NARAYANA IIT/PMT ACADEMY - iit jee coaching institute of delhi

NARAYANA IIT/PMT ACADEMY JEE ADVANCE - 2013 PAPER – I This booklet contains 33 printed pages XYZ

CODE: 0 Test Booklet Code

PAPER -1 : PHYSIC, CHEMISTY & MATHEMATICS Please Read the Instructions carefully You are allotted 5 minutes specifically for

0

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 The question paper CODE is printed on the right hand top corner of this sheet and on the back page (Page No 44) of this booklet Blank spaces and blank pages are provided in the question paper for your rough work No additional sheets will be provided for rough work Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers and electronic gadgets are NOT allowed inside the examination hall 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 when instructed 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 examination Using 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 DO NOT TAMPER WITH/MUTILATE THE ORS OR THE BOOKLET 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 each section and TOTAL MARKS 180. B, Filling the right part of the ORS The ORS also has a CODE printed on its left and right parts Check that the CODE printed on the ORS (on both sheets) is the same as that on this booklet and put your signature affirming that you have verified this IF THE CODES DO NOT MATCH, ASK FOR A CHANGE OF THE BOOKLET

1

NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS - INDIA

NARAYANA IIT/PMT ACADEMY JEE ADVANCE - 2013 CODE: 0

PAPER – I

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

x

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

x

2

y

2

3/ 2

ˆi

(x

2

y ˆj (K being a y 2 )3 / 2

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 2K K K (A) (B) (C) (D) 0 a a 2a Ans (D) y Sol Direction of Force tan x ds F

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 the other 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 in the configuration I The time to transport the same amount of heat in the configuration II is

(A) 20 s

Ans Sol

(B) 30 s

(C) 45 s

(D) 60 s

(A) In series, equivalent thermal conductivity 2



PAPER – I 1

2

K eq 2 K eq

1 K

K eq

4K 3

1

2

K1

K2

1 2K

In parallel

Keq (A1 A2 ) K1A1 K 2 A2 3K 2

K eq Heat Supplied

K1A1 T

Q1

t1

1

K2A2 T

Q2

t2

2

4K A . 9 3 2 6 = 3t2 t2 = 2 sec

3K 2A t2 2

(

Q1

Q2 )

3

Two non-reactive monatomic 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) Sol P1M1 = d1RT P2M2 = d2RT P1M1 d1 d1 4 2 8 P2 M 2 d 2 d2 3 3 9 4

A particle of mass m is projected from the ground with an initial speed u0 at 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 u0 The angle that the composite system makes with the horizontal immediately after the collision is (A)

Ans

4

(B)

4

(C)

4

(D)

2

(A)

3



PAPER – I o

u0

Sol

u 02 sin 2 2g

H max u o cos u o sin

u 02 sin 2 2g 2 2 u 0 cos

V2

u 02

V2

2g

v u 0 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 3 108 ms-1 The final momentum of the object is (A) 0.3 10 17 kg ms-1 (B) 1.0 10 17 kg ms-1 (C) 3.0 10 17 kg ms-1 (D) 9.0 10 17 kg ms-1

Ans (B) Sol

6

E = 30 mW 100 ns = 3000 10-12 J = 3 10-9 J P = E/C 3 10 9 = 10-17 kg m/s 8 3 10

In the Young’s 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

I0 2

I0 cos 2

2 1 cos 2 2 (2n 1) / 2

4



PAPER – I

2

. x

x (2n 1) / 4

7

Ans Sol

8

Ans

The image of an object, formed by a plano-convex lens at a distance of 8m behind the lens, is 2 real and is one-third the size of the object The wavelength of light inside the lens is times the 3 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 (C) 1 3/ 2 2/3 hi v m ho u 1 v 8 3 u u u = 24 1 1 1 ( 1) f R 1 1 1 v u R R=3m 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) 025 (B) 050 (C) 200 (D) 400

(C) R

Sol Y

F (2R) 2 L1 2L

R

L1

F L R 2 2Y

5



PAPER – I

Y L1 L2

9

F R2 L2 L

L2

2

A ray of light travelling in the direction

1 ˆ i 2 (B) 45°

it travels along the direction (A) 30°

Ans

(A)

Sol

Direction of Normal is

F L R2 Y

1 ˆ i 2

3jˆ is incident on a plane mirror After reflection,

3jˆ The angle of incidence is

1 1 (i 3j) (i 2 2 3jˆ

(C) 60°

(D) 75°

3) j

Angle between incident ray and normal 1 (i 3j).( 3j) 2 cos 1 ˆ (i 3jˆ 3jˆ 2 3 3 2 2 3 o 30 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 exactly coincides 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 2.45 VSD 0.049 cm 50 LC = 005 – 0049 = 0001 cm Reading = 510 + 24 0001 cm = 510 + 0024 cm = 5124 cm 6


NARAYANA IIT/PMT ACADEMY JEE ADVANCE - 2013 PAPER – I

CODE: 0

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 S1 is pressed first to fully charge the capacitor C1 and then released The switch S2 is then 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 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 –CV0 S1

S2 C1

2V0

Ans Sol:

12

Ans Sol:

S3 C2 V0

B, D When s1 is closed and then released the charge on upper plate of C1 is +2CVo When S2 is closed and then released charge will be distributed equally in C1 and C2 So the charge on the upper plate of C2 is +CVo When S3 is pressed the charge on upper plate of C2 will become (–CVo) ˆ 1 , enters A particle of mass M and positive charge Q, moving with a constant velocity u1 4ims 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 u 2 2 3iˆ ˆj ms 1. The correct statements (s) is (are) (A) The direction of the magnetic field is –z direction (B) The direction of the magnetic field is +z direction 50 M (C) The magnitude of the magnetic field units 3Q M (D) The magnitude of the magnetic field is units 3Q A, C

7


NARAYANA IIT/PMT ACADEMY JEE ADVANCE - 2013 PAPER – I y

x x x x x x x x x x x x x x x xo x x 30 x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x xx= x0 x x x x x x x x x x x x x x x x x x x x x x x x x xturns x x particle

x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x toward

As [use : F q v B ] Since u 2

2

x x x x x x x 30xo x x x x x x x x x x x xx= xL x x x x x x x x y axis the +ve

CODE: 0

direction of field must be along –z axis (as)

3iˆ ˆj

1 o 3 So particle emerges at an angle of 30o with initial direction T T Hence Time to turn by 30o is 2 6 12 1 2 m t 10 10 3 12 qB 50 M B 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 m–1) x] cos [(628 s–1)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 Hz Ans B, C Sol: y x, t 0.01m sin 62.8m 1 x cos 628s 1 t tan

Here, k And

v

2

62.8 20

628 200 s 10ms

0.1m

1

1

k Since string is vibrating in 5th harmonic so 5 5 0.1 L 0.25 m 2 2 8



PAPER – I

14

Ans Sol:

and fundamental frequency v 10 f 20Hz 2L 0.5 The midpoint is the antinode hence A = 001 m A solid sphere of radius R and density 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 3 The complete arrangement is placed in a liquid of density 2 and is allowed to reach equilibrium The correct statement (s) is (are) 4 R3 g (A) the net elongation of the spring is 3k R3 g (B) the net elongation of the spring is 3k (C) the light sphere is partially submerged (D) the light sphere is completely submerged A, D

A

4 3 R 2 g 3 kx

B B

15

4 3 R 3

g

If whole system is submerged then upthrust is equal to the weight of the system hence the sphere A 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 4 3 4 3 So for sphere B kx R 2 g R 3 g. 3 3 4 R3 g x 3 k Two non-conducting solid spheres of radii R and 2 R, having uniform volume charge densities 1 and 2 respectively, 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

can

2

be (A) –4

(B)

32 25 9



CODE: 0

32 (D) 4 25 B, D If 1 and 2 both are both positive or both negative then field will be zero at a point in between the line joining the centre R3 2 ie, 1 R 2 3 o 2R 3 o (C)

Ans Sol:

1

4

2

Case II: If one of the charge density is negative and other is positive the point will lies on the left side of smaller sphere 1

3

o

R3 2R 1 2

2 2

3

o

2R 5R

3 2

0

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 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 l required to complete a full circle in the vertical plane, the ratio 1 is l2 Ans 5 Sol: The speed of the mass attached to string of length l1 at the highest point will be gl1 which should be equal to the critical speed of mass attached to the string of length l 2 at the bottom-most point ie gl1 5gl2 l1 5 l2 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 ms–1) of the particle is zero, the speed (in ms–1) after 5 s is 5

or 17

Ans

10



PAPER – I Sol:

18 Ans Sol:

w

k 1 2 pt mv 2 v 5ms

1

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 1 kE max hf W and kE max evs h slope of either graph is which is constant e ratio of slopes = 1

vs

f

19

Ans Sol:

20

Ans Sol:

A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations 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 4 No 1 e t Required fraction = 1 e 0.04 No % fraction 4% A uniform circular disc of mass 50 kg and radius 04 m is rotating with an angular velocity of 10 rad s–1 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 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 s–1) of the system is 8 From conservation of angular momentum 1 50 R 2 10 2

1 R 50 R 2 2 6.25 2 2

2

R 6.25 2

2

8 rad / s

11



PAPER – I

PART II : CHEMISTRY 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 the which ONLY ONE is correct 21

Ans

22

Ans

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 spin –only magnetic moment values (in BM) is (A) R < Q < P (B) Q < R < P (C) R < P < Q (D) Q < P < R (B) [FeF6]3 , Fe3+ 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 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 (A) cation in octahedral voids r 0.414 (without distortion) r r 0.414 250

r+ = 0414 23 Ans

24

250 = 1035

104 (approx)

Sulfide ores are common for the metals (A) Ag, Cu and Pb (B) Ag, Cu and Sn (A) Common sulphide ore are given by Ag Ag2S : Silver glance Cu CuS : Copper glance Pb PbS : Galena

(C) Ag, Mg and Pb

(D) Al, Cu and Pb

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 12



PAPER – I 0

Ans

glucoses at 25 C is (A) +2900 kJ (B) -2900 kJ (C) I C + O2 CO2 II H2 +

1 O2 2

(C) -1611 kJ H = 400 kJ/mol

H2O

H = 300 kJ/mol

III 6C + 6H2 + 3O2 C6H12O6 C6H12O6 + 9O2 6CO2 + 6H2O H = (–400 6) + (–300 6) – (–1300) H = –2900 kJ / mole H of per gram of Glucose = 25 Ans

26

Ans

27

(D) +1611 kJ

2900 180

H = 1300 kJ/mol

16.11kJ / gram

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) (D) On reaction with H2S in presence of NH4Cl/NH4OH only Zn reacts and gives ZnS (white) Zn2+ + H2S ZnS (white ppt) (IV group) 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 250C (B) The adsorption is accompanied by a decreases in enthalpy (C) The adsorption increases with increases of temperate (D) The adsorption is irreversible (B) Adsorption of phenolphthalein is exothermic at 250C KI in acetone, undergoes SN2 reaction with each of P, Q, R and S The rate of the reaction vary as O Cl

H3C - Cl

Cl Cl

P

Q

(A) P > Q > R > S Ans

R

(B) S > P > R > Q

S

(C) P > R > Q > S

(D) R > P > S > Q

(B) S>P>R>Q SN2

13


NARAYANA IIT/PMT ACADEMY JEE ADVANCE - 2013 PAPER – I CODE: 0 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 Name – Advance Organic Chemistry) Average relative SN2 rates for some alkyl substrates Reactant 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 the concentration of Q varies with reaction time as shown in the figure The overall order of the reaction is [Q]0 [Q]

time

Ans

(A) 2 (B) 3 (C) 0 (D) 1 (D) In the reaction P + Q R+S time 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 t t

P

C0

C0/2

C0/4

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 Ans

Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of (A) NO (B) NO2 (C) N2O (D) N2O4 (B) On long standing concentrated HNO3 gives brown colored gas HNO3(con) NO2(g) + H2O + O2 (brown coloured) 14



PAPER – I 30

Ans

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) (D) Phenol does not reacts with NaHCO3 Because phenol is less acidic then HCO3 Thus according to acid-base concept PhO can not abstract the hydrogen from HCO3 and do not give CO2 gas SO 3H

COOH

COOH

OH OH

SECTION – 2 : (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

Ans

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 K a of HA is (A) 1 10-4 (B) 1 10-5 (C) 1 10-6 (D) 1 10-3 (A) r1 H 1

r2

H

r1 r2

H

100

2

H

H H

1 2

1

1 2

Ka C

Ka C = 10-4 Ka = 1 10-4 32

The hyperconjugative stabilities of tert-butyl cation and 2-butene, respectively, are due to * p (empty) and (A) electron delocalisations * (B) and electron delocalisations p (filled) and (C) electron delocalisations * * (D) p (filled) and electron delocalisations 15



PAPER – I Ans

(A) p overlapping t-butyl carbocation is stabilized by * 2-butene is stabilized by overlapping

33

The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are) (A) Cr NH3 5 Cl Cl2 and Cr NH3 4 Cl2 Cl (B) Co NH3 (C) CoBr2 Cl2 (D) Pt NH3

Ans

4 2

3

and Pt NH3

Cl2

and PtBr2Cl2 NO3

2

H 2O Cl

2

Cl and Pt NH3 3 Cl Br

(B), (D) (A) Cr NH3 5 Cl Cl2 does not exhibit isomerism, Cr NH3 4 Cl2 Cl exhibits geometrical isomerism (B) Co NH3

4

and Pt NH3

Cl2

2

H 2O Cl

Both can show geometrical isomerism 2 (C) CoBr2 Cl2 - tetrahedral so it does not exhibit isomerism PtBr2Cl2

2

- square planar so it exhibits geometrical isomerism

(D) Pt NH3

3

NO3

Cl and Pt NH3 3 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 Cl AlCl3

P

Q

NaH

NH 4

100

O O

(D) S

CO3 2 o 115 C

R

O HCl

Ans

S

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

16



PAPER – I Cl AlCl3

AlCl4 2 electron aromatic

H

Na H2

NaH

6 electron aromatic O

OH

O

O

OH

O

O

OH

O

enolization

H

H

6 electron aromatic O

OH

base

OH H

6 electron aromatic 35

Ans

Benzene and naphthalene form an ideal solution at room temperature For this process, the true statement(s) is(are) (A) G is positive (B) Ssystem is positive (C) Ssurroundings 0 (B), (C), (D) Gsystem Hsystem T Ssystem

Hsystem

Ssurrounding

Hsystem

0

0

As mixing is spontaneous so

Hsystem T Ssystem So as

H 0

T Ssurrounding

So as solution is ideal So

(D)

Hsystem = 0 ,

Gsystem

0

0 Ssystem must be greater than zero 17



CODE: 0

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) 36

Ans

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 5 He : m1 = 4, T1 = 200 K Ne : m2 = 20, T2 = 1000 K T h v He M m Ne Ne So

Ne

Ne

37

m Ne TNe

h m Ne

He

EDTA 4

mHe THe

h mHe

He

TNe m Ne THe mHe

1000 20 4 200

25

5

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

Co EDTA

1

complex ion is

Ans

8

38

The total number of carboxylic acid groups in the product P is 18



PAPER – I O

O O

1. H3O , 2. O 3

P

3. H 2 O 2

Ans

O

O

O

O

2

O O

O

O

O OH OH

H 3O

O

O

O

O

keto acid

O3 / H 2 O 2 O

O

O

O

HO HO

39

Ans

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 NH2 group attached to a chiral center is 4 CH3

CH 3 CH CH CO 2 H

Valine (V)

NH2

CH3 CH CO2 H Alanine (A) NH2

H 2 C CH CO 2 H Phenylalanine (P) NH2

H 2 N CH 2 CO 2 H

Glycine

(G)

As (A) is at the end so options possible with NH2 at beginning are (i) VPGA, (ii) VGPA, (iii) PVGA and (iv) PGVA 40 Ans

The total number of lone-pairs of electrons in melamine is 6

19



PAPER – I H2N

NH2

N N

N NH2

Melamine Total number of lone pairs = 6

PART – III : MATHEMATICS 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 the which ONLY ONE is correct 41

41

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

ax by c 0 and bx ay c 0 is less than 2 2. then (A) a b c 0 (B) a b c 0 (C) a b c 0 (A) c c Point of intersection , a b a b Distance of (1, 1) from point of 1

intersection

2

a b

c

2

a b

2 2

The area enclosed by the curves y sin x cos x and y is (A) 4

42

1

a b c a b

Acc to question a b c 2 a b a b c 0 42

2

c

2 1

(D) a b c 0

(B) 2 2

2 1

(C) 2

2 1

cos x sin x over the interval 0,

(D) 2 2

2

2 1

(B) Required Area /4

/2

2sin x dx 0

2 cos x

2 cos x dx /4

/4 0

2 sin x

/2 /4

20



PAPER – I

1 1 2

2

1 2

2 1

4 2 2 2 2 2 1 43 43

f' x

2x

f' x

x 2 cos x

When x

sin x x cos x

0 f' x

0

When x 0 f ' x

0

At

for which x 2 x sin x cos x (C) 2

, The number of points in (A) 6 (B) 4 (C) Let f x x 2 x sin x cos x

x=0

f' x

f 0

1

0, is

(D) 0

sin x

f x is sI function f x is sd function

0

When x

f x

f x will be zero at exactly two points 23

44

The value of cot cot

n 1

cot n 1

(A) 44

23 25

(B)

1

2k

is

k 1

25 23

(C)

23 24

(D)

24 23

(B) 23

n

cot

1

cot

1

n 1 n

n

2k

2k k 1

2

k 1

k k 1

2.n n 1 2

n n 1

23

cot

1

n2 n 1

n 1

Tr

cot

1

tan

1

tan

1

r2

r 1

1 1 r2 r r 1 r 1 r 1 r

21



PAPER – I Tr

tan

1

1

r 1

tan r

23

Tr

1

tan

24

4

n 1

cot tan

45

24

4

25 23

A curve passes through the point 1, y x

45

1

1 1 1 24 1 1 24

6

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

y , x 0. Then the equation of the curve is x y 1 y (A) sin (B) cos ec log x x 2 x 2y 2y (C) sec (D) cos log x 2 x x (A) Given dy y y …………(i) sec dx x x Let y vx dy dv …………(ii) v x dx dx Substituting (ii) in (i) dv v x v sec v dx dx cos v dx x sec

At x 1, y

v

6

v

x

cos v dv /6

1

sin v

v

y x

log x

1 2

at x 1

dx x x

log x 1

/6

sin v sin sin

6

log x 2

6

log x

log x

1 2

22



PAPER – I 46

1 ,1 2

Let f :

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

function such that f ' x

1 2 f x and f 2

1

1. Then the value of

f x dx lies in the 1/2

interval (A) 2e 1, 2e 46

(D) f' x

(B) e 1, 2e 1

2x

f' x

d e dx

Or

2x

2x

2e

f x

f x

(D) 0,

e 1 2

2x

f x is sd f n in

e

2x

f x

e

f x

e2 x

1

f x dx 1/2

1

0

1 ,1 2

1

1

0

0

0

e

0

f x dx 1/2

47

e 1 ,e 1 2

2f x e

47

(C)

e2 x 1 2

1

1/2

e 1 2

Let PR 3iˆ ˆj 2kˆ and SQ iˆ 3 ˆj 4kˆ determine diagonals of a parallelogram PQRS and PT iˆ 2 ˆj 3kˆ be another vector Then the volume of the parallelepiped determined by the vector PT , PQ and PS is (A) 5 (B) 20 (C) S R

(C) 10

(D) 30

b Q

P

a b

a b a b

a ˆ ˆ 3i j 2kˆ iˆ 3 ˆj 4kˆ

a b

c

iˆ 2 ˆj 3kˆ

10iˆ 10 ˆj 10kˆ 23



PAPER – I

10iˆ 10 ˆj 10kˆ

2 a b a b

5 iˆ

ˆj kˆ

Now volume of parallelepiped

cab

c. a b

51 2 3

10 48

Perpendiculars are drawn from points on the line feet of perpendiculars lie on the line x y 1 z 2 (A) 5 8 13 x y 1 z 2 (C) 4 3 7

x 2 2

x 2 x (D) 2 (B)

y 1 1

y 1 3 y 1 7

z to the plane x y z 3. The 3

z 2 5 z 2 5 P

48

(-2,-1, 0)

(D)

A

x 2 y 1 z The given line is r 2 1 3 General point on the line is P 2r 2, r 1, 3r Now the direction ratio of PQ 2r 2, r 1, 3r are which are proportional to (1, 1, 1) 2r 2 r 1 3r k 1 1 1 k 2r 2 k r 1 k 3r Point

, ,

lies on pane x y z 3 so

k 2r 2 k r 1 k 3r Now

Q

6 4r r 3 6 4r r 1 r 3 6 4r 3 3r r 3 5 3 3 3 3 6 2 7 5

3

k

6 4r 3

3 2

2r 2

3

3 7

and

6 x 2

y 1 7

z 2 5 24



PAPER – I 49

49

50

Four person independently solve a certain problem correctly with probabilities

the probability that the problem is solved correctly by at least one of them is 235 21 3 253 (A) (B) (C) (D) 256 256 256 256 (A) Prob that the problem is solved correctly = 1 – prob that problem is not solved correctly by none of them 1 1 3 7 1 . . . 2 4 4 8 21 1 256 235 256 1 2 2 y y0 r 2 and Let complex numbers and lie on circles x x0 2

x x0

(A) 50

1 3 1 1 , , , . Then 2 4 4 8

2

y y0

4r 2 , respectively If z0

1 2

(B)

1 2

x0 iy0 satisfies the equation 2 z0

1 7

(C)

(D)

2

r2

2, then

1 3

(c) …………(i)

z0

r

z0

2r

1 z0 (ii)/(i) gives 1 z0 z0

2r

1

1

Or

1

Or

1

From (i) 2 2 z0

z0

z0

…………(ii)

2

z0 1 2

2

z0 z0

z0 2

2

2

4

z0

z0

4

z0

z0 4

4

2

z0

4 2

z0

2

2

z0 2

z0

z0

2

z0

z0 1 4

z0 . …………(A) 2

…………(B)

r2

r2 1 r2 2 25



PAPER – I 2

2

1

r 2

…………(C)

From (B) & (C) 2

1

2

r2 1 2

4

r2 1 2

1 4

4

4

2

r2 1 2

2

1 7

SECTION – 2 : (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 51

A line l passing through the origin is perpendicular to the lines ˆ ˆ 1 2t ˆj 4 2t k, t 1: 3 t i ˆ : 3 2s ˆi 3 2s ˆj 2 s k, s 2

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

Key Sol

of

and

(A)

7 7 5 , , 3 3 3

1

2

at a distance of

17 from the point of intersection

is (are) (B)

1, 1, 0

(C) 1,1,1

(D)

7 7 8 , , 9 9 9

(B, D) Given equation of line 1 and 2 are x 3 y 1 z 4 … (1) 1 2 2 x 3 y 3 z 2 and … (2) 2 2 1 dr of line are 2,3, 2 Equation of line is x y z … (3) 2 3 2 2k,3k, 2k Any point on (3) is To find point of intersection of (1) and (3) We have 26



CODE: 0

3k 1 2k 4 2 2 4k 6 3k 1 2k 4 7k 7 k 1 Point of intersection of (1) and (3) is P 2, 3, 2 2k 3

Any point on line

PQ

2

is Q 3 2S,3 2S, 2 S

17 1 2S

2

6 2S

2

S2

17

2

9S 28S 20 0 9S2 10S 18S 20 0 9S 10 S 2

0

10 ,S 2 9 7 7 8 Q point is or 1, 1, 0 , , 9 9 9 Let f x x sin x, x 0 Then for all natural numbers n, f ' x vanishes at S

52

(A) a unique point in the interval n, n

1 2

1 ,n 1 2 (C) a unique point in the interval n, n 1

(B) a unique point in the interval n

Key Sol

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

27



PAPER – I y

y

x x 1, 0

0,0

x

f' x

x cos x.

f' x

x cos

Now f ' x

2,0

1 2

x

x

3 2

5 2

sin x , x > 0 sin x

0

x cos x sin x 0 x tan x Now taking y tan x, and y From graph,

x

1 , n 1 and a unique in the internal n, n 1 2

A unique point in the interval n

n N k k 1 4n

53

Let Sn

2

1

k 2 Then S n can take value(s)

k 1

Key

(A) 1056 (A, D)

(B) 1088

(C) 1120

(D) 1332

k k 1 4n

Sol

Sn

2

1

k2

k 1

Sn

Sn

1

1

2

2

5 32

2 2

2

9

r 1

4

2

7 2 112 .....

4n 1

24

2

.....4n terms

22

62 102 ......

4n 2

2

42

82 122 ......

4n

n

1 r 1

6

2

n

r 9 r 1

2

5

4n 3

n

r2

16

2

2

.....

n

Sn

3

n

r 2 16

16 r 1

2

n

r 4 r 1

2

1 r 1

28



PAPER – I n

n

r2 8

16 r 1

r 1 n

r 12 r 1

54

Sol

r

n

32

32

n

1

r2

16

r 1

1 r 1

n n 1 2

12n = 4n 4n 1

Therefore the required solutions are 1056, 1332 For 3 3 matrices M and N, which of the following statement(s) is (are) NOT correct? (A) N T 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 adj N adj MN for all invertible matrices M and N (C, D) (A) NT MN

T

T

NT MN

MN

T

NT

NT MT N NT MN NT MN

So option A is correct T MN (B) MN NM NT MT

T

NM

MT NT

T

(If M is symmetric M T M ) (If M is skew-symmetric MT

M)

T

NT

N

MT

M

NM MN MN NM

55

Key Sol

So option B is correct T (C) MN (M and N are symmetric ) NT MT NM So option (C) is wrong (D) adjM adjM adj MN Given So option D is wrong 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 (A, C)

29



PAPER – I a

a

a 8x

a

2a

8x

15x

2a

2a

Volume of the box V

8x 2a 15x 2a .a

V 4a3 46a 2 x 120x 2 .a For max volume dV d2V 0 and 2 0 da da dV 12a 2 92ax 120ax 0 da 5x d2V 24a 92x 0 at a 2 3 da Now given that sum of Area 4a 2 100 5x So from a x 3 3

a 5

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

aiˆ bjˆ ckˆ : a, b,c

1,1

There non-coplanar vectors

P

Ans Sol

57

can be chosen from V in 2 ways Then p is 5 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 4C3 .2.2.2 32 25 P 5 Of the three independent events E1, E2 and E3, the probability that only E1 occurs is , only E2 occurs is and only E3 occurs is Let the probability p that none of events E1, E2 or E3 occurs 2 p 3 p 2 satisfy the equations and All the given probabilities are assumed to lie in the interval (0, 1) 30



PAPER – I Then Ans Sol

Probability of occurrence of E1 Probability of occurrence of E3

6 Let P E1

x, P E 2

y, P E 3

z

x 1 y 1 z

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

Given equations are 2 p and

1

3 p

1 1 3 2 … (1) p p (1) x 2y and (2) y 3z x 6z x 6 z P E1 6 P E3 58 Ans Sol

2

1 x 1 y 1 z 2

… (2)

The coefficients of three consecutive terms of 1 x 6 n 5 Cr 1 n 5Cr n 5Cr 1 5:10 :14

n 5

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

n 5! n 5! n 5! : : 5 :10 :14 r 1 ! n r 6 ! r! n r 5 ! r 1 : n r 4 !

59

Ans

n n r 5 :1: 5 :10 :14 n r 6 r 1 r 1 2r n r 6 n r 6 2 … (I) n 3r 6 0 n r 5 7 5n 5r 25 7r 7 r 1 5 … (II) 5n 12r 18 0 … (III) 4n 12r 24 0 ____________ n=6 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 5 31



PAPER – I Sol

n

2

n

2k 1

2

1224

n2 n 4k 2450 0 n 2 n 2450 k 4 1 k n 1 But 4 n 2 n 2450 4n 4 n 2 3n 2459 0 3 9817 n 2 n 51 K = 25

1

9785 2 49, n N

n n

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

N

k 20 5 60

A vertical line passing through the point h, 0 intersects the ellipse and Q Let the tangents to the ellipse at P and Q meet at the point R If PQR,

Ans Sol

1

max 1/2 h 1

h and

min 1/2 h 1

2

h , then

8 5

1

8

x2 4

y2 3 h

1 at the points P area of the triangle

2

9 p 2cos , 3 sin

R 2sec ,0

(h, 0)

Q 2 cos ,

3 sin

1 2 3 sin 2sec 2 2 3 sin 3 cos Which is decreasing function of cos Area of

PQR

2cos

32


NARAYANA IIT/PMT ACADEMY JEE ADVANCE - 2013 1 2

PAPER – I 1 2cos 2

h 1

1 4

cos

CODE: 0 3

1 2

3 1

2 3

4

2 3 15. 15 16 45 5 8

8 5

1

2 3

2 3 3 3 4 8 2 36

1 2 15 4

3 2

2

9 2

So

8 5

1

8

2

9

45

33



PAPER – I

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

37

8

57

6

18

1

38

2

58

6

19

4

39

4

59

5

20

8

40

6

60

9

34
