JEE (Main + Advanced) : LEADER & ENTHUSIAST …s displacement law constant b = 2.89 × 10–3 m–K...

READ THE INSTRUCTIONS CAREFULLY

GENERAL :

1. This sealed booklet is your Question Paper. Do not break the seal till you are instructed to do so.

2. Use the Optical Response sheet (ORS) provided separately for answering the questions.

3. Blank spaces are provided within this booklet for rough work.

4. Write your name and form number in the space provided on the back cover of this booklet.

5. After breaking the seal of the booklet, verify that the booklet contains 28 pages and all the 20 questions in each

subject and along with the options are legible.

QUESTION PAPER FORMAT AND MARKING SCHEME :

6. The question paper has three parts : Physics, Chemistry and Mathematics. Each part has two sections.

7. Carefully read the instructions given at the beginning of each section.

8. Section-I :

(i) Section-I(i) contains 10 multiple choice questions with only one correct option.

Marking scheme : +3 for correct answer and 0 in all other cases.

(ii) Section-I(ii) contains 5 multiple choice questions with one or more than one correct option.

Marking scheme : +4 for correct answer, 0 if not attempted and –1 in all other cases.

9. There is no questions in SECTION-II & III

10. Section-IV contains 5 questions. The answer to each question is a single digit integer ranging from

0 to 9 (both inclusive)

Marking scheme : +4 for correct answer, 0 if not attempted and –1 in all other cases.

OPTICAL RESPONSE SHEET :

11. The ORS is machine-gradable and will be collected by the invigilator at the end of the examination.

12. Do not tamper with or mutilate the ORS.

13. Write your name, form number and sign with pen in the space provided for this purpose on the original. Do not write

any of these details anywhere else. Darken the appropriate bubble under each digit of your form number.

DO

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PAPER – 1

Test Type : FULL SYLLABUS Test Pattern : JEE-Advanced

TEST DATE : 24 - 04 - 2016

SCORE(ADVANCED)

JEE (Main + Advanced) : LEADER & ENTHUSIAST COURSE

Paper Code : 1001CT103115041

CLASSROOM CONTACT PROGRAMME(Academic Session : 2015 - 2016)

Please see the last page of this booklet for rest of the instructions

Time : 3 Hours Maximum Marks : 210

EN

GL

ISH

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Target : JEE (Main + Advanced) 2016/24-04-2016/Paper-1

1001CT103115041

Space for Rough Work

SOME USEFUL CONSTANTSAtomic No. H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16, Cl = 17,

Br = 35, Xe = 54, Ce = 58,

Atomic masse s : H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24,

Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127,

Xe = 131, Ba=137, Ce = 140,

Boltzmann constant k = 1.38 × 10–23 JK–1

Coulomb's law constant

9

0

1= 9×10

4

Universal gravitational constant G = 6.67259 × 10–11 N–m2 kg–2

Speed of light in vacuum c = 3 × 108 ms–1

Stefan–Boltzmann constant = 5.67 × 10–8 Wm–2–K–4

Wien's displacement law constant b = 2.89 × 10–3 m–K

Permeability of vacuum µ0 = 4 × 10–7 NA–2

Permittivity of vacuum 0 = 2

0

1

c

Planck constant h = 6.63 × 10–34 J–s

Leader & Enthusiast Course/Score(Advanced)/24-04-2016/Paper-1

PHYSICS

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PART-1 : PHYSICSSECTION–I(i) : (Maximum Marks : 30)

This section contains TEN questions.

Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four option(s) is correct.

For each question, darken the bubble corresponding to the correct option in the ORS

Marking scheme :

+3 If the bubble corresponding to the correct option is darkened

0 In all other cases

1. The magnetic field inside a solid conducting long wire at distance r from its axis is given as B = B0r3

where B0 is constant. Which of the following relations correctly represents current enclosed in the loop

of radius r shown in the figure:

r

(A)

30

0

2 B r

5

(B)

40

0

B r

2

(C)

20

0

2 B r

(D)

40

0

2 B r

2. Water is coming out from a tap of cross-section area A with speed v0. Area of cross-section of water

stream changes as it moves down. Which of the following graph can best represent cross-section area 'a'

of stream with depth h :

h

(A)

a

h

(B)

a

h

(C)

a

h

(D)

a

h

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS


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1001CT103115041


3. The vernier of a vernier scale is divided into 10 divisions which coincide with 9 divisions of the main

scale, each main scale division being 0.5 mm. When the two jaws of the instrument are in contact with

each other, the 4th division of the vernier scale coincides with a main scale division and the zero of the

vernier lies to the right of the zero of the main scale. When a sphere is inserted between the jaws, the

zero of vernier scale lies slightly to the right of 1.8 cm and the sixth vernier division coincides with a

main division. The diameter of sphere will be

(A) 1.850 cm (B) 1.810 cm (C) 1.750 cm (D) 1.710 cm

4. An open pipe 0.400 m in length is placed vertically in a cylindrical bucket and nearly touches the bottom

of the bucket which has an area of 0.100 m2. Water is slowly poured into the bucket until a sounding

tuning fork of frequency 440 Hz, held over the pipe, produces resonance. Find the mass of water in the

bucket at this moment. Take speed of sound in air as 330 m/s.

(A) 18.75 kg (B) 21.25 kg (C) 40.00 kg (D) 26.75 kg


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5. A circus acrobat of mass M leaps straight up with initial velocity V0 from a trampoline. As he rises up,

he takes a trained monkey of mass m hanging from a branch at a height h above the trampoline. What is

the maximum height attained by the pair (from the branch) ?

(A)

20VM

hM m 2g

(B)

2 20VM

hM m 2g

(C)

20Vm

hM m 2g

(D)

2 20Vm

hM m 2g

6. The piston is massless and the spring is ideal and initially stretched. The piston cylinder arrangement

encloses an ideal gas. If the gas is heated quasistatically, the PV graph is :-

(A)

P

V

2

1(B)

P

V

(C)

P

V

2

(D)

P

V

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1001CT103115041

7. A metallic ring (radius R) of negligible resistance has a resistance r connected across its diameter as

shown in the figure. It is moving with velocity v0 in a constant magnetic field B

0 acting perpendicular to

the plane of paper in inward direction. The current in the resistance is :

× × × × × ×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

B

D

v0

r

(A) 02BRv

r(B)

0BRv

r(C)

0BRv

2r(D) 0

8. Two scales on a voltmeter measure voltages up to 20.0 V and 30.0 V. The resistance connected in series

with the galvanometer is 1680 for the 20.0 V scale and 2930 for the 30.0 V scale. The resistance of

the galvanometer and the full scale current respectively are :-

(A) 320 , 8 mA (B) 70 , 10 mA (C) 820 , 10 mA (D) 820 , 8 mA



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9. If light of wavelength of maximum intensity emitted from surface at temperature T1 is used to cause

photoelectric emission from a metallic surface, the maximum kinetic energy of the emitted electron is

6eV, which is 3 times the work function of the metallic surface. If light of wavelength of maximum

intensity emitted from a surface at temperature T2 (T

2 = 2T

1) is used, the maximum kinetic energy of the

photoelectrons emitted is :-

(A) 2 eV (B) 4 eV (C) 14 eV (D) 18 eV

10. A triangular rigid wire frame 'AOB' is made, in which length of each wire is and mass is m. The whole

system is suspended from point O and free to perform SHM about x-axis or about z-axis. When it

performs SHM about x-axis it time period of oscillation is T1 and when it performs SHM about z-axis,

its time-period of oscillation is T2, then choose the CORRECT option.

y

O x

BA

(A) T1 < T

2(B) T

1 > T

2(C) T

1 = T

2(D) None


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1001CT103115041

SECTION–I(ii) : (Maximum Marks : 20)

This section contains FIVE questions.

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these

four option(s) is (are) correct.

For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS

Marking scheme :

+4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened

0 If none of the bubbles is darkened

–1 In all other cases

11. The continuous X - ray spectrum for the three potential differences between target and cathode, V1, V

2

and V3 is shown. Temperatures of filament in these cases are T

1, T

2 and T

3 respectively, then :

V1

V2

V3

T1

T2 T3

Inte

nsit

y(I)

(A) V1 > V

2 > V

3(B) T

1 > T

2 > T

3(C) V

1 = V

2 = V

3(D) T

1 < T

2 < T

3

12. A thin paper of thickness 0.02 mm having refractive index 1.45 is pasted across one of the slit in a

Young's double slit experiment. The paper transmits 4/9 of light falling on it. (Wavelength = 600 nm).

(A) Amplitude of light wave transmitted through the paper will be 2/3 time of incident wave.

(B) The ratio of maximum and minimum intensity in the fringe pattern will be 25.

(C) The total number of fringe crossing the centre if an identical paper is pasted on the other slit is 15.

(D) The ratio of maximum and minimum intensity in the pattern will be 5.



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13. In the circuit shown switch S is closed at t = 0. At time (t) which of the following is/are CORRECT:

S

a b

I1

I2

L

L

R

IV

R

(A)

4Rt

LV

I 1 e2R

(B)

4Rt

L1 2

VI I 1 e

2R

(C) At t = 0, abdV

0dt

(D) 2 21 2I I I


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1001CT103115041

14. Figure shows an arrangement of four identical rectangular plates A, B, C and D each of area S. Ignore

the separation between the plates in comparison to the plate dimensions.

A B C D

+Q1 +Q2

(A) Potential difference between plates A & B is independent of Q1.

(B) Potential difference between plates C & D is independent of Q1.

(C) Potential difference between plates A & B is independent of Q2.

(D) Potential difference between plates C & D is independent of Q2.

15.92

U235 is '' (Alpha) active. Then in a large quantity of the element :

(A) The probability of a nucleus disintegrating during one second is lower in the first half life and

greater in the fifth half life

(B) The probability of a nucleus disintegrating during one second remains constant for all time.

(C) More than half of U235 will remain even after the average life

(D) The energy of the emitted '' particle is less than the disintegration energy of the U235 nucleus.

SECTION –II : Matrix-Match Type & SECTION –III : Integer Value Correct Type

No question will be asked in section II and III



PHYSICS

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SECTION–IV : (Maximum Marks : 20)

This section contains FIVE questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive For each question, darken the bubble corresponding to the correct integer in the ORS Marking scheme :

+4 If the bubble corresponding to the answer is darkened



1. Electric field in a space is given as E = E0x(x – 1)(x – 2)(x – 3) N/C along X-axis. Where E

0 is a positive

constant. A positive charge q of mass m is in stable equilibrium at some specific position x, where x > 0.

If this charge is displaced slightly from equilibrium position to perform simple harmonic motion then

angular frequency (rad/s) of SHM will be n. Find n (assume q E0 = 2 m )

2. A uniform rod AB of length 4m and mass 12 kg is thrown vertically upwards such that just after the

projection the centre of mass of the rod moves vertically upwards with a velocity 10 m/s and at the same

time it is rotating with an angular velocity /2 rad/sec about a horizontal axis passing through its mid

point. Just after the rod is thrown it is horizontal as shown in the figure. Find the acceleration (in m/s2) of

the point A when the centre of mass is at the highest point. (Take g = 10m/s2 and 2 = 10)

10 m/s

A B/2 rad/sec


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1001CT103115041

3. A soap bubble of radius 4 cm has uniform surface charge density of 217.7 C / m . If the surface

tension of the liquid is 0.07 N/m. Find the air pressure difference between inside and outside the soap

bubble in N/m2.

4. A solid sphere of diameter 0.1 m is at 427°C and is kept in an enclosure at 27°C. Take Stefan's

constant = 820

103

W/m2K4, emissivity of the surface 0.84, specific heat 0.1 kcal/kg K,

density = 9280 kg/m3, J = 4200 J/kcal. If rate of decrease of temperature of the sphere is N × 10–1 °C/s,

find N.

5. A vessel, whose bottom is flat and perfectly reflecting, is filled with water (index = 4/3) upto a

height = 40cm. A point object in air above is moving towards the water surface with a constant

speed = 4 m/s. What is the relative speed of its final image (in m/s), as seen by the object itself, at a

moment when the object is 30 cm above the water surface ?



CHEM

ISTRY

E-13/281001CT103115041

PART-2 : CHEMISTRY

SECTION–I(i) : (Maximum Marks : 30)




Marking scheme :



1. For the given reaction :

H2(g) + S(s) H

2S (g)

at 300 K ; rS = 400 J/mol–K and

rG = – 20 kJ/mol–1

Temperature at which above reaction will occur reversibly is : [Assume rH &

rS as temperature

independent]

(A) 200 K (B) 250 K (C) 350 K (D) 400 K

2. An aqueous solution of 0.1M B+Cl–(aq.) solution has pH = 5 at 298 K. The equilibrium constant for

the reaction.

B+(aq.) + OH–(aq.) BOH (aq.) + H2O (l) is :

(A) 103 (B) 105 (C) 107 (D) 109

3. For the the galvanic cell operating at 298K as shown below -

Ag(s) | AgNO3(0.1M) || H

2SO

4 (0.05M) | H

2(0.1 bar) , Pt

Ecell

is [Given : SOP of Ag = – 0.8 V]

(A) – 0.57 V (B) – 0.77 V (C) – 0.83 V (D) 0.83 V

4. Which of the following molecule is diamagnetic and have last electron in sigma () B.M.O

(A) O2

(B) N2

(C) B2

(D) C2

5. Choose the pair of complex compounds which show stereoisomerism as well as structural isomerism

(A) [Co(H2O)

4Cl

2]Cl, [Cr(H

2O)

5Cl]Cl

2(B) [Co(NH

3)

4Cl

2], [Cr(H

2O)

3Cl

3]

(C) [Cr(NH3)

3Cl(en)] SO

4, [Fe(H

2O)

5NO]SO

4(D) [Ni(en)

2(NH

3)Cl]Br, [Zn(Pn)PyCl] Br


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CHEM

ISTRY


1001CT103115041


6. Which of the following reaction is INCORRECT ?

(A) SO2Cl

2 + P

4(white) PCl

5 + SO

2

(B) I– + H+ + KMnO4 IO

3– + Mn2+

(C) I2O

5 + CO CO

2 + I

2

(D) XeF6 hydrolysis

in alkaline medium HXeO

4–

OH

XeO64– + Xe + O

2

7. In castner-kellner cell if cathode is made up of graphite instead of mercury, then product that will be

obtained first at cathode will be -

(A) Na-amalgam (B) Na only (C) H2 gas (D) NaOH

8. What is the major product explected from the following reaction ?

CH3D–Cl

Product

(A) H

CH3

H

H

(B) D

CH3

Cl

H

(C) H

CH3

D

Cl

(D) Cl

D

CH3

H


CHEM

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9. If number of stereoisomers of compound CH–Ph

CH3

(A)

is Z & number of -H present in major product

of following reaction (B) is W.

OHconc. H SO2 4 (B)

(major product)

Then value of (W + Z) is :

(A) 13 (B) 11 (C) 12 (D) 15

10. For the preparation of compound, which of the following is used as best method :

O CH3

(A) Cl + CH ONa3 (B) ONa + CH –Cl3

(C) OH + CH –Cl3 (D) Cl + CH OH3

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CHEM

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1001CT103115041






Marking scheme :




11. For a Van der waals gas A(g) which of the following option(s) is/are correct -

(A) A(g) will always exert lesser pressure than an ideal gas

(B) A(g) may exert pressure greater than or lesser than an ideal gas

(C) A(g) is more difficult to be compressed than an ideal gas

(D) For A(g) Van der waals constant 'a' and 'b' will change with temperature.

12. For the reaction : A B. Differential rate law is : 1d[A]

k[A]dt

. If initial concentration of A is A

0M

and concentration at time 't' is At M then identify correct option (s)

(A) At2 = A

02 – 2kt (B) Half life , t

1/2 =

203A

4K

(C) Completion time for reaction : t100%

= 20A

2K (D) The reaction must be complex



CHEM

ISTRY

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13. Salt 'X' + conc. H SO Y (Heterodiatomic gas)2 4

FeCl3

(Neutral sol.)deep blue ppt.

Salt 'X' + CuSO4 solution — chocolate brown ppt.

Based on above observations select the correct option :-

(A) Central atom in anion of salt 'X' is sp3d2 hybridised

(B) Central atom in anion of salt 'X' belongs to '3d' series

(C) Salt 'X' gives HCN gas with dil. H2SO

4

(D) Gas 'Y' is acidic in nature

14. Which of the following reaction(s) is/are correctly matched with it's major product :

(A) HCl

Cl

(B)

F

NO2

H–N

NO2

N

(C) CH

ONH –NH2 2

KOH /CH3 (D)

F

Alc. KOH

(major)

15.Br /h2 v

(A) Mg/THF(B) 1. H–C–H

O

2. H+ (C)

conc. H SO2 4

170º C(D) HBr

CCl4

(E)

Which of the following is (are) true for above reaction sequence ?

(A) Compund (D) is (B) Compound (E) contains chiral atom

(C) Compound (C) is 1º alcohol (D) Compound (A) and (E) are identical




E-18/28

CHEM

ISTRY


1001CT103115041

SECTION–IV : (Maximum Marks : 20)


The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive

For each question, darken the bubble corresponding to the correct integer in the ORS

Marking scheme :




1. Element Mg [Atomic mass = 24 g mole–1]. Exist as a cubic lattice with edge length 4Å and has density

of 2.5 × 103 kg m–3. Calculate effective number of magnesium atoms per unit cell.

[Use : NA = 6 × 1023]

2. Total number of common metallurgical steps involved during extraction of Fe from Fe2O

3 and Cu

from CuFeS2 (high grade ore) -

Froth floatation, roasting (partial/complete), calcination, smelting, electro-refining

3. Z(Compound of potassium)

KIFeCl3

SO2

Y(Compound tosulphur)

SnCl2 X(Compound of iron)

Find the sum of oxidation states of Fe, S and K in X, Y, Z respectively -



CHEM

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E-19/281001CT103115041

4. A tetrapeptide has –NH2 group on glycine. This produces glycine (Gly), valine (Val), leucine (Leu)

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

(Primary structures) with valine (Val) never containing free –COOH group is :

5. Number of compounds which are more reactive towards nitration then benzene :

(i)

Cl

(ii)

DD

D

DD

D

(iii)

NO2

(iv)

CH3

(v)

OCH3

(vi)

O–C–CH3

O

(vii)

CC

O

OCH3 (viii)

NH–C–CH3

O

(ix)

CCl3

(x)

NH–Ph

:


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MATHEM

ATICS


1001CT103115041

PART-3 : MATHEMATICSSECTION–I(i) : (Maximum Marks : 30)




Marking scheme :



1.x

cos2xdx, x 0,

2(e cos x) 1 sin 2x

is equal to

(A) xn e cosx C (B) xn 1 e cos x C

(C) xn e sin x C (D) xn 1 e sin x C

(where C is constant of integration)

2. Number of solutions of the equation 2sinx + 5sin2x + 8sin3x + ...... = 1 for x [0, 2] is -

(A) 0 (B) 1 (C) 2 (D) 4

3. Area bounded by curves y = 2 – |x – 1| and y = |x – 2| – 2 is -

(A) 15

2 unit2 (B) 15 unit2 (C)

15

4 unit2 (D) 4 unit2



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4. A point is selected at random inside the circle x2 + y2 = 36, then the probability that it also lies inside

two parabolas y2 = 4x and x2 = 4y is

(A) 4

27(B)

4

27(C)

4

9(D)

4

95. In a triangle ABC the inradius (r) is half of its exradius (r

1), opposite to vertex A, then value of

A B Ctan tan tan

2 2 2

is

(A) 1 (B) 1

2(C)

1

3(D)

2

3

6. Solution of the differential equation e–y(2tan–1x + 1 + x2)dx + (1 + x2)dy = 0 is -

(A) x + ey + (tan–1x)2 = C (B) x + e–y + (tan–1x)2 = C

(C) e–y(x + (tan–1x)2) = C (D) ey(x + (tan–1x)2) = C

(where C is constant of integration)

7. Value of

1 2

1 2

1 2

1 (cosec(cosec x)) 2

cos (x ) 2 1

2 1 tan (x )

is equal to -

(A) 112

(B) 7

2

(C) –5 (D) 13

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8. Number of values of m I for which point (2m, 3m) lies inside both the circles x2 + y2 = 25 and

x2 + y2 – 2x – 2y + 1 = 0, is -

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

9. If a

and b

are two mutually perpendicular unit vectors such that a.c b.c

, where c

is a non zero

vector, then which of the following is always incorrect -

(A) a, b, c

are coplanar (B) a, b, c

are mutually perpendicular

(C) (a b).c 0

(D) vector c

is collinear with vector a

or vector b

10. Let a function ƒ(x) = (cosx)esinx, then -

(A) ƒ(x) is monotonic in x ,2 2

(B) ƒ(x) has a local maxima in x ,2 2

(C) ƒ(x) has a local minima in x ,2 2

(D) ƒ(x) has a local maxima and a local minima in x ,2 2


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Marking scheme :




11. Consider a function ƒ(x) = |x2 – 2mx + m2 – 1| x [m – 2, m + 2], where m = 4n + 1 and n I, then

(A) ƒ(x) is a periodic function

(B) Number of solutions of equation ƒ(x) = 1 in [0, 11] is 5.

(C) Number of local extremum points of ƒ(x) in (0, 11) is 10.

(D) Number of points where ƒ(x) is non-differentiable in [0, 11] is 10.

12. Consider a curve 23x 21

C : y7

and a line L of slope 1, which is tangent to C at point A, then

(A) Equation of L is y = x + 2

(B) Equation of L is y = x – 2

(C) Equation of circle passing through origin and touches the curve C at A is x2 + y2 + x – 41y = 0

(D) Equation of circle passing through origin and touches the curve C at A is 4x2 + 4y2 + x – 41y = 0


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1001CT103115041

13. Let x-y plane is Argand plane and z lies on the cross section of a solid sphere, centred at (2 2, 2 2, 0)

and radius 2, which cut by Argand plane, then

(A) Minimum value of |z – 2 – 2i| is zero.

(B) Minimum value of |z – 2 – 2i| is 2 2 2 .

(C) Area of quadrilateral formed by origin, (2 2,2 2,0) , z1 and z

2 is 4 3 is (where z

1 and z

2 are

positions of z which has minimum and maximum amplitude respectively).

(D) Area of quadrilateral formed by origin, (2 2,2 2,0) , z1 and z

2 is 2 3 is (where z

1 and z

2 are

positions of z which has minimum and maximum amplitude respectively).

14. Let a function 2{x}ƒ(x) e (where {.} is fractional part function), then

(A) ƒ(x) is a periodic function.

(B) ƒ(x) is an even function

(C) in x [0, 10], ƒ'(x) = 0 has 10 distinct solutions.

(D) x I, ƒ(x) has a local minima.

15. Let A and B are Involutary matrices of same order, then

(A) If AB = BA, then AB BA

2

will be involutary

(B) If AB = BA, then AB – BA will be a Nill potent matrix

(C) If A and B are inverse to each other, then A + B will be involutary matrix.

(D) Both A and B will be orthogonal matrix also.





MATHEM

ATICS

E-25/281001CT103115041

SECTION–IV : (Maximum Marks : 20) This section contains FIVE questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive For each question, darken the bubble corresponding to the correct integer in the ORS Marking scheme :




1. Let a function y = ƒ(x) is defined by x = esin and y = esin, where is a real parameter, then value

of lim ƒ '(x)

is (given that lim(sin cos ) 0

)

2. a and b are positive real numbers such that for constant value of a + b the value of ab is maximum,

then number of ordered pairs (a,b), for which lines x y 1 z 2

a 2 1

,

x 1 y 2 z 3

1 2 b

are

coplanar, is


E-26/28

MATHEM

ATICS


1001CT103115041

3. If coefficient of x12 in expansion of (1 – x2 + x4)3(1 – x)7 is equal to 10m –1, then value of m is

equal to

4. Let 3

ƒ(a) sin(x a).sin(x a)dx

, then value of ƒ'(4) is equal to

5. If total number of alpha-numeral linear arrangements of all letters and digits used in the word

"T20 WORLD CUP 2016" such that between each pair of digits there is exactly one letter and no three

letters are together is m(6!)2, then sum of digits in number m is



E-27/281001CT103115041



Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005

+91-744-5156100 [email protected] www.allen.ac.in

Your Target is to secure Good Rank in JEE 2016 1001CT103115041E-28/28

DARKENING THE BUBBLES ON THE ORS :

14. Use a BLACK BALL POINT PEN to darken the bubbles in the upper sheet.

15. Darken the bubble COMPLETELY.

16. Darken the bubbles ONLY if you are sure of the answer.

17. The correct way of darkening a bubble is as shown here :

18. There is NO way to erase or "un-darken" a darkened bubble.

19. The marking scheme given at the beginning of each section gives details of how darkened and not darkened

bubbles are evaluated.

20. Take g = 10 m/s2 unless otherwise stated.

I HAVE READ ALL THE INSTRUCTIONS

AND SHALL ABIDE BY THEM

____________________________

Signature of the Candidate

I have verified the identity, name and rollnumber of the candidate, and that questionpaper and ORS codes are the same.

____________________________

Signature of the invigilator

NAME OF THE CANDIDATE ................................................................................................

FORM NO. .............................................


1.

2. (ORS)

3.

4.

5. 28 20

6.

7.

8. -I :

(i) -I(i) 10

: +3 0

(ii) -I(ii) 5

: +4 0–1

9. –II III

10. -IV 5 0 9 ()

: +4 0–1

11.

12.

13.

PAPER – 1

Test Type : FULL SYLLABUS Test Pattern : JEE-Advanced

TEST DATE : 24 - 04 - 2016

Time : 3 Hours Maximum Marks : 210

SCORE(ADVANCED)

JEE (Main + Advanced) : LEADER & ENTHUSIAST COURSE

Paper Code : 1001CT103115041

CLASSROOM CONTACT PROGRAMME(Academic Session : 2015 - 2016)H

IND

I

H-2/28


1001CT103115041

SOME USEFUL CONSTANTSAtomic No. H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16, Cl = 17,

Br = 35, Xe = 54, Ce = 58,

Atomic masse s : H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24,

Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127,

Xe = 131, Ba=137, Ce = 140,

Boltzmann constant k = 1.38 × 10–23 JK–1

Coulomb's law constant

9

0

1= 9×10

4

Universal gravitational constant G = 6.67259 × 10–11 N–m2 kg–2

Speed of light in vacuum c = 3 × 108 ms–1

Stefan–Boltzmann constant = 5.67 × 10–8 Wm–2–K–4

Wien's displacement law constant b = 2.89 × 10–3 m–K

Permeability of vacuum µ0 = 4 × 10–7 NA–2

Permittivity of vacuum 0 = 2

0

1

c

Planck constant h = 6.63 × 10–34 J–s


PHYSICS

H-3/281001CT103115041

-1 : – I(i) : ( : 30)

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

:

+3

0

1. r B = B0r3 B

0

r

r

(A)

30

0

2 B r

5

(B)

40

0

B r

2

(C)

20

0

2 B r

(D)

40

0

2 B r

2. A v0

'a' h

h

(A)

a

h

(B)

a

h

(C)

a

h

(D)

a

h

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS

H-4/28

PHYSICS


1001CT103115041

3. 10 9

0.5 mm 4

1.8 cm

(A) 1.850 cm (B) 1.810 cm (C) 1.750 cm (D) 1.710 cm

4. 0.400 m

0.100 m2 440 Hz

330 m/s

(A) 18.75 kg (B) 21.25 kg (C) 40.00 kg (D) 26.75 kg


PHYSICS

H-5/281001CT103115041

5. M V0

h m

(A)

20VM

hM m 2g

(B)

2 20VM

hM m 2g

(C)

20Vm

hM m 2g

(D)

2 20Vm

hM m 2g

6.

(quasistatically)PV

(A)

P

V

2

1(B)

P

V

(C)

P

V

2

(D)

P

V

H-6/28

PHYSICS


1001CT103115041

7. R r

B0 v

0

× × × × × ×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

B

D

v0

r

(A) 02BRv

r(B)

0BRv

r(C)

0BRv

2r(D) 0

8. 20.0 V 30.0 V 20.0 V

1680 30.0 V 2930

(A) 320 , 8 mA (B) 70 , 10 mA (C) 820 , 10 mA (D) 820 , 8 mA


PHYSICS

H-7/281001CT103115041

9. T1

6eV

3 T2 (T

2 = 2T

1)

(A) 2 eV (B) 4 eV (C) 14 eV (D) 18 eV

10. 'AOB' m

O x-z-x-

T1 z-

T2

y

O x

BA

(A) T1 < T

2(B) T

1 > T

2(C) T

1 = T

2(D)

H-8/28

PHYSICS


1001CT103115041

–I(ii) : ( : 20)

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

:

+4 0

–1

11. V1, V

2 V

3 X-

T1, T

2 T

3

V1

V2

V3

T1

T2 T3

Inte

nsit

y(I)

(A) V1 > V

2 > V

3(B) T

1 > T

2 > T

3(C) V

1 = V

2 = V

3(D) T

1 < T

2 < T

3

12. 0.02 mm 1.45

4/9 ( = 600 nm).

(A) 2/3

(B) 25

(C) 15

(D) 5


PHYSICS

H-9/281001CT103115041

13. S t = 0 (t)

S

a b

I1

I2

L

L

R

IV

R

(A)

4Rt

LV

I 1 e2R

(B)

4Rt

L1 2

VI I 1 e

2R

(C) t = 0 abdV0

dt (D) 2 2

1 2I I I

H-10/28

PHYSICS


1001CT103115041

14. S A, B, C D

A B C D

+Q1 +Q2

(A) A B Q1

(B) C D Q1

(C) A B Q2

(D) C D Q2

15.92

U235 '' ()

(A)

(B)

(C) U235

(D) '' U235

– II : & – III :

II III


PHYSICS

H-11/281001CT103115041

–IV : ( : 20)

0 9

:

+4

0

–1

1. X-E = E0x(x – 1)(x – 2)(x – 3) N/C E

0

x x > 0 q m

(rad/s ) n n ( q E0 = 2 m )

2. AB 4m 12 kg

10 m/s

/2 rad/sec

A (m/s2 )

(g = 10m/s2 2 = 10)

10 m/s

A B/2 rad/sec

H-12/28

PHYSICS


1001CT103115041

3. 4 cm 217.7 C / m

0.07 N/m N/m2

4. 0.1 m 427°C 27°C = 820

103

W/m2K4

0.84, 0.1 kcal/kg K, = 9280 kg/m3, J = 4200 J/kcal

N × 10–1 °C/s N

5. (= 4/3) 40cm

4 m/s 30 cm

(m/s


CHEM

ISTRY

H-13/281001CT103115041

-2 : –I(i) : ( : 30)

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

:

+3

0

1.

H2(g) + S(s) H

2S (g) 300 K

rS = 400 J/mol–K

rG = – 20 kJ/mol–1

[rH

rS

]

(A) 200 K (B) 250 K (C) 350 K (D) 400 K

2. 298 K 0.1M B+Cl–(aq.) pH = 5

B+(aq.) + OH–(aq.) BOH (aq.) + H2O (l)

(A) 103 (B) 105 (C) 107 (D) 109

3. 298K

Ag(s) | AgNO3(0.1M) || H

2SO

4 (0.05M) | H

2(0.1 bar) , Pt

Ecell

[ Ag SOP = – 0.8 V]

(A) – 0.57 V (B) – 0.77 V (C) – 0.83 V (D) 0.83 V

4. () B.M.O (A) O

2(B) N

2(C) B

2(D) C

2

H-14/28

CHEM

ISTRY


1001CT103115041

5.

(A) [Co(H2O)

4Cl

2]Cl, [Cr(H

2O)

5Cl]Cl

2(B) [Co(NH

3)

4Cl

2], [Cr(H

2O)

3Cl

3]

(C) [Cr(NH3)

3Cl(en)] SO

4, [Fe(H

2O)

5NO]SO

4(D) [Ni(en)

2(NH

3)Cl]Br, [Zn(Pn)PyCl] Br

6.

(A) SO2Cl

2 + P

4() PCl

5 + SO

2

(B) I– + H+ + KMnO4 IO

3– + Mn2+

(C) I2O

5 + CO CO

2 + I

2

(D) XeF6

HXeO

4–

OH

XeO64– + Xe + O

2

7.

(A) Na- (B) Na (C) H2 (D) NaOH

8.

CH3D–Cl

(A) H

CH3

H

H

(B) D

CH3

Cl

H

(C) H

CH3

D

Cl

(D) Cl

D

CH3

H


CHEM

ISTRY

H-15/281001CT103115041

9. CH–Ph

CH3

(A)

Z (B)

-H W OH

conc. H SO2 4 (B)

(W + Z)

(A) 13 (B) 11 (C) 12 (D) 15

10.

O CH3

(A) Cl + CH ONa3 (B) ONa + CH –Cl3

(C) OH + CH –Cl3 (D) Cl + CH OH3

H-16/28

CHEM

ISTRY


1001CT103115041

–I(ii) : ( : 20)

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

:

+4

0

–1

11. A(g)

(A) A(g),

(B) A(g),

(C) A(g)

(D) A(g) 'a' 'b'

12. : A B 1d[A]k[A]

dt

A A0M

't' At M

(A) At2 = A

02 – 2kt (B) , t

1/2 =

203A

4K

(C) t100%

= 20A

2K (D)


CHEM

ISTRY

H-17/281001CT103115041

13. 'X' + H SO Y ( )2 4 FeCl3

( )

'X' + CuSO4 —

:-

(A) 'X' sp3d2

(B) 'X' '3d'

(C) 'X' H2SO

4 HCN

(D) 'Y'

14.

(A) HCl

Cl

(B)

F

NO2

H–N

NO2

N

(C) CH

ONH –NH2 2

KOH /CH3 (D)

F

Alc. KOH

(major)

15.Br /h2 v

(A) Mg/THF(B) 1. H–C–H

O

2. H+ (C)

conc. H SO2 4

170º C(D) HBr

CCl4

(E)

(A) (D) , (B) (E)

(C) (C) , 1º (D) (A) (E)

– II : & – III :

II III

H-18/28

CHEM

ISTRY


1001CT103115041

–IV : ( : 20)

0 9

:

+4

0

–1

1. Mg [= 24 g mole–1] 4Å

2.5 × 103 kg m–3

[ : NA = 6 × 1023]

2. Fe2O

3 Fe CuFeS

2 () Cu

, (/), ,

3. Z(

)

KIFeCl3

SO2

Y(

)

SnCl2 X(

)

X, Y, Z Fe, S K


CHEM

ISTRY

H-19/281001CT103115041

4. –NH2 (Gly), (Val),

(Leu) (Ala) (Val), –COOH

5.

(i)

Cl

(ii)

DD

D

DD

D

(iii)

NO2

(iv)

CH3

(v)

OCH3

(vi)

O–C–CH3

O

(vii)

CC

O

OCH3 (viii)

NH–C–CH3

O

(ix)

CCl3

(x)

NH–Ph

:

H-20/28

MATHEM

ATICS


1001CT103115041

-3 : – I(i) : ( : 30)

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

:

+3

0

1.x

cos2xdx, x 0,

2(e cos x) 1 sin 2x

(A) xn e cosx C (B) xn 1 e cos x C

(C) xn e sin x C (D) xn 1 e sin x C

(C )

2. x [0, 2]2sinx + 5sin2x + 8sin3x + ...... = 1 -

(A) 0 (B) 1 (C) 2 (D) 4

3. y = 2 – |x – 1| y = |x – 2| – 2

(A) 15

2 2 (B) 15 2 (C)

15

4 2 (D) 4 2

4. x2 + y2 = 36 y2 = 4x x2 = 4y,

(A) 4

27(B)

4

27(C)

4

9(D)

4

9

Enthusiast & Leader Course/Score(Advanced)/24-04-2016/Paper-1

MATHEM

ATICS

H-21/281001CT103115041

5. ABC (r) A (r1)

A B Ctan tan tan

2 2 2

(A) 1 (B) 1

2(C)

1

3(D)

2

3

6. e–y(2tan–1x + 1 + x2)dx + (1 + x2)dy = 0

(A) x + ey + (tan–1x)2 = C (B) x + e–y + (tan–1x)2 = C

(C) e–y(x + (tan–1x)2) = C (D) ey(x + (tan–1x)2) = C

(C )

7.

1 2

1 2

1 2

1 (cosec(cosec x)) 2

cos (x ) 2 1

2 1 tan (x )

(A) 112

(B) 7

2

(C) –5 (D) 13

H-22/28

MATHEM

ATICS


1001CT103115041

8. m I (2m, 3m) x2 + y2 = 25 x2 + y2 – 2x – 2y + 1 = 0

-

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

9. a

b

a.c b.c c

-

(A) a, b, c

(B) a, b, c

(C) (a b).c 0

(D) a

b c

10. ƒ(x) = (cosx)esinx

(A) x ,2 2

ƒ(x)

(B) x ,2 2

ƒ(x)

(C) x ,2 2

ƒ(x)

(D) x ,2 2

ƒ(x)


MATHEM

ATICS

H-23/281001CT103115041

–I(ii) : ( : 20)

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

:

+4

0

–1

11. ƒ(x) = |x2 – 2mx + m2 – 1| x [m – 2, m + 2] m = 4n + 1 n I

(A) ƒ(x)

(B) [0,11] ƒ(x) = 1 5

(C) (0, 11)ƒ(x) 10

(D) [0, 11]ƒ(x) 10

12. 23x 21

C : y7

L 1 A C

(A) L y = x + 2

(B) L y = x – 2

(C) C A x2 + y2 + x – 41y = 0

(D) C A 4x2 + 4y2 + x – 41y = 0

H-24/28

MATHEM

ATICS


1001CT103115041

13. x-y z, (2 2, 2 2, 0)

2 (A) |z – 2 – 2i|

(B) |z – 2 – 2i| 2 2 2

(C) (2 2,2 2,0) , z1 z

2 4 3 (z

1 z

2, z

)

(D) (2 2,2 2,0) , z1 z

2 2 3 (z

1 z

2, z

)

14. 2{x}ƒ(x) e ({.} ),

(A) ƒ(x)

(B) ƒ(x)

(C) x [0, 10] ƒ'(x) = 0 10

(D) x I, ƒ(x)

15. A B

(A) AB BA

2

AB = BA.

(B) AB – BA AB = BA.

(C) A B A + B

(D) A B

– II : & – III :

II III


MATHEM

ATICS

H-25/281001CT103115041

–IV : ( : 20)

0 9

:

+4

0

–1

1. y = ƒ(x), x = esin y = esin lim ƒ '(x)

( lim(sin cos ) 0

)

2. a b a + b ab

(a,b) x y 1 z 2

a 2 1

,

x 1 y 2 z 3

1 2 b

3. (1 – x2 + x4)3(1 – x)7 x12 10m –1 m

H-26/28

MATHEM

ATICS


1001CT103115041

4. 3

ƒ(a) sin(x a).sin(x a)dx

ƒ'(4)

5. "T20 WORLD CUP 2016"

,

m(6!)2m


H-27/281001CT103115041


Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005

+91-744-5156100 [email protected] www.allen.ac.in

Your Target is to secure Good Rank in JEE 2016 1001CT103115041H-28/28

:

14.

15.

16.

17. :

18.

19.

20. g = 10 m/s2

____________________________

____________________________

................................................................................................

.............................................

JEE (Main + Advanced) : LEADER & ENTHUSIAST …s displacement law constant b = 2.89 × 10–3 m–K...

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Transcript of JEE (Main + Advanced) : LEADER & ENTHUSIAST …s displacement law constant b = 2.89 × 10–3 m–K...