Test 8 Paper 1 JEE Main 2013

Roll No.:

Centre Code:

TEST No. 8

Test Date:

Test Booklet Code A

JEE (Main)-2013

for

ALL

IN

DIA

AAKASH TEST

SERIE

S

PAPER - I

INSTRUCTIONS FOR CANDIDATE1. Read each question carefully.

2. It is mandatory to use Black Ball point PEN to darken theappropriate circle in the answer sheet.

3. Mark should be dark and should completely fill the circle.

4. Rough work must not be done on the answer sheet.

5. Do not use white-fluid or any other rubbing material on answersheet.

6. Student cannot use log tables and calculators or any othermaterial in the examination hall.

7. Before attempting the question paper, student should ensure thatthe test paper contains all pages and no page is missing.

8. Before handing over the answer sheet to the invigilator,candidate should check that Roll No., Centre Code and Date ofBirth have been filled and marked correctly.

9. Immediately after the prescribed examination time is over, theanswer sheet to be returned to the invigilator.

There are parts in the question paperA, B and C consisting

of , and having 30 questions

in each part of equal weightage. Each question is allotted 4

marks for each correct response.

three

Physics Chemistry Mathematics

(four)

10.

11. The Test booklet consists of 90 questions. The maximum marks

are 360.

12. One fourth (¼) marks will be deducted for indicating incorrect

response of each question. No deduction from the total score

will be made if no response is indicated for any question in the

answer sheet.

13. Pattern of the questions are as under:

Section – I : Straight Objective Questions

Section – II : Assertion – Reason Type Questions

10/02/2013

TOPICS OF THE TEST

Test No. 8

PhysicsOptics, Modern Physics, Solids and Semiconductors, Principles of

communications, Units and Measurements (Including Experimental Physics)

Chemistry Functional group-II, Functional group-III, Polymers, Biomolecules

MathematicsDifferential Equation, Vectors, 3-D Geometry, Probability, Statistics, LinearProgramming

Paper - I

Space for Rough Work

Test - 8 (Paper-I) All India Aakash Test Series for JEE (Main)-2013

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PART - A (PHYSICS)

SECTION - I

Straight Objective QuestionsThis section contains 25 multiple choice questionsnumbered 1 to 25. Each question has 4 choices (1), (2),(3) and (4), out of which ONLY ONE is correct.

Choose the correct answer :

1. A light ray is made to incident on surface PQ of atriangular prism as shown. The face PR is silvered.The value of refractive index of prism so that light rayretraces its path is (μair = 1)

45º(air)

(Normal)

P

30º

Q R

(1) 1 (2) 2

(3) 3 (4)32

2. Let spherical aberration and chromatic aberration bedenoted by S and C respectively. In case of concavemirror

(1) S and C both may be present

(2) Both S and C do not exist

(3) S may be present but C does not exist

(4) C may be present but S does not exist

TEST - 8 (Paper-I)Time : 3 Hrs. MM : 360

3. Two radioactive nuclei P and Q, in a given sampledecay into a stable nucleus R. At time t = 0,number of P species are 4N0 and that of Q are N0.Half life of P is 1 minute whereas that of Q is2 minutes. If there are no nuclei of R initially, thenumber of R nuclei present in sample when numberof nuclei of P and Q are equal is

(1) 3N0 (2) 092N

(3) 052N

(4) 2N0

4. In the circuit shown, if each diode has a forwardbiased resistance of 75 Ω and infinite resistance inreverse bias, what will be the values of currents i1,i2, i3 and i4?

i4

i3

i2

i1

15 V

25 Ω

25 Ω

25 Ω

25 Ω

(1) i1 = 0.2 A, i2 = 0, i3 = 0.2 A, i4 = 0

(2) i1 = 0.26 A, i2 = 0, i3 = 0.26 A, i4 = 0

(3) i1 = 0.2 A, i2 = 0.07 A, i3 = 0.07 A, i4 = 0.07 A

(4) i1 = 0.2 A, i2 = 0.1 A, i3 = 0 A, i4 = 0.1 A


All India Aakash Test Series for JEE (Main)-2013 Test - 8 (Paper-I)

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5. A prism of angle 3º is made of glass havingrefractive index 1.64. Two thin prisms made of glasshaving refractive index 1.48 are intended to becoupled with the former prisms to yield acombination without an average deviation. Which ofthe following angles cannot correspond to the twoprisms?

μ = 1.48 μ = 1.48

θ1 θ2

μ = 1.64

3o

(1) 2º & 2º (2) 1.5º & 2.5º

(3) 1º & 3º (4) 5.5º & 3.5º

6. A kingfisher bird dives at a speed of 10 m/s to catcha fish moving vertically upward towards the surfaceof water at a speed of 5 m/s. Velocity of fish as seenby bird is (only magnitude)

water air4 , 13

⎛ ⎞μ = μ =⎜ ⎟⎝ ⎠

10 m/s

(1) 15 m/s

(2)50 m/s3 5 m/s

(3)55 m/s4

(4) 5 m/s

7. If the behaviour of light rays is as shown in figure,the correct relation between μ, μ1 and μ2 is

μ1 μ μ2

(1) μ > μ2 > μ1 (2) μ < μ2 < μ1

(3) μ < μ2 ; μ = μ1 (4) μ2 < μ1 ; μ = μ2

8. A uranium nucleus (atomic number 92, massnumber 238) emits an alpha particle and theresultant nucleus emits β– particle. The atomicnumber and mass number of final nucleus are

(1) 91, 234 (2) 90, 234

(3) 93, 242 (4) 91, 238

9. A particle of mass M at rest decays into two particlesof masses m1 and m2, having non-zero velocities. Theratio of the de Broglie wavelength of the particles is

(1)1

2

mm (2)

2

1

mm

(3) 1 (4)2

1

mm

10. In a Young's double slit experiment, the slits areilluminated by monochromatic light. The entire set upis immersed in pure water. Which of the followingact cannot restore the original fringe width?

(1) Bringing the slits close together

(2) Moving the screen away from the slit plane

(3) Replacing the incident light by that of a longerwavelength

(4) Introducing a thin transparent slab in front of oneof the slits

11. A man is walking under an inclined mirror at aconstant velocity 10 m/s along the x-axis. If theplane mirror is inclined at an angle θ with thehorizontal then what is the velocity of image?

(1) ˆ ˆ10sin 10cosθ + θi j

(2) ˆ ˆ10cos 10sinθ + θi j

y

xθ10 m/s(3) ˆ ˆ10sin2 10cos2θ + θi j

(4) ˆ ˆ10cos2 10sin2θ + θi j



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12. A beam of 400 nm light is incident on a metal havingwork function 2 eV and placed in a magnetic field B.The most energetic electron emitted perpendicular tothe field are bent in circular arcs of radius 0.1 m. Thevalue of B is

(1) 5.72 × 10–25 T (2) 5.72 × 10–5 T

(3) 35.8 × 10–6 T (4) 35.8 × 10–25 T

13. In the following circuit if A = 0 and B = 1 andA = 1, B = 1, then value of Y is (respectively)

A

B

Y

(1) 0, 1 (2) 1, 0

(3) 0, 0 (4) 1, 1

14. The equivalent resistance of the circuit across AB is

3 Ω

6 Ω 18 Ω

4 ΩA B

9 Ω

(1) 8 Ω (2) 8 Ω or 0 Ω

(3) 8 Ω or 19 Ω (4) 0 Ω

15. A T.V. tower has a height of 600 mπ

. If the average

population density around the tower is 1000 km–2.The population covered by T.V. broadcast is

(1) 76.8 lakhs (2) 38.4 lakhs

(3) 600 lakhs (4) 6 thousand

16. A vernier callipers has 1 mm marks on the mainscale. It has 20 equal divisions on the vernier scalewhich match with 18 main scale divisions. The leastcount for this vernier callipers is

(1) 0.2 mm (2) 0.1 mm

(3) 0.02 mm (4) 0.01 mm

17. The force acting on a particle as a function of

position (x) and time (t) is given by 2α

−β=

xtF Ae . The

dimensional formula for αβ is

(1) M0L0T0

(2) L–2T1

(3) L2T–1

(4) LT–1

18. In a standard YDSE with identical slits, the intensityof the central bright fringe is I0. If one of the slits iscovered in a manner that it allows one fourth of theintensity to pass through it which it was allowingwhen not covered, the intensity at the same point is

(1) I0 (2) 0

4I

(3) 0916I

(4) 0716I

19. An electron is in exited state in a hydrogen likeatom. It has a total energy of –3.4 eV. The kineticenergy of electron is E and its de Broglie wavelengthis λ. Then

(1) E = 6.8 eV, λ 6.6 × 10–10 m

(2) E = 3.4 eV, λ 6.6 × 10–10 m

(3) E = 3.4 eV, λ 6.6 × 10–11 m

(4) E = 6.8 eV, λ 6.6 × 10–11 m



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20. A ray of light is incident on a concave mirror. It isparallel to the principal axis and its distance fromprincipal axis is equal to the focal length of themirror. If C be the centre of curvature and B be thepoint at which refracted ray cuts the principal axis,then the distance BC is

(1)2

>f

(2)2

<f

(3)2

=f

(4)3

<f

21. A cubical block of glass of refractive index μ1 is incontact with the surface of the water of refractiveindex μ2. A beam of light is incident on vertical faceof the block as shown. After a total internal reflectionat the base and refraction at the opposite verticalface, a ray emerges out at an angle θ. The value ofθ is given by

(1) 2 21 2sinθ < μ −μ

(2) 2 21 2tanθ < μ −μ

(3)2 21 2

1sinθ <μ −μ

μ1 θair air

μ2

(4)2 21 2

1tanθ <μ − μ

22. A physical quantity Q is given by xyQz

= . If

percentage error in the measurement of x, y and zare 0.20%, 0.10% and 0.36% respectively, thenpercentage error in the measurement of Q is

(1) 0.48% (2) 0.12%

(3) 0.66% (4) 0.06%

23. Kα wavelength emitted by an atom of atomic numberZ = 11 is λ. The atomic number for an atom thatemits Kα radiation with wavelength 4λ is

(1) 6 (2) 4

(3) 11 (4) 44

24. A student performs an experiment for determination

of 2

24⎛ ⎞π

=⎜ ⎟⎝ ⎠

gT . The error in length is Δ and in time

T is ΔT and n is number of times the reading istaken. The measurement of g is most accurate for

Δ ΔT n

(1) 1 mm 0.15 50

(2) 1 mm 0.15 20

(3) 5 mm 0.25 20

(4) 1 mm 0.15 10

25. Figure shows graph of deviation δ versus angle ofincidence for a light ray striking a prism. Angle ofprism is

23º

15º 35º(1) 19º (2) 27º

(3) 15º (4) 35º

SECTION - IIAssertion – Reason Type Questions

Directions : Questions number 26 to 30 are Assertion-Reason type questions. Each of these questionscontains two statements. Statement-1 (Assertion) andStatement-2 (Reason). Each of these questions also hasfour alternative choices, only one of which is the correctanswer. You have to select the correct choice.

26. Statement-1 : A parallel beam of light travelling in aircan be displaced laterally by a transparent cuboidslab by a distance more than the thickness of theslab.and

Statement-2 : The lateral displacement of lighttravelling in air increases with rise in value ofrefractive index of slab.



5/14

(1) Statement-1 is True, Statement-2 is True;Statement-2 is a correct explanation forStatement-1

(2) Statement-1 is True, Statement-2 is True;Statement-2 is NOT a correct explanation forStatement-1

(3) Statement-1 is True, Statement-2 is False(4) Statement-1 is False, Statement-2 is True

27. Statement-1 : Quantity of characteristic x-raysincreases as the applied potential differenceincreases keeping number of incident electronsconstant.andStatement-2 : Kinetic energy of bombardingelectrons is directly proportional to the appliedpotential difference.(1) Statement-1 is True, Statement-2 is True;

Statement-2 is a correct explanation forStatement-1



28. Statement-1 : Figure shows graph of stoppingpotential and frequency of incident light inphotoelectric effect. For values of frequency less thanthreshold frequency (ν0) stopping potential isnegative.andStatement-2 : Lower the value of frequency ofincident light (for ν > ν0) the lower is the maxima ofkinetic energy of emitted photoelectrons.

ν0ν

Vs




29. Statement-1 : Keeping a point object fixed, if aplane mirror is moved, the image must move.andStatement-2 : In case of a plane mirror, distance ofa point object and its image from a given point onmirror is equal.(1) Statement-1 is True, Statement-2 is True;



(3) Statement-1 is True, Statement-2 is False

(4) Statement-1 is False, Statement-2 is True30. Statement-1 : When the temperature of a

semiconductor is increased, then its resistancedecreases.andStatement-2 : The energy gap between conductionband and valence band is small for semiconductorsas compared to insulators.(1) Statement-1 is True, Statement-2 is True;






6/14

SECTION - I

Straight Objective Questions

This section contains 25 multiple choice questionsnumbered 31 to 55. Each question has 4 choices (1), (2),(3) and (4), out of which ONLY ONE is correct.

31. How many geometrical isomers of aldehyde

H

O in its enol form are possible?

(1) 4 (2) 6

(3) 8 (4) 5

32. Which of the following compound will give positiveiodoform test?

(1)CH I2

O(2)

O

(3)CH 2I

O(4) All of these

33.

O

O

Δ(A)OH–

(major product).

Which one of the following statements is true forthe major product (A) formed in the above reaction?

(1) 'A' gives positive bromine water test

(2) 'A' absorbs two moles of H2 per mole ofcompound in hydrogenation

(3) 'A' gives positive DNP test

(4) All of these

PART - B (CHEMISTRY)

34.

OCH N2 2 CF CO3 3H(X) (Y). Product (Y) is

(major product)

(major product)

(1)

O

O (2) HC O CH3

O

(3)

O

O (4)O C CH3

O

35.

CHOCHO

NO2

O N2

OH–

Conc.Δ

(A) (major product)

Conc. H SO2 4 (B) (major product). Product (B) is

(1)

O

O

(2)

OO

NO2

O N2

(3)

O

O

NO2

NO2

(4)

O

O

NO2

NO2



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36.O

OH

OHOH

O

O

Δ

(major product)(A)

Which of the following statements is incorrect about(A)?

(1) (A) has molar mass 44 g lesser than molarmass of reactant

(2) (A) is bicyclic compound

(3) (A) in water acts as diprotic acid

(4) Cyclohexane retains its identity in formationof (A)

37. Two monosaccharides form a disaccharide by using

(1) Ether linkage

(2) Amide linkage

(3) Ester linkage

(4) Anhydride linkage

38. Amylose is a polymer of

(1) Maltose (2) Galactose

(3) Lactose (4) Sucrose

39. Glycine at its isoelectric point exists in solution as

(1) NH2CH2COO–

(2)3 2H NCH COOH+

(3) H N3 CH COO2–+

(4) NH2CH2COOH

40. Which of the following is true?

(1) Glucose and mannose are epimers

(2) Mannose and fructose are functional isomers

(3) α-mannose and β-mannose are anomers

(4) All of these

41.

O C NH2

KOH Br2, D O2

major product.

How many deuterium are there in major product?

(1) 0

(2) 1

(3) 2

(4) 4

42. The type/s of reaction involved in Aldol reaction is

(1) Nucleophilic addition reaction

(2) Acid-base reaction

(3) Redox reaction

(4) All of these

43. HCHO + PhCHO KOHConc. (A) + (B). (A) and (B) are

major product. Which of the following is true about(A) and (B) ?

(1) (A) is HCOOK formed by nucleophilic attack ofOH–

(2) (A) is PhCOOK, formed by nucleophilic attack ofOH–

(3) (B) is CH3OH, formed by hydride transfer

(4) (B) is PhCH2OK, formed by hydride transfer

44. Which one of the following compounds form solubleproduct in KOH on reaction with ArSO2Cl?

(1) PhNH2

(2) PhNHCH3

(3) PhNCH3

CH3

(4) All of these



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45.H CO3

O

OO O

dil.H SO2 4

(A). (A) is a mixture of

products. The number of organic products found inmixture (A) is

(1) 2 (2) 3

(3) 4 (4) 5

46.O

Odil.

H SO2 4(A)

(organic product). Product (A) does

not give

(1) Effervescence with NaHCO3

(2) Colour in victor meyer test

(3) DNP test

(4) Fruit smell on reaction with suitable alcohol inconc. H2SO4

47. The type of attraction force involved in polyamide is

(1) van der Waal attraction force

(2) Dipole-dipole attraction force

(3) H-bonding

(4) All of these

48. The intermediate involved in the formation ofpolyethene is

(1) Free radical (2) Carbocation

(3) Carbanion (4) Carbene

49. Which of the following is biodegradable polymer?

(1) Glyptal (2) Bakelite

(3) Nylon 2-Nylon-6 (4) Nylon-6

50. Vulcanised rubber has

(1) Thiol linkages (2) Thio ether linkages

(3) Sulphonic linkages (4) Alcohol linkages

51. In Stephen's reaction, the n-factor of SnCl2 is

(1) 1 (2) 2

(3) 4 (4) 3

52. Which one of the following compounds has thehighest boiling point?

(1) Methoxyethane (2) Acetone

(3) Propan-1-ol (4) Propan-2-ol

53. CH CH COOH3 2(i) Br /2 P lC 3

(ii) H O2(A)

(major product)

Compound (A) is

(1)

ClOH

O(2)

BrOH

O

(3)Br

Cl

O(4)

BrCl

O

54. R CX H

O+ R O R C O R

O

The best rate for esterification is observed when 'X'is

(1) OH (2) Cl

(3) O C R

O

(4) NH2

55. Zwitter ion formation is observed in

(1) Sulphonation of aniline at higher temperature

(2) Sulphonation of benzoic acid

(3) Sulphonation of phenol

(4) Sulphonation of chlorobenzene



9/14

SECTION - II

Assertion – Reason Type Questions


56. Statement-1 : Aniline on reaction with CH3COClforms a product, which is less reactive than anilinetowards electrophilic substitutions.

andStatement-2 : In amide lone pair of nitrogen is inresonance with C

O

group.




57. Statement-1 : HCOOH acid gives Tollen's reagenttest.

andStatement-2 : In Tollen's reagent test mirror of 'Ag' isformed.




58. Statement-1 : Phthalic acid on reaction withammonia at room temperature forms phthalamide.andStatement-2 : The ratio of molar mass to equivalentmass of phthalic acid in formation of phthalamideis 2.(1) Statement-1 is True, Statement-2 is True;




59. Statement-1 : Diazonium salt with aniline formscoloured dye.andStatement-2 : The coupled product has largeconjugation.(1) Statement-1 is True, Statement-2 is True;




60. Statement-1 : Quaternary structure is biologicallyactive form of protein.andStatement-2 : Sequence of amino acid is decided bysecondary structure of protein.(1) Statement-1 is True, Statement-2 is True;






10/14

SECTION - IStraight Objective Questions

This section contains 25 multiple choice questionsnumbered 61 to 85. Each question has 4 choices (1), (2),(3) and (4), out of which ONLY ONE is correct.

61. If the points A = (–6, 3, 2), B = (3, –2, 4),C = (5, 7, 3) and D = (–13, 17, k) are coplanar,then the value of k is(1) –1 (2) 1(3) –2 (4) 2

62. If ˆ ˆ ˆ2 3 5A i j k= + + and ˆ ˆ ˆB ai bj ck= + + where a, b,c are positive real numbers and 2a + 3b + 5c = 10,then the minimum value of a2 + b2 + c2 is

(1)1019 (2)

10019

(3)5019 (4)

2519

63. There are 6 members in a family then the probabilitythat exactly 3 of them have the birthday on monday is

(1)3

620 6

7×

(2)6

36120

7×C

(3)6

337

C(4) 3

17

64. The orthogonal trajectory of the family of ellipsesx2 + 2y2 – y = c is(1) Family of lines (2) Family of parabolas(3) Family of ellipses (4) Family of circles

65. If the S.D. of a set of observations is 10, and if eachobservation is divided by –2, the S.D. of new set ofobservations will be(where S.D. = standard deviation)(1) 20 (2) –20

(3) 5 (4) –5

PART - C (MATHEMATICS)66. Let P be a point (1, 2, 3) and A, B, C be its images

with respect to the xy, yz and zx planesrespectively. A plane passes through A, B, C andintersects the co-ordinate axes at L, M, N.Thevolume of the tetrahydron OLMN, O being origin, is

(1) 1 unit3 (2) 3 unit3

(3) 4 unit3 (4) 6 unit3

67. If ,a b are two non-zero vectors then

ˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) ( ) ( ) ( ) ( )a i b i a j b j a k b k× ⋅ × + × ⋅ × + × ⋅ × is

(1) 0 (2) a b⋅

(3) 2a b⋅ (4) 3a b⋅

68. A normal is drawn at a point P(x, y) of a curve. If theharmonic mean of the intercepts of the normalbetween the co-ordinate axes is constant and isequal to 2 units then the curve must be

(1) Circle (2) Ellipse

(3) Parabola (4) Hyperbola

69. The co-ordinates of the circumcentre of the triangleformed by the points A (3, 2, –5), B (–3, 8, –5) andC (–3, 2, 1) are

(1) (–1, 4, –3) (2) (1, 4, –3)

(3) (–1, 4, 3) (4) (–1, –4, –3)

70. Three unbiased dices are thrown together then theprobability that the sum of the numbers appearing onthem is a multiple of 4 under the condition that thefirst dice show the even number and the second diceshow the odd number is

(1)7

27(2)

13

(3)4

15 (4)8

27



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71. If the differential equation of the family of curveswhose equation is given by, y = aex + be–x + c.e3x

+ d.e–3x is

4 3 2

1 2 3 44 3 0d y d y d y dyk k k k ydx dxdx dx

+ + + + =

Then the value of 1 4

2 3

k kk k

++

is

(where a, b, c and d are arbitrary constants)

(1)7

8−

(2)8

7−

(3)87

(4)9

10−

72. If X1 = (x1, y1) and X2 = (x2, y2) are two optimal

solutions of an L.P.P., then

(1) λX1 + (1 – λ)X2, λ∈R, is also an optimalsolution

(2) λX1 + (1 + λ)X2, λ∈R, is also an optimalsolution

(3) λX1 + (1 + λ)X2, 0 ≤ λ ≤ 1, is also an optimalsolution

(4) λX1 + (1 – λ)X2, 0 ≤ λ ≤ 1, is also an optimalsolution

73. Bag A contains 6 red, 5 black balls and bag Bcontains 3 red, 4 black balls. One ball is drawn atrandom from bag A and is placed in bag B then oneball is again drawn from bag B and is placed in bagA. If now one ball is drawn from bag A, then theprobability that it is a black ball is

(1)12

(2)449968

(3)432968 (4)

107204

74. The mean deviation of first n natural numbers fromtheir mean is

(1) ,4n

if n is even

(2) ,2n

if n is odd

(3)( 1)( 2) ,

4n n

n+ +

if n is odd

(4)2 1,4

nn+

if n is odd

75. The projection of the line segment joining the pointsA = (1, 2, –3) and B = (3, 3, 3) on the planex + y + z = 3 is

(1) 14 units (2) 21 units

(3) 28 units (4) 35 units

76. If , ,a b c are three unit vectors then the maximum

value of 2 2 2| – | | | | |a b c b c a a c b+ + + − + + − is

(1) 6 (2) 9

(3) 10 (4) 12

77. The image of the plane x + 2y + 3z – 6 = 0w.r.t. the plane x + y – z + 1 = 0 is

(1) x + y – z + 1 = 0

(2) 2x + 3y + 5z + 1 = 0

(3) x + 2y + 3z – 6 = 0

(4) 61 2 3x y z+ + =

78. If the points P(a, 2a, 4) and Q(2a, – a, 3) lies on thesame side of the plane x + y + z = 6 then the rangeof a is

(1)2, (3, )3

⎛ ⎞−∞ ∪ ∞⎜ ⎟⎝ ⎠

(2)2,33

⎛ ⎞⎜ ⎟⎝ ⎠

(3)2,3

⎛ ⎞−∞⎜ ⎟⎝ ⎠

(4) (3, )∞



12/14

79. The solution of the differential equation3log loge ex x dy y dx x x dx+ = is

(1) 29 (log ) (2log 1)= + +e ey x x x c

(2) 39 (log ) (3log 1)e ey x x x c= − +

(3)3 3

log3 9ex xy x c= − +

(4) 29 (log ) (2log 1)e ey x x x c= − +

80. The equation of the plane passing through the line ofintersection of the planes 2x – 3y + 5z = 7 andx + y + 2z = 4 and perpendicular to the plane x +y + z = 5 is

(1) x – 4y + 3z = 3

(2) x + 2y – 3z = 7

(3) – x + 4y + 3z = 1

(4) 2x + y – 3z = 5

81. The mean deviation from the median is

(1) Equal to that measured from another value

(2) Maximum if all observations are positive

(3) Greater than that measured from any othervalue

(4) Less than that measured from any other value

82. A pair of dice is thrown 5 times, getting a doublet isconsidered as a success then the probability ofgetting at least three successes is

(1)23

648 (2)47

216

(3)53

1296 (4)1136

83. The equation of a line passing through the origin andintersecting the straight lines

L1 : 1 1 1

2 3 4x y z− + −

= =

and L2 : 3 9 / 2

1 2 1x y z− −

= = is

(1) 4 11 10x y z= = (2) 1 2 3

x y z= =

(3)1 1 1x y z= = (4) 4 7 5

x y z= =

84. If , ,a b c are three non-zero vectors, such that

| ( ) | | | | | | |a b c a b c× × = .

Then which of the following is not always true?

(1) . 0a b =

(2) [ ] 0abc =

(3) [ ] 0a b b c c a+ + + =

(4) 0b c× =

85. Let , ,u v w be three non-coplanar unit vectors such

that the angle between andu v is 60º, the angle

between andu w is 60º and the angle between

andv w is 90º. If , ,x y z are unit vectors along the

bisectors of the angles &u v , &u w and

&v w respectively then 2

[ ][ ]

x y y z z xu v w

× × × is

(1)29 (2)

49

(3)23 (4)

89



13/14

SECTION - II

Assertion – Reason Type Questions


86. Consider two lines L1, L2 as follows

L1: 1 2 3= =

x y z

and L2 : 1 2 3

1 2 3x y z− − −

= = .

Statement-1 : L1, L2 are two coincident lines.

and

Statement-2 : If the direction ratios of two lines aresame, then they must be coincident lines.




(4) Statement-1 is False, Statement-2 is True

87. Statement-1 : The solution of the differential equation

2 32 4 1

dy x ydx x y

+ +=

+ − is given by

– 4x + 8y = –7 loge |4x + 8y + 5| + k.

(where k is a constant)

and

Statement-2 : The above differential equation can bereduced to variable separable form by putting(x + 2y = t).





88. Consider two planes

P1 : x + 2y – 3z + 2 = 0

P2 : x – 2y + 3z + 7 = 0

Statement-1 : Origin and the point P(1, 2, 2) both liein the acute angle between the planes.

and

Statement-2 : Let a1x + b1y + c1z + d1 = 0, and a2x+ b2y + c2 z + d2 = 0 are two planes, where d1, d2> 0, then origin lies in acute angle if a1a2 + b1b2 +c1c2 < 0. Further both the points (x1, y1, z1) andorigin lie in acute angle if (a1x1 + b1y1 +c1z1 + d1)(a2x1 + b2y1 + c2z1 + d2) > 0.







14/14

89. Statement-1 : The order of differential equation of thefamily of curves, whose equation is given byy = (c1 + c2).e

x + c3.ex + c4 is 1.

and

Statement-2 : The order of a differential equation isalways equal to the number of independentparameters which is not necessarily equal to thenumber of all the parameters in the family of thecurves.





90. Statement-1 : The maximum value of f = 4x + 3ysubject to the constaints

x ≥ 0, y ≥ 0, 2x + 3y ≤ 18, x + y ≥ 10 is 40.

and

Statement-2 : The optimal value of the objectivefunction occurs at the corner points of the feasibleregion.





Test Booklet Code A

Test No. 7

Paper - I

JEE (Main)-2013

for

ALL

IN

DIA

AAKASH TEST

SERIE

S

Test - 7 (Paper-I) (Answers & Hints) All India Aakash Test Series for JEE (Main)-2013

1/9

TEST - 7 (Paper-I)

ANSWERS

1. (1)

2. (1)

3. (3)

4. (1)

5. (4)

6. (1)

7. (2)

8. (1)

9. (4)

10. (3)

11. (2)

12. (1)

13. (2)

14. (4)

15. (1)

16. (4)

17. (1)

18. (3)

19. (4)

20. (2)

21. (2)

22. (2)

23. (2)

24. (2)

25. (3)

26. (4)

27. (3)

28. (1)

29. (1)

30. (1)

PHYSICS CHEMISTRY MATHEMATICS31. (1)

32. (3)

33. (2)

34. (2)

35. (1)

36. (2)

37. (4)

38. (3)

39. (4)

40. (4)

41. (3)

42. (4)

43. (1)

44. (4)

45. (3)

46. (2)

47. (2)

48. (3)

49. (1)

50. (3)

51. (2)

52. (2)

53. (3)

54. (1)

55. (2)

56. (4)

57. (4)

58. (1)

59. (4)

60. (3)

61. (1)

62. (4)

63. (4)

64. (3)

65. (1)

66. (3)

67. (3)

68. (2)

69. (4)

70. (2)

71. (4)

72. (3)

73. (1)

74. (3)

75. (3)

76. (4)

77. (3)

78. (2)

79. (1)

80. (3)

81. (4)

82. (4)

83. (2)

84. (4)

85. (4)

86. (1)

87. (2)

88. (1)

89. (1)

90. (3)

All India Aakash Test Series for JEE (Main)-2013 Test - 7 (Paper-I) (Answers & Hints)

2/9

1. Answer (1)

Use conservation of energy

U1 + K1 = U2 + K2

210 02

mB I+ = − + ω

⇒2mB

Iω =

2. Answer (1)

For 100100 V dc 1 A (Given)EI

R R= = =

⇒ R = 100 Ω

For 2 2

100100 V ac 0.5 A (Given)L

EIZ R X

= = =+

⇒ 2 2 200LR X+ =

⇒ 100 3LX = Ω

⇒ 3 H 0.55 H2

LXL = = =πν π

3. Answer (3)

For X, 200 405

VRI

= = = Ω

For Y, I lags behind V by 90º

⇒ Y is pure inductance

⇒200 405LX = = Ω

In series combination of X and Y

2 2 40 2LZ R X= + =

⇒ Peak current = 200 5

40 2 2=

⇒ RMS current = 0 5 2.5 A22

I= =

4. Answer (1)

There is no change in flux across the bar.

5. Answer (4)

Since i2 is constant and positive, rate of change of i1must be constant and negative, because

2 12 2

2

andE dii E MR dt

= = −

PART - A (PHYSICS)

6. Answer (1)

F qv B= ×

7. Answer (2)

Current is same in each galvanometer

.NAB IK

θ =

⇒ 1 11

n A B IK

θ = and 2 22

n A B IK

θ =

⇒1 1 1

2 2 2

n An A

θ=

θ

8. Answer (1)

After a long connection with Y,

1

EIR

=

Now switch is shifted to Z, energy stored in L willdissipate as heat in R2.

⇒ Energy dissipated = 212

LI

= 2

212

E LR

9. Answer (4)

C

D

A

B

I

Y

X

See the figure, magnetic field at B and D are mutuallyopposite.

10. Answer (3)

QRΔφ

=

For θ = 180º

Δφ = φ – (–φ) = 2φ = 2BAn

⇒2BAnQ

R=

⇒ Q ≠ 0


3/9

11. Answer (2)

B1 = field due to segment (1)

= 0 01202 360 6Ι Ι

R Rμ μ°

× =°

60o

R/2

1 2

B2 = field due to segment (2)

= ( )0 sin60 sin604 /2

ΙRμ

° + °π

= 0 32Ι

Rμπ ⇒

0 3 36

ΙR

⎛ ⎞μ − π⎜ ⎟⎜ ⎟π⎝ ⎠

outward.

12. Answer (1)

Under steady state,

voltage across the capacitor = E ⇒ 212

=CU CE

Current through the inductor = 4ER

⇒21

2 4⎛ ⎞= ⎜ ⎟⎝ ⎠

LEU LR ⇒ 216

=L

C

U LU R C

13. Answer (2)

C Fx

FM

At any instant t,

=E B v

⇒ = =q CE CB v

⇒dvΙ CBdt

= ⋅ ⇒ 2 2M

dvF Ι B B Cdt

= =

Now − =MdvF F Mdt ⇒

dvdt

= constant.

14. Answer (4)

weber is unit of magnetic flux and ohm is unit ofresistance.

But RΔφ

= charge

⇒ coulomb

15. Answer (1)

1mvR

qB=

v1 is velocity perpendicular to magnetic field

i.e., net velocity in x-y plane

= 2 2 53 4 10+ × = 5 × 105 m/s

⇒5

125 1010 2

R ×=

×= 2.5 × 10–7 m = 2500 Å

16. Answer (4)2 2 2

2 ParabolaB l vP P vR

= ⇒ ∝ ⇒

17. Answer (1)

Fields of I1 and I2 should be equal and opposite

0 1 0 2

2 2 2I I

R Rμ μ

=π×

⇒1

2

2II= π

18. Answer (3)

By symmetry φ = θ

19. Answer (4)

Since there is no vertical component of earth's field atthe equator.

⇒ The needle is in neutral equilibrium in the givenplane

⇒ It stays at rest in all directions

20. Answer (2)

For hitting the wall r ≥ d

⇒mvdqB

=

⇒qBdvm

=

For α-particle, q′ = 2q

m′ = 4 m

⇒2vv ′ =


4/9

PART - B (CHEMISTRY)

21. Answer (2)

No magnetic force acts and electric force is oppositeto motion.

22. Answer (2)2

2 21. [ ]2

lll

r l

B rdr B r −=−

ε = ω = ω∫

= 232

B lω

B is at higher potential than A

⇒ 232A BV V B l− = − ω

23. Answer (2)

B BCircular 2 = B BStraight 1 =

2 2net 1 2B B B= +

0 01 2,

2 2I I

B BR R

μ μ= =

π

⇒ 0net 2

112

IBR

μ= +

π

24. Answer (2)

At t = 0 L behaves open circuited

⇒ Req = 2R

⇒2EIR

=

25. Answer (3)

Since V is same across L & C

⇒ IL and IC are π out of phase.

⇒ Their directions are different

Since 1LC

ω = A1

~ A3

IL

IC

I

V

A2⇒ XL = XC

⇒ IL = IC⇒ I = 0

⇒ Reading of A3 is zero.

26. Answer (4)

Direction of the currents are mutually opposite sincein first case magnet and loop are coming closer andin second case moving apart.

27. Answer (3)

Potential of a charge changes when it is moved alongthe field line.

28. Answer (1)

1 22= ⇒ >>M

E ME

F v v F FF c

29. Answer (1)

XL = ω.L

⇒ Statement 1 is correct

Statement-2 is correct explanation of statement-1

30. Answer (1)

The correct explanation is that the magnetic field dueto a magnet is non-uniform and a dipole experiencesa net force in non-uniform magnetic field.

31. Answer (1)

32. Answer (3)

2

4

4

2

2

4

2

33. Answer (2)

34. Answer (2)

Gauche is more stable due to intramolecularH-bonding.

35. Answer (1)

36. Answer (2)

37. Answer (4)

Both group present at equatorial position.

38. Answer (3)

39. Answer (4)

40. Answer (4)

41. Answer (3)

42. Answer (4)

43. Answer (1)

44. Answer (4)


5/9

45. Answer (3)

46. Answer (2)

47. Answer (2)

48. Answer (3)

Higher is the stability of carbocation, more is thereactivity of alcohol.

49. Answer (1)

50. Answer (3)

OH

H+

OH2

+

+

+

H+

–H+

51. Answer (2)

52. Answer (2)

53. Answer (3)

H

Anti-aromatic

54. Answer (1)

55. Answer (2)

56. Answer (4)

57. Answer (4)

58. Answer (1)

59. Answer (4)

60. Answer (3)

PART - C (MATHEMATICS)61. Answer (1)

Let f(x) = x4 – 7x + 2, since f(0) = 2;

f(1) = – 4; f(2) = 4.

So, by intermediate value theorem there isx1 ∈ (0, 1) and x2 ∈ (1, 2) such that f(x1) = f(x2) = 0.

f ′(x) = 4x3 – 7, so f(x) decreases for x < (7/4)1/3 andincreases for x > (7/4)1/3 and 1 < (7/4)1/3 < 2. Thus f(x)cannot be zero at any other point.

62. Answer (4)

f(x) = x3 – bx + b

f ′(x) = 3x2 – b

⇒ f ′(1) = 3 – b

Equation of tangent is

∴ (y – 1) = (3 – b) (x – 1)

∴ (3 – b)x – y = 2 – b

∴ Area of triangle formed by the straight line andthe coordinate axes is

1 2 (2 ) 22 3

b bb

−⎛ ⎞Δ = − =⎜ ⎟−⎝ ⎠

Solving for b, we have

b = 2

63. Answer (4)

Put tanx = t

I =2 2

24 2

(1 )(1 )1

t dttt t+

++∫

=4 2

4

1 2t tt

+ +∫

= 4 2

2dt dt dtt t

+ +∫ ∫ ∫

= 3 11 (tan ) tan 2(tan )3

x x x c− −− + − +

= 31 cot tan 2cot3

x x x c− + − +

∴13

α = − and β = 1, γ = –2

∴ α + β + γ = 43

−

64. Answer (3)

The graph is as shown

–2 O 2


6/9

And, area bounded by the curve and the x-axis is2

2

0

2 (2 – )x x dx∫

=23

2

0

2 –3xx

⎛ ⎞⎜ ⎟⎝ ⎠

=8 sq.units3

65. Answer (1)y

π2

xO

π2

–

f(x) = sin x

x

f ′(x) = 2

cos – sinx x xx

= 2

cos ( – tan ) 0x x xx

>

66. Answer (3)

Put ( )22 2 2x x t+ − − =

⇒ 2 4x x t− − =

⇒ (x – t)2 = x2 – 4

⇒ t2 + 4 = 2tx

⇒1 42

x tt

⎛ ⎞= +⎜ ⎟⎝ ⎠

Thus 2

1 412

dxt

⎛ ⎞= −⎜ ⎟⎝ ⎠

∴ I = 2

1 412

t dtt

⎛ ⎞−⎜ ⎟⎝ ⎠∫

=3/2 1/21 2 8

2 3t t c−⎛ ⎞+ +⎜ ⎟

⎝ ⎠

= ( )31 1 162 232 2 2 2

x x cx x

⎡ ⎤+ − − + +⎢ ⎥+ − −⎣ ⎦

= ( ) ( )31 1 2 – – 2 4 2 – 232 2

x x x x c⎡ ⎤+ + + + +⎢ ⎥⎣ ⎦

67. Answer (3)

If 2 2( ) ( ) ( ) ( )b b b

a a a

I f x g x dx f x dx g x dx= ≤∫ ∫ ∫

Put sin2x = t

∴1

3

0

1I t dt= +∫Put f(x) = 1 and g(x) = 31 x+

I = 1 1

3 3

0 0

1 1x dx x+ ≤ +∫ ∫

1 514 4

≤ + =

So, I < 2 and 7

2I <

68. Answer (2)

'( ) (3 3 )log 3ax axef x a −= −

So f is monotonically decreasing if 3ax – 3–ax > 0 i.e.

32ax > 1, which is true if 2ax > 0.

Since a < 0, we must have x < 0.

69. Answer (4)

Only this specific condition is insufficient to answerthat maxima or minima is at c.

70. Answer (2)

Put tan x = t

I = 3 4+ + +∫t dt

t t

I = 4 3t t t t dt⎡ ⎤+ − +⎣ ⎦∫

Let, In = t t n dt+∫Put t + n = u2

dt = 2u du

In = 2( ) (2 )−∫ u n u u du

= 5 3 5/2 3/22 2 2 2( ) ( )5 3 5 3

n nu u t n t n− = + − +

∴ I = 5/2 5/22 (tan 4) (tan 3)5

x x⎡ ⎤+ − +⎣ ⎦

( ) ( )3/ 2 3/22 4 tan 4 3 tan 33

x x c⎡ ⎤− + − + +⎣ ⎦


7/9

71. Answer (4)f(x) = 3sinx–[sinx]⋅ cosx

∵ f(π – x) = – f(x)

∴0

( ) 0f x dxπ

=∫72. Answer (3)

Consider the function φ(x) = f(x) – g(x) on theinterval [x0, x]. By Langrange's theorem we haveφ(x) – φ(x0) = φ′(z) (x – x0) for some z ∈ (x0, x)Since φ(x) = 0, φ′(z) = φ′(z) – g'(z) > 0

So φ(x) – φ(x0) = φ(x) = (f ′(z) – g'(z)) (x – x0) > 0⇒ φ(x) > 0Thus f(x) > g(x) for all x > x0

73. Answer (1)

1 | sin | | cos | 2x x≤ + ≤

∴ y = [ ]| sin | | cos | 1x x+ =

∴ Area bounded by the two curve can be repre-sented by

2

π/6

y = 1

∴ Area bounded is

= ( )22 / 3 1(2) 2 3 (1)2 2π⎛ ⎞ −⎜ ⎟

⎝ ⎠

4 33π

= −

74. Answer (3)f ′(x) = m (x – a)m–1 (x – b)n + n (x – a)m (x – b)n–1

= (x – a)m–1 (x – b)n–1 [(m+n)x – (mb + na)]∴ f ′(x) = 0 for three points.

75. Answer (3)g(x) = (f (x))2 – 2f(x) – 3(f (x))3

g′(x) = 2f (x).f ′(x) – 2f ′(x) – 9(f (x))2 f ′(x) = –f ′(x) (9(f(x)2 – 2f(x) + 2)

Quadratic in f in bracket is always positive asD = 4 – 72 < 0

∴ g(x) is increasing or decreasing according as f(x)is decreasing or increasing respectively.

76. Answer (4)2

cos2 sin 0x dxα

α

+ α =∫2sin2 sin 0

2x α

α

+ α =

sin4 sin2 sin 02 2α α− + α =

2cos3 sin sin 02α α

+ α =

∴ sinα = 0 or cos 3α = – 1 (α∈[–2π, 2π])

∴ α = –2π, –π, 0, π, 2π

3α = π, –π, 3π, –3π, 5π, –5π

⇒– 5 –5, , ,– , ,

3 3 3 3π π π π

α = π π

∴ Number of solution = 9

77. Answer (3)

3{x} – 2 [x] = 3 (x – [x]) –2[x]

= 3x – 5[x]

∴5/2

3/2

3 5[ ]−

= −∫I x x dx

1 0

3/2 1

3 10 | 3 5 |x dx x dx−

− −

= + + +∫ ∫

1 2

0 1

3 | 3 5 |x dx x dx+ + −∫ ∫

5/2

2

3 10 |x dx+ −∫

1 0

3/2 1

(3 10) (3 5)−

− −

= + + +∫ ∫x dx x dx

1 5/3

0 1

3 ( 3 5)+ + − +∫ ∫x dx x dx

2 5/2

5/3 2

(3 5) (10 3 )x dx x dx+ − + −∫ ∫

1 0 12 2 2

3/2 1 0

3 3 310 52 2 2x x xx x

−

− −

= + + + +

5/3 2 5/22 2 2

1 5/3 2

3 3 35 5 102 2 2x x xx x x−

+ + + − + −

12712

=


8/9

78. Answer (2)

Let f(x) = 4sin 2 cos

2 cosx x x x

x− −+

Differenting w.r.t. x, we have

2

(2 cos ) [4cos 2 cos cos ](4sin 2 cos )[ sin ]'( )

(2 cos )

x x x x x xx x x x xf x

x

+ − − +− − − −

=+

2 2

2

6cos 4 3cos 4sin'( )(2 cos )

x x xf xx

− + +=

+

⇒2

2

6cos cos'( )(2 cos )

x xf xx

−=

+

Now, f(x) is increasing

If 2

cos (6 cos ) 0(2 cos )

x xx

−>

+

⇒ cosx > 0 ; 30, ,2

2 2x π π⎛ ⎞ ⎛ ⎞∈ ∪ π⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

For, f(x) is decreasing, we have 3,

2 2x π π⎛ ⎞∈ ⎜ ⎟

⎝ ⎠

79. Answer (1)

I = 1

2 4 1/2

1

(1 2sin sin )x dx−

−

− α + α∫

12 4

21

2(1 2sin sin )2sin

x

−

− α + α=

− α

2 4 1/ 22

1 (1 2sin sin )sin

⎡= − − α + α −⎣α2 4 1/2(1 2sin sin ) ⎤+ α + α ⎦

( ) ( )2 2

2

– (1– sin – 1 sin )

sin

α + α=

α

= 2

80. Answer (3)

f(x) = 2x3 – 9ax2 + 12a2x + 1

⇒ f ′(x) = 6x2 – 18ax + 12a2

For point of maxima/minima,

f ′(x) = 0

∴ 6x2 – 18ax + 12a2 = 0

x = a, 2a

∵ β = α2

∴ a > 0

If a > 0, x = a is local maximum and x = 2a is localminimum

∴ α = a, β = 2a

∴ 2a = a2 ⇒ a = 2

or if x = 2a is local maximum and x = a is localminimum, then α = 2a, β = a

⇒ a = 4a2, a ≠ 0, a = 14

81. Answer (4)

16161

1 1

tan 14 1

x dxx xx x x

− − −−

∫

16

1

16 13 4 1

dx

x

π= −

−∫

3 22

0

16 1 4 (1 ) ( 1 )3 4π +

= − = +∫t t dt x t

t

33

0

163 3

tt⎛ ⎞π

= − +⎜ ⎟⎝ ⎠

( )16 163 3 2 33 3π π

= − + = −

82. Answer (4)

We have f ′(x) = 2ae2x + bex + c

31 = f ′(log2) = 8a + 2b + c

Also, –1 = f(0) = a + b, further

392

=log4

0

( ( ) )f x cx dx−∫

=log4

2

0

( )x xae be dx+∫

=log4

2

02x xa e be+

= 16 42 2× + − −

a ab b

=15 3

2a b+

∴ a = 5, b = – 6, c = 3


9/9

83. Answer (2)

1 4 9

0 1 4

0 1 2dx dx dx+ +∫ ∫ ∫

= 3 + 10 = 13

84. Answer (4)

g(x) is increasing and f(x) is decreasing function.

Now (1) x + 1 < x + 5 x R∀ ∈

∴ g(x + 1) < g(x + 5)

⇒ f(g(x + 1)) > f(g(x + 5)) ( (4) is correct)

(2) x < x +1

g(x) < g(x + 1)

⇒ f(g(x)) > f(g(x + 1)) ((2) is wrong)

(3) x < x + 2

f(x)) > f(x + 2)

∴ g(f(x)) > g(f(x + 2)) ((3) is wrong)

(4) x > x – 2 x R∀ ∈

f(x) < f(x – 2)

g(f(x) < g(f(x–2)) ((1) is wrong)

85. Answer (4)

mJ = 1

1 11

(ln ) (ln ) (ln )ee e

m m mmxx dx x x x x dxx

−= −∫ ∫

Jm = 1−− me mJ

∴ 9 89+ =J J e

86. Answer (1)f(x) = π/4 – tan–1x

f ′(x) = 2

1 01 x

− <+

for all x.

So the greatest value is f(0) = π/4.and the least value is f(1) = 0.

87. Answer (2)

2( 1)xxe x dx+∫ = 21 ( 1) ( )

2⎛ ⎞ + =⎜ ⎟⎝ ⎠ ∫

te t dt t x

= 12

te t c⋅ +

= 2

2

2x xe c⋅ +

88. Answer (1)20 cos 1,x≤ ≤ so

2

1 1 18 55 3cos x≤ ≤

+/2

2016 105 3cos

dxx

ππ π≤ ≤

+∫89. Answer (1)

x8 sin11x is an odd function

So 8 11sin 0

a

a

x x dx−

=∫

90. Answer (3)Set x3 + 1 = u

⇒2 92

3 21 2

10 103| 1|

=+∫ ∫x dudx

x u

3527

=

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Test 8 Paper 1 JEE Main 2013