Sol-PAPER-2

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AITS-FT-II (Paper-2)-PCM-Sol-JEE(Advanced)/17

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1

Q. No.

PHYSICS

CHEMISTRY

MATHEMATICS

1.

B

A

B

2.

D

C

B

3.

D

A

C

4.

C

B

D

5.

B

A

B

6.

D

B

B

7.

B

C

B

8.

A

B

D

9.

BCD

ABC

AD

10.

AD

ACD

ACD

11.

AC

ABC

BD

12.

ABC

AC

AD

13.

D

D

D

14.

C

B

A

15.

C

B

C

16.

A

C

A

17.

C

B

D

18.

A

D

D

1.

6

0

2

2.

5

4

4

3.

3

3

2

4.

2

6

7

5.

2

9

2

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2


Physics

PART – I SECTION – A

v 330 4   0.66m  n3 f 500 2n  1

1.

λ=

2.

x=2d= , Now, U=fx=2fd

3.

V1 =

4.

If another shell is kept upside down over it complete a sphere, net field should become zero

5.

F1 =FB -  πa 2  ρgh

kQ kQ ;V2 = 2b 5b

F2 =FB -  πa 2  ρg  2h  6.

2  1  3 v2  v1  v3 2  1  3 Frequency is independent of medium

7.

8.

Bw is magnetic field of wire B1  B0  Bw B2  B0  Bw B  B2  B0  1 2

fR 2 f is =3 for monoatomic but f is 5 for diatomic at normal temperature. But f>5 for diatomic gases. At high temperature energy of vibration also increases taken that into account C V increases since molecules of monoatomic gas do not vibrate, its C V remains same. CV >

9.

The loops is always attracted towards wire as the region part of loop getting attracted is experiencing stronger magnetic field

10.

The magnetic field due to the outer loop is into the paper. The force on an element of the inner







loop has the direction given by ds  B where ds is the element length in the direction of current. Thus the force is radially outward. Since the force is symmetric, i.e. radially outwards everywhere and the loops are concentric, there is no net force on the inner loop.


3

11.


Equivalent diagram is as shown in P is moved 2 cm right them R1=12, R3=3

R1 R 2 =  Hence wheat stone will be balanced  R2 R4 5 20 R R If s is moved left cm then R4  hence 1  2 (hence wheat stone will be balanced) 3 3 R3 R4 12.

I   t 1 2 t  Hence I   2 2I

15.

2

f  T

f increases by increasing T. i.e. f 2  f1  3  443 Hz So

16.

v f 340   0.94 360

17.

Pressure will increase, equally at all points

18.

The compression will be proportional to g eff

SECTION – C 1.

By linear momentum conservation impulse (J) = mV. By angular momentum conservation, angular impulse So

mv

J

  I 2

 mv mv 6v  I  or      6 rad / s 2 2 2I  m   2   12 

2.

2F M

Mg



For equilibrium

  2F 1 2 mg 100 F   25 N 4 4 Mg



Chemistry

4

PART – II SECTION – A

1.

log

P2 ΔH  1 1  =  -  P1 2.303R  T1 T2 

P2 9720  1 1  =  760 2.303×2  373 348  P2 =298torr

log

298-Psol. 0.1 0.1 = » 1000 1000 298 0.1+ 18 18 Psol =297.46torr Lowering in VP=(298-297.46)=0.54torr A

2.

54044'

C H3C

B

CH3

CB AC Terminal carbon carbon distance= 2  CB

Sin(540 44')=

=2  1.54  Sin(540 44') =2.51 A 0 3.

0.5.mol of CH3COONa and 0.5mol H2O is formed.

Tf = 4.

0.5  2  5.04 20.5 18 10 3

100g H2O2 per hr. (100/34 )mol H2O2 per hr. (100/34 )mol (NH4)2S2O8 per hr.

100×2 mole e - per hr 34 i =157.68amp current required=315 amp 5.

0.059 log Q 1 0.059 0=E 0 cell log K 1 E cell =E 0 cell -


5

0=(x-0.799) -


0.059 log(6  10 8 ) 1

x  0.37 6.

SO 3 2-  2H +   SO 2 + 2H +

7.

A, B, D is having plane of symmetry.

8.

Intercept=E 0 Cu 2+ /Cu =0.34

0.059 log[Cu 2+ ] 2 0.059 ECu/Cu 2+ = -0.34 log 0.1 2 ECu/Cu 2+ = -0.3105V ECu/Cu 2+ =E Cu/Cu 2+ -

9. 10.

12.

In isothermal process

ΔU=0 ,

35.6 mole Mg 24 35.6 (  2) mole of e 24 35.6 (  2  96500) coloumb 24 C4H10,CH3CH2OH, NH3 and I2 have high BP so low vapour pressure.

(13 – 14). heat FeSO 4 .7H 2 O   FeSO 4 (B)+H 2 O strong heat FeSO 4 .7H 2 O   Fe 2 O3 (D)+SO 2 (E) +SO3 ( F )

(15 – 16)

H 2 O 2 +2KI+ H 2 SO 4   K 2 SO 4 +I2 +2H 2 O I2 + Na 2 S2 O3   2NaI + Na 2 S4 O 6 4mmol of hypo means 2mmole of I 2 and 2 mmole of H 2 O 2 . so molarity=

2 4 , normality   0.16 N 25 25

And volume strength=

2  11.2V=0.896V 25



6

17.

C2 and C5 should having opposite configuration for S to be meso.

18.

C2 and C5 should having opposite configuration for S to be meso it is possible with D only. CH2OH O

(OH)H

H H H H O (OH)H CH2OH

(Q)

S,Meso(C2 and C5 have opposite configuration)

SECTION – C 1.

Pyrolusite   MnO 2 Iron Chromite   FeO.Cr2 O 3 Siderite   FeCO 3 Cassiterite   SnO 2 Calamine   ZnCO 3 Argentite   Ag 2S Lime stone   CaCO 3 Chalcopyrite   CuFeS 2

2.

NH 3 ,SO 2 ,Ca  OH 2 ,NaCl

3.

I,ii,and V are correct

4.

Sc3+ &Co3+ are diamagnetic Sol.

 2

1

1 CH2

1

1

CH3 CH2

 2

1


7

Mathematics


PART – III SECTION – A

9

1.

8

1 5      b 1  5   1  0 a   2   2   



 

29 a  28 b 5  1 







9

5 1  0

2a  b 5  1  76  34 5  2a  b  76, b  34  a  21 2.

3.

2ab  5a  b  55  2a  1 b  5   55  5 2 2   ( 2a  1)(2b  5)  105  3.5.7 Total number of positive solutions  2  2  2  8 Total number of solutions  16 . f n  2 n  3n n

 f n 1  2 f n  3 and f n  2  2 f n 1  3.3 So option ‘c’ is correct.

4.

Pn 

3!.3n  3!.(n !) 3

(n  1)!3 (3n )!

n

6 n3 n  3n  3n  13n  2 

 lim

2 9 ac   2 bx  2 y  7b  0   ac  2     2 bx  2 y  7b  0 b  7   x  y  b x    0 2  

5.

7 7 Line passes through  ,  maximum distance from origin 2 2  7  minimum is zero.  0,  2  6.

3

2

2

7 7 7 and      2 2 2

2

Let roots of the equation x  13 x  54 x  72  0 are, a, b, c

 a  13  ab  54


8


13 2 x 3  13 x 2  54 x  72  ( x  a )( x  b)( x  c )

abc  72 or

S

3

2

 13   13   13     13   54   72  ( S  a )( S  b)( S  c ) 2 2 2 35  ( S  a )( S  b)( S  c ) 8   S (S  a)(S  b)(S  c)

455 4

 7.

Clearly  is repeated root of f(x)=0  f(x)  f (x)  f (x)   f(x)  f (x)   lim   lim  1        = 1–0=1 x   f (x)  f(x) x   f (x)  f(x)    f(x)        f (x)  f(x)  0&  is bounded function   xlim     f (x)  f(x)   

8.

Clearly if f(x) = a(x–)(x–) then g(x) = b(x–)(x–)  h(x)  k(x  )2 (x   )2  h(x)h(x)  2k(x  )3 (x   )3 (2x    ) So distinct roots of 2

9.

d (h(x)h(x)) =0 are 4. dx

2

x  y  x  p      3  z z  z  To factories make   0

y  x  y      2   q  0 z z z 2

2

1 3 1  3 p  1  q  2   1   p  12  1     q   0 2 2 2  2 and also for perpendicularity p  q  1  0 

4 p 2  p  14  0 p  2, q  3,

10.

Let

 

( 4 p  7 ) ( p  2)  0 7 3 p ,q 4 4

dy  p, we have dx p 2  y  xp 2 pp' p  p  xp' 2 p  x or p'  0 2

which givens x  4 y or y  cx  ( 2c  1) or y  1  c( x  2) 11.

Fro more than one triplet satisfying all three equations we have


9

1 a 1 b 1 c


1 c 1 0 a 1 b  abc  ( x  y  z )  a satisfying the triplets 2 We get a  10, p  7    2  8  0   2 and 4. 12.

1 b 1 c 1 a





1 1   1 (Where lim t k  ). k   4 4

If f (x ) is continuous then lim f (t k )  f lim (t k )  f  k 

k 

1 0 4

Also as graph is smooth f  a

13.

h(n)   2 | sin nx | dx o

a

Let

 0

0

1 , n

kI



1/ n

k

k  , n

2 | sin nx | dx   2 | sin nx | dx 0

k 2(1  cos n)  n n h(n) 4( a  ) 2(1  cos n)   a a n 4 2(1  cos n) 4     na a h(n) is independent of n if   0 . a i.e. an  k  a  kN 4

a should be divisible by 2, 3, 4 i.e. minimum (a) = 12. a

14.

h(1)   sin 2 x. | cos x | dx 0

h(1) 1a  lim  sin 2 x. | cos x | dx a a a a 0 Let 0 a  n    lim

n  

 lim

n





sin 2 x. | cos x | dx  lim .

0

n 

   1   n  sin 2 x | cos x | dx   sin 2 x. | cos x | dx   n    0 0 

1 2 2 sin x | cos x | dx   0 3


10


15.

16.

17.

 5 –A

4 P 

A

 3

B

Either the stripes at odd number are painted with basic colour or at the even numbers. Total ways  3

18.

C

AP  3 cos  In APC 3 cos  5  sin  sin(   A) 25  cot   12 1 .PB.4. sin  PBC 4 2   . tan  PAB 1 3 .PB.3. sin( 90º  ) 2 4 12 16  .  3 25 25 n 1 2

4

n 1 2

3

n 1 2

4

n 1 2

 712

n 1 2

tn an



No of ways of painting n stripes



bn



No of way of painting n stripes when it ends with a particular basic colour (R, B or G) No of way of painting n stripes when it ends with a particular non basic colour ( R / B, B / G or R / G )

t n = 3(a n  bn ) a n = 2a n1  bn 1 bn = a n1

(i) (ii) (iii)

Equation (ii) and (iii) givens

bn1  2bn  bn 1 2

On solving characteristic equation x  2 x  1  0 we get



bn   1  2



n 1



 1 2



n 1

b2  1  1      2     n 1 1 1    bn   1  2  1 2 2 2 n n 3  tn   1  2  1  2  .  2





 







n 1

 



SECTION – C 1.

Required plane contains z-axis and normal of 3 x  4 y  z  1  0 hence its equation is

4 x  3 y  0 distance from (1, 2, 3) is 2.

46 16  9

 2 units.

On adding and subtracting the curves we get

( x  3) 2  4( y  1), y 2  20 y  59  0 respectively


11


Let

point of intersection be ( x1 , y1 )( x 2 , y1 )( x3 , y 2 )( x 4 , y 2 )



y1  y 2  20 and y1 , y 2  0 2

We required  ( x1  3)  ( y1  2)

2

  4( y1  1)  ( y1  2) 2

  y1  2( y1  y 2 )  40

3.

   A   

1 0 0  0 2 0 0  0  3 0 0     n 0 0  0 

   2 A    

1 0 2 0 3 0  n 0

   A2    

2 0 3 0  n 0

2

 2n 0  0 0  0 3

  0 0  0 0    0 0 0  0  0  0   2 1 2 2 2 3  2  n    0  0   0 0 0  0  0   0    0 0 0       0  0   0 0 0  0  0  0    i2 i 0 0 0  0     0  0   0 0 0 0  0   0    0      0  0   0 0 0 0  0   2

2

2

1 0 0  0   2 1 2 2 2 3  2 0 0  0   0 0 0 3 0 0   0 0 0    n 0 0  0   0 0 0 2 1 2 2 2 3  2  n   0 0 0  0  0 0 0  0    0 0 0  0 

     

 2n  0  0

      0 

  A   i.2  Trace A   i.2   a    i.2  1 2 3 lim Trace A   lim  i 2      ..... = 2 2 4 8 A 2   i.2 i A



 i n 1

n



i n

 i n 1

n



ii

n 1/ n



i

n 

1985

4.

1 0

      

1

1985

 x 1

n

1985

f ( x  1)   ( 1) x 1 x  2  f ( x) x 1

x 1

 As f (1)  f (1986) 1985

 3

x 1 1985



1985

 f ( x)   (1)

x 1

x

x 1

 f ( x)  331 . x 1

Sum of the digits = 5.

3  31  7 .

From the figure

A

  3  129 8 cos 2 120º   2



 = 43º

2



3



l3


Sol-PAPER-2

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