Aits Ft i (Paper 1) Pcm(Sol) Jee(Advanced)

1

FIITJEE

JEE(Advanced)-2016 ANSWERS, HINTS & SOLUTIONS FULL TEST – I (PAPER-1)

ALL INDIA TEST SERIES

FIITJEE Students From Classroom / Integrated School Programs have secured to 2 Zonal, 6 State & 18 City Topper Ranks. 33 in Top 100, 78 in Top 200 and 205 in Top 500 All India Ranks bagged by FIITJEE Students from All Programs have qualified in JEE (Advanced), 2015.

AITS-FT-I-(Paper-1)-PCM(Sol)-JEE(Advanced)/16

Q. No.

PHYSICS

CHEMISTRY

MATHEMATICS

1. 2.

C C

D D

B C

3.

B

A

D

4.

B

C

A

5.

B

C

D

6.

C

C

A

7. 8.

B C

A A, B, C

C A, B, D

9.

A, B, C, D

B

A, B, C

10.

A, B, C

B, C

B, C

11.

A, B, C

A, C

A, B

12. 13.

A B

C B

C D

14.

C

A

D

15.

D

A

B

16.

A (A)  s, (B)  pt, (C)  q, (D)  pr (A)  pqt, (B)  prt, (C)  s, (D)  qr 3

1. 2. 1. 2.

D (A)  r, (B)  pt, (C)  qs, (D)  qs (A)  pqs, (B)  q, (C)  rt, (D)  q 4

A (A)  prt, (B)  q, (C)  t, (D)  s (A)  qt, (B)  q, (C)  prs, (D)  t 1

3.

5 9

7 3

2 1

4.

9

3

5

5.

2

2

7

6.

4

3

2

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com

2 AITS-FT-I-(Paper-1)-PCM(Sol)-JEE(Advanced)/16

Physics

PART – I SECTION – A

1.

C 1  tan1  tan   e  vertical component after collision  eucos  & Horizontal component  u sin   tan  

2.

u sin  1     tan1  tan   eucos  e 

C Zero  T  f  2a

Ring

T a

2

TR  fR  2R 

f

a  R  f  0.

3.

B  I1 

V  t /RC1 V e & I2  e t /RC2 R R t

1 

1



    I1  eR  C1 C2  I2



I1  e RC1 I2

t

4.

B f 

1 2 LC

,

f  50 

1 2 LC / K

2

50   100   f  50   K    K  1  2    1   f   10 4   f    K  1.01

5.

B

B

 r1  2R sin30o , r2  2R sin 60o  R 3  R

A

 i  i BD  B AC  BCB   0 sin 60o  0 sin30 o  4  r 4  r  1 2  0 i  BD  4 R 3

r2

r1 30o 30o

30

30 o

C

6.

i

i o

R

R

D

C Beats 

V V V   4 4      4      

V 4 2



7.

B 3 sin60o   sin90o 2 3 3  , 2 2 3 3  4

8.

C   13.6  10.2  3.4eV   1st collision &  3.4  15  11.6eV  2nd collision.

9.

A, B, C, D V

1 n 1 r

P.E.     T  r 3/ 2 T.E.    

10.

1 r

A, B, C 2

2

 Vs2   3    8  4   25  Vs  5 volt Vs  5 I VR  3  1 R  R  3

Z

R 3   0.6 Z 5  VL  VC  I lags V. P.F. 

11.

A, B, C 1 1 1 v  v     1   f v u f  u since m is positive in the graph,



m 

v v    u u

v v   m    1     u f   int ercept on m  axis  1 & tan  

12.

1 f

A V , OP  x 2  y 2 R origin is ins tan taneous centre of rotation

 

 VP  r    OP  

V 2 x  y2 R



13.

B



 a  R

 a  R   2a

ac

2

V2 V ac  2R    R  R R   2a 

anet

ac 1 2a   45o tan  

14.

C Q 300   15J / o C T 20

Thermal heat capacity, C  15.

D Specific heat capacity, S 

16.

Q 300   600J / kg o C m T 0.025  20

A Molar heat capacity Q 300  nT  25     20   50   30J / mol o C



SECTION – B 1.

(A  s) (B  p, t) (C  q) (D  p, r) In wrist watch spring force is not dependent on gravity. T  2

m K

2.

(A  p, q, t) (B  p, r, t) (C  s) (D  q, r) In nuclear reactions energy, charge, mass and momentum remain conserved.

1.

3

SECTION – C for Rmax ,   45o  Hmax 

 20 

2



sin2 45o

2  10

  10m

 Hmax allowed  5m, 5

 20 

 Rmax 

2.

2

sin2 

2  10

 20 

2

   30o

sin60o

10

m  20 3m.

5 1 3R3  R2 2 2 22  Msphere  R3 , Isphere   R3   R2 3 53  29 5 Iremaining portion  Icylinder  2 Isphere  R . 30





 Mcylinder  R2  3R . Icylinder 







3.

9  18 sin30o  r    1    9 m / s.

18 sin 30o

18 m/s 18cos 30o

1m

60o 0.5m

4.

9 

1 1 1   F f1 f2

      1  1 1  1 1   2 1      1    1     24  4   20 20   4   20 20     3    3  5   1  2     . 9

5.

2 2

V 200 V 2  200  I    2.5A & P    500 watt R 80 R 80 When the wire is divided into two equal parts and connected in parallel we get maximum power.  Re sis tance of each part  40

 Imax  5A 2

 Pmax 

6.

 200  V2   2000W  2kw. Req  20 

4 Q Q , VPB  3 6 Q Q  12    Q  24C 3 6 24  VAP  8 3  VA  VP  8  VP  4volt.  VAP 



Chemistry

PART – II SECTION – A

1.

D H H

H B

B

H

H

H

H

CH3

B 2 H6

CH3 CH3

CH3

2.

B

B

methylation

H

D H OH

OH

+ +

H

CH2

H+

ring. exp

-H2 O

HO

HO

HO

HO OH

+

HO

OH

OH

+

H /excess similar repetation HO HO

3.

A At nodal surface,  = 0 r

4.

2 2  Å = 1Å z z

C +2

Co – ordination no of Fe in D is 6 -

+

Na

Fe

+2

(a sandwhich compound)

C -

D



5.

C O

O +

Zn(Hg) HCl

AlCl3

O O

A

SOCl2

HO

B

HO

Cl

O

O

C

O

AlCl3 Br

NaBH4

H2SO4

NBS/CCl4



F

G

D

E

OH

O

ONa OH heat

H

6.

C O

O H

(a)

O

-

OH

H

OH

-

+ OH O-D

D - OD

O

H

O

O

H

-

(b)

D - OD

DO-

O

-

D

D

-

DO

D

O

O

O

D

D

D

H

D - OD

O

D

D

D

D

D

D D

-

H

H DO - D

DO-

-

OD

O

O D

D D

D

D

O-D D H

-

D

D

O

D

D

D

D

D

D

D - OD

DO - D

O

O D

D

D

D

D

Repeat last two step

D

D

D D

D D

D

D

D


8 AITS-FT-I-(Paper-1)-PCM(Sol)-JEE(Advanced)/16 O

O

O

H H

(d) OH

-

OH

H

7.

-

OH

HO - H

A Synergic bonding.

8.

9.

A, B, C log

K2 Ho  T2  T1     K1 & K 2 are eq. cons tan t K1 2.303R  T1T2 

log

K2 Ea  T2  T1     K1 & K 2 are rate cons tan t K1 2.303R  T1T2 

B H3C

Me N

H3C H3C

+

Me

-

CH3

OH

CH3



H2C CH3 (Major product)

Due to greater probability of losing -H (9 : 3 ratio). 10.

B, C (b) conjugate base of H2PO4 is HPO24  (c) pH of 0.1 M NaCl (aqueous solution) =

11.

1 pK w 2

A, C A 3  3e  Al

Number of moles of Al produced 

4.5  0.166 27

Number of moles of electrons used  3  0.166 1 H  e   H2 2

Number of moles of H2 produced 

1  3  0.166 2

Volume of H2 produced  1.5  0.166  22.4  5.57  5.6 12.

C 2nd step is r.d.s. in sulphonation reaction.

13.

B 2nd step is fast step in nitration.



14.

A a   P    V  b   RT V  on converting into virial equation from, we get  a    b  2 3   2  RT   b  b  ....... to higher power of n PV  RT  2 4 V   V V    

…(i)

For Boyle’s temperature, 2nd virial co-efficient becomes zero b 

a

RT 2 a TB  Rb a  TB2  RB 2a  Ti   Ti  2TB2 Rb

15.

A Let Boyle’s temperature, TB'  TB  TB' 

16.

a Rb 8a a 2  2TB R  2b Rb

D b2 = 0.04 b = 0.2 lit/mol Vreal = 4  22.4 + 4  0.2 = 90.4 lit

SECTION – B 1.

 A  r  ,B  p,t  , C  q,s D  q,s  . (A) Sic covalent carbide (Not hydrolysed) (B) Al4 C3  12H2O  4Al  OH3  3CH4 (C) Cal2  2H2 O  Ca  OH2  H  C  C  H (D) Mg2 C3  4H2 O  2Mg  OH 2  CH3  C  CH



2.

 A  p,q,s  , B  q , C  r,t D  q . Al2  SO4 3  2Al3  3SO 4 0.2M

0.3M

3

AlPO 4  Al  PO 0.1M

3 4

0.1M

Urea   Urea 0.1M MgCl2  Mg2  2Cl 0.1M 0.2M

SECTION – C 1.

4 Based on fact

2.

7 OH B 2Na

+

-

O

O HO

B

O O B

-

B

OH 8H2O

O

OH

3.

3 OHC

CHO

CHO

,

, NO2

Cl

C N

4.

3 OCH3

, H3CO

H3CO

OCH3

NH2

OCH3

,

OCH3



5.

2 1 1 100 n100; t 90  n K K 10 n100 2   2 n10 1

t 99  t 99 t 90

hence x  2

6.

3 S

H

O

H Cl

H C

S H 0

C C

, Cl

,

H 0

O H 0



Mathematics

PART – III SECTION – A

1.

B

Q

  PQ  r 2  1  2r cos  r 2  1  r 3  & AT  r 2  1  2r cos  Where 'Q ' is the angle between   OA and OT.  2  2 now, PQ  4 AT





2



2

 1  r  r  4 r  1  2r cos  2

 3

P A





2

3r  r  3 3r  r  3  1   1; r  0 8r 8r  7  13 7  13  r , . 6  6  cos  

2.

C Here two balls are drawn and the result is known that the two balls are white, therefore Bayes’s theorem should be used. Required probability =

3.

T

1 .1 1 4  1 1 6 1 3 1 1 2 .1  .  .  . 4 4 10 4 10 4 10

D Let Q  x   f  x   g  x  in  x0 x  By LMVT, Q  x   Q  x 0   Q'  x   Q'  x 0  for some C   x 0 ,x   Q  x 0   0, Q'  C   f '  C   g'  C   0

 Q  x   Q  x0   Q  x   F'  C   g'  C   x  x0   0  Q  x   0  f  x   g  x .

4.

A A

AP = 23

(-3, 4)

5 5

13

P FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com


5.

D Sn 

5 2n  5  4 2 n  1 n  2 

K 

6.

5 4

A Equation of tan gent at 'P ' is Y  y  m X  x Now, 

y  mx

x

1  m2

y 2  x 2 dy  2xy dx

putting y  vx

Solving Differential equation x 2  y 2  Cx It passes through 1,2   C  5 Equation of curve  x 2  y 2  5x  0

7.

C x cos  y sin   1 3 2 x  0, y  2cos ec,

T

chord A 'P; y

2 sin  x  3 3  cos   1

x  0, OM 

2 sin  cos    1

Now, OQ2  MQ2  OQ2   OQ  OM

2

 2  OM  OQ   OM2  OM 2OQ  OM 2 sin 







 4



 cos   1  sin 

2 sin   1  cos  

2sin   4  4 cos   2 sin 2  

1  cos  

2

sin 

4 sin  2  2cos   1  cos2  

1  cos  

2

sin 

4

8.

A, B, D f  x   0  x 3  x 2  x  1   x 2  x  1  2  0  x 3  1 

2 x2  x  1 -2

-1



1 2

x3



9.

A, B, C Take f  x    x  and g  x   x and proceed.

10.

B, C  cos 2x    1  I   tan1   dx   tan tan   x  dx 4  1  sin 2x           x  dx 4  I

11.

 x2  x2 x  C  gx  x  . 4 2 4 2

A, B



,  







2  4  4 a2  1

 1  a, 1  a 2  f 1  a   0 & f 1  a   0

y



 1   a    ,1  4 

12.

C The value of K1 & K2 decide the opening of parabola  A max  Area  OBCO   A min  Area  OACO 

D

Y c A

X'

o

Y'

14.

1 4 1 K1  8

K1 

(1,0)

B

(2, 0)

K2  

X

1 K2 = -1 4

D Area  BFEC   Area  ABC  AFE 

A

 or  AOB  BOC  COA  AFE  1 2 R sin 2C  sin 2A  sin 2B  sin    2A   2  1  R2  sin2B  sin 2C  2

2R co sA

13.



  2A

F

E H

B

O C



15.

B AE  R, AF  2Rcos A  EF  R

16.

A  3  AFE is equilateral AFE 

2

R 3  9  sum of square of altitudes  3   R2 .  2  4  

SECTION – B 1.

(A   p, r, t) (B   q) (C   t) (D   s)       A  tan  1 2 3 4   0  1  2  3  4  2n 2  

B  x 2

 4  t 2  2, TR is diameter  6  2  TR   9 4

Area of circle 

C 6 D  y A  yB  yC  2.

2



 3

2

3

(A   q, t) (B   q) (C   p, r, s) (D   t)

SECTION – C 1.

1 D.C' s of OP are,

Z

 1 1  , ,  3 3  1  D.C' s of AL    , 3 

1   3 1 1  ,  3 3

C (o, o, a)

L

 cos   cosy 

 mn

3  mn 3

,cos  

,cos  

 P (a, a, a)

(o, a, a)

 1 1 1  D.C's of BM   , ,  3 3  3  1 1 1  D.C' s of CN   , ,  3 3 3  Let D.C' s of required line be  ,m,n 

M(a, o, a)

O

A (a, o, o)

X

B (o, a, o)

N (a, a, o)

Y

 m n

3  mn 3

4 2 4   m2  n 2  3 3 4  2 2 2 2   cos   cos   cos y  cos       1. 3  2

2

2

Hence, cos   cos   cos y  cos2  







2.

2 The gives equation is  x  1 y  2   4

A

XY  4 Equationof tan gent is,

Q

4 4   2  x  t t t  8  A   0,  & B   2t1 0   t

 4

P  t1 t  B

Y

1 8   2t  8 2 t 4K  8  k  2. Area 

3.

1 2

2

 2  3  25 CE  a  b     DE 16 16 C2  CE    1  DE  4.

5 10 1

1

1

1

  f K  1  x  dx   f  x  dx   f  x  1 dx.......... f  x  9  K 1 0

0

0

0

 putting x  1  t,  x  2   t............  x  9   t 1

2

3

10

  f  x  dx   f  x  dx   f  x  dx..........  f  x  dx 0

1

2

9

10



 f  x  dx  5. 0

5.

7 Every ball have two options  4 balls can be put in 24 ways. Arrangement among themselves in 2! 1   24  23 2! But, above count also includes the one case in which all the balls are put in one box,  Required number of ways = 23 – 1 = 7.

6.

2 3

3

3

K  1 K  2 K  3  K  1 K  2  K  3  1  K  2

1

0

1

 K 2


Aits Ft i (Paper 1) Pcm(Sol) Jee(Advanced)

Recommend Documents