jee (main + advanced) : enthusiast coursetarget : jee (main + advanced) 2016/07-02-2016/paper-1...

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

Marking scheme : +4 for correct answer and 0 in all other cases.

(ii) Section-I(ii) contains 3 ‘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 –1 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 – 1

Test Type : FULL SYLLABUS Test Pattern : JEE-Advanced

TEST DATE : 07 - 02 - 2016

Paper Code : 1001CT102115076

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

JEE (Main + Advanced) : ENTHUSIAST COURSE

SCORE-I

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Target : JEE (Main + Advanced) 2016/07-02-2016/Paper-1

1001CT102115076

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 =

20

1

c

Planck constant h = 6.63 × 10–34 J–s

Enthusiast Course/Score-I/07-02-2016/Paper-1

PHYSICS

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PART-1 : PHYSICSSECTION–I(i) : (Maximum Marks : 24)

This section contains SIX 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 In all other cases

1. In the figure shown, the uniform resistance wire between A and B has a total resistance Ro. The contact at C

can divide the wire into two resistors, one of which is fRo. Value of f for which the ammeter reading is :-

E Er

BA C

fR0

A

(A) Maximum if f = 0 (B) Maximum if f = 1 (C) Minimum if f = 2

1(D) Minimum if f =

3

1

2. A very long wire abcdef carrying current has the shape as shown in the figure. The magnetic field at

point P in terms of unit vectors.

a b

c

d e

f

x

y

z

(0, a, 0)

(a, 0, 0)

(0, 0, a) (a, 0, a)

(a/2, 0, 0)

P

(A) Due to wire ab is ka

Io ˆ5

11

4

(B) Due to wire bc is k

a

Io ˆ5

(C) Due to wire cd is ja

Io ˆ5

(D) Due to wire ef is j

a

Io ˆ5

21

2

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS


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3. A circuit consists of two capacitors of capacitances C each and ideal cell of emf as shown in figure.

Initially both the capacitors are uncharged. The key is pushed to position 1 and then to position 2. This

operation is repeated infinite times. Which of the following statement is/are CORRECT?

C

1 2

C

(A) The magnitude of net charge that crosses the cell is 2C

(B) The magnitude of net work done by the cell is 2C2

(C) The net heat produced in the circuit is C2

(D) None of the above is correct.

4. A body is rotated in the vertical plane by means of a thread of length with minimum possible speed.

At the highest point B the thread breaks and the body moves under gravity and strikes at C. Then

A

B

C

(A) the horizontal range AC is 2

(B) the horizontal range AC is 2

(C) tangential acceleration of the body at point C is 5

2g

(D) tangential acceleration of the body at point C will be 5

1g


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5. The density () of an ideal gas varies with temperature T as shown in figure. Then :-

(A) The product of P & V at A is equal to the product of P & V at B

(B) Pressure at B is greater than the pressure at A

(C) Work done by the gas during the process AB is negative

(D) The change in internal energy from A to B is zero

6. Here inductor is connected to a resistor as shown. When a magnet is moved towards the coil with a

constant speed till it comes out of coil completely. As seen by observer looking at the rectangular loop

as shown

S N

R

(A) Current first flows clockwise and then anticlockwise in loop.

(B) Current first flows anticlockwise and then clockwise in loop.

(C) Magnet first experiences a repulsive force and then an attractive force.

(D) When rate of change of flux due to magnet is d

dt

, current in coil is

d

Rdt

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SECTION–I(ii) : (Maximum Marks : 24)

This section contains THREE paragraphs.

Based on each paragraph, there will be TWO questions

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

four option(s) is(are) correct.


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

Paragraph for Questions 7 and 8

Spherical bob of a pendulum has massless thin walls. The bob is filled with water. Radius of this

sphere is R. This bob is suspended from a fixed point with the help of a massless rigid rod of circular

cross-section. Length from the fixed point to center of sphere is .

Case-1 : When water is in liquid state, the pendulum behaves like a simple pendulum. It performs

SHM when displaced from its mean position. Gravitational force provides the restoring force to the

pendulum.

Case-2 : When water freezes, the arrangement behaves like a rigid body. It behaves like a physical

pendulum. The gravitational force provides the restoring torque to this pendulum. Moment of inertia

of the sphere will decide the time period of the oscillations. Viscosity of water and volume changes on

freezing may be ignored. It is obvious that on freezing the time period of oscillations of the pendulum

will be more than the simple pendulum.

Time period of solidified sphere is x times that of liquid state i.e. T2 = xT

1 x =

1

2

T

T.



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7. Choose the CORRECT option(s) :-

(A) Square of angular frequency when water in the bob is in liquid state is given by l

g

(B) Square of angular frequency when water in the bob is in liquid state is given by l

g2

(C) Moment of inertia of the sphere about the fixed point when water is in solid state can be given as

22

5

2lRm

(D) Moment of inertia of the sphere about the fixed point when water is in solid state can be given as

22

5

2

2lR

m

8. Keeping the same angular amplitude, comparing case-1 & case-2 :-

(A) In case-1, water rotates about it's own centre of mass but in case-2, ice does not rotate about its

centre of mass

(B) In case-1, water does not rotate about it's centre of mass but in case-2, ice rotates about its centre

of mass.

(C) Energy of SHM in case-1 is more than in case-2

(D) Maximum angular velocity of the body in case-1 is less than maximum angular velocity in case-2.


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A truck is carrying two thin circular discs of mass m each and radius r joined by a rigid light rod of

length 2r. The axis of rod is connecting the centers of two discs. This arrangement is kept on the rough

floor of the truck such that its axis is perpendicular to the direction of motion of the vehicle. The floor

friction is sufficient to keep the object rolling without slipping. Let the direction of motion of the truck

be along x-axis and the direction normal to it be taken as z-axis. The truck accelerates with acceleration

a. In the reference frame of the truck the forces acting on a disc are shown. Take the mid-point of light

rod as the origin of the coordinate system.

z

y

x

a

Ff

mg

F

R

Here F is the psuedo force on system and Ff is the frictional force acting on one of the any discs. F is

acting opposite to the direction of motion of the truck. It is given that W1 is the work done by friction

in the reference frame of truck and W2 is the work done by friction in the reference frame of ground.



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(A) In vector form frictional force on the disc is ma

i3

(B) In vector form frictional force on the disc is ma

i3

(C) Position vectors of points of contacts is jrirjrir ˆ2ˆ2,ˆ2ˆ2

(D) Position vectors of points of contacts is krjrkrjr ˆˆ,ˆˆ


(A) Torques due to friction about centre of the rod are mar marˆ ˆ ˆ ˆ( j k), ( j k)

3 3

(B) Torques due to friction about centre of the rod are kjmarkjmar ˆˆ2

3,ˆˆ

2

3

(C) W1 = 0, W

2 0

(D) W2 = 0, W

1 0


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Various rules of thumb have been proposed by the scientific community to explain the mode of

radioactive decay by various radioisotopes. One of the major rules is called the n/p ratio. If all the

known isotopes of the elements are plotted on a graph of number of neutrons (n) versus number of

protons (p), it is observed that all isotopes lying outside of a “stable” n/p ratio region are radioactive as

shown in figure.

The graph exhibits straight line behavior with unit slope up to p = 25. Above p = 25, those isotopes

with an n/p ratio lying below the stable region usually undergo electron capture while those with n/p

ratios lying above the stable region usually undergo beta decay. Very heavy isotopes (p>83) are unstable

because of their relatively large nuclei and they undergo alpha decay.

For A 40, there is a preference for N > Z. Nuclear force is charge independent, i.e. we consider that

neutron-neutron, neutron-proton or proton-proton nuclear force is same. As the number of protons

increases, the coulombic repulsive force increases, so that excess of neutrons which produce only

attractive forces is required for stability, hence N vs Z curve departs from N = Z line. Gamma ray

emission does not involve the release of a particle. It represents a change in an atom from a higher

energy level to a lower energy level.

25 No of protons (p)

electron capture

-decayn=p line

stabl

e n/p

ratio

s

-decay

No.

of

neut

rons

(n)



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(A) The radioisotope of magnesium with mass number 27 and atomic number 12 undergo radioactive

decay by beta decay

(B) The radioisotope of magnesium with mass number 27 and atomic number 12 undergo radioactive

decay by alpha decay

(C) 23090

Th undergoes a series of radioactive decay processes, resulting in 21483

Bi being the final product.

The possible sequence of the processes that occurred is alpha, alpha, alpha, alpha, beta

(D) 23090

Th undergoes a series of radioactive decay processes, resulting in 21483

Bi being the final product.

The possible sequence of the processes is alpha, beta, beta, beta, gamma


(A) In a hypothetical radioactive series, Th21081 undergoes 3 beta decay processes, 1 alpha decay process

and 1 gamma ray emission to yield a stable product will be 20686

Rn

(B) In a hypothetical radioactive series, Th21081 undergoes 3 beta decay processes, 1 alpha decay process

and 1 gamma ray emission to yield a stable product will be 20682

Pb.

(C) For a hypothetical isotope of an element X17888 we may expect electron capture

(D) For a hypothetical isotope of an element 2412 X we may expect no decay

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 darkened


1. An air bubble of radius r = 3mm moves up in a viscous liquid with a constant velocity v = 0.7 cm/sec

and the density of the liquid is 1.75 gm/cc. If viscosity of liquid (in poise) is given by 10x then x will

be. Neglect the density of air in comparison to that of the liquid.

2. Two charges +2q and –3q are fixed at (4 m, 0, 0) and (9 m, 0, 0) respectively. If the distance between

2 points on x-axis (in m) where the potential is zero is given by 1.210x then the value of x will be.

Assume standard reference point.



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3. An ideal gas is heated from the temperature of 300 K to the temperature of 750 K at constant pressure,

resulting in its volume increased by V = 6 m3. Determine the initial volume of gas (in m3).

4. A 10 kg block B rests on smooth rollers of negligible mass and is in contact with a 2.0 kg solid

cylinder of the radius 0.1 m as shown. The rollers ensure that there is no friction between B and the

ground A below B. It is know that k =

s = 0.5 between B and C as well as between C and ground A.

Find the magnitude of the horizontal force P (in N) for which the cylinder will roll with an angular

acceleration of 20 rad/s2 on ground A without slipping. Fill P

16 in OMR sheet.

C B P

A


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5. A composite mirror is formed by arranging the upper half of a concave mirror above the lower half of

a convex mirror as shown. A real linear object AB is placed on the common principal axis and it is

found that images formed by the two halves of the composite mirror are equidistant from the point O.

If the radii of curvature of the concave mirror and convex mirror halves are 20 cm and 30 cm respectively.

The distance of the object AB from O is given by x cm then x

10 is.

////////

////

////

///

/////

///

/////

O

B

A

6. An arrangement consisting of slabs is shown in the figure. When system is placed in situation (a) with

temperature difference is maintained between (1) & (2), equivalent thermal conductivity of system is

K and when system is changed to situation (b) and temperature difference is maintained between (3)

& (4) then equivalent thermal conductivity increases by 20%. Find number of slabs in the arrangement.

d

12

k 2k nk

n

d

2 (a)nk

(b)

(4)

(3)

n

2kk

d

d2



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7. A vernier calliper has no zero error, one main scale division = 1 mm and 9 main scale division = 10

vernier scale divisions. A spring is held between jaws without exerting force on jaws. Main scale

reading is 3.4 cm and 5th vernier scale division coincides with a main scale division. When Jaws are

pressed by force of 10 N on both sides, main scale reading remains 3.4 cm, but 1st division of vernier

scale coincides with a main scale division. Now spring is removed, cut into two pieces and when

inserted between jaws without any force, main scale reading = 1.1 cm and 5th vernier scale division

coincides with a main scale. If we apply force of 30 N on both the jaws, main scale reading remains

1.1 cm and x vernier scale division coincides with a main scale division. Fill x in OMR sheet.

8. A block is released from the state of rest from a wedge, as shown. The block and the wedge are both

of the same mass. If friction is absent, the total horizontal displacement of the block with respect to

earth before it falls on the earth is given by 0.5R + x R. then the value of x will be.

Smooth

R

R


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PART-2 : CHEMISTRYSECTION–I(i) : (Maximum Marks : 24)





Marking scheme :



1. Which of the following methods can be used to produce phosphine gas ?

(A) Ca3P2 + H2O (B) PH4I + KOH

(C) Heating of H3PO2 (D) White P + NaOHinert

atmosphere

2. One molecule of which of the following ligand(s) formed only two ring towards a particular central

atom.

(A) en (B) bn (C) dien (D) trien

3. For an irreversible process, if P-V graphs are plotted simply to represent the work involved in the process

as the area under the P-V curve. Then which of the following options has/have the correct P-V curve

for the mentioned process on an ideal gas placed in a cylinder fitted with a piston.

(A) 3-step compression

V

P

(B) 3-step compression

V

P

(C) 3-step expansion

V

P

(D) 3-step expansion

V

P



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4. (I) NH OH2

O

(Product mixture-1) (II) Product mixture-1 H+

(Product mixture-2)

(A) Product mixture-1 contain diastereoisomer (B) Product mixture-2 contain diastereoisomer

(C) Product mixture-2 contain enantiomers (D) Product mixture-1 contain enantiomers

5. A sample of 23892 U when kept at 298 K in a closed vessel shows -decay and -decay to give 206

82 Pb .

After t hour the volume of the gas collected in the vessel at 1atm,273K is found to be 89.6 ml. Then

which of the following is/are correct for the nuclear process and the products of the process.

(A) Mass of Pb obtained in time t hour = 103 mg.

(B) If the temperature is increased by 10°C then the time take for the same amount of gas to be

produced will be t

2 hour..

(C) particles show more ionisation power than particles.

(D) radiations undergoes no deviation while traveling through electric or magnetic field.

6. M + NaCN + O2 + H2O Na[M(CN)2] + NaOH.

Which of the following elements gets dissolved in same manner in NaCN solution

(A) Ag (B) Cu (C) Au (D) Pt

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1001CT102115076








Marking scheme :





X Reddish brown

oxide

'A' Metal chloride

conc.HCl

KI/H+

B + I + H O2 2

B + NaOH Dirty green ppt. NH3

(excess) insoluble

7. Which of the following statements is/are correct regarding (B).

(A) It is soluble salt

(B) It is easily oxidisable by air

(C) It gets converted into higher oxidation state chloride with SO2 when pH of the medium is 1 - 2

(D) It's oxidation state is +3

8. Find the option where blue ppt is obtained

(A) When 'A' is treated with K4[Fe(CN)6] (B) When 'B' is treated with K4[Fe(CN)6]

(C) When 'A' is treated with K3[Fe(CN)6] (D) When 'B' is treated with K3[Fe(CN)6]


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We represent various reagents pictorially such that,

0.1 M strong acid; 0.1 M strong base

0.1 M weak acid (Ka = 10–5); 0.1 M weak base (K

b = 10–5)

The relative area of the squares represents the relative volume of the reagents used and the overlapping

region represents the relative amount of mixing of the reagents. Consider the temperature to be 298 K.

(use log 5 = 0.7, log 2 = 0.3, log 3 = 0.5, 2H Od = 1 g ml–1)

9. Which of the following leads to buffer formation ?

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

10. For a solution represented by

a2

a2

a

a

Which is/are correct ?

(A) pH = 1 (B) of weak acid = 5 × 10–4

(C) of H2O = 9 × 10–15 (D) pH = 1.7


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1001CT102115076


The letters P, Q, R, S, and T represent isomeric cyclobutane dicarboxylic acid, one of

them being a racemic form.

It was shown that :

(a) Only compound R forms a cyclic anhydride easily,

(b) Q yields an cyclic anhydride only at higher temperatures,

(c) Of all the acids under investigation, only P releases carbon dioxide when heated,

(d) S and T do not change at higher temperatures,

(e) Q and T are diastereomer

11. Which of the following represent correct match ?

(A) P is

COOH

COOH (B) Q on heating gives

O

O

O

(C) R is optically inactive (D) P is most acidic among all

12. Which of the following represent correct match ?

(A) S is optically active compound. (B) Q has plane of symmetry

(C) R is meso compound (D) R & S are diastereomer





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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. In how many of the following compounds 'd' orbital (which contains no nodal plane) is used in hy-

bridisation of central atom/ion.

Fe(CO)5, MnO

4–, Co(C

2O

4)

3–3, NiCl

2(PPh

3)

2, PtCl

4–2, Ni(CN)

4–2, Ni(CN)

4–4.

2. In how many of the following species % s - character of all bonds is not identical.

IF7, CH

4, SF

4, XeF

3+, OCl

2, ICl

2–, BF

3, SO

2

3. Number of crystal systems having, only 2 types of Bravais lattices = x,

Number of crystal systems having, atleast 2 interfacial angles equal = y,

all the three interfacial angles and all the three axes lenghts equal = z

Then find y – (x + z).


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1001CT102115076

4. How many of the following statements are correct ?

(i)HO

OH

O

O

OOH

OH

HO

HO

can show mutarotation

(ii) HOO

O

O

OH

HOHO OH

OH OHHO is reducing sugar

(iii)

O

HO

OHHO

OH

O

HOOH

O

HOC1-C4 Glycosidic linkage is present

(iv) — —C NH—

O

The marked group can give ninhydrin test.

(v) D–Glucose , D-Mannose and D-Fructose form same osazone.

(vi) Ala-Gly and Gly-Ala are enantiomer

(vii) Ala-Phe–Gly is having more M.w. than Ala-Gly-Ala

(viii) Iso-electric point (PI) of Glycine is more than aspartic acid

(ix) Monomer unit of starch is -D glucose.

5. For the cell reaction :

Hg2Cl

2(s) + 2Ag(s) 2Hg (l) + 2AgCl(s)

temperature coefficient of cell emf is found to be 0.02 VK–1. Find rSº for cell reaction in kJ mole–1

(Round off your answer to the next higher integer)



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6. How many of the following can be prepared by gabriel phthalimide synthesis in good yield.

NH2

, N

,

O

NH2 ,

NHMe

NO2

,

NH2

NH2

, NH2 ,

NH2

, NH2

7. 3–Amino–2,3–dimethyl butan–2–ol when treated with NaNO2/HCl gives product (X). (X) on treatment

with LiAlH4 forms (Y). (Y) on heating with conc. H

2SO

4 forms (Z). If -H of Z is (P) than find (P–3).

8. How many of the following statements are correct ?

(A) Delta at the mouth of a river is formed due to the coagulation of the colloidal sea water by the

electrolytes present in the river water.

(B) Artificial rain is caused by the coagulation of colloidal water droplets in the clouds by an electrolyte.

(C) Cottrell precipitator is used in the precipitation of the colloidal waste in the sewage water.

(D) Cheese is an example of solid foam.

(E) A colloid formed by adding small amount of AgNO3 to KI solution is negatively charged.

(F) Micelles are formed only below the Kraft temperature (Tk)

(G) Mixing of hydrated ferric oxide sol and arsenious sulphide s ol bring them in precipitated form.

(H) An oil in water emulsion can be diluted by addition of water.

(I) Emulsions show Brownian motion & tyndall effect.


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PART-3 : MATHEMATICSSECTION–I(i) : (Maximum Marks : 24)





Marking scheme :



1. Before I can open my bank locker, I must remember the combination. Two of the numbers of the three-

term sequence (-,-,-) are 17 and 24, but I have forgotten the third as well as the order of the numbers. Thepossibilities for the third number are any number between 1 and 40 (both inclusive). Assuming each trial

takes 10 sec, then (for example one of the possible password is 17245)

(A) Maximum possible time taken to open the locker is 40 mins

(B) Maximum possible time taken to open the locker is 39 mins

(C) Total possible password combination is 240

(D) Total possible password combination is 234

2. If from a point P(z1) on the curve |z| = 2, pair of tangents are drawn to the curve |z| = 1 meeting Q(z

2),

R(z3), then (where z

1,z

2, z

3 are complex numbers)

(A) complex number 1 2 3z z z

3

will lie on the curve |z| = 1

(B) 1 2 3 1 2 3

4 1 1 4 1 19

z z z z z z

(C) orthocentre and circumcentre of PQR will coincident

(D) 2

3

z 2arg 2n

z 3

, where n I



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ATICS

E-25/321001CT102115076


3. Let f(x) = x3 + 3x2 + 10x + 1 for x > 0 and 'g' be its inverse, then the value of 'k' such that kg'(15) = 1

is less than equal to

(A) –36 (B) 51 (C) 19 (D) 42

4. If tan(2| sin) = cot (2| cos) and y = (|sin| + |cos|)x, x > 1, then y can be equal to

(A) 2 (B) 3 (C) 4 (D) 1

5. If a quadrilateral formed by four tangents to the ellipse 3x2 + 4y2 = 12 is a square then

(A) The vertices of the square lie on y = + x (B) The vertices of the square lie on x2 + y2 = 7

(C) The area of all such squares is constant (D) Only two such squares are possible

6. Consider the parabola y2 = 4x. From a point A(–1,0) tangents AB and AC are drawn. Normals at B and

C intersect at D. A point is randomly chosen inside the quadrilateral ABCD. Let P denotes the probability

when the point lies inside the parabola as well as inside quadrilateral. Then P is greater than-

(A) 0.2 (B) 0.4 (C) 0.7 (D) 0.9

E-26/32

MATHEM

ATICS


1001CT102115076








Marking scheme :





If a chord of a parabola y2 = 4x subtends a right angle at centre of a circle x2 + y2 = 9 and the chordpasses through a fixed point P and from point P three normal are drawn to intersect the parabola atA, B & C. Then -

7. Area of ABC

(A) 8 2 (B) 4 2 (C) 2 2 (D) 6 2

8. Out of the 3 normal chord, the length of normal chord which is minimum is -

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


Consider three quadratic equations, x2 – 2rprx + r = 0; r = 1,2,3

Given that each pair has exactly one root common.

9. The common root between the equations obtained by r = 1 and r = 3 is

(A) 3 3

or2 2

(B) 2 2

or3 3

(C) 6 or 6 (D) 1

10. The number of the triplets (p1, p

2, p

3) for which such equations exist is

(A) 2 (B) 3 (C) 6C2

(D) infinite


MATHEM

ATICS

E-27/321001CT102115076


Consider a sequence I1, I

2, I

3, ...... I

n where the nth term I

n is given by

2n

n

0

I cos x cos nx dx

and another squence T1, T

2, T

3... ....T

n has general term given by T

n = 2(n 1)n! and

1 2 3 n

n

T T T ....... Ta

n!

11. The sequence I1, I

2, I

3,..... I

n represents

(A) A.P. (B) G.P.

(C) A.P. and G.P. both (D) Neither A.P. nor G.P.

12.5

n 1 n

1

a

equals to-

(A) 5/6 (B) (C) 1/6 (D) 4/5




E-28/32

MATHEM

ATICS


1001CT102115076

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. The lengths of two opposite edges of a tetrahedron are 2 cm and 3 cm. The angle between them is

6 and volume of tetrahedron is 2 cm3. Let shortest distance between the opposite edges is then

3

2

is

2. If the ratio of length of conjugate axis & transverse axis is 1

3 for the hyperbola

2 2

2 2

x y1

a b . Then the

eccentricity of conjugate hyperbola is



MATHEM

ATICS

E-29/321001CT102115076

3. The length of shortest distance between the two lines x 3 y 6 z

4 3 2

and

x 2 y z 7

4 1 1

is

4. Let f(x) be a function such that x 0

f(x)lim 1

x and

3x 0

x(1 a cos x) bsin xlim 1

f(x)

, then b – 3a is equal to

5. Let

2

nnn 1

rr 0

nS

r C

, then the value of S is


E-30/32

MATHEM

ATICS


1001CT102115076

6. x x/3

5

/3

e e1 xsin x n dx

1 x 2

is

7. If A be the area bounded by the curves |x| 1| y | e

2 and

| x | | y | | x | | y |2

2 2

then A – ln 4 – 7

equals to

8. If 2a2 – 9b2 – 3ab + 6b + a – 1 = 0, (where a,b are variable) then the line ax + by + 1 = 0 passes through

at least one of two fixed points A & B, let the area of triangle OAB is (where O is origin) then 2 is



E-31/321001CT102115076



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 1001CT102115076E-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) 6

: +4 0

(ii) -I(ii) 3 ‘’

: +4 0–1

9. –II III

10. -IV 8 0 9 ()

: +4 0

11.

12.

13.

PAPER – 1

Test Type : FULL SYLLABUS Test Pattern : JEE-Advanced

TEST DATE : 07 - 02 - 2016

Time : 3 Hours Maximum Marks : 240

JEE (Main + Advanced) : ENTHUSIAST COURSE

Paper Code : 1001CT102115076

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

IND

I

SCORE-I

H-2/32


1001CT102115076

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 =

20

1

c

Planck constant h = 6.63 × 10–34 J–s


PHYSICS

H-3/321001CT102115076

-1 : – I(i) : ( : 24)

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

+4 0

1. A B Ro C

fRo

E Er

BA C

fR0

A

(A) f = 0 (B) f = 1

(C) f = 2

1 (D) f =

3

1

2. abcdef P

a b

c

d e

f

x

y

z

(0, a, 0)

(a, 0, 0)

(0, 0, a) (a, 0, a)

(a/2, 0, 0)

P

(A) ab ka

Io ˆ5

11

4

(B) bc k

a

Io ˆ5

(C) cd ja

Io ˆ5

(D) ef j

a

Io ˆ5

21

2

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS

H-4/32

PHYSICS


1001CT102115076

3. C

1 2

C

1 2

C

(A) 2C

(B) 2C2

(C) C2

(D)

4. B

C

A

B

C

(A) AC 2

(B) AC 2

(C) C 5

2g

(D) C 5

1g


PHYSICS

H-5/321001CT102115076

5. () T

(A) A B P V

(B) B A

(C) AB

(D) A B

6.

S N

R

(A)

(B)

(C)

(D) d

dt

d

Rdt

H-6/32

PHYSICS


1001CT102115076

–I(ii) : ( : 24)

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

:

+4

0

–1

7 8

R

-1 :

-2 :

x T2 = xT

1 x =

1

2

T

T.


PHYSICS

H-7/321001CT102115076

7.

(A) l

g

(B) l

g2

(C)

22

5

2lRm

(D)

22

5

2

2lR

m

8. -1 -2

(A) -1 -2

(B) -1 -2

(C) -1 -2

(D) -1 -2

H-8/32

PHYSICS


1001CT102115076

9 10

m r 2r

x-z-

a

z

y

x

a

Ff

mg

F

R

F Ff F

W1

W2


PHYSICS

H-9/321001CT102115076

9.

(A) ma

i3

(B) ma

i3

(C) jrirjrir ˆ2ˆ2,ˆ2ˆ2

(D) krjrkrjr ˆˆ,ˆˆ

10.

(A) mar marˆ ˆ ˆ ˆ( j k), ( j k)

3 3

(B) kjmarkjmar ˆˆ2

3,ˆˆ

2

3

(C) W1 = 0, W

2 0

(D) W2 = 0, W

1 0

H-10/32

PHYSICS


1001CT102115076

11 12

n/p (n)

(p) n/p

p = 25 p = 25

n/p n/p

(p>83)

A 40 N > Z

N-Z N = Z

25 No of protons (p)

electron capture

-decayn=p line

stabl

e n/p

ratio

s

-decay

No.

of

neut

rons

(n)


PHYSICS

H-11/321001CT102115076

11.

(A) 27 12

(B) 27 12

(C) 23090

Th 21483

Bi

(D) 23090

Th 21483

Bi

12.

(A) Th21081 , 3 1 1

20686

Rn

(B) Th21081 , 3 1 1

20682

Pb

(C) X17888

(D) 2412 X

– II : & – III :

II III

H-12/32

PHYSICS


1001CT102115076

–IV : ( : 32)

0 9

:

+4

0

1. r = 3mm v = 0.7 cm/sec 1.75 gm/cc

10x x

2. +2q –3q (4 m, 0, 0) (9 m, 0, 0) x-2

1.210x x


PHYSICS

H-13/321001CT102115076

3. 300 K 750 K V = 6 m3

(m3 )

4. 10 kg B 2.0 kg 0.1 m

B B A

B C C A k =

s = 0.5 P (N )

A 20 rad/s2 P

16

C B P

A

H-14/32

PHYSICS


1001CT102115076

5.

AB

O

20 cm 30 cm AB O x cm x

10

////////

////

////

///

/////

///

/////

O

B

A

6. (a) (1) (2)

K (b)

(3) (4) 20%

d

12

k 2k nk

n

d

2 (a)nk

(b)

(4)

(3)

n

2kk

d

d2


PHYSICS

H-15/321001CT102115076

7. = 1 mm 9 =

10 3.4 cm

5 10 N

3.4 cm

1.1 cm 5

30 N 1.1 cm

x x

8.

0.5R + x R

x

Smooth

R

R

H-16/32

CHEM

ISTRY


1001CT102115076

-2 : –I(i) : ( : 24)

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

:

+4

0

1.

(A) Ca3P2 + H2O (B) PH4I + KOH

(C) H3PO2 (D) P + NaOH

2. (A) en (B) bn (C) dien (D) trien

3. P-V P-V

P-V

(A) 3-

V

P

(B) 3-

V

P

(C) 3-

V

P

(D) 3-

V

P


CHEM

ISTRY

H-17/321001CT102115076

4. (I) NH OH2

O

-1 (II) -1 H+

-2

(A) -1

(B) -2

(C) -2

(D) -1

5. 298 K 23892 U -- 206

82 Pb

t 1atm,273K 89.6 ml

(A) t Pb = 103 mg

(B) 10°C t

2

(C)

(D)

6. M + NaCN + O2 + H2O Na[M(CN)2] + NaOH.

NaCN (A) Ag (B) Cu (C) Au (D) Pt

H-18/32

CHEM

ISTRY


1001CT102115076

–I(ii) : ( : 24)

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

:

+4 0

–1

7 8

X

'A'

HCl

KI/H+

B + I + H O2 2

B + NaOH NH3

( )

7. (B)

(A)

(B)

(C) pH, 1-2 SO2

(D) +3

8.

(A) 'A' K4[Fe(CN)6]

(B) 'B' K4[Fe(CN)6]

(C) 'A' K3[Fe(CN)6]

(D) 'B' K3[Fe(CN)6]


CHEM

ISTRY

H-19/321001CT102115076

9 10

0.1 M ; 0.1 M

0.1 M (Ka = 10–5); 0.1 M (K

b = 10–5)

298 K

( log 5 = 0.7, log 2 = 0.3, log 3 = 0.5, 2H Od = 1 g ml–1)

9.

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

10.

a2

a2

a

a

(A) pH = 1 (B) = 5 × 10–4

(C) H2O = 9 × 10–15

(D) pH = 1.7

H-20/32

CHEM

ISTRY


1001CT102115076

11 12

P, Q, R, S T

(a) R

(b) Q,

(c) P

(d) S T,

(e) Q T,

11.

(A) P ,

COOH

COOH (B) Q

O

O

O

(C) R , (D) P

12.

(A) S , (B) Q

(C) R , (D) R S,

– II : & – III :

II III


CHEM

ISTRY

H-21/321001CT102115076

–IV : ( : 32)

0 9 :

+4

0

1. 'd' ()

Fe(CO)

5, MnO

4–, Co(C

2O

4)

3–3, NiCl

2(PPh

3)

2, PtCl

4–2, Ni(CN)

4–2, Ni(CN)

4–4.

2. % s- IF

7, CH

4, SF

4, XeF

3+, OCl

2, ICl

2–, BF

3, SO

2

3. = x,

2 = y,

= z

y – (x + z)

H-22/32

CHEM

ISTRY


1001CT102115076

4.

(i)HO

OH

O

O

OOH

OH

HO

HO

(ii) HOO

O

O

OH

HOHO OH

OH OHHO

(iii)

O

HO

OHHO

OH

O

HOOH

O

HO C1-C4

(iv) — —C NH—

O

(v) D– , D-D-

(vi) Ala-Gly Gly-Ala ,

(vii) Ala-Phe–Gly , Ala-Gly-Ala M.w.

(viii) (PI)

(ix) -D

5. :Hg

2Cl

2(s) + 2Ag(s) 2Hg (l) + 2AgCl(s)

emf 0.02 VK–1 kJ mole–1 rSº

( )


CHEM

ISTRY

H-23/321001CT102115076

6.

NH2

, N

,

O

NH2 ,

NHMe

NO2

,

NH2

NH2

, NH2 ,

NH2

, NH2

7. 3––2,3––2–NaNO2/HCl (X) (X), LiAlH

4

(Y) (Y) H2SO

4 (Z) Z -H , (P)

(P–3)

8.

(A)

(B)

(C) (Cottrell)

(D)

(E) KI AgNO3

(F) (Tk)

(G)

(H)

(I)

H-24/32

MATHEM

ATICS


1001CT102115076

-3 : –I(i) : ( : 24)

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

:

+4

0

1. (-,-,-), 17 24 1 40

()10 (17245 )

(A) 40

(B) 39

(C) 240

(D) 234

2. |z| = 2 P(z1) |z| = 1 Q(z

2), R(z

3)

(z1,z

2, z

3 )

(A) 1 2 3z z z

3

, |z| = 1

(B) 1 2 3 1 2 3

4 1 1 4 1 19

z z z z z z

(C) PQR

(D) 2

3

z 2arg 2n

z 3

, n I


MATHEM

ATICS

H-25/321001CT102115076

3. x > 0 f(x) = x3 + 3x2 + 10x + 1 'g' 'k' kg'(15) = 1

(A) –36 (B) 51 (C) 19 (D) 42

4. tan(2| sin) = cot (2| cos) y = (|sin| + |cos|)x, x > 1 y

(A) 2 (B) 3 (C) 4 (D) 1

5. 3x2 + 4y2 = 12

(A) y = + x

(B) x2 + y2 = 7

(C)

(D)

6. y2 = 4x A(–1,0) AB AC B C

D ABCD P,

P -

(A) 0.2 (B) 0.4 (C) 0.7 (D) 0.9

H-26/32

MATHEM

ATICS


1001CT102115076

–I(ii) : ( : 24)

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

:

+4

0 –1

7 8

y2 = 4x x2 + y2 = 9 P P A, B C -

7. ABC

(A) 8 2 (B) 4 2 (C) 2 2 (D) 6 2

8. 3 -

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

9 10

x2 – 2rprx + r = 0; r = 1,2,3

9. r = 1 r = 3 -

(A) 3 3

2 2 (B)

2 2

3 3 (C) 6 6 (D) 1

10. (p1, p

2, p

3)

(A) 2 (B) 3 (C) 6C2

(D)


MATHEM

ATICS

H-27/321001CT102115076

11 12

I1, I

2, I

3, ...... I

n, n I

n

2n

n

0

I cos x cos nx dx

T1, T

2, T

3.......T

n T

n = 2(n 1)n!

1 2 3 n

n

T T T ....... Ta

n!

11. I1, I

2, I

3,..... I

n -

(A) (B)

(C) (D)

12.5

n 1 n

1

a

-

(A) 5/6 (B) (C) 1/6 (D) 4/5

– II : & – III :

II III

H-28/32

MATHEM

ATICS


1001CT102115076

–IV : ( : 32)

0 9

:

+4

0

1. 2 3

6

2 3

2

2. 2 2

2 2

x y1

a b

1

3


MATHEM

ATICS

H-29/321001CT102115076

3. x 3 y 6 z

4 3 2

x 2 y z 7

4 1 1

4. f(x) x 0

f(x)lim 1

x

3x 0

x(1 a cos x) b sin xlim 1

f(x)

b – 3a

5.

2

nnn 1

rr 0

nS

r C

S

H-30/32

MATHEM

ATICS


1001CT102115076

6. x x/3

5

/3

e e1 xsin x n dx

1 x 2

7. |x| 1| y | e

2 | x | | y | | x | | y |

22 2

A A – ln 4 – 7

8. 2a2 – 9b2 – 3ab + 6b + a – 1 = 0, (a,b ), ax + by + 1 = 0 A B

OAB(O ) 2


H-31/321001CT102115076


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 1001CT102115076H-32/32

:

14.

15.

16.

17. :

18.

19.

20. g = 10 m/s2

____________________________

____________________________

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

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

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