JEE (Main + Advanced) : ENTHUSIAST COURSE

READ THE INSTRUCTIONS CAREFULLY

GENERAL :

1. This sealed booklet is your Question Paper. Do not break the seal till you are told

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, form number and sign in the space provided on the back cover of this

booklet.

5. After breaking the seal of the booklet, verify that the booklet contains 28 pages and

that all the 18 questions in each subject and along with the options are legible. If not,

contact the invigilator for replacement of the booklet.

6. You are allowed to take away the Question Paper at the end of the examination.

OPTICAL RESPONSE SHEET :

7. The ORS will be collected by the invigilator at the end of the examination.

8. Do not tamper with or mutilate the ORS. Do not use the ORS for rough work.

9. Write your name, form number and sign with pen in the space provided for this purpose

on the ORS. Do not write any of these details anywhere else on the ORS. Darken

the appropriate bubble under each digit of your form number.

DARKENING THE BUBBLES ON THE ORS :

10. Use a BLACK BALL POINT PEN to darken the bubbles on the ORS.

11. Darken the bubble COMPLETELY.

12. The correct way of darkening a bubble is as :

13. The ORS is machine-gradable. Ensure that the bubbles are darkened in the correct

way.

14. Darken the bubbles ONLY IF you are sure of the answer. There is NO WAY to erase

or "un-darken" a darkened bubble.

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

PAPER – 2

Test Type : MINOR Test Pattern : JEE-Advanced

TEST DATE : 26 - 06 - 2016

PHASE : I

JEE (Main + Advanced) : ENTHUSIAST COURSE

Paper Code : 1001CT102116010

CLASSROOM CONTACT PROGRAMME(Academic Session : 2016 - 2017)

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

Time : 3 Hours Maximum Marks : 186

EN

GL

ISH

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

E-2/28

Target : JEE (Main + Advanced) 2017/26-06-2016/Paper-2

1001CT102116010

Space for Rough Work

SOME USEFUL CONSTANTS

Atomic No. H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16,

Cl = 17, Br = 35, Xe = 54, Ce = 58,

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

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

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

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

Coulomb's law constant

9

0

1= 9×10

4

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

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

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

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

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

Permittivity of vacuum 0

= 2

0

1

c

Planck constant h = 6.63 × 10–34 J–s

Enthusiast Course/Phase-I/26-06-2016/Paper-2

PHYSICS

E-3/281001CT102116010

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS


PART-1 : PHYSICS

SECTION–I(i) : (Maximum Marks : 18)

This section contains SIX questions.

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

is correct.

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

For each question, marks will be awarded in one of the following categories :

Full Marks : +3 If only the bubble corresponding to the correct option is darkened.

Zero Marks : 0 If none of the bubbles is darkened.

Negative Marks : –1 In all other cases

1. A galvanometer of ig = 1mA & resistance = 10 is connected in a circuit as shown. This

combination (circuit) can be used as an ammeter or a voltmeter also. The range of ammeter(R

A) and voltmeter (R

v) are :-

5

40

80

G

(A) 9

mA8

, 15 mV (B) 1 mA, 12 mV (C) 8

mA3

, 20 mV (D) 4

mA3

, 10 mV

2. Three capacitors are connected as shown in the figure, where C1 = 3µF, C

2 = 6µF, C

3 = 2µF

and E = 10 V. Then the

C1 C2

C3

E

(1) Energy drawn from the source is 4 × 10–4 J

(2) Energy drawn from the source is 2 × 10–4 J

(3) Heat generated in the system during charging is 2 × 10–4 J

(4) Charges on the three capacitors are equal

Select most appropriate combination :-

(A) 1 & 3 (B) 2 & 3 (C) 1, 3 & 4 (D) 1 only

E-4/28

PHYSICS


1001CT102116010

3. A wire has resistance of 24 is bent in the following shape. The effective resistance between

A and B is-

60°

60°A

5cm10cm

B

(A) 24 (B) 10 (C) 163 (D) None of these

4. A standing wave with a frequency of 1100 Hz in a column of methane at 27.0°C produces

nodes that are 0.200 m apart. What is the approximate value of for methane?

(The molar mass of methane is 16.0 g/mol.)

(A) 7/5 (B) 4/3 (C) 6/5 (D) 5/3



PHYSICS

E-5/281001CT102116010

5. Consider a tug-of-war between two people who pull in opposite directions on the ends of a rope.

The first man has mass of 50 kg and friction coefficient between him and ground is 0.5. The

second man has mass of 60 kg and friction coefficient between him and ground is 0.4. Both are

equally powerful, that means that they can both exert the same force on the rope.

(A) Nobody can win because both exert the same force on each other.

(B) The first man will win because the ground exerts more limiting friction on him.

(C) The second man will win because he has more mass and hence more inertia.

(D) The first man will win because he can have more acceleration for same amount of force.

6. In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km

and the other at 64,000 km. Pluto already was known to have a large satellite Charon, orbiting

at 19,200 km with an orbital period of 8 days. Assuming that the satellites do not affect each

other, find the orbital periods of the two small satellites (in days) :-

(A) 8 × (2.5)3/2,

3/210

83

(B) 8 × 23/2 , 8 × 33/2

(C)

3/2 3/25 5

8 ,84 3

(D) None of these


E-6/28

PHYSICS


1001CT102116010

SECTION–I(ii) : (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


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

darkened.

Partial Marks : +1 For darkening a bubble corresponding to each correct option,

Provided NO incorrect option is darkened.


Negative Marks : –2 In all other cases.

for example, if (A), (C) and (D) are all the correct options for a question, darkening all these

three will result in +4 marks; darkening only (A) and (D) will result in +2 marks; and

darkening (A) and (B) will result in –2 marks, as a wrong option is also darkened

7. A capacitor of capacity C0 is connected to a battery of emf V

0. When steady state is attained a

dielectric slab of dielectric constant K is slowly introduced in the capacitor. Mark the Correct

statement(s), in final steady state. (Battery remains connected at all time) :-

(A) Magnitude of induced charge on the each surface of slab is C0V

0 (K – 1)

(B) Net electric force on one of the plate due to induced charges on dielectric is zero.

(C) Force of attraction between plates of capacitor is

2

0 0

0

K C V

2 A

(D) Net field due to induced charges in dielectric slab is

2

0

0

8V k 1

K A



PHYSICS

E-7/281001CT102116010

8. Pick the correct statements :-

(A) If a point charge is placed off-centre inside an electrically neutral spherical metal shell then

induced charge on its inner surface is uniformly distributed.

(B) If a point charge is placed off-centre inside an electrically neutral, isolated spherical metal

shell, then induced charge on its outer surface is uniformly distributed.

(C) A non metal spherical shell of uniform charge attracts or repels a charged particle that is

outside the shell such as all the shell's charge were concentrated at the centre of the

shell.

(D) If a charged particle is located inside a non metal spherical shell of uniform charge, there

is no electrostatic force on the particle due to the shell.

9. In a conductor of resistance R charge flows as Q = at – bt2. Then consider t from t = 0 to time

till current becames zero.

(A) i = 0 at a

tb

in the resistance

(B) Power dissipated is maximum at t = 0 in the resistance

(C) voltage across the resistance is maximum at a

t4b

(D) Power dissipated is maximum at a

t2b

in the resistance


E-8/28

PHYSICS


1001CT102116010

10. In each of the figures shown below, friction is negligible, mass of the strings is also negligible

and the pulleys are massless. The horizontal acceleration of M1 is g.

M1

M2

(i)

M1

M2

(ii)

(A) This is possible in case (i) if M2 is 4M

1. (B) This is possible in case (ii) if M

2 is 4M

1.

(C) This is not possible in case (i) (D) This is not possible in case (ii)

11. Two plates of mass m each are connected to mass M by two light inextensible strings. Thepulleys are ideal. To hold system in equilibrium, we need to charge the plates by a battery of atleast 8 volts. If distance between plates is 1µm, capacitance formed is 1µF, (g = 10 m/s2)

M

m

md

(A) M = 12.8 kg (B) m = 3.2 kg (C) M = 6.4 kg (D) m = 1.6 kg



PHYSICS

E-9/281001CT102116010

12. While making this paper, I stayed up till late at night working with candle light. I had two

candles of equal length L lit and placed at the same time, as shown in the figure. Immediately

thereafter I noticed that the shadow of the first candle on the left wall is not changing in

length, and the shadow of the second candle on the right wall is being shortened at a speed v.

(Take d1 = 10 cm, d

2 = 20 cm, d

3 = 30 cm, v = 1 mm/s, L = 20 cm.)

d1 d2 d3

1 2

(A) The left candle will be burning for longer time than right candle

(B) The candle to first extinguish will be extinguished at time t = 400 sec.

(C) The candle to extinguish second will be extinguished at time t = 800 sec.

(D) The right candle is decreasing in length at a faster rate.

13. A racing car has a siren and is speeding with the speed u = 306 km / h. The frequency of the

siren is 0 = 40 Hz. Ahead of it there is another car with exactly the same siren (this car is at

rest). Each of the drivers distinguishes the sound of siren of another car, because it is of

higher frequency than the sound of his own siren. (Take speed of sound = 340 m/s)

(A) The beat frequency heard by the speeding driver is 13.33 Hz

(B) The beat frequency heard by the speeding driver is 10 Hz

(C) The beat frequency heard by the driver at rest is 13.33 Hz

(D) The beat frequency heard by the driver at rest is 10 Hz


E-10/28

PHYSICS


1001CT102116010

14. At the equator on a spherical planet a body weighs one third of that at the pole. Radius of planet

is R. The time period of revolution of the planet around its axis is equal to T = 60 min.

(Take : 1120G 10

3 Nm2/kg2)

(A) The average density of the planet is 610192

kg/m3

(B) The height of geostationary satellite from the surface of planet is 3

31 R

2

(C) At a latitude of 60°, the weight of the body is half the weight at pole

(D) The geostationary satellite can be in a plane passing through the poles.



PHYSICS

E-11/281001CT102116010

SECTION–I(iii) : (Maximum Marks : 12)

This section contains TWO paragraphs.

Based on each paragraph, there are TWO questions.

Each question has FOUR options (A), (B), (C) and (D) ONLY ONE of these four options

is correct.



Full Marks : +3 If only the bubble corresponding to the correct answer is darkened.

Zero Marks : 0 In all other cases.

Paragraph for Questions 15 and 16

A potentiometer can also be used to calibrate an ammeter or a voltmeter. Here we require a

standard cell of which emf is known say E0. This cell is connected to secondary branch of

potentiometer and we balance this standard cell against a length 0 on potentiometer wire.

(by connecting 1 & 2). Now standard cell is removed and the galvanometer end 2 is connected

to point 3 on circuit and balanced against length 1. Current I can be obtained using principle

of potentiometer as well as from ammeter. I as given by potentiometer is true value and

ammeter reading may have error.

E0

I

1

32

G

E K

R

A BJ

A

+–

15. If resistance R is 15, E0 = 5V,

0 = 2.5m,

1 = 3m, current I as read by ammeter is 0.45 A

(A) The ammeter reads more than true value by 0.05 A

(B) The ammeter reads less than true value by 0.01A

(C) The ammeter reads less than true value by 0.05A

(D) The ammeter reads more than true value by 0.01 A

16. If the same error is obtained for another circuit where E0 = 10V,

0 = 3m,

1 = 1.5m, current I

in ammeter is 0.55A, the resistance R used is :

(A) 10 (B) 8 (C) 100

9 (D) 8.1


E-12/28

PHYSICS


1001CT102116010


An instructive example of a stationary current distribution occurs in the plane diode, a two-electrode

vacuum tube. One electrode, the cathode, is coated with a material that emits electrons when

heated. The other electrode, the anode, is simply a metal plate. By means of a battery the anode is

maintained at a positive potential with respect to the cathode. Electrons emerge from this hot

cathode with negligible velocities and then, being negatively charged, are accelerated toward the

positive anode by the electric field between cathode and anode. In the space between the cathode

and anode the electric current consists of these moving electrons. The circuit is completed by the

flow of electrons in external wires, possibly by the movement of ions in a battery, and so on, with

which we are not here concerned. In this diode the local density of charge in any region, , is

simply ne, where n is the local density of electrons, in electrons per cubic meter. The local current

density J is v, where v is the velocity of electrons in that region. In the plane-parallel diode we

may assume J has no y or z components. If conditions are steady, it follows then that J must be

independent of x. This is obvious; if we have a steady stream of electrons moving in the x direction

only, the same number per second have to cross any intermediate plane between cathode and

anode.

v

E

x

y

Anode

Cathode

Heater

But observe that v is not constant; it varies with x because the electrons are accelerated by

the field. Hence is not constant either. Instead, the negative charge density is high near the

cathode and low near the anode, just as the density of cars on an expressway is high near a

traffic slowdown and low where traffic is moving at high speed. Assume cathode to be at zero

potential.



PHYSICS

E-13/281001CT102116010

17. The potential at any point between the plates is V. The variation of V with x can be represented

by the differential equation: (C is constant)

(A)

2

2

d V C

dx V (B)

2

2

d VC V

dx

(C) 2

2

d VCV

dx (D)

2

2

d V C

Vdx

18. The constant current I mentioned in the paragraph depends on the applied potential

difference V0. This relation between current I and V

0 is given by: K is constant

(A) I = KV03/2 (B) I = KV

01/2 (C) I = KV

0–1/2 (D) I = KV

0–3/2


E-14/28

CHEM

ISTRY


1001CT102116010


PART-2 : CHEMISTRY

SECTION–I(i) : (Maximum Marks : 18)


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

is correct.






1. In a structure of a solid, atom A form fcc structure and atoms B are present in all octahedral

void. Now one atom of A is removed from one face center of unit cell then simplest formula

of compound will be -

(A) AB (B) A7B

8(C) A

5B

6(D) A

8B

7

2. Using IUPAC norms, the systematic name of the complex [Co(NH3)4Cl(NO

2)]Cl is :

(A) Tetraminenitrito-O-chloridocobalt(III) chloride

(B) Tetraamminenitrito-N-chloridocobalt(III) chloride

(C) Tetraamminechloridonitrito-N-cobalt(III) chloride

(D) Chloridonitrito-O-tetraamminecobalt(III) chloride

3. Which of the following has INCORRECT order of their boiling point ?

(A) H3BO

3 > B(OCH

3)

3(B) HF > HCl

(C) ICl > Br2

(D) NH3 > SbH

3

4. Which of following can be prepared by single organic reactant by wurtz reaction in good

yield :

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


CHEM

ISTRY

E-15/281001CT102116010

5. Which of following compound is most reactive towards SN1 reaction :

(A)

O Cl(B)

Cl

CH3

(C)

C1

(D) Cl

6. Which of following reaction(s) produce Saytzeff product as a major product :

(A)

F

NH2

–

(B)

Cl

alc. KOH /

(C)

Cl

Me CO K3– +

(D)

NMe3

OR–

+


E-16/28

CHEM

ISTRY


1001CT102116010




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



darkened.






three will result in +4 marks; darkening only (A) and (D) will result in +2 marks; and

darkening (A) and (B) will result in –2 marks, as a wrong option is also darkened

7. If 0formation 4H [CH (g)] = –100 kJ/mol

0formation 2H [CO (g)] = –400 kJ/mol

0combustion 2H [H (g)] = –300 kJ/mol

0combustion 4 10H [C H (g)] = –3000 kJ/mol

then, which of the following is/are true -

(A) 0formation 4 10H [C H (g)] = –100 kJ/mol (B) 0

combustion 4H [CH ( )] g = –900 kJ/mol

(C) 0formation 2H [H O( )] l –300 kJ/mol (D)

0combustion (diamond)H [C ] = –400 kJ/mol


CHEM

ISTRY

E-17/281001CT102116010

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

No question will be asked in section II and III


8. For a chemical reaction at 300K

R(g) P(g)

Hº = – 20 kJ ; Sº = – 100 J/K

Select the correct statement at 300 K

(A) R(g) will convert into P(g) if a container contains 1 mol R(g) and no P(g) at 300 K

(B) R(g) will not convert into P(g) if a container contains 1 mol R(g) and no P(g) at 300 K

(C) Gº is positive

(D) Equilibrium constant is more than one

9. For vander waal's gas equation

2

2

anP (V nb) nRT

V

(A) At very high pressure PV

nRT > 1

(B) If PV

nRT < 1 , repulsive forces dominates

(C) at low pressure z = 1 – m

a

V RT

(D) 'a' is measure of attractive forces among molecules

10. Assuming that if Hund's rule is violated, then the diamagnetic specie(s) is/are :

(A) B2

(B) O2

(C) NO

(D) O2

11. Which of the following compounds do not give free halogen acid (Hydra acid) on hydrolysis

with excess water as a final product ?

(A) NCl3

(B) PCl3

(C) SiCl4

(D) BF3

E-18/28

CHEM

ISTRY


1001CT102116010


12. Which of the following will produce immediate turbidity with Lucas reagent ?

(A)

OH

(B)

OH

(C) C

OH

CH3H3C

CH3

(D) CH3–OH

13. For the following reaction, choose the incorrect option(s) ?

CH –CH2 3

CH –OH2H

CH3

HCl (P)

( )-2-methylbutane-1-ol–

(A) Formation of major product (P) involves carbocation rearrangement

(B) Major product (P) is 2-methyl but-2-ene

(C) Reaction follows bimolecular nucleophilic substitution

(D) Major product (P) is 2-chloro-2-methyl butane

14. Which of following reactions are shown with incorrect major product :

(A) CH3–C—

CH3

CH3

ONa + CH3–Cl CH

3–C—

CH3

CH3

O–CH3

(B)

OAc

(C)

Me

Cl

H

D KCN

Me

CN

H

DDMF

(D) O18 H O3

+

OH +18

OH


CHEM

ISTRY

E-19/281001CT102116010






is correct.






One mole of an ideal monoatomic gas is processed in process ABCD then answer the following

question (Given : R = 2 cal./mol-K)

200K 300K T(Kelvin)

A

D

B C( )l –21V

2

15. Entropy change for gas in process BC

(A) 3

3ln cal.2

/K (B) 3 3

ln cal.2 2

/K (C) 3

4ln cal.2

/K (D) 5 3

ln cal.2 2

/K

16. Heat change in complete process ABCD ?

(A) 200 cal (B) 50 cal (C) 1200 cal (D) 100 cal

E-20/28

CHEM

ISTRY


1001CT102116010


Qualitatively, the formation of molecular orbitals can be understood in terms of the constructive

or destructive interference of the electron waves of the combining atoms. In the formation of

bonding molecular orbital, the two electron waves of the bonding atoms reinforce each other due

to constructive interference while in the formation of antibonding molecular orbital, the electron

waves cancel each other due to destructive interference.

17. Which of the following combinations give(s) antibonding sigma molecular orbital if z-axis

is the internuclear axis ?

(A)

2pz 2pz

–– + –+ (B)

2px 2px

+–

+

–

+

(C)

2pz 2pz

+– + –+ (D)

1s 1s

+ ++

18. Which of the following specie does not exist :-

(A) HeH+ (B) Be2

(C) C2

2– (D) NO+



MATHEM

ATICS

E-21/281001CT102116010


PART-3 : MATHEMATICSSECTION–I(i) : (Maximum Marks : 18)


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

correct.






1. The value of

k

3 2k 00

1 1 cos4xlim dx

2k x 2 is equal to -

(A) 0 (B) 1

3(C)

2

3(D) 1

2. If

x

2

a cos x bx sin x ce 2x; x 0

ƒ(x) x

0 ; x 0

is continuous at x = 0, then

(A) a + b + c = 2 (B) ab = –4

(C) a + c = 4 (D) a – b + c = 2

3. If number of points of non-derivability of ƒ(x) = (x – 1)(x – 2)|(x – 1)(x – 2)(x – 3).....(x – m)|

(where m N) in (–, ) are 9, then m is equal to -

(A) 7 (B) 8 (C) 10 (D) 11

4. If ƒ() = sin(sin) and ƒ''() + tanƒ'() + g() = 0, then g() is equal to -

(A) cos2sin(cos) (B) cos

2sin(sin)

(C) sin2cos(sin) (D) sin

2cos(cos)

5.3/ 4

2

1/ 4

nxdx

n(x x )

is equal to -

(A) 1

2(B)

1

4(C)

1

6(D)

1

8

6. If

n 2 2 2n

2n 4n 6n 3A lim ........

51 n 4 n 9 n, then nAe is equal to -

(A) n4 (B) 10 (C) n10 (D) 4

E-22/28

MATHEM

ATICS


1001CT102116010



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



darkened.






three will result in +4 marks; darkening only (A) and (D) will result in +2 marks; and darkening

(A) and (B) will result in –2 marks, as a wrong option is also darkened

7. If y = tan–1

(secx), then 1x cosec 2

dy a

dx b

, where a and b are coprime, then -

(A) a + b = 1 (B) a + 2b = 6 (C) 2a 3

b 4 (D) a2 + b2 = 13

8. If ƒ( ) 3cos sin 2cos cos where ,2

, then

(A)

/ 2

ƒ( )d 1 (B)

2

/2

ƒ ( )d4

(C) 3 1

ƒ '4 2

(D)

/2

3ƒ( )d

2



MATHEM

ATICS

E-23/281001CT102116010

9. Let ƒ(x – 3) = ƒ(x + 3) for all x R and x ; 0 x 3

ƒ(x)6 x ; 3 x 6

, then

(A) 36

0

ƒ(x)dx 54 (B) 48

0

ƒ(x)dx 64

(C) 54

0

ƒ(x)dx 81 (D) 18

0

ƒ(x)dx 9

10. Let ƒ be a derivable function on R and satisfying x

2 t

0

ƒ(x) x e ƒ(x t)dt , then

(A) 1

0

5ƒ(x)dx

6 (B)

1

0

5ƒ(x)dx

12

(C) 1

1

2ƒ(x)dx

3

(D) 4

ƒ(1)3

11. If 4 2

2

x x 4y

x 3x 2

and

dyx

dx , then

(A) ( ) cot12

(B) 2 2 21

tan4

(C) 5

sec6

(D)

2cos

3


E-24/28

MATHEM

ATICS


1001CT102116010

12. Consider y3 + py

2x

3 + qyx

6 – rx

9 = 0, where p, q, r R. If possible values of

3

3

d y

dx are –6, 12 and

–18, then

(A) p + q + r = 9 (B) pq + r = 5 (C) p – q – r = 1 (D) p q r

43

13. The equation of straight line passing through (4, 3) and making intercepts on the co-ordinate

axes whose sum is –1, is

(A) x y

12 1

(B)

x y1

2 1

(C)

x y1

2 3 (D)

x y1

2 3

14. Which of the following hold(s) good ?

(A) ƒ(x) = [x](x – 1) is discontinuous at x = 1. [Note : [.] denotes greatest integer function]

(B) If ƒ(a) and ƒ(b) are of opposite sign, then there must exist atleast one solution of equation

ƒ(x) = 0 in (a, b).

(C) If ƒ : R R and ƒ is an even degree polynomial, then ƒ(x) = 0 has atleast two real roots

if leading coefficient and absolute term are of opposite sign.

(D) The equation xex = 1 has exactly one positive root.



MATHEM

ATICS

E-25/281001CT102116010





is correct.






Let a, b, c, d be 4 distinct roots of the equation x4 – 4x + 3 = x(x

3 – ƒ'(1)x

2 + ƒ"(1)x – 4) + ƒ(1)

and ƒ(x) is a monic polynomial of degree 3.

15. The value of 3 3

5/3x 1

(x 1) sin (x 1)lim

(ƒ(x) 3)

is -

(A) 1

2(B) 2 (C)

1

6(D)

2

3

16.dx 1

g(x) Cƒ(x) 3 2

4x 1

, where C is constant of integration and g(3)4

, then the value of

g(5) + g(7) is -

(A) 2

(B) (C)

5

4

(D)

3

4


E-26/28

MATHEM

ATICS


1001CT102116010



Let ƒ be a differentiable function on R satisfying ƒ(x y) ƒ(x) ƒ(y) 2

xy x,y R2 2

and ƒ'(0) = 3.

17. The number of points where |ƒ(|x|)| is non-derivable

(A) 0 (B) 1 (C) 2 (D) 5

18.4 2

2

0

ƒ(x) x 2sin dx

3

is equal to

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


E-27/281001CT102116010



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 2017 1001CT102116010E-28/28

QUESTION PAPER FORMAT AND MARKING SCHEME :

16. The question paper has three parts : Physics, Chemistry and Mathematics.

17. Each part has one section as detailed in the following table.

I have read all the instructions

and shall abide by them.

____________________________

Signature of the Candidate

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

____________________________

Signature of the invigilator

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

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

Que. No. Category-wise Marks for Each Question Maximum

Section Type of Full Partial Zero Negative Marks of theQue. Marks Marks Marks Marks section

+3 0 –1 I(i) Single If only the bubble If none In all

correct option 6 corresponding to — of the other 18 the correct option bubbles is cases

is darkened darkened+4 +1 0 –2

One or more If only the bubble(s) For darkening a bubble If none In all I(ii) correct

8 corresponding corresponding to each of the other

32option(s) to all the correct correct option, provided bubbles is casesoption(s) is(are) NO incorrect option darkened

darkened darkenedParagraph +3 0

Based If only the bubble In all I(iii) (Single 4 corresponding to — other — 12

correct the correct option casesoption) is darkened

1.

2. (ORS)

3.

4.

5. 28 18

6.

7.

8.

9.

:

10.

11.

12. :

13.

14.

15. g = 10 m/s2

PAPER – 2

Test Type : MINOR Test Pattern : JEE-Advanced

TEST DATE : 26 - 06 - 2016

PHASE : I

JEE (Main + Advanced) : ENTHUSIAST COURSE

Paper Code : 1001CT102116010

CLASSROOM CONTACT PROGRAMME(Academic Session : 2016 - 2017)

Time : 3 Hours Maximum Marks : 186

HIN

DI

H-2/28


1001CT102116010

SOME USEFUL CONSTANTS

Atomic No. H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16,

Cl = 17, Br = 35, Xe = 54, Ce = 58,

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

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

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

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

Coulomb's law constant

9

0

1= 9×10

4

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

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

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

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

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

Permittivity of vacuum 0

= 2

0

1

c

Planck constant h = 6.63 × 10–34 J–s


PHYSICS

H-3/281001CT102116010

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS

-1 : –I(i) : ( : 18)

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

: +3

: 0

: –1

1. ig = 1mA 10

(RA)

(Rv)

5

40

80

G

(A) 9

mA8

, 15 mV (B) 1 mA, 12 mV (C) 8

mA3

, 20 mV (D) 4

mA3

, 10 mV

2. C1 = 3µF, C

2 = 6µF, C

3 = 2µF E = 10 V

C1 C2

C3

E

(1) 4 × 10–4 J (2) 2 × 10–4 J (3) 2 × 10–4 J (4) (A) 1 & 3 (B) 2 & 3 (C) 1, 3 & 4 (D) 1

H-4/28

PHYSICS


1001CT102116010

3. 24 A B -

60°

60°A

5cm10cm

B

(A) 24 (B) 10 (C) 163 (D)

4. 1100 Hz 27.0°C

0.200m (16.0 g/mol )

(A) 7/5 (B) 4/3 (C) 6/5 (D) 5/3


PHYSICS

H-5/281001CT102116010

5. 50 kg

0.5 60 kg

0.4

(A)

(B)

(C)

(D)

6. 2006 48,000 km 64,000 km

Charon 19,200 km 8

(A) 8 × (2.5)3/2,

3/210

83

(B) 8 × 23/2 , 8 × 33/2

(C)

3/2 3/25 5

8 ,84 3

(D)

H-6/28

PHYSICS


1001CT102116010

–I(ii) : ( : 32)

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

: +4

: +1

: 0

: –2

(A), (C) (D)

+4 (A) (D) +2

(A) (B) –2

7. C0 V

0

K

(A) C0V

0 (K – 1)

(B)

(C)

2

0 0

0

K C V

2 A

(D)

2

0

0

8V k 1

K A


PHYSICS

H-7/281001CT102116010

8.

(A)

(B)

(C)

(D)

9. R Q = at – bt2 t = 0 t

(A) a

tb

i = 0

(B) t = 0

(C) a

t4b

(D) a

t2b

H-8/28

PHYSICS


1001CT102116010

10. M1

g

M1

M2

(i)

M1

M2

(ii)

(A) (i) M2 4M

1 (B) (ii) M

2 4M

1

(C) (i) (D) (ii) 11. m M

8 1µm 1µF (g = 10 m/s2)

M

m

md

(A) M = 12.8 kg (B) m = 3.2 kg (C) M = 6.4 kg (D) m = 1.6 kg


PHYSICS

H-9/281001CT102116010

12. L v ( d

1 = 10 cm, d

2 = 20 cm, d

3 = 30 cm, v = 1 mm/s, L = 20 cm.)

d1 d2 d3

1 2

(A) (B) t = 400 sec (C) t = 800 sec (D)

13. u = 306 km/h 0 = 40 Hz

= 340 m/s)

(A) 13.33 Hz (B) 10 Hz (C) 13.33 Hz (D) 10 Hz

H-10/28

PHYSICS


1001CT102116010

14. R

T = 60 min ( : 1120G 10

3 Nm2/kg2)

(A) 610192

kg/m3

(B) 33

1 R2

(C) 60°

(D)


PHYSICS

H-11/281001CT102116010

–I(iii) : ( : 12)

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

: +3

: 0

15 16 E

0

0 1 2 )

2 3 1

I I

E0

I

1

32

G

E K

R

A BJ

A

+–

15. R 15, E0 = 5V,

0 = 2.5m,

1 = 3m I 0.45 A

(A) 0.05 A (B) 0.01 A (C) 0.05 A (D) 0.01 A

16. E0 = 10V,

0 = 3m,

1 = 1.5m,

I 0.55A R

(A) 10 (B) 8 (C) 100

9 (D) 8.1

H-12/28

PHYSICS


1001CT102116010

17 18

;

ne n

J v v

J y z

J x x

v

E

x

y

Anode

Cathode

Heater

v x


PHYSICS

H-13/281001CT102116010

17. V V x

(C )

(A)

2

2

d V C

dx V (B)

2

2

d VC V

dx

(C) 2

2

d VCV

dx (D)

2

2

d V C

Vdx

18. I V0 I V

0 K

(A) I = KV

03/2 (B) I = KV

01/2 (C) I = KV

0–1/2 (D) I = KV

0–3/2

H-14/28

CHEM

ISTRY


1001CT102116010

-2 : –I(i) : ( : 18)

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

: +3

: 0

: –1

1. A fcc B A (A) AB (B) A

7B

8(C) A

5B

6(D) A

8B

7

2. IUPAC [Co(NH3)4Cl(NO

2)]Cl :

(A) -O-(III)

(B) -N-(III)

(C) -N-(III)

(D) -O-(III)

3. ?

(A) H3BO

3 > B(OCH

3)3

(B) HF > HCl

(C) ICl > Br2

(D) NH3 > SbH

3


CHEM

ISTRY

H-15/281001CT102116010

4.

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

5. SN1

(A)

O Cl(B)

Cl

CH3

(C)

C1

(D) Cl

6.

(A)

F

NH2

–

(B)

Cl

alc. KOH /

(C)

Cl

Me CO K3– +

(D)

NMe3

OR–

+

H-16/28

CHEM

ISTRY


1001CT102116010

–I(ii) : ( : 32)

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

: +4

: +1

: 0

: –2

(A), (C) (D)

+4 (A) (D) +2

(A) (B) –2

7. 04H [CH (g)] = –100 kJ/mol

02H [CO (g)]

= –400 kJ/mol

02H [H (g)] = –300 kJ/mol

04 10H [C H (g)] = –3000 kJ/mol

(A) 04 10H [C H (g)] = –100 kJ/mol (B) 0

4H [CH ( )] g = –900 kJ/mol

(C) 02H [H O( )] l –300 kJ/mol (D)

0( )H [C ] = –400 kJ/mol

8. 300K

R(g) P(g)

Hº = – 20 kJ ; Sº = – 100 J/K

300 K (A) 300 K R(g) P(g) R(g) P(g)

(B) 300 K R(g) P(g) R(g) P(g)

(C) Gº (D)


CHEM

ISTRY

H-17/281001CT102116010

9. 2

2

anP (V nb) nRT

V

(A) PV

nRT > 1

(B) PV

nRT < 1 ,

(C) z = 1 – m

a

V RT

(D) 'a'

10. :

(A) B2

(B) O2

(C) NO (D) O2

11.

?

(A) NCl3

(B) PCl3

(C) SiCl4

(D) BF3

12.

(A)

OH

(B)

OH

(C) C

OH

CH3H3C

CH3

(D) CH3–OH

H-18/28

CHEM

ISTRY


1001CT102116010

13.

CH –CH2 3

CH –OH2H

CH3

HCl (P)

( )-2- -1-–

(A) (P)

(B) (P), 2--2-

(C)

(D) (P), 2--2-

14.

(A) CH3–C—

CH3

CH3

ONa + CH3–Cl CH

3–C—

CH3

CH3

O–CH3

(B)

OAc

(C)

Me

Cl

H

D KCN

Me

CN

H

DDMF

(D) O18 H O3

+

OH +18

OH


CHEM

ISTRY

H-19/281001CT102116010

–I(iii) : ( : 12)

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

: +3

: 0

15 16

ABCD : R = 2 cal./mol-K

200K 300K T(Kelvin)

A

D

B C( )l –21V

2

15. BC

(A) 3

3ln cal.2

/K (B) 3 3

ln cal.2 2

/K (C) 3

4ln cal.2

/K (D) 5 3

ln cal.2 2

/K

16. ABCD

(A) 200 cal (B) 50 cal (C) 1200 cal (D) 100 cal

H-20/28

CHEM

ISTRY


1001CT102116010

17 18

(constructive)

(destructive) (interference)

17. z- ?

(A)

2pz 2pz

–– + –+ (B)

2px 2px

+–

+

–

+

(C)

2pz 2pz

+– + –+ (D)

1s 1s

+ ++

18. :-

(A) HeH+ (B) Be2

(C) C2

2– (D) NO+


MATHEM

ATICS

H-21/281001CT102116010

-3 : –I(i) : ( : 18)

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

: +3 : 0 : –1

1.

k

3 2k 00

1 1 cos4xlim dx

2k x 2 -

(A) 0 (B) 1

3(C)

2

3(D) 1

2.

x

2

a cos x bx sin x ce 2x; x 0

ƒ(x) x

0 ; x 0

, x = 0

(A) a + b + c = 2 (B) ab = –4

(C) a + c = 4 (D) a – b + c = 2

3. (–, )ƒ(x) = (x – 1)(x – 2)|(x – 1)(x – 2)(x – 3).....(x – m)| (m N)

9 m -

(A) 7 (B) 8 (C) 10 (D) 11

4. ƒ() = sin(sin) ƒ''() + tanƒ'() + g() = 0 g() -

(A) cos2sin(cos) (B) cos

2sin(sin)

(C) sin2cos(sin) (D) sin

2cos(cos)

5.3/ 4

2

1/ 4

nxdx

n(x x )

-

(A) 1

2(B)

1

4(C)

1

6(D)

1

8

6.

n 2 2 2n

2n 4n 6n 3A lim ........

51 n 4 n 9 n nAe -

(A) n4 (B) 10 (C) n10 (D) 4

H-22/28

MATHEM

ATICS


1001CT102116010

–I(ii) : ( : 32)

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

: +4

: +1

: 0

: –2

(A), (C) (D)

+4 (A) (D) +2 (A)

(B) –2

7. y = tan–1(secx), then 1x cosec 2

dy a

dx b

a b -

(A) a + b = 1 (B) a + 2b = 6 (C) 2a 3

b 4 (D) a

2 + b

2 = 13

8. ƒ( ) 3cos sin 2cos cos , ,2

(A)

/ 2

ƒ( )d 1 (B)

2

/2

ƒ ( )d4

(C) 3 1

ƒ '4 2

(D)

/2

3ƒ( )d

2


MATHEM

ATICS

H-23/281001CT102116010

9. x R ƒ(x – 3) = ƒ(x + 3) x ; 0 x 3

ƒ(x)6 x ; 3 x 6

(A) 36

0

ƒ(x)dx 54 (B) 48

0

ƒ(x)dx 64

(C) 54

0

ƒ(x)dx 81 (D) 18

0

ƒ(x)dx 9

10. R ƒ x

2 t

0

ƒ(x) x e ƒ(x t)dt

(A) 1

0

5ƒ(x)dx

6 (B)

1

0

5ƒ(x)dx

12

(C) 1

1

2ƒ(x)dx

3

(D) 4

ƒ(1)3

11. 4 2

2

x x 4y

x 3x 2

dy

xdx

(A) ( ) cot12

(B) 2 2 21

tan4

(C) 5

sec6

(D)

2cos

3

H-24/28

MATHEM

ATICS


1001CT102116010

12. y3 + py

2x

3 + qyx

6 – rx

9 = 0 p, q, r R

3

3

d y

dx –6, 12 –18

(A) p + q + r = 9 (B) pq + r = 5

(C) p – q – r = 1 (D) p q r

43

13. (4, 3) –1

(A) x y

12 1

(B)

x y1

2 1

(C)

x y1

2 3 (D)

x y1

2 3

14.

(A) ƒ(x) = [x](x – 1), x = 1 [: [.] ]

(B) ƒ(a) ƒ(b) (a, b) ƒ(x) = 0

(C) ƒ : R R ƒ ƒ(x) = 0

(D) xex = 1


MATHEM

ATICS

H-25/281001CT102116010

–I(iii) : ( : 12)

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

: +3

: 0

15 16

a, b, c, d x4 – 4x + 3 = x(x

3 – ƒ'(1)x

2 + ƒ"(1)x – 4) + ƒ(1) ƒ(x)

15.3 3

5/3x 1

(x 1) sin (x 1)lim

(ƒ(x) 3)

-

(A) 1

2(B) 2 (C)

1

6(D)

2

3

16.dx 1

g(x) Cƒ(x) 3 2

4x 1

, C g(3)4

g(5) + g(7) -

(A) 2

(B) (C)

5

4

(D)

3

4

H-26/28

MATHEM

ATICS


1001CT102116010

17 18

R ƒ ƒ(x y) ƒ(x) ƒ(y) 2xy x,y R

2 2

ƒ'(0) = 3

17. |ƒ(|x|)| (A) 0 (B) 1 (C) 2 (D) 5

18.4 2

2

0

ƒ(x) x 2sin dx

3

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


H-27/281001CT102116010


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 2017 1001CT102116010H-28/28

16. 17.

____________________________

____________________________

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

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

I(i) +3 0 –1

6

—

18

+4 +1 0 –2

I(ii) 8 32

+3 0 I(iii)

4 — — 12

JEE (Main + Advanced) : ENTHUSIAST COURSE

Documents

Transcript of JEE (Main + Advanced) : ENTHUSIAST COURSE