Aits Ft i (Paper 1) Pcm(Sol) Jee(Advanced)

AITS-FT-I-(Paper-1)-PCM(Sol)-JEE(Advanced)/16

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1

ANSWERS, HINTS & SOLUTIONS

FULL TEST I (PAPER-1)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

1. C D B

2. C D C

3. B A D

4. B C A

5. B C D

6. C C A

7. B A C

8. C A, B, C A, B, D

9. A, B, C, D B A, B, C

10. A, B, C B, C B, C

11. A, B, C A, C A, B 12. A C C 13. B B D

14. C A D

15. D A B

16. A D A

1. (A) s, (B) pt, (C) q, (D) pr (A) r, (B) pt, (C) qs, (D) qs

(A) prt, (B) q, (C) t, (D) s

2. (A) pqt, (B) prt, (C) s, (D) qr (A) pqs, (B) q, (C) rt, (D) q

(A) qt, (B) q, (C) prs, (D) t

1. 3 4 1 2. 5 7 2 3. 9 3 1

4. 9 3 5

5. 2 2 7

6. 4 3 2

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2

PPhhyyssiiccss PART I

SECTION A 1. C

1

1

1tan tane

vertical component after collision eucos& Horizontal component usin

usin 1tan tan taneucos e

2. C

2

ZeroT f 2aTR fR 2Ra R

f 0.

Ring T

af

3. B

1 2

1 2

1

t /RC t /RC1 2

t 1 1R C C1

2t

RC1

2

V VI e & I eR R

I eI

Ie

I

4. B

2

4

1 1f , f 502 LC 2 LC / K

f 50 50 100K K 1 2 1f f 10

K 1.01

5. B

o o1 2

o o0 0D AC CB

1 2

0D

r 2Rsin30 , r 2Rsin60 R 3 R

i iB B B sin60 sin30

4 r 4 ri

B4 R 3

C R R D

A

B

30o

30o30o

30o

r1r2i i

6. C

2

V V VBeats4 4 4

V4



3

7. B

o o3 sin60 sin902

3 3,2 23 3

4

8. C

13.6 10.2 3.4eV 1st collision& 3.4 15 11.6eV 2nd collision.

9. A, B, C, D

3/2

1Vn

1P.E.r

T r1T.E.r

10. A, B, C

2 22s s

R

L C

V 3 8 4 25 V 5 voltVsZ 5I

V 3 1 R R 3R 3P.F. 0.6Z 5

V V I lags V.

11. A, B, C

1 1 1 v v1f v u f u

since m is positive in the graph,v vmu u

v vm 1f u

1intercept on m axis 1 & tanf

12. A

2 2

2 2P

V , OP x yR

origin is ins tan taneous centre of rotationVV r OP x yR



4

13. B

2 22

c

c

o

a R

V Va R RR R

2aa

tan 12a

45

acanet

a R 2a

14. C

Thermal heat capacity, oQ 300C 15J / CT 20

15. D

Specific heat capacity, oQ 300S 600J / kg Cm T 0.025 20

16. A Molar heat capacity

o

Q 300n T 25 20

5030J / mol C

SECTION B 1. (A s) (B p, t) (C q) (D p, r) In wrist watch spring force is not dependent on gravity.

mT 2K

2. (A p, q, t) (B p, r, t) (C s) (D q, r) In nuclear reactions energy, charge, mass and momentum remain conserved.

SECTION C 1. 3

2 2 omax o max

max

2 2o

2 omax

20 sin 45for R , 45 H 10m

2 10H allowed 5m,

20 sin5 30

2 1020 sin60

R m 20 3m.10

2. 5

2 3 2cylinder cylinder3 3 2

sphere sphere

5remaining portion cylinder sphere

1M R 3R . I 3 R R2

2 2 2M R , I R R3 5 3

29I I 2 I R .30



5

3. 9

o18 sin30 r 1

9 m / s.

18 m/so18cos30

o18sin30

1m

0.5m

o60

4. 9

1 2

1 2

1 2

1 1 1F f f

1 1 1 1 11 14 424 20 20 20 203 3

5 .9

5. 2

22 200V 200 VI 2.5A & P 500 watt

R 80 R 80

When the wire is divided into two equal parts and connected in parallel we get maximum power.

max

22

maxeq

Resis tance of each part 40I 5A

200VP 2000W 2kw.R 20

6. 4

AP PB

AP

A P P

Q QV , V3 6

Q Q12 Q 24 C3 6

24V 83

V V 8 V 4volt.



6

CChheemmiissttrryy PART II

SECTION A

1. D

B BCH3 CH3

CH3CH3 H

H

methylationB2H6

B B

H

HH

H

H

H

2. D

OH

OH

OH

OH

OH

OH+

H

OH

OH

CH2+

OH

OH

H

+H+

-H2Oring. exp

OH

OH

H+/excesssimilar repetation

3. A

At nodal surface, = 0

2 2rz z

= 1

4. C

Co ordination no of Fe+2 in D is 6

Na+

Fe+2

D

(a sandwhich compound)-

-

-C



7

5. C

+ O

O

O

AlCl3

O

OHO

Zn(Hg) HCl

OH O OCl

AlCl3

O

NaBH4

OH

H2SO4

Br

A B C

DEFG

H

ONaOH

heat

SOCl2

NBS/CCl4

6. C

(a)

OH

O O

+ O-H

O-H

-

H OH

(b)

OH

O OHD

O-D

-

D - OD DO- O

D -

D - OD

OD

D

H

D

D

O-

OD

D

HD

DD

D

D H

O

-OD

D - OD

DO-

DO - DDO-

ODD

D

D

OD

D

D D

H

D

-D D

D

DD

O

-

DO - D

O-D

D - OD

D

D

D

D

D D

O

D

D

DDD D

DD

O

Repeat last two step



8

O

H

H

OHO-H

O

H

HO - H

O-H

O

(d)

7. A

Synergic bonding. 8. A, B, C

o2 2 1

1 21 1 2

a2 2 11 2

1 1 2

K T THlog K &K are eq. constan tK 2.303R T T

EK T Tlog K &K are rate cons tant

K 2.303R T T

9. B

CH3

CH3 N+

CH3

Me

MeCH3

OH-

CH2

CH3

CH3(Major product)

Due to greater probability of losing -H (9 : 3 ratio).

10. B, C

(b) conjugate base of 2 4H PO is 24HPO

(c) pH of 0.1 M NaCl (aqueous solution) = w1 pK2

11. A, C

3A 3e Al

Number of moles of Al produced 4.5 0.16627

Number of moles of electrons used 3 0.166

21H e H2

Number of moles of H2 produced 1 3 0.1662

Volume of H2 produced 1.5 0.166 22.4 5.57 5.6

12. C 2nd step is r.d.s. in sulphonation reaction. 13. B

2nd step is fast step in nitration.



9

14. A

aP V b RTV

on converting into virial equation from, we get

2 32

2 4

abb bRTPV RT .......

V V V

to higher power of n (i)

For Boyles temperature, 2nd virial co-efficient becomes zero

2

2

22 2

B

B

i i B

abRT

aTRb

aTRBaT T T

Rb

15. A

Let Boyles temperature, 'BT

8 2 2

2

B

'B B

aTRba aT T

R b Rb

16. D

b2 = 0.04 b = 0.2 lit/mol

Vreal = 4 22.4 + 4 0.2 = 90.4 lit

SECTION B 1. A r , B p,t , C q,s D q,s .

(A) Sic covalent carbide (Not hydrolysed)

(B) 4 3 2 43Al C 12H O 4Al OH 3CH

(C) 2 2 2Cal 2H O Ca OH H C C H

(D) 2 3 2 32Mg C 4H O 2Mg OH CH C CH



10

2. A p,q,s , B q , C r,t D q .

32 4 43

3 34 4

22

Al SO 2Al 3SO

0.2M 0.3MAlPO Al PO

0.1M 0.1MUrea Urea

0.1MMgCl Mg 2Cl

0.1M 0.2M

SECTION C 1. 4

Based on fact

2. 7

OH BO

B-

OB

-

OO

BO

OH

OH

OH 8H2O2Na+

3. 3

OHC

Cl

CHO

NO2

CHO

C

N

, ,

4. 3

H3CO

OCH3OCH3

H3CO OCH3

NH2

OCH3, ,



11

5. 2

99 90

99

90

1 1 100t n100; t nK K 10

t n100 2 2t n10 1hence x 2

6. 3

S

S H

H

O

OH

H

, ,C C C

Cl

H

H

Cl

0 0

0



12

MMaatthheemmaattiiccss PART III

SECTION A 1. B

2 2

2

2 2

2 2

2 2

PQ r 1 2r cos r 1 r3

& AT r 1 2rcos

Where 'Q' is the angle between

OA and OT.

now, PQ 4 AT

1 r r 4 r 1 2rcos

3r r 3 3r r 3cos 1 1; r 08r 8r

7 13 7 13r , .6 6

P

Q

A

T

3

2. C Here two balls are drawn and the result is known that the two balls are white, therefore Bayess

theorem should be used.

Required probability = 1.1 14

1 1 6 1 3 1 1 2.1 . . .4 4 10 4 10 4 10

3. D

0

0 0 0

0

0

0

Let Q x f x g x in x xBy LMVT,Q x Q x Q' x Q' x for some C x ,x

Q x 0, Q' C f ' C g' C 0

Q x Q x Q x

F' C g' C x x 0

Q x 0 f x g x .

4. A

A

55

13

(-3, 4)AP = 23

P



13

5. D

n

5 2n 5S4 2 n 1 n 2

5K4

6. A

2

2 2

2 2

2 2

Equation of tangent at 'P ' isY y m X x

y mxNow, x1 m

y x dy putting y vx2xy dx

Solving Differential equationx y CxIt passes through 1,2 C 5

Equation of curve x y 5x 0

7. C

22 2 2

2

2

2

xcos ysinT 13 2

x 0, y 2cosec ,chord A 'P;

2siny x 33 cos 1

2sinx 0, OMcos 1

Now, OQ MQ OQ OQ OM

2 OM OQ OM

OM 2OQ OM

2sin 4 2sincos 1 sin 1 cos

2sin 4 4cos 2sin

1 cos sin

4sin 2 2cos

2

2

1 cos

1 cos sin4

8. A, B, D

3 2 2

32

f x 0 x x x 1 x x 1 2 0

2x 1x x 1

-2 -1 12

3x



14

9. A, B, C Take f x x and g x x and proceed. 10. B, C

1 1

2 2

cos2xI tan dx tan tan x dx1 sin2x 4

x dx4

x xI x C g x x .4 2 4 2

11. A, B

22 4 4 a 1, 1 a, 1 a

2f 1 a 0 & f 1 a 0

1a ,14

y

12. C The value of K1 & K2 decide the opening of parabola

max

min

A Area OBCO

A Area OACO

13. D

c

X'

Y' K2 = -1

(1,0) (2, 0) X

Y

A B

o

21K4

11K4

11K8

14. D

2

2

Area BFEC Area ABC AFE

or AOB BOC COA AFE1R sin2C sin2A sin2B sin 2A21R sin2B sin2C2

F

B C

E

A

H O

2A 2Rco

sA



15

15. B

AE R,AF 2RcosAEF R

16. A

2

2

AFE3

AFE is equilateral

R 3 9sum of square of altitudes 3 R .2 4

SECTION B

1. (A p, r, t) (B q) (C t) (D s)

1 2 3 4 1 2 3 4A tan 0 2n2

22

2

B x 4 t 2, TR is diameter 6

Area of circle TR 94

C 6

22A B CD y y y 3 3

2. (A q, t) (B q) (C p, r, s) (D t)

SECTION C 1. 1

2 2

D.C's of OP are,1 1 1, ,3 3 3

1 1 1D.C's of AL , ,3 3 3

1 1 1D.C's of BM , ,3 3 31 1 1D.C's of CN , ,3 3 3

Let D.C's of required line be ,m,nm n m ncos ,cos3 3

m n m ncosy ,cos3 3

Hence, cos cos co

2 2 2 2 2

2 2 2 2

4 4s y cos m n3 3

4cos cos cos y cos 1.3

L

Z

YN

XO

M(a, o, a)(o, o, a)

(a, o, o)

(o, a, o) (a, a, o)

(o, a, a) (a, a, a)

A

C

P

B



16

2. 2

2

1

The gives equation is x 1 y 2 4XY 4

Equationof tangent is,4 4Y x tt t

8A 0, & B 2t 0t

1 8Area 2t 82 t

4K 8 k 2.

A

Q B

P 14tt

3. 1

2 2

2

a b 2 3CE 25DE 16 16C

CE 1DE

4. 5

1 1 1 110

K 1 0 0 0 0

1 2 3 10

0 1 2 9

10

0

f K 1 x dx f x dx f x 1 dx.......... f x 9

putting x 1 t, x 2 t............ x 9 t

f x dx f x dx f x dx.......... f x dx

f x dx 5.

5. 7 Every ball have two options 4 balls can be put in 24 ways. Arrangement among themselves in 2!

4 31 2 22!

But, above count also includes the one case in which all the balls are put in one box, Required number of ways = 23 1 = 7. 6. 2

3 3 3K 1 K 2 K 3K 1 K 2 K 3 0

1 1 1

K 2 K 2

Aits Ft i (Paper 1) Pcm(Sol) Jee(Advanced)

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