ALLEN M 0 1 2 1 4 0 8 CLASSRO ONTAC OGRAMME...P S ALLEN CARESTI TUTE KOT( RAJASTHA) M CLASSRO ONTAC OGRAMME (CADEM SESSI 2014-2015) PAPE 0 1 2 1 4 0 8ANSW : PER-1. 1 2 3 4 5 6 7 8

Path to Success

ALLENCAREER INSTITUTEKOTA (RAJASTHAN)

T M

CLASSROOM CONTACT PROGRAMME(ACADEMIC SESSION 2014-2015)

PAPER CODE 0 1 C T 2 1 4 0 8 8

ANSWER KEY : PAPER-1Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20A. 4 2 3 3 3 2 1 2 2 1 3 2 3 3 2 1 2 1 3 2Q. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40A. 3 2 4 2 3 2 3 3 2 4 1 4 3 4 1 2 3 2 1 2Q. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60A. 1 4 2 3 1 2 1 2 4 4 3 1 4 1 4 1 1 3 4 1Q. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80A. 2 2 2 3 3 2 2 3 2 2 2 4 4 1 3 4 3 2 2 1Q. 81 82 83 84 85 86 87 88 89 90A. 1 3 4 4 4 1 3 2 2 4

PART-1 : PHYSICSANSWER KEY : PAPER-2

PART-2 : CHEMISTRY

PART-3 : MATHEMATICS

Path to Success



PAPER CODE 0 1 C T 2 1 4 0 8 9T M

Q. 1 2 3 4 5 6 7 8 9 10A. A,B,C A,C,D B,C A,B,C A,C,D B,C,D A,B,C,D A,B B CQ. 11 12A. A DQ. 1 2 3 4A. 006 020 090 008Q. 1 2 3 4A. 1 8 4 7

SECTION-I

SECTION-III

SECTION-IV

Q. 1 2 3 4 5 6 7 8 9 10A. A, B, C A,B,C,D A,B,C,D B, D A, B, D A, B, D B, C, D A, B, C B BQ. 11 12A. A DQ. 1 2 3 4A. 006 004 003 008Q. 1 2 3 4A. 2 4 2 1

SECTION-I

SECTION-III

SECTION-IV

Q. 1 2 3 4 5 6 7 8 9 10A. B,C A,B,C B,C A,C A,B,D A,B,C B,C B,C,D B AQ. 11 12A. B BQ. 1 2 3 4A. 150 010 006 013Q. 1 2 3 4A. 1 1 9 7

SECTION-I

SECTION-III

SECTION-IV

DATE : 22 - 02 - 2015ALLEN JEE (Main) TEST

TARGET : JEE (Main + Advanced) 2015 LEADER COURSE (PHASE-TLV, TLX & TVX) : SCORE-I

DATE : 22 - 02 - 2015ALLEN JEE (Main) TEST


Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 +91-744-2436001 [email protected] HS-1/7

1. Ans. (4)

Sol. input reject

input input

Q Qwork done0.6Q Q

-= = = r

i

Q1Q

-

i

200.6 1Q

= -

Qi = 50

W = Qi – Q

r = 30 J

2. Ans. (2)Sol. Truth table of AND gate

A B C0 0 00 1 01 0 01 1 1

3. Ans. (3)Sol. s = 4t = 2t + 5t2

5t2 = 2tt = 0.4v = 2 + gt = 6 m/sJ = mv – mv

1

–J = mv – mv2

v = 5J = 4 (v – 4)

4. Ans. (3)Sol. F = mg sin q

q

» mg tan q

dy 2xmg mgdx 40

= = - ´

a = x m2

-

a = x

2-

-

w = 12


PAPER CODE 0 1 C T 2 1 4 0 8 8

Path to Success


T M

DATE : 22 - 02 - 2015

Ans. (3)

Sol. v = kw

. Also Tv =m

20 Mgv10 1 0.01

= =´

M = 0.0046. Ans. (2)Sol. For constant pressure process.

Work done (W) = nRDT.

\25TnR

æ öD = ç ÷è ø

Now, for above process f = 6,

So CP

f1 R 4R2

æ ö+ =ç ÷è ø

\ Heat absorbed = nCPDT

= n × 4R × 25 100 JnR

=

7. Ans. (1)Sol. 6000 × 1 × (40 – 20)

= m × 540 + m × 1 × (100 – 40)12 × 104 = m × 600Þ m = 2 × 102 = 200 gm

8. Ans. (2)

Sol. At equilibrium dU 0dr

=

13 712A 6B 0r r

-+ =

1/ 62ArB

æ ö= ç ÷è ø

9. Ans. (2)

Sol.

x2 2

02 0

x0

x 4 x dxE 4 x

r ´ p´ p =

Î

ò

x=

504 x

5 0

prÎ

SOLUTIONPAPER-1

ALLEN JEE (Main) TEST


Kota/01CT214088HS-2/7

Target : JEE (Main + Advanced) 2015/22-02-2015/Paper-1

R2 2

0 500

y 2 20

x 4 x dxRE

4 y 5 y0

r ´ pr ´

= =Î ´ p Î

ò

3 50 0

2

x R5 5 y0 0

r r=Î Î

x3y2 = R5

10. Ans. (1)

Sol.R

RR

R R

RR R =3R0

Þ 7R3 = Req

11. Ans. (3)

Sol. In time T m2 qB

p=

V^ gets reversed

ˆ ˆ ˆv 2i 3j 4k= - -r

( )q E v B 0+ ´ =r rr

E (v B)= - ´r rr = ˆ ˆ ˆ ˆ(2i 3j 4k) 2i- - - ´ = ˆ ˆ6k 8j- +

12. Ans. (2)

Sol. 600

C

Vi 300 2 100 10X

-= = ´ ´ = 30 2 mA

irms

= 0i2

irms

= 30 mA

13. Ans. (3)

Sol. O PI

14. Ans. (3)

Sol. h = 21V(10) g(10)2

+

V/2V/2

h

h = 2V 1(20) g(20)2 2

- +

adding we get h = 1250 m15. Ans. (2)

Sol. x3 = t3 + 1 Þ 2 2dx3x 3tdt

=

x2v = t2 Þ 2dx2x v x a 2tdt

+ =

42

4t2x x a 2tx

+ = Þ 4

23

2tx a 2tx

= -

3 3

52t[x t ]a

x-= Þ 5

2tax

=

16. Ans. (1)

Sol. 3 2 11 1E V E E Ek k

´ Þ D ´ Þ < <

and DV3 < DV

2 < DV

1

17. Ans. (2)

Sol.0.2T20

D =

25T 1.2520

= =

T 0.2 20 100 0.8%T 1.25

D= ´ =

18. Ans. (1)

Sol.34

19 31

h h 6.63 10mv 2Km 2 54 1.6 10 9.1 10

-

- -

´l = = =

´ ´ ´ ´ ´19. Ans. (3)

Sol. I µ A2, I µ 2

1

r, 1 2

2 1

A rA r

=

20. Ans. (2)

Sol. -

æ ö= -ç ÷´ ¥è ø10 2 2

1 1 1R

2700 10 2

-

æ ö= - -ç ÷´ è ø

2

10 2 2

1 1 1R(Z 1)

1 10 1 2

Leader Course/Phase-TLV, TLX & TVX/Score-I/22-02-2015/Paper-1

HS-3/7Kota/01CT214088

21. Ans. (3)Sol. J = m(V – u) ...(1)

J4

l =

2m

12

lw ...(2)

VP = 2u = V +

4

wl...(3)

m(V – u)2

m4 12

= wl l

...(4)

from (3) and (4)

2u – 4

wl= u +

3

wl

l

4J P

uV =2u(after impulse)

P

V =uC

vc= +u

Vp = +2u (after impulse)

u = 7

12wl

4u 11uV u u

3 7 7w

= + = + =l

22. Ans. (2)Sol. mv

A cos q = mv'

Ay

mvA sin q = mv'

Bx

23. Ans. (4)Sol. From Newton's law of cooling

ms 1 2 1 20t 2

q - q q + qæ ö é ùµ - qç ÷ ê úè ø ë û

in the first case ms50 40

5

-æ öç ÷è ø µ 0

50 40

2

+é ù- qê úë û

......(i)

in second case 0

45 41.5 45 41.5

5 2

- +æ ö é ùµ - qç ÷ ê úè ø ë û

......(ii)

From (i) and (ii)

q0 = 33.3ºC

24. Ans. (2)Sol. Pascal's law

1 2

1 2

F FA A

= Þ F1A

2 = F

2A

1 ;

as A1 < A

2

So F1 < F

2

25. Ans. (3)

Sol. Flux going in pyramid = 0

Q2e

Which is divided equally among all 4 faces

\ Flux through one face = 0

Q

8e26. Ans. (2)Sol. Behaviour of lens is opp. If outside medium is

denser than lens medium27. Ans. (3)Sol. V = fl

f = 1

2 LCp28. Ans. (3)

Sol.1 1 1

2 2 2

A NA N

l=

l

( )( )

n22

20

n22

40

n2 1N e 2

2 2 21n2

N e24

-´

-´

´= =

l

l

l

l

29. Ans. (2)

Sol. t = 2h

g ....(1)

x = vt ....(2)

30. Ans. (4)Sol. L

a = L

b

mvar = 3mv

br

Þ a

b

vv = 3 : 1

31. Ans. (1)

Sol. Energy of one photon = 310012400

= 4ev = 4 x 96

= 384 kJ mol–1

\ % of energy converted to K.E. = 384288384 -

= 38496

x 100 = 25%

32. Ans. (4)

Sol. pOH = pKb + log ]Base[

]Salt[= pK

b + log ]Base[

]Cation[

[ +4NH ] = 2 × mole of (NH

4)

2SO

4

\ pOH = 5 + log 2 = 5.3

or pH = 8.7

Kota/01CT214088HS-4/7


33. Ans. (3)

PCl5(g) �� PCl

3(g) + Cl

2(g)

Kp = 3 2

5

PCl Cl

PCl

P P (0.4 2)(0.4 2)P 2 0.2

´ ´=

´

= 0.4 × 4 = 1.634. Ans. (4)

Sol. H2C

2O

4(s) +

21

O2(g) ® H

2O(l) + 2CO

2(g);

Dng = 3/2

DUc = –

175.8312.0 ´

× 90= – 245.7 kJ/mol

DH = DU + DngRT = – 245.7 +

23

× 1000

300314.8 ´

= –241.96 kJ/mol

35. Ans. (1)

Sol. Anode ïþ

ïýü

ïî

ïíì

++®++®-+

-+

e4H4O OH2e2H2OSHSOH2

22

82242

Cathode {2H2O ® H

2 + 2OH– – 2e–} × 3.

Net : 2H2SO

4 + 8H

2O ® H

2S

2O

8 + O

2 + 3H

2 +

6H+ + 6OH–

Hence ratio of 2On and 2Hn is 1 : 3.

36. Ans. (2)

(M ¾®¾ Mn+(0.02M)

+ ne–) × 2

(2H+ + 2e– ¾®¾ H2) × n

2M + 2nH+ ¾®¾ 2Mn+ + nH2

0.81 = (0.76 + 0) – 0.062n log

( )n2

2

)1(

02.0

(0.81 – 0.76) = – 0.062n log (4 × 10–4)

n = – 0.06

2 0.05´ × log (4 × 10–4)

(– 4 + 0.6) = 2.37. Ans.(3)

200.7 1 19.8100 2106 18x 10 1000

´´ = ´

+106 + 18x = 141.41 = 18x = 35.41x = 2

38. Ans. (2)Sol. Given DT

b = 1.080C , i = 2 at

boiling pt. of solution. and DTf = 1.800C , and

f

b

kk

= 0.3

sof

b

TT

DD

= mkimki

ff

bb so i

f = 1

i.e., AB behaves as non–electrolyte at the f.pof the solution.

39. Ans. (1)40. Ans. (2)Sol. Adsorption

DH = –veDS = –veDG = –veDSsurr = 0

41. Ans. (1)42. Ans. (4)43. Ans. (2)44. Ans. (3)45. Ans. (1)46. Ans. (2)47. Ans. (1)48. Ans. (2)49. Ans. (4)50. Ans. (4)51. Ans. (3)

More the a–H more will be reactivity forAromatic electrophilic substitution

52. Ans. (1)

53. Ans. (4)

Cl2

hv

Cl2

54. Ans. (1)Rate of SN1 a stability of carbocation

55. Ans. (4)


HS-5/7Kota/01CT214088

56. Ans. (1)

OH

CH3

BrBrBr ,H O2 2

OH OH

CH3 CH3

Br2

CS2

Br

57. Ans. (1)CH3

CHMe2

3 Ha

EÅ

1 aH 58. Ans. (3)

CHCl3 n

¾¾®¾hO2 COCl

2(A) EtOH¾¾¾®

EtO

COEt

O

2MeMgBr¾¾¾¾®CH3

CCH3

O¾¾¾¾¾ ®¾ .)aq(CaOCl2

CHCl3 + (CH

3COO)

2Ca (D)

CH –C–CH3 3

O59. Ans. (4)60. Ans. (1)

Thymine base of DNA is replaced by uracil inRNA

61. Ans. (2)

· x

yx = 2

(0,0)

c

D < 0

f(x) = y = ax2 + 2bx + c

f(2) = 4a + 4b + c < 0

so c < 0

62. Ans. (2)

( )2 2 2 2 20 1 2 3 100S 1.C 2C 3C 4C ...... 101 C= - + - + +

2 2 2 2 20 1 2 3 100S 101C 100C 99C 98C ...... 1.C= - + - + +

2 2 2 20 1 2 1002S 102 C C C .......... Cé ù= - + +ë û

100502S 102. C=

10050S 51. C=

63. Ans. (2)

|z| £ 4 and Arg z =3

p

So it represent radius of the circle.

p/34

Im(z)

Re(z)

64. Ans. (3)c

1 ® c

1 – c

2

a b cb c ac a b

D = = –[a3 + b3 + c3–3abc]

Or D = 1

2- (a+b+c)[(a–b)2 + (b–c)2 + (c–a)2]

So D < 0Also D = 0 Þ a = b = c (\ a + b + c > 0)Since a + b + c ¹ 0

65. Ans. (3)4x + y – 2z = 0x – 2y + z = 0x + y – z = 0Here D = D

1 = D

2 = D

3 = 0

so non-trivial solution.66. Ans. (2)

Sum of last eight coefficient are

S = 8 9 10 15C C C C15 15 15 ...... 15+ + + + ....(i)

S = 7 6 5 0C C C C15 15 15 ...... 15+ + + + ....(ii)

(We know that 8 7C C15 15= )

equation (i) + (ii)

2S =0 1 2 15C C C C15 15 15 ...... 15+ + + +

Þ 2S = 215 Þ S = 214

67. Ans. (2)MYNK4C

2 × 2! × 3C

2 × 1 = 36 (x – x – x)

68. Ans. (3)

P = 1 1 1 1 1 1

1 1 1 1 1 1

C C C C C C

C C C C C C

13 13 13 13 26 26 6!

52 52 52 52 52 52 2!2!2!´ ´ ´ ´ ´ ´

= 10

190

2æ ö´ ç ÷è ø

Kota/01CT214088HS-6/7

Target : JEE (Main + Advanced) 2015/22-02-2015/Paper-169. Ans. (2)

Arrange 3H and 3V in a row6C

3 × 3C

3 × 1 = 20 ways

70. Ans. (2)

2

4xy

(x y)+ £ 1 Þ (x – y)2 < 0

Þ x = y but x ¹ 0 (cosec2q ¹ 0)71. Ans. (2)

s2 = 2 2

1 1 2 2

1 2

n n

n n

s + s+ +

1 22

1 2

n n

(n n )+ 2

1 2(x x )-

s2 = 6 24 3 36

6 3

´ + ´+

+ 2

6 3

(6 3)

´+ (11 –14)2

s2 = 144 108

9

+ +

18

81 × 9 = 28 + 2 = 30

72. Ans. (4)73. Ans. (4)

If lines ˆ ˆ ˆr (2 )i (1 2 ) j k= + l + - l +r

& ˆ ˆ ˆr i (1 ) j ( 3 2 )k= + + m + - + mr

intersects each other, then2 + l = 1, 1 – 2l = 1 + m & 1 = –3 + 2mÞ l =–1, m = 2

74. Ans. (1)Required plane is

ˆ ˆ ˆ ˆ ˆ ˆ[r (2i j 3k)].(2i j 3k) 0- + - + - =75. Ans. (3)

1[a bc] 3

6=

rr r (Given)

[a bc] 18=r r

Vol. of parallelopiped= [ a b+rr b c+

r r c a+r

]

2[a b c ]=rr r Þ 2 × 18 = 36

76. Ans. (4)(3i + 4j + 12k). (ai + bj + ck) < |3i + 4j + 12k|

|ai + bj + ck| 2 2 213 a b c= + +77. Ans. (3)

b (a.b)a (a.a)b (a.b)a a (b a)- = - = ´ ´r r r r rr r r r r r r r

here a b a (b a)´ ^ ´ ´r rr r r

ÐABC = 90°

3 a btan 3

a b a

´q = =

´

rr

rr r

a × ( b × a )CB

A

3( a × b )

q

q = p/3Ð ABC = p/2 : ÐBAC = p/6Ð ACB = p/3

78. Ans. (2)The equation of the chord of contact of tangentsdrawn from the origin to the circle

x2 + y2 + 2gx + 2fy + c = 0 is

gx + fy + c = 0 ...(i)

The required circle passes through theintersection of the given circle and line (i).Therefore, its equation is

(x2 + y2 + 2gx + 2fy + c) + l (gx + fy + c) =0...(ii)

this passes through (0, 0)

\ c + lc = 0 Þ l = – 1

putting l = –1 in (ii), the eq. of the req. circleis x2 + y2 + gx + fy = 0.

79. Ans. (2)The equations of the lines are

2x – y + 4 = 0 and – x + 2y + 1 = 0

We have, 2 × – 1 + (–1) × 2 < 0 i.e. a1a2+b1b2<0

Therefore, the equation of the bisector of acuteangles is

2 2

2x y 4 x 2y 1

1 4 ( 1) 2

- + - + +=

+ - +

Þ 2x – y + 4 = – x + 2y + 1

Þ 3x – 3y + 3 = 0

Þ x – y + 1 = 0

80. Ans. (1)81. Ans. (1)

Eq. of focal chord is y = m(x – 4).......(1)

(1) is tangent to (x – 6)2 + y2 = 2

\ 2

2m

1 m+ = 2

Þ 4m2 = 2 + 2m2

Þ 2m2 = 2

Þ m = ±1.

82. Ans. (3)ƒ(x) = 5x – 3 cosx – 4sinx

ƒ'(x) = 5 + 3sinx – 4 cosx

–5 < 3sinx – 4 cosx < 5

0 < 5 + 3sinx – 4cosx < 10

function is stricitly increasing Þ Range = (–¥,¥)

Þ function is one-one onto.


HS-7/7Kota/01CT214088

83. Ans. (4)

(–1, )p

y= /2p

(–1,0) (1,0)

abvious from figure

sec–1(1) < sec–1(2) < sec–1(–2) < sec–1(–1)

84. Ans. (4)

85. Ans. (4)

86. Ans. (1)

Let AB be the lamp - post & CD be the man ata particular time t.

2m

x y EA

B

6mD

C

Let AC = x & CE = y (Shadow of the man)

dx

dt = 5 km/hr

dy?

dt=

DABE ~ CDE

\AB AE

CD CE=

Þ6 x y

2 y

+= Þ 2y = x

Þdy dx

2 5dt dt

= =

Þdy 5

2.5 km / hr.dt 2

= =

Hence ; the length of the shadow increases atthe rate 2.5 km/hr.

87. Ans. (3)

As the point (2 + 13 cosq, 3 + 13sinq) alwayslies on the direction circle of the given ellipse

( ) ( )2 2x 2 y 3

125 144

- -+ = , so angle between

tangents 2

p=

88. Ans. (2)

{ } { }1/3

0

1 2 4 101x x x x 1 x ... x dx

3 3 3 3

æ öì ü ì ü ì ü ì ü+ + + + + + + + + +í ý í ý í ý í ýç ÷î þ î þ î þ î þè ø

ò

{ }1/ 3

0

1 234 x x x dx

3 3

æ öì ü ì ü= + + + +í ý í ýç ÷î þ î þè ø

ò

( )1/3 1 /3

0 0

1 234 x x x dx 34 3x 1 dx

3 3æ ö= + + + + = +ç ÷è øò ò

1/ 32

0

3x 1 134 x 34 17

2 6 3

é ù é ù= + = + =ê ú ê úë ûë û

89. Ans. (2)

90. Ans. (4)

2n

r 1

r r 1sin cos .

n n n=

æ öæ öç ÷ç ÷è øè ø

å

12

0

sin x cos xdxò

Let cosx = t Þ –sinx dx = dt

( )cos1

12 3

cos11

1t dt t

3- =ò ( )31

1 cos 13

= -

Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 +91-744-2436001 [email protected]

PART-1 : PHYSICS SOLUTION

Path to Success

ALLENCAREER INSTITUTE

T M

CLASSROOM CONTACT PROGRAMME

HS-1/6

PAPER CODE 0 1 C T 2 1 4 0 8 9

KOTA (RAJASTHAN)

DATE : 22 - 02 - 2015

(ACADEMIC SESSION 2014-2015)

PAPER-2

SECTION-I1. Ans. (A,B,C)

Sol. Acceleration

a ; x 0m2bx ; x 0m

ì <ïïí-ï >ïî

2. Ans. (A,C,D)

Sol. xdVEdx

-=

Work done per unit charge = DV

3. Ans. (B,C)

Sol.

P

1 2

QB1 B1 B1

B1

B2

B2

4. Ans. (A,B,C)

Sol. (A)0

3

2E 1fh n

æ ö= ç ÷è ø

;

Frequency of revolution.

Frequency of photon emitted

( )0

2 2

E 1 1vh nn p

é ù= -ê ú

-ê úë û; where p = 1, 2, 3, ......

n >> p

(D) Electrons de broglie wavelength is morethan the size of the nucleus.

5. Ans. (A,C,D)

Sol. minhceV

l =

6. Ans. (B,C,D)

Sol. 2l

b =mq

7. Ans. (A,B,C,D)Sol. Angular impulse-angular momentum theorem

t = Dòrrf

idt L

Area under the curve gives linear impulse.8. Ans. (A,B)

Sol.

N1

N2

f

NMg

N

fmg

9. Ans. (B)

Sol.dA constantdt

=

1 2

1 2

A At t

=

a

b

FS D

E

ab Area of ESFA 440 12

p+ D

=

A Area of ESFA 440 12

+ D=

Area of DESF = 12 A 12 10 AA A40 4 40 20

-æ ö- = =ç ÷è ø

FS

CBA Area of SFCB Area of SCF40 1

- D=

A40 = Area of SFCB – Area of DESF

ALLEN JEE (Main) TEST


HS-2/6 Kota/01CT214089


Area of SFCB = A40 + Area of DESF

Total area of shaded region = Area of ESF +Area of SFCB

= A40 + 2 [Area of ESF]

= A8

10. Ans. (C)

Sol.A L L 2T 2m mA T

= Þ =

11. Ans. (A)

Sol.( )

= aird M Vmg

dt

mg = rpr2v2 ....(i)

P = ( )d K.E.

dt

P = rp 2 31r V

2....(ii)

12. Ans. (D)

Sol.Fa gm

= -2 2r v ' g

mrp

= -

v' = 21/3 v = (41/3 – 1) g

SECTION-III

1. Ans. 006

Sol. Power = 2V ;R

R Ar

=l

2. Ans. 020

Sol. AB = BC & tAB

= tBC

Þ Velocity of m at A = velocity of m at B = 0

Þ m = m0

2M m ga g 5m / sM m 2

-æ ö= = =ç ÷+è ø

20

HV 2 a 5 12

= ´ ´ = ´

Velocity of m just before collision is 0V

Velocity of m0 just after collision is V

0

Distance travelled by both masses m0; along the

length of string is same.

2 20 0 0

1 1gt V t gt2 2

= -

20 0 0gt V t=

00

Vtg

=

22 00 2

V 5 1 1tg 100 20 2 10

= = = =´

20

1tX

=

X = 203. Ans. 090

Sol. 0ˆ ˆv 5i 5 3 j= +

r

omˆ ˆv 10 i 5 3 j= +

r

imˆ ˆv 10 i 5 3 j= - +

r

iˆ ˆv 15 i 5 3 j= - +

r

0 iv v 0× =r r

4. Ans. 008

Sol. (3m)g(2l) = w21I

2....(i)

J(3l) = Iw ....(ii)SECTION-IV

1. Ans. 1Sol. m

1C(T – T

1) = m

1C(T

1 – T);

v1 = 1 2

1 20 0

m mv , v= =r r

( ) ( )1 1 11 1

1 0

m m mv ' 1 T 1 T T= = + gD = + g -é ùë ûr r r

( ) ( )2 12 2

0

m mv ' 1 T 1 T T= + gD = + g -é ùë ûr r

( ) ( )2 12 2 2 1 1 1

0

m mv ' v T T ;v ' v T T- = g - - g -r r

2 2 1 1v ' v v ' v- = -

2 1 2 1v v v ' v '+ = +

HS-3/6


Kota/01CT214089

2. Ans. 8

Sol. Time period Ia

3. Ans. 4

Sol. f × R + f × R – 100 = 0 ....(i)

f = µN ....(ii)

N = Force applied by brakes

4. Ans. 7

Sol. =2mv '

mga

....(i)

a

v'

u

2a

( )= +2 21 1mu mg 3a mv '

2 2....(ii)

PART–2 : CHEMISTRY SOLUTIONSECTION-I

1. Ans.(A,B,C)Sol. Facts to remember.

2. Ans.(A,B,C,D)3. Ans.(A,B,C,D)Sol.

900 mm Hg 860 mm Hg

X XA BX' X'

A

B

=0.3=0.7

X'' X''

A

B

=0.6=0.4

Solution Residue Condensate

+

Let the mole of A & B be 'a' & 'b' respectivelyin the initial mixture.Initial moles of A in the solution = Final molesof A in the residue and condensate

a = 0.3 × 3)ba( +

+ 0.6 × 3)ba(2 +

a = 0.1 a + 0.1 b + 0.4 a + 0.4 ba = 0.5 a = 0.5 b 0.5 a = 0.5 b a = b

4. Ans. (B, D)5. Ans. (A, B, D)6. Ans. (A, B, D)7. Ans. (B,C,D)8. Ans. (A,B,C)9. Ans. (B)Sol. No. of octahedral voids in hcp = 6; so formula

A6B

4 = A

3B

2

10. Ans. (B)Sol. no. of tetrahedral voids = 8

no of octahedral voids = 4so formula = AB

2O

4

11. Ans. (A)

12. Ans. (D)

SECTION-III

1. Ans. 006

Sol. Dm(g) l mD(g)

1 —

After reaction 0.4 0.6 m

1m6.04.0

nn

ectedexp

actual += Þ 0.4 + 0.6 m = 3.0

2.1

Þ 0.4 + 0.6 m = 4Þ 0.6 m = 3.6 Þ m = 6

2. Ans. 004

3. Ans. 003

4. Ans. 008

SECTION-IV

1. Ans. 2

G = al

= 10010

= 0.1 ; G = 0.0001 S ;

V = 100 × 10 = 1000 cm3 = 1 litrek = G G = 0.1 × 0.0001 = 10–5

^m= M100´k

= 5.01000)0001.01.0( ´´

=0.02Scm2 mol–1

2. Ans. 4

3. Ans. 2

4. Ans. 1

HS-4/6 Kota/01CT214089


PART-3 : MATHEMATICS SOLUTIONSECTION-I

1. Ans. (B,C)ƒ(x) is discontinuous at,

1 11 12015 2015x , tan (2) , tan (3) .......

4- -p

=

1 11 12015 2015tan (2014) , tan (2015)- -

hence ƒ(x) is discontinuous at1 12015 2015x tan 2014, tan 2015-=

1 12015 2015tan 4 / 4 tan 2

0 0 / 4

ƒ(x)dx 0 1dx- -p

p

= +ò ò ò1 1 20152015

1 12015 2015

tan 3 tan 4

tan 2 tan 3

2dx 3dx- -

- -

+ +ò ò2. Ans. (A,B,C)

3ƒ2(x)ƒ'(x) =1

ƒ(x)

Þ 33ƒ (x)ƒ '(x)dx dx=ò ò43 ƒ (x) x c

4= +

since ƒ(1) = 1

Þ1c4

= -

\1/ 4

4 1ƒ(x) x3 4

é ùæ ö= -ç ÷ê úè øë û3. Ans. (B,C)

Curve through the intersection of S1 and S2 is

given by S1 + lS2 = 0

Þ x2(tan2q + lcot2q) + y2(cot2q + ltan2q)

+ xy(2hsinq – 2n'cosq) + x(22 + l11)

+ y(4 + l1) + 15 + 7l = 0

it represent a circle, if

\ tan2q + lcot2q = cot2q + ltan2q

Þ l = 1 or 4p

q =

4. Ans. (A,C)2

2

2ƒ( )1

aa =

- a

\n 2

n 2n times

2ƒ(ƒ.......ƒ( )....)1 (2 1)

aa =

- - a

2 22

2 2

3b 2a14b 3a

-- a =

- Þ

2 22

2 2

3b 2a14b 3a

-a = -

-

Þ2 2

22 2

b a4b 3a

-a =

-

Þ

2

2

2

2

b 1a

4b 3a

-

-, Let

2b ta

æ ö =ç ÷è ø

2 t 14t 3

-a =

-, here t > 1

\ range of a is 10,2

æ öç ÷è ø

5. Ans. (A,B,D)

Let ƒ(x) = (x – log23)(x – log34)

+ (x – log34)(x – log42)

+ (x – log42)(x – log23)

ƒ(log23) = (log23 – log34)(log23 – log42) > 0

ƒ(log34) = (log34 – log42)(log34 – log23) < 0

ƒ(log42) = (log42 – log23)(log42 – log34) > 0

6. Ans. (A,B,C)

Arrangement of n2 different things into n equal

groups is equal to 2

n 1

n !(n!) +

Arrangement of 4n2 different things into

2n equal groups is equal to 2

2n 1

4n !(2n!) +

7. Ans. (B,C)

2 1 4 1 43

1 4 1 4

| z | | z z | | z z || z | 1| z z 1| | z z 1|

- -= = £

- -

|z1 – z4| < 1 4| z z 1|-

Þ |z1 – z4|2 < 2

1 4| z z 1|-

(z1 – z4) 1 4 1 4 1 4(z z ) (z z 1)(z z 1)- £ - -

|z1|2 + |z4|

2 – |z1|2|z4|

2 – 1 < 0

(|z1| – 1)(|z4|2 – 1) > 0

\ |z4| < 1

HS-5/6


Kota/01CT214089

8. Ans. (B,C,D)ƒ(x) = Asinx + Bcosx,A = (cos2a1 + cos2a2 + .... + cos2an)B = (sina1cosa1

+ sina2cosa2 +....+ sinancosan)ƒ(x1) = Asinx1 + Bcosx1 = 0

Þ 1

1

A cos xB sin x

= -

ƒ(x2) = Asinx2 + Bcosx2 = 0

Þ1

1

A cos xB sin x

-=

Þ tanx1 = tanx2

x1 = np + x2 Þ |x1 – x2| = np,

n = 0, 1, 2 .....

Paragraph for Question 9 & 10Let c (a b) µ(a b)= l + + ´

r rr r r

take dot product with ar

a.c (a.a a.b)= l +rr r r r r

3 35 5 5 5 5 52 2

æ ö´ ´ = l ´ + ´ +ç ÷ç ÷

è ø

3 312 2

æ ö= l +ç ÷ç ÷

è ø

3 2 3 32 3

l = Þ l = -+

take dot product with 1cr

2| c | (a.c b.c) µ[(a b).c]= l + + ´r rr r r r r r

3 325 25 25 µ[abc]2 2

æ ö= l + +ç ÷ç ÷

è ø

rr r

take dot product with (a b)´rr

2 2 2(a b).c µ | a | | b | sin´ = qr rr r r

1µ25 254

= ´ ´

\ 2 25 2525 (2 3 3)(25 3) µ4´

= - +

2 2µ (25)75 3 1754

- =

2 253 3 5 µ4

- =

2µ 3 3 55

= -

\ volume of parallelopiped

( ) 125[abc] 3 3 52

æ ö= -ç ÷è ø

rr r

Volume of tetrahedron

( )1 125[abc] (3 3 5)6 12

= -rr r

If height of parallelopiped = harea × height = volume

| a b | h V´ ´ =rr

( )1 1255 5 h 3 3 52 2

æ ö´ ´ = -ç ÷è ø

h 5 3 3 5= -height of tetrahedron is same as parallelopiped.

9. Ans. (B)10. Ans. (A)11. Ans. (B)

1DEF a cos A.bcos Bsin( 2C)2

D = p -

12cos A cos Bcos C absin C2

æ ö= ç ÷è ø

1 2 31 B C AI I I 4R cos , 4R cos ,sin 902 2 2 2

æ ö æ öD = ° -ç ÷ ç ÷è ø è ø

12. Ans. (B)

DDFRR2

=

DEFDEF

DEP

1(2cos Acos Bcos C) absin C2r a cos A bcos B ccos CS

2

D= =

+ +

2R cosA cos BcosC=

DEF

R 1r 4cos A cos Bcos C

æ ö =ç ÷è ø

SECTION – III1. Ans. 150

Three b's and four a's can arrange abababa

frouth 'b' can take 5 places,

babababaabbababaababbabaabababbaabababab

ìïïïíïïïî

take 8 place from 12 places and arrange lettersin same order, cc dd can take remaining 4 place

\12

84!C 5

2! 2!´ ´

HS-6/6 Kota/01CT214089


2. Ans. 010

ƒ(e) =2

ee 1-

ƒ(ƒ(ƒ........ ƒ(e))) = e, n is even

= 2

ee 1-

, n is odd

\6

22

e dee 1-

òLet e2 – 1 = t2

2ede = 2tdt

ede = tdt35

3

dt 35 3 k= - =ò\ 76k2 – k4 = 1024 = 2p

p = 10

3. Ans. 006

30°11

1

11

1

C

A

B1D

M

in DOBD : 1cos30

OB° =

2OB3

=

OM = OB + BM

2 13

= +

2 33

+

a = 2, b = 34. Ans. 013

|z1|2 3=

3

3z|z|

p3

Oz1

SECTION – IV1. Ans. 1

ƒ(1) < ƒ(2) < ƒ(3) < ƒ(4) < ƒ(5) < ƒ(6) < ƒ(7) <ƒ(8)< ƒ(9) < ƒ(10) < ƒ(11) < ƒ(12) < ƒ(13)< ƒ(15)

equality needed = 5total number of ways of selecting 5 equality from14 inequality = 14C5

Now, ƒ(5) = 3 and ƒ(10) = 64 5 5

2 2 114

5

C C CC

´ ´

150 p1001 q

=

2. Ans. 11 2 1sin(cos x) 1 x cos(sin x)- -= - =

\ 1 1 1 1tan (cos(sin x)) cot (sin(cos x))2

- - - - p+ =

3. Ans. 9cos(9x) + cos(7x) = 0

2cos8x.cos2x = 0

8x (2n 1)2p

= + = or x(2k 1)2p

+ =

(2n 1)x16+ p

=(2k 1)x

2+ p

=

3 5 7 9 11 13 15x , , , , , , , ,16 16 16 16 16 16 16 16 2p p p p p p p p p

=

No solution = 94. Ans. 7

PQ PRPD2+

=uuur uuur

uuur

Q R

P

Dˆ ˆ î j 6kPD2

- - l +=

uuur

21| PD | 9 244 4

l= + + =

uuur

21 9 244

+ l + =

59l =

\ˆ ˆ î 59j 6kPD

2- - +

=uuur

\ QD QP PD= +uuur uuur uuur

ˆ î 59 6kˆ ˆ ˆ4i 5k i2 2 2

æ ö= - + - - +ç ÷ç ÷

è øˆ7i 59 ˆ ˆj 2k

2 2= - -

49 59QD 4 31, QR 1244+

= + = =uuur uuur

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