FIITJEE AITS Solutions 1 Ft 5

1

FIITJEE

JEE(Advanced)-2015 ANSWERS, HINTS & SOLUTIONS FULL TEST– V (Paper-1)

Q. No.

ALL INDIA TEST SERIES

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AITS-FT-V-(Paper-1) –PCM(Sol)-JEE(Advanced)/15

1.

PHYSICS A

CHEMISTRY B

MATHEMATICS B

2.

A

C

A

3.

D

B

C

4.

B

A

D

5.

A

B

C

6.

B

A

C

7.

A

C

B

8.

C

A

D

A, C

B, C

A, C

A, D

B, C

A, D

A, B, C, D

A, C

A, B, D

A, B, C

C, D

13.

C

D

B

14.

C

A

A

15.

B

C

D

16.

D

A

D

17.

A

A

C

18.

C

B

B

1.

3

2

8

2.

2

4

4

3.

6

0

4

4.

6

1

6

5.

2

4

1

9. 10. 11. 12.

B, C

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2 AITS-FT-V-(Paper-1)–PCM(Sol)-JEE(Advanced)/15

Physics

PART – I SECTION – A

1.

2.

m1v1  m 2 v 2  1m / s m1  m2 At maximum extension both blocks will move in same direction with VCM now use energy y conservation. VCM 

 M 2   2 12 F – N = Ma N = Ma1  M 2 N  1 2 12 Acceleration of hinge O

 F  N

a1 1 N N



    a1  1 2 2  Nx = – F/4  a = 5F/4M and  = 9F/2M

=

a

F

a

   F   ao   a    ˆi   ˆi 2  M  9 L 4

3.

y  y0 sin kx cos t and

4.

d KA   T1  T2   equation of heat flow through the box. dt L

5.

Just after collision. a  b  v e ucos   L2  m2  v  ucos   a  b   m1   a2   12  

6. 7.



u sin v

1 2 2 L   constant 2 2C Using COM; vcos = u sin  2v sin  and e   2 tan2  ucos   2tan2 < 1

u v

v  

 1   tan < 1/2    tan1    2

8.

Power factor = VRMS RMS cos 


3 AITS-FT-V-(Paper-1) –PCM(Sol)-JEE(Advanced)/15

9. 10. 11. 12.

np p for minima y   2n  1 d 2d y y  A sin  t  kx   A  cos  t  kx  t C Vmedium  

For maxima y 

Isothermal PV  cons tan t adiabatic PV r  constant

13-14. h    eV0 15-16. E  mc 2 17-18. f 

C  V0  f0 C  VS

SECTION – C 1.

3 2 km/hr

2.

The centre of mass of system will not move horizontally.

3.

F  URel

4.

In steady state the current in branch containing capacitor will become zero.

5.

   F  q E  V  B 

dM dt



Chemistry

PART – II SECTION – A 

1

1.

1

3

    NH2OH  H  N2 O  2NH4  2H  3H3O

S

2.

P S S S

S

P

P

S

S

S

S

P Has 3 p - d P – S bond Has 3 P = S bonds Has 3 tetrahedral units of P Has 6 P – S – P linkages. 3.

KI  2KMnO4  H2O  KIO3  2KOH  MnO2 (nf = 6) (nf = 3) Moles of KI × 6 = 3 × 0.2 × 10 × 10-3 Moles of KI = 10-3 = 1 milli mole = x 

S  C  N  I2  H  SO42  HCN  I  H2O

To produce 10-3 moles I , 

Moles of S CN  6  10 3  1  10 3 Moles S CN   1.6  10 4 moles 6 4.

O

CH3

OH  CH3OH

H H2 O

 

OH

 Tautomerises

H 5 C6 C

H

O

O C6H5 CHO

  OH

A

N2H4 / OH

OH H 5C 6 C

H

D 2 enantiomeric pairs possible for (D).



5.

CoCl2  2KCN   Co  CN2  2KCl Buff coloured Co  CN 2  4KCN   K 4 Co  CN 6 

6.

o

L.H.S.

Half cell H2  2H  2e

Reaction E Eoxo  0.00 V

R.H.S.

Zn2  2e   Zn

Eored  0.76 V

Net  zn2  H2   Zn  2H , Eocell  0.76 V 2

H  0.0591 K   2  log K  Zn  n o Ecell  Ecell 

0.0591 log K n

 0.0591 log10 H  0.46  0.76  2 0.3  6 H   4.60  10 M

  HSO3  



0.4 M

2

H



4.6  10 6

SO32 6.4  10 3

H  SO23  4.6  10 6  6.4  103 Ka         7.3  108 HSO3 0.4 pKa = 7.13

7.

O

O C

OMe

MeOH

 O 

NMe

Cl O

O OMe

MeNH

2  

OH O

O OMe

PCl

3  

C O

O

NHMe O

O

1 1 1   RH   2    1  R H 1 1  

8.

Shortest  of Lyman series    1 for H-atom

9.

  (2) B  OH3  NaOH   NaBO2  Na B OH 4   H2O can be made to proceed forward by

adding cis-diol to stabilization by chelation. (3) The solution turning milky on passing H2S through group II solution indicates the presence of oxidizing agent where H2S is oxidised to 8O2. 10.

(A) The heat of formation for NO2 is calculated as – 45 kJ mol-1. (B) For reaction N2  2O2   2NO2 , H  310 kJ / mole. i.e. it is exothermic. 1 1 N2  O2  NO, Ho  55 kJ / mol 2 2 1 (D) NO  O2  NO2 , Ho  100 kJ / mol, it is exothermic. 2

(C)



11.

(A) No H-bonding can occur in chlorobenzene, so +I effect increases acidic strength. (B) Due to both steric and polar effect has no ortho effect is observed in case of o-MeOC6H4NH2. MeOC6H5NH2. p-MeOC6H4NH2 > o-MeOC6H4NH2 > m-MeOC6H4NH2 pKa = 5.24 (pKa = 4.45) (pKa = 4.2) N N >

NH2 >

(C)

N

pKa = 4.63

pKa = 5.25

pKa = 1.3

NH > (CH3 CH2) 2NH

(D)

12.

Pyrrolidine (pKa = 11.27)

(pKa = 10.98)

(A) is correct, S sys  nRn

V2 , V 2 > V 1. V1

 S sys   ve

(B) PV = constant, as  is high, slope is higher. (D) At boiling point, the process is at equilibrium  G  0 . 14.

x = 6, y = 3, z = 2 Solution for the Q. No. 13 & 14. IO3  6OH  Cl2  IO65  3H2O  2Cl 

H  IO56  H5IO6 o

o

100 C 200 C  2H5IO6  2HIO 4  I2O5 . 4H2O

17.

V. D. = 28.75 Dd   0.6 n  1 d     2NO2 At equilibrium : 1  0.6  1.2 N2O 4

 0.4

PN2O 4 PNO2

0.4  1.2

If moles of N2O4 escaping out is x, then for NO2 it is (1 – x)  x  0.4 46  x  0.188    1  x  1.2 92 XN2O4 escaping out   0.188, XNO2  escaping out   0.812 2

18.

 1.2  PNO2  1.6  KP    2.2 PN2O4  0.4     1.6   KP does not change.



SECTION – C 1.

Cl

Cl

Cl As

As Cl

Cl

Cl Cl

Cl

2.

H CrO  C H N

OH ArO

3 5 5 2   B  CH2Cl2

 Collins reagent 

O

O ArO

O

O O

CN

3.

CN

copolymerisation nCH2  CH  CH  CH2  nCH2  CH  

CH2  CH  CH  CH2  CH2  CH

n



Mathematics

PART – III SECTION – A

1.

If [x3 + x2 + x + 1] = [x 3 + x2 + 1] + x, then x is integer  log |[x]| = 2 – |[x]| has same solutions as log |x| = 2 – |x|, x is integer  No integral solution  0

2.

Each element has 5 choices, as it may be in any one of A1, A2, A3, A4 or may not be in any one  The required number of ways = 57

3.

(1 + x)n = C0 + C1x + C2x2 ..... Cnxn 

1  x n 1  1 n1

 C0 x 

C1x 2 C2 x3 C x n 1  ..... n 2 3 n 1

n2



1  x  C x 2 C x3 Cn xn2 x 1    0  1 ..... n  1n  2  n  1 n  1n  2  2  1 3  2 n  1n  2 

Put x = 2, we get 

C0  2 C1  22 Cn  2n1 1 1 3n 2   .....     2 23 n  1n  2  n  1 2  n  1n  2  2 n  1n  2 

 Sn  

C  22 C1  23 Cn  2n2 3n 2  1 2   0  ..... 23 n  1n  2  n  1 2  1 n  1n  2 

3n 2 2  n  1n  2 

S7 3 9  2  7  8 7   S 6 2  8  9  38 3

4.

It is given that |z| = 1 z = cos  + i sin  (let),   (0, 2) z z Now,   1 gives 2 |cos 2| = 1 z z 1 1  cos2  ,  2 2  5 7 11  2 4  5   , , , , , , , i.e. 8 values 6 6 6 6 3 3 3 3

5.

As cot  – tan  = 2 cot 2  tan  = cot  – 2 cot 2 1  1  1   tan  tan  cot  cot  2 2 4 4 4 4

6.

Latus rectum, 4a  2 2 1  a 2  1 1 Vertex =  ,   2 2 Axis = x – y = 0 As origin lies outside the parabola  Focus = (1, 1), Directrix = x + y = 0 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com


 Equation of parabola is

xy

2

2  x2 + y2 – 2xy – 4x – 4y + 4 = 0

2

  x  1   y  1

2

7.

Tangent at (0, 0) will be same (3 + sin B)x + (2 cos )y = 0 and 2 cos  x + 2cy = 0 are same 2cos2   c  cmax = 1 where sin  = –1 and  = 0 3  sin 

8.

Consider g(x) = xf(x) g is continuous on [0, 1] and differentiable on (0, 1)  f(1) = 0  g(0) = 0 = g(1) By Rolles’ Theorem g(c) = 0 some for c  (0, 1)  cf(c) + f(c) = 0

9.

At x = 2, RHD = 5 + |1 – x| = 5 + 1 = 6 whereas LHD = 5

11.

Let the coordinates of point A are (ct, c/t) So, the slope of normal at A will be t2. And normal will be parallel to BC. So, t will be  2  c =  2.

12.

‘a’ must be negative and x2 – a = x should have no solution 1 Now, D < 0  1 + 4a < 0  a   4

y=x 0

13.



ABC b  PQR a

 ABC 

14.

b 3 3 2 3 3  a  ab a 4 4

Slope of OP and OQ will be

am1 am2 and respectively b b

a m1  m2  ab m1  m2  If POQ =  than tan  = b 2  2 a b  a2m1m2 1  2 m1m2 b

Now, area of sector OPQ of circle x 2 + y2 = a2 = Area of sector OAB =

a2 1   a2  2 2

ab  m  m2  b1 2  1 a     ab where  = tan1 2 12 a  2 2 b  a m1m2 



15.-16.  LCM = 90º and CL = CM  CLM = CML = 4 Let ACB = 2  3  LAC =   and LCB =  4 4 By sine Rule in ABC     3  a2 + b2 = 4R 2  sin2      sin2    4   4        a2 + b2 = 4R 2  sin2      cos2      = 4R2 4 4      Let ADC =   ABC =  –   CBM =       4    CAD =      2 4  BAC = CAD  BC = CD 17.-18. We want to have an = k if

k  k  1 2

n

 n is an integer, this is equivalent to

A   4

L

 4 B

D  

C

 4 M

k k  1

2 k  k  1 2



k  k  1 1 1 n  8 2 8

1 1  2n  k 2  k  4 4 1 1 1  k   2n  k   k  2n   k  1 2 2 2 1  Hence, an   2n     = 2 =  2 

 k2  k 

17.

Now, a = 2, b = 3, c = 5 Let A = number of numbers which are divisible by 2 B = number of numbers which are divisible by 3 C = number of numbers which are divisible by 5 Required number = A + B + C – A  B – B  C – C  A + A  B  C  1000  1000  1000  1000  1000  1000  1000  =          734  2   3   5   6   15   10   30 

18.

a = 2, b = 3, c = 5, d = 7 Hence, the given number is 25 · 35 · 53 · 73  4n + 1 is odd number therefore the factor 2 will not occur in divisor. 3 and 7 are of 4n + 3 form, odd powers of 3 and 7 will be of 4n + 3 form and even powers will be 4n + 1 form 5 is 4n + 1 form and any power of 5 will be of 4n + 1 form  Number of divisors of 4n + 1 type = number of terms in the product (1 + 32 + 34)(1 + 5 + 52 + 53)(1 + 72) 2 5 2 3 3 + number of terms in the product (3 + 3 + 3 )(1 + 5 + 5 + 5 )(7 + 7 ) = 48



SECTION – C 1.

As g(x) is inverse of f(x)  Domain of g(x) is range of f(x)  a = 3, b = 11

2.

Required number of ways = co-efficient of x30 in (x2 + x3 + x4 + x5 + x6 + x7 + x8)5





5

 x 2 1  x7   = co-efficient of x in   1 x    20 7 5 –5 = co-efficient of x in (1 – x ) (1 – x) 24 17 10 = C20 – 5 C13 + 10 C6 = 826  A = 8, B = 2, C = 6 A+B–C=4 30

3.





x3  1  1

2



For x  [0, 2] 1   f(x) = 2



x3  1  3



2

=

x3  1  1 

x3  1  3

x3  1  3

2



 f  x  dx  4 0

4.

Orthocentre is foot of perpendicular drawn from origin on the plane


FIITJEE AITS Solutions 1 Ft 5

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