PAPER-1 · 01CT313082 KOTA / HS - 5/12 LEADER & ENTHUSIAST COURSE 23-04-2014 TM Path to success...

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SOLUTION

PART-1 : PHYSICS ANSWER KEYPAPER-1

PAPER CODETM

Path to success KOTA (RAJASTHAN)0 1 C T 3 1 3 0 8 2

PATTERN : JEE (Advanced)LEADER & ENTHUSIAST COURSE

Date : 23 - 04 - 2014TARGET : JEE 2014

SCORE-II : TEST # 03

SECTION–I

1. Ans. (D)

Sol. vcm = v

2 2sys Ac Bc

1 1k (2m)v mv 0

2 2

P = 2mvAB = 0 [vAB = 0]

L = (2m)vAB = 0

2. Ans. (B)

Sol. Conserving moment of system along rod.

mv cos45 = 2mv1

v1= v/22

By work energy theorem:

2

21 1 vkx 2m

2 2 2 2

x = 1 m

v2 k

3. Ans. (B)

4. Ans. (B)

5. Ans. (C)

6. Ans. (A)

7. Ans. (A, C, D)

8. Ans. (B, D)

Sol. Momentum along Y-axis will remain conserved

So, mu cos = 2m Vy

Q. 1 2 3 4 5 6 7 8 9 10

A. D B B B C A A,C,D B,D A,C,D A,B,C

Q. 11 12 13 14 15 16

A. C C B B A B

A B C D A B C D

Q,R,S R,S R,S P P,Q,S Q,S P,Q P,Q,R

Q. 1 2 3 4

A. 2 3 3 5

SECTION-I

SECTION-IV

SECTION-II Q.1 Q.2

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PAPER – 1

y

4uV

10 ……(i)

Conserving angular momentum about Y-axis during collision

22mv

mu sin m3

3u sin 9u

4 20

……(ii)

Velocity of particle just after impact is

p P

9u 4uˆ ˆ ˆ ˆV V j ( i) i j20 10

for completing full rotation about Y-axis

gW k

2 21 4

mg(2 ) mg 0 m2 3

min

9 g

2

min

20 9 gu

9 2

2 1010 g 2g

9 3

9. Ans. (A,C,D)

Sol. 020

2 22

Vi

R X

50 2 A

10i 50 A

V0

20 Li X

20i R

Xc=2

~ V=100 sin t

R=1 XL=1

i2

i1

i

i10

V0

i10Xc=V0

Phase difference between v and i is = 0°

1 2i i i

0i 50

0 0av

i vP cos 2500 W

2

i

45°

i10

i20 0

0

Vz 2 .

i

10. Ans. (A, B, C)11. Ans. (C)12. Ans. (C)13. Ans. (B)

14. Ans. (B)Sol. PB – PA = (2a) 2g – (2ag) = 2ag15. Ans. (A)16. Ans. (B)

Sol. S(Px – PA) =5a /3

2 2 2

a

8Sxdx S a

9 . . . (1)

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similarly S(PB – Px) = 2 256

S a9

. . . (2)

solving (1) and (2)

Fx = 9

Sag4 And =

9g

32a

SECTION–II

1. Ans. (A) (Q,R,S) ; (B) (R,S) ; (C) (R,S) ; (D) (P)

2. Ans. (A) (P, Q, S) ; (B) (Q, S) ; (C) (P, Q) ; (D) (P, Q, R)

SECTION-IV

1. Ans. 2

Sol. Stress, = Y(T) = (1.21 10-5 )(2 1011 ) (20) = 4.84 107 N/m2

Tension, T = A = 48.4 N

Since the point plucked would be an antinode, therefore, the string vibrate in its second harmonics.

f = 21

2l

T

0.75m

NA A NN

0.25m

or f = 22 Hz.

K = 2

2. Ans. 3Sol. Frame will complete vertical circular motion if it crosses position where line joining O and centre of

mass C makes an angle

= 1 oatan 37

g with vertical

By work energy theorem

2 21 3 32m 2mg 1 cos 2ma sin

2 2 2

2ma

2mg

C

C

3/2

O

3. Ans. 3

Sol. Using Moseley’s law

= A(Z - 1) , we have

A

c

= A(ZA - 1)

and xx

ca Z 1

Dividing yields A x

x A

Z 1

Z 1

Zx = 29.

4. Ans. 5

Sol. 2

22 2

B 0

1 1a dt 2H g dt v dt

2 2

AB = 2g = 20 m/s2

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PAPER – 1

SOLUTION

PART-2 : CHEMISTRY ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10

A. C C A D B C A,D A,B A,B,D B,C,D

Q. 11 12 13 14 15 16

A. B C B C D C

A B C D A B C D

Q,S Q R,S P,Q S Q P R

Q. 1 2 3 4

A. 1 2 2 4

SECTION-I

SECTION-IV

SECTION-II Q.1 Q.2

SECTION-I

1. Ans. (C)

2. Ans. (C)

3. Ans. (A)

Sol. General mechanism will be

In water 3º halide dissociates & shows SN

1

mechanism whereas CH3–X shows (S

N2)

mechanism.

(CH3)

3C – X (CH ) C3 3

+ + X-

X- + CH

3Y X – CH

3 + Y

-

(CH3)

3C + H

2O (CH

3)

3C–OH

Since I- is a good leaving group and a good

nucleophile hence reaction ‘a’ will be the fastest.

4. Ans. (D)

5. Ans. (B)

Sol.

has no plane of symmetry or centre of symme-try. It is optically active.

has 3 chiral centres but it also has a plane ofsymmetry it is a meso compound and opticallyinactive.

has 3 chiral centres but no plane or centre ofsymmetry it will be optically active.

has no plane or centre of symmetry. It is

optically active.

6. Ans. (C)

7. Ans. (A, D)

8. Ans. (A, B)

Sol.N N

H H

H

Possible Two Isomers of N H22

H

–I –I

N N

N N

F F

F F

Possible Two Isomers of N F22

+I +I

N N

N NHO HO

HO

Possible two isomersof H N O22 2

OH

+I +I

N N

Due to having different substituents N–

N bond strength is not equal for all.

9. Ans. (A,B,D)

Sol.

(A) Cl

X /OH2 CHX + CH CH COO2 3 2

X /OH2

OH

(B) X/OH2 COOCH3

O

CHX +3

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PAPER – 1

(C) C

CHCH – CH3 2 3

OHCH – CH3 2

* Not a methyl carbinol

* Can't be oxidised by OX

* So not show Iodoform test.

(D) CHX + OOC – COO2

X /OH2

O||

COOHCCH3

10. Ans. (B,C,D)

11. Ans. (B)

12. Ans. (C)

13. Ans. (B)

14. Ans. (C)

15. Ans. (D)

16. Ans. (C)

SECTION-II

1. (A) Q, S ; (B) Q ; (C) R, S ; (D)

P, Q

2. (A) S ; (B) Q ; (C) P ; (D) R

SECTION-IV

1. Ans. (19) [OMR Ans. 1]

Kt = ln 0[A]

[A]t

K (10) = ln100

90

K(10) = ln 10

9

..... (1)

Any fraction of completion of first order reactionis independent of its initial concentraton.

K(20) = ln0

t

[A]

[A] ..... (2)

Putting the value of (2) in (1) we get

2ln0

t

10 [A]ln

9 [A]

ln 0

t

100 [A]ln

81 [A]

[A]t = 81

the reaction between is 19% completed

2. Ans. 2

Sol. [Be(gly)2], Hybridisation of Be2+ in [Be(gly)2]is sp3

as. Be++

3sp.Hyb

p2s2

CH2

H2 H2

O = C

N N

BeII

O O– –

CH2

C = O

(optically active)

Asymmetric configuration due to absence ofelements of symmetry, hence optically active.Therefore, number of space/stereoisomers = 2

3. Ans. 2Sol. PCl5 PCl3 + Cl2

In 1st vessel at equilibrium 1 atm 1atm1atmhence Kp = 1 atm

In 2nd vessel initially 3 atm 3atm3atmat equilibrium 3+x 3–x 3–x

Kp = )x3(

)x3( 2

= 1

hence x = 1

atm2P2Cl

4. Ans. 4

Monochloro alkenes EtOH)ii(

Mg)i( CH2 = CH–CH3

so possible alkenes

Cl|CHCHCH 22

Cl|

CHCCH 32

C = C C = C

Cl ClH

HH HCH3

CH3

Ans. 4

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PAPER – 1

SECTION-I

1. Ans. (A)

Sol. (x – y – 1)2 + (x + z – 3)2 + (z – 5)2 = 0

2. Ans. (B)

Sol.( ) ( )

( ) ( )

2

2

( )

( )

1

3. Ans. (B)

Sol. 3y (ydx xdy) xd(xy) 0

2 2

y y d(xy)d 0

x x x y

4. Ans. (C)

Sol.

1

tan x1

2 x

x 0lim (1 x )

x

2 tan x x

x 0lim (1 x )

3

2

1 x

2 tan x xx

x 0lim (1 x )

3

3x 0

xlim

x(x .... x)

3e

= e3

5. Ans. (D)

Sol.x / 2 1

I

dy 1 ye

dx 2 2

1II

dy dy 22y 4

dx dx y

m1m2 = –1

6. Ans. (B)

Sol. f(x) = x + 1

7. Ans. (A, B, C, D)

Sol. || = a + b

– () = ab

Hence, a is negative b is positive

Now || < b – 1

a < –1

8. Ans. (A, C)

Sol.

1 1

1

sin x cos x 0 x 1

f(x)2sin x 1 x 0

2

9. Ans. (A, D)

Sol. If the sign of f'(x) changes from –ve to +ve at

x = 2

, then x =

2

is a point of minima.

10. Ans. (A, B)

Sol. As circle can intersect a praobola in four points,

so quadilateral may be cyclic and the

quadilateral may be isoscleles trapezium.

11. Ans. (B)

12. Ans. (D)

13. Ans. (C)

Sol. (G3A)2 = (G3M) (G3C)

a2 + b2 = 2c2

sin CAM sin CBM

In CAG1

A

G3

BG1

C

G2M

SOLUTION

PART-3 : MATHEMATICS ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10

A. A B B C D B A,B,C,D A,C A,D A,B

Q. 11 12 13 14 15 16

A. B D C C D A

A B C D A B C D

R P S R P S R P,Q

Q. 1 2 3 4

A. 4 4 1 2

SECTION-I

SECTION-IV

SECTION-II Q.1 Q.2

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PAPER – 1

1 1CG AG

sin CAM sin C

1

1

CG sin Csin CAM

AG

In CBG2

2 2CG BG

sin CBM sin C

2

2

CG sin Csin CBM

BG

AG1 = 2 2 21 3

2(b c ) a b2 2

BG2 = 2 2 21 3

2(a c ) b a2 2

sin CAM sin CBM = a b

sin C3 b 3 a

= 2 2sin C[a b ]

3 ab =

22c sin C

3 ab

cos C = 2 2 2 2a b c c

2ab 2ab

2cos C = 2c

ab

2

sin CAM sin CBM sin 2C3

C = 45°22 2c 1

3 3 2 ab

2c2

ab

14. Ans. (C)

15. Ans. (D)

16. Ans. (A)

Sol. 2 7 11 1

SECTION-II

1. Ans. (A)(R); (B)(P); (C)(S); (D)(R)

Sol. (C)

The points of discontinuity are , 4

where

cot = 3, cot = 2, cot 4

= 1

2. Ans. (A) (P); (B) (S); (C) (R);

(D) (P,Q)

Sol. (A) B P(B A)

PA P(A)

= P(A B) P(A)

P(A)

=

1/ 4 1

3 / 4 3

(C) Ladies can be seated in 6 ways then men

can be placed in t wo ways only.

(D) Total ways of forming 5 letter word from

the word CALCULUS is 1110.

SECTION-IV

1. Ans. 4Sol. The equation of given line is symmetrical

form is

3 9x y

z 02 21 1 2

2. Ans. 4

Sol. Distance of p(x, y) from origin 2 2x y ,

Also perpendicular distance of P is 2

x2 + y2 4

3. Ans. 1

Sol. Both equations hold for x = 1

4. Ans. 2

Sol. 2

f ' (x) 1 sin 2x

f(x) x cos x

PAPER-2

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Path to success KOTA (RAJASTHAN)0 1 C T 3 1 3 0 8 3

PATTERN : JEE (Advanced)LEADER & ENTHUSIAST COURSE

Date : 23 - 04 - 2014TARGET : JEE 2014

SCORE-II : TEST # 03

Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, INDRA VIHAR, KOTA-324005

PHONE : +91 - 744 - 2436001, Fax : +91-744-2435003, E-mail: [email protected] Website: www.allen.ac.inKOTA / HS - 8/12

SOLUTION

PART-1 : MATHEMATICS ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10

A. B D D B B D C A D A,D

Q. 11 12 13 14 15 16 17 18 19

A. B A,D A,C A D A A D A

Q. 1 2 3

A. 2 3 9SECTION-IV

SECTION-I

SECTION-I

1. Ans. (B)

For perimeter to be minimum AM + MB should

be least

2. Ans. (D)

2 4 6

3 5

1 3 5

x x x dx1 1 1

1x x x

3. Ans. (D)

n 2 n 1 n 1 1 n

1 1 1 2.....

a, a a a a a (a a )

1 2 n

1 1 1.....

a a a

4. Ans. (B)

Let l(x) = loge(xx + 1) for 0 < x < 1

x = 1

e is a point of minimum

5. Ans. (B)

f(x) = ln|x| + 1

6. Ans. (D)

f(x) = 21(x x )

2

7. Ans. (C)

OB = tan

COA =

AB = sec

Equation of lien OC is y = x tan

8. Ans. (A)

Use the property a a

0 0

f(x)dx f(a x)dx

9. Ans. (D)

A B C

B C A

( 1) C ( 1)A

= –1

10. Ans. (A, D)

tan 3 =3

2

3 tan tank (say)

1 3 tan

(3 tan – tan3)2 = k2 (1 – 3 tan2 )2

tan6 + 3(3 + 2k2) tan2

= k2 + 3(3k2 + 2) tan4 )

On comparing

k = 3

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PAPER – 2

11. Ans. (B)

We have

A2 = cos 2 sin 2

sin 2 cos 2

A3 = cos 3 sin 3

sin 3 cos 3

A4 = cos 4 sin4

sin4 cos 4

12. Ans. (A,D)

If 5

x , 32

1cos (cosx) x 2

13. Ans. (A, C)

Area of triangle PQR with eccentric angles ,,

is given as 2absin sin sin2 2 2

Paragraph for Question 14 to 16

14. Ans. (A)

15. Ans. (D)

16. Ans. (A)

1 1(A ') (A )'

If A is a non singular Matrix

As A and B are invertible matrices, An and Bn

are invertible for for every positive integer n

Paragraph for Question 17 to 19

17. Ans. (A)

18. Ans. (D)

19. Ans. (A)

1, 4, 25

2, 5, 23

3, 6, 24

Required probability

=8 8 9 8 2 8 8 9 9

625

3530.5648

625

SECTION - IV

1. Ans. 2

Let a < b

f(x) = |x – a| + |x –b| =

a b 2x x a

b a a x b

2x (a b) x b

Given that f(0) = f(1) = f(–1)

{–1, 0, 1} [a, b]

(a –b)min = 2

2. Ans. 3

a2 + b2 + c2 + 2abc = 1

(b + ca)2 = (1–a2) (1– c2)

c2 1

similarly b2 1

3. Ans. 9

Let y = m1x

1

21

| 2 m | 5

21 m

2 2

1 2

82m m

9

01CT313083KOTA / HS - 10/12

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PAPER – 2

SECTION-I

1. Ans. (B)

2. Ans. (B)

Sol. Ndt = m2 1

L Lm

2 2

2 2

2 1

L mL mLN dt

2 12 12

Ndt = 2 1

mL

6

2 = 1/2

3. Ans. (A)

4. Ans. (A)

Sol.

N

mg

a

5. Ans. A

Sol. Before charging the bubble pressure inside the bubble P1 = P

0 + 4T/r

After charging electrostatic pressure will be 2

02

Which will pull the bubble outward.

When radius becomes 2r then 2

10

0

P 4TP

8 2 2r

, where =

2

q

4 2r6. Ans. (D) ; 7. Ans. (B) ; 8. Ans. (C) ; 9. Ans. (A)

10. Ans. (A, B, C)

Sol. For any point on the surface of paraboloid, (x2 + z2) = 4ay

(A) I =

3

1i

21

21 )zx(m (distance of mi from y-axis is 2

i2i zx ) = 4ma (y1 + y2 + y3).

(B) mg (x1 + x2 + x3) = mg(y1 + y2 + y3)

(C) mg x1 = 2

1mv1

2 v1 = 1gy2

(D) Distance y-mass mp from y-axis, ri = i

2i

2i ay4zx

KE = 2

1m2 (r1

2 + r22 + r3

2) = 2

1m2 4a (y1 + y2 + y3)

SOLUTION

Q. 1 2 3 4 5 6 7 8 9 10

A. B B A A A D B C A A,B,C

Q. 11 12 13 14 15 16 17 18 19

A. A,C A,B A,C C B A D B A

Q. 1 2 3

A. 2 2 2SECTION-IV

SECTION-I

PART-2 : PHYSICS ANSWER KEY

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PAPER – 2

11. Ans. (A, C)

Sol. Q value for fusion = 75 30 45(B / A) 75 B/ A 30 B / A 45 = 99 MeV.

12. Ans. (A, B) ; 13. Ans. (A,C)

14. Ans. (C)

Sol. Immediately after collision centre of disc and particle will rotate about common centre of mass in circular

path of radius R/3 and R/6 respectively with common angular velocity ’.

By conservation of angular momentum about vertical axis passing through combined centre of mass.

22 2

o

mR 11 2mR mR '

2 18 36

o

3'

4

R/3

m 2m C

R/6

15. Ans. (B)

Sol. Friction on particle = f = 2R 752m ' N

6 4

16. Ans. (A)

Sol.

2

2 2 RI m u m ' 20.2 Ns

6

17. Ans. (D)

Sol. 0mg 2K X X i B ma ……(i)

qB V

C ……(ii)

v

Mean position 2K(x+x0)

i

mg

i B x

from (i) and (ii)

2 2

2Kxa

(B C m)

2 2B C mT 2

2K

18. Ans. (B)

19. Ans. (A)

SECTION - IV

1. Ans. 2

Sol. 2 3 2

(8 2) (6 3) (10 6) 7ˆ(E n)A

472 3 6

2. Ans. 2

Sol. = 1

0

2

hd

h

3. Ans. 2

Sol.1

1 1 5 21

f 10 2 R

2

1 1 21

f 15 R

µ = 2.

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PAPER – 2

SECTION-I

1. Ans. (D)

2. Ans. (A)

3. Ans. (A)

Sol. I & IV are enantiomers they have same physi-cal & chemical solubilities. They have identicalsolubility in methanol.

4. Ans. (B)

Sol.2

Water

Cl BrBr2 2 CCl4 layer becomes red or

reddish-Brown.]

5. Ans. (A)

6. Ans. (C)

Sol. Compound given in option C do not have anyhemi Acetal group so can't be oxidised bytollen's regent and will not give tollen's test.

7. Ans. (D)

Sol. As PtF6 is a powerful oxidizing agent hence.

Na + PtF6 Na+[PtF6]–;

NO + PtF6 NO+[PtF6]–

Xe + PtF6 Xe+ [PtF6]–

8. Ans. (C)

9. Ans. (C)

10. Ans. (A,D)

11. Ans. (C)

12. Ans. (B, D)

Sol. The reaction is SN2 reaction in which bromide

rather than Fluoride is replaced.

13. Ans. (A, B)

Sol. Product is

14. Ans. (B)

15. Ans. (A)

16. Ans. (C)

17. Ans. (C)

SOLUTION

PART-3 : CHEMISTRY ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10

A. D A A B A C D C C A,D

Q. 11 12 13 14 15 16 17 18 19

A. C B,D A,B B A C C D B

Q. 1 2 3

A. 8 7 9SECTION-IV

SECTION-I

18. Ans. (D)

19. Ans. (B)

Sol.

SECTION-IV

1. Ans. (728)

[OMR Ans. 7 + 2 + 8 = 17 = 1 + 7 = 8]

Sol. 8e– + 2 Ph – NO2 Ph – N = N – Ph

OH¯ 4

1O2 +

2

1H2O + e–

C60 + 60 O2 60 CO2

96 160 4 182

60 12 8 728 gm

2. Ans. 7

3. Ans. 9

Sol. Total moles in layer M = x

Total moles in layer N = y

In layer M Moles of A = 0.2 x

Moles of B = 0.8 x

and

In layer N Moles of A = 0.6 y

Moles of B = 0.4 y

As per question

0.2 x + 0.6 y = 1

0.8 x + 0.4 y = 3

x = 3.5

y = 0.5Ratio of masses of layer M to N

= 404.05.0206.05.0

408.05.3202.05.3

= 9 Ans.

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