JEE Advanced Full Test II Paper 1 Answers

JEE(Advanced)-2014 ANSWERS, HINTS & SOLUTIONS FULL TEST – II PAPER-1

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PHYSICS

CHEMISTRY

MATHEMATICS

1.

A

A

C

2.

C

A

D

3.

B

A

A

4.

C

A

B

5.

C

A

B

6.

A

A

C

7.

A, B, C, D

A, B, C, D

A, C, D

8.

A, B

A, C, D

A, C

9.

C, D

A, B, C

A, B

10.

B, C

A, D

A, D

11.

B

B

C

12.

B

A

B

13.

B

A

A

14.

C

C

A

15.

A

B

C

16.

B

C

D

17.

A

C

C

18.

A

B

B

19.

C

D

C

20.

D

C

D

21.

C

D

B

22.

D

B

D

23.

C

A

A

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2 AITS-FT-II (Paper-1)-PCM(Sol)-JEE(Advanced)/14

Physics 2.

3.

4.

PART – I

(v 1cos)t = (v2cos)t = 30 3 1 1 (v 1sin)t  gt 2 +h = (v 2sin)t  gt 2 2 2 from (1) and (2) h = 60 m

. . . (1) . . . (2)

1 2  =  tan   g 2 2 u cos2  1 2  =  tan   g 1  tan2   2 2g 4   tan2  4 tan  +  1  =0    If particle will not hit the target. (b2  4ac) < 0 4   16  4  1  <0    4 > 3 mv B2 At B, mg sin  = r Using energy considerations 1 mv B2 = mgr(cos   sin ) 2 From (1) and (2) mg sin  = 2mg(cos  sin ) 2cos = 3 sin 

(, )

u 2g

 

Smooth

. . . (1)

A

. . . (2)

Here the position y on the screen will correspond to maxima. n D y d 30,000  when n = 1, 2, 3, 4, ………….. n

8.

a = kx 1 da  2kx  tan 60 = 2k 3  k  dx 2 x2 a  2 dv x2 v 2 x3 v     C1 at x = 0, v = 3 m/s dx 2 2 6 9 C1  2 x3 v2  9 3

2

3 9


N

B

 

7.

hence a = 1.5 and v 



u

vB


2T cos   h = 59.6 mm rg here 59.6 mm is greater than the protruding part hence water will rise in the capillary of insufficient height 25 mm. 2T Now, R   0.6 mm hrg

9.

Capillary height h 

10.

10. Fluid particles passing through the bend are in circular motion. The centripetal acceleration is provided by the variation in pressure. In the section as shown P1A  P2 A  maC

maC

P2A

2 P1A

 P1  P2 applying Bernoulli theorem at 1 and 2 1 1 i.e., P1  1v12  gh1  P2  v 22  gh2 2 2 1 1 2 2 P1  1v1  P2  v 2 as h1  h2 2 2  P1  P2

1

v1  v 2 15.

mAg sin  = 5  10  sin 37 = 5  10 

3 = 30 N 5

fmax   AB (mC  mB )gcos   48 N hence f = 30 N 16.

mC gsin   2  10 

3  12N 5

fmax  mc gcos   0.1 2  10 

4  1.6 N 5

hence f = 1.6 N 17.

T

T  fA  fC  mB gsin37 = 91.6 N

fC

fa

18.

mBg sin 37

Optical path difference between the beans arriving at P x  ( 2   1 )  dsin  for maxima ( 2   1 )  dsin   n 1  n  (  2   1 )  d   n    sin1 2   1    40   sin  



19.

sin   1 n  1  2   1  1 40   or 20  n  60 Hence number of maxima = 60 – 20 = 40

20.

At C, phase difference 2  (  2  1 ) = 80   Hence for maximum intensity will appear at C. Now for minimum intensity at C  (  1)t  2 t t  500 nm 2(  1)



Chemistry

PART – II

3.

Neither N nor F contain d-orbitals. Further in N2F4 N—N bond is shorter than in N2H4 due to s-character (Bent’s Rule).

4.

Due to poor shielding effect of 3d electron the size of Br  ion nearly same as that of Cl ion.

5. C8 H7OCl

N 2O 287 m

Molar mass of C8H7OCl = 8  12 + 7  1 + 16 + 35.5 = 154.5 g Molar mass of N2O = 2  14 + 16 = 44g According to Graham’s Law of diffusion rN2 O MC8H7OCl 154.5    3.5  1.87 : 1 rC8H7 OCl MN2O 44 1.87 th  287 = 187 row from N2O side 2.87 1.0 th dC8H7OCl   287 = 100 row from weeping gas side 2.87 Therefore, the spectator from the side of N2O in the 187th row will be laughing and weeping simultaneously Alternatively, the spectator from the side of weeping gas in 100th row will laugh and weep. dN2O 

6.

OH attacks at the more reactive (C=O) group, (containing more EWG or less EDG) Et is more EDG than Me (due to +I effect here no H.C.). Therefore OH attack (C=O) with (Me) group. OH

Me

Me

Me O

O C



OH  

O

C

H O

OH

3 

OH

OH

Et

10.

O

O

Et

Et

NH4NO3  aq   N2 O  H2 O NH2 OH  HNO2  N2 O  2H2O Hence, (A) and (D) are the correct answers.

13.

Velocity = 2.18  106 

14.

(i)

Z n

EBrO   M / 5 3

(ii) EBrO   M / 6 3

 ratio = 6/5



17.

Me

Me HNO2

 

Me H2N

N2

 

Me

2 C

N

OH

C

 bond

2 4    breaks Ring exp ansion

+

N

OH

Me

Me 1 OH

3

A 

Me

1

Me

2

Me



 

Me 4 O

3 H

Me O

Me

RCO3H    Me Baeyer  Villiger

H

O

i LiAlH4   ii H2O Me

OH OH

O

(B)

(C) (Cyclic ester)

(D)

(Lactone)

18.

2KIO3

  Pb  NO3 2   Pb IO3 2  2KNO3



0.815  12

350.15 5.25

1 mol of Pb(NO3) reacts with 2 mol of KIO3 5.25 m mol of Pb(NO3)2 reacts = 2  5.25 = 10.5 m mol m mol of KIO3 left = 12 – 10.5 = 1.5 Hence Pb(NO3)2 is the limiting reagent 19.

IO3   1.5 m mol  0.03 M   15  35  mL

20.

Due to common ion IO3

  left in the solution the solubility Pb IO

S=

K sp

0.03 2





3 2

decreases.

2.7  1013  104 9

= 0.30  10–9 = 3.0  10–10 21.

pH range of all these indicators lies in between 2-12. Hence, all are suitable indicators.

22.

Since curve is given for strong acid and strong base hence, pH at the end point is 7.



Mathematics 1.

PART – III

y = ax2 + bx + c, vertex is (4, 2) 4=–

b 4ac  b2 b2 , b = –8a, = 2 + 16a  2 c=2+ 2a 4a 4a 2

2

3

Now  = abc = –8a (2 + 16a) = –16(a + 8a ) 

d 2 = –16(2a + 24a ) < 0  a  [1, 3] da

 |max = –144, |min = –3600  Difference = 3456 2.

x – [x] = {x} x – [x + 1] = {x} – 1 4

 f x dx

2

= 6.

1 1. 1 = 3 2

1

-2

x x

3.

lim

x 0

 x  sin x  0



t 2dt

at

= lim

x 0

-1

1

2

3

4

t 2dt

at x2 = lim x  sin x  x 0 a  x 1  cos x  0

x2 2 = = 1 (given) x0 x 2 a 2 sin . ax 2

= lim

a=4 4.

(3)6 = 729 < 900 and (3)7 = 2187 > 900

5.

Now by property of triangles 1 a RQ = BC = 2 2 c b Similarly PQ = , PR = 2 2 abc Area of ABC = 4R abc That of PQR = 2 2 2 4R ' Also area of ABC = 4 (area of PQR) abc abc  R 2      32R' 16R  R' 1 

A

R

B

Q

P


C


6.

Clearly 2 solutions

B y = 2x –3

y =logx 3

7.

f(x) = cos (|x| + [x]) f(0) =1, f(0 – 0) = cos (–1) = –1

 1 2

1   1  0   cos   = 0 2  2

f(1/2) = cos    = 0, f 

 1  f   0   cos  0 2 2  for x  (0, 1) f(x) = cos x for x  (–1, 0) f(x) = cos  (–x – 1) = –cos x 8.

OP = 5 2 sec 

P1

OP1 = 5 2 cosec  PP1P2 =

100 sin 2

 O

( | PP1P2)min = 100   = /4  OP =10  P = (10, 0), (–10, 0) 9.

P2

Let P(2, –1) goes 2 units along x + y = 1 upto A and 5 units along x – 2y = 4 up to B Slope of PA = –1 = tan 135°, slope of PB =  sin  =

1 5

, cos  =

1 = tan  2

2 5

 A  (x1 + r cos 135°, y1 = r sin 135°) = (2 + 2x  B  (x1 + r cos , y1 + r sin ) = (2 + 5 

10.

P

2 5

, –1 +

1 2 5 5

, –1 + 2 

1 2

) = (2 – 2, 2 – 1)

) = (25 + 2, 5 – 1)

1 b Slope of tangent = a   1 a b



dy dy  dt  dx dx dt b So, 0 a 12.

4 t2   1 < 0 4 t2 

b >0 a

 A  1 P  A  B 1 x P    2 1 P B  x B 2 x +x–10

5 1 1  5 or x  2 2  x is positive 5 1 x 2 x

2

13.

a2  b2  c 2  d2  a b  c  d    4  4 

 8  e 2 16  e2  16 4 2 64 + e – 16e  64 – 4e2 5e2 – 16e  0 0e

16 5

14.

Let A(a, P(a)), B(b, P(b)), then slope of AB = P(a) = P(b) from LMVT  c  (a, b) Where P(c) = slope of AB

15.

Given QT = QA = 1 Let PQ = x, then PT  x 2  1 Then  TQP and  APO are similar triangles x 1 Then OT  OA  x2  1 2(x  1)  1 x   x2  1  8 2 x 1 5 x= 3

P T

Q

O

A

16.

From above, OA = 2 and AQ = 1  coordinate of Q  (2, 1)  equation is (x – 2)2 + (y – 1)2 = 1

17.

ORQ is a right angled triangle. Then O and R are the extremities of the diameter then the coordinates of R(2 3, 2)  Equation of circle (x  0)(x  2 3)  (y  0)(y  2)  0  x 2  y2  2 3x  2y  0



18.

Perpendicular tangents intersect on the director circle of hyperbola and director circle of rectangular hyperbola is a point circle. Hence centre of hyperbola is (1, 1) and equation of asymptotes are (x – 1) = 0 and y – 1 = 0

19.

Equation of hyperbola is xy – x – y + 1 +  = 0 It passes through (3, 2) hence  = –2 Equation of hyperbola is xy = x + y + 1

20.

From the centre of hyperbola we can draw two real normals to the rectangular hyperbola

21.

Taking point (r cos , r sin ), we get 3 2 r2 =   3  2 sin  2   4   rmax =

3 2 3 2

, rmin = 1

Max exist when 2 

22.

Required equation is

     4 2 8

x2 3 2

 y2  1

3 2

23.

Centre of circle be origin and its radius is the length of semi minor axis Hence area = 


JEE Advanced Full Test II Paper 1 Answers

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