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JEE(Advanced)-2014 ANSWERS, HINTS & SOLUTIONS FULL TEST –I (Paper-1)

ALL INDIA TEST SERIES

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FIITJEE

Q. No.

PHYSICS

CHEMISTRY

MATHEMATICS

1.

B

B

B

2.

B

C

A

3.

A

D

A

4.

B

D

A

5.

A

A

C

6.

A

D

B

7.

C

B

A

8.

D

B

C

9.

B

B

A

10.

A

A

A

11.

D

B

C

12.

B

B

D

13.

B

A

C

14.

A

B

A

15.

B

C

C

16.

C

A

C

17.

A

B

A

18.

C

A

B

19.

C

C

C

1.

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

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

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

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

A → (r, s) B → (r, s)

2.

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

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

A → (q) B → (r)

3.

C → (p, r) D → (p, q)

C → (s) D → (p)

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

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2

AITS-FT-I-(Paper-1)-PCM(S)-JEE(Advanced)/14

Physics 1.

PART – I

Now downward force on the right block is more.

T

T

mg

2. 3.

mg

IC = I0 + M(OC)2 = I0 + M(OB2 + BC2) = IB + M(BC)2 Potential difference across AC is zero as IAC = 0 5  2I = 0 I = 2.5 A Let the resistance of part BC be r Applying KVL 10 + 5  2I  Ir  I = 0 2.5r = 7.5  r = 3  As resistance of part AB = 9  Length AC = 66.7 cm

E1 = 10V r1 = 1

A

G

C

E2 = 5V r2 = 2

5.

Initial energy of electron = 2eV Energy after formation of hydrogen atom in the ground state = 13.6 eV. Energy released = 2  (13.6) = 15.6 eV hc   793 Å 15.6

6.

Threshold wavelength = 5000Å hc Work function = = 2.48 eV  K.E. = eV = 3 eV hC   2258 Å 5.48

7.

B

V 2 3002 V2 R= 2   900  R P 100 V 300 1  current i   = A R 900 3 Let L be the required inductance, then 500 500 1 1  or = 2 2 z 3 3 (900)  x 2 P

XL  1200 L

1200 1200   W 2f

1200  4H 150 2  


3


8.

In steady state potential difference across each capacitor = E

9.

H=

4

14.

15.

E2 0 R dt 

4

 0

 6t 

2

12

dt

= 64 J

q1 + q2 = 0 kq kQ kq2 vA = 1   R 2R 4R kq1 kQ kq2 vC =   4R 4R 4R vA = vC  q1 = Q/3 and q2 = Q/3

q2 Q q1 A

C B

Q  Q  Q Q vA = k       3R 2R 12R  160R

Q  5Q  Q Q vB = k      48 R 6R 2R 12R   0 di 17-19. When current is maximum 0 dt  emf across L = 0 and potential difference across the capacitor will be same. V From conservation of charge V  + +  3CV + CV = 6CV0  CV0 C 3C 5CV0 L  V= 4 Loss in energy of capacitor = energy stored in inductor 3V0 3C  Imax = 2 L

16.

S


4


Chemistry 1.

PART – II

Ag  I  AgI  e 

Eo  0.152 V

Ag  e   Ag

Eo  0.800 V

Ag  I  AgI

Ecell = 0. 952 V

At eqm, Eo 

2.303RT 1     Ag  I  K SP

2.303RT logK SP  0.059 logK SP F  16.13

 0.952 

 logK SP 2.

3.

A 2  t   on differentiating the equation we get: dA   1 2AdA  dt or   A dt 2A 2 Hence order is -1. rA  1   0.5 . As it lies in the range 0.414 to 0.732. AB has octahedral structure like that of NaCl. rB 2





 a  2 rA   rB  2 1  2   6 pm 3

4.

Volume = a = (6 pm)3 = 216 pm3 2    BaF2  s    Ba  aq   2F aq  K sp  Ba2   F 

2

 F 

Again K sp  2  Ba2   F   F  

5.

Co3 

K sp Ba2  

2

K sp 2 Ba2  

1s2 2s2 2p6 3s2 3p6 3d6 3d6

4p

4s

4d

sp3 hybridisation

6.

Bond order of N2 ,N2 ,NO  ,NO,CN and CN are 3, 2.5, 3, 2.5, 3, 2.5 respectively. Higher is bond order smaller is bond length. Bond order of CO and CO+ are 3 and 3.5. Br

7. MgCl

Br2 /h 

Cl2 /h CH4  CH3 Cl  

8.

C KCN  

N

COOH 

H3 O  

O CH3 CH2 O Br  

H

C6H5


5

O


O O Br 

Br

C6H5

CH3

 

C6H5

C6H5

9. O O3 / Zn  H2O

 

O

15.

Y

H

O K

+

1   3  Ho  H CaSO 4 . H2 O  s    H  H2O  g  2   2  1 H CaSO 4 .2H2 O  s    833 kJ mol = + 484 kJ for 1 kg Ho  Ho  So = 17920 J mol-1 Go  2.303RTlogK p

Go  7.22  10 4 pH2O 2.303RT  8.1 10 3 atm

logK p  pH2 O

16.

OH

KOH   Cannizzaro reaction



O

14.

H

O





3 2

pH2O  1, K p  1 Go  0 at eqm. Ho  TSo  T 

Ho  380 K So

= 107oC


6


Mathematics 1.

PART – III

x2 + y2 – 2x – 2y – 1 = 0 x 2 + y2 –1=0 Common chord is –2x – 2y = 0  2x + 2y = 0 y = –x 1   1 Points of intersection of circles are   ,   2 2  4

k k 1 8

10  2 5  8  7 14

2.

P(E) 

3.

The line can be written as y = mx and curve as x 2 + y2 = 4



C2

Let, C(h, k) be a point on the circles and A  3, 1 be given point then,

B 1

h2 3  3



3, 1 A

 h  3  2 3 k2  m 3  k = 3m – 2 Now, this point (h, k) lies on the circle 2

 

2

C(h, k)

(, m) (0, 0)

y = mx

  3  2 3    3m  2   4 2

9 2  12  12 3  9m2  2  4  12m  4  9 1  m2   2  12  3  m  12  0

3 1  m2   2  4  3  m   4  0 2

16  3  m   4  3 1  m2   4   0



2

3  m   3 1  m2   0

3  m2  2 3m  3  3m2  0 2 3m  2m2  0 2m2  2 3m  0 m   0,

4.

3

Suppose m is an integer root of x4 – ax3 – bx2 – cx – d = 0 as d  0 m0 (I) m > 0 m4 – am3 – bm2 – cm = d  d = m(m3 – am2 – bm – c) dm Also, m4 – am3 = bm2 + cm + d 3 2 m (m – a) = bm + cm + d  m > a  contradiction (II) m < 0 m = –n  n4 + an3 – bn2 + cn – d = 0


7


n4 + n2(an – b) + (cn – d) > 0 So, contradiction Hence, equation has no integral solution 5.

(I)

+

+

1 3  2 2 + + 1

(II) x – 3x + 2

+ – 1

–

1

– – 1

1 3  2 2

1

1 3  2 2

+

1

2

+

1

1 3  2 2

2

1 1  x    1  2 4

 x

–

1

1 2

x2  x 

2

– + 1

2

+ + 1

1 3  2 2

2 –2x –1

1 (III) x  1  x  1  2 1 3   set x can be  ,      4 4

1 2 1  x 4

 

/2

6.

Required area, A  4



y

0

/2

= 4

 0

1 1 1 2 3  x 4

2x 

2x 

y

dx dt dt

(0, a) 0

x

 a sin t   a cos2 t  dt  sin t 

(0, –a)

/2

= 4a

2

 cos

2

t dt  a

2

y

0

7.

Distance from origin, D  x2  y2  z2 where P(x, y, z) is any point on the curve D2 = x2 + y2 + z2 = x2 + y2 +

2 2  2xy  4 xy xy

 D = 2 and occurs at point(s) 1, 1, 2  and  1,  1, 2  , 1, 1,  2  ,  1, -1,  2 

8.

Differentiating w.r.t. ‘r’, 2r dr = 2a cos 2 d, a   rdr   r

r2 sin 2

r2 cos 2 d sin 2

d  tan2 dr


x

8


So, differential equation of orthogonal trajectory, r 

d   cot 2 dr

dr r  ln(cos 2) = 2 ln r + k  r2 = c cos 2

Solving  tan 2 d 

9.

Let f  x   x 3  bx 2  cx  1.

f  0   1  0, f  1  b  c  0



So,    1,0  . So, 2 tan1  cosec   tan1 2sin  sec 2 



 1  1    1  1  2sin   1  2 tan1    tan    2  tan    tan  sin    2 sin  sin     1  sin          2       as sin   0   2

1

2

10.

is continuous every where except where 1 + z = 0 1  z2  z = i so when |z| < 1 then above points are excluded so f(z) is continuous

11.

Continuity : for any point z, lim f  z   lim x  x 0  f  z0  . Non differentiable : for any point z, z  z0

f '  z   lim

f  z  z   f  z 

z  0

12.

z

y1  cos  m sin 1 x  

 lim

z  z0

z  0

x x x  0 and lim 1 now lim x 0 x  iy y 0 x  iy x  iy y 0

x 0

m 1  x2

1  x2 y1  mcos msin1 x   1  x 2  y 2  xy1  m2 y  0

 n  n  1    2 2 yn    xyn1  n.1.yn   m2 yn  0  1  x  yn 2  n  2x  yn1  2  2    Simplifying we get, 1  x y n 2  2n  1 yn1  n2  m2  yn  0 13.

dy 3at  2  t 3   dt 1  t 3 2 1 3  t  dx   6a  2 2 dt 3 1  t 

Now

dy 1/3   at t  (2) dx

       14.-16. Here, Pv  P2  PK so Pv will make acute angle with all the vectors from P2 , P3 , ....., Pk Pn–2 Pn–1 Pn

Pk

Pm P3 P2 P1


9


    So, OPv   OP2  OP3  .....  OPk 1   0

Again, if n is odd  n = 2k – 1    OP1  OP2k 1  OPk     Now OPk will make acute angle with all the vectors from OP1, OP2 ..... OP2k 1         OPk   OP1  OP2  .....  OPk  OPk 1  .....  OP2k 1   OPk  OPk  1           Now, OPk   OP1  OP2  .....  OP2k 1   OPk OP1  .....  OP2k 1 = OP1  OP2  .....  OPn     Again, V makes acute angle with all the vectors OP1, OP2 , ....... OP7        So, V  S  OP1 + OP2 +.......+ OP7  1  V  S  1 17.

Eigen values are roots of the equation A – X = 0  (A – I)X = 0  |A – I| = 0 8   4  0 2 2  ( – 4)( – 6) = 0   = 4, 6

18.

When  = 6 8  6 4   x1  2 4   x1  0       2 2  6   x 2  2 4   x2  0    2x1 – 4x2 = 0  x1 = 2x2  2  X  C   1

19.

|A – I| = 0 1  0 1  1 2 1 0 2 2 3 3 2   – 6 + 11 – 6 = 0   = 1, 2, 3, For  = 3 1 X  C  1  which is orthogonal  2 SECTION – B

1.

2

2

2

(A) Assume sphere as, x + y + z + 2ux + 2vy + 2wz = 0 Now P, Q, R are (–2u, 0, 0), (0, –2v, 0), (0, 0, –2w) x y z So, equation of plane is   1 2u 2v 2w 1 2 3 Since it passes through (1, 2, 3)   1 2u 2v 2w If centre is (x, y, z)  (–u, –v, –w) 1 2 3  locus is    2 x y z 2 2 (B) Assume equation of sphere as, x + y + 2ux + 2vy + 2wz = 0


10


It passes through (, 0, 0), (0, , 0), (0, 0 )   r  u   , v=  ,    2 2 2 2

2

2

           1  2  2 2 2 2 2  +  +  = 4 if (x, y, z) are coordinates of centroid    4  x  , y  , z   x 2  y 2  z2  3 3 3 9 (C) P  (, 0, 0), Q  (0, , 0), OPQ = 15º   tan 15º = ….. (1)  Now if sphere is made with PQ as diameter (x – )x + (y – )y + z2 = 0 2 2 2  x + y + z = ax + by ….. (2) A plane through PQ parallel to z–axis is x y  1 ….. (3)    x y Using (1), (2), (3) we get x 2 + y2 + z2 = (x + y)     

Radius = 1 

z2 0 2 (D) Using family of planes any plane through line of intersection is x + y – 6 + (x – 2z – 3) = 0 1  6  3 Now, its distance from centre of sphere is radius  3 ,  = 1,  2 2 2 1     1  4  z2 = xy(tan 15º + cot 15º)  2xy 

 Equation is x + 2y + 2z = 9; 2x + y – 2z = 9 2.

  y  z , sin (y – z)  0 2 1 1 p  cos z sin  x  y   sin  x  y    cos2 z 2 2 As, sin (x – y)  0 and sin (x + y) = cos z 1 1  2 3 p  1  cos 2z    1  cos   4 4 6 8

(A) Let p = cos x sin y cos z. As,

(B) The equation can be written as 3u2  8u  3  0  u1  Clearly, u1, u2 are negative so

8  2 7 8  2 7  u2  6 6

  x1, x2   2

   3    < x1 + x2 < 2 as cot x tan x = 1 = cot x cot   x  = cot x  cot   x 2   2  3 7  x1  x 2  and another pair, x '1  x '2     x '1, x'2  2 2 2  x1 + x '1 + x2 + x '2 = 5


11


(C) |EF|  |E1F1| = a – [|BF| cos B + |CF| cos C] |DE|  C – [|AE| cos A + |BD| cos B] |FD|  b – [|CD| cos C + |AF| cos A] 1 |DC| + |CE| = |EA| + |AF| = |FB| + |BD| =  a  b  c  3  |DE| + |EF| + |FD| 1 1  a  b  c    a  b  c  cos A  cosB  cosC     a  b  c  3 2 1 So, minimum value is 2 2 2 2 (D) sin x1 + sin x2 + ….. + sin x10 = 1 cos xi 

2

 sin

A E F B

F1

D

E1

xj

j 1

 sin x For each 1  i  10 we have cos xi 

2

 sin

xj 

j i

10



 i1

 3.

10

cos xi 

 i1 j i

sin x j 3

10



9 i1

j

j i

3

sin x i 3

cos x1  cos x 2  .....  cos x10 3 sin x1  sin x 2  .....  sin x10

(A) Property of ellipse (B) Normal chord is made between points A  2, 2 2  and B  8,  4 2  so its length is 6 3 (C) Q  (4, 2), S : (x – 4)2 + (y – 2)2 = 16 then required circle is (x – 1)2 + (y – 1)2 + (x – y) = 0 162 Here,  = 9 and radius = 2 2 3y (D) x 2  c 2 5 5 e (c > 0) and e  (c < 0) 3 2


C

fiitjee

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