Solutions to JEE Mains/Test Series - 9/IITJEE - 2013

Vidyamandir Classes

VMC/JEE-2013/Solutions 1 JEE Mains/Test Series-9/ACEG

Solutions to JEE Mains/Test Series - 9/IITJEE - 2013

[CHEMISTRY]

1.(C) By conservation of moles

1 1 2 2PV P V P V

RT RT RT= +

P × 4 = 2 × 1 + 3 × 2

P = 2 atm.

2.(C) Statement-II violates the inert pair effect.

Hence it is false.

3.(A) KMnO4 will react with FeSO4 only.

1 × moles = 100

5 21000

× ×

moles = 1

4FeSO 1/ 3χ =

4.(D)

5.(D) Only Cl2 is oxidsing among the given options

∴ 2Cl2 4 4K MnO KMnO→

6.(D) 2 2v

1 1

T VS nC n nR n

T V∆ = +� �

= v

1nC n2 nR n

2+� �

= v(C R) n 2− �

7.(D) P-anthranilic acid has less acidic nature due to the +M effect of 2NH− group.

8.(B) 2 2 3N (g) 3H (g) 2NH (g)

t 0 a 3a

+

=

��⇀↽��

eq

a 3at t a

2 2=

3NH

a PP P

3a 3= = .

9.(C) 3 2 3 3 44BF 3H O H BO 3HBF+ → +

10.(D) 1 1 2 2[H ] Ka C Ka C+ = + (Learn as a result)

= 5 51.8 10 0.01 6.3 10 0.01− −× × + × ×

= 7 48.1 10 9 10− −× = × .

11.(B) 22 7Cr O − on reduction converts to Cr

+3 which is green in colour.

Vidyamandir Classes


12.(C) ( )n n 2 1.73 3+ = =

n (n + 2) = 3

n = no. of unpaired e s 1− =

v = 4s2 3d3.

Hence Vanadium must exist in the form V4+ compound is VCl4

13.(C) 3AgNO2 2 2 2CH CH CH Cl AgCl CH CH CH

⊕= − − → ↓ + = − Allylic carbocation.

14.(A) ( )4 4 2 4 4 4 2 4 2MgSO NH OH Na HPO Mg NH PO Na SO H O+ + → + + .

15.(C) 2 2P I Mg /Ether HCHO H O3 2CH CH OH A B C D

+− − → → → →

(A) is 2 2CH CH I− − (B) is 3 2CH CH Mg I− − −

(C) is 3 2 2CH CH CH O MgI⊕

− − − � (D) is 3 2 2CH CH CH OH− − −

16.(B)

17.(B) 3 2 2 3

O O|| ||

CH C CH C CH CH− − − − − Most acidic hydrogen (Active methylene group)

18.(D) NF3 dipole > NH3 dipole.

19.(D) R.D.S. of cannizaro involves transfer of H ion−

to the carbonyl group.

20.(A)

O O|| ||

R C Z Nu R C Nu Z− −− − + → − − +

Leaving group capacity of Cl− is maximum among the given options hence & fastest reaction when Z = Cl.

21.(C) Both (1) & (2) contain chiral carbon will undergo SN2 reaction with alc. KCN, hence inversion.

22.(B) iCRTπ =

moles

0.75 2.47 0.0821 3002.5

= × × ×

Moles = 1

0.0330

=

23.(A) 2

2

CZE

n=

3

3 3

22 2BeH H

2 2 2 2HBe Be

nZ 1 2 1

4n Z 1 4

+

+ +

ε= = × =

ε

24.(D) Graphite: free e− are spread out between the structure & thus graphite is conducting.

25.(D) V1 = volume for complete neutralization of Na2CO3.

V2 = Volume for complete neutralization of NaHCO3.

1V x2

=

V1 + V2 = x + y

Hence 2V y x= −

Vidyamandir Classes


26.(D) (g) (g)Cl e Cl− −+ → is exothermic, rest all endothermic.

27.(D)

28.(C) Metal oxides an basic (or more ionic character ≈ more basic).

∴ TiO > VO > CrO > FeO

29.(C) 2dxk[A]

dt=

dx

log log k 2 log[A]dt

= +

y 2x logk= +

On comparing with y = mx + c.

Where m slope≡

c constant≡

30.(A) 4r a 2=

2

r a4

=


[PHYSICS]

31.(C) A crystal structure is composed of a unit cell, a set of atoms arranged in a particular way; which is periodically

repeated in three dimensions on a lattice. The spacing between unit cell in various directions is called its lattice

parameters or constants. Increasing these lattice constant will increase or widen the band-gap (Eg), which means

more energy would be required by electrons to reach the conduction band from the valence band. Automatically Ec

and Ev decreases.

32.(D) 3 41 2

3 4

60 3050 30 100

90

×= + + = + + =

+K K

K K K N / mK K

0 012 2

100 50= = =

m .T Hz

K

ππ π .

33.(B) 1

22 2

1

2

0 18 20 20 08

10 2

− −= ⇒ = ⇒ =

− −t

tt t

v d .v . m / s

v d v

ρρ

34.(A) × =C ||E B & CB E in magnitude.

As 8 63 10 and 720 2 4 10−= × = ⇒ = × ˆˆ ˆC i E j B . k

35.(C) 300 0

10 000 9500300

−=

− ,

v ⇒ 300 285 15− = ⇒ =v v m / s

36.(B) Only electric force acts on the particle.

37.(C)

38.(A) Lowest frequency = HCF of 420 Hz & 315 Hz = 105 Hz.

Vidyamandir Classes


39.(D) = BAN sin tφ ω

= − = − ⇒ =m

dBAN cos t BAN

dt

φε ω ω ε ω

40.(B) ( )20 50 10 3= − = − − = − =d

t v at t sdt

φε

41.(B) D1 is not conducting but D2 is conducting

⇒ 12

24 2

= =+

i A

42.(B) Energy of proton = 8 7 06 7 5 6× − ×. .

= 17.28 MeV

43.(A) K.E. is maximum when P.E. is minimum ⇒ 0=dv

dx ⇒ 3 0 0 1− = ⇒ = = ±x x x or x

2

2

23 1 0 0 0= − < = ⇒ =

d vx for x x

dx is a point of maximum for V

and 2

2

23 1 0 1 1= − > = ± ⇒ = ±

d vx for x x

dx are points of minima for V

⇒ minimum potential energy = 1 1 1

J4 2 4

−− = at x 1= ±

⇒ maximum 1 9

24 4

− = − =

KE ⇒ 29 11

4 2= =maxmv & m kg ⇒

3

2=maxv m / s

44.(C) Let Q be total charge on the two spheres & let q1 & q2 be charges on A & B respectively in equilibrium.

(rA = 1mm, rB = 2mm)

⇒ q1 + q2 = Q and 1 2

0 04 4=

∈ ∈A B

q q

r rπ π ⇒ 1 2

2and

3 3= =Q Q

q q .

( )

2 21 2 1 2 2

2 3: 1 3 2 1

2= = =A B

/E / E q / r q / r / : :

45.(A) 1 22

2

+ < ⇒ < =

∵R B

A Dsin

D Dsin A /

µ µ µ

⇒ Ans. (A) or for small angled prism ( )1= −D A µ

46.(D) 47.(C)

48.(A) ( )

2

1 2

1 1

−∝ ⇒ ∝

n n /

vv

R R R

1 11

2 22

− ++

= ∝ ⇒ ∝n n

RT R T R

V

π

49.(D) ⇒ 2 1

4

2

−=

+

T TH

x x

KA KA

= ( )2 1

3

−KA T T

x

⇒ f = 1/3 ⇒ Ans. (D)

50.(B) ( ) 2| P | E / C E / C E / C∆ = − − = ⇒ Ans. (B)

H

K 2K T1 T2

x 4x

Vidyamandir Classes


51.(A) u = v ⇒ the object is placed at the centre of the equivalent mirror.

1 1 5 1 1 1

2 1 101 30 30 2

=− − − + ⇒ = − ∞ − − m

m

.F cm

F /.

2 20= =mR | F | cm

52.(C) ( )

64

2 6 1 5 3= =

+ i A

|| . ||

53.(A) 20 4

5080 100

= = ⇒ =−

X X x; x cm

Y Y x

54.(B) Let � : length of the wire ; R : radius in first case; r : radius in second case.

⇒ 2 (2 )= =� �R & n rπ π

0 0

2= =

�

i iB

R

µ µ π

2

20 0

2= = =

�

n i n iB' n B

r

µ µ π

55.(B) 2=I

TMB

π where I: moment of Inertia = 2

12

�m

M : pole strength × �

( ) ( )

2

9 pole strength remain same 3 312

= = = × ×�

�m / s

I ' I / ; M ' / M

⇒ 9

2 2 33

= = =I / T

T ' / secM .B

π

56.(D) As P.d across L & C are out of phase.

57.(B) induced current = ( )change influx × no of turns

total resistance × time− =

( )( )

2 1

4

− − × ÷

+

W Wn t

R R =

( )2 1

5

−−n W W

RT

58.(C) 1 1

2(2 )

= = ⇒ =L' L /LC L' C

ω

59.(B) ( ) ( )22 4 51 10 2 10 5 1 5 10

2 2

− −= = × × × × = ×� . Vε ∆ω

= 50 vµ

60.(A)

⇒ truth table of OR gate.

Vidyamandir Classes



[MATHEMATICS]

61.(D) For the graph to touch x-axis, the polynomial 2 2 1y x px p= − + + would have to be a perfect square. Therefore,

2 1p p= + , or 2 1 0p p− − = , ( )1 5 2p /= ± .

62.(C) ( ) ( )1 8 7 4Z sin cos i sin cosθ θ θ θ= + + +

( ) ( )2 4 8 7= + + +Z sin cos i sin cosθ θ θ θ

Hence, 1

2

Z x iy

Z y ix

= + = + where ( )8 7x sin cosθ θ= + and ( )4y sin cosθ θ= +

( ) ( )2 21 2 = − + +Z .Z xy xy i x y

2 20a ib a ; b x y= + ⇒ = = +

Now, ( ) ( )2 22 2 8 7 4x y sin cos sin cosθ θ θ θ+ = + + +

2 265 65 120sin cos sin cosθ θ θ θ= + +

65 60 2sin θ= +

Hence, ( ) 125+ =max

a b

63.(C) ( ) ( )2 3× +| a b | a .b

2 3 where= + ,|a||b|sin |a||b|cosθ θ θ is angle between anda b

2 212 18 12 18 6 13= + ≤ + =sin cosθ θ

64.(B) Line 0ax y+ = and 0x by+ = intersect at O(0, 0).

If AB subtends right angle at O(0, 0), then 0ax y+ = and 0x by+ = are perpendicular.

⇒ ( ) 11a

b

− − = −

⇒ 0a b+ =

65.(D) The first two columns of first determinant are same as first two rows of second. Hence, transpose the second.

Add the two determinants and use 1 1 3C C C→ + . It gives D = 0.

66.(D) dr

cdt

=

and h ar b= + ; also 3

dh dr

dt dt

=

(given)

ax + y = 0

Vidyamandir Classes


∴ 3 3dr dra adt dt

= ⇒ =

Hence, 3h r b= +

When 1 6 6 3 3r ; h b b= = ⇒ = + ⇒ =

∴ ( )3 1h r= +

( ) ( )2 2 3 23 1 3V r h r r r rπ π π= = + = +

( )23 3 2

dV drr r

dt dtπ= +

where 6 1dV

r ; cc / secdt

= =

Therefore, ( )1 3 108 12 360 1dr dr

dt dtπ π

= + ⇒ =

Again when 36dV

r , ndt

= =

( )( )23 3 36 2 36

= +

drn .

dtπ

( ) ( )3 36 110 1 360n . . /π π=

33n =

67.(A) 3 20

3

x

sin x aL lim b

x x→

= + +

3

30

3

x

sin x ax bxlim

x→

+ +=

2

20

33

3

x

sin xa bx

xlimx→

+ +=

For existence of limit 3 0a+ =

⇒ 3a = −

∴ 3

30

3 3

x

sin x x bxL lim

x→

− += ( )

3027 0 3t

sint tlim b x t

t→

−= + = =

27

06

b= − + = (Apply LH rule to get 30

1

6→

− −=

t

sin t tlim

t ⇒

9

2b =

68.(C) Given integral

( ) ( )

1

2 20

1

dx

x cos cosα α=

+ + −∫

( )

1

2 20

dx

x cos sinα α=

+ +∫

11

0

1 x costan

sin sin

αα α

− +=

1 11 1 cos cos

tan tansin sin sin

α αα α α

− −+ = −

Vidyamandir Classes


( )1 11

2tan cot tan cot

sin

αα

α− −

= −

1 11

2 2 2tan tan tan tan

sin

π α πα

α− −

= − − −

1

2 2 2 2sin sin

π α π αα

α α

= − − − =

69.(B) ( )2 2

31 2 0 0

2log x log x x− − + > >

⇒ ( )2 2

3 11 1 0

2 2log x log x− − − + >

Let 2 1 0log x t− = ≥ . . . .(i)

⇒ 2 1 2log x x≥ ⇒ ≥

∴ 23 1

02 2

t t− + >

⇒ 2 22 3 1 0 3 2 1 0t t t t− + > ⇒ − − <

⇒ 1

13t− < < . . . .(ii)

From (i) and (ii), 0 1t≤ <

20 1 1log x≤ − <

20 1 1log x≤ − <

21 2log x≤ <

2 4x≤ <

70.(A) Let H be the midpoint of BC since 2 2 290TBH , TH BT BH∠ = ° = +

2 2

5 5 50= + =

Also 2 2 290 50 25THG , TG TH GH∠ = ° = + = +

75=

Let θ be the required angle of elevation of G at T.

Then 5 1

5 3 3

GHsin

TGθ = = =

⇒ 1 1

3sinθ −=

71.(D) ( )[ ]

[ ] [ ]

[ ]

[ ][ ] ( ) [ )

[ )

0 0 1

0 0 0 0 1

tan x tan xx x

x xf x

x x

, ≠ , ∈ − ∞, ∪ , ∞

= , = , ∈ ,

L.H.L = ( ) [ ][ ]

[ ][ ]

( )( )0 0 0

11

1x x h

tan x tan h tanlim f x lim lim tan

x h→ − → →

− −= = = =

− −

R.H.L. = ( )0 0

0 0x xlim f x lim→ + →

= =

Since L.H.L. ≠ R.H.L. Therefore ( )0x

lim f x→

does not exist.

Vidyamandir Classes


72.(B) ( ) ( ) ( ) ( )10 10 10 10 10 10 10 10 10 100 0 1 0 1 2 0 1 2 9C C C C C C . . . . C C C C+ + + + + + + + +

10 10 10 100 1 2 910 9 8C C C . . . C= + + + +

10 10 10 101 2 3 102 3 10C C C . . . C= + + + +

10 1010 9 9

1

1 1

10 10 2r r

r r

r C C −= =

= = = ⋅∑ ∑

73.(D) ( )

( )22

2 20 2

4 5

xy x

x x

− +′ = = ⇒ = −

+ +;

( )( )1

2 2

3 33 2

4 5 2 1dx dx tan x C

x x x

−= = + ++ + + +∫ ∫

Hence A = 3 and B = 2

Hence A B≠

Area ( ) ( ) ( )3 2

3 21 1

22

3 3 2 3 3f x dx tan x tan π−

−− −

−−

= = + = =∫

Range is (0, 1]

Graph of ( )3y f x= is as shown. Turning point is 2x = −

74.(B) Circles S1,and S2 with centres C1 and C2 have radius r1 and r2, respectively

Let S3 be the variable circle which touches the given two circles as explained in the question and which has centre

C and radius r.

Now 2 2CC r r= + and 1 1CC r r= +

Hence 1 2 1 2CC CC r r− = − (= constant)

The locus of C is hyperbola whose focii are C1 and C2.

75.(B) Let the tangents interest at P(h, k) then

According to the question

hcos k sin aθ θ+ =

3 3

hcos k sin aπ π

θ θ + + + =

Eliminating θ from the above equations we get

2

2 2 4

3

ah k+ =

76.(B) 11 1

4tan

π− = <

1 Radian 57 17 44 8' . ′′= ° for which 1 1sin cos> but both values are less than 1.

Obviously, tan 1 > 1

Hence greatest value is tan 1

77.(D) We know that ( )( )

( ) ( ) ( )( )

P A B P A P B P A BAPB P B P B

∩ + − ∪ = =

Since ( ) 1P A B∪ < , therefore ( ) 1P A B− ∪ > −

⇒ ( ) ( ) ( ) ( ) ( ) 1P A P B P A B P A P B+ − ∪ > + −

S1 S2

S3

Vidyamandir Classes


⇒ ( ) ( ) ( )

( )( ) ( )

( )1P A P B P A B P A P B

P B P B

+ − ∪ + −> ⇒

( ) ( )( )

1P A P BAPB P B

+ − >

Thus, choice (A) is correct.

Choice (B) holds good.

Choice (C) is also correct since

( ) ( ) ( ) ( )P A B P A P B P A B∪ = + − ∩

( ) ( ) ( ) ( )P A P B P A P B= + − If A, B are independent

( )( ) ( )( ) ( ) ( )1 1 1 1P A P B P A P B= − − − = −

Choice (D) is obviously not true because ( ) ( ) ( )P A B P A P B∪ = +

78.(B) ( )( )( )

2

2 21 1 1

x dxf x

x x

=+ + +

∫

Let 2x tan dx sec dθ θ θ= ⇒ =

⇒ ( )( ) ( )

2

2 21 1 1

x dxf x

x x

=+ + +

∫

( )

2 2 2

2 11

tan sec d tan d

secsec sec

θ θ θ θ θθθ θ

= =++∫ ∫

( )

2

1

sin d

cos cos

θ θθ θ

=+∫

( )

21

1

cos d

cos cos

θ θθ θ−

=+∫

( )1 cos d

cos

θ θ

θ

−= ∫

sec d dθ θ θ= −∫ ∫

( )2 11log x x tan x c

−= + + − +

Given ( )0 0f =

⇒ 0 1 0log c= − + ⇒ c = 0

⇒ ( ) ( ) ( )2 11 1 1 1 1f log tan

−= + + −

( )1 24

logπ

= + −

79.(D) Let 2

2

1

1e

x lnxI dx

x x ln x

+=

+∫

( ) ( )

( )

22

2

11

1

x ln x x ln x x ln xx ln x

x x ln xx x ln x

+ − −+=

++

( ) ( )1 11 1

1 1

ln x x ln x x ln x ln x

x x ln x x x ln x

− + − −= − = +

+ +

1 1

11

ln x

x x ln x

+ = + − +

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Solutions to JEE Mains/Test Series - 9/IITJEE - 2013