Aiits 2016 Hct Vii Jeem Jeea Advanced Paper 1 Solutions Solutions

1

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AIITS-HCT-VII (Paper-1)-PCM (Sol)-JEE(Advanced)/16

JEE(Advanced)-2016 ANSWERS, HINTS & SOLUTIONS HALF COURSE TEST–VII (Paper - 1)

Q. No.

PHYSICS

CHEMISTRY

MATHEMATICS

1.

B

A

C

2.

B

A

C

3.

A

A

A

4.

C

A

B

5.

B

A

A

6.

A

D

C

7.

A

D

A

8.

A, C

A, B, D

A, B, C

9.

A, C

A, B, C, D

A, C

10.

B, C, D

A, B, C, D

A, B, C, D

11.

A, B, C, D

A, B, C, D

A, B

12.

B

C

C

13.

A

A

A

14.

C

A

C

15.

D

B

B

16.

A

A

D

1.

(A  p); (B q); (C  s); (D  p) (A  s); (B  p); (C  q); (D  r) 7

(A  t); (B  s); (C  p); (D  q, r) (A  q, r); (B  p); (C  s); (D  t) 7

(A  q); (B  r); (C  p) (D  s) (A  s) (B  p); (C  q); (D  r) 4

2.

1

2

6

3.

3

1

1

4.

4

4

3

5.

2

4

0

6.

6

2

1

1. 2.

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2 AIITS-HCT-VII (Paper-1)-PCM (Sol)-JEE(Advanced)/16

Physics

PART – I

Hints & Solution SECTION – A 1.

2.

Pressure in the air inside the column of mercury is equal to the weight of mercury over the air divided by the internal cross sectional area of the tube. When the temperature increases, the weight of the upper part of the mercury column does not change. That is why the pressure in the air is also constant. For the isobaric process, the change in volume is proportional to the change in temperature. The same is true for the lengths of the air column. lT l T   l = 0 = 11 cm. l0 T0 T0

v  w u Frequency received by car B is f1   f v  w u Now the car B will be treated as a source of frequency f1  wavelength of reflected sound received by the driver of car A is

 v  w  u  v  w  u v  w  u  f1  v  w  u f  330  5  25 330  5  25   31 m 36  330  5  25  300

' 

3.

Hence (B) is correct vmax F F  a  0 e Kx   v dv  0  e Kx dx m m0 0 v max 2 F F   0  K max  0 2 mK K

4.

Let initial location of center of mass is 0 then find location of cm  (90 cm)(5 cm)   (10 cm)(45) Xcm = = 9 cm  (90)  (10)

5.

As escape velocity is

6.

The power developed by the man = power required in the reference frame of the man = Force  velocity = m(g + a).(h/t) Hence, (A) is correct.

7.

x + y = 100 Heat lost = Heat gained 5 5 y × 3.36 × 10 = x × 21 × 10 y 21 = x 3.36 336y = 21x × 100 Using (1) & (2) 210000 y = = 86.21 2436

2 times of orbital velocity.

…(1)

…(2)



8.

9.

10.

For a plane wave intensity (energy crossing per unit area per unit time) is constant at all point For spherical wave p 1  I=   2  2  r  4r 2 Acceleration of block is = 10 m/s 1 2 1 4  Displacement s = at = × 10 × = 2m 2 2 10 Tension in the string is 40 N N 20 N

N = 50 – 20 = 30 N Limiting friction force = µN = 12 N and applied force in horizontal direction is less than the limiting friction force, therefore the block will not slide. For equilibrium in horizontal direction, friction force must be equal to 5 N.

5N

mg = 50 N

From the top view, it is clear that  = 37° i.e. 127° from the x-axis that is the direction of the friction force. It is opposite to the applied force. Contact force =

2

N f

2

=

925 N

f  53 3iˆ  4ˆj

11.

P = 3 V3 , for ideal gas PV = nRT (A) relation between V and T

 nR  V × (3V3 ) = nRT  V4 =  T  3  (B) Relation between P and T PV = nRT

V4  T

1/ 3

12. So





P  P   = nRT  P4/3  T 3 (C) For expansion V  increases work-done : positive internal energy : increases Hence, heat will have to supplied to the gas. (D) As T  V4 with increase in temperature, volume increases Hence work done is positive. 2 2 So  = 0.8 and effective length of air column = 0.99 + 0.01 = 1m.  0.8   5   4  l = 5 , so five half loops will be formed 4   = 5   so second overtone  4



13.

Pex = – B

d dx

2 2 = (5 × 105) × (0.1 × 10–3 ) sin (y + 1cm) cos 2(400) t 0.8 0.8 2 = (125  N/m2) sin (y + 1cm) cos 2(400t) 0.8

14- 16.

m 2 m 2 m 2 . .v 0    v1 4 3 3 4 3 3v 0 8  2v 0   3 4 2 v1   v0 3 v v1  0 2 mg 1 1  cos    I2 2 2 SECTION – C

1.

2.

3.

P

 0.16  10 6   0.16  106  3 6 =    1000  4.2  80  10     1000  2.3  10 = 7 watt. 60 60    

h h T    N  > 0 2 4 N T >0 2 Mg Tmin = 2 2 a = g/2 = 5 m/s . mg – T = m(g/2) Mg g = m  mg  2 2 Mg  mg  =    2  2 m =1 M

The acceleration of center of mass of the system    m a  m2 a 2  m3 a 3 acm = 1 1 m1  m2  m3 

The net force acting on the system    = m1a1  m2 a2  m3 a3

a3

a2

a1 A

B

C

m1

m2

m3

 Fnet = (m 1 a1 + m 2 a 2 – m 3 a 3) 1   = (1)(1)  (2)(2)  (4) N = 3 N. 2  



4.

Rmax 

u2 g

2 So, maximum area covered = Rmax  .

5.

u4  4m2 g2

Velocity of block w.r.t. ground 

V1G = 20iˆ  15ˆj 

 V1P = 15jˆ Velocity of block w.r.t. ground after 10 sec. is zero (since block will stop w.r.t. board after 5 sec.) 



V '1P = V1P   gt ˆj  0 (t = 5 sec) 

 6.

V '1G = 20 ˆi

Area under graph =

1

 v dx  t



Chemistry 1.

2.

E

PART – II

Molecular weight = 98/1 = 98 Re placed H

He

B :

He

C : +

2+

Sr , Ba are relatively easily formed in order to achieve completely filled valence shell orbitals. 2

3.

 Ag  NH3  K c 1    Ag NH3   2 

2

K sp NH3  1 K sp   Ag  Cl    K c  Ag  NH3    Cl   2   



F

4. XeOF 2 :

Xe

O

F

Hence, or

5.



h 6.626  10 34  = 3.321  1010 m 31 6 mv  9.1 10  2.19  10  CH3

6. H3C

CH2 C

CH3

  HC CH CH3   3 HBr / H2 O2 or HO _ Br H  H2O  Br 

CH2 C

CH CH3 Br

HBr

CH3 H3C

CH2 C H

7.

CH CH3 Br Br

1 O2  g   CO2  g 2 1   G0  G0  CO2   G0  CO   G0  O2  = 394.4   137.2  0  =  257.2 kJ 2  

CO  g 

 Spontaneous (ve sign) But G0  H0  TS0  257.2 = H0  298 K  0.094 



0  H  285.2 kJ  exothermic ( ve sign)

8.

Allyl carbocation is less stable than benzyl carbocation due to less number of resonating structures.

9.

In (A) Cl and O In (B) S and O In (C) only O In (D) P and O

10.

OH

Br (A)

11.

(B)

(C)

(D) CHI3

(E) CH3COO

(E) CH3COOH

The balanced equation is K 2Cr2O7  14HCl   2KCl  2CrCl3  7H2O 3Cl2 14 36.5 g

294 g

3 71g

Mass of K2Cr2O7 =

96  61.3  58.848g 100

30  320  1.15  110.4g 100 Hence, K2Cr2O7 is limiting reagent.  Mass of Cl2 produced = 42.63 g M Also equivalent of K2Cr2O7 = 6 7M Equivalent of HCl = 3

Mass of HCl =

16.

All the given molecules are cyclic isomers with molecular formula = C5H10 Number of pairs of diastereomers = total number of pairs of stereomerms – number of pairs of enantiomers = 3C2 – 1 = 2 CH3 H C 3

is th missing cyclic isomer.

SECTION – C 1.

OH OH OH OH

HO

HO

HO

OH

OH

OH

OH

HO

OH

OH

x = 14 so x/2 = 7



2.

O

O Pure

O

Cr

Cr O

O

O O As these two does not participate in resonance.

3.

It will produce most stable free radical.

4.

meq of KOH = 20 meq = meq of acid 20 0.45   n  factor 1000 90 n-factor for acid = 4 = number of replaceable proton.

5.

[H2CO3] in blood = 4M [NaHCO3] = 10 M Volume of blood = 10 mL Let the volume of NaHCO3 used = x mL 4  10 H2 CO3  in mixture  x  10  

NaHCO3  in mixture  pH  pK a  log

10  x  x  10 

salt acid

7.4  7  log 4  log

10  x  x  10   x  10  4  10 

 7.4  6.4  log

10x 40

10x 10x  1 so  10 40 40 x = 40 mL Hence x/10 = 4   2NH3  g  N2  g   3H2  g   log

6.

1

3

0 0 2  Ratio of initial and final volumes = 4/2 = 2



Mathematics

PART – III SECTION – A

1.

2. 3.

4.

C Since O(0, 0) does not lie on 11x + 7y = 9 hence it is the equation of AC,  2 7 M 7x+2y=0 solving OC with AC we get C    ,   3 3 5 4 A solving OA with AC we get A   ,   3 3 A (0, 0)  1 1 4x+5y=0 hence M = mid point of C & A   ,   2 2 hence equation of OB  y = x. The curve remains unchanged when y & x are interchanged.  1 2 2 Let the circle be x + y + 2gx + 2fy + c = 0, if  mi ,  lies on it then  mi  4 3 2 m + 2gm + 2fm + cm + 1 = 0 and m 1 m 2, m 3, m 4 are its roots.  1 1   1    2    2  1 ...   2  r  1 r     2 r Coefficient of x is    r! r

5.

6.

7.

8.

B

11x+7y=9

r

1.3.5...  2r  1  1  1 2r 1.3.5...  2r  1  2r ! r = .   2 r! r! r!r! 2r 1 1 3 9 27 2     ...   2. 3 2 8 32 128 1 4 x 1 1 a  1  a  1  a2 1  a4 1 a







i.e. 1  ax 1  1  x 8 (multiply by (1 – a) on both the sides)  x = 7. ABC can speak in 10C3  1 ways other 7 in 7! ways 10! so totally 10C3 . 7! ways  . 6 Putting x =  in the equation, 0 = a0 + a1  +a22 +a3 + . . . Putting x = 2 in the equation, 2 0 = a0 + a1 +a2 + a3 + . . . Putting x = 1 in the equation, 3n = a0 + a 1 +a2 + a 3 + ... adding (1), (2) and (3), n 3 = 3(a0 +a3 +a6 + . . . .)  a0 + a3 +a6 + . . . . = 3 n-1 (option C) subtracting (2) from (1), 2 0 = ( -  ) (a1 –a2 + a4 –a5 + . . . ) Since  - 2  0 , a1 +a 4 + a7 +. . . = a2 + a5 +a8 +

.... (1) .... (2) .... (3) .... (a)

...

. . . (4)



Also from (3) – (a), a 1 + a2 +a4 +a5 + . . . . = 3n – 3 n-1 = 2.3 n-1 . . . .(5) n-1 From (4) and (5) , a1 +a 4 + a7 +. . . = a2 + a5 + a 8 + . . . = 3 = a0 + a3 + a6 + . . 9.

Clearly the inscried triangle is equilateral. 2

z2

2

i i z  z0 z  z0  2 e 3, 3 e 3 z1  z0 z1  z0

 z2 = -1 +i(2 +

3 ) and z 3 = -1 + i(2 -

3)

z0(2i) z3

10.

b is the H.M. of a and c and their G.M. = ac G.M > H.M  ac > b n n n n And A.M. of a and c > G.M. of a and c an  c n  an c n 2 an  c n  2

11.

12.



ac

n



 2bn

Put n = 100, 3, 5, 2 A A A A A A =sin – cos if sin cos 1  sin A = sin  cos 2 2 2 2 2 2  A 5 A  ie    0    . 4 2 4 2 2 6 can be written in the following ways using 1’s and 2’s only. Six 1’s in 1 way 5! Four 1’s and one 2’s in = 5 ways 4! Two 1’s and two 2’s in

13.

z1(2 +2i)

4! = 6 ways 2! 2!

Three 2’s in 1 way. So f(6) = 1 + 5 + 6 + 1 = 13. We have f(6) = 13, so f{f(6)} = f(13). Now 13 can be expressed in the following ways: Thirteen 1’s in 1 way 12! Eleven 1’s and one 2’s in = 12 ways 11! Nine 1’s and two 2’s in

11! = 55 ways 9! 2!

Seven 1’s and three 2’s in Five 1’s and four 2’s in

9! = 126 ways 5! 4!

Three 1’s and five 2’s in One 1’s and six 2’s in

10! = 120 ways 7! 3!

8! = 56 ways 3! 5!

7! = 7 ways. 6!

 f(13) = 377. 14–16. x – y + 2(1 + ) = 0  (x + 2) – (y – 2) = 0 pass through (– 2, 2) x – y + 2(1 – ) = 0  (x – 2) – (y – 2) = 0 pass through (2, 2).



Clearly these represent the foci of ellipse, so 2ae = 4. 2

2

2

2

2

The circle x + y – 4y – 5 = 0  x + (y – 2) = 9 represents auxiliary circle thus a = 9  e =

2 and 3

2

b = 5.

1

SECTION – B (A). |a + ib| |z| = |z| |(a – 1) + ib| 1 1 2 2 2    a  1  b2 and a + b = 2 2 1  1 – 2a = 0  a = 2 1 1 2 and b = b= 4 2  a – b = 0. (B) We must have either 6a > 2a . 3b or 6b > 2a . 3b. If b = 0, then 6a > 2a  a = 1, 2, 3, 4. b If a = 0, then 6b > 3  b = 1, 2 suppose a > 0 and b > 0  if 6a > 2a . 3b thus 6 a > 2a . 3 so 2a > 2a (not possible) a b b similarly if 6b > 2 . 3 than 3b > 3 not possible  only solutions are 2, 4, 8, 76, 3, 9. z z (C). Re     0, 1 , Im    0, 1 4 4 means that if z = a + ib than a, b  (0, 4) 4 4a 4bi Now   a  ib a2  b 2 a2  b 2 a 2  b2  0 < a, b < 4  (a – 2)2 + b 2 > 4 and a2 + (b – 2)2 > 4 so we want area inside the square and outside the two circles  area = 16 – 4 + (2 – 4) 12 – 2. (D). 3x + 4y + z = 5  let a  3iˆ  4ˆj  kˆ  b  xiˆ  yjˆ  zkˆ      a  b  a b 5

9  16  1  x 2  y 2  z2

 26  x 2  y 2  z 2   2

25 26

(A)

sin  cos x  = sin(cosx)  0 2n  cosx  (2n + 1) 0  cosx  1    x   2n  , 2n   .  2 2

(B)



cos  sin x   cos(sinx) > 0

1



   sin x   x  R. 2 2 (C) tan( sinx)   sinx   2 1 sinx   2   x  n  6  x  R – n  . 6 (D) ln(tanx) tanx > 0  0 < x< 2  n < x < n – . 2 





SECTION – C 1.

2.

4.

 1 log 5   2 1

4  1/ 4 1   5 log5 4     1 2 5  1 2 7 c 7      ,     2  2   2 2 4 7 7 49 7 i.e.            i.e.   2x    c = 6. 4 2 4 4 The given expression can be written as k3 = k 3  sin3 A cos  B  C    sin2 A  sin 2B  sin2C  2 = k 3  sin A sinB  sin A cosB  cos A sinB  x



5. 6.

1



= k 3  sin A sinB sinC  k 3 3 sin A sinB sinC  3  ab c  . Line ‘L’ passes through the centre of second circle. a 1 1 1 b 1  0  a(b – 1)(c – 1) – (b – 1)(1 – a) – (c – 1)(1 – a) = 0 1 1 c



a 1 1    0. 1 a 1 b 1  c


Aiits 2016 Hct Vii Jeem Jeea Advanced Paper 1 Solutions Solutions

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