figure has a partition which can slide without friction along the length of the box. Initially each of the two chambers of the box has one mole of a monatomic ideal gas at a pressure P0, volume V0 and temperature T0. The chamber on the left is slowly heated by an electric heater. The walls of box and partition are thermally insulated. Heat loss through lead wire of heater is negligible. The gas in left chamber expands, pushing the partition until final

pressure in both chambers becomes 24332

P0. (Take R = 8.3 J/mol K) [TH-FL](104)

fp=k esa iznf'kZr ?kukdkj fMCcs esa fLFkr fiLVu fcuk ?k"kZ.k ds fMCcs dh yEckbZ ds vuqfn'k fQly ldrk gSA izkjEHk esa fMCcs ds nksuksa Hkkxksa esa ,d eksy ,dijek.kqd vkn'kZ xSl] nkc P0, vk;ru V0 rFkk rkieku T0 ij Hkjh gqbZ gSA cka;s okys Hkkx esa fLFkr xSl dks fo|qr ghVj }kjk /khjs&/khjs xeZ fd;k tkrk gSA fMCcs dh nhokjs rFkk fiLVu Å "eh; dqpkyd gS rFkk ghVj ds rkj }kjk Å "eh; gkfu ux.; gSA cka;s okys Hkkx esa fLFkr xSl QSyrh gS

vkSj fiLVu dks rc rd nckrh gS tc rd fd nksuksa Hkkxksa dk nkc 24332

P0 u gks tk;sA (R = 8.3 J/mol K ysa)

29. Final temperature of gas in left chamber is : [TH-FL](104) cka;s Hkkx esa xSl dk vfUre rkieku gksxk & (A) 2.25 T0 (B) 4.5 T0 (C) 8.75 T0 (D*) 12.93 T0

Sol. 0

right

T

T

= 1

0

0

P 32

243P

0

right

T

T =

223

Tright = 2.25 T0

For the left chamber, the process of heating is slow. Initial parameters are P0, V0, T0. nka;s Hkkx ds fy,, izØe vuqlkj /khsjs&/khjs xeZ djrs gSA izkjfEHkd izkapy P0, V0, T0 gSA

Final parameters are vfUre izkapy 0243

P32

, V1, T1

0 0

0

P V

T = 0243P

32

1

1

VT

T1 = 24332

1

0

VV

T0 ��.(1)

Adiabatic compression occurs in right chamber. Initial parameters are P0, V0, T0. nka;s Hkkx esa :)ks"e lEihM+u gksxkA blds izkjfEHkd izkapy P0, V0, T0 gSA



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Final parameters are vfUre izkapy 24332

P0, V2, T2 gSA

P0V0 = P0

24332

V2

5 / 3

2

0

V

V

= 32243

2

0

VV

= 827

V1 + V2 = 2V0

V1 = 2V0 � 827

V0

V1 = 4627

V0 ��.(2)

by equation lehdj.k (1) & o (2) ls T1 = 12.93 T0 30. The work done by the gas in the left chamber is [TH-FL](104) cka;s Hkkx esa xSl }kjk fd;k x;k dk;Z gksxk (A) 5.58 T0 J (B) 10.58 T0 J (C) 25.58 T0 J (D*) 15.58 T0 J Sol. Work done by gas on right chamber. nka;s Hkkx esa fLFkr xSl }kjk fd;k x;k dk;ZA

W = 2 2 0 0P V P V

1

= 32

9

14

P0V0

= 158

× 8.3 T0

= (�15.58 T0) J

Paragraph for Question Nos. 31 to 32

iz'u 31 ls 32 ds fy, vuqPNsn Comprehension # 1 In standard YDSE the intensity on screen varies with position according to following graph as we

move away from the central maxima : ] ekud YDSE iz;ksx esa dsfUnz; mfPp"B ls nwj tkus ij fLFkfr ds lkFk insZ ij izkIr rhozrk fuEu vkjs[k ds

vuqlkj ifjofrZr gksrh gSA [YE-YE](104)

0

2 4 6 8 10 y (in mm)

0

Wave length of light used is 400 nm and distance between the two slits is 1 mm. iz;qDr izdk'k dh rjaxnS/;Z 400 nm rFkk nksuksa fLyVksa ds e/; nwjh 1 mm. gSA



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31. If the distance between two slits is reduce to 0.5 mm then the intensity : [YE-YE](104)

;fn nksuksa fLyVksa ds e/; nwjh 0.5 mm rd de gks tk;s rks] izkIr rhozrk

(A) will become zero at y = 4 mm and y = 8 mm

y = 4 mm rFkk y = 8 mm ij 'kwU; gks tk;sxhA

(B) will remain same 0 at y = 4 mm and y = 8 mm

y = 4 mm rFkk y = 8 mm ij rhozrk 0 ds leku jgsxhA

(C*) will become zero at y = 4 mm and will remain 0 at y = 8 mm

y = 4 mm ij 'kwU; gks tk;sxh rFkk y = 8 mm ij eku 0 ds leku jgsxhA

(D) will become zero at y = 8 mm and will remain 0 at y = 4 mm

y = 8 mm ij 'kwU; gks tk;sxh rFkk y = 4 mm ij eku 0 ds leku jgsxhA

Sol. Fringe width will become double. fÝUt pkSM+kbZ nqxuh gks tk;sxhA

32. If the distance between the two slits is kept 1 mm and then a thin film of refractive index 1.5 and

thickness 2.2 m is used to cover the one of the slit. What will be intensity at y = 2 mm.

;fn nksuksa fLyVksa ds e/; nwjh 1 mm j[kh tk;s rFkk fdlh ,d fLyV ds lkeus 1.5 viorZukad dh 2.2 m

eksVkbZ okyh iryh fQYe j[kus ij y = 2 mm ij rhozrk D;k gksxh ? [YE-YE](104)

(A*) /2 (B) (C) /4 (D) zero

Sol. Shift foLFkkiu = t( 1)

= �6

7

2.2 10 0.5

4 10

= 114

So, intensity will become 0

2

vr% rhozrk 0

2

gks tk;sxhA

Paragraph for Question Nos. 33 to 34 iz'u 33 ls 34 ds fy, vuqPNsn

Passage The common prism telescope is shown in figure is generally made up of three parts, viz., a prism and two

telescopes. One of these telescopes is used to render the rays which fall on the prism parallel to each other, and is therefore called the collimator, the other telescope is made movable so that it can be turned into the proper direction for observing any desired color, and is, therefore, called the view-telescope. A circular scale is also provided to major angular position of telescope.

On the circular scale 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half-a-degree (0.5°). [GO-OI](104)

First telescope is focused to collect rays directly from collimator and position of telescope is measured. This is position (1). Now a prism is placed as shown in the figure and telescope is focused for red, yellow and violet. The respective positions of telescope are (2), (3) and (4).



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fp=k esa iznf'kZr ,d la;qDr nwjn'khZ lk/kkj.kr;k rhu Hkkxksa ls feydj cuk gksrk gSA tSls fizTe rFkk nks nwjn'khZA buesa ls ,d nwjn'khZ dk iz;ksx fizTe ij fdj.kksa dks lekUrj izsf{kr djuk gksrk gS] blfy, bldks dkWyhehVj dgrs gS rFkk blh izdkj nwljk nwjn'khZ pyk;eku :i esa cuk gksrk gSA ftlds dkj.k bldks ,d fuf'pr fn'kk esa O;ofLFkr djds bPNkuqlkj jax izsf{kr dj ldrs gSA blfy, bldks n'; nwjn'khZ dgrs gSA ofÙk; iSekuk nwjn'khZ ds vyx&vyx ogn dks.kh; fLFkfr dks iznf'kZr djrk gSA ofÙk; iSekus ij eq[; iSekus ds 29 Hkkx] ofuZ;j iSekus ds 30 Hkkxksa ls laikrh gSA eq[; iSekus dk ,d Hkkx vk/kk fMxzh (0.5°) dks iznf'kZr djrk gSA

izFke nwjn'khZ dks dkWyhehVj ls fdj.ksa izkIr djus ds fy, Qksd'khr fd;k tkrk gS] rFkk nwjn'khZ dh fLFkfr ekih tkrh gSA ;g fLFkfr (1) gSA vc fp=kkuqlkj ,d fizTe j[kk tkrk gS rFkk ;g fdj.kksa dks yky] ihys rFkk cSaxuh jax esa Qksdflr djrk gS] vksj buds fy, nwjn'khZ dh fLFkfr;k¡ Øe'k% (2), (3) rFkk (4) ls iznf'kZr gSA

Reading of main scale Reading of veriner scale Position (1) 20.0° 20 Position (2) 30.0° 16 Position (3) 32.5° 9 Position (4) 33.5° 11

eq[; iSekus dk ikB~;kad ofuZ;j iSekus dk ikB~;kad fLFkfr (1) 20.0° 20

fLFkfr (2) 30.0° 16

fLFkfr (3) 32.5° 9

fLFkfr (4) 33.5° 11

Telescope ¼nwjn'khZ½

Prism

¼fizTe½

Collimator ¼dksyhehVj½

F2

L2

P L1 S

F1

EYE ¼vk¡[k½

33. What is deviation for violet light : [GO-OI](104) cSaxuh jax ds fy, fopyu D;k gksxk % (A) 12°22 (B) 13°19 (C) 33°5 (D*) 13°21 Sol. 29 division of main scale coincides with 30 divisions of vernier scale Hence one division of vernier

scale eq[; iSekus ds 29 Hkkx ofuZ;j iSekus ds 30 Hkkx ds lEikrh gSA vr% ofuZ;j iSekus dk ,d Hkkx

= 30

29�30 of main scale eq[; iSekuk

= º5.0301

= 605.0301

min

= 1 min. Position fLFkfr (1) 20° 20

Position fLFkfr (2) 30° 16 Position fLFkfr (3) 32° 39

Position fLFkfr (4) 33° 41



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DV = 33°41� 20°20= 13°21 34. If we find angular derivation for different colours then percentage error in derivation will be

maximum for [GO-OI](104) (A*) Red (B) Yellow (C) Violet (D) Same for all ;fn ge vyx jaxksa ds fy, dks.kh; fopyu Kkr djs rks izfr'kr fopyu =kqfV fdl jax ds fy, T;knk gksxhA (A) yky (B) ihyk (C) cSaxuh (D) lHkh ds fy, leku gSA

Sol. Percentage error izfr'kr =kqfV =

× 100

is minimum for red is same for all, so percentage error will be maximum for Red. yky jax ds fy, U;wure gS lHkh ds fy, leku gS, vr% yky jax ds fy, izfr'kr =kqfV vf/kdre gksxhA

SECTION � 3 : (Integer value correct Type) [k.M � 3 : (iw.kk±d eku lgh çdkj)

This section contains 6 questions. The answer to each question is a Two digit integer, ranging from 00 to 99 (both inclusive).

bl [k.M esa 6 ç'u gSaA çR;sd ç'u dk mÙkj 00 ls 99 rd ¼nksuksa 'kkfey½ ds chp dk nks vadksa okyk iw.kk±d gSA

35. A particle starts from rest and moves on a circular path with an angular acceleration which is proportional to square root of instantaneous angular velocity. Find the mean angular velocity (in rad/sec) averaged over the time it takes to acquire an angular velocity of 12 rad sec�1

,d d.k fLFkj voLFkk ls oÙkkdkj iFk ij dks.kh; Roj.k ds lkFk xfr izkjEHk djrk gSA bldk dks.kh; Roj.k

blds rkR{kf.kd dks.kh; osx ds oxZewy ds lekuqikrh gSA 'kwU; ls 12 rad sec�1 dks.kh; osx izkIr djus esa yxs le;kUrjky ds nkSjku vkSlr :i ls bldk ek/; dks.kh; osx (rad/sec esa) Kkr djksA [CM-KN](104)

Ans. 4 Sol. = k

d

kd

0 0

d k d

3 / 2

3 / 2

= K

= 3 / 22

3k

d

kdt

t

0 0

dk dt

1/2 = kt

t = 2

1 kk

2

wavg = 4t 3 rad/sec.



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36. A piece of metal weighs 46 g in air. When it is immersed in a liquid of specific gravity 1.24 at 37C it

weighs 30 g. When the temperature of the liquid is raised to 42C, the metal piece weighs 30.5 g. The specific gravity of the liquid at 42C is 1.20. Coefficient of linear expansion of the metal is nearly 69 × 10

�N, find N. [FL-BY](104)

/kkrq ds ,d VqdM+s dk gok esa Hkkj 46 g gSA tc bldks 37C rkieku ij 1.24 fof'k"V xq:Ro ds nzo esa Mqcks;k

tkrk gS] rc bldk Hkkj 30 g gSA tc nzo dk rkieku 42C rd c<+k;k tkrk gS] rks /kkrq ds VqdM+s dk Hkkj 30.5 g gks tkrk gSA 42C rkieku ij nzo dk fof'k"V xq:Ro 1.20 gSA ;fn /kkrq dk js[kh; izlkj xq.kkad yxHkx 69 ×

10�N, gks rks N dk eku Kkr djksA

Ans. 06 Sol. We have ge tkurs gS, Upthrust mRiykod cy , FB = Vg Apparent weight vkHkklh Hkkj = mg � 0Vg = 30g �(i)

When temperature is increased tc rkieku c<+k;k tkrk gS FB = Vg Apparent weight vkHkklh Hkkj = mg � 0Vg = 30.5g = mg � 0V(1 + 3 )g = 30.5g �(ii) on solving gy djus ij

and rFkk = 6.9 × 10�5 C�1

37. A conducting tube is passing through a bath. A liquid at temperature 90°C and specific heat s is

entering at one end of tube. Rate of flow of liquid is 1 kg/s and exit temperature is 50°C. In bath

another liquid having specific heat 2s and inlet temperature 20°C is entering at a rate of 2 kg/s.

Find the exit temperature of the liquid coming out of the bath. (Assume steady state condition) [CT-CL](104)

,d pkyd uyh LukuVc ls xqqtjrh gSA uyh ds ,d fljs ls 90°C ij s fof'k"V Å "ek dk nzo izokfgr fd;k tkrk gSA bl nzo dh izokg nj 1 kg/s rFkk ckgj fudyrs le; rkieku 50°C gSA LukuVc esa Hkjs gq, vU; nzo dh fof'k"V Å "ek 2s rFkk izos'k djrs le; rkieku 20°C gS rFkk ;g 2 kg/s dh nj ls izos'k djrk gS rks LukuVc ls ckgj fudyus okys nzo dk rkieku Kkr djksA (LFkkbZ voLFkk ekurs gq,)

2s 20°C

90°C 50°C

[Ans 30°C ] 38. A conductor carrying current 60A is in the form of a semicircle AB of radius R and lying in xy-plane

with its centre toat origin as shown in the figure. The magnitude of

d.B for the circle x2 + z2 =

3R2 in xz-plane due to current in curve AB is n0. Find the value of n. [EM-AL](104)



PH

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PHYSICS





,d pkyd esa 60A /kkjk izokfgr gSA pkyd R f=kT;k dk v)ZoÙk AB gS tks fd xy ry esa fLFkr gS ,oa bldk

dsUnz fp=kkuqlkj ewy fcUnq ij gSA xz ry esa oÙk x2 + z2 = 3R2 ds fy,

d.B dk ifjek.k ywi AB esa izokfgr

/kkjk ds dkj.k n0 gS] rks n dk eku Kkr djksA Y

B

O

Z A

X

Ans. 30

Sol.

d.Bd.Bd.B 21net

1B

magnetic field due to straight part lh/ks Hkkx }kjk mRiUu pqEcdh; {ks=k

2B

magnetic field due to curved part oØ kdkj Hkkx }kjk mRiUu pqEcdh; {ks=k

R 3

0 = B1 2R 3 + 2B .d

0 = � 0

4 R 3

2R 3 + 2B .d

0 = � 0

2

+ 2B .d

0

2

= 2B .d

39. The energy that can be obtained from 1 kg of water through the fusion reaction 2H + 2H 3H + p is

P × 108J. Assume that 1.5 × 10

�2% of natural water is heavy water D2O (by number of molecules ) and all the deuterium is used for fusion. Find the value of P.

Atomic masses are m( 21 H ) = 2.014102 u , m ( 3

1 H ) = 3.016049 u [NP-FF](104)

Mass of proton mp = 1.007276 u ; 1u = 2

MeV931

c

fuEu lay;u vfHkfØ;k }kjk 1 fdxzk ikuh esa mRiUu Å "ek P × 108J gks rks P ds eku dh x.kuk dhft;s %

2H + 2H 3H + p ;g eku yhft;s fd izkÑfrd ty esa 1.5 × 10 � 2 % Hkkjh ikuh D2O gS (v.kqvksa dh la[;k ds vk/kkj ij) vkSj

leLr M~;wfVfj;e lay;u esa iz;qDr gks tkrk gSA ijek.kq Hkkj m( 2

1 H ) = 2.014102 u , m ( 31 H ) = 3.016049 u

izksVksu dk nzO;eku mp = 1.007276 u ; 1u = 2

MeV931

c

Ans : 37 Sol. Energy released in one fusion ,d lay;u esa mRiUu Å tkZ



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Q = [2m(2H) � m(1H) � m(3H)]c2

Number of fusion = �4

A

mass of waterN 1.5 10

moleculer weight

lay;uksa dh la[;k = �4

AN 1.5 10 ikuh dk nzO;eku

v.kqHkkj

40. A capillary tube of radius 0.50 mm is dipped vertically in a pot of water. The difference between the

pressure of the water in the tube 5.0 cm below the surface and the atmospheric pressure is 10x N/m2. Surface tension of water = 0.075 N/m contact angle between glass and water is zero degree. Find the value of x. (g = 9.8 m/s2) [SF-EX](103)

0.50 feeh f=kT;k dh ds'kuyh dks ty ds ik=k esa m/okZ/kj Mqck;k x;k gSA uyh esa p<s gq, ty dh lrg ls 5.0 lseh uhps nkc ,oa ok;qe.Myh; nkc dk vUrj 10x N/m2 gks rks x dk eku Kkr dhft;sA ty dk i"B rukao =

0.075 N/m rFkk dkWp o ikuh ds e/; Li'kZ dks.k 'kwU; fMxzh gSA (g = 9.8 m/s2) Ans. 19 Sol.

A

PA = P0 � 2S

pghR

= P0 �300 + 490 = P0 +190



Page # 1

Course : JR AIOT (JEE ADVANCE) - 2 Test Date : 03-05-2015

Test Type : JEE ADVANCED -2 (ELPD)

Page # 1

PAPER-2 MCQ (6) 41. A 5g sample containing Fe3O4 (FeO + Fe2O3) and an inert impurity is treated with excess of KI solution in

the presence of dilute H2SO4. The entire iron converted to Ferrous ion along with liberation of Iodine. The resulting solution is diluted to 100 ml. 20 ml of the diluted solution requires 10 ml of 0.5M Na2S2O3 solution to reduce the iodine present. Amongs the following select correct statements.

(A*) % of Fe2O3 in sample is 40% (B) % of FeO in sample is 28% (C*) % of inert impurity in sample is 42% (D) % of inert impurity in sample is 32% Fe3O4 (FeO + Fe2O3) rFkk ,d vfØ; v'kqf);qDr] ,d 5g izkn'kZ dk ruq H2SO4 dh mifLFkfr esa KI foy;u ds vkf/kD;

ds lkFk mipkfjr fd;k tkrk gSA lEiw.kZ vk;ju] Qsjl vk;u esa ifjofrZr gksrk gS rFkk lkFk gh vk;ksMhu eqDr gksrh gSA ifj.kkeh foy;u dks 100ml rd ruq fd;k x;kA 20ml ruq foy;u esa mifLFkr vk;ksMhu dks vipf;r djus ds fy,

0.5M Na2S2O3 dk 10ml vko';d gksrk gSA fuEu esa ls lgh dFkuksa dk p;u dhft,A (A*) izkn'kZ esa Fe2O3 dk 40% gSA (B) izkn'kZ esa FeO dk 28% gSA (C*) izkn'kZ esa vfØ; v'kqf) 42% gSA (D) izkn'kZ esa vfØ; v'kqf) 32% gSA Sol. 40 Let the moles of Fe3O4 is x. So mole of Fe2O3 (in Fe3O4) = x eq of Fe2O3 (in Fe3O4) = eq of K = eq of 2 = eq of Na2S2O3. or 2x = 0.025 or x = 0.0125 mol of Fe2O3 So mass of Fe2O3 = 0.0125 × 160 = 2g

% of Fe2O3 = 25

× 100 = 40%.

Sol. 40 ekuk fd Fe3O4 ds eksy x gSA vr% Fe2O3 ds eksy (Fe3O4 esa) = x

Fe2O3 ds rqY;kad (Fe3O4 esa) = K ds rqY;kad = 2 ds rqY;kad = Na2S2O3 ds rqY;kad vFkok 2x = 0.025

vFkok x = 0.0125 mol of Fe2O3

vr% Fe2O3 ds eksy = 0.0125 × 160 = 2g

Fe2O3 dk izfr'kr =25

× 100 = 40%.

42. -bond results due to overlap of : (CBO-VBT)_203 (A*) dxy and py along x�axis (B) dx2�y2 and py along x�axis (C*) dxy and px along y�axis (D) dx2�y2 and py along y�axis

-cU/k] fuEu ds vfrO;kiu dk ifj.kke gksrk gS % (A*) x-v{k dh vksj dxy rFkk py ds vfrO;kiu ds dkj.k (B) x-v{k dh vksj dx2�y2 rFkk py ds vfrO;kiu ds dkj.k (C*) y-v{k dh vksj dxy rFkk px ds vfrO;kiu ds dkj.k (D) y-v{k dh vksj dx2�y2 rFkk py ds vfrO;kiu ds dkj.k Sol. -bond are formed by parallel overlaping of same orbitals. So, correct options are : (A) dxy and py along x�axis (C) dxy and px along y�axis

gy leku d{kdksa ds lekukUrj ¼ikf'oZd½ vfrO;kiu ds dkj.k - ca/k dk fuekZ.k gksrk gSA vr% lgh fodYi gSa % (A) x-v{k dh vksj dxy rFkk py ds vfrO;kiu ds dkj.k (C) y-v{k dh vksj dxy rFkk px ds vfrO;kiu ds dkj.k 43. The vapour pressure of two miscible liquids A and B are 300 and 500 mm of Hg respectively. In a flask,

2 moles of A are mixed with 6 moles of B. Further to the mixture, 32 g of an ionic non-volatile solute MCl (partially ionised, mol. mass = 70 u) were also added. Thus, the final vapour pressure of solution was found to be 420 mm of Hg. Then, identify the correct statement(s) : (Assume the liquid mixture of A and B to behave ideally). (SCP-RLVP)_206

(A*) The numerical value of relative lowering in vapour pressure upon addition of solute MCl is 1/15.

Page # 2

(B*) The solute MCl is 25% ionised in the above question. (C) The solute MCl is 23.33% ionised in the above question. (D*) Upon addition of excess Pb(NO3)2, the number of moles of PbCl2 precipitated is 2/35.

nks feJ.kh; nzoksa A rFkk B ds ok"i nkc Øe'k% 300 rFkk 500 mm Hg gSaA ,d ¶ykLd esa] 2 eksy A dks B ds 6 eksy ds lkFk fefJr fd;k tkrk gSA blds ckn bl feJ.k esa] ,d vok"i'khy vk;fud foys; MCl ¼vkaf'kd :i ls vk;fur] vkf.od nzO;eku = 70 u½ dk 32 g Hkh feyk;k tkrk gSA bl çdkj foy;u dk vfUre ok"i nkc 420 mm Hg ik;k tkrk gSA rc lgh dFku@dFkuksa dh igpku dhft, % (A rFkk B ds nzo feJ.k dk O;ogkj vkn'kZ ekusaA)

(A*) MCl foys; feykus ij ok"i nkc esa vkisf{kd voueu dk vkafdd eku 1/15 gSA (B*) mijksDr iz'u esa foys; MCl, 25% vk;fur gksrk gSA (C) mijksDr iz'u esa foys; MCl 23.33% vk;fur gksrk gSA (D*) Pb(NO3)2 dk vkf/kD; feykus ij] vo{ksfir PbCl2 ds eksyksa dh la[;k 2/35 gSA

Sol. P0 = XAPAº + XBPBº =28

× 300 + 68

× 500 = 450 mm of Hg.

Now, ¼vc]½ RLVP = 0 S

0

P � P

P=

450 � 420

450 =

115

Also, ¼lkFk gh½ RLVP = in

in N

1

15 =

32i70

32i8

70

i = 1.25 = (1 + (2 � 1)). So, = 0.25 (or 25%)

�Cln produced ¼mRikfnr �Cl

n ) =25

100×

3270

=4

35

2PbCln precipitated ¼vo{ksfir

2PbCln ½ =12

×4

35=

235

44. Which of the following properties is/are common to both K2Cr2O7 & KMnO4 ? (DBC-IDM)_203 (A*) Both on heating form an oxosalt, a metal oxide & oxygen gas. (B*) Both have their colour because of charge transfer spectrum. (C) Both behave as primary titrants as well as self-indicators in redox titrations. (D*) Both form a slightly coloured precipitate with hydrogen sulphide in acidic medium. fuEu esa ls dkSulk@dkSuls xq.k/keZ K2Cr2O7 rFkk KMnO4 nksuks ds fy, leku gS@gSa \

(A*) nksuksa dks xeZ djus ij vkDlks yo.k] /kkrq vkWDlkbM o vkWDlhtu xSl curh gSA

(B*) nksuks dk jax vkos'k LFkkukUrj.k LisDVªe ds dkj.k gksrk gSA

(C) nksuksa jsMkWDl vuqekiuksa esa izkFkfed vuqekid o Lor% lwpd dh rjg O;ogkj djrs gSA

(D*) nksuks vEyh; ek/;e es gkbMªkstu lYQkbM ds lkFk gYdk jaxhu vo{ksi cukrk gSaA

Sol. (A) Both on heating form an oxosalt (K2CrO4 & K2MnO4), a metal oxide (Cr2O3 & MnO2) & oxygen gas.

(B) Both have their colour (orange and purple respectively) because of charge transfer spectrum (since no

unpaired electrons).

(C) Only K2Cr2O7 behaves as primary titrant, while KMnO4 behave as self-indicator in redox titrations.

K2Cr2O7 requires diphenyl amine indicator for end point detection while KMnO4 behaves as secondary

titrant.

(D) Both form a slightly coloured precipitate (S) with hydrogen sulphide in acidic medium.

Page # 3

gy- (A) nksuksa dks xeZ djus ij vkDlks yo.k (K2CrO4 & K2MnO4)] /kkrq vkWDlkbM (Cr2O3 & MnO2) o vkWDlhtu xSl

curh gSA

(B) nksuks dk jax ¼Øe'k% ukjaxh o cSaxuh½ vkos'k LFkkukUrj.k LisDVªe ¼pqafd dksbZ v;qfXer bysDVªkWu ugha gksrk gS½ ds dkj.k

gksrk gSA

(C) jsMkWDl vuqekiu esa dsoy K2Cr2O7 esa vfUre fcUnq fu/kkZj.k ds fy, MkbZQsfuy,ehu lqpd dh vko';drk gksrh gS

tcfd KMnO4 f}rh;d vuqekid dh rjg O;ogkj djrk gSA

(D) nksuks vEyh; ek/;e es gkbMªkstu lYQkbM ds lkFk gYdk jaxhu vo{ksi (S) cukrk gSaA

45. F2 + dil. NaOH A + B + C (PBC-XVII, SBC-CAM)_204

F2 + conc. NaOH A + D + C (A*) A is NaF (B) B is O2 (C*) C is H2O (D) D is OF2

F2 + ruq NaOH A + B + C (PBC-XVII, SBC-CAM)_204

F2 + lkanz NaOH A + D + C

(A*) A, NaF gS (B) B, O2 gS (C*) C, H2O gS (D) D, OF2 gS Sol. F2 + dil. NaOH NaF + OF2 + H2O

F2 + conc. NaOH NaF + O2 + H2O

gy- F2 + ruq NaOH NaF + OF2 + H2O

F2 + lkanz NaOH NaF + O2 + H2O

46. (CHK-PRFO)_205

All reactions are of Ist order.

At time t = t1 (t1 > 0), [B][C]

= ; At time t = t2 (where t2 t1), [C][D]

=

Which of the following is/are correct ?

lHkh vfHkfØ;k,¡ izFke dksfV dh gSaA

le; t = t1 ij (t1 > 0), [B][C]

= ; le; t = t2 ij (tgk¡ t2 t1), [C][D]

=

fuEu esa ls dkSulk¼ls½ lEcU/k lgh gS¼gSa½ \ (A) = 0.5 (B) = 4/3 (C*) = 0.375 (D*) = 16/15

Page # 4

Sol. 1

1

3k[B] 3[C] 8k 8

1

1

[C] 8k 8[D] 7.5k 7.5

MCQ (2) 47. Observed the following reaction sequence and identify the correct options ?

fuEu vfHkfØ;k vuqØe dk voyksdu dj lgh fodYi dk p;u dhft, \ (NAV Sir) (AC) (M) (MCQ)

+

CH2�C

O

O CH2�C

O

3AlCl (A) HCl

Hg�Zn (B) 5PCl (C) 3AlCl (D)

HClHg�Zn (E)

(F)

Pd�C/

(A*) F reacts with KMnO4/OH gives phthalic acid which upon heating gives phthalic anhydride.

(B*) E upon monochlorination (Cl2/h) gives total four products. (C*) F upon complete catalytic hydrogenation gives saturated compound which show geometrical

isomerism. (D*) Compound (A) gives the NaHCO3 test.

(A*) ;kSfxd F, KMnO4/OH ds lkFk fØ;k djds FkSfyd vEy cukrk gS ftls xeZ djus ij FkSfyd ,ugkbMªkbM curk gSA

(B*) ;kSfxd E ds ,dyDyksjksuhdj.k (Cl2/h) ij dqy pkj mRikn curs gSA

(C*) ;kSfxd F ds iw.kZ mRizsjdh; gkbMªkstuhdj.k ij larIr ;kSfxd curk gS tks T;kferh; leko;ork n'kkZrk gSA

(D*) ;kSfxd (A) NaHCO3 ijh{k.k nsrk gSA

Sol.

+

CH2�C

O

O CH2�C

O

3AlCl

C CH2

CH2 HO�C

O

O (A)

HCl

Hg�Zn

C

O (B)

H

5PCl

C

O (C)

Cl

AlCl3

O (D)

Zn�Hg + HCl

(E)

Pd�C/

Cl2/h

(d)

Cl H *

+

(d)

Cl H

*

(F)

KMnO4

Pd + H2 /

H

H + H

H

OH

COOH

COOH

�H2O

O

O

O

cis.

trans. (G)

Page # 5

48.

Product (X) Conc. H2SO4

80ºC Reaction-I

Conc. H2SO4

160ºC

Product (Y) Conc. H2SO4

160ºC Reaction-II

Which of the following option(s) is/are correct for above give reactions �

(SSS Sir) (AC-EAS) (MCQ) (204) (M)

(A*) In the reaction-I product is formed by most stable intermediate and product (X) is converted into (Y) by

conc. H2SO4 at 160ºC

(B) In the reaction-II product is -naphthalene sulphonic acid

(C*) Product 'X' is converted into product Y by conc.H2SO4/ and it is a thermodynamically controlled

process.

(D) In the reaction I & II electrophile attacks on naphthalene with same regioselectivity.

mRikn (X)

lkUnz H2SO4

80ºC vfHkfØ;k-I

lkUnz H2SO4

160ºC

mRikn (Y) lkUnz H2SO4

160ºC vfHkfØ;k-II

mijksDr nh xbZ vfHkfØ;kvksa ds fy, fuEu esa ls dkSuls fodYi lgh gS@gSa& (SSS Sir) (AC-EAS) (MCQ) (204) (M)

(A*) vfHkfØ;k-I esa mRikn lokZf/kd LFkk;h e/;orhZ ls curk gS rFkk mRikn (X) 160ºC ij lkUnz H2SO4 dh mifLFkfr esa

;kSfxd (Y) esa ifjofrZr gks tkrk gSA

(B) vfHkfØ;k-II esa mRikn -uS¶FkSyhu lY¶;wfjd vEy gksrk gSA

(C*) mRikn 'X' lkUnz H2SO4/ dh mifLFkfr esa mRikn Y esa ifjofrZr gks tkrk gS rFkk ;g vfHkfØ;k Å "ekxfrdh;

:i ls fu;af=kr izØe gSA

(D) vfHkfØ;k I o II esa bysDVªkWu Lusgh leku fjft;ksp;ukRedrk ds lkFk us¶Fksyhu ij vkØe.k djrk gSA

Sol.

SO3/80ºC

H SO3 �

�H

SO3H

SO3/160ºC

H

SO3 �

�H SO3H

Page # 6

Comp.(3 x 2Q.) (2) Paragraph for Question Nos. 49 to 50

iz'u 49 ls 50 ds fy, vuqPNsn Calomel electrode : It consists of mercury at the bottom over which a paste of mercury-mercurous

chloride is placed. A saturated solution of potassium chloride is then placed over the paste. A platinum wire sealed in a glass tube helps in making the electrical contact. The electrode is connected with the help of the side tube on the left through a salt bridge with the other electrode to make a complete cell.

Fig. Calomel electrode

The potential of the calomel electrode depends upon the concentration of the potassium chloride solution.

If potassium chloride solution is saturated, the electrode is known as saturated calomel electrode (SCE) and if the potassium chloride solution is 1 N, the electrode is known as normal calomel electrode (NCE) while for 0.1 N potassium chloride solution, the electrode is referred to as decinormal calomel electrode (DNCE). The electrode reaction when the electrode acts as cathode is :

12

Hg2Cl2 + e� Hg + Cl�

(Given : 2.303 RT / F = 0.06) Now answer the following questions. dSyksey bysDVªksM : blds vUrxZr edZjh rys ij gksrh gS ftlds Å ij edZjhµ ejD;wjl DyksjkbZM dk isLV j[kk tkrk gSA

bl isLV ds Å ij iksVsf'k;e DyksjkbM dk larIr foy;u Mkyk tkrk gSA fo|qr lEidZ dks cukus ds fy, dk¡p dh ufydk esa ,d IysfVue dk rkj izR;kjksfir fd;k tkrk gSA lEiw.kZ lSy cukus ds fy, bysDVªksM dks ck¡;h vksj fLFkr ik'oZ ufydk dh lgk;rk ls vU; bysDVªksM ds lkFk yo.k lsrq ls tksM+k tkrk gSA

fp=k % dSyksey bysDVªksM

dSyksey bysDVªksM dk foHko] iksVsf'k;e DyksjkbM foy;u dh lkanzrk ij fuHkZj djrk gSA ;fn iksVsf'k;e DyksjkbM dk

foy;u larIr gS] rks bysDVªksM dks larIr dSyksey bysDVªksM (SCE) dgk tkrk gS rFkk ;fn 1 N iksVsf'k;e DyksjkbM foy;u gS, rks bysDVªksM dks ukeZy dSyksey bysDVªksM (NCE) dgk tkrk gS] tcfd 0.1 N iksVsf'k;e DyksjkbM foy;u ds fy, bysDVªksM dks MslhukWeZy dSyksey bysDVªksM (DNCE) dgk tkrk gSA tc bysDVªksM ,d dSFkksM dh rjg dk;Z djrk gS rks bysDVªksM vfHkfØ;k fuEu gS %

12

Hg2Cl2 + e� Hg + Cl�

(fn;k x;k gS : 2.303 RT / F = 0.06)

Page # 7

vc fuEu iz'uksa ds mÙkj nhft,A 49. For the calomel half-cell Hg, Hg2Cl2 | Cl� (aq), values of electrode oxidation potential are plotted at different

log [Cl�]. Variation is represented by : ( 22

0

Hg /HgE

= 0.79 V ; Ksp(Hg2Cl2) = 10�18) (ECH-NE)_205

dSyksey v)Z lsy] Hg, Hg2Cl2 | Cl� ¼tyh;½ ds bysDVªksM vkWDlhdj.k foHko ds ekuksa dks log [Cl�] ds fofHkUu ekuksa ij

vkjsf[kr fd;k tkrk gSA rc bldk ifjorZu fuEu }kjk n'kkZ;k tkrk gS % ( 22

0

Hg /HgE

= 0.79 V ; Ksp(Hg2Cl2) = 10�18

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

Sol. EOP = 0SOPE � 0.06 log

1

[Cl ]

EOP = �(0.79 + 0.06

2log1010�18) + 0.06 log [Cl�]

�

OP E � 0.25 0.06 log [Cl ]y c m x

50. E.M.F. of the following cell is 0.67 V at 298 K. (ECH-ANE)_205

2 2 2Pt(s) | H (g, 1 atm) | H (aq, pH 6.5) || 1N KCl(aq) | Hg Cl (s) |Hg( ) | Pt(s) l

Calculate �2 2

º

Cl /Hg Cl /HgE (calomel electrode).

298 K ij fuEu lsy dk E.M.F. 0.67 V gSA

2 2 2Pt(s) | H (g, 1 atm) | H (aq, pH 6.5) || 1N KCl(aq) | Hg Cl (s) |Hg( ) | Pt(s) l

�2 2

º

Cl /Hg Cl /HgE ¼dSyksey bysDVªksM½ dh x.kuk dhft;sA

(A*) 0.28 V (B) 0.30 V (C) 0.26 V (D) 0.4 V

Sol. H2 + Hg2Cl2 2Hg + 2Cl� + 2H+

Ecell = �2 2Cl /Hg Cl /Hg

Eº � 2H /H

Eº � 0.06

2log [H+]2 (1)2

0.67 = �2 2Cl /Hg Cl /Hg

Eº + 0.06 pH

0.67 = �2 2Cl /Hg Cl /Hg

Eº + 0.06 × 6.5

0.28 = �2 2Cl /Hg Cl /Hg

Eº


iz'u 51 ls 52 ds fy, vuqPNsn

The coordination number of iron in both the complexes (A) and (B) is six. (QUA-IIG)_204 Now answer the following questions.

Page # 8

nksuksa ladqyksa (A) o (B) esa vk;ju dh leUo; la[;k N% gSA vc fuEu iz'uksa ds mÙkj nhft,A 51. Which of the following options is incorrect ? (COR-ACFT)_203 (A) A + Fe2+ = Prussion blue colour (B) A + Fe3+ = Brown colour (C*) B + Fe2+ = Light green colour (D) B + Fe3+ = Prussion blue colour fuEu esa ls dkSulk fodYi lgh ugha gS \ (A) A + Fe2+ = izqf'k;u uhyk jax (B) A + Fe3+ = Hkwjk jax (C*) B + Fe2+ = gYdk gjk jax (D) B + Fe3+ = izqf'k;u uhyk jax Sol. B + Fe2+ = white colour B + Fe2+ = lQsn jax 52. Which statement is incorrect ? (COR-ACFT)_203 (A) Complex A absorbs radiation of greater frequency. (B) Complex B has smaller spin magnetic moment. (C) Both complexes have same hybridisation of central atom/ion. (D*) Complex B is thermodynamically more stable. fuEu esa ls dkSulk dFku xyr gS \ (A) ladqy A mPp vkorhZ dh fofdj.ksa vo'kksf"kr djrk gSA (B) ladqy B de pØ.k pqEcdh; vk?kw.kZ j[krk gSA (C) nksuksa ladqyksa esa dsUnzh; ijek.kq@vk;u dk ladj.k leku gksrk gSA

(D*) ladqy B m"ekxfrdh; :i ls vf/kd LFkk;h gksrk gSA Sol. (A) Complex A - K3[Fe(CN)6] absorbs radiation of greater frequency because it contains Fe3+, so greater

O. (B) Complex B - K4[Fe(CN)6] has zero spin magnetic moment. (C) Both complexes have same hybridisation (d2sp3) of central atom/ion. (D) Complex B is thermodynamically less stable (O.N. stability). (A) ladqy A - K3[Fe(CN)6] mPp vkorZ dh fofdj.ksa vo'kksf"kr djrk gSA D;ksafd ;g Fe3+ vk;u j[krk gSA blfy, O

mPp gksrk gSA (B) ladqy B - K4[Fe(CN)6] dk 'kwU; pØ.k pqEcdh; vk?kzw.kZ gksrk gSA (C) nksuksa ladqy ds dsUnzh; ijek.kq@vk;u dk ladj.k (d2sp3) gksrk gSA (D) ladqy B m"ekxfrdh; :i ls de LFkk;h gksrk gSA (vkWDlhdj.k vad LFkkf;Ro).


iz'u 53 ls 54 ds fy, vuqPNsn Each of these reactions is a substitution of the leaving group (�OTs or �Cl) by solvent.

PhS

Cl 2

1

H O

r (r1 > r2)

Cl 2

2

H O

r

The mechanism by which they speed up the reactions is known as neighbouring group participation.

PhS

Cl Ph�S

H2O

Three membered ring intermediate

OH S

Ph

In this, ionization of the starting material is assisted by the lone pair of an electron-rich functional group.

mailto:ijek.kq@vk

mailto:ijek.kq@vk

Page # 9

In neighbouring group participation relation of configuration is observed. Observe the following reaction also

OAc

OTs

AcOH

OAc

OAc

Which follows through the mechanism.

O C

O

OTs

O C

O

CH3

AcOH

OAc

OAc

CH3

Neighbouring group participation is only possible when leaving group and neighbouring group are anti to

each other. (RSS Sir) (RM-MX) (207) (NGP) (Tough)

uhps nh xbZ izR;sd vfHkfØ;ksa esa mi;qDr foyk;d dh mifLFkfr esa fu"dklu lewg (�OTs ;k �Cl) dk izfrLFkkiu gksrk

gSA

PhS

Cl 2

1

H O

r (r1 > r2)

Cl 2

2

H O

r

fØ;kfof/k ftlds }kjk vfHkfØ;k dh nj c<+ tkrh gS mls iM+kSlh lewg dk ;ksxnku dgrs gSA

PhS

Cl Ph�S

H2O

rhu lnL;h oy; e/;orhZ

OH S

Ph

mijksDr vfHkfØ;k es izkjfEHkd inkFkZ dk vk;uu ,d bysDVªkWu /kuh fØ;kRed lewg ds ,dkdh ;qXe }kjk lEiUu gksrk gSA

iM+kSlh lewg ;ksxnku esa foU;kl dk lEcU/k izsf{kr gksrk gSA fuEu vfHkfØ;k Hkh izsf{kr gksrh gS %

OAc

OTs

AcOH

OAc

OAc

tks fuEu fØ;kfo/kh ds ek/;e ls gksrh gSA

O C

O

OTs

O C

O

CH3

AcOH

OAc

OAc

CH3

iM+kSlh lewg ;ksxnku dsoy rHkh lEHko gS tc fu"dklu lewg o iM+kSlh lewg ,d&nwljs ds foifjr gksrs gSA.

(RSS Sir) (RM) (NGP) (Tough) 53. Which of the following are true statement -

(A)

OAc

OTs

reacts with acetic acid faster than

OAc

OTs

Page # 10

(B) O

Cl reacts with H2O slower than Cl

(C*)

SMe OSO2Ph Me reacts with ROH raster than

Me OSO2Ph Me

(D)

I HO reacts faster than

I HO with H2O.

fuEu esa ls dkSulk dFku lR; gS &

(A) ,flfVd vEy ds lkFk

OAc

OTs

dh fØ;k

OAc

OTs

dh rqyuk es vf/kd rsth ls gksrh gSA

(B) H2O ds lkFk O

Cl dh fØ;k Cl dh rqyuk es /kheh gksrh gSA

(C*) ROH ds lkFk

SMe OSO2Ph Me dh fØ;k

Me OSO2Ph Me dh rqyuk esa vf/kd rst gksrh gSA

(D)

I HO ,

I HO fd rqyuk esa H2O ds lkFk rsth ls vfHkfØ;k djrk gSA

Sol.

SMe OSO2Ph Me show neighbouring group participation.

SMe OSO2Ph Me iM+kSlh lewg dk ;ksxnku iznf'kZr djrk gSA

54.

OAc

OTs

AcOH Product, ¼mRikn½

Product is ¼mRikn gS½ �

(A)

OAc

OAc

(B*)

OAc

OAc

(C)

OAc

OTs

(D)

OAc

OAc

Sol. In

OAc

OAc

neighbouring group participation is not possible.

OAc

OAc

esa iM+kSlh lewg dk ;ksxnku lEHko ugh gSA

Integer (Double digit) (4)

55. [Au(CN)x]� is a very stable complex (though EAN = 82) under certain conditions and Kf of [Au(CN)x]

� is 4

1028. z 10�y M concentration of cyanide ion is required to maintain the equilibrium at which 99 mole % of the gold is in the form of cyanide complex. Then determine (y + z). Atomic number of Au = 79. z is a natural number & 1 z 9. (CEQ-HOEL, COR-BICC)_205

fuf'pr fLFkfr;ksa ds vUrxZr [Au(CN)x]� ,d vR;f/kd LFkk;h ladqy gS (;|fi EAN = 82) rFkk [Au(CN)x]

� dk Kf 4

1028 gSA lk;ukbM vk;u dh vko';d lkUnzrk z 10�y M gS] tks fd lkE;koLFkk cuk;s j[krs gq, 99 eksy % lksus dks lk;ukbM ladqy ds :i esa cuk;s j[kus ds fy, vko';d gSA rc (y + z) Kkr dhft,A Au dk ijek.kq Øekad = 79. z

,d izkd r la[;k gS rFkk 1 z 9. Ans. 19

Page # 11

Sol. Au+ + 2CN� [Au(CN)2]�

99 mol% [Au(CN)2]�

It means 2

2

[Au(CN )]

[Au ] [Au(CN )]

=

99100

bldk vFkZ gS 2

2

[Au(CN )]

[Au ] [Au(CN )]

=

99100

100[Au(CN)2]� = 99[Au+] + 99[Au(CN)2]

�

[Au+] = 2[Au(CN )]

99

� (I)

K� = 2

2

[Au(CN )]

[Au ] [CN ]

Putting the [Au+] from (I) in K�

K� esa (I) ls [Au+] dk eku j[kus ij

4 × 1028 = 2

99

[CN ] [CN�] = 5 × 10�14 M

56. A sparingly soluble salt MX is dissolved in water to prepare 1 L saturated solution. Now 10�6 mole NaX

(assume 100% dissociation) is added into this. Conductivity of this solution is 29 10�6 S/m. If Ksp of MX is

a 10�b, then find value of (a + b). a is a natural number & 1 a 9. (ECH-EC)_206

Given : �

0 �3

x4 10 S m2 mol�

0 �3

Na5 10

S m2 mol�

0 �3

M6 10

S m2 mol� (Made by AKK SIR ON Jan.2014)

,d vYi foys;h yo.k MX dks ty esa ?kksydj 1 yhVj larIr foy;u cuk;k tkrk gSA vc blesa 10�6 eksy NaX

(100% fo?kVu ekusa) feyk;k tkrk gSA bl foy;u dh fof'k"V pkydrk 29 10�6 S/m gSA ;fn MX dk Ksp a 10�b

gS] rks (a + b) dk eku Kkr dhft,A a ,d izkd r la[;k gS rFkk 1 a 9.

fn;k x;k gS % �

0 �3

x4 10 S m2 mol�

0 �3

Na5 10

S m2 mol�

0 �3

M6 10

S m2 mol� (Made by AKK SIR ON Jan.2014)

Ans. 18

Sol. MX M + X+ �

x x + 10�6

[Na+] = 10�6 M

KSol = M

K + �XK +

NaK

Page # 12

29 10�6 = 103[6 10�3x + (4 10�3 (x + 10�6) + (5 10�3 10�6)]

x = 2 10�6

Ksp = 2 10�6 × 3 10�6 = 6 10�12

57. Two moles of an ideal gas is changed from their initial state (16 atm, 6L) to final state (4 atm, 15L) in such

a way that this change can be represented by a straight line in P�V curve. Let 'y' L atm be the maximum

average translational kinetic energy possessed by the gas sample during the above change. Report your

answer as y/2. (Take R =1

12L atm K�1 mol�1) (GST-KTG)_207

,d vkn'kZ xSl ds nks eksy dks izkjfEHkd voLFkk (16 atm, 6L) ls vfUre voLFkk (4 atm, 15L) rd bl izdkj ifjofrZr

fd;k tkrk gS] fd P�V oØ esa bl ifjorZu dks ,d lh/kh js[kk }kjk iznf'kZr fd;k tk ldrk gSA ekusa fd mijksDr

ifjorZu ds nkSjku xSl izkn'kZ }kjk izkIr vf/kdre vkSlr LFkkukUrj.k xfrt Å tkZ 'y' L atm gSA viuk mÙkj y/2 ds :i

esa nhft,A

(yhft, % R =1

12L atm K�1 mol�1)

Ans. 81

Sol. Equation of straight line (lh/kh js[kk dk lehdj.k esa) :

(y � y1) =2 1

2 1

y y

x x

(x � x1)

(P � 16) =4 1615 6

(V � 6)

3P + 4V = 72

Tmax =max(PV)

nR

For (PV)max ((PV)max ds fy,), 3P =722

and (rFkk) 4V =722

P = 12, V = 9

Tmax =12 9

2 (1/ 12)

= 648K.

KEmax = 32

nRTmax = 32

× 2 ×1

12× 648 = 162 L atm.

58. [H+] concentration in 0.01 M H2O2 solution (1aK = 3 × 10�12 and

2aK 0) is x M. Fill first two digits of 108 x

as answer. (IEQ-PPAB)_205

0.01 M H2O2 ( 1aK = 3 × 10�12 rFkk 2aK 0) foy;u esa [H+] lkanzrk x M gSA 108 x ds igys nks vadksa dks mÙkj ds :i

esa Hkfj,A

Ans. 20

Sol. [H+] = 12 2 14a 0 wK C K 3 10 10 10

= 2 × 10�7 M = x 108 x = 20

Page # 13

Integer (Double digit) (2)

59. How many cyclic structural isomer of C6H10 possible which after ozonolysis give diketones that show Intra molecular Aldol condensation. (Only between ketone).

C6H10 ds fdrus pfØ; lapukRed leko;oh lEHko gS tks vkstksuh vi?kVu ds i'pkr~ dsoy MkbZdhVksu nsrk gS tks

vUr%vkf.od ,YMksy la?kuu iznf'kZr djrk gSA ¼dsoy fdVksu ds e/;½ Ans. 03

Sol.

3

2

O

Zn�H O

O O

�OH Intra molecule aldol ¼vUr%vkf.od ,YMksy½

3

2

O

Zn�H O

O O

3

2

O

Zn�H O

O O

60. How many reactions will proceed through free radical addition mechanism.

fuEu esa ls fdruh vfHkfØ;k eqDr ewyd ;kSxkRed fØ;kfof/k ds ek/;e ls lEiUu gksrh gSA

(i)

NBS (PS Mam) (RM-MX) (M) (206) (Integer)

(ii)

2 2

HCl

R O

(iii) CH3�C=CH�CH3

CH3

HBr

h

(iv) CH3�CC�CH2�CH3 I

2 2

HR O

(v) CH3�CH=CH2 2 2

HBrR O

(vi)

CH=CH�CH3

NBS

Ans. 02

Sol. Only (iii) & (v)

Sol. dsoy (iii) o (v)

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