AITS 2016 FT I JEEM JEEAAdvancedPAPER-2SolutionsSolutions

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

JEE(Advanced)-2016 ANSWERS, HINTS & SOLUTIONS FULL TEST – I (PAPER-2)

ALL INDIA TEST SERIES

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1

Q. No.

PHYSICS

CHEMISTRY

MATHEMATICS

1.

A

C

B

2.

A

B

A

3.

B

C

C

4.

C

A

A

5.

C

B

B

6.

C

A

C

7.

D

A

A

8.

A, B

A, B

A, B

9.

B, C

B, D

A, C

10.

C, D

B, C

B, D

11.

A, D

A, B, C

A, B

12.

D

D

B

13.

C

C

C

14.

B

D

A

15.

C

A

B

16.

D

C

C

1.

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

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

(A)  r, (B)  t, (C)  q, (D)  r 4

(A)p,s, (B)  q,r,t, (C)  p,s, (D) q,r,t 3

2. 1.

(A)  p, q, r, s (B)  p, r, s (C)  t (D)  p, s (A)  q, (B)  p, (C)  r, s, (D)  r 8

2.

8

2

6

3.

1

6

4

4.

3

2

7

5.

1

6

3

6.

0

4

2

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Physics

2

PART – I SECTION – A

 charge in flux  R resis tance B  A  n   0  2I  1     A n R 4  r  R   IAn   0   2rR 

1.

Q

2.

Relative acceleration of rod with respect to wedge is parallel to the inclined plane.

3.

4.

 v  vD  fapp    f0  v  fs  Simply use the above relation

for B to C part A h1 v = gT B

liquid h2 C

u = gT a

 v    v  0  g v0

T 2    0 O  gt    0 v0 a

1

 T g 2 

  0  20    0 20

 3 0

5.

Calculate the initial and final distance of object from the concave mirror.

6.

To avoid t toppling


3


F N

O

fx mg

 net 0

0

F  a  mg 

a 0 2

mg 2 Now apply Newton’s laws of motion. F

7.

As we know L0 = L1 + L2 L1 = due to rotational motion L2 = due to linear motion.

8.

If temperature of a body is constant that means amount of hear energy received and amount of hear energy emitted are equal.

9.

If direction of field is along the x-axis then its magnitude will be 20 V/cm other wise greatest then hat.

10.

Apply conservation of linear momentum 2 KE  2  mass

11.

Calculate the time constant

R

R a b

Battery is replaced by a wire L L   Rab R 12 – 13. Now apply newton’s law of motion for the particle B. T

N

T

B 60

o

r



m r

mg mg



4

14 – 16. Initially current will pass through the resistance only. I R1 = 4 k 

C1 E R2

7 k

C2

Power in R2 = 7  (I2) = 2.8 W 2 Power in R1 = 4  I = 4  (4) = 1.6 -6

2.8 7  103

3

Charge in C1 = 3  10  8  10  = 240 C.

SECTION – B 1.

In such type of situation, we have to use these basic facts. (a) charge will flow between the two body if there is potential difference between the two. (b) during this process energy will lose due to sparking.

2.

At equilibrium Fnet will be zero and at the extreme point, particle will be in state of rest.

SECTION – C 1.

in this situation block B will not move at all. N fk 2a1 

T 2 = 2T 1

T2

mag Now apply newton’s laws of motion. T1

C

a1 mg

mg 2.

a sin30o 

v 02 R

v = 8 m/sec 30

o

2

a sin30o

a = 2 m/sec Where R is radius of curvature.


5

3.


apply newton’s laws of motion for the body A and body B N M A Mgsin 

Mgcos 

T Mg  m

B 4.

length of rod is R 2 Hollow sphere

2v O v P Velocity of point P 2v  R  v KE of rod = rotational KE t translatory KE 5.

simply use this concept After earthing a conductor its potential will become zero. If conductor are connected together both will be at the same potential.

6.

from P – T diagram, write the P – T equation. By using PV = nRT Find P in terms of volume w   Pdv .


6


Chemistry

PART – II SECTION – A

1.

C Reimer – Tiemann’s reaction

2.

B +   The cell reaction is: H2(g) + I2(s)   2H (aq) + 2I (aq) 2

Ecell = E

o cell

H  I  0.0591  log     2 PH2

2

 

2

H   0.1 0.0591 0.7714 = 0.535  log   2 1 pH = 3

3.

2

C Mosley’s equation is :

  a  z  b

= az  ab Given that, ab = 1, a = tan45o = 1

   51  1  50    2500 Hz 4.

A H2O S  2Cl2   SCl4  

OH

HO S

OH

HO Hybridisation of S in H2SO3 = 5.

OH

1  6  2  0   4  sp3  3  1  2

B H

H 6.



 H2O   HO  S  O

H

H

up

down

A 10 3  1000  0.01 M 50  50 0.1 50 [Cl-] initial just after mixing   0.05 M 50  50 I.P = [Ag+] [Cl-] = 5  104 I.P  K sp

[Ag+]initial just after mixing =


7

Ag  aq


Cl  aq  AgCl  s 



0.01M

0.05

initial conc.

x

0.04 M

conc. after precipitation





K sp   Ag  Cl  1010   Ag   0.04   Ag   2.5  109 M

7.

A OCH 3

OCH 3

OCH 3

OCH 3

H+

O2/Pt

Ph - CH = CH2 H+

Ph

O

O O

OH H3C

Ph +

O

H3C

H

1. -H +

2

-H 2O

OCH 3

+

H3C

Ph O

H2 O

+

OCH 3

+O H

H

+ OH2 Ph O

2. H

CH3

Ph

CH3

Ph

CH3

rearrangement

Ph

OCH 3

+

CH3

O

OCH 3

OH +

HO

O

H

+

+

C Ph

CH3

O

-H +

C Ph

CH3

C CH3

Ph (Y)

CH3 (X)

8.

A, B All reducing sugars are muatrotating. Although (IV) is an   hydroxy ketone and hence reducing but it can’t mutarotate as it not a carbohydrate, can’t form ring. In (II) the glycosidic linkage is in between two anomeric carbons and hence ring opening can’t occur, thus non – reducing as well as non – mutarotating.

9.

B, D In B and D compound after mixing undergoes H – bonding and hence boiling point will be raised and hence vapour pressure lowered.

10.

B, C Fact



11.

8

A, B, C Re dhot Fe tube CH3  C  CH   

Al2O 3 /Cr2 O3 /  n  C6H14  

+

+

4H2 Cl

t - Buo-

CHCl3

N

N

H

12.

D  NH4NO3   N2 O  H2 O

A

(B )

(C)

N2 O  P4  P4 O10  N2 (D)

13.

C P4 O10  HClO 4   Cl2 O7  H3PO4 P4 O10  6H2O  4H3PO 4

14.

D If k1, k2 and k3 be the rate constants of the reaction along path I, II and III respectively, then overall rate constant of consumption of A will be k1 + k2 + k3. So, 86.625 min 

0.693 k1  k 2  k 3

or k1  k 2  k 3 

0.693  8  10 3 min1 86.625

0.693  4  103 min1 173.25 0.693 k2   2  103 min1 346.5 k 3  8  103   4  2   103  2  103 min1 k1 

% distribution of C 

15.

2k 2 2  2  103  100   40 k1  2k 2  k 3 10  10 3

A d  A  dt

 k1  k 2  k 3   A   8  103  0.25  2  103 mol L1 min1 dC  dD   dt dt dt  k1  A   2k 2  A   k 3  A  

dB 

 k1  2k 2  k 3   A   10  103  0.25  2.5  103 mol L1 min1 .

16.

C


9

Ea overall 

k 1E a1 k1  k 2  k 3



k 2Ea2 k1  k 2  k 3




k 3E a3 k1  k 2  k 3

 0.5  40  0.25  60  0.25  80  55 kJ / mol

SECTION – B 1.

(A  p, r, s) (B  p, r, s, t) (C  q, r) (D  p, r, s) 3 O 2  2HF 2 P4  3OH  3H2 O   PH3  3H2PO 2 3XeF4  6H2 O   XeO3  2Xe 

B2H6  6H2O   2H3BO3  6H2 

B  OH3  2H2 O   B  OH 4   H3 O  3Br2  3Na2 CO 3   5NaBr  NaBrO3  3CO2

2.

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

SECTION – C 1.

3 92

2.

  U238  o n1  92 U239  Np239  Pu239 93 94

2 1  3.5 8 Effective no. of atoms of Y = 4 – 1 = 3 Effective no. of atoms Z = 8 – 4 = 4 34 Therefore, required answer = 2 7 2

Effective no. of atoms of X = 4  4 

3.

6 H3C

CH2

C

+

H

H

HC

O

CHO

OH-

CH3

+

C

H

CH2

C

CH2 CH3

O

OH-

+

H

CH

C

CH3

O

CH3

H2C CH3 H3C

C

CH2

HO

H2C CH3 C

CH CH CHO

O

O

H3C

CH2

HO

H2C CH3 H3C

H3C

CH3

OH-

C

CH2 CH3

O

C2H5 H3C

C HO

HC CH3

C

CH3

O



H3C

CH2

C

H

+

H

CH2

CH2

C

CH2

C

H

+

H3C

CH2 HC

O

+

CH2

C

CH2

CH3

O

OH

OHH

CH

C

CH3

O

H3C

CH3

CH2 HC

H

CH

CH

OH CH3

CHO

C

CH3

O

OH

OH-

CH3

4.

OH-

O

O

H3C

CH2 CH3

C

O

H3C

10

H3C

CH3

CH2

C

CH

CHO

CH3 CH3

2 U  nC v T 5.6  C v  10 22.4 50 Cv   5 cal 10 Cp  Cv  R  Cp  7 cal 12.5 

 

CP 7   1.4 CV 5

The gas is thus diatomic

5.

6 No. of eq. of NaOH that can be produced theoretically for 100% current efficiency = no of equivalents of NaCl decomposed = no of eq. of Cu deposited 

wt. of Cu 31.75  1 Eq. wt. of Cu 63.5 / 2

No. of eq. of NaOH produced experimentally = 

 % yield 

6.

meq. of NaOH 1000

N  V 1 60   0.06 1000 1000

0.06  100  6% 1

4 Hoffmann’s mustard oil reaction is a test of 1o amine. SH R  NH2  S  C  S  S

C

NHR



HgCl2

RNCS

(Alkyl isothiocyanote) smells like mustard oil

+ HgS + 2HCl


11

Mathematics


PART – III SECTION – A

n

1.

f  x 

  x  dx

0 n

 x dx 0

  f  x  n  1 

2.

 y y   3  y  



n  n  1 2

n 

n2  n 2

2

y

y 

 y dx  3 ydx

 x  A 1y 2  A 2 y  A 3

3.

d  AC  d

A

 bcosec .cot   asec  tan 

a sec 

P

a

y

bcosec 

b B

C

x 2

AC 

2

b a3  b3 b

1 3

2



2

a a3  b3 1

a3 3

2 2  2    a3  b3   

4.

Let , , ,  be the four roots Then         4 and   1 AM = GM               1  x 4  4x3  ax 2  5x  1   x  1

4

 a  6, b  4

5.

Let APB =  PA 2  PB 2  AB2 cos   2PA.PB



12

 PA 2  PB 2  AB 2  2PA.PB cos   2PA.PB  as cos  1 2

 PA  PB   AB 2

Thus the max value of |PA – PB| is AB This is possible only when P lies on AB but P lies on AB  P is the point of intersection of x + y = 2 and 2x + 3y + 1 = 0. 6.

r 6! P q  2    x  3x 2 P!q!r!  P  q  r  6 & q 2r  5





 coeff  5052

7.

tan3  tan2  tan2  tan  sin  3  2  sin  2      cos 3 cos 2 cos 2 cos   sin   2 sin  sin2   0

 sin   0 or sin2  0

n is not for possible as n is odd tan is not define. 2 Hence   n, n  z is the only solution.    n, n  z or 2  n   

8.

 AB  c, BC  a, AC  b

In OBD A

z O

E y

A x B

D

C

OD OB x cos A  R x  R cos A cos A 

a a b c   cos A     2R  2 sin A  sin A sinB sinC  a a x  cot A  tan A  2 2x Similarly b tanB  2y x

c 2z Also tan A  tanB  tanC  tanA .tanB.tanC (when A+ B + C = )

And tanC 


13

9.



a b c abc    2x 2y 2z 8xyz



a b c abc    x y z 4xyz


 4 tan x  cot x

(A) 0  x 

also ln  sin x   0 ln sin x

 tan x 

ln sin x 

  cot x 

  (B) in  0,   2 cosecx  1 4ln cos ecx  5ln cosec x   (C) in  0,  , cos x   0,1  2 lncos x  0 also  1    2

10

1 1  2 3

ln  cosx 

 1   3

ln cos x 

  = 5m – 3n  7(5m – 3n)2 + 5m2 – 3n2 = 0  6m2 – 7mn + 2n2 = 0 2 n m  n or m  3 2 If m = 2/3n from (1)  = 1/3 n If m = n/2 …..  = -n/2  cos  

1 1  2  1  3  2 12  22  33

11.

 1

2

7



14 6

 12  22



   2 2      a  b  c  d   a  2 a.b  4  2d. a  b  c     Let d  a  b  c    Then d.a  d.b  d.c  cos         cos 



Also  2  2   2  1  3cos2  or cos      2 3  a  b  c  d  4  2. 3

7 12

 a,b,c are mutually  r 

1 3

 42 3

12.

end point of latus rectum P(1, 2) & Q(1, -2)  PAQ is isosceles right angled


14


 slope of PA is – 1 In equation y – 2 = - (x – 1) x+y–3=0 similarly equation of line QB x–y–3=0 2 solving x + y – 3 = 0 with parabola y = 4x 2  3  x   4x  x  1,9  co-ordinate of B & C are (9, -6) & (9, 6) respectively Area of trapezium 1   12  4  x8 2  64 sq units

13.

let the circumcenter of trapezium be T(n, 0) Then PT = PB

n  1

2

n  9 

4 

2

 36

or n = 7 radius 40  2 10

14. – 16 z     2  1 Case I when 1    1 We get 2 < 1 z    i 1   2 or x = -   y2 = 1 – x2 or x2 + y2 = 1 case II  > 1 2  1  0 z     2  1 or x     2  1, y  0





Roots are    2  1,0 ,    2  1,0



One root lies inside the unit circle and other will lies outside the unit circle Case III where  is very large then z     2  1  2 z     2  1 

1 2

    1



1 2

SECTION – B 1.

6

total number of function = 3 = 729 (a) total number of onto function = 3 C116  3 C2 26  3 C3 36  3  192  729  540


15


(b) since f(ai)  bi It means that a1, a2, a3 cannot be assigned images b1, b2, b3 Number of function = 2333 = 216 (c) number of invertible function = 0 as function is not one – one (d) total many one function = 729 – 0 = 729 2.

(a) sin1 sin  3  10   3  10 (b) cos cos  4  10  1

 4  10 1  (c) tan1 x  tan1  if x  0 x 2    if x  0 2  n  n 1  n  n 1  (d) sin1  tan   1  n  n  1  n  n  1    n  n 1   n  n 1     sin1   tan   1  n n  1  n 1 n 1 n  n  1  

 tan1 n   2

SECTION – C 1.

let each friend has n sons.  3 tickets can be distributed among 2n sons in 2n C3 ways. The number of ways distributing 3 tickets such that two tickets go to the sons of one and one tickets goes to sons of the other.  n C2  n C1  n C1  n C2  2  n C1  n C2  probability that two tickets go to the sons of one and one tickets goes the sons of the other 2  n C1  n C2  2n C3 

6n2  n  1 2n  2n  1 2n  2 



3n 2  2n  1

But from question 6 3n  7 2  2n  1 n  4 Hence total number of boys = 8.

2.

2

2

2

2

2

2

the equation of the tangents to the circle x + y = a at P and the hyperbola x – y = a at Q are x  my  a 1  m2 or x  my  a 1  m2 respectively Where y = mx is intersecting line through (2, 0) Let (h, k) be the point of intersection of these two lines



16

 h  mk  a 1  m 2 are h  mk  a 1  m2 2



2





  h  mK   a2 1  m2 and  h  mK   a2 1  m2

 K

   2mhK  h



 m2 K 2  a2  2mhK  h2  a2  0  m2

2

 a2

2

 a2  0

Elimination m from these two equation and we get   a 4  4y 4  x 2  a6 3.

A BC  

Since ei  1, eiB  C  ei  A   eiA  e 

i A B

ei2A    e 

i A B 

e 

i A C

1

ei2B

e 

i A C 

e 

i B  C

e 

i B C 

ei2C

1 1

  1 1 1 1 1 1  4    4  4

x 1 t x2

4.

put

5.

OP  HP

 x2  y2  3  x 3  or y 2  6  x   2  3

Area of region =

9  y2 0 6 dy

= 3 sq unit 1

6.

 1 2  3...  n  n f  x   lim  n n  n  ....  n    1 1 2 n log f  x   lim  log  log  ...  log  n n n n n   n 1 r  lim  log n n r 1 n 1

  log x dx  1 0

f  x   e1 

1 e

Hence e2f(x) = e2 

1 e e

 e2 f  x     e  2


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