TEST TYPE : MAJOR PATTERN : JEE (Adaced) · part-1 : physics answ : per-1 part-2 : chemistry part-3...

PART-1 : PHYSICSANSWER KEY : PAPER-1

PART-2 : CHEMISTRY

PART-3 : MATHEMATICS

Path to Success

ALLENCAREER INSTITUTEKOTA (RAJASTHAN)

CLASSROOM CONTACT PROGRAMME(ACADEMIC SESSION 2014-2015)

PAPER CODE 0 0 C T 2 1 4 0 0 5

PATTERN : JEE (Advanced)TEST TYPE : MAJOR

ENTHUSIAST & LEADER COURSE

Date : 10 - 05 - 2015TARGET : JEE (Advanced) 2015

ALL INDIA OPEN TEST #02

T M

Q. 1 2 3 4 5 6 7 8 9 10A. C B B B C B C D D CQ. 11 12A. C A

A B C D A B C DP S,T R,T Q P,R Q Q S,T

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

SECTION-I

SECTION-IV

SECTION-II Q.1 Q.2

Q. 1 2 3 4 5 6 7 8 9 10A. B D A A D D C C B DQ. 11 12A. D A

A B C D A B C DQ,S R,T S P Q,R,S,T P,Q,R,S,T P Q,S

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

Q.1 Q.2

SECTION-I

SECTION-IV

SECTION-II

Q. 1 2 3 4 5 6 7 8 9 10A. B D D C A D D B A DQ. 11 12A. B C

A B C D A B C DQ P S R P,Q,S P,Q,S P,Q,S R,T

Q. 1 2 3 4 5 6 7 8A. 2 7 1 7 6 1 2 2

Q.1 Q.2

SECTION-I

SECTION-IV

SECTION-II

PART-1 : PHYSICSANSWER KEY : PAPER-2

PART-2 : CHEMISTRY

PART-3 : MATHEMATICS

Path to Success



PAPER CODE 0 0 C T 2 1 4 0 0 6





T M

Q. 1 2 3 4 5 6 7 8 9 10A. A,B,C B,D A,C,D B,D B,C,D A,B,C,D B,C B,C,D D AQ. 11 12 13 14 15 16A. C A A B D CQ. 1 2 3 4A. 7 5 1 9

SECTION-IV

SECTION-I

Q. 1 2 3 4 5 6 7 8 9 10A. A,D B,C B,C,D A,B,D A,B,D A,C A,B,C,D B,C C CQ. 11 12 13 14 15 16A. D B C A D AQ. 1 2 3 4A. 6 2 1 6

SECTION-I

SECTION-IV

Q. 1 2 3 4 5 6 7 8 9 10A. A,C,D A,B,C A,C B A,C, A,B A,B B,C D AQ. 11 12 13 14 15 16A. B D A C B AQ. 1 2 3 4A. 6 2 6 1

SECTION-I

SECTION-IV

Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 +91-744-2436001 [email protected] HS-1/14

PART-1 : PHYSICS SOLUTIONPAPER-1

Path to Success


T M


PAPER CODE 0 0 C T 2 1 4 0 0 5





SECTION-I1. Ans. (B)Sol. As msinq = constant so 1 × sin 60°

= ( )max2 3 y sin90- °

60°

y

x

max

3 3y

2Þ =

2. Ans. (D)Sol.

6C 3C

3C 6C

V

S

net charge zero

–q +q

O

Initial condition

6C 3C

3C 6C

V

–(3CV/2) +3CV

Þ

6C 3C

3C 6C

V

Final conditionNet charge in initial condition on the boundedsystem is zero.

Final charge on the same system

is +3CV –3CV 3CV

2 2=

Therefore 3CV

2charge will flow through the

switch.

3. Ans. (D)

Sol.y

y60cmd

x

d

Deviation produced by prism

d = (m–1)A = (1.5–1)0

4 2 1rad

90æ ö æ ö= =ç ÷ ç ÷è ø è øp p

y = fd = (60) 190

æ öç ÷è ø

=23

cm

4. Ans. (C)

Sol. x

y

BB

BABA

5. Ans. (A)Sol. T

1 = 60g

\\\\\\\\\\\\\\\\\

T1

T

T – T1 – 50g

= 50 × 2Þ T = 110 g + 100

= 1200 N6. Ans. (D)

Sol.1 1F q(v B) aq j q(ai B)= ´ Þ - = ´r r rr $ $

$0B B kÞ =

r

Now $1 2v v (ai) (ai b j) abk´ = ´ + =r r $ $ $

Therefore 1 2F q (v v ) B 0é ù= ´ ´ =ë ûr rr r

HS-2/14

ALL INDIA OPEN TEST/JEE (Advanced)/10-05-2015/PAPER-1

Kota/00CT214005

7. Ans. (D)Sol. Let frequencies of A and B are n1 and n2 then

n1 = 1

v v v4 4 15 60

= =´l

and 21

v v vn

2 2 30.5 61= = =

´l

But n1 – n2 = 6 so v v

60 61- = 6

Þ v = 21960 cm/s

1

21960n

60Þ = = 366 hz, n2

21960

66= = 360 Hz

8. Ans. (B)

Sol. Least count = pitch 0.5

No.of divisions 50=

= 0.01 mm ;Diameter of wire = 6 × 0.5 + 46 × 0.01= 3.46 mm

9. Ans. (A)

Sol.

Rsin /np

p/n

I = 2

2 2 2m.(2R sin )

nn m.R cos MR12 n

pé ùê úp

+ +ê úë û

Þ I =

2

2 2 2sin

nnmR cos MR3 n

pé ùê úp

+ +ê úë û10. Ans. (D)

Sol.

where 2

2 2 2sin

nI nmR cos MR3 n

pé ùê úp

= + +ê úë û

(M + nm) g sin q – f = (M + nm) a ........(i)fR = Ia/R ........(ii)

form (i) & (ii) f = 2

(M nm)gsin I[I (M nm)R ]

+ q+ +

Þ µ(M + nm) g cosq = 2

(M nm)gsin I[I (M nm)R ]

+ q+ +

µ = 2

I tanI (M nm)R

q+ +

11. Ans. (B)

12. Ans. (C)

Sol. For Q. 11 & 12

For resonance condition w of electric field must

be equal to the angular frequency of electrons

rotation

\ w = qB

*mAt low temperature the lattice vibration

decreases and increasing w will increase the path

length between two collisions.

Holes are positively charged thus direction of

electric field must be reversed

SECTION-II

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

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

Sol. (A) It represents induce electric field for time

varying magnetic field upto some

distance R.

For magnetic field it may be field of a

cylinder with varing current density

(B) Induce electric field lines or magnetic field

lines of a current carring wire.

(C) Induce electric field lines or magnetic field

lines of a solenoid.

(D) Only electrostatics field of a dipole

SECTION-IV

1. Ans. 2

Sol. Assumeing initial angular velocity of the disc

to be zero, we can assume it to be performing

the pure translational motion as the net torque is

also zero.

We can treat this as a smiple pendulum.

T 2g

Þ = pl

= 0.993

29.8

p

T ; 2 sec

HS-3/14


Kota/00CT214005

2. Ans. 7

+ x

P(x,y)

(3,0)( 3,0)-

y

q -2q

Potential at

P = 0 = ( )

( )( )2 22 2

K 2qKq0

x 3 y ) 3 x y- =

+ + - +

Þ ( ) ( )2 22 23 x y 4 x 3 4y- + = + +

Þ 9-6x + x2 + y2 = 4x2 + 24 x + 36 + 4y2

Þ 3x2 + 30x + 27 + 3y2 =0

Þ (x+5)2 + y2 = 42 Þ a =5, b=2 Þ a + b =7

3. Ans. 1

Sol. p = charge × (distance between mass centres of

both the portions) = 2

2 2 4l læ ö´ =ç ÷è øl l l

= 1 mC–m

4. Ans. 7

Sol.34 8

min 19 3

hc 6.63 10 3 100.62Å

eV 1.6 10 20 10

-

-´ ´ ´

l = = =´ ´ ´

minhc 12400

OR 0.62ÅeV 20000

æ öl = = =ç ÷è ø

Also, ( )2 2 2K 1 2

1 1 1R Z 1

n n

é ù= - -ê ú

l ê úë û

Þ ( )27K

K

1 11.09 10 41 1 1 0.76Å

4é ù= ´ - - Þ l =ê úl ë û

Now, l - l = - =K min 0.76 0.62 0.14Å

- -= ´ = ´10 80.14 10 m 14 10 m

5. Ans. 6

Sol. From PV = NkT

Þ (10–5 × 13.6 × 103 × 9.81) × 7.4

= Nk × 293 \ Nk = 0.0338

Energy discharged through capacitor

= 2 51CV 12.0 10 J

2= ´

Energy transformed to plasma K.E. = 24 × 104 J

As average kinetic energy associated with the

gas molecules is 32

NkT, but each deuterium

molecule produces two ions and two electrons,

have 4 × 32

NkT = 24 × 104

Þ T = 1.18 × 106 Þ So a = 6

6. Ans. 1

Sol. By symmetry circuit can be reduced to

5W

10V

I 5W I

10

I 1A5 5

= =+

7. Ans. 2

Sol. Electric field at

P : 2 2

2 2 5/2 2 2 5/2

kp(2x y ) 3kpxyˆ ˆE i j(x y ) (x y )

-= +

+ +

r.

At P xE 0=r

Þ y x 2 2m= =

8. Ans. 2

Sol. For power to be minimum current should be

zero Þ n n 2

n 2n 1 n 4

+= Þ =

+ +

HS-4/14


Kota/00CT214005

PART–2 : CHEMISTRY SOLUTIONSECTION - I

1. Ans. (B)

Kh = w

a

KK

1410

1410 1010

--

- =

2W

Ha

K CKK (1 )

a= =

- aa << 1, hence10–10 = 0.01 a2

a = 10–4

% a = 10–2

2. Ans. (D)3. Ans. (A)

Specific conductance , K = 2 11

. 10 ScmR A

- -=l

\ molar conductance eqL = ´K 1000

M

= 2

31

1010

10

-

- ´

= 102 S cm2 mol–1

4. Ans. (A)5. Ans. (D)6. Ans. (D)7. Ans. (C)8. Ans. (C)

CH COONa2Electrolysis CH2

CH2CH COONa2

Br2

CCl4CH –Br2

CH2

Alc K

OH

CH CH

(A) (B)

(C)

Br–

9. Ans. (B)

C(graphite) + 2H2(g) ® CH4(g)

DrHº = {– 400 + 2 (–300)} – {–900}

DrHº = – 100

DrHº = – 100 = 700 + 880 – 4EC–H

E(C–H) = 420kJ

mole

10. Ans. (D)11. Ans. (D)12. Ans. (A)

SECTION - II

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

2. (A)®(Q,R,S,T); (B) ® (P,Q,R,S,T);

(C) ® (P); (D)®(Q,S)

SECTION - IV1. Ans. 2

2 = hc

- fl

........(i)

6 = 2hc

- fl

........(ii)

Solving : f = 2eV Ans.

2. Ans. 2

2 NO (g) + O2 (g) ® 2NO

2 (g)

t = 0 moles 2 0.5 0

t = t moles 2–2 × 0.5 0.5 – 0.5 2 × 0.5

= 1 = 0 = 1

nf = 1 + 1 = 2

Dn = 2.5 - 2 = 0.5 moles

\ change in pressure

nRT 1 300P 0.5 2atmV 12 6.25

DD = = ´ ´ =

3. Ans. 4

3A

3 3 2 M 3 3N 64r2

´= ´

´

3A

3A

1 2M64 r N2

4 M 2 2dN 64r

=´ ´

´ ´=

´

1 22 d 4 2 2

=´ ´

d = 4

HS-5/14


Kota/00CT214005

4. Ans. 9

5. Ans. 36. Ans. 7

Ph–C + NaHCO 3 ® Ph–C + H O + CO2 2

||O

OH

||O

ONa(A) PhOH + NaNH2 ® PhONa + NH3

(B) CH3OH + NaH ® CH3ONa + H2

(C) CH3SH + ELLi ® CH3SLi + CH3–CH3

(D) 2PhSO3H + 2Na ® 2PhSO3Na + H2

(E) (y + z + w + v) – (x)

(17 + 2 + 30 + 2) – ( 44) = 7

7. Ans. 2

COOEtCOOEt

(i) NaOEt/D(ii) H O(iii) (iv) Zn-Hg/HCl

3+

D

NBS/hv

Br

(two)product are possible

8. Ans. 5

PART-3 : MATHEMATICS SOLUTIONSECTION-I

1. Ans. (C)

Let ( )5x

ƒ x ......... x 15!

= + + +

( )4 3x x

ƒ ' x ......... x 14! 3!

= + + + +

( )3 2x x

ƒ" x x 13! 2!

= + + +

( )2x

ƒ"' x x 1 0 x R2!

= + + > " Î

Þ ƒ" is increasing with only 1 real root

say a so 3 2

1 03! 2!

a a+ + a + =

Þ ƒ'(x) have only one point of minima x = aÞ ƒ'(a) = +ve Þ ƒ'(x) > 0 " xÎR

Þ ƒ'(x) > 0 Þ ƒ(x) = 0 have only one real root

2. Ans. (B)

Let M = cosx cos2x........cos999x

( ) ( )( )

999

999

2 sin xsin 2x.........sin 999x cosx cos2x....cos999x

2 sin x sin 2x......sin 999x=

999

sin 2x sin 4x.......sin1998x

2 sin xsin 2x.....sin 999x=

( )999

sin1000xsin 1002x .......sin1998x

2 sin xsin 3x.........sin 999x=

999

1

2=

Q sinx = sin1998x sin3x = –sin1996x and so on.

3. Ans. (B)z1 + z3 = z2 + z4 Þ PQRS is a parallelogramand set A contain points on the following circle

P31

R C A Q B(–4,0) (–2,0) (–1,0) (2,0)

(–5/2,0)

parallelogram inscribed in a circle is rectangleand area of rectangle will be maximum when it

is a square so area ( )21 93

2 2= =

4. Ans. (B)

x ycos sin 1

a bq + q = , OP

bm tan

a= q

aP(a cos ,bsin )q q

O

T

bm cot

a= - q

HS-6/14


Kota/00CT214005

2

2

b btan cot

a atanb

1a

q + qa =

-2 2

abtan cot

a b= q + q

-

)( )

( )22 2least

2 2 32abtan

a b 2 3 1

+a = =

- + -

least

1

63

p= Þ a =

5. Ans. (C)

1 2x x 90S =x1x2.....x5 = 243

as 1 2x x9

10

S= & ( )1/10

1 2x x 9P =

Þ x1x2 = x2x3=...... = 9Þ x1 = x2 =...... = x5 = 3so a = –5C13b = –5C33

3

c = 5C434

a > b, a + b < 0, b + c > 0, c > 06. Ans. (B)

( )/ 2

13

1

0

I cosx cosxdxp

= ò

( ) ) ( ) ( )/ 2/2

13 13 1

0 0

cosx sin x 13 cosx sin x sin xdxpp

-= - -ò

( ) ( )/ 2

13 1 2

0

13 cos x 1 cos x dxp

-= -ò

( )1 2 1I 13 I I= -

2 21 2

1 1

13 I 1 II I 1 1

I I13 1 13

é ù= Þ = + Þ =ê ú+ ë û

7. Ans. (C) ƒ(x,y) > x2 + 2y " x,yÎR

and ƒ(x,y) > y2 + 4x " x,yÎR

2ƒ(x,y) > x2 + 4x + y2 + 2y

ƒ(x,y) > ( ) ( )2 21x 2 y 1 5

2é ù+ + + -ë û

5

2³ -

so least value of ƒ(x,y) is 5

2æ ö-ç ÷è ø

obtained when

x =–2 & y = –1

8. Ans. (D)

x2 = 1002 + 1002 – 2(100)2cosq

= 2.1002[1 –cosq]

4dx d2x 2.10 sin

dt dt

q= q ...(1)

Qx

S100

100

Also s =100q Þ ds d d 1

100dt dt dt 10

q q= Þ = ...(2)

(2) & 2

pq = in (1)

Þ 4dx 2.10 .1 1

.dt 102.100 2

= 5 2=

Paragraph for Question 9 & 10

x : Judge see a plus sign

P(x) =æ ö æ ö æ ö æ ö+ + =ç ÷ ç ÷ ç ÷ ç ÷è ø è ø è ø è ø

4 2 2 44

2

1 1 2 2 41C

3 3 3 3 81

y : A originally wrote a plus sign

4 21 1 2 1

3 .P(x y) 133 3 3 3

P(y / x)41P(x) 4181

æ ö æ ö+ç ÷ ç ÷Ç è ø è ø= = =

9. Ans. (D)

10. Ans. (C)

11. Ans. (C)

12. Ans. (A)

Q Reflection of (x, y) in

L is 3x 4y 4x 3y,

5 5- + +æ ö

ç ÷è ø

\ 2 23x 4y 3x 4y 4x 3y 4x 3y12 7 12 25 0

5 5 5 5- + - + + +æ ö æ ö æ ö æ ö- - + =ç ÷ ç ÷ ç ÷ ç ÷

è ø è ø è ø è ø

Þ 12(9x2 + 16y2 – 24xy) – 7(–12x2 + 12y2 +7xy) – 12(16x2 + 9y2 + 24xy) + 625 = 0

Þ xy = 1

T.A. of xy = 1 is y = x and its reflection iny = 2x gives T.A. of S1 which is

4x 3y 3x 4y5 5+ - +

= Þ y = 7x

HS-7/14


Kota/00CT214005

SECTION – II1. A ns. (A)®(P); (B)®(S,T); (C)®(R,T);

(D)®(Q)

( )

sin x , x 0,2

ƒ x 1 , x ,2

cosx , x ( ,2 ]

ì pé ùÎï ê úë ûïï pï æ ù= Î pí ç úè ûïï- Î p pïïî

which is differentiable in [0,2p]

(A) ( )( ) ( )10

0

I x 1 x 2 .... x 9 dx= - - -òI = – I using king.Þ I = 0

(B) ( )( )1

1/ 44 3 5 4

0

x x 4x 5x dx+ +ò4x5 + 5x4 = tdt = 20(x4 + x3)dx

9 5/ 41/ 4

0

dt tt c

520 20.4

= +ò 19 3

25=

(C) ƒ(t) = 16t3 – 4t2 – 16t + 8, t Î [–1,1]ƒ'(t) = 48t2 – 8t – 16= 8[6t2 – t – 2]= 8(6t2 – 4t + 3t – 2)= 8[2t(3t – 2) + (3t – 2)]

= 8(2t + 1) (3t – 2) = 0 Þ 1 2

t ,2 3

= -

ƒ(–1) = 4, ƒ(1) = 4, 1

ƒ 132

æ ö- =ç ÷è ø

,

2 8ƒ

3 27æ ö =ç ÷è ø

Þ ƒmax = 13

(S) sin–1x = x + c(1, /2)p

(–1,– /2)p

y = x + c Þ c = y – x

c 1, 12 2

p pé ùÎ - + -ê úë ûc = 0 only value

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

(A) ( ) 22tan

7a + b =

B D C3 17

h

A

a b

2

3 1722h h

51 71h

+=

-

2

2

20 h 22.

h h 51 7=

- Þ 70h = 11(h2 – 51)

Þ h = 11

(B) ah 1 2 s 12

r r a a

S D= = D

Då å

= (a + b + c) 1 1 1

9a b c

æ ö+ + ³ç ÷è ø

(by AM > GM)

(C) Let ( ) ( ) ( ) ( )D 0 . A a , B b ,C cr rr r

Þ a 2b b 2c 2a c

E F G3 3 3

æ ö æ ö+ + +æ öç ÷ ç ÷ ç ÷

è øè ø è ø

r rr r r r

D

AG

CF

BE

1V DE DF DG

6é ù= ë ûuuur uuur uuur

3

1a 2b b 2c c 2a

6.3é ù= + + +ë û

r rr r r r

1 2 01

0 1 2 a b c 96.27

2 0 1

é ù= =ë ûrr r

(D)

1 2 3

2 3 1 a b c

3 1 2

é ùë ûrr r

( )18 a b .c= - ´rr r

218c= -r

so max. value of 218c-r

is 18

{as c a b£rr r

}

HS-8/14


Kota/00CT214005

SECTION-IV1. Ans. 6

Multiply gh in(1), ƒh in(2) and gh in (3)and add(ƒgh)' = 6(ƒgh)2 + 6

Þ 2

dt6dx

t 1=

+Þ tan–1t = 6x + c

tan–1(ƒ(x) g(x) h(x)) = 6x + c

by x = 0 Þ c4

p=

Þ ( ) ( ) ( )ƒ x g x h x tan 6x4

pæ ö= +ç ÷è ø

2. Ans. 41

2 x 1 x2

± - aa + a = ± Þ =

a

put in 2 1x x x

24= a + a +

Þ3 2

,2 3

a = , 13 601

12

- ±-

3. Ans. 9by x ® 3 – x & x ® 8 – x

ƒ(6 – x) = ƒ(x) = ƒ(16 – x)

Þ period = 10 ...(1)

by x = 3, ƒ(6) = ƒ(0) ...(2)

also ƒ'(3 + x) = –ƒ'(3 – x) Þ ƒ'(3) = 0 by x = 0& ƒ'(6) = 0 by x = 3

and ƒ'(8 + x) = –ƒ'(8 – x) Þ ƒ'(8) = 0 by x = 0

so roots are 0,3,6,8,10,–10,–7,–4,–2

4. Ans. 9The curve is(x – 2) + 4(y + 1)2 = 4

P A B(0,–1)

(2,–1)(–2,–1) (4,–1) so

2

2

Max PB

Min PA=

5. Ans. 6(0,40)

(40,0)

Let z = x + iy Þ [ ]x y, 0,1

40 40Î

Þ x,y Î [0,40]

also [ ]2 2 2 2 2

40 40z 40x 40y, 0,1

z | z | x y x y= Þ Î

+ +Þ 40x < x2 + y2 & 40y < x2 + y2

Þ (x – 20)2 + y2 > 202 & x2 + (y – 20)2 > 202

( ) ( )2 22 1 1A 40 40 20

4 2= - - p

( )A2 6

100= - p Þ Least integer greater then

A

100 is equal to 6.

6. Ans. 6

x+y=p

x+y=0

x+y=p

x + y Î ......[–p,0]

Area 1 12

62

æ öp =ç ÷pè ø

7. Ans. 8

D1 = D2 = D3 = 0 Þ a + c = 2b

Þ max (b) = 8

8. Ans. 4

b c b cM N

2 2

æ ö æ öl + + mç ÷ ç ÷è ø è ø

r rr r

( ) ( )1 1 1AMN b c b c

2 2 2D = l + ´ + m

r rr r

1b c b c

8é ù= lm ´ - ´ë û

r rr r

D

C

N

M

A O E B bl b

mC

c

( )1BCDE CE BD

2D = ´

uuur uuur

( ) ( )1b c c b

2= l - ´ m -

r rr r

1b c b c

2= lm ´ - ´

r rr r

BCDE4

AMN

D=

D

Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005 +91-744-2436001 [email protected] HS-9/14

PART-1 : PHYSICS SOLUTIONPAPER-2

Path to Success


T M


PAPER CODE 0 0 C T 2 1 4 0 0 6





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

Sol. For image of B BB

1 1 1v 60cm

v 30 20- = Þ =

- ;

For image of A AA

1 1 1v 40cm

v 40 20- = Þ =

-

length of image = vB – vA = 20 cm2. Ans. (A,B,C)Sol. Let velocity of dog be

1 2v i v j+$ $ such that2 21 2v v 89+ = .

To catch the rabbit

1 2 0(v i v j)t (5i 10j) (3i 4j)+ = + + +$ $ $ $ $ $ t0Þ v1t0 = 5 + 3 to & v2 t0 = 10 + 4 t0Þ 2v1 = v2 + 2 Þ 5v1

2 – 8v1 – 85 = 0

1

8 64 1700 8 1764 8 42v 5

10 10 10± + ± ±

Þ = = = =

Þ v2 = 8ms–1 and 01

5 5 5t s

v 3 5 3 2= = =

- -3. Ans. (A,C)Sol. For (A) : Power consumed P = I2R

But 0

q= f

Î. so q = aÎ0t2

0

dqI 2 t

dtÞ = = a Î

Þ P = a2Î02t4R

For (B) , Assuming initial charge in resevoir beq0 then electric flux through a closed spherical

surface around S2 will be 2

20 0

s0

q t- a Îf =

Î

For (C) 2sd

2 tdt

f= - a

4. Ans. (B)

Sol. vsound = Er Here E is same and rsea > rfresh

5. Ans. (A, C)We add a resistance R in series to achieve avoltmeter of range 50V.

Þ 50 × 10–6 = 50

100 R+ Þ R = 106 W

Now, when a current of 10 mA is passed throughthe ammeter, 50 mA should go through the coil.We add a resistance S in parallel to the ammeter,50 mA should go through the galvanometer is :

50 × 10–6 = (10 × 10–3) S

S 100+Þ S = 0.5 W

6. Ans. (A,B)

Sol. Average rate of rise of temperature( ) ( ) ( )

( ) ( )-1 1

P 0.99 50kV 20mA2 C/ sec

t ms 1kg 495 J kg C-

Dq= = = °

D °

The minimum wavelength of the X-rays emitted

min 3

hc 124000.248Å

eV 50 10l = = =

´

7. Ans. (A,B)Try yourself by using BE/nucleon curve

8. Ans. (B, C)9. Ans. (D)Sol. The block will move updown periodically

\ centre of mass moves periodically in vertical.(In horizontal both will have same vertical andacceleration in magnitive but opposite indistance)

10. Ans. (A)Sol. When smaller block will be at point C it will

has zero xE but max potential energy \ byconservation principle of mechanical energy wecan say that velocity of bigger block will bezero.

11. Ans. (B)

Sol. Emax = hcl

+ evA = f (work function)

12400124

= ev + 10,000 ev – 10 ev = 10,090 ev

Þ min

hcl

= 10,090 ev Þ min

1240010,090

l = =1.24 Å

HS-10/14


Kota/00CT214006

SECTION - I1. Ans. (A, D)2. Ans. (B,C)

Since 'A' is more volatile than 'B' & initial molesare equal : Y

A2 > Y

B2 , Y

B2 < X

B2but on vapourisation, composition changes andvapour pressure decreases P

T2 > P

T33. Ans. (B,C,D)4. Ans. (A,B,D)

3I2 + 6NaOH ¾¾®NaIO

3 + 5NaI + 3H

2O

5. Ans. (A, B, D)6. Ans. (A, C)7. Ans. (A,B,C,D)

(A)I show poor rate of SN2 due

to steric hindrane

(B) Ph—CH—CH—Me

OH D

Optically pure

HIS 1N

CH—Me + Ph CH—Me

H

I

Ph

I

H

D D

Diastercomer Not Racemic mixture

(C) Me—Cl Me—OH Not Me—O—MeMoistAg O2

(It is obtained in dry Ag2O)

(D) MeI + NH 3 Me NI4dÅ

(Excess)8. Ans. (B,C)9. Ans. (C)10. Ans. (C)11. Ans. (D)12. Ans. (B)13. Ans. (C)14. Ans. (A)15. Ans. (D)16. Ans. (A)

SECTION - IV1. Ans. 62. Ans. 23. Ans. 1

O O ¾¾¾®

2

KOHI

CHI3 + 2CH3 – COOK

4. Ans. 6

12. Ans. (D)

Sol.Power outputPower input

=ò3 6

0.5 100 100= =

´Þ 6%

13. Ans. (A)Sol. n = a (z–b) (Mosley's law)

b is less for K so compared to L due to lessshielding effect in K.Energykb > Energy ka Þ nkb > nka \ ab > aa

14. Ans. (C)Sol. For (A) By symmetry E = 0

For (B) By superposition principle

E = 20

q4 ap Î

For (C)0

6qV

4 a=

p Î

For (D)0

5qV

4 a=

p Î15. Ans. (B)

Sol. dF Idl B; B® ® ® ® ®

= ´ t = m´ò ò

16. Ans. (A)

Sol.P RQ S

= \ S = RQP

In option (a), P = 100, Q = 10 \ S = R10

Since correct R lies between 400 and 500 W,correct value of S will lie between 40 and 50.Similarly for other options.

SECTION-IV1. Ans. 6

Number of electric field lines are drawn in pro-portion to charge magnitudes.

2. Ans. 2

Sol. n0 =

1 T

2 ml

T = y A a D T (D T = 80)µ = r An0 = 200

3. Ans. 6

( )2

2 d I RI R mgy mgv v

dt mg= = Þ =

4. Ans. 1Sol. 30VS = 29MS and 1 MS = (1/2)°; least count

= 1MS – 1VS = 1'

PART–2 : CHEMISTRY SOLUTION

HS-11/14


Kota/00CT214006

PART-3 : MATHEMATICS SOLUTIONSECTION-I

1. Ans. (A,B,C)

cos2x = –1 Þ x = (2n + 1)2

p

2. Ans. (B,D)

For min value a = 0

–10 10

ƒ(0) = 50

and for maximum value a = 11 –10 10 11

( ) 1ƒ 11 22.20 220

2= =

3. Ans. (A,C,D)

1234

1235

43214322

Shaded area = Q

1234

1235

43214322

Shaded area = P

P < R < Q

Q – R < R – P Þ P + Q < 2R

4. Ans. (B,D)

ƒ2(x) = x Þ ƒ4(x) = ƒ6(x)........ = x

Þ fn(x) = nx

(n + 1)x > nx Þ x > 0 so A false, B true.

for (D) 4x = 5 sinx,

Þ 3 solutions

5. Ans. (B,C,D)

as ƒ(x) is a continuous, odd function in closed

interval therefore its range will be [-a,a], where

a Î R+ so a = –a & b = a

Now check options

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

Let tn is k, 122333........ k k.......k

k times

tn

Þ ( ) ( )k k 1 k k 1

1 n2 2

- ++ £ £

Þ 4k2 – 4k + 8 < 8n < 4k2 + 4k

Þ (2k – 1)2 + 7 < 8n < (2k + 1)2 – 1

Þ (2k – 1)2 < 8n – 7 < (2k + 1)2 – 8 <(2k + 1)2

Þ 2k 1 8n 7 2k 1- £ - < +

Þ8n 7 1

k k 12

- +£ < +

Þ n

8n 7 1t

2

é ù- += ê ú

ë û

7. Ans. (B,C)

by x = 7 Þ ƒ(6) = 0

by x = 4 Þ ƒ(4) = 0

Þ ƒ(x) = (x – 4) (x – 6) g(x) ...(1)

(1) in given relation

(x – 4) (x – 5) (x – 7) g(x – 1)

= (x – 7)(x – 4) (x – 6) g(x)

Þ (x – 5) g(x – 1) = (x – 6) g(x) ...(2)

by x = 5 Þ g(5) = 0 Þ g(x) = (x – 5) h(x) ..(3)

(3) in (2) gives (x – 5) (x – 6) h(x – 1)

= (x – 6) (x – 5) h(x)

Þ h(x) = h(x – 1) Þ h(x) = l ...(4)

by (4), (3) in (1) ƒ(x) = (x – 4) (x – 6) (x – 5)lby ƒ(7) = 6, l = 1

8. Ans. (B,C,D)

( ) ( )2

2ƒ ' x x 2

x= -

y

2 xx y® ±¥ Þ ® ¥

x 0 y-® Þ ® -¥

x 0 y+® Þ ® ¥

HS-12/14


Kota/00CT214006

Paragraph for Question 9 to 10Locus of Q is a circle

C A F D

B(2,3,4)

Let BC is maximum value of

QB and BD is minimum value

Line x 2 y 3 z 4

BF :1 1 1

- - -= = = l

F(l + 2, l + 3, l + 4) Þ 3l = –6 Þ l = –2

Þ F(0,1,2)

so BC2 = BF2 + FC2

= 12 + (FA + AC)2

= 12 + ( )2

2 2 2 30+ =

BC 30= so 'D' is correct.

Now F is mid point of

( )1 1 1AD , , 0,1,2

2 2 2

a + b + g +æ öÞ ºç ÷è ø

Þ D(–1,1,3)

9. Ans. (D)10. Ans. (A)

Paragraph for Question 11 to 12

( )1

n

0

ƒ n 1 x sin x dx2

pæ ö+ = ç ÷è øò

1 1n n 1

00

2 2x cos x nx cos x dx

2 2-ùp pæ ö æ ö= - +ç ÷ ç ÷úp pè ø è øû

ò

Þ ( ) ( )2nƒ n 1 g n+ =

p...(1)

Now ( )1

n 1

0

nƒ n nx sin x dx2

- pæ ö= ç ÷è øò

1 1n n

0 0

x sin x x cos dx2 2 2

p p pæ ö= -ç ÷è ø ò

( )1 g n 12

p= - +

Þ ( )nlim nƒ n 1

®¥=

Q ( ) ( )nlim ƒ n g n 0

®¥= = ...(2)

Þ ( ) ( )nlim n 1 ƒ n 1

®¥+ =

Þ ( ) ( )2

n nlim n g n lim nƒ n 1

2®¥ ®¥

p= +

( ) ( ) ( )nlim n 1 ƒ n 1 ƒ n 1

2®¥

pé ù= + + - +ë û 2

p= ...(3)

by (1) (2) & (3)

( ) ( )( ) ( )2n

3n 1 ƒ n 3lim

24n 4n 1 g n®¥

+=

p+ +

11. Ans. (C)

12. Ans. (A)

13. Ans. (A)

( )2cos t 1 0

A t 1 2cos t 1

0 1 2 cos t

é ùê ú= ê úê úë û

|A(t)| = 8cos3t – 4cost

= 4 cost cos2t sin 4t

sin t=

|A(4t)| = 4cos4t cos8t = sin16t

sin 4t

(P) t 0

sin 4t sin16tlim 16

sin t sin 4t®=

(Q) |A(t)| |A(3t)| = 4cost cos2t.

4cos3t cos6t < 16

which is maximum at t = 0

(R)t

17

4 sin16tA A 1

17 17 sin t p=

p pæ ö æ ö ö= =ç ÷ ç ÷ ÷è ø è ø ø

(S)0

sin16tdt 0

sin t

p

=ò

HS-13/14


Kota/00CT214006

14. Ans. (B)

by x = i

(1 + i)62 = (C0 – C

2 + C

4–.......–C

62)

+ i(C1 – C

3 + C

5–....+ C

61)

Þ ( )62i / 42e p = (C

0 – C

2....... –C

62)

+ i(C1 – C

3 + C

5..... +C

61)

Þ comparing real parts

0 2 4 62

610 2 4 62

600 4 4 60

602 6 62

C C C ...... C 0

C C C ....... C 2

C C C ..........C 2

C C ..............C 2

- + - =ìïí

+ + + =ïîÞ + + =

Þ + + =

comparing imaginary parts

311 3 5 61

611 3 61

C C C ......... C 2

C C ................. C 2

- + - + = -

+ + + =

by adding C1 + C

5 + C

9.... + C

61 = 230(230–1)

by subtraction C3 + C

7+.....+C

59 = 230(230+1)

15. Ans. (D)2 4 34

i i i18 18 18A 1,e ,e , .....e

p p pì üï ï= í ýï ïî þ

2 4 94i i i18 18 48B 1,e ,e , .....e

p p pì üï ï= í ýï ïî þ

LCM of 2 2

,18 48 3

p p pæ ö =ç ÷è ø

so

2 4 5i i i i

i3 3 3 3A B 1,e ,e ,e ,e ,ep p p p

pì üï ïÇ = í ýï ïî þ

( )1 22k 2k 2

i i i18 48 144

1 2zw e e e 8k 3kp p p

= = +

by k1= 0 & k

2Î{0,1,.....47} ® 48 values of zw

by k1= 1 & k

2Î{0,1,.....47} ® 48 values of zw

by k1= 2 & k

2Î{0,1,.....47} ® 48 values of zw

for rest of the values of k1 Î {3,........17} the

values of zw repeats. \ 144 differentiablevalues will be there so n(C) = 144

n(A Ç B) = 4. {Plot 18, sided polygon insidecircle of unit radius}

n(B Ç D) = 9. {Plot 48, sided polygon insidecircle of unit radius}

16. Ans. (C)

( )n

210

k 1

cos k log k 3k 2 1=

p + + =å

( ) ( )( )n

k 1

cos k log k 1 log k 2 1=

p + + + =å

I case : n even

Þ |– log2 – log3 + log3 + log4 – .....

.....+ log(n + 1) + log(n + 2)| = 1

Þ |log (n + 2) – log2| = 1

Þ n 2 1

10, n 182 10

+= Þ =

II case : n odd

|– log2 – log3 + log3 + log4.... – log(n+1)

– log(n+2)| =1

|– log(n + 2) – log2| = 1 Þ 2(n + 2) = 10, 1

10Þ n = 3.

for equation (2)

I case : n even

|log (n + 2) – log2| = 2 Þ 2

2

n 2 110 ,

2 10

+=

n = 98 one solution.

II case : n odd.

|– log(n + 2) – log2| = 2 Þ 2(n + 2) = 102, 2

1

10n = 48 rejected as n is odd.

SECTION-IV

1. Ans. 7

( ) ( ) ( )3 3 3

1 1 1

ƒ x ƒ x ƒ xdx dx dx

x x 2= -ò ò ò

( )3

1

1 1ƒ x dx

x 2æ ö= -ç ÷è øò

( )3

1

1 1ƒ x dx

x 2æ ö£ -ç ÷è øò

3

1

1 1dx

x 2£ -ò

HS-14/14


Kota/00CT214006

2 3

1 2

1 1 1 1dx dx

x 2 2 xæ ö æ ö= - + -ç ÷ ç ÷è ø è øò ò

1 1 3n2 n

2 2 2= - + -l l

4n

3= l

Note : Maximum can be achieved by letting

( ) [ ]1, x 1,2ƒ x

1, x (2,3]

ì Î= í

- Îî

2. Ans. 5

Power of R with respect to 2nd circle gives x.2x= 10.26

Þ x2 = 130

2PQ 526

=

3. Ans. 1

ex(1 + yex)(dy + ydx) = 2x dx

Þ (1 + yex) (exdy + yexdx) = 2xdx

Þ ( )2x

21 ye

x c2

+= + Þ x 2ye 1 2x 4+ = +

\ ( ) ( )ƒ 1 6 1 e- = -

4. Ans. 9

R'' R'

R Q

C

B Q' Q'' A

P

R 'Q PQCA AB

= Þ102 dPQ .85

102-

=

R'' R PRBC AB

= Þ90 dPR .85

90-

=

QR = d =102 d 90 d.85 .85

102 90- -

+

Þ85 85d 1 85 85

102 90æ ö+ + = +ç ÷è ø

Þ5 17d 1 1706 18

æ ö+ + =ç ÷è ø

Þ 170.18d

50=

Þ5d 934

=

TEST TYPE : MAJOR PATTERN : JEE (Adaced) · part-1 : physics answ : per-1 part-2 : chemistry part-3...

Documents

Transcript of TEST TYPE : MAJOR PATTERN : JEE (Adaced) · part-1 : physics answ : per-1 part-2 : chemistry part-3...