JEE (Main + Advanced) : ENTHUSIAST & LEADER COURSE · 2016. 2. 29. · TARGET : JEE (ADVANCED) 2016...

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 32 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 : 14 - 02 - 2016

TARGET : JEE (ADVANCED) 2016

JEE (Main + Advanced) : ENTHUSIAST & LEADER COURSE

Paper Code : 0000CT103115003

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)/14-02-2016/PAPER-2

0000CT103115003

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,

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


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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 certain mass of oxygen, treated like an ideal gas undergoes a cycle shown on P-T diagram here.

The maximum volume of gas during the cycle is 16 litres. Select the CORRECT statement(s) :-

(Take : 25

R3

J/mol K)

2

311

2

P(in 10 Pa)5

300 400T(in K)

(A) The mass of gas is 15.36 gm

(B) The mass of gas is 30.72 gm

(C) The minimum volume of gas during cycle is 12 litres

(D) The minimum volume of gas during cycle is 8 litres.

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS


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2. A fluid of density is rotated with a certain angular velocity in a cylindrical vessel as shown :-

h2h

R

(A) The angular velocity is 2

2gh

R (B) The angular velocity is 2

4gh

R

(C) The kinetic energy of fluid is 2 216 gh R

15

(D) The kinetic energy of fluid is

2 25 gh R

6

3. In the given figure m1 & m

2 are attached to springs on left & right respectively and kept on smooth level

ground. The left block is pushed to the left to compress spring 1 by 10 cm and right block is pushed to

right to compress spring 2 by 10 cm. Both the blocks are released simultaneously. Springs are initially

relaxed as shown in the diagram:

m =m =1kg1 2

k =100N/m1 k =400N/m2

m1 m21 2

(A) They will collide at time when both springs come to the natural length.

(B) They will collide 5cm to left of position shown.

(C) They will first collide at t = 30

sec.

(D) They will first collide at t = 10

sec.


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4. The ideal inductor is connected in circuit as shown :-

V

R

A

R L

R

(A) Just after switch is closed, potential difference across L is V

2

(B) Long time after switch is closed, potential difference across A is V

3.

(C) After a long time, switch is opened again just after switch is opened, potential difference across L is

2V

3.

(D) Potential difference across A just after switch is opened (as in option C) is V

2.

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5. If the effective voltage of the AC source is 200 V and it's frequency is variable :

10H 100 40µF

(A) The frequency at which power dissipated in resistor is maximum is 50 Hz.

(B) The frequency at which power dissipated in resistor is maximum is 25

Hz.

(C) The frequency at which power dissipated across the resistor drops to half it's maximum value is

45

2Hz.

(D) The frequency at which power dissipated across the resistor drops to half it's maximum value is

20

Hz.

6. A square wire frame is hinged about one of it's sides AB. It carries a current of 0.5A. Which of the

following magnetic fields can hold it in equilibrium in horizontal position shown. It has a mass of

500gm :-

A B

CD

x

z

y

side=2m

(A) 5ˆ ˆ3i k2

(B) 5 ˆ ˆ ˆi 3 j 2k2

(C) 5ˆ ˆ2 j k2

(D) 5 5ˆ ˆ ˆi j k2 2



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7. In which of the following cases can the body execute SHM for small displacement from the equilibrium

position?

(A)–q

+Q

A charged bead of charge –q is confined to move on the axis of a

uniformly fixed charged ring.

(B)I2

I1

A rod carrying current I2 is hanging below an infinite current

carrying wire. The rod can move only in vertical plane.

(C) On surface of earth we kept a smooth table. On the horizontal surface,

a small block is kept as shown.

(D) A bead is threaded on a horizontal smooth rod & connected to a spring.

When the spring is vertical, it is at it's natural length.


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8. Two tubes with cross sections S1 and S

2 are interconnected and closed by the pistons, whose masses are

equal m1 and m

2. An explosion occurs in the gas enclosed. Some time after explosion first piston was

moving with speed v to the left and v2 is the velocity of other piston. Friction of the piston tube wall can

be neglected then :-

m1

m2

(A) if the pipe is fixed and cannot move; v2 = m

1v/m

2

(B) if the pipe is not fixed and its mass is M; v2 = m

1 v/m

2 + m

1 + M

(C) if the pipe is fixed and cannot move; v2 = m

1S

2v/m

2S

1

(D) if the pipe is not fixed and its mass is M; v2 = (m

1+ M)v/m

2



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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

A photo cell is kept in dark. It is connected to two cells and resistances as shown. Now monochromatic

light of wavelength 310 nm is switched on. If the anode potential of photo cell is higher than the potential

of the cathode, the current I = 1A flows (Saturation current). The work function of the cathode plate is

2.5eV. Neglecting the internal resistance of batteries, all resistance are listed in ohm.

+

–

anodecathode

x

200

50+

– 100V

9. Find the voltage x such that saturation current flows in photo cell.

(A) 80 V (B) 30 V (C) 40 V (D) 61.5 V

10. If x = 100 V, what can be the incident wavelength for saturation current to flow? (Assume isat

& work

function are same)

(A) 500 nm (B) 600 nm (C) 310 nm (D) 400 nm


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Earth is in a circular orbit of angular frequency about the Sun. The Sun is so much more massive than

the Earth that, for our purposes, it may be taken to sit at rest at the center of our coordinate system.

Lagrange discovered that there exist a certain number of equilibrium points at which an artificial satellite

of negligible mass can orbit the Sun with the same frequency as the Earth (while maintaining a fixed

distance from the Earth and the Sun). We will explore some of the properties of the ‘Lagrange points’ in

this problem. If we draw a straight line from Sun to earth, such “Lagrange points” lie on this straight

line. If the distance of the Lagrange Point from Earth is very small compared to Earth Sun distance, a

Lagrange Point is Located inside the earth’s orbit (called L1) and another Lagrange Point is outside the

earth’s orbit (called L2). The distance of L1 from earth is denoted by x1 and distance of L2 from earth is

denoted by x2. Such orbits are ideally suited for space-based observatories of various kinds. The Wilkinson

Microwave Anisotropy Probe (WMAP) is stationed at one of the Lagrange Points. Take mass of earth

as Me, mass of sun as M

s, Radius of earth’s orbit as R. For this problem, neglect the rotation of earth

around its axis.

11. Their respective distances from earth are:

(A) e

31

s

Mx R

3M (B)

e3

2

s

3Mx R

M (C)

e3

1

s

Mx R

M (D)

e3

2

s

Mx R

M

12. Choose CORRECT statement(s) :-

(A) As seen from satellite in L1, Earth and Sun move in opposite direction

(B) As seen from satellite in L2, Earth and Sun move in opposite direction

(C) As seen from satellite in L1, Earth and Sun move in same direction

(D) As seen from satellite in L2, Earth and Sun move in same direction

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 energy E required (in MeV) to remove an alpha particle from 147 N . The excess of mass of atom

(in amu) is given below. (Take : 1 amu 931 MeV)

atom mass of atom – A amu

42 He

105 B

147 N

0.00260

0.01294

0.00307

Fill E

2 in OMR sheet after rounding off to nearest integer.

2. A heavy mass is hung from ceiling as shown. Both wires are identical. A pulse is produced in both the

strings simultaneously at the point of contact with mass. If it reaches the ceiling in 33 3 10 sec for

string on the right, in what time will it reach the ceiling for the left string (in 10–3 sec).

53°37°

m


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3. For the inversion of the image we often use a Dove prism (see figure), representing truncated regular

rectangular isosceles prism. Find approximately the length of side AB (in cm) at which the beam of light

falling on side AD will come out of side BC completely. The height of the trapezium ADCB is equal to

h = 1.05 cm. Refractive index of the prism is 1.41. Angle A and B are 45° each. (Given : 2 = 1.41,

3 1.73 )

A B

CD

h

4. Two bodies of charge –Q and mass m each orbit a body of charge Q at rest in a circle of radius r. What

is the total mechanical energy of the system? If answer is 23kQ

nr , fill n in OMR sheet.

Q

–Q

–Q

r

mass m

mass m



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5. A beam balance is used to suspend a Capillary tube with very thin walls vertically in a cup. The other

side of the beam balance is balanced with very accurate weights. The system is carefully lowered into

water, so that the surface of water touches the capillary. To restore the balance, we have to increase the

load on the other side by 0.056gm. Determine the radius r (in mm) of the capillary. Take surface

tension of water as 0.07 N/m and contact angle 0º.

6. The efficiency of a thermodynamic cycle 1-2-3-1 (see picture) is 20% and for another thermodynamic

cycle 1-3-4-1 efficiency is equal to 10%. Determine the efficiency (in %) of the thermodynamic cycle

1-2-3-4-1. The gas is assumed to be ideal. Fill /4 in OMR Sheet.

1

2 3

4

V

P


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7. To measure the resistance of the resistor Rx we use a circuit shown in Fig. Ammeter shows current

I = 2.0 A, voltmeter shows a voltage V = 120 V. The resistance of the voltmeter RV = 960 Ohm. What

is the error in Rx (in ohms) if we use the approximate formula R

x = V/I?

Rx

A

V

8. Circular coils are placed very close to each other so that they make up an infinite solenoid.The current in

coils is 1A and number of coils/length of solenoid is 4000 turns/m. The diameter of the solenoid is 2m.

A wire of length equal to diameter of circular coil and carrying the same current is kept along the

diameter as shown. Its mass is 4 gm. This wire is near the centre of solenoid. Find the velocity of the

wire (in m/s) when it comes out of the solenoid. (Neglect friction & assume that the rod is moving in the

horizontal plane) (Take : 2 = 10) (Neglect gravity)

i

i



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PART-2 : CHEMISTRY






Marking scheme :




1. A 1kg real gas is liquified at temperature 300K as shown in diagram. Select the correct statement-

2P(atm)

1 10 20

T=300K

V(Litre)

1

(A) 2 atm is vapour pressure of liquid at 300 K

(B) Density of gas when V equals to 10 litre, is 0.1 gm / ml

(C) Density of gas when V equals to 10 litre, is 0.05 gm / ml

(D) Density of liquid when V equals to 10 litre, is 1 gm / ml

2. In a FCC unit cell

x = distance between two nearest OV

y = distance between two nearest T.V.

z = distance between nearest O.V. and T.V.

Select the correct order of distance

(A) x = y = z (B) x < y < z (C) x > y < z (D) x > y > z

3. Which of the following option is CORRECT regarding borax bead test.

(A) Iron : Green colour in hot oxidising flame

(B) Chromium : Green colour in hot reducing flame

(C) Nickel : Reddish-brown in hot oxidising flame

(D) Mn : Amethyst (pale violet) colour in hot oxidising flame


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CHEM

ISTRY


0000CT103115003


4. The lewis acid strength of following molecule is in the order

(A) SiF4 < SiCl

4 < SiBr

4(B) BF

3 > BCl

3 > BBr

3

(C) SiF4 > SiCl

4 > SiBr

4(D) BF

3 < BCl

3 < BBr

3

5. In acidic medium the reaction of H2O

2 with potassium permanganate produces a compound in which

the oxidation state of Mn is not.

(A) 0 (B) +2 (C) +3 (D) +4

6. N–H

O

O

NaOH

NaOH

Br2

P

Q + Na CO2 3

CH –CH –I3 2 RH O3

+

S + T

If T can evolve effervescence of CO2

with aq. NaHCO3, then correct statement(s) is/are :

(A) S & Q can be distinguished by dye azo test

(B) T is most acidic among all isomeric benzenoid dicarboxylic acid

(C) Q & S can be distinguished by mustered oil test

(D) P, Q & T all are soluble in aq. NaHCO3


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7. Possible X in following reaction scheme is :

OH (i) CCl / NaOH4

(ii) H+

P

'X'

(A) Zn-dust (B) NaOH, CaO, (C) CHCl3/NaOH (D) LiAlH

4

8. Which of the following is/are present in mixture of product :

C–O–CH –CH2 3

O

HOH / H+

C–O–Ph18

O

Mixture of product

(A) CH3–CH

2–OH (B)

CO OH

C–OH

O

18(C) Ph–OH

18

(D)

CO OH

C–OH

O

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Marking scheme :





Radioactive elements can be categorised into

(i) Neutron rich nuclide

(ii) Neutron poor nuclide

In order of achieve stable n/p ratio, these nuclide either emits – , + (positron) or -particle depending

upon the n/p ratio of unstable nuclei and stable nuclei.

53P135 Q + a

Q R + b

[Note : only stable isotope of element P and Q are 53

P133 and 54

Q137 respectively]

Elements P, Q, R does not show resemblence to any known element

9. Particles a and b can be respectively

(A) 2He4 ,

–1eº (B)

–1eº ,

2He4 (C)

–1eº ,

+1eº , (D)

2He4 ,

2He4

10.53

P135 Q + a ; t1/2

= 1000 hrs

Q R + b ; t1/2

= 10 min

Number of nuclei of Q and R respectively after 1000 hr if we start with 2 mol P.

[NA = 6 × 1023]

(A) 6 × 1023 , 6 × 1023 (B) 6 × 1022 , 6 × 1023 (C) 2 × 1020 , 6 × 1023 (D) 1020 , 6 × 1023



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ONH OH2 (P) + (Q)

PCl 5

SOCl2

(R)

(S)

H O3+

(T) + (U)

H O3+

(V) + (W)

(V) & (U) will give CO2

gas with aq. NaHCO3 :

11. Choose the correct option(s) ?

(A) (P) & (Q) Stereoisomers (B) (R) & (S) Metamers

(C) (W) & (T) Identical (D) (U) & (V) Homologous

12. Choose the incorrect option(s) :

(A) Reaction sequence involve beckmann rearrangement

(B) Hydrolysis of (R) is faster than CH –C–O–CH3 3

O

(C) Formation P & Q involve nucleophilic addition followed by elimination

(D) Reaction sequence involve migration at electron deficient nitrogen




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CHEM

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0000CT103115003

SECTION–IV : (Maximum Marks : 32)


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

0 In all other cases

1. Find freezing point temperature (in Kelvin)of liquid if vapour pressure of solid & liquid are given by

following expression.

lnps = 10 –

3000(K)

Twhere P

s = vapour pressure of solid

lnp = 5 –

2000(K)

TP = vapour pressure of liquid

Note : All terms are in SI unit

Fill your answer as sum of digits (excluding decimal places) till you get the single digit answer.

2. A reversible cyclic process involves 6 steps. In step -1, 3 system absorb 500 J, 800 J of heat from a

heat reservoir at temperature 250K & 200K respectively. Step 2, 4, 6, are adiabatic such that the temperature

of one reservoir changes to that of next. Total work done by the system in whole cycle is 700 J. Find

the temperature during step 5 if it exchange heat from a reservoir at temperature T5

Fill your answer as sum of digits (excluding decimal places) till you get the single digit answer.

3. Count the number of correct statements :

(i) Minimum potential required for electrophoresis is called zeta potential

(ii) Tyndall effect is increases with increase in difference in size of particle and wavelength of light used.

(iii) Minimum amount of electrolyte in millimoles per 100 ml required to cause precipitate in two

hours is called coagulating value.

(iv) Both physical and chemical adsorption decreases with temperature

(v) Chemical bonds may be covalent or ionic in chemical adsorption

(vi) Zeolites are shape -selective catalyst.



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4. FeO.Cr2O

3 + Na

2O

2 A + B + Na

2O

A + H2O NaOH + (C)

red-brown ppt.

B + H2O (D) + Na+

yellow

(D) + 2H+ E + H2O

orange

Find the sum of peroxylinkanges in A, B, C, D and E.

5. Find the number of CORRECT statements about NH3

(i) basic gas

(ii) turns red litmus to blue litmus

(iii) gives white dense fumes with HCl

(iv) gives brown ppt. with nessler's reagent

(v) gives deep blue colour with CuSO4 solution when passed in excess

(vi) gives deep blue colour with NiCl2 solution when passed in excess

6. Find the number of 3C– 2e– bonds in following molecule.

H

BB

H

HH

Be

HH

H H


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CHEM

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0000CT103115003

7. Number of oxidation reactions in which organic reactant gets oxidised & one of the major product

is carboxylic acid/salt :

(i)

C

O

CH3 NaOI(ii)

CH OH2

CHO

OHHH IO5 6

(iii)

CHOFehling solution

(iv) D-Glucose Br2

H O2

(v)

C

O

OCH3H O3

+

(vi)

CHOconc. NaOH

(vii)

NCH O3

+(viii) L-Ribose HNO3

(ix) Lactose Tollen's reagent

8. If X, Y & Z is number of mole of HIO4 consumed by 1 mole of sucrose, 1 mole of Glucose & 1 mole

of fructose respectively :

Then value of Y Z 1

X

is :



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PART-3 : MATHEMATICS






Marking scheme :




1. Let ƒ : R R be a continuous function such that 1

0

ƒ xt dt 0 for all x , then-

(A) There exists three integral values of P such that the graph of y = ƒ(x) and y = x3 – 3x2 + P intersects

at three distinct points.

(B) y = ƒ(x) is a periodic function.

(C) If A dentoes the minimum area bounded by the curves y = ƒ(x), y = x4 – 4x – a and the ordinates

x = 2 & x = 4, then a = 69

(D) ƒ(x) is neither even function nor odd function.

2. Consider the function / 2

20

dxƒ k

1 k cos x

where k [0,1), then which of the following is/are

correct-

(A) 1 1 1

ƒ ƒ ƒ2 3 4

(B)

1 1 1ƒ ƒ ƒ

4 3 2

(C) 1 1 3

ƒ ƒ ƒ4 2 4

(D)

1 1 3ƒ ƒ ƒ

4 2 4


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3. Let B is an invertible square matrix and B is the adjoint of matrix A such that AB = BT, then-

(A) A is an identity matrix. (B) B is a symmetric matrix.

(C) A is not an identity matrix. (D) B is a skew symmetric matrix.

4. If ƒ(x) is a polynomial of degree 4 with rational coefficients and touches x-axis at 2,0 , then for the

equation ƒ(x) = 0 -

(A) sum of roots is 0 (B) sum of roots is 4 2

(C) product of roots is 4 (D) product of roots is –4.

5. Let

12016

n2

2016

I n x cos x 2016 dx

, then which of the following is/are correct ?

(A) 20161

I2

(B) 40321

I4

(C) nlim n.I n 2016

(D) nlim n.I n 2017

6. Let z1 and z

2 are the roots of the equation z2 + z + 1 = 0, then-

(A) z12016 + z

22016 = 2 (B) z

12016 + z

22016 = –2

(C) z12015 + z

22015 = 1 (D) z

12015 + z

22015 = –1


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7. A tangent is drawn at any point P(t) on the parabola y2 = 8x and a point Q() is taken on it from

which pair of tangents QA and QB are drawn to the circle x2 + y2 = 4, where A & B are points of

contact. Now if point P(t) varies on the parabola then the locus of all such point of concurrency of the

chord of contact AB is denoted by S = 0, then-

(A) Equation of S = 0 is 2x + y2 = 0

(B) Equation of S = 0 is x + y2 = 0

(C) Area bounded by S = 0 and the line 2x + 1 = 0 is 2

3 sq. units

(D) Tangents drawn at the extremities of intersection of S = 0 and the line 2x – y + 1 = 0 to the curve

S = 0 includes an angle 75º.

8. Let two planes P1 : 2x – y + z – 2 = 0 and P

2 : x + 2y – z – 3 = 0 are given then-

(A) The equation of the plane through line of intersection of P1 = 0 and P

2 = 0 and the point (3,2,1) is

x – 3y + 2z + 1 = 0

(B) The equation of the plane through line of intersection of P1 = 0 and P

2 = 0 and the point (3,2,1) is

3x – y + 2z – 9 = 0

(C) The equation of acute angle bisector plane of P1 = 0 and P

2 = 0 is x – 3y + 2z + 1 = 0

(D) The equation of acute angle bisector plane of P1 = 0 and P

2 = 0 is x + 3y + 2z + 2 = 0

E-26/32

MATHEM

ATICS


0000CT103115003







Marking scheme :





Let A be the area of region in the first quadrant bounded by the line x

y2

, the x-axis and the ellipse

22x

y 19 . If m (m > 0) is such that A is equal to the area of the region in the first quadrant bounded

by the line y = mx, the y-axis and the ellipse 2

2xy 1

9 , then

9. Value of A is -

(A) 13 2

sin2 13

(B) 1 3

3sin13

(C) 13 2

cos2 13

(D) 1 2

3cos13

10. Value of m is -

(A) 2

3(B)

2

9(C)

2

7(D)

1

9



MATHEM

ATICS

E-27/320000CT103115003


Let ABC be a triangle (with usual notations) such that a = 3 and b = 2c, then

11. Maximum area of ABC is (in sq. units)-

(A) 2 (B) 6 (C) 3 (D) 4

12. Value of c corresponding to maximum area of triangle is-

(A) 3 (B) 2 5 (C) 5 (D) None of these




E-28/32

MATHEM

ATICS


0000CT103115003

SECTION–IV : (Maximum Marks : 32)


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

0 In all other cases

1. In isosceles triangle ABC with AB = AC, if coordinates of A are (2,2) and medians through B & C

are 3y + 2x = 6 and 3y – 2x = –2 respectively, then slope of BC is

2. If coefficient of xr in the the product of (1 – x + x2 – x3 +......–x99 + x100) (1 + x + x2+......+x100) is

denoted by Tr, then value of (T99 + T101 + T103) is

3. Triangle ABC is right angled at C and BAC = . A point D is

E

C

A F D B

chosen on AB such that AC = AD = 1 and a point E is chosen on

BC such that CDE = as shown in figure. The perpendicular to

BC at E meets AB at F. If 0

lim EF

can be expressed as p

q (where

p & q are coprime), then value of (p + q) is



MATHEM

ATICS

E-29/320000CT103115003

4. A box contains 6 cards. Three of the cards are black on both sides, one card is black on one side and

red on the other and two of the cards are red on both sides. A card is randomly picked from box and

one side is randomly seen. Given that the side seen was red, the probability that the other side is also

red can be expressed as a

b

(where a & b are coprime), then (a + b) is

5. Let a,b,c are the roots of the equation x3 – 2x2 + 3x – 4 = 0, then the value of 1 1 1

a b ca b c

is

6. As shown in diagram, the volume of tetrahedron ABCD is 1

6 cu. units.

Also ACB = 45º, BC = 1, AC

AD BC 32

, then (CD)2 is

D

C45º

B

A


E-30/32

MATHEM

ATICS


0000CT103115003

7. Consider the region S of complex numbers 'a' such that |z2 + az + 1| = 1 where complex number z

is moving on the locus |z| = 1. If A denotes the area of S in argand plane, then A

2

is

(where [.] denotes greatest integer function)

8. If 2015

n 2015 2012 kn 0

1 2016S

2 2 2

(where k N), then the value of k is



E-31/320000CT103115003



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 0000CT103115003E-32/32

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. 32 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 : 14 - 02 - 2016Time : 3 Hours Maximum Marks : 240

Paper Code : 0000CT103115003

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

IND

I

TARGET : JEE (ADVANCED) 2016

JEE (Main + Advanced) : ENTHUSIAST & LEADER COURSE

H-2/32


0000CT103115003

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-3/320000CT103115003

-1 : – I(i) : ( : 32)

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

:

+4 0 –2

1. P-T

16 (25

R3

J/mol K)

2

311

2

P(in 10 Pa)5

300 400T(in K)

(A) 15.36 gm

(B) 30.72 gm

(C) 12

(D) 8

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS

H-4/32

PHYSICS


0000CT103115003

2.

h2h

R

(A) 2

2gh

R (B) 2

4gh

R

(C) 2 216 gh R

15

(D)

2 25 gh R

6

3. m1 m

2

1 10 cm 2 10 cm

m =m =1kg1 2

k =100N/m1 k =400N/m2

m1 m21 2

(A)

(B) 5cm

(C) t = 30

sec

(D) t = 10

sec


PHYSICS

H-5/320000CT103115003

4.

V

R

A

R L

R

(A) L V

2

(B) A V

3

(C) L 2V

3

(D) (C) A V

2

H-6/32

PHYSICS


0000CT103115003

5. AC 200 V

10H 100 40µF

(A) 50 Hz

(B) 25

Hz

(C) 45

2Hz

(D) 20

Hz

6. AB 0.5A

500gm

A B

CD

x

z

y

side=2m

(A) 5ˆ ˆ3i k2

(B) 5 ˆ ˆ ˆi 3 j 2k2

(C) 5ˆ ˆ2 j k2

(D) 5 5ˆ ˆ ˆi j k2 2


PHYSICS

H-7/320000CT103115003

7.

(A)–q

+Q

–q

(B)I2

I1 I2

(C)

(D)

H-8/32

PHYSICS


0000CT103115003

8. S1 S

2 m

1 m

2

v

v2

m1

m2

(A) v2 = m

1v/m

2

(B) M v2 = m

1 v/m

2 +m

1 +M

(C) v2 = m

1S

2v/m

2S

1

(D) M v2 = (m

1+M)v/m

2


PHYSICS

H-9/320000CT103115003

–I(ii) : ( : 16)

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

:

+4 0

–2

9 10

310 nm

I = 1A 2.5eV

+

–

anodecathode

x

200

50+

– 100V

9. x (A) 80 V (B) 30 V (C) 40 V (D) 61.5 V

10. x = 100 V isat

(A) 500 nm (B) 600 nm (C) 310 nm (D) 400 nm

H-10/32

PHYSICS


0000CT103115003

11 12

()

L1

L2 L1 L2 x1 x

2

The Wilkinson Microwave

Anisotropy Probe (WMAP) Me

Ms R

11.

(A) e

31

s

Mx R

3M (B)

e3

2

s

3Mx R

M (C)

e3

1

s

Mx R

M (D)

e3

2

s

Mx R

M

12.

(A) L1

(B) L2

(C) L1

(D) L2

– II : & – III :

II III


PHYSICS

H-11/320000CT103115003

–IV : ( : 32)

0 9

:

+4

0

1. 147 N E (MeV )

(amu ) (1 amu 931 MeV)

atom mass of atom – A amu

42 He

105 B

147 N

0.00260

0.01294

0.00307

E

2

2.

33 3 10 sec (10–3 )

53°37°

m

H-12/32

PHYSICS


0000CT103115003

3. Dove prism

AB (cm ) AD

BC ADCB h = 1.05 cm

1.41 A B 45° (: 2 = 1.41, 3 1.73 )

A B

CD

h

4. –Q m Q r

23kQ

nr n

Q

–Q

–Q

r

mass m

mass m


PHYSICS

H-13/320000CT103115003

5.

0.056gm

r (mm ) 0.07 N/m 0º

6. 1-2-3-1 20% 1-3-4-1 10%

1-2-3-4-1 (% ) /4

1

2 3

4

V

P

H-14/32

PHYSICS


0000CT103115003

7. Rx I = 2.0 A

V = 120 V RV = 960 Ohm R

x = V/I

Rx () ?

Rx

A

V

8.

1A 4000 /m

2m

4 gm

(m/s ) (

) (2 = 10 )i

i


CHEM

ISTRY

H-15/320000CT103115003

-2 : – I(i) : ( : 32)

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

:

+4 0 –2

1. 300K 1 kg

2P(atm)

1 10 20

T=300K

V(Litre)

1

(A) 300 K 2 atm

(B) V, 10 0.1 gm / ml

(C) V, 10 0.05 gm / ml

(D) V, 10 1 gm / ml

2. FCC

x = OV

y = T.V.

z = O.V. T.V.

(A) x = y = z (B) x < y < z (C) x > y < z (D) x > y > z

H-16/32

CHEM

ISTRY


0000CT103115003

3.

(A) :

(B) :

(C) :

(D) Mn : Amethyst

4.

(A) SiF4 < SiCl

4 < SiBr

4(B) BF

3 > BCl

3 > BBr

3

(C) SiF4 > SiCl

4 > SiBr

4(D) BF

3 < BCl

3 < BBr

3

5. H2O

2 Mn

(A) 0 (B) +2 (C) +3 (D) +4

6. N–H

O

O

NaOH

NaOH

Br2

P

Q + Na CO2 3

CH –CH –I3 2 RH O3

+

S + T

T NaHCO3 CO

2

(A) S Q

(B) T,

(C) Q S

(D) P, Q T NaHCO3


CHEM

ISTRY

H-17/320000CT103115003

7. X

OH (i) CCl / NaOH4

(ii) H+

P

'X'

(A) Zn- (B) NaOH, CaO, (C) CHCl3/NaOH (D) LiAlH

4

8.

C–O–CH –CH2 3

O

HOH / H+

C–O–Ph18

O

(A) CH3–CH

2–OH (B)

CO OH

C–OH

O

18(C) Ph–OH

18

(D)

CO OH

C–OH

O

H-18/32

CHEM

ISTRY


0000CT103115003

–I(ii) : ( : 16)

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

:

+4

0

–2

9 10

(i)

(ii) n/p

n/p – , + () -

53P135 Q + a

Q R + b

[ : P Q 53

P133 54

Q137 ]

P, Q, R

9. a b

(A) 2He4 ,

–1eº (B)

–1eº ,

2He4 (C)

–1eº ,

+1eº , (D)

2He4 ,

2He4

10.53

P135 Q + a ; t1/2

= 1000 hrs

Q R + b ; t1/2

= 10 min

2 P 1000 hr Q R

[NA = 6 × 1023]

(A) 6 × 1023 , 6 × 1023 (B) 6 × 1022 , 6 × 1023 (C) 2 × 1020 , 6 × 1023 (D) 1020 , 6 × 1023


CHEM

ISTRY

H-19/320000CT103115003

11 12

ONH OH2 (P) + (Q)

PCl 5

SOCl2

(R)

(S)

H O3+

(T) + (U)

H O3+

(V) + (W)

(V) (U), NaHCO3 CO

2

11.

(A) (P) (Q) (B) (R) (S)

(C) (W) (T) (D) (U) (V)

12.

(A)

(B) (R) CH –C–O–CH3 3

O

(C) P Q

(D)

– II : & – III :

II III

H-20/32

CHEM

ISTRY


0000CT103115003

–IV : ( : 32)

0 9

:

+4

0

1.

lnps = 10 –

3000(K)

T P

s =

lnp = 5 –

2000(K)

TP =

Note : SI

()

2. 6 -1, 3 250K 200K

500 J, 800 J 2, 4, 6,

700 J

5 T5 5

()

3.

(i)

(ii)

(iii) 100 ml

(iv)

(v)

(vi)


CHEM

ISTRY

H-21/320000CT103115003

4. FeO.Cr2O

3 + Na

2O

2 A + B + Na

2O

A + H2O NaOH + (C)

B + H2O (D) + Na+

(D) + 2H+ E + H2O

A, B, C, D E

5. NH3

(i)

(ii)

(iii) HCl

(iv)

(v) CuSO4

(vi) NiCl2

6. 3C– 2e–

H

BB

H

HH

Be

HH

H H

H-22/32

CHEM

ISTRY


0000CT103115003

7.

(i)

C

O

CH3 NaOI(ii)

CH OH2

CHO

OHHH IO5 6

(iii)

CHOFehling solution

(iv) D-Glucose Br2

H O2

(v)

C

O

OCH3H O3

+

(vi)

CHOconc. NaOH

(vii)

NCH O3

+(viii) L-Ribose HNO3

(ix) Lactose Tollen's reagent

8. X, Y Z 1 1 1 HIO4

Y Z 1

X


MATHEM

ATICS

H-23/320000CT103115003

-3 : – I(i) : ( : 32)

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

:

+4

0

–2

1. ƒ : R R x 1

0

ƒ xt dt 0 -

(A) P y = ƒ(x) y = x3 – 3x2 + P (B) y = ƒ(x) (C) y = ƒ(x), y = x4 – 4x – a x = 2 x = 4 A a = 69 (D) ƒ(x)

2. / 2

20

dxƒ k

1 k cos x

, k [0,1) -

(A) 1 1 1

ƒ ƒ ƒ2 3 4

(B)

1 1 1ƒ ƒ ƒ

4 3 2

(C) 1 1 3

ƒ ƒ ƒ4 2 4

(D)

1 1 3ƒ ƒ ƒ

4 2 4

H-24/32

MATHEM

ATICS


0000CT103115003

3. B B, A AB = BT -(A) A (B) B (C) A (D) B

4. ƒ(x) 2,0 x-

ƒ(x) = 0 -

(A) (B) 4 2

(C) 4 (D) –4

5.

12016

n2

2016

I n x cos x 2016 dx

(A) 20161

I2

(B) 40321

I4

(C) nlim n.I n 2016

(D) nlim n.I n 2017

6. z1 z

2, z2 + z + 1 = 0 -

(A) z12016 + z

22016 = 2 (B) z

12016 + z

22016 = –2

(C) z12015 + z

22015 = 1 (D) z

12015 + z

22015 = –1


MATHEM

ATICS

H-25/320000CT103115003

7. y2 = 8x P(t) Q(), x2 + y2 = 4 QA QB A B P(t)

AB S = 0 -

(A) S = 0 2x + y2 = 0

(B) S = 0 x + y2 = 0

(C) S = 0 2x + 1 = 0 2

3

(D) S = 0 2x – y + 1 = 0 S = 0 75º

8. P1 : 2x – y + z – 2 = 0 P

2 : x + 2y – z – 3 = 0

(A) P1 = 0 P

2 = 0 (3,2,1)x – 3y + 2z + 1 = 0

(B) P1 = 0 P

2 = 0 (3,2,1) 3x – y + 2z – 9 = 0

(C) P1 = 0 P

2 = 0 x – 3y + 2z + 1 = 0

(D) P1 = 0 P

2 = 0 x + 3y + 2z + 2 = 0

H-26/32

MATHEM

ATICS


0000CT103115003

–I(ii) : ( : 16)

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

:

+4

0

–2

9 10

xy

2 , x-

22x

y 19 A

m (m > 0) A, y = mx, y-2

2xy 1

9

9. A -

(A) 13 2

sin2 13

(B) 1 3

3sin13

(C) 13 2

cos2 13

(D) 1 2

3cos13

10. m -

(A) 2

3(B)

2

9(C)

2

7(D)

1

9


MATHEM

ATICS

H-27/320000CT103115003

11 12

ABC () a = 3 b = 2c

11. ABC () -

(A) 2 (B) 6 (C) 3 (D) 4

12. c -

(A) 3 (B) 2 5 (C) 5 (D)

– II : & – III :

II III

H-28/32

MATHEM

ATICS


0000CT103115003

–IV : ( : 32)

0 9

:

+4

0

1. ABC AB = AC A (2,2) B C 3y + 2x = 6 3y – 2x = –2 BC

2. (1 – x + x2 – x3 +......–x99 + x100)(1 + x + x2+......+x100) xr Tr (T99 + T101 + T103)

3. ABC, C BAC = AB D

E

C

A F D B

AC = AD = 1 BC E CDE = BC E

AB F 0

lim EF

p

q

(p q ) (p + q)


MATHEM

ATICS

H-29/320000CT103115003

4. 6

a

b

(a b ) (a + b)

5. a,b,c x3 – 2x2 + 3x – 4 = 0 1 1 1

a b ca b c

6. ABCD 1

6

ACB = 45º, BC = 1, AC

AD BC 32

(CD)2

D

C45º

B

A

H-30/32

MATHEM

ATICS


0000CT103115003

7. 'a' S |z2 + az + 1| = 1 z, |z| = 1

A, S A

2

([.]

)

8. 2015

n 2015 2012 kn 0

1 2016S

2 2 2

(k N) k


H-31/320000CT103115003


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 0000CT103115003H-32/32

:

14.

15.

16.

17. :

18.

19.

20. g = 10 m/s2

____________________________

____________________________

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

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

JEE (Main + Advanced) : ENTHUSIAST & LEADER COURSE · 2016. 2. 29. · TARGET : JEE (ADVANCED) 2016...

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Transcript of JEE (Main + Advanced) : ENTHUSIAST & LEADER COURSE · 2016. 2. 29. · TARGET : JEE (ADVANCED) 2016...