JEE-MAIN 2016 (PEN-PAPER MODE)

HINTS & SOLUTIONS FOR CODE-F

CHEMISTRY

1. [3]

p

h

mvph

2mv

2

1E.K

m2

p

m2

vm 222

e.k.m2p

q.v.m2p

eV.m2p

meV2ph

2. [4]

Cl–C–CH–CH–CH 223

3CH

3CH

CCH–CH–CH 23

3CH

3CH

The reaction undergoes via E1 mechanism so the methanol can be attacks on the formed 3

0 carbocation so all

3 products are possible.

3. [1]

CrO2 is a ferromagnetic substance and it is used in radio cassets

4. [3]

LDPE It low grade polythein and it not used in formation of buckets, dust bins it is used in caring bags.

5. [2]n/1p

m

x

n/1p.k

m

x

n/1plogklogm

xlog

plogn

1klog

m

xlog

Y C ‘m’ ‘x’

Only n

1 appears as the slope

6. [3]

C + O2 CO2(g) H = –393.5 (i)

)g(COO2

1CO 22 H = –283.5 (ii)

Asked COO2

1C 2 H = ? (iii)

By taking inverse of egn (ii) and adding in eq

n (i)

= –eqn (ii) + eq

n (i)

= –(–283) + (–393.5)

= –110 kJ

7. [2]

In the Bunsen flame the hottest region is blue which is region

8. [1]

NaO–S–)CH(–CH 23

O

O (sodium lauryl sulphate)

Long hydrocarbon has negative change

9. [2]

C6H12O6 = 12 × 6 + 12 + 16 × 6

= 72 + 12 + 96

= 180 gm mol–1

(wt) H2O = 178.2 gm

(moles)H2O = mole9.918

2.178

(moles)C6H12O6 = 1.0180

18

100

1

10

1.0

1.09.9

1.0

P

P–P0

S

0

P0–PS =

0P10

1

76010

1P–760 S

6.7P–760 S

6.7–760PS

= 752.4

10. [3]

The separation of glycerol from spent lye is taken by distillation under reduced pressure because glycerol at

its boiling point decompose

11. [4]

2NO has SP hybridization.

12. [1]

2H2O2 2H2O + O2

1st order reaction

Kt = Ct

Coln

Ct

Colog303.2Kt

T = 50 min

Co = 0.5 ; Ct = 0.125m

min50k125.0

5.0ln

25

100ln

50k4ln

2ln250

1k

2log303.2250

1k

Rate of reaction = K [H2O2]1

05.0693.0250

1

Rate of formation

of O2 = dt

]O[d 2

dt

]OH[d–

2

1 22

50

693.005.02

2

1

4–1093.6 1–1– mntLmol

13. [1]

[Cr(H2O6]2+ and [Fe(H2O)6]2+

Both Cr2+

and Fe2+

has 4 unpaired es. So they have same magnetic moment

14. [1]

Accn to CIP rule the 2 S and 3 R

15. [2]

A + B C + D Keqb = 100

T = 0 ; 1m 1m 1m 1m

T = teqb ; 1–x 1–x 1+x 1+x

Keqb = 100 = 2

2

)x–1(

)x1(

)x–1(

)x1(= 10

10 – 10x = 1+x

11x = 9

11

9x

[D] = 1+x = 11

20

11

91

= 1.818

16. [2] Froth flotation method used for sulphide ore.

17. [1]

CxHy + nO2 xCO2 + OH2

y2

Give 15mL 375100

20

Volume = 15mL 75mL

Moles = 0.00066 0.0033

Accn = 1 mole 5 mole

Accn to options

C3H8 + 5O2 3CO2 + 4H2O

1mole 5mole

18. [4]

H3PO3 and OH–P–H

OH

O

(orthphosphorous acid)

(pyrophosphorous acids)

19. [2]

Cis-form, forms Non-super imposable mirror images

20. [4]

Zn + 4. HNO3 Zn (NO3)2 + 2NO2 + 2H2O

(conc.)

4Zn + 10. HNO3 4Zn (NO3)2 + 5H2O + 2N2O

(dil.)

21. [2]

Water has intermolecular hydrogen bonds not intra-molecular H-bonds

22. [2]

ppm02.0

pm100ppm4.0

ppm10

Fe

NOPb

F

2

–

3

2

–

So–

3NO is correct

23. [3] In excess O2

Li + O2 (excess) (Not form, peroxide and super oxide)

Na + O2 (excess) Na2O + Na2O2 (both normal oxide and peroxide)

24. [2] This group is present in cysteine amino acid.

25. [3] In Galvanisation the coating is taken by Zn

26. [3] Sc has more ionization energy by high zeff.

27. [3] NaOH4BrNH–C–R 22

O R–NH2 + Na2CO3+2NaBr

28. [2]

1

1i1

RT

VPn

1

i

1

i21

RT

VP

RT

VPnn

1

2i2

RT

VPn )1.....(

RT

VP2

1

i

Moles are constant before and after the temp increased

(n1 + n2)

2

f

1

f

RT

VP

RT

VP

)2.....(T

1

T

1

R

VP

21

f

Eq (1) and (2)

21

f

1

j

T

1

T

1

R

VP

RT

VP2

21

21

1

j

fTT

T.T

T

P2P

21

2jf

TT

TP2P

29. [1]

ClHC–CHClHOCHCH–CH 323

OH

Cl–CH–CH–CH 23OH

30. [1]

1. NBS/hv ResonanceBr

KOH

OH

Sterically hinderdgroup (bulky group)

MATHEMATICS

31. [2]

Solving two lines

06–6

––

01–

05––7

x

yx

yx

x = 1

7(1) –y –5 = 0

Y = 2

Hence (1, 2) is one vertex,

Let (h, R) be another vertex

2–2

2,1–

2

1 hh

( Diagonals bisect each other)

h = –3, R = –6 Hence (–3,–6) is a vertex

Now BO1OC 1–OCOB mm

1–1

2

13–

26–

a

b

2(b + 2) = –a –1

a + 2b = –5 (i)

Also (a,b) his on 7x –y –5 = 0

7a – b = 5 ……….(ii)

Solving (i) and (ii)

A = 3

8–,

3

1ba

Hence vertex 3

8–,

3

1bc

32. [1]

Let 2nd

, 5th, 9

th terms of A.P. are

A +d , a + 4d, a + 8d

(a+4d)2 = (a + d) (a +8d)

[ terms are in G.P.]

a2 + 16d

2 + 8ad = a

2 + 9ad + 8d

2

8d2 = ad

(as A.P. is non constant so d 0)

a = 8d

33. [4]

Let coordinates of P = )4,2( 2 tt

CP = 222 )64()2( tt

CP = 3648164 24 ttt

If CP is minimum, then (CP)2

will be minimum Let (CP)

2 = z

z = 4t4 + 16t

2 + 48t + 36

For z to be minimum, 0dt

dz

and 483216 3 ttdt

dz

16t3 + 32t + 48 = 0

T3 + 2t + 3 = 0

(t+1) (t2–t+

3) = 0

t = –1 as t2 –t + 3 = 0 has non-real roots

Now 3248 2

2

2

tdt

zd

0

1–

2

2

tdt

zd z is minimum at t = –1

P = (2, –4)

Eqn. of circle having centre at P and passing through is

(x–2)2 + (y + 4)

2 = 8

(CP will be radius)

x2 + y

2 –4x + 8y + 12 = 0

34. [3]

0

–11

1–1–

1–1

0)1(1–)1(––)1(1 2

01––1 3

0)1–( 2

1,–1,0

35. [2]

Given f(x) + 2f xx

31

, ).....(0 ix

Replacing ‘x’ by x

'1'we get

)(..........3

)(21

iix

xfx

f

)(..........6

)(41

2 iiix

xfx

f

Subtracting (i) form (iii)

xx

xf 3–6

)(3

xx

xf –2

)(

Now )(–)( xfxf xx

xx

2––

2

xx

24

22

x 22x

2x

36. [2]

xx

xP 2/12 )tan1(

0

lim

)1–tan1log(0

lim22

1

xex

x

x

xe

x 2

tan

0

lim 2

2

1

)(

)(tan

0

lim2

2

x

xe

x

P = e1/2

P =

2

1

37. [3]

Let sin2–1

sin32

i

iz

Ration ling the given complex number, we have

)sin41(

)sin21)(sin32(2i

iiz

Now Re(z) 0 (z is purely imagining)

z + 6i2sin

2 = 0

z – 6 sin2

= 0

z = 6sin2

= 1/3

sin = 3

1

sin–1

3

1

38. [2]

Let hyperbola be 1–2

2

2

2

b

y

a

x

Latus rectum 82 2

a

b 4

2

a

b

)........(4

2

iib

a and also given

)2(2

12 aeb

2b = ae

).......(2

iie

a

b

Also 1–2

2

2

ea

b

1–4

22

ee

Using (ii)

4

–12

2 ee

14

3 2e

3

42e

3

2e

39. [1]

Given S.D is 3.5

variance = (3.5)2 = 4

49

5

72

Now numbers are 2, 3,a, 11

4

16

4

1132 aax

Deviations from mean are

4

16–11,

4

16–,

4

16–3,

4

16–2

aaa

aa

I.E. 4

–28,

4

16–3,

4

4––,

4

–8– aaaa

Variance =

4

4

–28

4

16–3

4

4––

4

–8–2222

aaaa

4

49

416

)–28()16–3()4–(–)–8(– 2222 aaaa

49 × 16 = 12a2 –125a + 112 = 0

49 ×4 = 3a2 –32a + 280

3a2 –32a + 84 = 0

40. [1]

dxxx

xx335

912

)1(

52

Let I = dx

xxx

xx3

52

15

912

111

52

= dxxx

xx35–2–

6–3–

)1(

52

Let txx 5–2–1

dtdxxx )52(– 6–3–

dtdxxx –)52(– 6–3–

I = dttt

dt 3–

3––

cxx

I25–2– )1(2

1

cxx

x235

10

)1(2

ct

2–

– 2–

ct22

1

41. [3]

Time lieszyx

3

4

1–

2

2

3–

In the plan ln + my – z = 9

If line lies in two plan so points of lim (3,–2,–4) satisfy plane

3 l –2m + 4 = 9

3 l –2m = 5 ………(i)

Drs of line & plane are 1

2, –1, 3 l, m, –1

2l –m –3 = 0

2l –m –3 = 3 ………(2)

Sol Eq (i) & (ii)

l = 1, m = 1

So l2 + m

2 = 1

2 + 1

2

= 2

42. [2]

2cos2x cosx + 2cos 3x cosx = 0

(using cosC + cosD = 2cos 2

cos2

DCDC

02

cos2

5cos2cos2

xxx

02

cosx

or cosx = 0 or 02

5cos

x

cosx = 0 2

3,

2x

02

5cos

x

5

3,

5

9,

5

7,,

2

3,

5x

02

cosx

x

5

3,

5

9,

5

7,,

2

3,

2,

5x

43. [1]

y2 >, 2x

x2 + y

2 = 1x < 0

x >, 0 , y >, 0

Requred area is shaded area

Solving y2 = 2x & x

2 +y2 –4x = 0

x2 + 2x –4x = 0 x = 0 or x = 2

When x = 0, y = 0

When x = 2, y = ± 2

Required area =

2

0

)–( dxyy porabolauncle

=

2

0

2 )2––4( dxxxx

2

0

2

0

22 2–)2–(–2 dxxdxx

3

2

3

2–2

2–sin2

2

1)2–(–2

2

)2–(

2

02

2

0

1–222

xx

xx

3

8–22

3

22–)1(–1sin–2–

44. [3]

ba

, and c

unit vector

1cba

cbcba

2

3

2

3)(

cbcbabca

2

3

2

3).(–).(

Coppery

ccba

2

3).(

2

3–).( ba

2

3–cosba

2

3–cos

2

3–cos 1–

6/–

6/5

45. [2]

4x + 2πr + 2

2x + πr = 1

2x = 1 –πr

x = )........(2

–2

1i

rx

Also A = x2 + πr

2

2

2

2–

2

1r

rA

A = 2

22

42–

4

1r

rr

A is minimum

0dr

dA

022

–

2

– 2

rr

022

–

2

1–r

r

041– rr

1)4( r

).(..........)4(

1iir

Also 022

2

2

2

dr

Ad

A is minimum

Using (ii) in (i)

4

1

2–

2

1x

)4(2

–4x

4

2x

x = 2r

46. [1]

1

9–

1

5

1

1– zyx

Point on

(r + 1, r –5, r + 9)

Satis/es plane x –4 + z = 5

r + 1 – r + 5 + r + 9 = 5

r = –10

Ponts

(–9, –15, –1)

So distance between

(1, –5, 9) and (–9, –15, –1)

222 )19()155(–)91(

100100100

300

= 310

47. [3]

y = f(x)

y = (1 +xy) dx = xdy

y = vx

dx

dvxv

dx

dy

vx (1 + x vx) = x (v +x dx

dv)

dx

dvxvxvv 2

2

2

v

dvvxdx

cv

x 1–

2

2

cy

xx–

2

2

(1, –1) y2

1

8

5

c12

1

10

8y

c2

1–

5

4

So

cy

xx–

2

2

2

1–

1

2

2

y

x

sol (–1/2)

2

1–

24

1

8

1

y2

1

8

5

48. [Bonus]

A per the question, none of the option is correct, so this should be awarded as bonus to students.

49. [1]

f(x) = 2

21–

)2/sin–2/(cos

)2/sin2/(costan

xx

xx

= tan–1

2/sin–2/cos

2/sin2/cos

xx

xx

= 2/tan–1

2/tan1tan 1–

x

x

24

tantan 1– x

Y = t(x) 24

x

4,0

2

x

2

1)(xf

dx

dy

Normal so

2–dx

dy

)6

–(2–3

– xy –

42x 33

3

2

So passing point

3

2,0

50. [1] Near x = 0

f(x) = log 2 –sin x

f[f(x)] = log2 –sin[f(x)]

f[f(x) = log2 –sin (log2 –sinx)

g’(x) = log2 –sin(log2 –sinx)

g’(x) = –[cos(log2–sinx)] [–cosx]

g’(0) = 1 × cos (log2) = cos(log2)

51. [3]

E1 : Die A shows 4

E2 : Die B shows 2

E3 : sum of number on both dice is odd

P(E1) 6

1, P(E2)

6

1, P(E3)

2

1

36

18

Now P(E1 E2) = 36

1 (Cases are (4,2) only

P(E1) P(E2) = P(E1 E2)

P(E1 E2) = 12

1

36

3 cases are (4,1),(4,3)(4,5)

So is P(E1) P(E3)

P(E1 E3) = P(E1 E3)

and P(E2 E3) = 12

1

36

3 as cases are

(2,1),(2,3)(2,5)

Also P(E2) P(E3) = 12

1

P(E2 E3) = P(E2 E3)

Hence (E1,E2) and & (E2,E3) (E1,E3) are

Pair wise independent. Hence 1,2,4 are correct

Obviously (3) is not true

52. [1]

23

–5 baA

A adj A = a

bbaA

53–

2

23

–5

ab

ba

1030

0310

2–

35

23

–5

b

aba

132–15

2–1525 22

ba

baba

A(adj.A) = AAT

10a + 3b = 25a2 + b

2 and 15a –2b = 0

and 3b + 10a = 13

from (ii) egn. b = 2

15a

putting (in) (iii)

13102

45a

a

5

213

2

65a

a

3b

5

32

5 ba

53. [2]

By truth table

p q –q p

T T F F T

T F T T T

F T F F T

F F T F F

qp qqp )(

54. [4]

Given equation is

60–42 2

)55–( xxxx = ?

Taking log both sides

log log)55–( 60–42 2 xxxx

(x2 + 4x –60) log |x

2 –5x + 5| = 0

x2 + 4x –60 = 0 or log |x

2 –5x + 5| = 0

(x+10) (x–6) = 0 or x2 –5x + 5 = ± 1

x2 –5x + 4 = 0, x

2 –5x +6 = 1

(x–2) (x–3) = 0

x = 2,3

(x –4) ( x–1) = 0

x = –10,6,4,1,2,3

x = 3 is rejected as it does not satisfy the eqn.

sum of solutions = –10 + 6 + 4 + 1 + 2

3

55. [3] Let eqn of circle be x

2 + y

2 + 2gx + 2fy + c = 0

It touches x = a is C = g2

eqn. of circle becomes x

2 + y

2 + 2gx + 2fy + g

2 = 0

It also touches x2 + y

2 –8x –8y –4 = 0

Externally

C1C2 = r1 + r2

ffg 6)4()4( 22

(g+4)2 + (f+4)

2 = 3b + f

2 + 12|f|

(g+4)2 = 20+12|f| –8f

(g+4)2 = 20 +4f or (g+4)

4 = 20–2f

Locus of (–g,–f) will be parabola

Hence (3) is correct

56. [3]

S M A L L

A, L, L, M, S

Words starting with A = 1221

41

words starting with L = 41 + 24

words starting with SA 321

31

words starting with SL = 31 = 6

next words is SMALL = 58th

word

57. [1]

Let A n

nn

nnn

n

Lim1

2

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

A n

n

nn

n

n

n

n

n

Lim1

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

)2()1(

log A = n

nn

n

n

n

n

nn

Lim )2(.....................

)2()1(log

1

Log A = nn

Lim 1

n

nn

n

n

n

n )2(log........

)2(log

)1(log

log A =

n

r n

rn

nn

Lim 2

1

log1

log A =

2

0

)1log( dxx

log A = 3 log3 – 2

log A = log 27 – 2

log 2–27

A

2–

27e

A A =

2–

27

e

58. [1]

Given series is

......5

444

5

13

5

22

5

31

2

2

222

....5

24

5

20

5

16

5

12

5

822222

=

10

1

2

5

44

r

r

=

10

1

2)1(25

16

r

r

=

10

1

2 )12(25

16

r

rr

= 5

1610

2

11102

6

21)11)(10(

25

16in

= 77 + 22 + 2 = in

M = 101

59. [1]

Centre of x2 + y

2 –4x + 6y –12 = 0 is

(2, –3), it will be on diameter, say AB n

Now AB is chord of circle S.

for circle ‘s’

Learly AC = 52512)3()2(– 22

(Radius of given circle )

and OC = 55055 22

Radius of circle S = OA =

35755025

60. [3]

CD is pillar

).(..........60tan 0 ix

h

).(..........30tan 0 iiyx

h

)(3 ifromxh and

)()(3

1iifromyxh

)(3

13 yxx

yxx3

)......(..........2 iiixy

Speed is uniform

10

yv and

t

xv

t

xy

10

t

xx

10

2

t = 5

PHYSICS

61. [2]

l = <v>. t

tVV

l BA .2

tl .lg2

2210

202/2 glt s

62. [3] m g h

(10) (g) (1000) = (m) (3.8 × 107) (0.2)

= 12.89 × 10–3

kg

63. [2]

4302

1nmglosmgh

42

32

2

1N

29.0

73.12

1

32

1N

and Nmg x = mgh2

1

29.0

2

2

1x

= 3.5 m

0.29 and 3.5 m

64. [3]

For ring

Magnetic field at the center of a circular ring

R

I

2

0

har 2πR = l

2

lR

So. BA =

22

0

l

I

l

I.0

…..(1)

For square

Magnetic field at conter ‘B’ ,

)cos(cos4

4 210

R

IBB

)45cos45(cos

84

4 000

L

I

2

12

24 0

l

I

l

IBB

028 …………..(i)

(i) (ii)

2828

2

0

0

I

l

l

I

B

B

B

A

65. [4]

0

1

0

1

0

1

– g

gR

10010–10

103–

3–

= 0.01

66. [3] Magnetizing tower of telescope is 20.

So image of tree will be 20 time toller .

67. [2]

Cu is conductor so with procreating temperature the resistance will also increase linearly but Si is

semiconductor so resistance increases exponentially.

68. [4] By definition.

69. [3]

After 420

80 half liter the remaining nuclei of A is

1624

NoNoA

But after 240

80 half lives the remaining nuclei of B is

422

NoNoB

Thus ration of remaining nuclei

4:1: BA

Decayed ratio = 4: 1

70. [4]

nR

VPnR

PVT

1

2

3

2

300

nR

VPT 00

4

9

71. [3]

810

80

i

VR

But in AC

I = 22 )(WLR

V

z

V rmsrms

10 =22 )314()8(

22010

L

22 )22()314(64 L

420314L

49.20314L

314

49.20L

= 0.065 H

72. [3]

l2

Tenancy l

vvf

2

Now

l21

An gain l

vvf

21

1

ff 1

73. [2]

a

nU

1n

a

U

l

aQ

la2

la

lbm 4

74. [2]

Potentials at 4MF (Combination of 3MF & 9MF

Are 6Volt &I 2volt respectively;

This charge on 4MF is

Q1= C1V1

= 4 × 10–6

× 6

= 24 × 10–6

Coulomb

The charge on 9MF is

Q2= C2V2

= 9× 10–6

× 2

= 18× 10–6

Coulomb

Next charge Q = Q1 + Q2

The charge on 9MF is

Q = 24 × 10–6

+ 18 + 10–6

= 42 × 10

–6 coulomb

The field intensity

2r

kgE

2

6–9

)30(

1042109

mV /420

1042 1

75. [4]

Radio wave < Yellow light < Blue light < X-rays

Thus D < B < A < C

76. [3] Conceptual Question

77. [4]

Conceptual Question

78. [4]

r

axdxrQ

b

a

21

Q π B –a2

Q

)–(2 22 ab

QA

79. [4]

)–()–( 21 reri

)–79()–35(40 21 rr

)(–11440 21 rr

40 = 114 – A

= 74

But A)1–(

74)1–(40

174

40

= 1.5

80. [3] Conceptual Question

81. [4] Conceptual Question

82. [4]

Conceptual Question

83. [3]

sin5 1wAAw

cos3

2 1AA

3

7

9

49

9

45

222 AAA

A

84. [Bonus]

According to definition of angular momentum lrL

85. [1]

stn

QC

.

cantPV n

WUQ

n

TRTRfQ

–1

..2/

= n

fTR

–1

1

2.

T

n

fTR

C–1

1

2.

n

RCV

–1

R

n

CC V

–1

–

1

nCC

R

V

–1–

= n =

VCC

R

––1

V

P

CC

Ccn

–

– =

V

V

V

V

CC

RCC

CC

RCC

–

)(–

–

––

86. [Bonus] A per the question, none of the option is correct, so this should be awarded as bonus to students.

87. [3]

Conceptual Question

88. [2]

Ans. 8

Hint : Do your self

89. [Bonus]

A per the question, none of the option is correct, so this should be awarded as bonus to students.

90. [3]

RRVVV e 9–29– 0

)1–2(9R