Jee Advance-2013 (Paper-i)

5/26/2018 Jee Advance-2013 (Paper-i)

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NARAYANA IIT PMT ACADEMYJEE ADVANCE - 2013

PAPER I CODE: 0

1

NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS - INDIA

This booklet contains 33printed pages XYZ Test Booklet Code

PAPER -1 : PHYSIC, CHEMISTY & MATHEMATICS

Please Read the Instructions carefully You are allotted 5 minutes specifically for

this purpose

Important Instructions :

INSTRUCTIONS

A General :

This booklet is your Questions Paper Do not break the seals of this booklet before being instructed to

do so by the invigilators

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Blank spaces and blank pages are provided in the question paper for your rough work No additionalsheets will be provided for rough work

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Write your name and roll number in the space provided on the back cover of this booklet

Answers to the questions and personal details are to be filled on a two part carbon less paper,

which is provided separately These parts should only be separated at the end of the examination wheninstructed by the invigilator The upper sheet is machine gradable objective Response Sheet (ORS)

which will be retained by the invigilator You will be allowed to take away the bottom sheet at the end

of the examinationUsing a black ball point pen darken the bubbles on the upper original sheet Apply sufficient pressure

so that the impression is created on the bottom duplicate sheet

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On breaking the seals of the booklet check that it contains 44 pages and all the 60 questions and

corresponding answers choices are legible Read carefully the instruction printed at the beginning of

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0


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PART I : 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 2 2 3/ 22 2

x y K 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 (B)

K

a (C)

K

2a (D) 0

Ans (D)

Sol Direction of Forcey

tanx

F

ds

Force is radially outwards and displacement is tangential Hence work done = 0

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 k and theother 2k 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 inthe configuration I The time to transport the same amount of heat in the configuration II is

(A) 20 s (B) 30 s (C) 45 s (D) 60 s

Ans (A)Sol In series, equivalent thermal conductivity


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1 2 1 2

eq 1 2K K K

eq

2 1 1

K K 2K

eq

4KK

3

In parallel

eq 1 2 1 1 2 2K (A A ) K A K A

eq

3KK

2

Heat Supplied

1 1

1 1

1

K A TQ t

2 2

2 2

2

K A TQ t

2

4K A 3K 2A. 9 t

3 2 2

1 2( Q Q )

6 = 3t2t2 = 2 sec

3 Two non-reactive monatomic ideal gases have their atomic masses in the ratio 2 : 3 The ratio oftheir 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)Sol P1M1= d1RT

P2M2= d2RT

1 1 1 1

2 2 2 2

PM d d 4 2 8

P M d d 3 3 9

4 A particle of mass m is projected from the ground with an initial speed u0at an angle

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 u0The angle that the composite system makes with the horizontal immediately afterthe collision is

(A)4

(B)4

(C)4

(D)2

Ans (A)


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Sol

u0 o

2 20

max

u sinH2g

ou cos

ou sin

2 2

2 2 0

0

u sinV u 2g

2g

2 2 2

0V u cos

0v u cos

tan 1

4

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 83 10

ms-1

The final momentum of the

object is

(A) 170.3 10

kg ms-1

(B) 171.0 10

kg ms-1

(C) 173.0 10

kg ms-1

(D) 179.0 10

kg ms-1

Ans (B)

Sol E = 30 mW 100 ns

= 3000 10-12

J= 3 10-9J

P = E/C9

8

3 10

3 10= 10

-17kg m/s

6 In the Youngs double slit experiment using a monochromatic light of wavelength , the path

difference (in terms of an integer n) corresponding to any point having half the peak intensity is

(A) (2n 1)2

(B) (2n 1)4

(C) (2n 1)8

(D) (2n 1)16

Ans (B)

Sol 200

II cos

2 2

1cos

2 2

(2n 1) / 2


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2. x

x (2n 1) / 4

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

real and is one-third the size of the object The wavelength of light inside the lens is2

3times the

wavelength in free space The radius of the curved surface of the lens is

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

Ans (C)

Sol1

3/ 22 / 3

i

o

h vmh u

1 v 8

3 u u

u = 24

1 1 1( 1)

f R

1 1 1

v u R

R = 3 m

8 One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end ofanother 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 wireis

(A) 025 (B) 050 (C) 200 (D) 400

Ans (C)

SolR

R

2

1

F(2R)

YL

2L

1 2

F LL

R 2Y


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2

2

F

RYL

L

2 2

F LL

R Y

1

2

L2

L

9 A ray of light travelling in the direction1

i 3j2

is incident on a plane mirror After reflection,

it travels along the direction1

i 3j2

The angle of incidence is

(A) 30 (B) 45 (C) 60 (D) 75

Ans (A)

Sol Direction of Normal is1 1

(i 3j) (i 3) j2 2

3j

Angle between incident ray and normal

1(i

3j).( 3j)2cos1

(i 3j 3j2

332

23

o30

10 The diameter of a cylinder is measured using a Vernier calipers with no zero error It is found

that the zero of the Vernier scale lies between 510 cm and 515 cm of the main scale The Vernier

scale has 50 divisions equivalent to 245 cm The 24th

division of the Vernier scale exactlycoincides with one of the main scale divisions The diameter of the cylinder is

(A) 5112 cm (B) 5124 cm (C) 5136 cm (D) 5148 cm

Ans (B)

Sol MSD = 005 cm

2.45VSD 0.049

50cm

LC = 0050049 = 0001 cm

Reading = 510 + 24 0001 cm

= 510 + 0024 cm = 5124 cm


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SECTION -2 : (One or more options correct Type)

The section contains 5 multiple choice questions Each question has four choices (A), (B), (C) and (D)out of which ONE and MORE are correct

11 In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C

The switch S1is pressed first to fully charge the capacitor C1and then released The switch S2isthen pressed to charge the capacitor C2After some time, S2 is released and then S3 is pressed

After some time,

(A) the charge on the upper plate of C1is 2CV0

(B) the charge on the upper plate of C1is CV0(C) the charge on the upper plate of C2is 0

(D) the charge on the upper plate of C2isCV0S1 S2 S3

C1 C2

V02V0

Ans B, D

Sol: When s1is closed and then released the charge on upper plate of C 1is +2CVoWhen S2is closed

and then released charge will be distributed equally in C1 and C2 So the charge on the upperplate of C2is +CVoWhen S3is pressed the charge on upper plate of C2will become (CVo)

12 A particle of mass M and positive charge Q, moving with a constant velocity 11u 4ims , enters

a region of uniform static magnetic field normal to the x-y plane The region of the magnetic field

extends form 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

u 2 3i j ms . The correct

statements (s) is (are)(A) The direction of the magnetic field isz direction

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

(C) The magnitude of the magnetic field50 M

3Qunits

(D) The magnitude of the magnetic field isM

3Qunits

Ans A, C

Sol:


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x x x x x x x x

x x x x x x x x

x x x x x x x xx x x x x x x x

x x x x x x x xx x x x x x x x

x x x x x x x x

x x x x x x x xx x x x x x x xx x x x x x x x

x = 0 x = L

30o30

o

x

y

As particle turns toward +ve y axis the direction of field must be alongz axis (as)

[use : F q v B ]

Since2

u 2 3i j

o1tan3

So particle emerges at an angle of 30owith initial direction

Hence Time to turn by 30ois

T T

2 6 12

31 2 mt 10 1012 qB

50 MB

3Q

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

equation, y(x, t) = (001 m) sin[(628 m1) x] cos [(628 s1)t]

Assuming

= 314, the correct statement (s) is (are)(A) The number of nodes is 5(B) The length of the string is 025 m

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

001 m

(D) The fundamental frequency is 100 HzAns B, C

Sol: 1 1y x, t 0.01m sin 62.8m xcos 628s t

Here,

2

k 62.8 20 0.1m

And 1628 200 s

1v 10msk

Since string is vibrating in 5th

harmonic so

5 5 0.1L 0.25

2 2

m


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and fundamental frequency

v 10f 20Hz

2L 0.5

The midpoint is the antinode hence A = 001 m14 A solid sphere of radius R and density

is attached to one end of a mass-less spring of forceconstant k The other end of the spring is connected to another solid sphere of radius R and

density 3 The complete arrangement is placed in a liquid of density 2 and is allowed to reach

equilibrium The correct statement (s) is (are)

(A) the net elongation of the spring is34 R g

3k

(B) the net elongation of the spring is3R g

3k

(C) the light sphere is partially submerged

(D) the light sphere is completely submergedAns A, DSol:

A

B

B

34 R 2 g3

kx

34 R g3

If whole system is submerged then upthrust is equal to the weight of the system hence the sphereA must be completely submerged On sphere B net force must be zero as weight of sphere B is

greater than upthrust, hence there must be elongation in spring

So for sphere B 3 34 4

kx R 2 g R 3 g.3 3

34 R g3x

k

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

1

and 2respectively, touch each other The net electric field at a distance 2 R and from the centre

of the smaller sphere, along the line joining the centres of the spheres, is zero The ratio 1

2

can

be

(A)4 (B)32

25


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(C)32

25 (D) 4

Ans B, D

Sol: If 1and 2both are both positive or both negative then field will be zero at a point in betweenthe line joining the centre

ie,3

1 2

2

o o

R

3 32RR

1

2

4

Case II: If one of the charge density is negative and other is positive the point will lies on the left

side of smaller sphere33

1 2

2 2

o o

2RR0

3 32R 5R

1

2

32

25

SECTION-3: (Integer value correct Type)

The section contains 5 questions The answer to each question is a single digit integer, ranging from 0 to

9 (both inclusive)

Integer

16 A bob of mass m, suspended by a string of length l1, is given a minimum velocity reqired tocomplete a full circle in the vertical plane At the highest point, it collides elastically with anotherbob 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

Sol: The speed of the mass attached to string of length l1 at the highest point will be 1gl which

should be equal to the critical speed of mass attached to the string of length l 2at the bottom-mostpoint ie

1 25gl

gl

or 1

2

5l

l

17 A particle of mass 02 kg is moving in one dimension under a force that delivers a constant power

05 W to the particle If the initial speed (in ms1

) of the particle is zero, the speed (in ms1

) after 5s is

Ans 5


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Sol: w k

21pt mv2

1v 5ms

18 The work functions of Silver and Sodium are 46 and 23 eV, respectively The ratio of the slope of

the stopping potential versus frequency plot for Silver to that of Sodium is

Ans 1

Sol: maxkE hf W and max skE ev

slope of either graph ish

ewhich is constant

ratio of slopes = 1

sv

f

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

per second Given that ln 2 = 0693, 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

Sol: Required fraction =

t

o

o

N 1 e

1 e 0.04N

% fraction 4%

20 A uniform circular disc of mass 50 kg and radius 04 m is rotating with an angular velocity of 10

rad s1

about its own axis, which is vertical Two uniform circular rings, each of mass 625 kg and

radius 02 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 enoughsuch that the rings are at rest relative to the disc and the system rotates about the original axis

The new angular velocity (in rad s1

) of the system is

Ans 8

Sol: From conservation of angular momentum

2 22 21 1 R R 50 R 10 50 R 2 6.25 6.25

2 2 2 2

8rad/s


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PART II : CHEMISTRY

SECTION1 (Only One option correct Type)

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

out of the which ONLY ONEis correct

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

P = [FeF6]3-

, Q = [V(H2O)6]2+

and R = [Fe(H2O)6]2+

The correct order of the complex ions, according to their spinonly magnetic moment values (in

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

Ans (B)

[FeF6]

3

, Fe

3+

has 5 unpaired electron= 35 BM

[V(H2O)6]2+

, V2+

has 3 unpaired electrons

= 15 BM

[Fe(H2O)6]2+

, Fe2+

has 4 unpaired electrons

= 24 BM

P > R > Q

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 pmAns (A)

cation in octahedral voids

r0.414

r (without distortion)

r0.414

250

r+= 0414

250 = 1035 104 (approx)

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)Common sulphide ore are given by

Ag Ag2S : Silver glance

Cu CuS : Copper glance

Pb PbS : Galena

24 The standard enthalpies of formation of CO2(g), H2O(l) and glucose (s) at 250C are -400 kJ/mol,

-300 kJ/mol and -1300 kJ/mol, respectively The standard enthalpy of combustion per gram of


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glucoses at 250C is

(A) +2900 kJ (B) -2900 kJ (C) -1611 kJ (D) +1611 kJAns (C)

I C + O2 CO2 H = 400 kJ/mol

II H2 + 21

O2

H2O H = 300 kJ/mol

III 6C + 6H2+ 3O2 C6H12O6 H = 1300 kJ/mol

C6H12O6 + 9O2 6CO2 + 6H2O

H = (400 6) + (300 6)(1300)

H =2900 kJ / mole

H of per gram of Glucose = 2900

16.11kJ / gram180

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)On reaction with H2S in presence of NH4Cl/NH4OH only Zn reacts and gives ZnS (white)

Zn2+

+ H2S ZnS (white ppt)

(IV group)

26 Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at 250C For this

process, the correct statement is(A) The adsorption requires activation at 25

0C

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

(C) The adsorption increases with increases of temperate

(D) The adsorption is irreversibleAns (B)

Adsorption of phenolphthalein is exothermic at 250C

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

as

H3C - Cl ClCl

Cl

O

P Q RS

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

Ans (B)

S > P > R > Q

SN2


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Note :Methyl halide has lower rate of SN2 then allyl halide according to the data given in Jerry

March (4th

Edition, Page No 339, Book NameAdvance Organic Chemistry)

Average relative SN2 rates for some alkyl substratesReactant Relative rate

Methyl 30

Isopropyl 0025

Allyl 40

But considering the different options given in question the most appropriate order

S > P > R > Q

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 theconcentration of Q varies with reaction time as shown in the figure The overall order of the

reaction is

timetime

[Q]

[Q]0

(A) 2 (B) 3 (C) 0 (D) 1

Ans (D)

In the reaction

P + Q R + Stime taken for 75% of composition is two times of the decomposition of 50% of P then the order

of reaction with respect to P is 1

for

P C0 C0/2 C0/4

t t

according to graph concentration of Q completely decomposed in given time then order of

reaction with respect to Q is 0

overall order of reaction = 1 + 0 = 1

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

(A) NO (B) NO2 (C) N2O (D) N2O4Ans (B)

On long standing concentrated HNO3 gives brown colored gas

HNO3(con) NO2(g) + H2O + O2(brown coloured)


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30 The compound that does NOT liberate CO2, on treatment with aqueous sodium bicarbonatesolution is

(A) Benzoic acid (B) Benzenesulphonic acid(C) Salicylic acid (D) Carbolic acid (Phenol)

Ans (D)

Phenol does not reacts with NaHCO3Because phenol is less acidic then HCO3 Thus according

to acid-base concept PhO can not abstract the hydrogen from HCO3 and do not give CO2gas

COOH SO3H COOH

OH

OH

SECTION2 : (One or more options correct Type)

This section contains 5 multiple choice questions Each question has four choices (A), (B), (C) and

(D) out of which ONE or MORE are correct

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

of that

of a strong acid (HX, 1 M), at 25oC The aK of HA is

(A) 1 10-4

(B) 1 10-5

(C) 1 10-6

(D) 1 10-3

Ans (A)

1 1r H

2 2r H

1 1

22

Hr

r H

1

a2

H 1100

H

K C

Ka C = 10-4

Ka= 1 10-4

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

(A) p (empty) and * electron delocalisations

(B) * and electron delocalisations

(C) p (filled) and electron delocalisations

(D) p (filled) * and * electron delocalisations


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Ans (A)

t-butyl carbocation is stabilized by p overlapping

2-butene is stabilized by * overlapping

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

(A) 3 25Cr NH Cl Cl and 3 24Cr NH Cl Cl

(B)3 24

Co NH Cl and 3 22Pt NH H O Cl

(C)2

2 2CoBr Cl and2

2 2PtBr Cl

(D) 3 33Pt NH NO Cl and 3 3Pt NH Cl Br

Ans (B), (D)

(A) 3 25Cr NH Cl Cl does not exhibit isomerism, 3 24Cr NH Cl Cl exhibits geometrical

isomerism

(B)3 24

Co NH Cl and 3 22Pt NH H O Cl

Both can show geometrical isomerism

(C)2

2 2CoBr Cl - tetrahedral so it does not exhibit isomerism

2

2 2PtBr Cl - square planar so it exhibits geometrical isomerism

(D) 3 33Pt NH NO Cl and 3 3Pt NH Cl Br

Both can show isonisation isomerism

34 Among P, Q, R and S, the aromatic compound(s) is/are(A) P (B) Q (C) R (D) SCl

3AlCl

P

NaH Q

O O

4 32NH CO

o100 115 C

R

OHCl

S

Ans (A), (B), (C), (D)


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Cl

3AlCl4

AlCl

2 electronaromatic

H

NaH

Na

2H

6 electronaromatic

O

O

OH

O

OO

OH

O

H

enolizationO

H

H

baseO

H

6 electronaromatic

O

H

OH

6 electronaromatic

35 Benzene and naphthalene form an ideal solution at room temperature For this process, the true

statement(s) is(are)

(A) G is positive (B) systemS is positive

(C) surroundingsS 0 (D) H 0

Ans (B), (C), (D)

system system systemG H T S

system surroundingH T S So as solution is ideal

systemH 0

Sosurrounding

S 0

As mixing is spontaneous so systemG 0

system systemH T S 0

So assystemH = 0 , systemS must be greater than zero


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SECTION3 : (Integer value correct Type)

This section contains 5 questions The answer to each question is a single digit integer, ranging from 0 to9 (both inclusive)

36 The atomic masses of He and Ne are 4 and 20 amu, respectively The value of the de Broglie

wavelength of He gas at 73oC is M times that of the de Broglie wavelength of Ne at 727

oC M

is

Ans 5He : m1= 4, T1= 200 K

Ne : m2= 20, T2= 1000 K

He

Ne Ne

h

m

T

v

M

So HeHe

He He

mh

m T

NeNe

Ne Ne

mh

m T

He Ne Ne

Ne He He

T m

1000 2025 5

T m 4 200

374

EDTA is ethylenediaminetetraacetate ion The total number of N Co O bond angles in1

Co EDTA complex ion is

Ans 8

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


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O

O

O O

O

31. H O ,

2. O3

3. H O2 2

P

Ans 2

O

O

O O

O

O

O

3H O

O

O O

O

OH

OH

keto acid

3 2 2O / H O

O

OO

O

OH

OH

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 NH2group attached to a chiral center isAns 4

3 2CH CH CH CO H

CH3

NH2

Valine (V)

3 2CH CH CO H

NH2

Alanine (A)

2 2H C CH CO H

NH2

Phenylalanine (P)

2 2 2H N CH CO H Glycine (G)

As (A) is at the end so options possible with NH2at beginning are(i) VPGA, (ii) VGPA, (iii) PVGA and (iv) PGVA

40 The total number of lone-pairs of electrons in melamine isAns 6


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N

N

N

NH2NH2

NH2 Melamine

Total number of lone pairs = 6

PART III : MATHEMATICS

SECTION1 (Only One option correct Type)

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

out of the which ONLY ONEis correct

41 For 0,a b c the distance between 1,1 and the point of intersection of the lines

0ax by c and 0bx ay c

is less than 2 2. then

(A) 0a b c (B) 0a b c (C) 0a b c (D) 0a b c 41 (A)

Point of intersection ,c c

a b a b

Distance of (1, 1) from point of

intersection

2 2

1 1c c

a b a b

2 a b c

a b

Acc to question

2 2 2a b c

a b

0a b c

42 The area enclosed by the curves sin cosy x x and cos siny x x over the interval 0,2

is

(A) 4 2 1 (B) 2 2 2 1 (C) 2 2 1 (D) 2 2 2 1

42 (B)

Required Area/4 /2

0 /4

2sin 2cosx dx x dx

/4 /2

0 /42 cos 2 sinx x


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1 12 1 2 1

2 2

4 2 2 2 2 2 1

43 The number of points in , for which2 sin cos 0,x x x x is

(A) 6 (B) 4 (C) 2 (D) 0

43 (C)

Let2 sin cos

' 2 sin cos sin

' 2 cos

f x x x x x

f x x x x x x

f x x x

When 0 ' 0x f x f x is sI function

When 0 ' 0x f x f x is sd function

At x = 0 ' 0f x

0 1f

When x f x

f x will be zero at exactly two points

44 The value of cot23

1

1 1

cot cot 1 2n

n k

k is

(A)23

25 (B)

25

23 (C)

23

24 (D)

24

23

44 (B)23

1

1 1

cot cot 1 2n

n k

k

1 1

2. 12 2 1

2

n n

k k

n nk k n n

231 2

1

cot 1n

n n

1 2cot 1rT r r

1

2

1tan

1 r r

1 1

tan1 1

r r

r r


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1 1tan 1 tanrT r r

231

1

tan 24

4

r

n

T

1

11 1

2524cot tan 2414 23

124

45 A curve passes through the point 1, .6

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

sec , 0.y y

xx x

Then the equation of the curve is

(A)

1

sin log 2

y

xx (B) cos log 2

y

ec xx

(C)2

sec log 2y

xx

(D)2 1

cos log2

yx

x

45 (A)

Given

secdy y y

dx x x (i)

Let y vx

dy dvv x

dx dx

(ii)

Substituting (ii) in (i)

secdv

v x v vdx

cos

dxv dx

x

At 1,6 6

x y v at 1x

/6 1

cos

v xdx

v dvx

/6 1sin logv xv x

sin sin log6

v x

1sin log

2

yx

x


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46 Let1

: ,12

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

function such that ' 2f x f x and1

1.2f Then the value of

1

1/2

f x dx lies in the

interval

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

, 12

ee (D)

10,

2

e

46 (D)

' 2f x

f x

2 2' 2 0

x xe f x e f x

Or 2 0xd

e f x

dx

2xe f x is sd nf in1

,12

2 1xe f x e

2 10 xf x e

11 2 1

1/2 1/2

02

xef x dx

1

1/2

10

2

ef x dx

47 Let 3 2PR i j kand 3 4SQ i j k determine diagonals of a parallelogram PQRS and

2 3PT i j kbe another vector Then the volume of the parallelepiped determined by the

vector ,PT PQ and PS is

(A) 5 (B) 20 (C) 10 (D) 30

47 (C)S R

P Q

a

b

3 2a b i j k 2 3c i j k

3 4a b i j k

10 10 10a b a b i j k


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2 10 10 10a b i j k

5a b i j k

Now volume of parallelepiped

. 5 1 2 3c a b c a b

10

48 Perpendiculars are drawn from points on the line2 1

2 1 3

x y zto the plane 3.x y z The

feet of perpendiculars 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

48 (D)

The given line is2 1

2 1 3

x y zr

General point on the line is

2 2, 1, 3P

r r r

Now the direction ratio of PQ

are 2 2, 1, 3r r r

which are proportional to (1, 1, 1)

2 2 1 31 1 1

r r r k

2 2

1

3

k r

k r

k r

Point , , lies on pane 3x y z so

6 42 2 1 3 3

3

rk r k r k r k

Now6 4 3

2 2

3 2

rr r

6 4 3 31

3 7

rr r and

6 4r 3 63r r

3 5

3 3 3 3 6 1 2

2 7 5 2 7 5

x y z

P

A

(-2,-1, 0)

Q


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49 Four person independently solve a certain problem correctly with probabilities1 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

49 (A)

Prob that the problem is solved correctly = 1prob that problem is not solved correctly bynone of them

1 1 3 71 . . .

2 4 4 8

211

256

235

256

50 Let complex numbers

and1

lie on circles2 2 2

0 0x x y y r and

2 2 2

0 0 4 ,x x y y r respectively If 0 0 0z x iy satisfies the equation2 2

02 2,z r then

(A)1

2 (B)

1

2 (C)

1

7 (D)

1

3

50 (c)

0z r (i)

0

12z r

01 2z r (ii)

(ii)/(i) gives

0

0

12

z

z

2

0 0 0 01 1 4z z z z

Or2 2 2 2 2

0 0 0 0 0 01 4 .z z z z z z (A)

Or2 2 4 2 2 2

0 0 0 01 4 4 1 4z z z z (B)

From (i)2 2 2

0 0 0z z z r

22 2

0 0 12

rz z r


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22

12

r (C)

From (B) & (C)2 2

2 4 21 1 4 4 1

2 2

r r

22 2

1 1 42

r

1

7

SECTION2 : (One or more options correct Type)

This section contains 5 multiple choice questionsEach question has four choices (A), (B), (C) and (D)

out of which ONE or MOREare correct

51 A line lpassing through the origin is perpendicular to the lines

1 : 3 t i 1 2t j 4 2t k, t

2 : 3 2s i 3 2s j 2 s k, s

Then, the coordinate(s) of the point(s) on2

at a distance of

17 from the point of intersection

of and 1 is (are)

(A)7 7 5

, ,3 3 3

(B) 1, 1,0 (C) 1,1,1 (D)7 7 8

, ,9 9 9

Key (B, D)

Sol Given equation of line 1 and 2 are

x 3 y 1 z 4

1 2 2 (1)

andx 3 y 3 z 2

2 2 1 (2)

dr of line are 2,3, 2

Equation of line

is

x y z

2 3 2 (3)

Any point on (3) is 2k,3k, 2k

To find point of intersection of (1) and (3)

We have


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3k 1 2k 42k 3

2 2

4k 6 3k 1 2k 4

7k 7 k 1

Point of intersection of (1) and (3) is P 2, 3,2

Any point on line 2 is Q 3 2S,3 2S,2 S

PQ

17 2 2 21 2S 6 2S S 17

29S 28S 20 029S 10S 18S 20 0

9S 10 S 2 0

10S ,S 2

9

Q point is7 7 8

, ,9 9 9

or 1, 1,0

52 Let f x x sin x, x 0 Then for all natural numbers n,f ' x vanishes at

(A) a unique point in the interval1

n, n2

(B) a unique point in the interval1

n , n 1

2

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

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

Key (B, C)

Sol


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y x

y

x

1x 2

3

x 2

5x

2

2,01, 00,0

f ' x x cos x. sin x , x > 0

f ' x x cos sin x

Now f ' x 0

xcos x sin x 0 x tan x

Now taking

y tan x, and y x

From graph,

A unique point in the interval 1n , n 12

and a unique in the internal n, n 1

n N

53 Let

k k 14n 2

2

n

k 1

S 1 k Then nS can take value(s)

(A) 1056 (B) 1088 (C) 1120 (D) 1332

Key (A, D)

Sol

k k 14n 2

2

n

k 1

S 1 k

2 2 2 2 2 2

nS 1 2 3 4 5 6 .....4n terms

2 2 2 2

nS 1 5 9 ..... 4n 3 22 2 22 6 10 ...... 4n 2

22 2 23 7 11 ..... 4n 1 22 2 24 8 12 ...... 4n

n n n n n n2 2

n

r 1 r 1 r 1 r 1 r 1 r 1

S 16 r 24 r 9 1 16 r 16 r 4 1


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n n n2

r 1 r 1 r 1

16 r 8 r 1 216 r

n n

r 1 r 1

32 r 12 1

n n 132 12n

2= 4n 4n 1

Therefore the required solutions are 1056, 1332

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

(A)T

N M N is symmetric or skew symmetric, according as M is symmetric or skew symmetric

(B) MN NM is skew symmetric for all symmetric matrices M and N(C) MN is symmetric for all symmetric matrices M and N

(D) adjM adjN adj MN for all invertible matrices M and N

Sol (C, D)(A)

T T TTT T TN MN N MN MN N

T TN M N TN MN (If M is symmetric TM M

)TN MN (If M is skew-symmetric TM M

)

So option A is correct

(B)T T T

MN NM MN NM

T

T T T T

T

N NN M M N

M M

NM MN

MN NM

So option B is correct

(C)T T TMN N M NM (M and N are symmetric )

So option (C) is wrong

(D) adjM adjM adj MN Given

So option D is wrong55 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 volumeThen the lengths of the sides of the rectangular sheet are

(A) 24 (B) 32 (C) 45 (D) 60

Key (A, C)

Sol


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a a

aa

8x 2a 8x 2a

15x 2a Volume of the box V 8x 2a 15x 2a .a

3 2 2V 4a 46a x 120x .a For max volume

dV0

daand

2

2

d V0

da

2dV

12a 92ax 120ax 0da 2

2

d V24a 92x 0

daat

5xa

3

Now given that sum of Area 24a 100 a 5

So from5x

a3

x 3

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)

56 Consider the set of eight vectors V ai bj ck : a, b,c 1,1 There non-coplanar vectors

can be chosen from V in 2Pways Then p is

Ans 5Sol Consider a cube of side length 2 unit, whose centre is at origin (assumed sides parallel to

coordinate axis)

Required no of Triplets of non-coplanar vectors 4 53C .2.2.2 32 2

P 5 57 Of the three independent events E1, E2and E3, the probability that only E1occurs is , only E2

occurs is and only E3occurs is Let the probability p that none of events E1, E2or E3occurs

satisfy the equations 2 p and 3 p 2 All the given probabilities are

assumed to lie in the interval (0, 1)


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Then 1

3

Probability of occurrence of E

Probability of occurrence of E

Ans 6Sol Let 1 2 3P E x,P E y,P E z

x 1 y 1 z

y 1 x 1 z

z 1 x 1 y and p 1 x 1 y 1 z

Given equations are

2 p and 3 p 2

1 2 1

p (1)

1 3 2

p (2)

(1) x 2y and (2) y 3z x 6z

x6

z

1

3

P E6

P E

58 The coefficients of three consecutive terms ofn 5

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

Ans 6

Sol n 5 n 5 n 5r 1 r r 1

C C C 5:10:14

n 5 ! n 5 ! n 5 !: : 5 :10 :14

r 1 ! n r 6 ! r! n r 5 ! r 1 : n r 4 !

n n r 5:1: 5 :10 :14

n r 6 r 1 r 1

2r n r 6n r 6 2

n 3r 6 0 (I)

n r 5 7

r 1 5 5n 5r 25 7r 7

5n 12r 18 0 (II)4n 12r 24 0 (III)____________

n = 659 A pack contains 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 5


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Sol

2n n2k 1 1224

2

2n n 4k 2450 02n n 2450

k4

But 1 k n 124 n n 2450 4n 4

1

9785n

2

n 49, n N

But for n 49,51, k NSo n = 50,

2n 3n 2459 0

3

9817n

2

n 51

K = 25

k 20 5

60 A vertical line passing through the point h,0 intersects the ellipse2 2x y

14 3

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, max min1 1/2 h 1 2 1/2 h 1h and h , then 1 28

85

Ans 9

Sol

p 2 cos , 3 sin

R 2 sec , 0

Q 2 cos , 3 sin

(h, 0)

Area of 1PQR 2 3 sin 2sec 2cos2

32 3sin

cos

Which is decreasing function of cos


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1 1h 1 2cos

2 2

1 1cos

4 2

3

1

152 3 4

4

2 3 15.

15

16

45 5

8

1

845

5

3

2

32 3 2

2

2

3 3 3 9

4 2

28 36

So

1 2

88 9

5


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ANSWER KEY

PHYSICS CHEMISTRY MATHEMATICS

Q.No. Answer Q.No. Answer Q.No. Answer

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 5 378

57 6

18 1 38 2 58 6

19 4 39 4 59 5

20 8 40 6 60 9

Jee Advance-2013 (Paper-i)

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