72
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 36 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 three sections. 7. Carefully read the instructions given at the beginning of each section. 8. Section-I contains 10 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. 9. Section-II contains 2 ‘match the following’ type questions and you will have to match entries in Column-I with the entries in Column-II. Marking scheme : for each entry in column-I . +2 for correct answer, 0 if not attempted and –1 in all other cases. 10. There is no questions in SECTION-III 11. 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 : 12. The ORS is machine-gradable and will be collected by the invigilator at the end of the examination. 13. Do not tamper with or mutilate the ORS. 14. Write your name, form number and sign with pen in the space provided for this purpose on the original. Do not write any of these details anywhere else. Darken the appropriate bubble under each digit of your form number. DO NOT BREAK THE SEALS WITHOUT BEING INSTRUCTED TO DO SO BY THE INVIGILATOR PAPER – 1 Test Type : ALL INDIA OPEN TEST Test Pattern : JEE-Advanced TEST DATE : 07 - 02 - 2016 TARGET : JEE (ADVANCED) 2017 JEE (Main + Advanced) : NURTURE COURSE Paper Code : 0000CT100115002 CLASSROOM CONTACT PROGRAMME (Academic Session : 2015 - 2016) Time : 3 Hours Maximum Marks : 264 ENGLISH Please see the last page of this booklet for rest of the instructions

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Page 1: JEE (Main + Advanced) : NURTURE COURSE · ALL INDIA OPEN TEST/NURTURE COURSE/JEE (Advanced)/07-02-2016/PAPER-1 PHYSICS 0000CT100115002 E-5/36 Space for Rough Work 4. Figure shows

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 36 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 three sections.

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

8. Section-I contains 10 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.

9. Section-II contains 2 ‘match the following’ type questions and you will have to match entries in Column-I with the

entries in Column-II.

Marking scheme : for each entry in column-I. +2 for correct answer, 0 if not attempted and –1 in all other cases.

10. There is no questions in SECTION-III

11. 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 :

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

13. Do not tamper with or mutilate the ORS.

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

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

DO

NO

T B

REA

K T

HE

SEA

LS W

ITH

OU

T B

EIN

G IN

ST

RU

CT

ED T

O D

O S

O B

Y T

HE

INV

IGIL

ATO

R

PAPER – 1

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

TEST DATE : 07 - 02 - 2016

TARGET : JEE (ADVANCED) 2017

JEE (Main + Advanced) : NURTURE COURSE

Paper Code : 0000CT100115002

CLASSROOM CONTACT PROGRAMME(Academic Session : 2015 - 2016)

Time : 3 Hours Maximum Marks : 264

EN

GL

ISH

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

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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,

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PART-1 : PHYSICS

SECTION–I : (Maximum Marks : 40)

This  section  contains  TEN  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. Figure shows two point sources S1 and S2 that emit sound of wavelength  = 2.0 m. The emissions are

isotropic and inphase and the separation between the sources is d = 16m. At any point P on the axis, the

wave from S1 and the wave from S2 interfere. When P is very far away ( x ), mark the CORRECT

statement(s):-

(A) The phase difference between the arriving waves from S1 and S2 is zero.

(B) Interference they produce is approximately fully constructive.

(C) As we then move P along the x axis towards S1,  the phase difference between the waves from

S1 and S2 increases

(D) As we then move P along the x axis towards S1,  the phase difference between the waves from

S1 and S2 decreases

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS

Space for Rough Work

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2. A small sphere of radius R is arranged to pulsate so that its radius varies in simple harmonic motionbetween a minimum of R – R and a maximum of R + R with frequency f. This produces soundwaves in the surrounding air of density , take atomspheric pressure to be p

atm and ratio of specific

heats constant pressure (CP) and constant volume (C

V) to be . (Assume the amplitude of oscillation of

the sphere is the same as that of the air at the surface of the sphere.) Mark the CORRECT statement(s):-

(A) The intensity of sound waves at the surface of sphere is  22 2

atmI 2 f p R

(B) The total acoustic power radiated by the sphere is  23 2 2

atmp 8 R f p R

(C) At a distance d >> R from the center of the sphere, the amplitude is R

A Rd

(D)  At  a  distance  d  >>  R  from  the  center  of  the  sphere,  the  pressure  amplitude  is

max atm

Rfp 2 p R

d

3. Two blocks B and C each of mass m are connected by a light spring of force constant k and naturallength L. The whole system rests on a frictionless table such that x

B = 0 and x

C = L, where x

B and x

C are

coordinates of the block B and C respectively. Another block of mass M, which is travelling at speed

V0 collides head on elastically with the block B at t = 0. Mass ratio is 

m

M . For t > 0 the positions of

the blocks are given byx

B = t +  sin t

xC = L + t –  sin t

Mark the CORRECT option(s) :

A B Cv0

(A) 2k

m (B) 

0v

1

(C) 

0v

1

(D) Maximum speed of C is 

02v

1

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4. Figure shows a double chambered vessel with thermally insulated walls and partitions. On each side

there are n moles of an ideal monoatomic gas. Initially the pressure, volume and temperature in each

side is P0, V

0, T

0. The heater in first chamber supplies heat very slowly till the gas in the first chamber

expands  such  that  the  pressure,  volume  and  temperature  of  the  gas  on  the  left  side  is  P1,  V

1,  T

1

respectively and on right chamber is 0

2

27PP

8 , VV

2 and T

2 respectively :-

P ,V ,T0 0 0 P ,V ,T0 0 0

(A) Volume of first chamber is 

3 / 5

0

82 V

27

(B) Temperature in second chamber is 

2 / 5

0

27T

8

(C) Work done on the gas in second chamber in terms of molar heat capacity at constant volume and

T0 is 

2 / 5

V 0

27nC T 1

8

(D) Work done on the gas in first chamber in terms of molar heat capacity at constant volume and T0

is 

2 / 5

V 0

27nC T 1

8

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5. A projectile is projected on the inclined plane as shown. V1 & V2 are components of it's initial velocity

along the incline and perpendicular to incline and V3 & V4 are components of it's final velocity along

the incline and perpendicular to incline. {Here we are comparing the magnitudes only}

  V1

V2

V3

V4

(A) V1 > V3 (B) V1 = V3 (C) V2 = V4 (D) V2 > V4

6. Let n1 and n

2 moles of two different ideal gases be mixed. If ratio of specific heats of the two gases are

1 and 

2 respectively, then the ratio of specific heats  of the mixture is given through the relation :

(A) 

1 1 2 2

1 2

1 2

1 2

n n

1 1

n n

1 1

(B)  1 2 1 2

1 2

n n n n

1 1 1

(C)  1 21 2 1 2

1 2

n n n n1 1 1

(D) (n1 + n

2) ( – 1) = n

1(

1 – 1) + n

2(

2 – 1)

Space for Rough Work

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7. The rate of change of angular momentum of a system of particle about the centre of mass is equal to

the sum of external torques about the centre of mass when the centre of mass is :

(A) fixed with respect to an inertial frame (B) in linear acceleration

(C) in rotational motion (D) is in a translational motion

8. A particle moves in one dimension in a conservative force field. The potential energy is depicted in the

graph below.  If the particle starts to move from rest from the point A, then :

A B C D Ex

V(x)

Potential energy

(A) The speed is zero at the points A and E

(B) The acceleration vanishes at the points A,B,C,D,E

(C) The acceleration vanishes at the points B,C,D

(D) The speed is maximum at the point D

Space for Rough Work

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9. Consider a sinusoidal wave moving in the +x direction.

1

2 xy x, t Asin t

T v

A wall located at x = L reflects this wave. Mark the CORRECT statement(s) :

(A) If the wall is a fixed end the expression for reflected wave is 2 x 2L

Asin tT v

(B) If the wall is a free end the expression for reflected  waves is 2 x 2L

Asin tT v

(C) If the wall is a fixed end the resultant wave is 2 L x 2 L

2Asin cos tT v T v

(D) If the wall is a free end the resultant wave is 2 L x 2 L

2A cos sin tT v T v

Space for Rough Work

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10. A "hot reservoir" at 100.0 °C is connected to a "cold reservoir" at 0° C via two separate rods each of

length 10 cm which are insulated perfectly on their sides as shown in the figure(i). Rod A is made of

steel, has a cross sectional area of 4.0 cm2. Rod B is made of copper. Take thermal conductivites of

steel and copper respectively as 50 W m–1 K–1 and 400 W m–1 K–1. The areas are such that in figure (i)

rate of heat flow through both rods are equal. Mark the CORRECT statement(s) :-

(A) The cross–sectional area of the copper rod is 0.5 cm2

(B) In figure (i)  the time it takes for a total  of 1.0 × 103 J of heat to flow from the hot to the cold

reservoir via the steel rod is 50 sec

(C) In figure (ii) suppose the two rods of above specifications were welded end to end and the free

ends  connected  to  the  reservoirs.  The  rate  of  flow  of  heat  from  hot  to  cold  reservoirs  in  this

situation is 15 J/sec.

(D) In figure (ii) suppose the two rods of above specifications were welded end to end and the free

ends  connected  to  the  reservoirs.  The  rate  of  flow  of  heat  from  hot  to  cold  reservoirs  in  this

situation is 10 J/sec.

Space for Rough Work

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SECTION–II : (Maximum Marks : 16) This  section  contains  TWO questions. Each  question  contains  two  columns,  Column-I  and  Column-II. Column-I has  four entries  (A),  (B),  (C)  and  (D) Column-II has  five entries  (P),  (Q),  (R),  (S)  and  (T) Match  the  entries  in  Column-I with  the  entries  in  column-II. One  or  more  entries  in  Column-I  may  match  with  one  or  more  entries  in  Column-II. The  ORS  contains  a  4  ×  5  matrix  whose  layout  will  be  similar  to  the  one  shown  below  :

(A)  (P) (Q) (R) (S) (T)

(B)  (P) (Q) (R) (S) (T)

(C)  (P) (Q) (R) (S) (T)

(D)  (P) (Q) (R) (S) (T)

For  each entry  in  column-I,  darken  the  bubbles  of  all  the  matching  entries.  For  example,  if  entry(A) in Column-I matches with entries (Q), (R) and (T), then darken these three bubbles in the ORS.Similarly,  for  entries  (B),  (C)  and  (D).

Marking  scheme  :For  each  entry  in  Column-I+2 If  only  the  bubble(s)  corresponding  to  all  the  correct  match(es)  is  (are)  darkened0 In  none  of  the  bubbles  is  darkened–1 In  all  other  cases

1. A calorimeter of water equivalent 1 kg contains 10 kg of ice & 10 kg of water in thermal equilibrium.

The atmospheric  temperature  is  15° below freezing point due  to which  the calorimeter  loses heat.

As a result ice is formed inside the calorimeter at a rate of 10.8 gm per second. To try to compensate

for  this heat  loss,  steam at  100°C  is  supplied  to  the  calorimeter  at  a  rate  of  r.  (LV  =  540  cal/gm,

Lf = 80 cal/gm, sp heat of water 1 cal/gm °C.) Column-I gives the value of r and column-II gives the

situation just after the introduction of steam.

Column-I Column-II

(A) r = 1.6 gm/sec (P) Amount of ice in calorimeter increases.

(B) r = 1.35 gm/sec (Q) Amount of water in calorimeter increases.

(C) r = 1.2 gm/sec (R) Amount of ice remains constant at 10 kg

(D) r = 1 gm/sec (S) Amount of water remains constant at 10 kg

(T) Amount of ice in calorimeter decreases.

Space for Rough Work

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2. Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especiallyuseful  in  studying  the  changes  in  motion  as  initial  position  and  momentum  are  changed.  Here  weconsider some simple dynamical systems for which phase space is a plane in which position is plottedalong horizontal axis and momentum is plotted along vertical axis. The phase space diagram is x(t) vs.p(t) curve in this plane. The arrow on the curve indicates the time flow. For example, the phase spacediagram for a particle moving with constant velocity is a straight line as shown in the figure. Similarlywe may also plot momentum of a pendulum versus  (with sign convention shown in figure (b)).

Mom

entu

mPosition

Figure (a)Figure (b) shows phase diagram of motion of simple pendulum (momentum P versus angle ). Choosepotential energy level at the lowest point of the pendulum. E represents total energy of simple pendulum.Pendulum has a point mass connected with light rod.

a bc

d

P

+–

      

P

O

D

=– =+

+–

Figure (b)

Column-I Column-II

(A) Phase diagram a (P) E < 2 mg(B) Phase diagram b (Q) E  2mg(C) Phase diagram c (R) May perform periodic and oscillatory motion(D) Phase diagram d (S) May represent SHM

(T) May represent angular velocity  versus  for apendulum bob

SECTION –III : Integer Value Correct Type

No question will be asked in section III

Space for Rough Work

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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. The block has mass M and rests on a surface for which the coefficient of friction µ. If a force F = kt is

applied to the cable (see figure). Find the power developed by the force at t = t2 in watt. If your answer

is N fill value of N

320. (Given : M = 20 kg, µ = 0.4, k = 40 N/s, t2 = 3 sec.)

F

2. A rod of length R and mass M is free to rotate about a horizontal axis passing through hinge P as in

figure. First it is taken aside such that it becomes horizontal and then released. At the lowest point the

rod hits the block B of mass m and stops. Find the ratio of masses M

m

 such that the block B completes

the circle. Neglect any friction. If your answer is N fill value of 2N

3.

Space for Rough Work

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3. A vessel A filled with water (Mariotte vessel) communicates with the atmosphere through a glass tube

passing through the throat of the vessel as shown in figure. A faucet F is h2 = 2cm from the bottom of

the vessel. Find the velocity (in m/s) with which the water flows out of the faucet F when the distance

between the end of the tube and the bottom of the vessel is h1 = 10 cm. If your answer is 

1N

10 fill

value of N.

Fh2

h1

A

a

4. A sphere made of polymer is floating on the surface of water which is at a temperature of 20°C and 1%

of the volume of the sphere is above the water. To what final temperature (in °C) must the water be

heated in order to submerge the polymer sphere completely. Take coefficient of linear expansion of the

polymer 2.30 × 10–5/°C, and coefficient of volume expansion of  water 3.19×10–4/°C. If your answer

is N fill value of N

20.

Space for Rough Work

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5. A pan with a set of weights is attached to a spring. The period of vertical oscillations is 0.3 s. After

additional weights are placed on the pan, the period of vertical oscillations becomes 0.4 s. By how

much does the spring stretch (in cm) owing to the additional weight? If your answer is n fill value of

4n. (g = 2)

6. Consider a uniform square plate of  side 2 = 6m made of wood. A semicircular portion is cut and

attached to the right as shown. Determine the displacement of centre of mass of the redesigned plate.

2L

Space for Rough Work

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7. Shin Chan and his mother have a tin whistle each. The pipe length of  Shin chan's tin whistle is 52 cm

long while the pipe  length of mother's tin whistle  is 50 cm  long. They both play at  the same  time,

sounding the whistles at their fundamental resonant frequencies. They note that they are not in tune

with each other. The velocity of sound in air is 325 m/s. Assume the whistle  is a pipe with one end

closed find the beat frequency (in Hz) that is heard when both whistles are playing simultaneously. If

n beats are heards in 4 sec fill value of  n .

8. Two light wires A and B of breaking stress 8 × 108 Pa and 3 × 108 Pa are used to support a light bar

horizontally as shown. The area of cross-section of A & B are 1 mm2 and 2 mm2  respectively. An

increasing external force directed vertically downward is applied as shown. If the angle made by wire

B with horizontal  is such that both wires break simultaneously. What  is the value of external force

(in N) at which wires break? If your answer is X fill value of X

250.

Space for Rough Work

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PART-2 : CHEMISTRY

SECTION–I : (Maximum Marks : 40)

This section contains TEN 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. Comment about the fraction of molecules moving between 400 to 500 m/sec for a gas

(molecular mass = 20 gm/mol) if its temperature increases from 300 K to 400 K [R = 25/3 J/mol/K]

(A) Fraction of molecules increases (B) Fraction of molecules decreases

(C) Fraction of molecules remains constant (D) Fraction of molecules may increase of decrease

2. 0.1M AgNO3 solution was slowly added to a solution containing salt NaX(aq.), MgY2(aq.), AlZ3(aq.)

If it is known that silver forms precipitate with X, Y, Z one by one (order may be different). Find

the element which was precipitated first using following information. (Consider complete dissociation

of salt).

Given : Atomic mass of X = 120

Atomic mass of Y = 80

Atomic mass of Z = 60

vol. of AgNO3

Mas

s of

pre

cipit

ate

(A) X (B) Y (C) Z (D) Incomplete information

Space for Rough Work

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3. For s-orbital , 4r2r2 vs r is plotted. Select correct statements (s)

Note : S = spherical surface at given distance

P = point (very small volume) at given distance

4 r 2 2

r

r

s1 s2 s3 s4

p1

p2

p3

p4

s-orbitalP6P5

(A) Order of probability of finding an electron at points is P4 > P3 > P2 > P1

(B) Order of probability of finding an electron at spherical surface is S4 > S3 > S2 > S1

(C) Order of probability of finding an electron at points is P6 = P5 = P4

(D) Order of probability of finding an electron at points is P4 = P3 = P2 = P1

4. A sample containing 1 mol KHC2O4. H2C2O4 is titrated with different reagent select correct statement-

(A) 1 mol of KOH are used

(B) 3/2 moles of Ba(OH)2 are used

(C) 4/5 mol of KMnO4 are used in alkaline medium

(D) 2/3 mol of K2Cr2O7 are used in acidic medium

5. Which of the following pair of orbital has electron density along the axes.

(A) dxy

,dyz

(B) 2 2 xyx yd , d

(C) 2xz zd , d (D) 2 2 2x y z

d , d

6. Which of the following pair of elements are not of same group of periodic table ?

(A) Li, Na (B) Be, B (C) N, As (D) O, At

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7. Among the following most acidic compound of molecular formula C4H

4O

4 is :

(A)

COOH

COOH

(B)

OHO O

OH

(C) HOOC–CC–COOH (D)

O

HO

O

OH

8. In which of the following pair(s), (II) has more heat of hydrogenation as well as heat of combustion

than (I) ?

(A) & (B) &

(C) & (D) &

9. Which of the following reactions will not undergo in forward direction ?

(A) HCCH + NaOH (B)

OH

COOH

+ NaHCO3

(C)

OH

+ CH –C–ONa3

O– + (D)

C

+ NaOHH

10. In which of the following pair(s), (II) has more resonance energy than (I) ?

(A) NH ,NH2

(B) ,–

(C)+

,

+

(D)

OH

,

O–

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SECTION–II : (Maximum Marks : 16) This section contains TWO questions. Each question contains two columns, Column-I and Column-II. Column-I has four entries (A), (B), (C) and (D) Column-II has five entries (P), (Q), (R), (S) and (T) Match the entries in Column-I with the entries in column-II. One or more entries in Column-I may match with one or more entries in Column-II. The ORS contains a 4 × 5 matrix whose layout will be similar to the one shown below :

(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)

(D) (P) (Q) (R) (S) (T)

For each entry in column-I, darken the bubbles of all the matching entries. For example, if entry (A)in Column-I matches with entries (Q), (R) and (T), then darken these three bubbles in the ORS. Similarly,for entries (B), (C) and (D).

Marking scheme :For each entry in Column-I+2 If only the bubble(s) corresponding to all the correct match(es) is (are) darkened0 In none of the bubbles is darkened–1 In all other cases

1. Column-I Column-II

(A) H2O(l) H

2O(g) (P) Favours forward reaction

Addition of H2O(l) at equilibrium

(B) I3

–(aq) I2(aq.) + I–(aq.) (Q) Does not shift equilibrium state

Addition of H2O(l) at equilibrium

(C) 2AB(g) A2(g) + B

2(g) (R) Concentration of product decrease

on increasing the volume at equilibrium

(D) A2(g) 2A(g) (S) Concentration of product remain

addition of catalyst at equilibrium constant

(T) Concentration of reactant decreases

Space for Rough Work

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2. Column-I Column-II

(A) SO3

(P) Paramagnetic

(B) SO2

(Q) d-p

bond

(C) CO2

(R) linear

(D) NO2

(S) planar

(T) angular

SECTION –III : Integer Value Correct Type

No question will be asked in section III

Space for Rough Work

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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. Calculate the wavelength of a rested electron (in Å) after it absorbs a photon of wavelength 9 nm.

[Given h = 6 × 10–34 J-sec , m = 9 × 10–31 kg].

2. An ideal gas occupy 2 litre volume at 300K & 1atm. Calculate the volume occupied by equal moles

of real gas at same temperature and pressure.

Given : b = 0.05 litre/mol

R = 0.08 atm

Z = 1.5 at given condition

Space for Rough Work

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3. -D-glucopyranose reacts with periodate ion as follows :

C6H12O6(aq.) + IO4– (aq.) HCOOH(aq.) + HCHO(aq.) + IO3

– (aq.)

In a typical experiment, a 1 ml solution of -D-Glucopyranose required 80 ml 0.25 M periodate solution

to reach the equivalence point. The solution is made free from formic acid and iodate ion by extraction

and then treated with H2O2, an oxidizing agent, oxidizing all formaldehyde into formic acid and finally

titrated against 0.1M NaOH solution. Titration required 40 mL of alkali to reach the equivalence point.

Determine molarity of -D-glucopyranose solution.

4. Maximum moles of Na4[Cu6(S2O3)5] which can be produced by 6 moles of CuSO4 and 10 moles of

Na2S2O3 using following series of reaction -

CuSO4 + Na2S2O3 CuS2O3 + Na2SO4

2CuS2O3 + Na2S2O3 Cu2S2O3 + Na2S4O6

3Cu2S2O3 + 2Na2S2O3 Na4 [Cu6(S2O3)5]

Fill your answer to nearest integer.

Space for Rough Work

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5. XOn(OH)

m is a formula of oxyacid then find the value of (n + m) if oxyacid has basicity 5 and central

atom has covalency seven.

6. Find the ratio of sp3 and sp2 hybridised atoms which are present in ring in following molecule.

N N

PP

N Cl

Cl

PClCl

Cl

Cl

hexachlorotriphosphazene

Space for Rough Work

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7. How many of the following compound(s) will show geometrical isomerism ?

(i) (ii) (iii) (iv)

(v) (vi) C=C=CCH3

HH

H C3 (vii) (viii)

Br

ICl

F

8. How many of the following compounds will have tautomerism phenomenon ?

(i)

O

(ii)

O

(iii)

O

OH

(iv)

O

(v)

O

O

(vi)

O

(vii)

O

(viii)

O

(ix)

O

(x) H

O

Space for Rough Work

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PART-3 : MATHEMATICS

SECTION–I : (Maximum Marks : 40)

This section contains TEN 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. The terms in the expansion of n

4

1x n 1, n N

2 x

, are arranged in descending powers of x. If

the coefficients of first three terms form an arithmetic progression, then-

(A) Number of terms with integer powers of x is 3

(B) Total number of terms in above expansion is 10

(C) Coefficient of term independent of x is 256

(D) Coefficient of middle term in above expansion is 35

8

2. Consider the equation of circles

S1 : x2 + y2 + 24x – 10y + a = 0

S2 : x2 + y2 = 36

(A) Number of non-negative integral values of a such that S1 = 0 represents a real circle 170

(B) If S1 = 0 and S

2 = 0 has no point in common, then number of integral values of a is 49

(C) If S1 = 0 andS

2 = 0 intersect orthogonally, then a = 36

(D) If a = 0, then number of common tangents to the circles S1 = 0 and S

2 = 0 are 3.

Space for Rough Work

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3. If sinx + siny = 96

65 and cosx + cosy =

72

65, then-

(A) 24

sin x y25

(B) x y 12

cos2 13

(C) x y 4

tan2 3

(D)

7cos x y

25

4. Let

2014

2014k

k 0

S C .k and E denotes the quotient when S is divided by 19, then-

(A) Highest exponent of 2 in E is 2014(B) Number of divisors of E is 4030(C) (E + 1) is an even number(D) Sum of coefficients in the expansion of (a – 2b + 3c)2016 is equal to E

5. For the equation 240 160 200 3206x 27x

x 1 x 4 x 5 x 8

(A) Number of real solutions of above equation is 3

(B) If E denotes the product of non-zero real or complex roots of the equation, then sum of divisors of

E is 2904

(C) If S denotes the set of all real roots of the equation then, sum of elements of S taken two at a time

is 81

(D) If

R be two roots of the equation such that

2 1log 2 is defined then it must be 1.

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6. Consider the trigonometric equation

63 3

3 36 3 3

cos x cosec x 2 2 sin x1 3

2tan x cot xcot x 2 2 cos x sec x 2 2

,

which of the following is correct ?

(A) Number of solutions of the equation in [0,4] is 8

(B) Number of solutions of the equation in [0,4] is 16

(C) If ƒ : A B be a function where A is set of solutions of above equation in [0,3] and B is set of

solutions of above equation in [0,5], then number of such function(s) is/are 106

(D) Sum of first four positive solution of equation is 13.

7. Given 3 3 32 2 2

1ƒ x

x 2x 1 x 1 x 2x 1

and E = ƒ(1) + ƒ(3) + ƒ(5) +........+ ƒ(999), then-

(A) Value of E is an irrational number

(B) Value of E is 5

(C) In the expansion of

7E

115

, greatest term are T

2 and T

3

(D) The value of ECk is maximum for k = 3 only.

8. Consider the function ƒ : (0, ) R defined as | nx|e , 0 x 3

ƒ xx 6 , x 3

, then which of the

following is correct ?(A) If a,b,c and d are four distinct positive number such that ƒ(a) = ƒ(b) = ƒ(c) = ƒ(d) then number of possible integral values of abcd is 7

(B) ƒ(x) is a surjective function

(C) Area bounded by the line x = 1, x = 3, y = ƒ(x) and x-axis is 4 sq. units

(D) There exists only one circle touching ƒ(x) at (7,1) and radius 2 which is x2 + y2 – 16x + 62 = 0

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9. Which of the following statements is/are correct ?(A) If ƒ(x) = log((x2 + 1)1/2 + x) + sinx and ƒ(a + 2) = 5 and ƒ(b – 7) = –5, then possible value of (a + b) is equal to 5.

(B) If 2

2

1ƒ x sin x x

1 x

(where [.] denotes greatest integer function), then ƒ(x) is an even

function.

(C) If 2015 4030 2015ƒ x 1 x x 1 , then the sum of the coefficients of 2015ƒ x 1 is 1.

(D) The number of solution of 1 11sin tan 2x

x

is 2.

10. Let A,B,C be angles of ABC and 5 A 5 B

D , E32 32

&

5 CF

32

, then

(where D,E,F n

2

& n I)

(A) cotD cotE + cotE cotF + cotD cotF = 1

(B) cotD + cotE + cotF = cotD.cotE.cotF

(C) tanD tanE + tanE tanF + tanF tanD = 1

(D) tanD + tanE + tanF = tanD tanE tanF

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SECTION–II : (Maximum Marks : 16) This section contains TWO questions. Each question contains two columns, Column-I and Column-II. Column-I has four entries (A), (B), (C) and (D) Column-II has five entries (P), (Q), (R), (S) and (T) Match the entries in Column-I with the entries in column-II. One or more entries in Column-I may match with one or more entries in Column-II. The ORS contains a 4 × 5 matrix whose layout will be similar to the one shown below :

(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)

(D) (P) (Q) (R) (S) (T)

For each entry in column-I, darken the bubbles of all the matching entries. For example, if entry (A)in Column-I matches with entries (Q), (R) and (T), then darken these three bubbles in the ORS. Similarly,for entries (B), (C) and (D).

Marking scheme :For each entry in Column-I+2 If only the bubble(s) corresponding to all the correct match(es) is (are) darkened0 In none of the bubbles is darkened–1 In all other cases

1. Column-I Column-II

(A) The value of n

n 1

n 0

2

is (P) 1

(B) Number of solutions of equation (Q) 2

3 316sin x 14 sin x 7 in [0,4] is

(C) Number of solutions of equation (R) 3

tan–1(2 sinx) = cot–1(cosx) in [0,10] is

(D) If the sum of roots of the quadratic equation (S) 4

(–a + 1)x2 + (a2 – a + 4)x – 2a + 3 = 0 is

minimum, then the value of a (a > 1) is (T) 5

Space for Rough Work

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2. Column-I Column-II

(A) The value of

1

n 0

2tan

n 2 n n 2 n n n 2

(P) 1

(where [.] denotes greatest integer function) is

(B) If a,b,c are the sides of a triangle, then 2 2 22 a b c

ab bc ca

(Q) 2

is always less than

(C) If a,b,c are distinct real numbers such that (R) 3

a2(b + c) = b2(a + c) = 2, then the value of c2(a + b) is

(D) If P = sinA sinB, Q = sinC cosA, R = sinA cosB, (S) 4

S = cosAcosC,then the value of 5(P2+Q2 + R2 + S2) is (T) 5

SECTION –III : Integer Value Correct Type

No question will be asked in section III

Space for Rough Work

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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. If the value of [sin(sin–1(0.5) + sin–1(0.4)) × sin(sin–1(0.5) – sin–1(0.4))] can be expressed as p

q

(where p, q N are relatively prime), then q p

13

is

2. For positive integer n, let Sn denotes the minimum value of the sum n

2 2k

k 1

2k 1 a

where a1,a2,....,an

are positive real numbers whose sum is 17. If there exist a unique positive integer n for which Sn is also

an integer, then n

2

is

Space for Rough Work

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3. Triangle ABC has right angle at B and contains a point P for which PA = 10, PB = 6 and

APB = BPC = CPA, then PC

11

is

4. If

22015n

n 1

n n 1 b1 a

n ! c!

(where a,b,c N), then the minimum value of

a b c

576

is

Space for Rough Work

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5. 6 lecturers A,B,C,D,E,F want to deliver lecture in a particular batch (one by one), such that A wants

to take class before B and B wants to take class before C (not necessarily consecutive), if total number

of ways are 2n

n, then n is equal to (where n N)

6. Let x and y be positive real numbers and an angle such that n

2

for any integer n. Suppose

sin cos

x y

and

4 4

4 4 3 3

cos sin 97sin 2

x y x y xy

, then the value of

x y

y x is

Space for Rough Work

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7. Let C1 and C

2 be externally tangent circles with radius 2 and 3 respectively. Let C

1 and C

2 both touch

circle C3 internally at points A and B respectively. The tangents to C

3 at A and B meet at T and

TA = 4, then radius of circle C3 is

8. If n n

n mm p

p 1 m p

C . C 19

, then the value of n is

Space for Rough Work

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E-35/360000CT100115002

Space for Rough Work

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ALL INDIA OPEN TEST/NURTURE COURSE/JEE (Advanced)/07-02-2016/PAPER-1

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

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

Your Hard Work Leads to Strong Foundation 0000CT100115002E-36/36

DARKENING THE BUBBLES ON THE ORS :

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

16. Darken the bubble COMPLETELY.

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

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

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

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

bubbles are evaluated.

21. 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. .............................................

Space for Rough Work

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1.

2. (ORS)

3.

4.

5. 36 20

6.

7.

8. -I 10

: +4 0–2

9. -II 2 ‘’ -I -II

: -I +2 0 –1

10. –III

11. -IV 8 0 9 ()

: +4 0

12.

13.

14.

PAPER – 1

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

TEST DATE : 07 - 02 - 2016

Time : 3 Hours Maximum Marks : 264

TARGET : JEE (ADVANCED) 2017

JEE (Main + Advanced) : NURTURE COURSE

Paper Code : 0000CT100115002

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

IND

I

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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,

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-1 : – I : ( : 40)

(A), (B), (C)  (D) 

  :

+4 0

–2

1. S1  S2  = 2.0 m 

d = 16m P S1 S2 

P ( x ) 

(A) S1  S2 

(B) 

(C) P x S1 S1 S2 

(D) P x S1 S1 S2 

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS

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2. R  R – R R + R f  p

atm (C

P) 

(CV) () 

(A)  22 2

atmI 2 f p R  

(B)  23 2 2

atmp 8 R f p R  

(C) d >> R R

A Rd

 

(D) d >> R  max atm

Rfp 2 p R

d

 

3. m B  Ck L x

B = 0  x

C = L x

B  x

C B  C 

M V0  t = 0 B 

m

M  t > 0 

xB = t +  sin t

xC = L + t –  sin t

A B Cv0

(A) 2k

m (B) 

0v

1

(C) 

0v

1

(D) C  

02v

1  

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4. n 

P0, V

0, T

P1, V

1, T

1  0

2

27PP

8 , V

2 T

P ,V ,T0 0 0 P ,V ,T0 0 0

(A)  

3 / 5

0

82 V

27

 

(B) 2 / 5

0

27T

8

 

(C) T0 

2 / 5

V 0

27nC T 1

8

(D) T0 

2 / 5

V 0

27nC T 1

8

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5.

V1  V2 V3

 V4 

  V1

V2

V3

V4

(A) V1 > V3 (B) V1 = V3 (C) V2 = V4 (D) V2 > V4

6. n1 n

(A) 

1 1 2 2

1 2

1 2

1 2

n n

1 1

n n

1 1

(B)  1 2 1 2

1 2

n n n n

1 1 1

(C)  1 21 2 1 2

1 2

n n n n1 1 1

(D) (n1 + n

2) ( – 1) = n

1(

1 – 1) + n

2(

2 – 1)

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7.

(A)  (B) 

(C)  (D) 

8.

A B C D Ex

V(x)

Potential energy

(A) A E 

(B) A,B,C,D,E 

(C) B,C,D 

(D) D 

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9. x 

1

2 xy x, t Asin t

T v

x = L 

(A) 2 x 2L

Asin tT v

 

(B) 2 x 2L

Asin tT v

 

(C) 2 L x 2 L

2Asin cos tT v T v

 

(D) 2 L x 2 L

2A cos sin tT v T v

 

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10. 100.0 °C 0° C 10 cm 

(i) A 

4.0  cm2  B 50 W m–1 K–1 

400 W m–1 K–1 (i) 

(A) 0.5 cm2 

(B) (i) 1.0 × 103 J 50 sec

(C) (ii) 

15 J/sec

(D) (ii) 

10 J/sec

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–II : ( : 16)

-I -II

-I (A),  (B),  (C)    (D)  -II (P),  (Q),  (R),  (S)    (T)  -I -II -I -II

4  ×  5    :

(A)  (P) (Q) (R) (S) (T)

(B)  (P) (Q) (R) (S) (T)

(C)  (P) (Q) (R) (S) (T)

(D)  (P) (Q) (R) (S) (T)

-I -I (A) (Q),  (R)  (T)  (B),  (C)    (D) 

  :

-I +2 0 –1

1. 1 kg 10 kg 10 kg 15° 10.8 100°C r (LV = 540 cal/gm, Lf = 80 cal/gm,  =1 cal/gm°C.)

I r II I II

(A) r = 1.6 gm/sec (P) (B) r = 1.35 gm/sec (Q) (C) r = 1.2 gm/sec (R) 10 kg (D) r = 1 gm/sec (S) 10 kg 

(T)

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2. (Phase space diagrams) x(t) vs. p(t)(b)  

Mom

entu

mPosition

Figure (a)

(b)P ) E 

a bc

d

P

+–

      

P

O

D

=– =+

+–

Figure (b)

-I -II

(A) a (P) E < 2 mg

(B)  b (Q) E  2mg

(C)  c (R) (D)  d (S)

(T)    

– III :

III

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–IV : ( : 32)

0 9 

  :

+4

0

1. M µ F = kt 

t = t2 N N

320

 ( : M = 20 kg, µ = 0.4, k = 40 N/s, t2 = 3 sec.)

F

2. R M P 

m B M

m

 

B N 2N

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3. A Mariotte vessel) 

h2  = 2cm F 

h1 = 10cm F (m/s 

1N

10 N 

Fh2

h1

A

a

4. 20°C 1% 

(°C ) 

2.30 × 10–5/°C 3.19×10–4/°C N N

20

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5. 0.3 s 

0.4  s 

nwj h (cm ) n 4n  (g = 2)

6. 2 = 6m 

2L

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7.

52 cm 50 cm 

325 m/s 

(Hz ) 4 sec n 

n  

8. A B 8 × 108 Pa 3 × 108 Pa 

A  B 1 mm2 2mm2

(N )X

X

250

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0000CT100115002

-2 : –I : ( : 40)

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

:

+4

0

–2

1. (= 20gm/mol) 300K 400K 400 500 m/sec

[R = 25/3 J/mol/K]

(A) (B)

(C) (D)

2. 0.1M AgNO3 NaX(aq.), MgY2(aq.), AlZ3(aq.)

X, Y, Z

( )

: X = 120

Y = 80

Z = 60

AgNO3

(A) X (B) Y (C) Z (D)

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H-17/360000CT100115002

3. s- , 4r2r2 vs r

: S =

P = ()

4 r 2 2

r

r

s1 s2 s3 s4

p1

p2

p3

p4

s-orbitalP6P5

(A) P4 > P3 > P2 > P1

(B) S4 > S3 > S2 > S1

(C) P6 = P5 = P4

(D) P4 = P3 = P2 = P1

4. 1 mol KHC2O4. H2C2O4

(A) 1 mol KOH

(B) 3/2 moles Ba(OH)2

(C) 4/5 mol KMnO4

(D) 2/3 mol K2Cr2O7

5.

(A) dxy

,dyz

(B) 2 2 xyx yd , d

(C) 2xz zd , d (D) 2 2 2x y z

d , d

6.

(A) Li, Na (B) Be, B (C) N, As (D) O, At

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0000CT100115002

7. C4H

4O

4

(A)

COOH

COOH

(B)

OHO O

OH

(C) HOOC–CC–COOH (D)

O

HO

O

OH

8. (II) (I)

(A) (B)

(C) (D)

9.

(A) HCCH + NaOH (B)

OH

COOH

+ NaHCO3

(C)

OH

+ CH –C–ONa3

O– + (D)

C

+ NaOHH

10. (II) (I)

(A) NH ,NH2

(B) ,–

(C)+

,

+

(D)

OH

,

O–

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H-19/360000CT100115002

–II : ( : 16)

-I -II

-I (A), (B), (C) (D) -II (P), (Q), (R), (S) (T)

-I -II

-I -II

4 × 5 :

(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)

(D) (P) (Q) (R) (S) (T)

-I

-I (A) (Q), (R) (T)

(B), (C) (D)

:

-I

+2

0

–1

1. -I -II

(A) H2O(l) H

2O(g) (P)

H2O(l)

(B) I3

–(aq) I2(aq.) + I–(aq.) (Q)

H2O(l)

(C) 2AB(g) A2(g) + B

2(g) (R)

(D) A2(g) 2A(g) (S)

(T)

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0000CT100115002

2. -I -II

(A) SO3

(P)

(B) SO2

(Q) d-p

(C) CO2

(R)

(D) NO2

(S)

(T)

– III :

III

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CHEM

ISTRY

H-21/360000CT100115002

–IV : ( : 32)

0 9

:

+4

0

1. 9 nm (Å )

[ h = 6 × 10–34 J-sec , m = 9 × 10–31 kg].

2. 300K 1 atm 2

: b = 0.05 litre/mol

R = 0.08 atm

Z = 1.5

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H-22/36

CHEM

ISTRY

ALL INDIA OPEN TEST/NURTURE COURSE/JEE (Advanced)/07-02-2016/PAPER-1

0000CT100115002

3. -D-

C6H12O6(aq.) + IO4– (aq.) HCOOH(aq.) + HCHO(aq.) + IO3

– (aq.)

-D-1ml 80 ml 0.25 M

H2O2

0.1M NaOH

40 mL -D-

4. 6 CuSO4 10 Na2S2O3

Na4[Cu6(S2O3)5]

CuSO4 + Na2S2O3 CuS2O3 + Na2SO4

2CuS2O3 + Na2S2O3 Cu2S2O3 + Na2S4O6

3Cu2S2O3 + 2Na2S2O3 Na4 [Cu6(S2O3)5]

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CHEM

ISTRY

H-23/360000CT100115002

5. XOn(OH)

m (n + m) 5

7

6. sp3 sp2

N N

PP

N Cl

Cl

PClCl

Cl

Cl

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H-24/36

CHEM

ISTRY

ALL INDIA OPEN TEST/NURTURE COURSE/JEE (Advanced)/07-02-2016/PAPER-1

0000CT100115002

7.

(i) (ii) (iii) (iv)

(v) (vi) C=C=CCH3

HH

H C3(vii) (viii)

Br

ICl

F

8.

(i)

O

(ii)

O

(iii)

O

OH

(iv)

O

(v)

O

O

(vi)

O

(vii)

O

(viii)

O

(ix)

O

(x) H

O

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MATHEM

ATICS

H-25/360000CT100115002

-3 : –I : ( : 40)

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

:

+4

0

–2

1. n

4

1x n 1, n N

2 x

x

-

(A) x 3

(B) 10

(C) x 256

(D) 35

8

2. S

1 : x2 + y2 + 24x – 10y + a = 0

S2 : x2 + y2 = 36

(A) a S1 = 0 170

(B) S1 = 0 S

2 = 0a 49

(C) S1 = 0 S

2 = 0 a = 36

(D) a = 0 S1 = 0 S

2 = 0 3

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H-26/36

MATHEM

ATICS

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0000CT100115002

3. sinx + siny = 96

65 cosx + cosy =

72

65 -

(A) 24

sin x y25

(B) x y 12

cos2 13

(C) x y 4

tan2 3

(D)

7cos x y

25

4.

2014

2014k

k 0

S C .k S 19 E -

(A) E 2 2014

(B) E 4030

(C) (E + 1)

(D) (a – 2b + 3c)2016 E

5. 240 160 200 3206x 27x

x 1 x 4 x 5 x 8

(A) 3

(B) E E 2904

(C) S S 81

(D)

R

2 1log 2 1

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MATHEM

ATICS

H-27/360000CT100115002

6.

63 3

3 36 3 3

cos x cosec x 2 2 sin x1 3

2tan x cot xcot x 2 2 cos x sec x 2 2

?

(A) [0,4] 8

(B) [0,4] 16

(C) ƒ : A B A [0,3] B

[0,5] 106

(D) 13

7. 3 3 32 2 2

1ƒ x

x 2x 1 x 1 x 2x 1

E = ƒ(1) + ƒ(3) + ƒ(5) +........+ ƒ(999) -

(A) E (B) E 5

(C)

7E

115

T

2 T

3

(D) k = 3 ECk

8. ƒ : (0, ) R, | nx|e , 0 x 3

ƒ xx 6 , x 3

?

(A) a,b,c d ƒ(a) = ƒ(b) = ƒ(c) = ƒ(d) abcd

7

(B) ƒ(x)

(C) x = 1, x = 3, y = ƒ(x) x-4

(D) 2 ƒ(x) (7,1) x2 + y2 – 16x + 62 = 0

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MATHEM

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0000CT100115002

9. ?

(A) ƒ(x) = log((x2 + 1)1/2 + x) + sinx ƒ(a + 2) = 5 ƒ(b – 7) = –5 (a + b) 5

(B) 2

2

1ƒ x sin x x

1 x

([.] ) ƒ(x)

(C) 2015 4030 2015ƒ x 1 x x 1 2015ƒ x 1 1

(D) 1 11sin tan 2x

x

2

10. A,B,C ABC 5 A 5 B

D , E32 32

5 CF

32

(D,E,F n

2

n I)

(A) cotD cotE + cotE cotF + cotD cotF = 1

(B) cotD + cotE + cotF = cotD.cotE.cotF

(C) tanD tanE + tanE tanF + tanF tanD = 1

(D) tanD + tanE + tanF = tanD tanE tanF

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MATHEM

ATICS

H-29/360000CT100115002

–II : ( : 16)

-I -II

-I (A), (B), (C) (D)

-II (P), (Q), (R), (S) (T)

-I -II

-I -II

4 × 5 :

(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)

(D) (P) (Q) (R) (S) (T)

-I

-I (A) (Q), (R) (T)

(B), (C) (D)

:

-I

+2 0

–1

1. -I -II

(A) n

n 1

n 0

2

(P) 1

(B) [0,4] 3 316sin x 14 sin x 7 (Q) 2

(C) [0,10] (R) 3

tan–1(2 sinx) = cot–1(cosx)

(D) (S) 4

(–a + 1)x2 + (a2 – a + 4)x – 2a + 3 = 0

a (a > 1) (T) 5

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0000CT100115002

2. -I -II

(A)

1

n 0

2tan

n 2 n n 2 n n n 2

(P) 1

([.] )

(B) a,b,c 2 2 22 a b c

ab bc ca

(Q) 2

(C) a,b,c (R) 3

a2(b + c) = b2(a + c) = 2 c2(a + b)

(D) P = sinA sinB, Q = sinC cosA, R = sinA cosB, (S) 4

S = cosAcosC 5(P2+Q2 + R2 + S2) (T) 5

– III : III

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MATHEM

ATICS

H-31/360000CT100115002

–IV : ( : 32)

0 9

:

+4

0

1. [sin(sin–1(0.5) + sin–1(0.4)) × sin(sin–1(0.5) – sin–1(0.4))] p

q

(p, q N ), q p

13

2. n Sn, n

2 2k

k 1

2k 1 a

a1,a2,....,an

17 n

Sn n

2

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MATHEM

ATICS

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0000CT100115002

3. ABC, B P PA = 10, PB = 6

APB = BPC = CPA PC

11

4.

22015n

n 1

n n 1 b1 a

n ! c!

(a,b,c N)

a b c

576

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MATHEM

ATICS

H-33/360000CT100115002

5. 6 A,B,C,D,E,F ()

A, B B, C (),

2n

n n (n N)

6. x y n n

2

sin cos

x y

4 4

4 4 3 3

cos sin 97sin 2

x y x y xy

x y

y x

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0000CT100115002

7. C1 C

2 2 3 C

1 C

2

C3 A B C

3 A B T

TA = 4 C3

8. n n

n mm p

p 1 m p

C . C 19

n

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H-35/360000CT100115002

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

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

Your Hard Work Leads to Strong Foundation 0000CT100115002H-36/36

:

15.

16.

17.

18. :

19.

20.

21. g = 10 m/s2

____________________________

____________________________

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