JEE-Advanced-2013 Paper 1 Code 0 - Amazon Web...

Vidyamandir Classes

VMC/Solutions 1 Paper 1 - Code 0 – JEE Advanced - 2013

JEE-Advanced-2013

Paper 1

Code 0

2 June, 2013 | 9:00 AM – 12 Noon

Detailed solutions to the paper by Vidyamandir Classes

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Part 1 : PHYSICS

Section – 1: (Only One option correct Type)

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which

ONLY ONE is correct

1. The work done on a particle of mass m by a force,

( ) ( )3 2 3 22 2 2 2

+ + +

x yˆ ˆK i j

x y x y

(K being a constant of

appropriate dimensions), when the particle is taken from the point (a, 0) to the point (0, a) along a circular path of

radius a about the origin in the x-y plane is :

(A) 2K

a

π (B)

K

a

π (C)

2

K

a

π (D) 0

Ans.(D) ( )= = +∫ ∫��

x yW F .dr F dx F dy

( ) ( )3 2 3 22 2 2 2

= ++ +

∫ / /

xdx ydyW K

x y x y

=

( )

00

2 2 2 20 0

1 10

− = = + +

∫,a

( , a )

a, a,

K d

x y x y

2. Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or in configuration

II as shown in the figure. One of the blocks has thermal conductivity κ and the other 2κ . The temperature

difference between the ends along the x-axis is the same in both the configurations. It takes 9s to transport a certain

amount of heat from the hot end to the cold end in the configuration I. The time to transport the same amount of

heat in the configuration II is :

(A) 2.0s (B) 3.0s (C) 4.5s (D) 6.0s

Ans.(A) ∆ ∆

= = ⇒ =dQ T T

H Q tdt R R

Since Q and ∆T is same.

1 2

1 2

=t t

R R …(i)

12

= +� �

RKA KA

(configuration 1)

⇒ 1

3

2=

�R

KA

2

1 2 3= + =� � �

KA KA KA

R (configuration 2)

⇒ 23

=�

RKA

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Putting in (i)

2

2

92

3 2 3= ⇒ =

� �

tt sec

/ KA / KA

3. Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partial

pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their densities is :

(A) 1 : 4 (B) 1 : 2 (C) 6 : 9 (D) 8 : 9

Ans.(D) 1

2

2

3=

M

M

1

2

4

3=

P

P

Using = ρPM RT

1 1 2 2 1 1 1

1 2 2 2 2

4 2 8

3 3 9

ρ= ⇒ = = × =

ρ ρ ρ

PM P M PM

P M

4. A particle of mass m is projected from the ground with an initial speed u0 at an angle α with the horizontal. At the

highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was

thrown vertically upward from the ground with the same initial speed u0. The angle that the composite system

makes with the horizontal immediately after the collision is :

(A) 4

π (B)

4

πα+ (C)

2

πα− (D)

2

π

Ans.(A) From energy conservation both will have equal speeds at the point of collision.

0 cos=V u α

For perfectly inelastic collision, to conserve linear momentum

Final velocity of combined mass will be along the Resultant ⇒ 45θ = �

5. A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is

30 mW and the speed of light is 8 13 10 ms−× . The final momentum of the object is :

(A) 17 10 3 10. kg ms− −× (B) 17 11 0 10. kg ms− −×

(C) 17 13 0 10. kg ms− −× (D) 17 19 0 10. kg ms− −×

Ans.(B)

39

8

30 10100 10

3 10

−−× ∆ ×

∆ = = = × ××

E power tP

C C

= 1710−

kgm / s

6. In the Young's double slit experiment using a monochromatic light of wavelength λ , the path difference (in terms

of an integer n) corresponding to any point having half the peak intensity is :

Vidyamandir Classes


(A) ( )2 12

nλ

+ (B) ( )2 14

nλ

+ (C) ( )2 18

nλ

+ (D) ( )2 116

nλ

+

Ans.(B) 204

2

∆φ=I I cos

⇒ 20 02 4

2

∆φ=I I cos

⇒ 2 1

2 2

∆φ=cos

⇒ ( )2 12 4

∆φ π= +n

⇒ ( ) ( )1 2

2 1 2 12 4 4

π π λ= + ⇒ = +

λp.d . n p.d n

7. The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one-third

the size of the object. The wavelength of light inside the lens is 2

3 times the wavelength in free space. The radius

of the curved surface of the lens is :

(A) 1m (B) 2m (C) 3m (D) 6m

Ans.(C) 1 8 1

243 3

= ⇒ = ⇒ = −−

Vu

u u

1 1 1 1 1 1

68 24

− = ⇒ = − ⇒ =−

fv u f f

0 0

0

23 2

3

λ λλ = ⇒ λ = ⇒ µ =

µ µ/

1 1 1 1

1

µ − = −

∞ f R

1 3 2 1 1

36 1

− = ⇒ =

/R

R

8. One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin

copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio

of the elongation in the thin wire to that in the thick wire is :

(A) 0.25 (B) 0.50 (C) 2.00 (D) 4.00

Ans.(C) Force in both the wires will be same.

2

∆F F

xAy r yπ

= =� �

2 221 1 2 1 2

22 2 2 11

∆ 2 1

∆ 2 2

x r r L R.

x r L Rr

⇒ = × = = =

� �

� �

2

1

∆2 (ratio of elongation of thin to thick wire)

∆

x

x=

O I

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9. A ray of light travelling in the direction ( )13

2ˆ ˆi j+ is incident on a plane mirror. After reflection, it travels along

the direction ( )13

2

ˆ ˆi j− . The angle of incidence is :

(A) 30° (B) 45° (C) 60° (D) 75°

Ans.(A) 1

2

−= =

a.bCos

a bθ

120θ = °

180

302

iθ−

⇒ = = °

10. The diameter of a cylinder is measured using a Vernier callipers with no zero errors. It is found that the zero of the

Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to

2.45cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions.

The diameter of the cylinder is :

(A) 5.112 cm (B) 5.124 cm (D) 5.136 cm (D) 5.148 cm

Ans.(B) 2 45

1 1 0 0550

.L.C MSD VSD . cm cm= − = −

0 05 0 049 0 001. . . cm= − =

5 10 24 0 001D . cm . cm= + ×

5 10 0 024 5 124. cm . cm . cm= + =

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Section – 2: (One or more options correct Type)


ONE or MORE are correct

11. In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch S1 is

pressed first to fully charge the capacitor C1 and then released. The switch S2 is then pressed to charge

the capacitor C2. After some time, S2 is released and then S3 is pressed. After some time,

(A) the charge on the upper plate of C1 is 2CV0 (B) the charge on the upper plate of C1 is CV0

(C) the charge on the upper plate of C2 is 0 (D) the charge on the upper plate of C2 is 0CV−

Ans.(BD)

12. A particle of mass M and positive charge Q, moving with a constant velocity 11 4û i ms ,−=�

enters a region of

uniform static magnetic field normal to the x y− plane. The region of the magnetic field extends from x = 0 to

x = L for all values of y. After passing through this region, the particle emerges on the other side after 10

milliseconds with a velocity ( ) 12 2 3ˆ û i j ms .

−= +�

The correct statement(s) is (are)

(A) The direction of the magnetic field is -z direction.

(B) The direction of the magnetic field is +z direction

(C) The magnitude of the magnetic field 50

3

M

Q

π units.

(D) The magnitude of the magnetic field is 100

3

M

Q

π units.

Ans.(AC) 1 4û i=�

2 2( 3 )ˆ û i j= +�

2

302 3

Tanθ θ= ⇒ = °

B should be along directionz .−

M

tQB

θ=

3 610 10

.M

QB

π−⇒ × =

50

1006 3

M MB

Q Q

π π⇒ = × = units

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13. A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation,

1 1( ) (0 01 ) [(62 8 ) ] [628 ) ]y x,t . m sin . m x cos s t .− −= Assuming 3 14. ,π = the correct statements(s) is (are)

(A) The number of nodes is 5.

(B) The length of the string is 0.25 m.

(C) The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 m.

(D) The fundamental frequency is 100 Hz.

Ans.(BC) ( ) ( ) ( )1 10.01 sin 62.8 cos 628y m m x s t

− − =

For 5th harmonic

52

2 2 262.8

62.8

l

k m

λ

π π πλ

λ λ

=

= ⇒ = ⇒ =

5 2 5 2 3.14 10.25

2 62.8 2 62.8 4l m m

π ×⇒ = × = × = =

Number of nodes = 6

At midpoint of string there is an antinode

maxy at midpoint = 0.01 m

0

/ 628 / 62.8 1020

2 2 2 0.25 0.5

v kf Hz

l l

ω= = = = =

×

14. A solid sphere of radius R and density ρ is attached to one end of a mass-less spring of force constant k. The other

end of the spring is connected to another solid sphere of radius R and density 3ρ . the complete arrangement is

placed in a liquid of density 2ρ and is allowed to reach equilibrium. The correct statement(s) is (are)

(A) the net elongation of the spring is 34

3

R g

k

π ρ

(B) the net elongation of the spring is 38

3

R g

k

π ρ

(C) the light sphere is partially submerged.

(D) the light sphere is completely submerged.

Ans.(AD) In equilibrium let elongation of spring is x

Then

34

3

vg R gkx vg x

k k

ρ ρπρ= ⇒ = =

In equilibrium net force of weight and spring force on light sphere in

downward direction is

2 Bkx vg vg Fρ ρ+ = =

So it is fully submerged.

15. Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities 1 2andρ ρ

respectively, touch each other, The net electric field at a distance 2R from the centre of the smaller sphere, along

the line joining the centres of the spheres, is zero. The ratio 1

2

ρ

ρ can be.

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(A) 4− (B) 32

25− (C)

32

25 (D) 4

Ans.(BD)

0=PFor E

( )

( )( )

31

1 2 2

2 20 00 0

4

3

3 34 2 4 2

⋅= ⇒ =

RQ R R

R R

ρ πρ ρ

ε επε πε

1

2

4=ρ

ρ

Also for QE = 0

( ) ( )1 2 1 1

2 22 20 0

4 320

25 254 2 4 5

Q Q Q

QR R

ρ

ρπε πε+ = ⇒ = − ⇒ = −

Section – 3: (Integer value correct Type)

This section contains 5 questions. The answer to each question is a single digit integer,

ranging from 0 to 9 (both inclusive)

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16. A bob of mass m, suspended by a string of length l1, is given a minimum velocity required to complete a full circle

in the vertical plane. At the highest point, it collides elastically with another bob of mass m suspended by a string

of length l2, which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after

collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio 1

2

l

lis

Ans.(5) 1 15=u gl

At highest point 1 1v gl= (from energy conservation)

After collision (identical mass and elastic collision)

2 1 1u v gl= =

To complete vertical circle

11 2

2

5 5l

gl gll

= ⇒ =

17. A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W to the

particle. If the initial speed 1(in ms )− of the particle is zero, the speed 1(in )ms− after 5 s is

Ans.(5) 0.5=P W

0.5 5 2.5= = = × =∫W Pdt Pt J

2 2

2

1 10.2 2.5

2 2

25 5 /

∆ =

⇒ = ⇒ × × =

⇒ = ⇒ =

K W

mv W v

v v m s

18. The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stopping

potential versus frequency plot for Silver to that of Sodium is

Ans.(1) 0

= +C

h w eVυ

0⇒ = −C

whV

e eυ

Slope of VC versus υ is Constanth

e=

Hence ratio = 1

19. A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second. Given

that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will

decay in the first 80 s after preparation of the sample is

Ans.(4) 31/ 2

0.693 0.693 1 11386

1386 20002 10= = ⇒ = = =

×T λ

λ

0

0 0

0 0

(1 )1

−

−−

=

− −= = = −

t

t

t

N N e

N N N ef e

N N

λ

λλ

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0.04

0.04

1 180 0.04

2000 25

1 (0.04) 0.96(1 ) 1 0.96 0.04

Using e 1 for 1

−−

−

= × = =

= − == − = − =

− << �x

t

ef e

x x

λ

Percentage fraction = 4

20. A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad 1s− about its

own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are gently placed

symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are

horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system

rotates about the original axis. The new angular velocity (in rad 1

s−) of the system is

Ans.(8) From conservation of angular momentum

1 1

2 22 2

1

( 2 )

22 2 2 2

= +

= + +

I I I

MR MR R Rm m

ω ω

ω ω

2 22

1

1

1

2 2

50 5010 6.25

2 2

8 rad/s

= +

⇒ × = +

⇒ =

MR MRmRω ω

ω

ω

Part 2 : CHEMISTRY


Vidyamandir Classes



ONLY ONE is correct

21. Consider the following complex ions, P, Q and R.

[ ] ( ) ( )2 23

6 2 26 6

+ +− = = = P FeF , Q V H O and R Fe H O .

The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is.

(A) R < Q < P (B) Q < R < P (C) R < P < Q (D) Q < P < R

Ans. (B) 36P [FeF ] −=

3 0 5Fe 4s 3d n 5+ ⇒ ⇒ =

22 6Q [V(H O) ] +=

2 0 3V 4s 3d n 3+ ⇒ ⇒ =

22 6R [Fe(H O) ] +=

0 6Fe 4s 3d n 42+ ⇒ ⇒ =

Magnetic moment will be in order Q < R < P

22. The arrangement of X− ions around A+ ion in solid AX is given in the figure (not drawn to scale). If the radius of

X− is 250 pm, the radius of A

+ is.

(A) 104 pm (B) 125 pm (C) 183 pm (D) 57 pm

Ans. (A) Using radius ratio for octahedral void (FCC)

r

0.414r

+

−=

∴ r 0.414 250 104+ = × =

23. Sulfide ores are common for the metals

(A) Ag, Cu and Pb (B) Ag, Cu and Sn

(C) Ag, Mg and Pb (D) Al, Cu and Pb

Ans. (A) Ag2S Silver glance

Cu2S, CuFeS2 Copper glance & Copper pyrite

PbS Galena

24. The standard enthalpies of formation of CO2(g), H2O(l) and glucose (s) at 25 C�

are

400 300 1300− − −kJ / mol, kJ / mol and kJ / mol , respectively. The standard enthalpy of combustion per gram

of glucose at 25 C�

is.

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(A) + 2900 kJ (B) 2900− kJ (C) 16.11− kJ (D) + 16.11 kJ

Ans. (C) 6 12 6 2 2 2C H O 6O 6CO 6H O+ → +

As we know

f (Product) f (Reactants)H H H∆ = ∑ −∑

2 2 6 12 6f (CO ) f (H O) f (C H O )6H 6H H= + − 6 ( 400) 6( 300) ( 1300)= × − + − − −

2400 1800 1300 2900 kJ= − − + = −

To calculate H∆ for one gram.

H∆ per gram =2900

16.11 kJ180

−= − .

25. Upon treatment with ammoniacal H2S, the metal ion that precipitates as a sulfide is.

(A) Fe (III) (B) Al (III) (C) Mg (II) (D) Zn (II)

Ans. (D) 2

2H S Zn ZnS 2H+ ++ → +

26. Methylene blue, from its aqueous solution, is absorbed on activated charcoal at 25 C� . For this process, the correct

statement is.

(A) The adsorption requires activation at 25 C�

(B) The adsorption is accompanied by a decrease in enthalpy

(C) The adsorption increases with increase of temperature

(D) The adsorption is irreversible

Ans. (B) Methylene blue on activated charcoal is an example of physiosorption. Since entropy decreases adsorption

therefore enthalpy has to be negative to make G 0∆ < .

27. KI in acetone, undergoes SN2 reaction with each of P, Q, R and S. The rates of the reaction vary as.

(A) P > Q > R > S (B) S > P > R > Q

(C) P > R > Q > S (D) R > P > S > Q

Ans. (B) Rate of SN2 :

28. In the reaction,

P Q R S+ → +

the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies

with reaction time as shown in the figure. The overall order of the reaction is.

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(A) 2 (B) 3

(C) 0 (D) 1

Ans. (D) Since time taken for 75% reaction of P is twice the time taken for 50% reaction of P. Hence reaction is first order

w.r.t. P.

Analysis of graph of Q w.r.t. time shows, reaction of Q is of zero order.

Hence overall order of the reaction is one.

29. Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of

(A) NO (B) NO2 (C) N2O (D) N2O4

Ans. (B) HNO3 on standing decomposes as 3 2 2 2

12HNO 2NO O H O

2→ + +

30. The compound that does NOT liberate CO2, on treatment with aqueous sodium bicarbonate solution, is.

(A) Benzoic acid (B) Benzenesulphonic acid

(C) Salicylic acid (D) Carbolic acid (Phenol)

Ans. (D) Phenol is too weak to liberate CO2 with NaHCO3.




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31. The initial rate ofr hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is 1/100th of that of a strong acid

(HX, 1M), at 25o C. The Ka of HA is

(A) 41 10−× (B)

51 10−×

(C) 61 10−× (D)

31 10−×

Ans. (A) 2R K[RCOOR][H O]=

Clearly w w

s s

R K 1

R K 100= = (Given) [Since (RCOOR) is common]

But Ea / RtK Ae

−= [∵ Ea will not change as catalyst is same (H+)]

So w w

s s

K A 1

K A 100= = . . . .(i)

Also, A depends on no. of particles i.e. A [H ]+∝

So for 2 3C C C C

HA H O H O A+ −

− α − α α+ +��⇀↽��

In equation (i)

C 1

0.01 ( C 1M)1 100

α= ⇒ α = =∵ and

24

a

CK 10

1

−α= =−α

32. The hyperconjugative stabilities of tert-butyl cation and 2-butene, respectively, are due to

(A) ( ) * .p empty and electron deloaclisationsσ σ π→ →

(B) * .and electron deloaclisationsσ σ σ π→ →

(C) ( ) .p filled and electron deloaclisationsσ σ π→ →

(D) ( ) * * .p filled and electron deloaclisationsσ σ π→ →

Ans. (A) Hyperconjugation in tert-butyl carbocation refers to delocalisation of σ -electrons over the empty p-orbitals of C+.

In 2-butene, it refers to delocalisation of σ -electrons over the *π orbitals of C = C.

33. The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are)

(A) [ ] [ ]3 5 2 3 4 2( ) ( )Cr NH Cl Cl and Cr NH Cl Cl

(B) [ ] [ ]3 4 2 3 2 2( ) ( ) ( )Co NH Cl and Pt NH H O Cl+ +

(C) [ ] [ ]2 2

2 2 2 2CoBr Cl and PtBr Cl− −

(D) [ ] [ ]3 3 3 3 3( ) ( ) ( )Pt NH NO Cl and Pt NH Cl Br

Ans. (BD)(B) 3 4 2 3 2 2[Co(NH ) Cl ] and [Pt(NH ) (H O) Cl]+ +

Both can show geometrical isomerism.

(D) 3 3 3 3 3[Pt(NH ) NO ]Cl and [Pt(NH ) Cl] Br

Both can show ionisation isomerism.

34. Among P, Q, R AND S, the aromatic compound(s) is/are

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(A) P (B) Q

(C) R (D) S

Ans. (ABCD) (A)

(B)

(C) 110 115 C

4 2 3 3 2 2(NH ) CO 2NH CO H O°− °

→ + +��

(D)

35. Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is (are)

(a) is positiveG∆

(b) is positivesystem

S∆

(c) 0surroundings

S∆ =

(d) 0H∆ =

Ans. (BCD) Benzene and Naphthalene form an ideal solution at room temperature.

⇒ H 0∆ =

For mixing of liquids, systemS∆ is positive. (Entropy increase for mixing).

( )sysSurr mix

qS 0 Since for ideal solution H 0

T

∆ = − = ∆ =


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36. The atomic masses of He and Ne are 4 and 20 a.m.u., respectively. The value of the de Broglie wavelength of He

gas at -73 0C is “M” times that of the de Broglie wavelength of Ne at 727

0C. M is

Ans. (5) As we know 2 ( )

h

m KEλ =

As we know KEavg =3/2 KbT

20 10005

4 200

He Ne Ne

Ne He He

m TM

m T

λλ

×= = = =

×

So M = 5

37. 4

EDTA− is ethylenediaminetetraacetate ion. The total number of N-Co-O bond angles in [ ]1( )Co EDTA

−

complex ion is

Ans. (8)

Total number of N Co O− − bond angles = (4 2)× = 8

38. The total number of carboxylic acid groups in the product P is

Ans. (2)

39. A tetrapeptide has COOH− group on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe)

and alanine (Ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primary

structures) with 2NH− group attached to a chiral centre is

Co

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Ans. (4) (gly/valine/Ph-al)

3

O||

NH C C OH|CH

− − − −

Primary structure

If glycine is at left corner, then NH2 will be at if valine at corner

2

O|| gly

NH CH C NH alaninePhenyl alanine|

i pr

− − − −

−

2 arrangements

If Phenyl alanine is at left corner

2

O|| gly

NH CH C NH alaninePhenyl alanine|

Ph

− − − −

2 arrangements

⇒ 4 possible structures.

40. The total number of lone-pairs of electrons in melamine is

Ans. (6) The structure of Melamine is

Hence number of lone pairs = 6

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Part 3 : MATHS



ONLY ONE is correct

41. For 0a b c ,> > > the distance between (1, 1) and the point of intersection of the lines 0ax by c+ + = and

bx + ay + c = 0 is less than 2 2 . Then

(A) 0a b c+ − > (B) 0a b c− + <

(C) 0a b c− + > (D) 0a b c+ − <

Ans.

42. The area enclosed by the curves y sin x cos x= + and y | cos x sin x |= − over the interval 02

π,

is

(A) 4( 2 1)− (B) 2 2( 2 1)−

(C) 2( 2 1)+ (D) 2 2( 2 1)+

Ans.

43. The number of points in ( , ),−∞ ∞ for which 2

0x x sin x cos x ,− − = is

(A) 6 (B) 4

(C) 2 (D) 0

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Ans.

44. The value of

+∑ ∑

= =

−23

1 1

121

n

n

k

kcotcot is:

(A) 23

25 (B)

25

23

(C) 23

24 (D)

24

23

Ans.

45. A curve passes through the point 16

π, .

Let the slope of the curve at each point x, y be 0y y

sec ,xx x

+ >

. Then

the equation of the curve is

(A) 1

2

ysin log x

x

= +

(B) 2

ycos ec log x

x

= +

(C) 2

2y

sec log xx

= +

(D)

2 1

2

ycos log x

x

= +

Ans.

Vidyamandir Classes


46. Let 11

2f : , R

→ (the set of all real numbers) be a positive, non-constant and differentiable function such

that 2f '(x) f ( x )< and 1

12

f .

=

Then the value of

1

1 2/

f ( x )dx∫ lies in the interval

(A) (2 1 2 )e , e− (B) ( 1 2 1)e , e− −

(C) 1

12

e,e

− −

(D)

10

2

e,−

Ans.

47. Let 3 2 ˆˆ ˆPR i j k= + −��

and 3 4 ˆˆ ˆSQ i j k= − −��

determine diagonals of a parallelogram PQRS and 2 3ˆˆ ˆPT i j k= + +��

be

another vector. Then the volume of the parallelepiped determined by the vectors PT ,PQ��

and PS��

is :

(A) 5 (B) 20

(C) 10 (D) 30

Ans.

Vidyamandir Classes


48. Perpendiculars are drawn from points on the line 2 1

2 1 3

x y z+ += =−

to the plane 3x y z .+ + = The feet of

perpendicular lie on the line.

(A) 1 2

5 8 13

x y z− −= =

− (B)

1 2

2 3 5

x y z− −= =

−

(C) 1 2

4 3 7

x y z− −= =

− (D)

1 2

2 7 5

x y z− −= =−

Ans.

49. Four persons independently solve a certain problem correctly with probabilities 1 3 1 1

2 4 4 8, , , . Then the probability

that the problem is solved correctly by at least one of them is

(A) 235

256 (B)

21

256

(C) 3

256 (D)

253

256

Ans.

Vidyamandir Classes


50. Let complex numbers α and 1

αlie on circles 0

2 2 20( ) ( )x x y y r− + − = and 0

2 2 20( ) ( ) 4x x y y r ,− + − =

respectively. If 0 0 0z x iy .= + satisfies the equation

0

2 22 2| z | r ,= + then | |α =

(A) 1

2 (B)

1

2

(C) 1

7 (D)

1

3

Ans.

Vidyamandir Classes





51. A line l passing through the origin is perpendicular to the lines

( ) ( ) ( )1ˆˆ ˆ: 3 1 2 4 2 ,l t i t j t k t+ + − + + + −∞ < < ∞

( ) ( ) ( )2ˆˆ ˆ: 3 2 3 2 2 ,l s i s j s k s+ + + + + −∞ < < ∞

Then, the coordinate(s) of the point(s) on l2 at a distance of 17 from the point of intersection of l and l1 is (are)

(A) 7 7, ,

3 3 3

5

(B) ( )1, 1, 0− −

(C) ( )1, 1, 1 (D) 7 7 8, ,

9 9 9

Ans.

52. Let ( ) sin , 0f x x x xπ= > . Then for all natural numbers n, ( )f x′ vanishes at

(A) a unique point in the interval 1,

2n n +

(B) a unique point in the interval 1, 1

2n n + +

(C) a unique point in the interval ( ), 1n n+

(D) two points in the interval ( ), 1n n+

Vidyamandir Classes


Ans.

53. Let ( )( 1)

4 22

1

1 .

k kn

n

k

S k

+

=

= −∑ Then Sn can take value(s)

(A) 1056 (B) 1088

(C) 1120 (D) 1332

Ans.

54. For 3x3 matrices M and N, which of the following statement(s) is(are) NOT correct?

(A) NT M N is symmetric or skew symmetric, according as M is symmetric or skew symmetric

(B) M N – N M is skew symmetric for all symmetric matrices M and N

(C) M N is symmetric for all symmetric matrices M and N

(D) (adj M) (adj N) = adj (M N) for all invertible matrices M and N

Ans.

Vidyamandir Classes


55. A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8:15 is converted into an open

rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed

squares is 100, the resulting box has maximum volume. Then the lengths of the sides of the rectangular sheet are

(A) 24 (B) 32

(C) 45 (D) 60

Ans.

Vidyamandir Classes





56. Consider the set of eight vectors { }{ }ˆˆ ˆ : , , 1,1 .V ai bj ck a b c= + + ∈ − Three non-coplanar vectors can be

chosen from V in 2p ways. Then p is

Ans.

57. Of the independent events E1, E2 and E3, the probability that only E1 occurs is α, only E2 occurs is β and only E3

occurs is γ. Let the probability p that none of events E1, E2 or E3 occurs satisfy the equations

( ) and ( 3 ) 2p pα β αβ β γ βγ− 2 = − = . All the given probabilities are assumed to lie in the interval (0,1).

1

3

Probability of occurrence of EThen =

Probability of occurrence of E

Ans.

Vidyamandir Classes


58. The coefficients of three consecutive terms of 5

(1 )n

x++ are in the ratio 5 : 10 : 14. Then n =

Ans.

59. A pack contain n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and

the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is k,

then k – 20 =

Ans.

60. A vertical line passing through the point (h, 0) intersects the ellipse

2 2

14 3

x y+ = at the points P and Q. Let the

tangents to the ellipse at P and Q meet at the point R. If ( )h∆ =area of the triangle PQR, 1 1/2 1

max ( )h

h≤ ≤

∆ = ∆ and

2 1/2 1min ( ),

hh

≤ ≤∆ = ∆ then

1 2

88

5∆ − ∆ =

Ans.

Vidyamandir Classes


JEE Advanced 2013

Answer Key

Paper 1 - Code 0

PHYSICS

CHEMISTRY

MATHS

1 D 21 B 41 A

2 A 22 A 42 B

3 D 23 A 43 C

4 A 24 C 44 B

5 B 25 D 45 A

6 B 26 B 46 D

7 C 27 B 47 C

8 C 28 D 48 D

9 A 29 B 49 A

10 B 30 D 50 C

11 B,D 31 A 51 B,D

12 A,C 32 A 52 B,C

13 B,C 33 B,D 53 A,D

14 A,D 34 A,B,C,D 54 C,D

15 B,D 35 B,C,D 55 A,C

16 5 36 5 56 5

17 5 37 8 57 6

18 1 38 2 58 6

19 4 39 4 59 5

20 8 40 6 60 9

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