JEE (Main + Advanced) : ENTHUSIAST COURSE...E-4/36 PHYSICS Target : JEE (Main + Advanced)...

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 36 pages and that

all the 20 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 – 1

Test Type : MINOR Test Pattern : JEE-Advanced

TEST DATE : 21 - 08 - 2016

PHASE : I

JEE (Main + Advanced) : ENTHUSIAST COURSE

Paper Code : 1001CT102116017

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 : 264

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Target : JEE (Main + Advanced) 2017/21-08-2016/Paper-1

1001CT102116017

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/21-08-2016/Paper-1

PHYSICS

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BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS


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

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

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

darkened.

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

Negative Marks : –2 In all other cases.

1. In uniform magnetic field B, a conducting resistance less rod OA is made to rotate with

uniform angular velocity on a fixed resistance less circular loop. The length of rod and

radius of loop both are ''. An external battrey having cmf E and internal resistance r is also

joined as shown in diagram. Which of the following is INCORRECT regarding potential

difference (V) across rod OA.

A

O

rE

(A) for E > 21

B2

, V > 21

B2

(B) for E = 21

B2

, V = 0

(C) for E < 21

B2

, V > 21

B2

(D) for any value of E, V can never be zero

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PHYSICS


1001CT102116017

2. The network has C1, C

2, C

3, C

4 four capacitors each of capacitance C = 1 µF, all resistance are

identical having r = 1 . The battery has emf E = 30 V. The circuit is analyzed at steady state.

Mark the CORRECT options.

r

r

r

r

r

C2

C3

C4

E

C1

(A) ratio of steady state charge of C1 and C2 is 1 : 1

(B) ratio of steady state charge of C2 and C4 is 1 : 1

(C) ratio of steady state charge of C1 and C

3 is 1 : 1

(D) ratio of steady state charge of C2 and C

3 is 1 : 1

3. A U shaped conducting wire frame having capacitor C is coplanar with an infinite wire having

current I. On frame, a wire PQ having resistance R makes sliding contact as shown. Frame

and wire are moving with speed V and 2V respectively as shown. (There is no friction

anywhere) Mark the CORRECT options.

IP Q

R

2V

V

Ca

2a

(A) Charge on capacitor at time t is 0 t / RCCµ Iv n2

q (1 e )2

(B) Charge on capacitor at time t is 0 t / RCCµ Iv n2

q (1 e )

(C) Current through resistor at time t is 0 t / RCµ Iv n2

i (e )2 R

(D) Current through capacitor at time t is 0 t / RCµ Iv n2

i (e )R



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4. Two long wires parallel to Z-axis, having current I each in same direction are at distance 4a

from each other. A cylinder of radius a and length L has its axis as Z-axis and centre the

origin of coordinates.If the net upward magnetic flux through half of the curved cylindrical

surface above X-Z plane be . Then mark the INCORRECT options.

L

(2a, 0, 0)

(–2a, 0, 0)

I

I

Z

X

Y

(A) 0µ IL n2

(B)

0µ IL n3

(C)

0µ IL n2

(D)

0µ IL n3


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1001CT102116017

5. A cylindrical space is having uniform magnetic field. The field varies with time as

B = B0 + t. From point A to B there are three different paths as shown in figure.

Path 1 is a semi circle of radius a, with AB as diameter.

Path 2 contain two straight lines AO and OB.

Path 3 is a single straight line AB.

If E.d from A to B along paths 1, 2 and 3 are E1, E2 and E3 respectively, then mark the

INCORRECT options. :-

A B2a

1

2

3

O

(A) E1 = E2 = E3 (B) E1 = E3 > E2 (C) E1 > E3 > E2 (D) E3 > E2 > E1

6. Three thin concentric conducting spherical shells A, B, C of radius R, 2R and 3R are given

charges Q, 2Q, 3Q respectively as shown. The inner most and outermost can be joined through

switch 'S' which is initially opened. Choose the INCORRECT options :-

3R

3Q

2QQ

2R

R S r

B

A

C

(A) After closing switch S, potential of shell C does not change.

(B) After closing switch S, charge on shell A becomes zero.

(C) After closing switch S, potential of shell B does not change.

(D) After closing switch S, charge on shell C does not change.



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7. A wire frame is in the shape of a quadrant of a circle of radius a, having resistance R and is

free to rotate about O about axis perpendicular to plane of paper. Above line PQ a uniform

magnetic field exist having magnitude B and direction out of plane for region I and inside

plane for region II. If frame rotates with constant . Mark the CORRECT options :-

× × × × ×

× × × × ×

× × × × ×

× × × × ×

× × × × ×

× × × ×

/2 a

QP

Region I Region II

O

a

(A) As frame goes from region I to region II, the thermal energy dissipated is 2 4B a

2R

(B) As frame goes from region I to region II, the thermal energy dissipated is 2 4B a

4R

(C) Total thermal energy dissipated in one cycle is 2 43B a

8R

(D) Average power is 2 2 43B a

8R


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8. A disc of mass M, radius R has massless string wrapped over it, the other end of the string connects

a block of mass M as shown. Initially the system was at rest. If released, no slipping takes place at

any contact surface. Imediately after system is released. Mark the CORRECT options :-

M

y

x

(A) Tension in the string is 3Mg

5

(B) Acceleration of the block is 2 2ˆ ˆgi gj5 5

(C) The disc moves forward

(D) Acceleration of the block from centre of mass frame of the system is 2 ˆgj5

9. A uniform rod AB has mass 2m, length 2kept on a smooth horizontal plane. Rod AB is

orthogonaly connected to a taut string whose other end has a particle of mass m. If particle is

given initial velocity v0, perpendicular to string then just after giving velocity. Mark the

INCORRECT options :-

A B

2m

2

m v0

(A) Tension in string is 202mv

9

(B) Acceleration of rod is 20v

6

(C) Angular acceleration of rod is 20

2

2v

3

(D) From the frame of centre of mass of rod, particle moves in a circular path.



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10. A rolling ball slows down because the normal force does not pass exactly through the CM of

the ball, but passes in front of the CM. This is rolling friction. v & both decrease such that

v = r is maintained. Assume that deformation is small so that friction acts at r0 from centre.

(r0

is radius of ball).

v

FN

F

Mg

(A) The friction force opposes translation as well as rotation.

(B) The normal force produces a torque which is greater in magnitude than torque of friction.

(C) Friction force opposes translation but aids rotation.

(D) = 0

7r

5 where µ is the coefficient of static friction.


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1001CT102116017

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).


For each entry in Column-I

Full Marks : +2 If only the bubble(s) corresponding to all the correct match(es) is (are)

darkened

Zero Marks : 0 In none of the bubbles is darkened

Negative Marks : –1 In all other cases

1. A square loop ABCD of side , carries a current i as shown. The loop lies in x-y plane withcentre as origin. for a point P having coordinate (0, 0, z).

D C

A B

y

xi

Column–I Column–II

(A) For z > 0, unit vector along (P)

20

3/22

2

i

2 z4

magnetic field at P is



PHYSICS

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(B) For z > 0, unit vector along (Q)2

2

ˆ ˆk zj2

z4

magnetic field at P only due towire AB is.

(C) For z > 0, unit vector along (R)2

0

2 22 2

i

2 z z4 2

magnetic field at P only due towire BC is

(D) For z > 0, the magnitude of (S)2

2

ˆ ˆk zi2

z4

magnitude field at P is

(T) k̂


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1001CT102116017

2. Match the column-I with column-II. Take mutual inductance = 0 in all cases.Column-I Column-II

(A) 1

k = 1V

3mH 3mH

3mH3mH

(P) at t = 3 millisecond current through

the battery is e 1

amp.e

(B)

6 18 mH

k

= 3V

6 18 mH

(Q) at t = 3 millisecond current through any

inductor is 1 e 1

amp.2 e

(C)

30mH

10

1010V

k(R) at t = 3 millisecond current through any

inductor is e 1

amp.e

(D)

2

2

2

2

3mH

2V k

(S) Just after closing switch current throughany inductor is zero.

(T) Current through battery after verylongtime is 1 amp.

SECTION –III : Integer Value Correct Type

No question will be asked in section III



PHYSICS

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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.


Full Marks : +4 If only the bubble corresponding to the correct answer is darkened.

Zero Marks : 0 In all other cases.

1. A disk is rotating in a horizontal plane with constant 2

rad/s. A small particle A of mass m

is moved with in a frictionless radial groove, towards centre with uniform radial speed of 1 m/s

relative to platform. If particle goes from r = 2m to r = 1m, the net impulse on particle A is

expressed as

22

netI m 1 12

Find value of .

A


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1001CT102116017

2. A, B, C are three identical neutral conducting plates, A and C are rigidly attached to fixed

supports whereas plate B is attached to a spring having constant K. The supports and spring

are very small in size and their effect on induced charges on plates A, B, C is negligible. The

seperations between plates are d as shown. If plate B is displaced by x (x << d) in the plane and

released it is found that plate B under goes SHM. If K = 2 N/m, mass of plate B is 8 kg, emf of

battery = 2V and system is in gravity free space, the time period of B is found to be

T = . Find value of .

A B C

d d

E

3. A charge q is kept just outside a cube on extension of digonal GA. The magnitude of electric

flux of side ABCD is 0

q

168

. Find value of .

qA

B C

D

E

F

H

G



PHYSICS

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4. A 1 m long rod is dropped (keeping it horizontal) from a height of 20 m. The end A of rod collides

with end B of a stationary horizontal table. The impact is perfectly elastic. The post impact

angular velocity of rod is 10 (in rad/sec.), find value of .

A

B

20m

5. A solid sphere of radius r is inside a horizontal ring of slightly larger radius than r. The ring

is rigidly fixed with a horizontal rod of length L. The rod is hinged at O and is rotated about

a vertical axis passing through O with uniform . Centre of ring and sphere coincides. A point is

marked as A on the sphere close to ring. The horizontal plane below is smooth. If

r = 1 cm, L = 2 cm = 1 rad/s. The distance travelled by point A in one complete rotation of rod

is cm find .

A (point on sphere)

horizontal ringL

O


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1001CT102116017

6. A solid sphere of mass m radius r undergoes pure rolling inside a fixed vertical stationary circular

track of radius R. The sphere just completes vertical circle. The normal at bottom point B is

given by 34mg

. The find .

R

r

B



PHYSICS

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7. The magnetic dipole moment of a uniformly charged lamina in shape of equilateral triangle of

side a is M, when rotated about axis AB with uniform (see figure). If a hexagonal lamina of

side a and having same charge density is rotated about axis A'B' (with same angular speed ). The

diopole moment generated is 5M. Find .

a a

a

A

B

a

a

a

a

a

a

A'

B'

8. A solid body starts rotating about a stationary axis with angular acceleration = at2, a is a

positive constant and t is time in seconds. After few moments at time t =

1/4tan

a

, it has

been found that for a general point on body the velocity vector and acceleration vector form

an angle with each other. Find the value of .


E-18/36

CHEM

ISTRY


1001CT102116017


PART-2 : CHEMISTRY





For each question, darken the bubble(s) corresponding to all the correct option(s) in the

ORS



darkened.



1. Select the correct statements(s)

(A) Those nucleus which have high n

p may undergo – decay

(B) Those nucleus which have low n

pmay undergo – decay

(C) Those nucleus which have high n

pmay undergo K-electron capture

(D) Those nucleus which have low n

pmay undergo either + decay or K-electron capture]

2. For the gase phase reaction : R – H + X2 R – X + HX

Following mechanism has been proposed :

(i) X2

1

2

k

k 2X•

(ii) X• + R –H 3kR•+ H–X (slowest)

(iii) R•+ X2 4k R –X + X•

Based on this select the correst option (s)

(A) Effective rate constant for the formation of RX is 1

3 4

2

kk k

k

(B) 2

d[RX]X

dt

(C) Overall order of the reaction is 3/2

(D) d[RX]

RHdt

1


CHEM

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3. For the reaction R(g) 2P(g)

Following observation is made at 300K.

P (P.Pr of P)P

P (P.Pr of R)R

Time(min)

P Pr.(bar)

2

1

10 20

Select the correct statements (s)

(A) (rG) at 10 min = 0 (B) (

rG) at 20 min = 0

(C) rG0 at 10 min = –ve (D)

rG0 at 20 min = 0

4. From the following phase diagram of sulphur, select the correct statements(s)

liquid (liq.)

Rhombic (rh)

15411911495Temperature (ºC)

10–5

10–4

103

Pre

ssu

re (

atm

)

Gas

100

Monoclinic

(A) At 1 atm and 25ºC GS(rh)

> GS(monoclinic)

[G : Gibb's function]

(B) At 103 atm and 154ºC GS(rh)

= GS(monoclinic)

= GS(liq.)

(C) At 1 atm and 114ºC GS(rh)

> GS(liq.)

(D) Melting point of monoclinic 'S' increases on increasing pressure

5. Select in which atleast two d-orbitals are used in bonding.

(A) XeO4

(B) XeO3F

2(C) XeO

6

4– (D) XeO2F

4


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CHEM

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1001CT102116017


6. The paramagnetic molecule(s) which have non fractional bond order :-

(A) O2

(B) O2

(C) N2

2– (D) B2

7. Molecule(s) in which maximum number of atoms in a plane may be ten.

(A) B2Me

4H

2(B) Al

2(CH

3)

6(C) N(SiH

3)

3(D) [Co(NH

3)

6]3+

8. Major product of which of the following reaction(s) can show iodoform test :

(A) (CH3COO)

2Ca (B)

H / KMnO+

4CH –CH –CH–CH3 2 3

OH

(C) CH3–CH

2–CHO

OH / H O2

–

(D) CH =CH2 2

(i) BH –THF3

(ii) CH COOH3

9. Which statement(s) is/are correct about following reaction :

2HCHO Conc. NaOH CH3OH + HCOONa

(A) It's a disproportionation reaction

(B) Hydride transfer is involved in rate determining step

(C) Increase in concentration of base can increase the rate of reaction

(D) It is a unimolecular reaction

10. Which of the following reaction(s) are correctly matched with their major product ?

(A)

O

O (i) LiAlH4HO OH

(ii) H O2

(B) CH3–CHO + 3Br

2 + 4NaOH HCOONa + CHBr

3 + 3NaBr + 3H

2O

(C)

O ONH –NH2 2

N

H

NpH (4.5-6.0)

(D)

COOH

O O


CHEM

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(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).




darkened



1. Column-I (Structure) Column-II

(A) Rock salt (P)r

0.414r

(B) Caesium chloride (Q) Anion forms fcc lattice

(C) Antifluorite (R) r+ + r

– =

3 a

2

(D) Zinc blende (sphalerite) (S) Co-ordination no. of cation is 6

(T) r+

+ r– = a/2

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CHEM

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1001CT102116017


2. Column - I Column - II

(One of the organic product can)

(A) But-2-ene

(i) O , (ii) Zn, H O3 2

(iii) dil. NaOH

(iv) CrO , H SO3 2 4

(P) Evolve CO2 on heating

(B) CH –CH=O +3

HO

HO

H+

(Q) Cyclic gemdiether

(C) CH CHO3

(i) KCN/HCl

(ii) H O3

+ (R) Give cyclic ester product on heating

(D)(i) O3

(ii) H O2 2

(S) Evolve CO2

with aq. NaHCO3

(T) Can give diol on reduction with LiAlH4




CHEM

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inclusive.





1. In order to determine volume of blood in a person, radioactive sample of activity 30 Ci

(t1/2

= 4 hrs) is injected into patient. After 8 hrs. , 10 ml blood sample withdrawn shows

activity of 0.015 Ci. What is total volume (in litre) of blood in the person ?

2. For any elementary reaction following observation is made.

3.2 3.8

–5.2

–8.8

lnk

1T

×103

If the above reaction is carried out in the presence of catalyst at 300K the reaction

proceeds with same rate as without catalyst at 400K. By what amount (in kcal) catalyst

has decreased Ea.

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CHEM

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1001CT102116017


3. Total number of enantiomeric pair in complex compound [Co(NH3)2Cl

2BrI]

4. Total number of ores which contain two ions of different metal :

Ilmenite , Chromite ore , Haematite , Magnesite , Copperpyraties Ironpyraties,

Azurite


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5. Total number of molecules in which bridge bond formed by sp3–s–sp3 type overlap

B2H

6 , Al

2(CH

3)

6 , I

2Cl

6 , Al

2H

6 , Si

2Cl

6

6. Which of the following low spin complex as well as paramagnetic complex :

[Co(H2O)

6]3 , [Co(NH

3)6]2 , [Fe(CN)

6]3– , [IrF

6]3–

[Ni(en)3]2 , [Cr(H

2O)

6]3 , [Cr(CN)

6]4–


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CHEM

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1001CT102116017

7. Total number of compound which on reaction with alkaline solution of iodine gives yellow crystals.

(i) CH –C–CH3 3

O

(ii) CH3CHO (iii)

O

(iv) Me2CH–CHO

(v) HCHO (vi)

CH–CH3

HO

(vii) Ph–C–CH3

O

(viii) CH –CH –C–CH –CH3 2 2 3

O

8. Number of moles of CrO3 required to oxidise 6 moles of 2-heptanol :



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PART-3: MATHEMATICS





For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS



darkened.



1. If xdyy e

dx and y(0) = 1, then-

(A) 5

y n24

(B) 9

y n24

(C) minimum value of y is 2 (D) n2

0

3ydx

4

2. Let x,y,z are roots of equation t3 – 3t2 + t – 1 = 0, then which can be true -

(A)

3

3

3

1 x x

1 y y

1 z z3

x y y z z x(B)

2 3

2 3

2 3

1 x x

1 y y

1 z z1

x y y z z x

(C)

1 x 1 1

1 1 y 1 2

1 1 1 z

(D)

2

2

2

1 x x

1 y y

1 z z1

x y y z z x

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1001CT102116017

3. Let x = ,y = , z = where are non-zero real numbers,

satisfies [xsin3 – 1 xcos2 2x] – [y –4y –7y] + [z 3z

–

2 7z] = [–1 –2 0], 0,4 .

If N denotes number of all possible values of and S is sum of all possible values of then(A) N > 8 (B) N < 8 (C) S < 14 (D) S > 14

4. Let

1 0 0

M 2 1 0

4 2 1

and I be the identity matrix of order 3. If Q = [qij] is a matrix such that

M100–Q = I, then-(A) Q is triangular matrix (B) q11 + q21 + q31 = q31+q32+q33

(C)

31 32

21

q q102

q(D) Q is nilpotent matrix of index 3.

5. Let P,Q is circumcentre of ABC and PBC respectively. In ABC, AB = 3, AC 2 , S is

perimeter of ABC and Q is midpoint of BC, then which can be true-

(A) circumradius of ABC5

2(B) circumradius of ABC 5

(C) [S] = 6 (D) [S] = 8(where [S] denotes greatest integer less than or equal to S)

6. Let

n

nr 0

1P lim

2r 1 2r 2 2r 3

, then value of 10P lie in-

(A)

17 9,

10 5(B)

9 19,

5 10(C)

19,2

10(D)

212,

10

7. Let g(x) is a real valued function which satisfies the equation

g()g() = g()g( + ) + g() R and is independent from and . If g(0) = 1,

then value of

3 1

1

g 2 t dt is/are-

(A) less than 1

2(B) less than 1 (C) greater than

1

2(D) 2

1 1

2 3



MATHEM

ATICS

E-29/361001CT102116017


8. Let y = ƒ(x) is the solution of the differential equation

22

2

d y dyy 2 0

dxdx, ƒ(1) = 1. S is area

bounded by curve y = ƒ(x), y = ƒ–1(x) and xy = 0. If ƒ–1(2) = 0, then which of the followingstatement(s) can be true ?

(A) Angle of intersection of y = ƒ(x) and y = ƒ–1(x) is

1 3tan

4

(B) Angle of intersection of y = ƒ(x) and y = ƒ–1(x) is

2

(C) [S] = 1

(D) [S] = 2

(where [S] denotes greatest integer less than or equal to S)

9. For any real number 'm' Let F(m) and ƒ(m) denotes the maximum and minimum of the function

2t 3 m t 2 3m

3 t t [–1,1], respectively, then select the incorrect option(s)

(A) ƒ(m) is non differentiable at one point

(B) Area bounded by y = F(m), y = ƒ(m), y = 2 is 87

16 sq. units

(C) Area bounded by y = ƒ(F(m)), y = ƒ(m), y = 2 is 87

16 sq. units

(D) Minimum distance between y = ƒ(m) and y = F(m) is 3

2 2

10. Consider curve C1 : y2 = 7x

curve C2 : x2 = 7y

and equation x3 – ax2 + bx – c = 0, a,b,c are rational numbers. Let N denotes maximum numberof possible irrational roots of given cubic equation. Straight line x = N divides the area enclosedby curves C1

and C2 into two parts R

1 and R

2. If area of region R

1 > area of region R

2, then

select the correct option(s) -

(A) sum of area of region R1 and R2 is 49

3sq. units

(B) area of region R2 is

28 14 8

21 sq. units

(C) area of region R1 is 351 28 14

21sq. units

(D) sum of area of region R1 and R

2 is not equal to

49

3sq. units

E-30/36

MATHEM

ATICS


1001CT102116017










(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 threebubbles in the ORS. Similarly, for entries (B), (C) and (D).




darkened



1. Column-I Column-II

(A) The maximum value of 8sin + 15cos – 18 is (P) 2

(B) Let ƒ : R R be defined by (Q) –2

k x 3 , x 1ƒ x

2x 3 , x 1. If ƒ(x) has a local maximum

at x = –1, then k can be

(C) Value of

4

2

log x 1I dx

log x 1 log 7 x is (R) –1

(D) Let ƒ : (–1,1) R be a continuous function (S) 1

If sinx

2

0

tƒ t dt 6x , then value of (T) 4

ƒ sin x .sin 2x

6x, x 0 is


MATHEM

ATICS

E-31/361001CT102116017


2. Let ƒ(x) = x3 – 6x2 – 36x + 2

Column-I Column-II

(A) ƒ(x) is increasing function in (P)

30,

2

(B) x = C is point of inflection for curve y = ƒ(x), then (Q)

3,4

2

C lie in

(C) Let N denotes number of real root(s) of equation (R) (5,7)

ƒ(x) = 0, then N lie in

(D) Let Rolle's theorem holds for ƒ(x) in [0,a]. If k is (S) (8,10)

sum of all possible values of a, then k lie in (T) (9,11)



E-32/36

MATHEM

ATICS


1001CT102116017





inclusive.





1. If 4 3 a c

b 5 1 5

, then the value of

c

b

a dx , is

2. Area bounded by ellipse 2 2x y

116 25

is k sq. units, then value of k

4 is

3. The number of arbitrary constant in the particular solution of a differential equation of oneorder is


MATHEM

ATICS

E-33/361001CT102116017


4. Area bounded by 12x – y2 > 0 and y > 2x is

5. In a triangle XYZ, let a,b,c be the length of the sides opposite to the angles X,Y and Z

(where Z = c radian) respectively and dentoes s(s – a) (s – b) (s – c). If

a b c 3b a b c

a b c 3b a b c 6

a b a 3b c b c

,

b2 + c2 > a2, a2 + b2 > c2, a2 + c2 > b2, then sum of all possible integral values of

1, where

a b cs

2 is

6. Let sin cosx – tan sinx + 2sin2x = 0 and

22 2

x

2 2

0

tan sincos3x cosx tan 1 sin3x sin x 1 sin sin 2x

2ƒ x dx

cosxsin sin x tan 2sin 2x

, where

cosx sin2 – sinx tan2 + 2 sin2x 0. If M is maximum value of (ƒ(x) + cos2x)2,

x 0,2

, then

value of M is

E-34/36

MATHEM

ATICS


1001CT102116017

7. The system of equations

2tanx + pcosy + 6sinz = 8

tanx + 2cosy+qsinz = 5

tanx + cosy + 3sinz = 4, x,y,z [0,2]

has exactly n ordered triplet (x,y,z) in x,y,z [0,2], n N, then minimum possible value of n

8. Number of solution(s) of equation n(sinx)secx = (cosecx)n(cosx),

x 0,2

is



E-35/361001CT102116017



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 1001CT102116017E-36/36

QUESTION PAPER FORMAT AND MARKING SCHEME :

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

17. Each part has three sections 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

+4 0 –2One or more If only the bubble(s) If none In all

I correct10

corresponding —

of the other40option(s) to all the correct bubbles is cases

option(s) is(are) darkeneddarkened

+8 +2 0 –1 II Matrix If only the bubble(s) For darkeninga bubble If none In all

Match Type 2 corresponding to corresponding to each of the other 16all the correct correct match is bubbles is cases

match(es) is(are) darkened darkeneddarkened

+4 0Single digit If only the bubble In all

IV Integer 8 corresponding — other — 32(0-9) to correct answer cases

is darkened

1.

2. (ORS)

3.

4.

5. 36 20

6.

7.

8.

9.

:

10.

11.

12. :

13.

14.

15. g = 10 m/s2

PAPER – 1

Test Type : MINOR Test Pattern : JEE-Advanced

TEST DATE : 21 - 08 - 2016

PHASE : I

JEE (Main + Advanced) : ENTHUSIAST COURSE

Paper Code : 1001CT102116017

CLASSROOM CONTACT PROGRAMME(Academic Session : 2016 - 2017)

Time : 3 Hours Maximum Marks : 264

HIN

DI

H-2/36


1001CT102116017

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-3/361001CT102116017

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS

-1 : –I : ( : 40)

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

: +4

: 0

: –2

1. B OA

'' E r

OA (V)

A

O

rE

(A) E > 21

B2

V > 21

B2

(B) E = 21

B2

V = 0

(C) E < 21

B2

V > 21

B2

(D) E V

H-4/36

PHYSICS


1001CT102116017

2. C = 1 µF C1, C

2, C

3, C

4 r = 1

E = 30 V

r

r

r

r

r

C2

C3

C4

E

C1

(A) C1 C2 1 : 1

(B) C2 C4

1 : 1

(C) C1 C3 1 : 1

(D) C2 C3

1 : 1

3. C U I R

PQ V 2V

IP Q

R

2V

V

Ca

2a

(A) t 0 t / RCCµ Iv n2

q (1 e )2

(B) t 0 t / RCCµ Iv n2

q (1 e )

(C) t 0 t / RCµ Iv n2

i (e )2 R

(D) t 0 t / RCµ Iv n2

i (e )R


PHYSICS

H-5/361001CT102116017

4. Z-I 4a a

L Z-X-Z

L

(2a, 0, 0)

(–2a, 0, 0)

I

I

Z

X

Y

(A) 0µ IL n2

(B)

0µ IL n3

(C)

0µ IL n2

(D)

0µ IL n3

H-6/36

PHYSICS


1001CT102116017

5. B = B0 + t

A B

1 a AB

2 AO OB

3 AB

1, 2 3 A B E.d E1, E2 E3

A B2a

1

2

3

O

(A) E1 = E

2 = E

3(B) E

1 = E

3 > E

2(C) E

1 > E

3 > E

2(D) E

3 > E

2 > E

1

6. R, 2R 3R A, B, C Q, 2Q 3Q

'S'

3R

3Q

2QQ

2R

R S r

B

A

C

(A) S C

(B) S A

(C) S B

(D) S C


PHYSICS

H-7/361001CT102116017

7. a R

O PQ B

I II

× × × × ×

× × × × ×

× × × × ×

× × × × ×

× × × × ×

× × × ×

/2 a

QP

Region I Region II

O

a

(A) I II 2 4B a

2R

(B) I II 2 4B a

4R

(C) 2 43B a

8R

(D) 2 2 43B a

8R

H-8/36

PHYSICS


1001CT102116017

8. M R M

:-

M

y

x

(A) 3Mg

5

(B) 2 2ˆ ˆgi gj5 5

(C)

(D) 2 ˆgj5

9. 2m 2AB AB

m

v0

A B

2m

2

m v0

(A) 202mv

9

(B) 20v

6

(C) 20

2

2v

3

(D)


PHYSICS

H-9/361001CT102116017

10.

v v = r

r0 (r0 )

v

FN

F

Mg

(A)

(B)

(C)

(D) = 0

7r

5 µ

H-10/36

PHYSICS


1001CT102116017

–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. ABCD i x-y (0, 0, z) P

D C

A B

y

xi


PHYSICS

H-11/361001CT102116017

–I –II

(A) z > 0 P (P)

20

3/22

2

i

2 z4

(B) z > 0 (Q)2

2

ˆ ˆk zj2

z4

AB P

(C) z > 0 BC (R)2

0

2 22 2

i

2 z z4 2

P

(D) z > 0 P (S)2

2

ˆ ˆk zi2

z4

(T) k̂

H-12/36

PHYSICS


1001CT102116017

2. -I -II = 0 -I -II

(A) 1

k = 1V

3mH 3mH

3mH3mH

(P) t = 3

e 1

amp.e

(B)

6 18 mH

k

= 3V

6 18 mH

(Q) t = 3

1 e 1

amp.2 e

(C)

30mH

10

1010V

k(R) t = 3

e 1

amp.e

(D)

2

2

2

2

3mH

2V k

(S)

(T) 1 amp

–III : III


PHYSICS

H-13/361001CT102116017

–IV : (: 32)

0 9

: +4

: 0

1. 2

rad/s m A

1 m/s

r = 2m r = 1m A

22

netI m 1 12

A

H-14/36

PHYSICS


1001CT102116017

2. A, B, C A C B

K A, B, C

d B x (x << d)

B K = 2 N/m, B 8 kg,

2V B T =

A B C

d d

E

3. q GA ABCD

0

q

168

qA

B C

D

E

F

H

G


PHYSICS

H-15/361001CT102116017

4. 1 m 20 m A B

10 (rad/sec.)

A

B

20m

5. r r L

O O

A

r = 1 cm, L = 2 cm = 1 rad/s A

cm

A (point on sphere)

horizontal ringL

O

H-16/36

PHYSICS


1001CT102116017

6. m r R

B 34mg

R

r

B


PHYSICS

H-17/361001CT102116017

7. a M

AB a

A'B' 5M

a a

a

A

B

a

a

a

a

a

a

A'

B'

8. = at2 a

t t =

1/4tan

a

H-18/36

CHEM

ISTRY


1001CT102116017

-2 : –I : ( : 40)

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

: +4

: 0

: –2

1.

(A) n

p –

(B) n

p –

(C) n

p K-

(D) n

p + K-

2. : R – H + X2 R – X + HX,

(i) X2

1

2

k

k 2X•

(ii) X• + R –H 3kR•+ H–X ()

(iii) R•+ X2 4k R –X + X•

(A) RX 13 4

2

kk k

k

(B) 2

d[RX]X

dt

(C) 3/2

(D) d[RX]

RHdt

1


CHEM

ISTRY

H-19/361001CT102116017

3. R(g) 2P(g)

300K

P (P.Pr of P)P

P (P.Pr of R)R

Time(min)

P Pr.(bar)

2

1

10 20

(A) 10 (rG) = 0 (B) 20 (

rG) = 0

(C) 10 rG0 = –ve (D) 20

rG0 = 0

4.

liquid (liq.)

Rhombic (rh)

15411911495Temperature (ºC)

10–5

10–4

103

Pre

ssu

re (

atm

)

Gas

100

Monoclinic

(A) 1 atm 25ºC GS(rh)

> GS()

[G : ]

(B) 103 atm 154ºC GS(rh)

= GS()

= GS()

(C) 1 atm 114ºC GS(rh)

> GS()

(D) 'S'

5. d-

(A) XeO4

(B) XeO3F

2(C) XeO

6

4– (D) XeO2F

4

H-20/36

CHEM

ISTRY


1001CT102116017

6. (non fractional) :-

(A) O2

(B) O2

(C) N2

2– (D) B2

7. 10

(A) B2Me

4H

2(B) Al

2(CH

3)

6(C) N(SiH

3)

3(D) [Co(NH

3)

6]3+

8.

(A) (CH3COO)

2Ca (B)

H / KMnO+

4CH –CH –CH–CH3 2 3

OH

(C) CH3–CH

2–CHO

OH / H O2

–

(D) CH =CH2 2

(i) BH –THF3

(ii) CH COOH3

9.

2HCHO Conc. NaOH CH3OH + HCOONa

(A)

(B)

(C)

(D)

10.

(A)

O

O (i) LiAlH4HO OH

(ii) H O2

(B) CH3–CHO + 3Br

2 + 4NaOH HCOONa + CHBr

3 + 3NaBr + 3H

2O

(C)

O ONH –NH2 2

N

H

NpH (4.5-6.0)

(D)

COOH

O O


CHEM

ISTRY

H-21/361001CT102116017

–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) (P)r

0.414r

(B) (Q) fcc

(C) (R) r+ + r

– =

3 a

2

(D) () (S) 6

(T) r+

+ r– = a/2

H-22/36

CHEM

ISTRY


1001CT102116017

2. - I - II

(A) -2-

(i) O , (ii) Zn, H O3 2

(iii) dil. NaOH

(iv) CrO , H SO3 2 4

(P) CO2

(B) CH –CH=O +3

HO

HO

H+

(Q)

(C) CH CHO3

(i) KCN/HCl

(ii) H O3

+ (R)

(D)(i) O3

(ii) H O2 2

(S) NaHCO3 CO

2

(T) LiAlH4

–III :

III


CHEM

ISTRY

H-23/361001CT102116017

–IV : (: 32)

0 9

: +4

: 0

1. 30 Ci (t

1/2 = 4 hrs)

8 10 ml 0.015 Ci

( ) ?

2.

3.2 3.8

–5.2

–8.8

lnk

1T

×103

300K 400K

Ea (kcal )

H-24/36

CHEM

ISTRY


1001CT102116017

3. [Co(NH3)2Cl

2BrI]

4. :

, , , , (Copperpyraties)

,


CHEM

ISTRY

H-25/361001CT102116017

5. sp3–s–sp3

B2H

6 , Al

2(CH

3)6 , I

2Cl

6 , Al

2H

6 , Si

2Cl

6

6. :

[Co(H2O)

6]3 , [Co(NH

3)

6]2 , [Fe(CN)

6]3– , [IrF

6]3–

[Ni(en)3]2 , [Cr(H

2O)

6]3 , [Cr(CN)

6]4–

H-26/36

CHEM

ISTRY


1001CT102116017

7.

(i) CH –C–CH3 3

O

(ii) CH3CHO (iii)

O

(iv) Me2CH–CHO

(v) HCHO (vi)

CH–CH3

HO

(vii) Ph–C–CH3

O

(viii) CH –CH –C–CH –CH3 2 2 3

O

8. 2-6 CrO3


MATHEM

ATICS

H-27/361001CT102116017

-3 : –I : ( : 40)

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

: +4

: 0

: –2

1. xdyy e

dx y(0) = 1 -

(A) 5

y n24

(B) 9

y n24

(C) y 2 (D) n2

0

3ydx

4

2. x,y,z t3 – 3t2 + t – 1 = 0 -

(A)

3

3

3

1 x x

1 y y

1 z z3

x y y z z x(B)

2 3

2 3

2 3

1 x x

1 y y

1 z z1

x y y z z x

(C)

1 x 1 1

1 1 y 1 2

1 1 1 z

(D)

2

2

2

1 x x

1 y y

1 z z1

x y y z z x

H-28/36

MATHEM

ATICS


1001CT102116017

3. x = ,y = , z = [xsin3 – 1

xcos2 2x] – [y –4y –7y] + [z 3z

–

2 7z] = [–1 –2 0], 0,4

N, S, (A) N > 8 (B) N < 8 (C) S < 14 (D) S > 14

4.

1 0 0

M 2 1 0

4 2 1

I, 3 Q = [qij]

M100–Q = I -

(A) Q (B) q11 + q21 + q31 = q31+q32+q33

(C)

31 32

21

q q102

q(D) Q, 3

5. P,Q ABC PBC ABC AB=3, AC 2 S ABC Q, BC -

(A) ABC 5

2 (B) ABC 5

(C) [S] = 6 (D) [S] = 8

([S], S )

6.

n

nr 0

1P lim

2r 1 2r 2 2r 3

10P -

(A)

17 9,

10 5(B)

9 19,

5 10(C)

19,2

10(D)

212,

10

7. g(x) g()g() = g()g( + ) + g() R

g(0) = 1

3 1

1

g 2 t dt -

(A) 1

2 (B) 1 (C)

1

2 (D) 2

1 1

2 3


MATHEM

ATICS

H-29/361001CT102116017

8. y = ƒ(x)

22

2

d y dyy 2 0

dxdx ƒ(1) = 1 S, y = ƒ(x), y = ƒ–1(x)

xy = 0 ƒ–1(2) = 0

(A) y = ƒ(x) y = ƒ–1(x)

1 3tan

4

(B) y = ƒ(x) y = ƒ–1(x)

2

(C) [S] = 1

(D) [S] = 2

([S], S )

9. m F(m) ƒ(m)

2t 3 m t 2 3m

3 t t [–1,1]

-(A) ƒ(m)

(B) y = F(m), y = ƒ(m), y = 2 87

16

(C) y = ƒ(F(m)), y = ƒ(m), y = 2 87

16

(D) y = ƒ(m) y = F(m) 3

2 2

10. C1 : y2 = 7x, C2 : x

2 = 7y x3 – ax2 + bx – c = 0 a,b,c Nx = N, C1 C2

R1 R

2 R

1 > R

2

-

(A) R1 R

2

49

3

(B) R2

28 14 8

21

(C) R1

351 28 14

21

(D) R1 R2

49

3

H-30/36

MATHEM

ATICS


1001CT102116017

–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) 8sin + 15cos – 18 (P) 2

(B) ƒ : R R,

k x 3 , x 1ƒ x

2x 3 , x 1 (Q) –2

ƒ(x), x = –1 k

(C)

4

2

log x 1I dx

log x 1 log 7 x (R) –1

(D) ƒ : (–1,1) R (S) 1

sinx

2

0

tƒ t dt 6x ƒ sin x .sin 2x

6x, x

0 (T) 4


MATHEM

ATICS

H-31/361001CT102116017

2. ƒ(x) = x3 – 6x2 – 36x + 2

-I -II

(A) ƒ(x), (P)

30,

2

(B) x = C y = ƒ(x) (Q)

3,4

2

C (C) N ƒ(x) = 0 (R) (5,7)

N

(D) [0,a] ƒ(x) k, a (S) (8,10)

k (T) (9,11)

–III :

III

H-32/36

MATHEM

ATICS


1001CT102116017

–IV : (: 32)

0 9

: +4

: 0

1. 4 3 a c

b 5 1 5

c

b

a dx

2. 2 2x y

116 25

k k

4

3.


MATHEM

ATICS

H-33/361001CT102116017

4. 12x – y2 > 0 y > 2x 5. XYZ a,b,c X, Y Z ( Z = c ) ,

s(s – a)(s – b)(s – c)

a b c 3b a b c

a b c 3b a b c 6

a b a 3b c b c

, b2 + c2 > a2, a2 + b2 > c2,

a2 + c2 > b2

1

a b cs

2

6. sin cosx – tan sinx + 2sin2x = 0

22 2

x

2 2

0

tan sincos3x cosx tan 1 sin3x sin x 1 sin sin 2x

2ƒ x dx

cosxsin sin x tan 2sin 2x

cosx sin2 – sinx tan2 + 2 sin2x 0 M, (ƒ(x) + cos2x)2,

x 0,2

M

H-34/36

MATHEM

ATICS


1001CT102116017

7. 2tanx + pcosy + 6sinz = 8

tanx + 2cosy+qsinz = 5

tanx + cosy + 3sinz = 4, x,y,z [0,2]

x,y,z [0,2], n (x, y, z), n N n

8. n(sinx)secx = (cosecx)n(cosx),

x 0,2


H-35/361001CT102116017


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 1001CT102116017H-36/36

16. 17.

____________________________

____________________________

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

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

+4 0 –2

I 10 — 40

+8 +2 0 –1 II

2 16

+4 0

IV (0-9)

8 — — 32

JEE (Main + Advanced) : ENTHUSIAST COURSE...E-4/36 PHYSICS Target : JEE (Main + Advanced)...

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