JEE (Main + Advanced) : LEADER & ENTHUSIAST COURSE

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 8 multiple choice questions with one or more than one correct option.

Marking scheme : +4 for correct answer, 0 if not attempted and –2 in all other cases.

(ii) Section-I(ii) contains 2 ‘paragraph’ type questions. Each paragraph describes an experiment, a situation or a

problem. Two multiple choice questions will be asked based on each paragraph. One or more than one option

can be correct.

Marking scheme : +4 for correct answer, 0 if not attempted and –2 in all other cases.

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

10. Section-IV contains 8 questions. The answer to each question is a single digit integer ranging from

0 to 9 (both inclusive)

Marking scheme : +4 for correct answer and 0 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.

PAPER – 2

Test Type : ALL INDIA OPEN TEST Test Pattern : JEE-Advanced

TEST DATE : 08 - 05 - 2016

TARGET : JEE (ADVANCED) 2016

JEE (Main + Advanced) : LEADER & ENTHUSIAST COURSE

Paper Code : 0000CT103115007

CLASSROOM CONTACT PROGRAMME(Academic Session : 2015 - 2016)

Time : 3 Hours Maximum Marks : 240

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

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ALL INDIA OPEN TEST/JEE (Advanced)/08-05-2016/PAPER-2

0000CT103115007

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 masses : 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,

Space for Rough Work

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

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PART-1 : PHYSICS

SECTION–I(i) : (Maximum Marks : 32)

This section contains EIGHT 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

–2 In all other cases

1. A stationary observer receives a sound of frequency f0 = 2000 Hz when the source is at rest. When the

source starts moving with a constant velocity starting from a large distance, the apparent frequency f

varies with time as shown in figure. Speed of sound = 300 m/s. Choose the CORRECT alternative (s):-

1800

2000

fm

f(Hz)

t(s)

(A) Speed of source is 66.7 m/s. (B) fm shown in figure can be 2500 Hz.

(C) Speed of source is 33.33 m/s. (D) fm shown in figure cannot be greater than 2250 Hz.

2. A projectile is projected from the level ground in vertical x-y plane (vertical is y axis). The point of

projection is considered as the origin. For which of the following initial velocities of projection u and

angle of projection with the horizontal will the projectile pass through the point (4, 2) (all in m) :-

(A) u = 4 5 m/s, = 45° (B) u = 4 5 m/s, tan–1(3)

(C) u = 2 5 m/s, = 45° (D) u = 8 5 m/s, tan–1(3)

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS


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PHYSICS


0000CT103115007


3. A block of mass 1 kg is pushed towards another block of mass 2 kg from 6 m distance as shown in

figure. Just after collision velocity of 2 kg block becomes 4 m/s :-

6m/s

2kg1kg

6m

(A) coefficient of restitution between two blocks is 1

(B) coefficient of restitution between two blocks is 1/2

(C) velocity of centre of mass after 2 s is 2 m/s

(D) velocity of centre of mass after 2 s is 1 m/s

4. Consider a solid sphere placed on an inclined plane. The friction coefficient between the plane and the

sphere is . For which of the following cases will the friction acting on the sphere be dissipative?

(A) = 0.2 , = 30 (B) = 0.25, = 45 (C) = 0.2, = 37 (D) =0.6, = 53.


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5. A tunnel is dug along a chord of the earth at a perpendicular distance R/2 from the earth’s centre. The

wall of the tunnel may be assumed to be rough with friction coefficient µ. A particle is released from

one end of the tunnel. Choose the CORRECT statement(s) :

(A) The normal force exerted by the wall is constant in magnitude as well as direction.

(B) The friction force exerted by the wall is constant in magnitude as well as direction.

(C) The normal force exerted by the wall is constant in magnitude but not in direction.

(D) The friction force exerted by the wall is constant in magnitude but not in direction.

6. Two point charges q1 and q

2 are fixed with a finite distance d between them. It is desired to put a third

charge q3 in between these two charges on the line joining them so that the charge q

3 is in equilibrium.

This is possible if :-

(A) q1 is positive, q

2 is negative and q

3 is negative

(B) q1 is positive, q

2 is positive and q

3 is positive

(C) q1 is positive, q

2 is positive and q

3 is negative

(D) q1 is positive, q

2 is negative and q

3 is positive

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7. A body of mass m is suspended from two light springs of force constants k1 and k

2 separately.

The periods of vertical oscillations are T1 and T

2 respectively. Now the same body is suspended from

the same two springs which are first connected in series and then in parallel. The period of vertical

oscillations are Ts and T

p respectively :-

(A) Tp < T

1 < T

2 < T

s for k

1 > k

2(B)

2 2 2p 1 2

1 1 1

T T T

(C) Ts2 = T

12 + T

22 (D)

s 1 2T T T

8. All capacitors were initially uncharged :-

5µF

10µF 15

15

12

10

50VS

(A) Battery current just after closing of switch S is 3.42 A

(B) Battery current just after closing of switch S is 0.962 A

(C) Battery current after long time of closing of switch S is 3.42 A

(D) Battery current after long time of closing of switch S is 0.962 A



PHYSICS

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SECTION–I(ii) : (Maximum Marks : 16)

This section contains TWO paragraphs.

Based on each paragraph, there will be TWO questions




Marking scheme :

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



Paragraph for Questions 9 and 10

This problem investigates the Planck blackbody spectrum in one dimension.

Consider a one-dimensional electromagnetic cavity of length L. At the ends of cavity, there are perfect

reflectors. The waves form standing wave pattern in cavity analogous to a string fixed at both ends.

Thus in frequency range to + d, we have number of modes as dN which can be found by formulae

for standing waves in a string. Power stored in interval to + d is given by E dN . We can

assume

2

x

0

xdx

e 1 6

. Theoretically all frequencies from 0 to are possible.

The average energy in a single mode of frequency v is given by h / kT

hE

e 1

. Use this to write

down the one-dimensional blackbody spectrum. This is the energy stored in the frequency interval

(v, v + dv).

In three dimensions, Stefan found that the total power radiated by a blackbody was proportional to T4.

9. The number of modes in frequency range to + d :-

(A) Increase with increasing length L

(B) Decrease with increasing length L

(C) Increase with increase in number of harmonic

(D) Decrease with increase in number of harmonic

10. Power P is proportional to :-

(A) T4 (B) T2 (C) L–3 (D) L


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Two long, thin concentric hollow cylindrical shells are each free to rotate around the z-axis. A mechanical

attachment (not shown) keeps them concentric. The two cylinders have the same length , but different

radii a and b. Each cylinder is an insulator, with a fixed charge per unit area, given by a and

b,

respectively. Suppose that the both cylinders rotate at angular frequency . [Take |a| = |

b| = ,

>> a, b and a = b

3]

a

b

z

11. What is the correct value of electric field & magnetic field in 3 regions ?

(A) For 0 < r < a, E = 0 & B = µ0 (a + b) along +ve z direction.

(B) For a < r < b, E = 0

a

r

in radially outward direction and B = µ0(b) along – ve z direction.

(C) For r > b, E = 0, B = 0

(D) For r > b, E =

0

b a

r

in radially inward direction & B = 0.

12. Due to electric field and magnetic field, pressure acts on the surfaces. The pressure due to magnetic

field is pB and pressure due to electric field is p

E :-

(A) On inner surface pE is outward and p

B is inwards.

(B) On outer surface pE is inward and p

B is outwards

(C) On inner surface pE as well as p

B is outwards.

(D) On outer surface pE as well as p

B is inwards.

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

No question will be asked in section II and III



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SECTION–IV : (Maximum Marks : 32) This section contains EIGHT 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 darkened0 In all other cases

1. Find the quantum number ‘n’ corresponding to exciting state of He+ ion if on transition to the ground

state that ion emits two photons in succession with wavelength 1026.7 Å and 304 Å.

(R = 1.096 × 107 / m)

2. Five rods with identical geometries are arranged as shown. Their thermal conductivity are shown.

Only A and C are maintained at 100°C and 0°C respectively. In other words, heat flows in from A and

flows out of C. If temperature difference between ends D and B can be written as 10x °C where x is an

integer. Then find x.

E K F

B

3K

K/2

CD2K

K/2A


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0000CT103115007

3. A 30 V storage battery is charged from 120 V direct current supply mains with a resistor being connected

in series with battery to limit the charging current to 15 amp. If all the heat produced in circuit, could

be made available in heating water, the time it would take to bring 1 kg of water from 15ºC to the

100ºC is.......... minute [Neglect the internal resistance of the battery]. Round off to nearest integer.

4. In alternating current circuit with a voltage of 200 V and angular frequency of 500 rad/s is connected

to two capacitors of capacitance C = 1 F each. Parallel to one of the capacitor resistor of 2000 is

connected. Find the thermal power (in W) dissipated in the chain.

C

C

R



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5. To anticipate the dip and hump in the road the driver of a car applies her brakes to produce a uniform

deceleration. Her speed is 20m/s at the bottom A of the dip and 10m/s at the top C of the hump, which

is 50m along the road from A. The radius of curvature of the hump at A is 100m. Calculate the total

acceleration at A (in m/s2).

|||

|||

|||

||

||

|||

|||

|||

||

||

||

|||

|||

||

|||||

||||

||||| |||||||||||||||

||| |||||||||||60m 60m C

BA

150m

x

6. Two fixed, equal, positive charges, each of magnitude q = 5 × 10–5C are located at points A and B

separated by a distance of 6m. An equal and opposite charge moves towards them along the line COD,

the perpendicular bisector of the line AB. The moving charge, when it reaches the point C at a distance

of 4m from O, has a kinetic energy of 1.5 joules. Calculate the distance of the farthest point D from O

(in m) which the negative charge will reach before returning towards C. Round off to nearest integer.

–qC

D O

+q

+q B

A


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PHYSICS


0000CT103115007

7. The half life of a certain radioactive specimen is to be measured. We have a geiger mueller counter

which measures the activity rate. We start the counter. At t = 0, the activity rate is 1000 ± 5 dps. At

t = 100 sec, the activity drops to 500 ± 5 dps. If the least count of stop watch is 1 sec, find the

percentage error in the half life. (Round off to nearest integer)

8. An infinite wire carrying current i = i0 sin t is kept along the axis of a toroid of mean radius a. The

cross section area of the toroid is A ( << a2). If the resistance of toroid is R, find the mean power

dissipated P in the toroid. Number of turns in toroid = N. i0 = 10A, = 100 rad/s, N = 1000, A = 1 cm2,

a = 10 cm, R = 1µ. Fill 100 P in OMR sheet.



CHEM

ISTRY

E-13/280000CT103115007

PART-2 : CHEMISTRY






Marking scheme :




1. Select the correct statement for an ideal gas

(A) If all gas molecules are assumed to be rigid sphere of negligible volume, the only possible molecular

motion is translation

(B) When temperature increases at constant pressure, collision frequency increases

(C) Mean free path increases by increasing temperature at constant pressure

(D) Kinetic energy decreases by increasing pressure at constant temperature

2. The nature of electric charge on collidal particle can be determined by -

(A) Ultramicroscope (B) Dialysis (C) Electrophoresis (D) Electroosmosis

3. Which of the following reactions are correctly represented ?

(A) 5525Mn (n,) 56

25Mn (B) 105 B (n) 13

7 N (C) 2713 Al (n, p) 27

12 Mg (D) 28 2914 15Si(D,n) P

4. Roasted silver ore + CN + H O–

2

O (air)2

'X' + OH–

Zn

'Y' + Ag

Which of the following statement is CORRECT for the above process.(A) Co-ordination number of X is two(B) 'X' and 'Y' both are diamagnetic(C) Zn acts as a catalyst(D) CN– ion is a complex forming agent as well as acts as a reducing agent.

5. The complex that is expected to show optical isomerism is(A) [Ni(en)

3]2+ (B) [Fe(ox)Cl

4]3–

(C) [Cr(NH3)

3Cl

2(NO

3)] (D) [Zn(NO

3)

2(NH

3)(CO)]


E-14/28

CHEM

ISTRY


0000CT103115007


6. Select incorrect option(s) :

(A)

OH

Cl &

Cl

HO can be differentiated by victor mayor's test

(B) Monomer of orlon is vinyl acetate

(C) Structure of tripeptide Gla-Ala-Ala is HOOC–CH –NH–C–CH–NH–C–CH–NH2 2

O OMe Me

(D) D-glucose, D-fructose & L-mannose give same osazone with Ph–NH–NH2

7. Select option(s) with correctly mentioned major product in following reaction :

(A) CH –CHO3NaOH CH –COONa + CH –CH OH3 3 2

(B) Me–C–MeNaOH C=CH–C–Me

O OMe

Me

(C) Ph–CHO + Me–C–O–C–Me(i) AcONa,

O O

(ii) H O ,3

+ Ph–CH=CH–CH=O

(D)

O

O

MeOK

OH

COOMe

8. Select correct statement(s) :

(A) Me–CH=CH–C

O

–Me will replace maximum of 4 H from D on prolonged keeping in D2O under

acidic or basic condition

(B)Cl

Cl

can show optical as well as geometrical isomerism

(C) Methane can't be prepared by Wurtz reaction in good yield

(D) With HCl , ClCH2–CH=CH

2 will give 1,2-dichloropropane as major product


CHEM

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Marking scheme :





A voltaic cell consist of an electrode of solid metal 'A' immersed in a 0.1M ANO3 solution and an

electrode of metal 'B' immersed in a 0.1M B(NO3)

2 solution. A porous barrier seperates the two half

cells. Current flows from electrode 'A' to 'B' in the outer circuit.

Given : 0

A / AE 0.4V ; 0

CellE 0.7V

9. Select the correct statement regarding the cell -

(A) 'B' is better reducing agent than hydrogen

(B) 'A' is better reducing agent than 'B'

(C) Ecell

< E0cell

(D) As the cell operate concentration of B2+ increases

10. What is the cell emf if in the above cell electrode 'A' is replaced by hydrogen electrode having pH one.

2.303RT0.06

F

(A) 0.3 V (B) 0.27 V (C) – 0.3 V (D) –0.33 V

E-16/28

CHEM

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0000CT103115007



Salt solution of 'A'dil. H SO2 4

C( ) + Gas 'B' White

turbidity

SCN–

(aq)

+KCN

+FeCl3

Blood redcolour of 'D'

Cr O /H2 7

2– +

(E) Green colour compound

11. The molecular formula of 'A' is

(A) Na2S

2O

3.5H

2O(s) (B) Na

2S (C) Na

2SO

3(D) Na

2CO

3

12. Which of the following statements is CORRECT for the above flow diagram ?

(A) Gas 'B' is SO2

(B) The molecular formula of 'D' is [Fe(H2O)

5(SCN)]+2

(C) The green colour compound (E) contains three unpaired electrons on the metal ion.

(D) The white turbidity of C is elemental sulphur




CHEM

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1. Calculate relative lowering in vapour pressure when 15 gm urea (Mw = 60) is dissolved in 18 gm of

water.

Fill your answer by multiplying it with 10.

2. In the II order reaction : 2A A2. The rate of formation of A

2 is 10–5 M sec–1 at 0.01M concentration

of A. Calculate the rate constant in the rate of disappearance of 'A'.

If your answer in scientific notation is x × 10–y then fill ‘X’ in OMR.

3. Find out total reaction(s) with their given condition(s), which converts Mn2+ into MnO4–.

(i) Mn2+ + BiO3–

H (ii) Mn2+ + KNO

3 + Na

2CO

3

(iii) MnO2 + KOH + O

2 fused (iv) Mn2+ + PbO

2

H

(v) Mn2+ + NaOH air

(vi) Mn2+ + S2O

82– + H

2O


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CHEM

ISTRY


0000CT103115007


4. The total number of optical active isomers for the complex ion [Co(NH3)

3Cl(NO

2)

2]+ is

5. Consider MOT and find the bond order of Be2 .

6. How many out of following reagents will change 1-cyclohexylmethanol into cyclohexane carbaldehyde

(i) Pyridinium chloro chromate (ii) H/ KMnO4 (iii) Tollen's reagent (iv) NBS /

(v) Red hot Cu tube (vi) TsCl / DMSO / NaHCO3 (vii) LiAlH

4(viii) NaBH

4


CHEM

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E-19/280000CT103115007

7. Find out total number of 1,2-shifts in this reaction :

OH

Conc. H SO2 4

8. How many N-atom are present in final product of following sequence :

HC CHRed hotFe tube

AConc. HNO3 BConc. H SO2 4

25ºC

Sn/HCl CNaNO /HCl,2 D0º–5ºC

C Final product


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MATHEM

ATICS


0000CT103115007

PART-3 : MATHEMATICS






Marking scheme :




1. If 3 32

3 3

37 3 91

1 x dx 1a b

12x 2 3x x

where a,b N, then-

(A) a = 36 (B) a = 16 (C) b = 6 (D) b = 4

2. Let 3

2

xƒ x

9 x

and

= number of integers in the domain of ƒ(x) where ƒ(x) is increasing.

m = number of solutions of ƒ(x) = sinx

n = least positive integral value of k for which ƒ(x) = k posses exactly 3 different solutions then-

(A) m2 = (B) 5m = n2 – 2 (C) n + + m = 20 (D) n + m = 75

3. Let 1 1 1

A 1 ......2 3 120

and 1 1 1

B ....2 3 121

, then-

(A) A > 20 (B) B < 20 (C) A + B > 40 (D) A + B < 40

4. Let n is the number of complex number(s) satisfying |z + 3| = |z – 3| + 6 and |z – 4| = r, where r R+, then

identify the correct statement -

(A) if r = 1, then n = 2 (B) if r = 1, then n = 1

(C) if r = 8, then n = 2 (D) if r = 8, then n = 1



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5. Let x, y, z are positive real numbers and 1 is the least value of 2x4 + 2y4 + 4z4 – 8xyz and

2 is the least value

of 4 4

2 3 3 2

4 1x y xy 8

x y x y , then -

(A) 1 = –1 (B)

1 > –1 (C)

2 = 10 (D)

2 > 10

6. Curve y = ƒ(x) satisfy the differential equation ysinx dx = cos4x dx – cosx dy and passes through origin,then

(A) y = ƒ(x) never intersect x

y cosx 12

(B) y = ƒ(x) passes through ,2 4

(C) 5 1

ƒ '2 2

(D) the equation of tangent at origin is y = x.

7. Let L is a line and A is foot of perpendicular from O (origin) on L. P and Q are two other points on line

L. Position vectors of A, P, Q are a,x, y

respectively. If | a | 1

, centroid of OPQ is c

and

b a (x y)

, then

(A) b 3c

x2

(B) 3c b

y2

(C) b

is parallel to line L

(D) a 2b , a 3b,a 5b

are position vectors of collinear points.

8. If z is a complex number satisfying |z2 – 1| = |z| + 2, then-

(A) Maximum value of |z| is 1 13

2

(B) Maximum value of |z| is

1 5

2

(C) Minimum value of |z| is 1 5

2

(D) Minimum value of |z| is

1 2

2

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MATHEM

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0000CT103115007








Marking scheme :





Let S1 is a hyperbola and S

2 be an ellipse whose centers are at origin. The transverse axis of S

1 and major

axis of S2 lie along the x-axis and y-axis respectively. Let H be the hyperbola x2 – y2 = 7. The straight

line 4x – 3y = 7 touches the curves H,S1 and S

2 at P,Q and R respectively. Suppose

5PQ QR

2 .

9. If e1 and e

2 are eccentricities of S

1 and S

2 respectively. Then-

(A) 1

23e

15 (B) 2

1e

2

(C) 2 21 215e 4e 24 (D) e

1 is an irrational number

10. Identify the incorrect statement(s) about S1 and S

2-

(A) S1, S

2 intersect at 4 distinct points (B) No common point in S

1 and S

2

(C) 4 common tangents in S1 and S

2(D) 2 common tangents of S

1 and S

2


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There are n urns each containing 2n balls such that the ith urn contains i white and 2n – i black balls.

Let Ei be the event of selecting ith urn, i = 1, 2, ...... n and w denotes the event of getting a white ball.

P(E) denotes probability of occurence of event E.

11. If P(Ei) i2, where i = 1, 2, ........ n, then -

(A) 3(2n 1)

P(w)16(n 1)

(B)

3(n 1)P(w)

4(2n 1)

(C) n

3lim P(w)

8 (D)

nn

3lim nP(E )

2

12. If P(Ei) i, where i = 1, 2, ..... n, then

(A) 1

6P(E / w)

n(n 1)(2n 1)

(B) P(E

i/w) increases as i increases from 1 to n

(C) P(Ei/w) decreases as i increases from 1 to n (D)

nnlim nP(E / w) 3




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MATHEM

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0000CT103115007



1. Let P and Q be distinct points on the parabola y2 = 4x such that circle, for which PQ is diameter passes

through vertex of the parabola. If area of OPQ (O is origin) is 20 and the diameter of circumcircle

of OPQ is d, then 2d

65 is equal to

2. The line of intersection of a plane P : 2x + y + z = 2 with xy plane is L1, with zx plane is L2 and with

yz plane is L3 and area of triangle made by lines L1,L2,L3 is , then 2 is equal to

3. If y = y(x) and it follows the relation xsiny + x = y, then absolute value of ( + 1)2 y''() is



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4. The sum of all in [0, 2] for which 3tan2 + 8tan + 3 = 0 is n , then n is equal to

5. Let R is the radius of the circle passing through the origin and cutting orthogonally each of the circles

x2 + y2 – 8y + 12 = 0 and x2 + y2 – 4x – 6y – 3 = 0, then 2R

5 is equal to

6. If 2n cos20º = sin40º + n cos40º, then 2

1

n is equal to


E-26/28

MATHEM

ATICS


0000CT103115007

7. A unit vector r

has equal length of projections along the vectors ˆ ˆ ˆ ˆ2i j 2i j

,5 5

and k̂ .

If ˆ ˆ ˆr. i j k

is positive, then number of such vectors r

will be

8. Let x is the number of diagonal matrices A of real entries and order 3 × 3 such that A7 + 3A5 + 7A = 11I and

y is the number of diagonal matrices B of complex entries with at least one non zero real entry and

order 3 × 3 such that B5 = I, then y – 55x is equal to

(where I is an identity matrix of order 3 × 3)



E-27/280000CT103115007



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 0000CT103115007E-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) 8

: +4 0–2

(ii) -I(ii) 2 ‘’

: +4 0–2

9. –II III

10. -IV 8 0 9 ()

: +4 0

11.

12.

13.

PAPER – 2

Test Type : ALL INDIA OPEN TEST Test Pattern : JEE-Advanced

TEST DATE : 08 - 05 - 2016Time : 3 Hours Maximum Marks : 240

Paper Code : 0000CT103115007

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

IND

I

TARGET : JEE (ADVANCED) 2016

JEE (Main + Advanced) : LEADER & ENTHUSIAST COURSE

H-2/28


0000CT103115007

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 masses : 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-3/280000CT103115007

-1 : – I(i) : ( : 32)

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

:

+4 0 –2

1. f0 = 2000 Hz

f

300 m/s

1800

2000

fm

f(Hz)

t(s)

(A) 66.7 m/s (B) fm 2500 Hz

(C) 33.33 m/s (D) fm 2250 Hz

2. x-y y

u (4, 2)

() :-

(A) u = 4 5 m/s, = 45° (B) u = 4 5 m/s, tan–1(3)

(C) u = 2 5 m/s, = 45° (D) u = 8 5 m/s, tan–1(3)

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS

H-4/28

PHYSICS


0000CT103115007

3. 1 kg 6 m 2 kg

2 kg 4 m/s

6m/s

2kg1kg

6m

(A) 1 (B) 1/2

(C) 2 s 2 m/s (D) 2 s 1 m/s

4.

(A) = 0.2 , = 30 (B) = 0.25, = 45 (C) = 0.2, = 37 (D) =0.6, = 53.


PHYSICS

H-5/280000CT103115007

5. R/2 µ

(A)

(B)

(C)

(D)

6. q1 q

2 d

q3 q

3

(A) q1 , q

2 q

3

(B) q1 , q

2 q

3

(C) q1 , q

2 q

3

(D) q1 , q

2 q

3

H-6/28

PHYSICS


0000CT103115007

7. m k1 k

2

T1 T

2

Ts T

p

(A) k1 > k

2 T

p < T

1 < T

2 < T

s(B) 2 2 2

p 1 2

1 1 1

T T T

(C) Ts2 = T

12 + T

22 (D) s 1 2T T T

8.

5µF

10µF 15

15

12

10

50VS

(A) S 3.42 A

(B) S 0.962 A

(C) S 3.42 A

(D) S 0.962 A


PHYSICS

H-7/280000CT103115007

–I(ii) : ( : 16)

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

:

+4

0

–2

9 10

L

+ ddN

+ d

E dN 2

x

0

xdx

e 1 6

0

v h / kT

hE

e 1

(v, v + dv)

T4

9. + d

(A) L

(B) L

(C)

(D)

10. P (A) T4 (B) T2 (C) L–3 (D) L

H-8/28

PHYSICS


0000CT103115007

11 12

z-

a b

a

b

[|a| = |

b| = , >> a, b a =

b

3]

a

b

z

11.

(A) 0 < r < a E = 0 B = µ0 (a + b); z

(B) a < r < b E = 0

a

r

B = µ

0(b); z

(C) r > b E = 0, B = 0

(D) r > b E =

0

b a

r

B = 0.

12.

pB p

E

(A) pE p

B

(B) pE p

B

(C) pE p

B

(D) pE p

B

– II : & – III :

II III


PHYSICS

H-9/280000CT103115007

–IV : ( : 32)

0 9

:

+4

0

1. He+ 'n'

1026.7 Å 304 Å (R = 1.096 × 107 / m)

2.

A C 100°C 0°C A C

D B 10x °C x x

E K F

B

3K

K/2

CD2K

K/2A

H-10/28

PHYSICS


0000CT103115007

3. 30 V 120 V 15 amp

1 kg 15ºC 100ºC [

]

4. 200 V 500 rad/s C = 1 F

2000

(W )

C

C

R


PHYSICS

H-11/280000CT103115007

5.

A 20m/s C 10m/s A

50m A 100m A (m/s2 )

|||

|||

|||

||

||

|||

|||

|||

||

||

||

|||

|||

||

|||||

||||

||||| |||||||||||||||

||| |||||||||||60m 60m C

BA

150m

x

6. q = 5 × 10–5C 6m A B

COD, AB

O 4m C 1.5 DO

C

–qC

D O

+q

+q B

A

H-12/28

PHYSICS


0000CT103115007

7.

t = 0 1000 ± 5 dps t = 100 sec

500 ± 5 dps 1 sec

8. i = i0 sin t a

A ( << a2) R P

= N i0 = 10A, = 100 rad/s, N = 1000, A = 1 cm2, a = 10 cm,

R = 1µ100 P


CHEM

ISTRY

H-13/280000CT103115007

-2 : – I(i) : ( : 32)

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

+4 0 –2

1. -(A) (B) (C) (D)

2. -(A) (B) (C) (D)

3. ?

(A) 5525Mn (n,) 56

25Mn (B) 105 B (n) 13

7 N

(C) 2713 Al (n, p) 27

12 Mg (D) 28 2914 15Si(D,n) P

4. + CN + H O–

2

O ( )2

'X' + OH–

Zn

'Y' + Ag

(A) X 2 (B) 'X' 'Y' , (C) Zn (D) CN–

5. -(A) [Ni(en)

3]2+ (B) [Fe(ox)Cl

4]3–

(C) [Cr(NH3)

3Cl

2(NO

3)] (D) [Zn(NO

3)

2(NH

3)(CO)]

H-14/28

CHEM

ISTRY


0000CT103115007

6.

(A)

OH

Cl

Cl

HO

(B)

(C) Gla-Ala-Ala HOOC–CH –NH–C–CH–NH–C–CH–NH2 2

O OMe Me

(D) D- , D- L-Ph–NH–NH2

7.

(A) CH –CHO3NaOH CH –COONa + CH –CH OH3 3 2

(B) Me–C–MeNaOH C=CH–C–Me

O OMe

Me

(C) Ph–CHO + Me–C–O–C–Me(i) AcONa,

O O

(ii) H O ,3

+ Ph–CH=CH–CH=O

(D)

O

O

MeOK

OH

COOMe

8.

(A) Me–CH=CH–C

O

–Me D2O

4 H D

(B)Cl

Cl

(C)

(D) HCl ClCH2–CH=CH

2 1,2-


CHEM

ISTRY

H-15/280000CT103115007

–I(ii) : ( : 16)

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

:

+4

0

–2

9 10

0.1M ANO3 'A' 0.1M B(NO

3)

2

B

'A' 'B'

: 0

A / AE 0.4V ; 0

CellE 0.7V

9. -

(A) 'B',

(B) 'A' , 'B'

(C) Ecell

< E0cell

(D) B2+

10. 'A' pH

emf 2.303RT

0.06F

(A) 0.3 V (B) 0.27 V (C) – 0.3 V (D) –0.33 V

H-16/28

CHEM

ISTRY


0000CT103115007

11 12

dil. H SO2 4

C( ) + 'B' SCN

–

(aq)

+KCN

+FeCl3

Cr O /H2 7

2– +

(E)

'A'

'D'

11. 'A'

(A) Na2S

2O

3.5H

2O(s) (B) Na

2S (C) Na

2SO

3(D) Na

2CO

3

12. (flow)

(A) 'B' , SO2

(B) 'D' [Fe(H2O)

5(SCN)]+2

(C) (E)

(D) C

– II : & – III :

II III


CHEM

ISTRY

H-17/280000CT103115007

–IV : ( : 32)

0 9

:

+4

0

1. 15 gm (Mw = 60) 18 gm

10

2. II : 2A A2. A 0.01M A

2 10–5 M sec–1 'A'

x × 10–y OMR ‘X’

3. Mn2+ MnO4–

(i) Mn2+ + BiO3–

H (ii) Mn2+ + KNO

3 + Na

2CO

3

(iii) MnO2 + KOH + O

2 fused (iv) Mn2+ + PbO

2

H

(v) Mn2+ + NaOH air

(vi) Mn2+ + S2O

82– + H

2O

H-18/28

CHEM

ISTRY


0000CT103115007

4. [Co(NH3)

3Cl(NO

2)

2]+

5. MOT Be2

6. 1-

(i) (ii) H / KMnO4 (iii) (iv) NBS /

(v) Cu (vi) TsCl / DMSO / NaHCO3 (vii) LiAlH

4(viii) NaBH

4


CHEM

ISTRY

H-19/280000CT103115007

7. 1,2-

OH

Conc. H SO2 4

8. N-

HC CHRed hotFe tube

AConc. HNO3 BConc. H SO2 4

25ºC

Sn/HCl CNaNO /HCl,2 D0º–5ºC

C

H-20/28

MATHEM

ATICS


0000CT103115007

-3 : – I(i) : ( : 32)

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

:

+4 0 –2

1. 3 32

3 3

37 3 91

1 x dx 1a b

12x 2 3x x

a,b N -

(A) a = 36 (B) a = 16 (C) b = 6 (D) b = 4

2. 3

2

xƒ x

9 x

ƒ(x)ƒ(x) = sinx m nk ƒ(x) = k 3 (A) m2 = (B) 5m = n2 – 2 (C) n + + m = 20 (D) n + m = 75

3. 1 1 1

A 1 ......2 3 120

1 1 1

B ....2 3 121

-

(A) A > 20 (B) B < 20 (C) A + B > 40 (D) A + B < 40

4. |z + 3| = |z – 3| + 6 |z – 4| = r n r R+

-

(A) r = 1 n = 2 (B) r = 1 n = 1

(C) r = 8 n = 2 (D) r = 8 n = 1


MATHEM

ATICS

H-21/280000CT103115007

5. x, y, z 2x4 + 2y4 + 4z4 – 8xyz 1

4 4

2 3 3 2

4 1x y xy 8

x y x y

2 -

(A) 1 = –1 (B)

1 > –1 (C)

2 = 10 (D)

2 > 10

6. y = ƒ(x) ysinx dx = cos4x dx – cosx dy

(A) y = ƒ(x) xy cosx 1

2

(B) y = ƒ(x), ,2 4

(C) 5 1

ƒ '2 2

(D) y = x 7. L A, L O () P Q L A,

P, Q a,x, y

| a | 1

, OPQ c

b a (x y)

(A) b 3c

x2

(B) 3c b

y2

(C) b

, L

(D) a 2b

, a 3b,a 5b

8. z |z2 – 1| = |z| + 2 -

(A) |z| 1 13

2

(B) |z|

1 5

2

(C) |z| 1 5

2

(D) |z|

1 2

2

H-22/28

MATHEM

ATICS


0000CT103115007

–I(ii) : ( : 16)

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

:

+4 0

–2

9 10

S1 S

2 S

1 S

2

x-y-H : x2 – y2 = 7 4x – 3y = 7 H,S1

S2 P,Q R 5

PQ QR2

9. e1 e

2 S

1 S

2 -

(A) 1

23e

15 (B) 2

1e

2

(C) 2 21 215e 4e 24 (D) e

1

10. S1 S

2

(A) S1, S

2 4

(B) S1 S

2

(C) S1 S

2 4

(D) S1 S

2 2


MATHEM

ATICS

H-23/280000CT103115007

11 12

n 2n i i 2n – i

Ei, i (i = 1, 2, ...... n)w

P(E), E

11. P(Ei) i2, i = 1, 2, ........ n -

(A) 3(2n 1)

P(w)16(n 1)

(B)

3(n 1)P(w)

4(2n 1)

(C) n

3lim P(w)

8 (D) n

n

3lim nP(E )

2

12. P(Ei) i, i = 1, 2, ..... n

(A) 1

6P(E / w)

n(n 1)(2n 1)

(B) i 1 n P(Ei/w)

(C) i 1 n P(Ei/w)

(D) n

nlim nP(E / w) 3

– II : & – III :

II III

H-24/28

MATHEM

ATICS


0000CT103115007

–IV : ( : 32)

0 9

:

+4

0

1. P Q y2 = 4x PQ

OPQ (O ) 20 OPQ d 2d

65

2. P : 2x + y + z = 2 xy L1, zx L2

yz L3 L1,L2,L3 2

3. y = y(x), xsiny + x = y ( + 1)2 y''()


MATHEM

ATICS

H-25/280000CT103115007

4. [0, 2] 3tan2 + 8tan + 3 = 0 n n

5. R x2 + y2 – 8y + 12 = 0

x2 + y2 – 4x – 6y – 3 = 0 2R

5

6. 2n cos20º = sin40º + n cos40º 2

1

n

H-26/28

MATHEM

ATICS


0000CT103115007

7. r

ˆ ˆ ˆ ˆ2i j 2i j,

5 5

k̂

ˆ ˆ ˆr. i j k

r

8. 3 × 3A x A7 + 3A5 + 7A = 11I

3 × 3 B y,

B5 = I y – 55x (I, 3 × 3)


H-27/280000CT103115007


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 0000CT103115007H-28/28

:

14.

15.

16.

17. :

18.

19.

20. g = 10 m/s2

____________________________

____________________________

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

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

JEE (Main + Advanced) : LEADER & ENTHUSIAST COURSE

Documents

Transcript of JEE (Main + Advanced) : LEADER & ENTHUSIAST COURSE