Answer Solution (3)

7/27/2019 Answer Solution (3)

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FIITJEE L t d . , F I I T JEE Hou se, 2 9 -A , Ka l u Sa r a i , Sa r va p r i y a V i h a r , New De l h i -1 1 00 16 , Ph 461 06000 , 2 65694 93 , Fa x 265139 42 websi t e : w ww . f i i t j e e .c om

ANSWERS, HINTS & SOLUTIONS

FULL TEST – II (Main)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. A A B

2. B C A

3. C B A

4. B D A

5. D B B 6. C C B

7. A C B

8. C C C

9. A. B B

10. C B B

11. D D D

12. C B B

13. C C A

14. A B A

15. B C B 16. D B A

17. A C A

18. C B A

19. A C A

20. C B C

21. A B D

22. D B C

23. A D D

24.C B B

25. D D C

26. B B C

27. B D C

28. D D A

29. C C A

30. D D A

A

L L

I N D I A

T E S T

S E

R I E S

FIITJEE JEE (Main)-2014

F r o m C

l a s s r o o m / I n t e g r a t e d S c h o o l P r o g r a m s 7 i n T o p

2 0 , 2 3 i n T o p 1 0 0 , 5 4 i n T o p 3 0 0 ,

1 0 6 i n T o p 5 0 0 A l l I n d i a R a n k s &

2 3 1 4 S t u d e n t s

f r o m C

l a s s r o o m /

I n t e g r a t e d S c h o o l P r o g r a m s &

3 7 2 3 S

t u d e n t s f r o m A

l l P r o g r a m s h a v e

b e e n A w a r d e d a R a n k i n J E E ( A d

v a n c e d ) , 2 0 1 3



AITS-FT-II-PCM(S)-JEE(Main)/14


2

P P h h y y s s i i c c s s PART – I

1. Avg. speed2 2

r mr (v t) (v t)3

t

2

r v 5 9 r v 2m/ s

2. avg

1 t 1v v 3v2 2 2vt 4

3.m m

g T a2 2

…(i)

T cos 60 =ma

cos 60

…(ii)

Solving (i) and (ii) acceleration of ring =2g9

4. Work done by all the forces on the block equal to change in kinetic energy.

6. No effect of ‘a’ and ‘g’ on time period of spring pendulum.

7.3 1

1

= R

2 2

1 1 8R1 3 9

2 1

1

= R 2 2

1 1 3R

1 2 4

3 1

2 1

2732

8. Work done by frictionx

0

dxF ds mgcos

cos

= mg x = 20 J

N

f

mg dxcos

ds

dx

dyds

9. There will be no change. h = h h

A

10. Conservation of energy 21 3mv mg mg m 3 2

2 2 2 2

g

v 8g 60





3

11. Potential across capacitor is zero, hence energy stored is zero.

12. vA – vB = E. dx = 10 (1)= 10 volt.

13. Z = 2 2R X = 29 X

but cos =R 3Z 5

X = 4 .

14. =0

vA – vB =0 0

(b) (a)

W = Q(vA – vB) =

0

Q(b a)

15. = 0 + 1 10 = 10 rad/sec2 v = r = 1 10 = 10 m/s

03

q(v r )B

r

| B |= 02

qv4 r

B =7

2

10 0.1 10(1)

= 10-7 T

16. P = AT4

4

A B B

B A A

P A T

P A T

16 =4

B

A

T2

1 T

TB = TA(8)1/4

Since mT = constant, 1 4A B

B A

T8

T

B =

A1 4 1 4

5000

8 8

Å

17. For adiabatic process, TV1 = constant

T1

m

= constant

1

T

= constant

1 1 1T 31

= 4/3

f =2 2

641 13

18. P(x, y), A(3a, 0), B = (3a, 0)

VPA =0

1 2Q4 PA

, VPB =

0

1 Q4 PB

According to equation





4

0 0

1 2Q 1 Q4 PA 4 PB

= 0

2 1PA PB

4PB2 = PA2

(x 5a)2 + y2 = (4a)2

19. i = 5A2 2

2mq q 1Li

2C 2C 2 qmax = 6 C

21. f = ma2

=2f

I mR

a2 = F/4m, f = F/4

F

a1

f

f

a2

22. The process is equivalent to TV1/2 = C

Compare with TVx1 = C x = 3/2

C =R R

1 1 X

=

R R2 3 1 3 2

=3 1

R 2R R2 2

23. Intensity will be highest at the nearest point.

28. tan =2 2u u 100 10

RRg gtan 10 3 3





5

C C h h e e m m i i s s t t r r y y PART – II

14. 19hcE 2.9 10 J

Total energy of 10 quanta = 10 × 2.9 × 10-19J

Energy stored for process3

1923

112 4.18 107.8 10 J

6 10

% efficiency19

19

7.8 10100

29 10

= 26.9%

15. a b

b a

r M2r 1 M

rms

TV

M (As

rms

3RTV

M )

rms a a b

b arms b

V T M 2 2 2 2

V T M 1 1 1

16. 1 1

1 1 2

k E 1 1n

k R T T

... (i)

2 2

2 1 2

k E 1 1n

k R T T

... (ii)

Eqn. (ii) – Eqn.(i) 2 12 1

2 1 1 2

E Ek k 1 1n

k k R T T

(For equimolar formation of B and C, 2 1k k )

1 2

2 2k T 3008314n k 8.314 300 T

n2 2

2

T 30083148.314 300 T

T2 = 329.77 K.

17.

3 2 22SO g 2SO g O g

at equilibrium 1 2x 2x x

Only SO2 will oxidized.Equivalent of SO2 = Equivalent of KMnO4 2x × 2 = 0.2 × 52x = 0.5

2

c 2

0.5 0.252 2

K 0.1250.52

18. b bT i k m 2.08 = 0.52 × 1 × ii = 4It means salt on dissociation gives 4 ions. Thus the salt that gives 4 ions is K3[Fe(CN)6].





6

19. 2 o o1 1 1Cu e Cu ; E X V G FX ... (i)

3 1 o o2 2 2In 2e In E X V G 2FX ...(ii)

2 o o3 3 3In e In E X V G FX ... (iii)

From Eqn. (i) + (iii) – (ii)2 2 3 o o

In Cu In Cu G FE

o o

1 3 2G F X X 2X FE

o1 3 2E X X 2X

20. Due to Fajan’s rule.23. Green ppt. is – Cr(OH)3

25.

CH3 C

O

OH2

HVZBr /P C OH

O

2BrCH3

KCN

H O

Y

||

2

O

OH C CH COOH

X Malonic acid

26. 1o aromatic amine on diazotisation followed by coupling with -napthol gives azo dye test.

28. (A) Viscosity intermolecular attraction intermolecualr H-bonding tendency

(B) No intramoecular H-bonding in the cis-isomer due to loss in planarity as a result of steric andelectronic repulsion. stability symmetry.

(C) Even though o-nitrophenol shows intramolecular H-bonding but has greater – R and – I effectthan Cl.

(D) H3BO3 + 3 C2H5OH B(OC2H5)3 + 3H2O. No cage lattice structure of H3BO3.

29. Because phenol is less acidic than H2CO3.30. S is less electronegative and therefore more reactive towards SN2.





7

M M a a t t h h e e m m a a t t i i c c s s PART – III

1. Given equation can be written as |x – 1|2 + |x – 1| – 6 = 0t2 + t – 6 = 0 t = 2 or –3 |x – 1| = 2

x – 1 = 2 or x = 3, –1

2. |z1 – z2|2 + |z2 – z3|

2 + |z3 – z4|2 + |z4 – z1|

2 = 2(|z1|

2 + |z2|2 + |z3|

2 + |z4|2) + |z1 + z3|

2 + |z2 + z4|2 8 + |z1 + z3|

2 + |z2 + z4|2

3.

r 4 22 2

r r T

r 4 r 2 4r

=

2 2

r

r 2r 2 r 2r 2

=2 2

1 1 14 r 2r 2 r 2r 2

= 2 2

1 1 14 r 1 1 r 1 1

n 2 2

1 1 1 1S 1

4 2 n 1 n 1 1

51 1 1 1 1 38 37 26 1

S 14 2 26 37 4 26 37

=

1 1380 3454 26 37 26 37

57 338

37S 1326 26

4. Number of terms = Number of non–negative integral solution of the equation p + q + r + s = 50= 50 + 4 – 1C50 =

53C50 =53C3

5. If a function is symmetric about two mutually perpendicular lines, it must be symmetric about theirpoint of intersection.

6.5 5

2 26 6x y dy dya x ydx dx1 x 1 y

5 5

2 2

6 6

y dy xay ax

dx1 y 1 x

6 2 6 3

6 2 6 3

1 ydy x a 1 x xdx 1 x y a 1 y y

..... (1)

Also,

6 6

3 3

6 6

1 x 1 ya x y

1 x 1 y

or 3 3 6 6x y a 1 y 1 x

6 3 6 3a 1 y y a 1 x x

Hence, from equation (1),6 2

26

1 ydy xdx y1 x

7. For a triangle formed by joining any three co–normal points, orthocentre consides with the fourthpoint.

8. HM of SP and SQ = 2 3 6

43 6

= semi latus rectum.





8

9. If a circle cuts two circles orthogonally, radical axis of the two circles pass through the centre ofthe first circleHere radical axis of given circles is 6x – 4y + 4 = 0 or 3x – 2y + 2 = 0

Hence 3 – 2 + 2 = 0 3 14 2 2

10. sin =21

21

r r r r

= 2 sin –1

21

21

r r r r

O1 O2

r 2

r 1

11. Let jbiar a2 + b2 = 1 …(1)Also,

cos 45° =| ji||r |

) ji(r

a + b = 1 …(2)

cos 60° =| j4i3||r |

) j4i3(r

3a – 4b =

25

…(3)

There exists no real values of a and b satisfying (1), (2) and (3)Hence no such unit vector exists

12. (1 + x)n = ....3240380383 43 = 1 + nx + 2 x)1n(n2

+ . . . .

On comparison, n = –3 and x =23

13. From the given conditions,abxy

1,

bcyz

1,

acxz

1

x + a y + b z + c The determinant is a symmetric one. The determinant will be equal to zero ifx + a + y + b + z + c = 0

but a + b + c = 0 (given) x + y + z = 0

x + z = 2 2y x, z,2y are in A.P.

14. For x =1, y = a + b + c

Tangent at (1, a + b + c) is c1x2b

axcbay21

Comparing with y = 2x, c = a, b = 2(1 – a)





9

15. x (2, 3) –1 < x2 – 4x + 3 < 0, so f(x) is increasing in (–1, 0)

f(sin x) is increasing onn I

(4n 1) ,2n2

16. Let point P on the straight line x + y = 4 be (m, 4 – m), this will be nearest to the parabola if at

this point to the straight line becomes normal to the parabola.Let it is normal at x – 10 = t2, y = 2tPerpendicular to x + y = 4 at (m, 4 – m) is y – (4 – m) = (x – m) …..(1)Normal at parabola at (t2 + 10, 2t) is y + t(x –10) = 12t + t3 …..(2)

(1) and (2) are same t = –1, m =172

so required point is17 9

,2 2

17. a a c 3b 3 b

a . a c sin 3.12

3 = a . a c sin

3 = 3.2 sin

sin =

1

2

cos

2

=

3

4

18. (x) =

x 2 2

2 x 2

e 2cos2x 2xsec x

ln 1 x cosx sinx

cosx e 1 sinx

+

x 2

2 x 2

e sin2x tanx

1sin x cosx

1 x

cosx e 1 sinx

+

x 2

2 x 2

e sin2x tanx

ln 1 x cosx sinx

2xsinx e 2xcosx

= B + 2Cx + ….

Put x = 0, B =

1 2 0 1 0 0 1 0 0

0 1 0 1 0 1 0 1 0

1 0 0 1 0 0 0 1 0

= 0

19. Let g(x) be the inverse of f, then f(g(x)) = x 3g (x)(g(x)–2) = x (g(x))2 – 2g(x) – log3 x = 0

g(x) = 33

2 4 4log x1 1 log x

2

Since g : [1, ] [2, ]So g(x) = 31 1 log x

20. Since 2x 3x 2 0 0 1 2tan x 3x 2 <2

Since 24x x 3 0 0 < 1 2cos 4x x 3 2

0 < L.H.S < The given equation has no solution

21. Let f = (5 2 – 7)19 x – f = an integer [x] + {x} – f = an integer {x} – f = an integer, but – 1 < {x} – f < 1 {x} = fSo, x{x} = x.f = 119 = 1





10

23.1x

= 2e1 2 3

...3 ! 5 ! 7 !

= 2eS

S =r 1 r 1 r 1

r 1 (2 r 1) 1 1 1 1(2 r 1)! 2 (2 r 1)! 2 (2 r)! (2 r 1)!

= 1 1 1 1 1 ...2 2 ! 3 ! 4 ! 5 !

=12

e1 =1

2 e

1x

= 2eS =2 e2 e

= 1

So,1

0

f (y) logy x dy = 0

24. Let any point on second line be (, 2, 3)

cos = 642

sin = 642

OAB =12

(OA).OB sin =12

3 . 14 6

42 = 6

= 2So, B is (2, 4, 6) B(, 2, 3)O

A (1, 1, 1)

25. Centroid of triangle will bea b c

3

Now line joining the orthocentre and the circumcentre is divided by centroid in 2 : 1 ratio

internally, so orthocentre will be a b c .26. If we draw the graph of tan x and cot x, we

observes that range of f (x) is [–1, 0) [1, )

So f (x) =1

3 does not have any root.

27.300

1SS n

300

1

31

1

3

11

31

1

1 n

001

3

111

23

n

32

3001

3

1n

450

1

3

1n

3n > 450

Least value of n = 6





11

28. f(x) =x 4

|t x| |t x|

0 x

e dt e dt =x 4

x t t x

0 x

e dt e dt =x 4x t t x x 4 x

0 xe e e e 2

f (x) = ex – e4–x = 0 x = 4 – x x = 2f(0) = f(4) = e4 – 1, f(2) = 2(e2 – 1), so maximum value of f(x) is e4 – 1.

29. Given inequation can be rewritten as2sec x 2 2y 23 y

3 9 1

2

2sec x 1 1

3 y3 9

1

Now,2

2sec x 1 1 1

3 3 and y3 9 3

So, we should have sec2 x = 1, y =13

x = 0, , 2, 3

30. arc(AC) = 3, arc(AB) = 4, arc(BC) = 5Let radius of the circle be ‘r’ r = 3 = 3 /r, = 4/r, = 5/r

Now, 3 + 4 + 5 = 2r r = 6/ 6r

1

ABC = OAC + OAB + OBC

=

r 5

sinr 4

sinr 3

sinr 21 2

=

65

sin3

2sin

2sin

36.

21

2

B

A

C

O

= 22

313921

23

118

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Answer Solution (3)