(acade ses : 2019 - 2020) jee(m+aance) : leader

READ THE INSTRUCTIONS CAREFULLY

GENERAL :

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

5. After breaking the seal of the booklet, verify that the booklet contains 28 pages andthat 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 purposeon the ORS. Do not write any of these details anywhere else on the ORS. Darkenthe 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 correctway.

14. Darken the bubbles ONLY IF you are sure of the answer. There is NO WAY to eraseor "un-darken" a darkened bubble.

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

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

Time : 3 Hours Maximum Marks : 186

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)0000CJA102119027)(0000CJA102119027) Test Pattern

CLASSROOM CONTACT PROGRAMME(Academic Session : 2019 - 2020)

JEE(Main+Advanced) : LEADER & ENTHUSIAST COURSE

JEE(Advanced)AIOT

09-02-2020

PAPER-1

En

glis

h

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All India Open Test/Leader & Enthusiast Course/JEE(Advanced)/09-02-2020/Paper-1

Space for Rough Work

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 = 58Atomic 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,

0000CJA102119027

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

· Coulomb's law constant pe9

0

1 = 9×104

· 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 s = 5.67 × 10–8 Wm–2–K–4

· Wien's displacement law constant b = 2.89 × 10–3 m–K· Permeability of vacuum µ0 = 4p × 10–7 NA–2

· Permittivity of vacuum Î0 = 20

1

cm· Planck constant h = 6.63 × 10–34 J–s

ALLEN

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All India Open Test/Leader & Enthusiast Course/JEE(Advanced)/09-02-2020/Paper-1ALLEN

0000CJA102119027


PART-1 : PHYSICSSECTION-I(i) : (Maximum Marks : 12)

� This section contains FOUR 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 massless inextensible string is wrapped over a cylinder of radius 20 cm and mass of 2kg.It is connected to a hanging block of mass 1 kg. If the surface is smooth, the cylinder reachesa distance l in time t1. If the surface is sufficiently rough, the cylinder reaches the distance l

in time t2. Then 1

2

tt

=

m

M

(A) 1 (B) 7

10 (C) 107 (D)

12

BEWARE OF NEGATIVE MARKINGHAVE CONTROL ¾® HAVE PATIENCE ¾® HAVE CONFIDENCE Þ 100% SUCCESS

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2. An equi-convex lens (µ = 1.5) is combined with an equi concave lens (µ = 1.3). The radii ofcurvature of all surfaces is the same. (Assume thin lenses)(A) The combination behaves like a converging lens in medium of refractive index 1.4 and

diverging lens in a medium of refractive index 1.6.(B) The combination behaves like a converging lens in a medium of refractive index 1.2 as well as

1.4.(C) The combination behaves like a diverging lens in medium of refractive index 1.2 as well as

1.6.(D) The combination behaves like a diverging lens in medium of refractive index 1.4 and converging

lens in a medium of refractive index 1.6.3. A certain substance (not an ideal gas) has a PV graph as a straight line AB when subjected to an

adiabatic process. If the same substance goes through process A ® C ® B, what is the heat givento it ?

A C

B

2×10 Pa5

1×10 Pa5

1 litre 2 litreV

P

(A) 50 J (B) 200 J (C) 150 J (D) 100 J

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4. A charge of +10µC and mass of 1gm is moving with velocity of 106 m/s at an angle of 45° tox & y axis as shown. It experiences a magnetic force along –ve z-direction. When it is projectedwith 106 m/s along +ve z-direction, it experiences a magnetic force of 10–2 N in +ve x-direction.

The magnetic field Br

is :

x

y

45°

v

(A) ( )3 ˆ ˆ10 T i j-- + (B) ( )3 ˆˆ10 j k- + (C) 3 ˆ10 i- (D) 3 ˆ10 T j--

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0000CJA102119027

SECTION-I(ii) : (Maximum Marks: 32)� This section contains EIGHT questions.� Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of these

four option(s) is (are) correct option(s).� For each question, choose the correct option(s) to answer the question.� Answer to each question will be evaluated according to the following marking scheme:

Full Marks : +4 If only (all) the correct option(s) is (are) chosen.Partial Marks : +3 If all the four options are correct but ONLY three options are chosen.Partial Marks : +2 If three or more options are correct but ONLY two options are chosen,

both of which are correct options.Partial Marks : +1 If two or more options are correct but ONLY one option is chosen

and it is a correct option.Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).Negative Marks : –1 In all other cases.

� For Example : If first, third and fourth are the ONLY three correct options for a question withsecond option being an incorrect option; selecting only all the three correct options will resultin +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options),without selecting any incorrect option (second option in this case), will result in +2 marks.Selecting only one of the three correct options (either first or third or fourth option), withoutselecting any incorrect option (second option in this case), will result in +1 marks. Selectingany incorrect option(s) (second option in this case), with or without selection of any correct option(s)will result in –1 marks.

5. In the given circuit the potential difference between a & b is 300 V. The resistors are of 100 W each.

M1

M2

R R R

ba

(A) If the upper meter M1 is ideal voltmeter & lower is ideal ammeter M2, reading of voltmeteris 300 V.

(B) If both meters are ideal voltmeters, they show 200 V each.(C) If both meters are ideal ammeters, they read 6A each.(D) If the upper meter M1 is ideal voltmeter and lower is ideal ammeter M2, reading of ammeter

is 3A.Space for Rough Work

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0000CJA102119027

6. A plastic body of density r is to be held in air of density ra first and then in water ofdensity rw. For holding, we need to apply a force on it.

(A) It is easier to hold it in water if r > ra & w a

2r + r

r < .

(B) It is easier to hold it in water if r > rw.

(C) It is easier to hold it in air if r < rw & w a

2r + r

r < .

(D) It is easier to hold it in water if r < rw & w a

2r + r

r > .

7. A long metallic rod of mass m is being moved on two smooth horizontal metallic rails with constantvelocity v. The angle between the rails is a. A uniform magnetic field (B) in vertical direction ispresent as shown. The capacitor is initially uncharged. The resistance of circuit is zero.Choose CORRECT graph(s) :-

A

v

a

C

x0

(A)

time

icurrent in therail

(B)

time

External force to be

appliedon the rod

(C)

time

Distance ofpoint of

application of force

from lower end

(D)

time

Energystored incapacitor


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0000CJA102119027

8. A radioactive nucleus A decays to C after emitting two a and three b– particles. Anothernucleus B decays to C by emitting one a and five b– particles. Half lives of A & B are 1 hr and2hr respectively and intermediaries have negligible half lives. At time t = 0; population of Ais 4N0, of B is N0 and of C is zero.(A) Atomic mass number of A– Atomic mass number of B = 4(B) Atomic number of A – Atomic number of B = – 2

(C) When A & B have equal population, population of C = 09N2 .

(D) The decay rate of A & B are equal at t = 6hr.9. When a narrow resonance column apparatus is filled with ozone gas (g = 4/3), it resonates in

1st resonance with a tuning fork. When it is filled with nitrogen gas (N2), it resonates with the

same tuning fork in 1st resonance when water column is shifted down by 10 cm. (Take 5 = 2.25)(A) Wavelength of sound in Ozone is approximately 1m.(B) Wavelength of sound in Nitrogen is approximately 1.6 m.(C) For second resonance in Ozone gas, the water column should be shifted down by 60 cm.(D) For second resonance in Nitrogen gas, the water column should be shifted down by 80 cm.


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0000CJA102119027

10. In an AC circuit, box may contain an inductor or a capacitor.

Box

The voltage across source is shown by given graph

V

t

Here Vbox versus time & Vresistance versus time graphs are plotted. Choose the possible CORRECTalternative(s) :

(A) If box contains inductor,

Vbox

t

(B) If box contains inductor,

Vresistance

t

(C) If box contains capacitor,

Vresistace

t

(D) If box contains capacitor,

Vbox

t


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0000CJA102119027

11. A body of mass m experiences a central force F kr= -r r . Here rr is the position from centre of

force and k is a positive constant. It moves in a closed circular path such that it's angular

momentum about centre of force is quantized nhL2

=p

. Take potential energy = 0 at r = 0.

(A) It's possible radius is quantized such that r µ n1/2.(B) It's possible mechanical energy is quantized such that E µ n.(C) It's speed is quantized such that v µ n1/2.(D) It's angular velocity is quantized such that w µ n.

12. 5 elastic balls A,B, C, D and E of same mass are moving on a long frictionless horizontal wire. Agraph is drawn for their position x against time. Select the possible CORRECT statement(s) onbasis of graph.

1m

2m

3m

4m

1 sec. 1.8 sec.0ED

C

B

A

1.5m

(A) There are a total of 10 collisions.(B) Ball A finally moves with a velocity of +1.5 m/s.(C) Ball C finally moves with a velocity of 0.5 m/s.

(D) Ball D finally moves with a velocity of 10 m / s9

-.


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0000CJA102119027

SECTION-II : (Maximum Marks: 18)� This section contains SIX questions.� The answer to each question is a NUMERICAL VALUE.� For each question, enter the correct numerical value (in decimal notation, truncated/rounded-

off to the second decimal place; e.g. 6.25, 7.00, –0.33, –.30, 30.27, –127.30, if answer is 11.36777.....then both 11.36 and 11.37 will be correct) by darken the corresponding bubbles in the ORS.For Example : If answer is –77.25, 5.2 then fill the bubbles as follows.

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

� Answer to each question will be evaluated according to the following marking scheme:Full Marks : +3 If ONLY the correct numerical value is entered as answer.Zero Marks : 0 In all other cases.

1. Electromagnetic waves of wavelength l = 500 nm are incident on a YDSE apparatus. Both slitsare of same width. Here d = 3 mm, D = 2 m. Intensity at central maxima is 1 W/m2, what is theintensity (in W/m2) at a point 1 mm above maxima of order 1?

2. A point source and a receiver are stationary . Source sends a wave of frequency 256 Hz, whichdirectly reaches to receiver and also after being reflected from a wall. The wall is moving at speedof 1 m/s. Find beat frequency (in Hz) heard by receiver. Velocity of sound = 321 m/s.

1m/s

S R


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0000CJA102119027

3. Surface tension of a liquid is being estimated by measuring height of liquid rises in a capillary.The travelling microscope has vernier scale whose reading for water level B is 2.3 cm on mainscale and 4th vernier scale division coinciding. The reading for lower level of meniscus A is 4.7 cmon main scale and 9th vernier scale division coinciding. For measuring the diameter, the left endof the capillary C measures 3.1 cm on main scale and 1st vernier scale division coinciding. Theright end of capillary D measures 3.2 cm on main scale and 4th vernier scale division coinciding.Here angle of contact = 0°, density is 103 kg/m3 and acceleration due to gravity = 10 m/s2 and10 vernier scale divisions = 9 main scale divisions. What is value of surface tension (in N/m) ?

A

B

C D

4. A charge of 8µC is given to point A and point B is earthed. If each capacitance is 1 µF, what is thetotal energy (in µJ) of the system ?

A

BSpace for Rough Work

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0000CJA102119027

5. The charge density inside a nucleus is given by 2

0 2r1a

æ ör = r -ç ÷

è ø where r £ a, a = radius of

nucleus = 3 fm and r0 = 5 × 1025 C/m3. The electric field is maximum at r = r0. Find r0 (in fm)and fill it in OMR sheet.

6. Two rigid walls at a distance of 1m act as a heat reservoir at 100 °C and 0°C. A well laggeduniform rod having length of 1m at 0°C is held between both the walls. It is found that when thesystem reaches steady state, the supporting force can be removed and limiting friction acts on therod at both the ends. Here a for rod = 10–5/°C, Y = 2 × 1011 N/m2, µwalls = 0.3 and area = 4mm2.What is the mass of the rod (in Kg) ?

100°C 0°C


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0000CJA102119027


PART-2 : CHEMISTRYSECTION-I(i) : (Maximum Marks : 12)




1. Select reaction(s) which will favour in forward direction

(A) O

O OH

OH+ NaHCO3

O

O ONa

ONa+ CO + H O2 2

(excess)

(B) Me –C º C – Me + NaNH2 Δ��

MeCH2 – C 3C Na NHÅº +1

(C) EtOH + NaOH � EtONa + H2O(D) None of these

2. Select an option having incorrect major product?

(A) Me3C – Cl + MeONa MeOH¾¾¾¾® Me3C – OMe

(B)

Cl

+ Me2NH ¾¾®

NMe2

(C) OAc

Δ¾¾®

(D) 2 2 4Conc. H SOCatalytic Δ¾¾¾¾¾¾®

3. Choose CORRECT outer electronic configuration of Cu, Cr & Pd are respectively.(A) 3d104s1, 3d54s1, 4d105s0 (B) 3d94s2, 3d44s2, 4d85s2

(C) 3d104s1, 3d54s1, 4d85s2 (D) 3d94s1, 3d54s2, 4d105s0

4. The solubility of Ag2CO3 in water is 5520 mg/L. Ksp of Ag2CO3 is (Ag = 108)(A) 4.0 × 10–10 (B) 8.0 × 10–15 (C) 3.2 × 10–14 (D) 1.6 × 10–11

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0000CJA102119027


SECTION-I(ii) : (Maximum Marks: 32)� This section contains EIGHT questions.� Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of

these four option(s) is (are) correct option(s).� For each question, choose the correct option(s) to answer the question.� Answer to each question will be evaluated according to the following marking scheme:




� For Example : If first, third and fourth are the ONLY three correct options for a question withsecond option being an incorrect option; selecting only all the three correct options will resultin +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options),without selecting any incorrect option (second option in this case), will result in +2 marks.Selecting only one of the three correct options (either first or third or fourth option), withoutselecting any incorrect option (second option in this case), will result in +1 marks. Selectingany incorrect option(s) (second option in this case), with or without selection of any correctoption(s) will result in –1 marks.

5. Which of the following atom/s has low ionization energy than Nitrogen atom.(A) Li (B) F (C) O (D) Ne

6. Choose CORRECT statements.(A) The bond dissociation energy of C2 decreases on forming 2CÅ and increases on forming 2C1

(B) Formation of 2OÅ from O2 electron is removed from gerade bonding molecular orbital(C) HOMO of F2 is gerade while LUMO of Cl2 is ungerade(D) In Halogen colour intensity increases F2 to I2

7. Which of the following statement(s) is/are correct about coagulation ?(A) Tanning of leather is an example of mutual coagulation.(B) Delta formation is an example of coagulation by electrolyte.(C) Cottrell precipitator coagulate by electrophoresis.(D) Lyophilic colloids may be coagulated by adding other suitable solvent.

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8. The molar conductance of H+ ion and CH3COO– ion at 298K and at infinite dilution are3.474 × 10–2 ohm–1 m2 mol–1 and 1.351 × 10–2 ohm–1 m2 mol–1 , respectively. The resistance of0.1 M CH3COOH solution is 2000 ohm at 298K in a conductivity cell of cell constant 20m–1.The correct information(s) related with the given parameters at 298K is/are-(1F = 96500 C/mol)

(A) 0m 3(CH COOH)L = 4.825 × 10–2 ohm–1 m2 mol–1

(B) k = 1.0 × 10–2 ohm–1m–1

(C) mL = 1.0 × 10–4 ohm–1 m2 mol–1

(D) [H+] = 2.07 × 10–3 M9. Which of the following will give alkane as product with RMgCl?

(A) PhOMe (B) Me3N (C) Me3C – Cl (D)

O

O10. Neopentane can be prepared in good yield by

(A) Catalytic reduction (B) Wolf-Kischner's reduction(C) Wurtz reaction (D) Decarboxylation using soda lime

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11.

Cl KO

CH3

CH3

CH3

D(X)

Major product

O3Zn / H2O

(A) + (B)

Product A and B are seprated and than heated with NaOH. A gave C where as B gaveD and ECorrect statement(s) is/are :

(A) Product C is

O

(B) Product D is H – C – OO

(C) Product E is CH3 – OH

(D) Major product (X) is

12. Choose CORRECT statements.(A) The dark blue colour of the solution formed when excess of ammonia is added to a solution

of copper(II) sulphate is due the formation of diamagnetic [Cu(NH3)4]2Å ion(B) [Cr(H2O)6]3Å have more paramagnetism than [Ni(H2O)6]2Å

(C) [Ni(CO)4] is tetrahedral and paramagnetic complex(D) [Ni(DMG)2] is low spin square planar complex

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0000CJA102119027



off to the second decimal place; e.g. 6.25, 7.00, –0.33, –.30, 30.27, –127.30, if answer is11.36777..... then both 11.36 and 11.37 will be correct) by darken the corresponding bubblesin the ORS.For Example : If answer is –77.25, 5.2 then fill the bubbles as follows.

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••


1. The normal boiling point of pure water is 373.15 K. Calculate the boiling point (in K) of anaqueous solution containing 11.7 gm NaCl and 34.2 gm sugar, C12H22O11 , in 260 gm water. Forwater, Kb = 0.52 K–Kg/mol. [Atomic mass of Na = 23, Cl = 35.5]

2. The reaction : A(g) ® 2B(g) + C(l) follows first order kinetics. The reaction is started withA(g) only. Calculate the rate constant of reaction ( in min–1) from the following data :(ln2 = 0.6931)

Time (min.) 0 10 ¥

Total pressure of system (bar) 0.080 0.125 0.165

3. One mole of non-ideal gas undergoes a process (10 L, 2 atm, 300 K) to (10L , 5 atm, 800K).The molar enthalpy change of gas in this process (in L-atm/mole) is(Given : Cv,m= 3.2R, Cp,m= 4.1 R , R = 0.08 L-atm/K-mole).

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0000CJA102119027


4. How many of the following are more reactive than benzene for electrophile substitutionreaction?(1) Anisole (2) Acetanilide (3) Carbolic acid (4) TNT(5) Mesitylene (6) Benzaldehyde (7) Acetophenone (8) Phenyl benzoate(9) Meta-Xylene (10) Ethyl benzoate

5. Total number of plot(s) in which slope is not correctly represented.

0°C273K

400°C673K

800°C1073K

1200°C1473K

1600°C1873K

2000°C2273K

– 1200

– 1100

– 1000

– 900

– 800

– 700

– 600

– 500

– 400

– 300

– 200

– 100

0

4Cu + O 2Cu O2 2®

C + OCO

2

2

®

2C + O 2CO2 ®

4/3Al + O 2/3Al O2

2 3®

2Mg + O 2MgO2 ®

Temperature

DG/k

Jmol

orO

–12

A

2Zn + O 2ZnO2®

2CO+O 2CO2

2®

6. Total number of anion which give white ppt. with aqueous solution of AgNO3.S2–, 3HSO1 , 2

3SO - , 22 3S O - , 3CH COO1 , Cl1 , I1 , 3

3BO -

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0000CJA102119027


PART-3 : MATHEMATICSSECTION-I(i) : (Maximum Marks : 12)




1. The sum of the series sinx sec3x + sin3x sec32x + sin32x sec33x+....upto n terms is

(A) n1 cot3 x cot x2é ù-ë û (B) n1 tan3 x tan x

2é ù-ë û

(C) [tan3nx – tanx] (D) cot3nx – cotx

2. If ax + y – 16 = 0 and x + ay – 6 = 0 represent two adjacent sides of a rhombus whose diagonalsintersect at (3,5) then sum of possible value of a is

(A) 4 (B) 2 (C) –2 (D) 94

-

3. Locus of a variable point P(x,y) is given by x3 + y3 + 3xy = 1 where (x,y) ¹ (–1,–1). The equationof circle touching the locus of P at (–1,2) and passing through (1,–2) is given by

(A) x2 + y2 + 2x + y – 5 = 0 (B) x2 + y2 + 12x + 6y – 5 = 0

(C) x2 + y2 + 4x + 2y – 5 = 0 (D) x2 + y2 – 4x – 2y – 5 = 0

4. If ( ) ( ) ( )2`

sin x 1ƒ : R R,ƒ x 2x 1 x x 14x 2x 3

pé ùë û® = + - + - ++ +

(where [x] denotes greatest integral value

less than or equal to x) denotes a function, then number of real solutions of equation

ƒ(x) = ƒ–1(x) is

(A) 0 (B) 1 (C) 2 (D) 3

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SECTION-I(ii) : (Maximum Marks: 32)� This section contains EIGHT questions.� Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of

these four option(s) is (are) correct option(s).� For each question, choose the correct option(s) to answer the question.� Answer to each question will be evaluated according to the following marking scheme:




� For Example : If first, third and fourth are the ONLY three correct options for a question withsecond option being an incorrect option; selecting only all the three correct options will resultin +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options),without selecting any incorrect option (second option in this case), will result in +2 marks.Selecting only one of the three correct options (either first or third or fourth option), withoutselecting any incorrect option (second option in this case), will result in +1 marks. Selectingany incorrect option(s) (second option in this case), with or without selection of any correctoption(s) will result in –1 marks.

5. Let ƒ be a twice differentiable function defined in [–4,4] such that ƒ'(–4) = 10, ƒ'(4) = –14 and

ƒ"(x) > – 3 " x Î [–4,4]. If ( ) ( )x

0

g x ƒ t dt= ò and ƒ(0) = 0, then

(A) g(x) increases in (–4,0)(B) g(x) decreases in (0,4)(C) graph of g(x) is concave down in (0,4)(D) ƒ(x) has maximum value at x = 2

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6. Which of the following can be antiderivative of ( )4

6x 1ƒ xx 1

+=

+ ? (Where 'C' is the constant of

integration)

(A) ( )1 1 31tan x tan x C3

- -+ + (B) ( )1 1 31 2tan x tan x Cx 3

- -æ ö- - +ç ÷è ø

(C) ( )1 1 31 tan x tan x C3

- -+ + (D) ( )1 1 32 1tan x tan x C3 x

- -æ ö- - +ç ÷è ø

7. Let 1 itz c1 it+æ ö- = rç ÷-è ø

where c is a complex number, r and t are real parameters then

(A) if z = c and r = 0 then t is undefined(B) if z – c Î ¡ then r = z – c and t = 0

(C) if z ¹ c then r = ± |z – c| and ( )( )

Im z ct

z c Re z c-

=± - + -

(D) If z ¹ c then r = ± |z – c| and ( )( )

Im z ct

z c Re z c-

=± - - -

E-23/28


0000CJA102119027

8. Let |r| < 1 and n kn

k 1S k r

¥

=

=å , then which of the following options is/are correct ?

(A) 01S

1 r=

-(B) ( )21 0rS S= (C) ( )n 1 n

dS r Sdr+ = (D) 2 3

2 0 0rS S 2S= +

9. In a game show a contestant selects one of three doors. Behind one of the doors there is aprize and behind other two there is no prize. Each door has equal probability of having aprize. After contestant selects a door, the game show host who knows what's behind each ofthe doors, deliberately opens one of the remaining doors and reveals there is no prize behindit. The host then asks the contestant whether he want to SWITCH his choice to other unopeneddoor or STICK to his original choice. Which of the following options is/are correct ?

(A) Probability of winning the prize is same whether contestant STICK or SWITCH

(B) Probability of winning the prize when contestant STICK, is 13

(C) Probability of winning the prize when contestant SWITCH, is 23

(D) It is advantageous for contestant to SWITCH than to STICK


E-24/28


0000CJA102119027

10. Consider a system of linear equations in three variables

–x + y + z = a

x – y + z = b

x + y – z = c

then

(A) a bx y c

2+

+ = + (B) b cy z a

2+

+ = + (C) a cx z b

2+

+ = + (D) x + y + z = a + b + c

11. Two tangents are drawn from point P(a,b) to the hyperbola. 3x2 – 2y2 = 6 and are inclined at

angles A and B to positive direction of x-axis. Let tanA.tanB = k, then

(A) if k = 2 then b2 = 2a2 + 7 (B) if k = 2 then b2 = 2a2 – 7

(C) if k = 3 then b2 = 3a2 – 9 (D) if k = 3 then b2 = 3a2 + 9

12. If 2n–1Tn = 2nC0 + 3(2nC2) + 32 (2nC4) + ....+3n(2nC2n), n Î N, then

(A) Tn, Tn+1, Tn+2 are in AP (B) Tn, 2Tn+1, Tn+2 are in AP

(C) T4 = 192 (D) T4 = 194


E-25/28


0000CJA102119027


off to the second decimal place; e.g. 6.25, 7.00, –0.33, –.30, 30.27, –127.30, if answer is11.36777..... then both 11.36 and 11.37 will be correct) by darken the corresponding bubblesin the ORS.For Example : If answer is –77.25, 5.2 then fill the bubbles as follows.

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••


1. Evaluate tanx

x tan x

x 0

tan xlimx

æ öç ÷-è ø

®

æ öç ÷è ø

(consider e = 2.7)

2. If ( ) n 11ƒ x

2 -= for n n 11 1x2 2 -< £ , x Î [0,1], n Î N and ƒ(0) = 0. Find the value of ( )

1

0

ƒ x dxò

3. If a curve represented by differential equation (tan–1y – x)dy = (1 + y2)dx, passes through (0,tan1)

and 1,b3pæ ö-ç ÷

è ø, then value of b is


E-26/28


0000CJA102119027

4. If ar and br

are perpendicular vectors with a 2, b 3= =rr and c a b´ =

rr r , then the least value

of c a-r r

5. If lx + 13y + mz + n = 0 is a plane through intersection of planes 2x + 3y – z + 1 = 0 and

x + y – 2z + 3 = 0 and is perpendicular to plane 3x – y – 2z = 4, then n

m-l

equals

6. Define permutation of set {1,2,3,.....,n} to be sortable if upon cancelling one appropriate termof such permutation remaining n – 1 terms are in increasing order. If ƒ(n) is the number of

such sortable permutations then ( )( )

ƒ 5ƒ 4 equals



E-27/280000CJA102119027

ALLENSpace for Rough Work

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

+91-744-2757575 [email protected] www.allen.ac.in

QUESTION PAPER FORMAT AND MARKING SCHEME :16. The question paper has three parts : Physics, Chemistry and Mathematics.17. Each part has two sections as detailed in the following table.

I have read all the instructionsand 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. .............................................


Your Target is to secure Good Rank in JEE 2020 0000CJA102119027E-28/28

ALLEN

Que. No. Category-wise Marks for Each Question MaximumSection Type of Full Partial Zero Negative Marks of the

Que. Marks Marks Marks Marks section+3 0 –1

I(i) Single If only the bubble If none In allcorrect option 4 corresponding to — of the other 12

the correct option bubbles is casesis darkened darkened

+4 +1 0 –1One 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 darkened+3 0

Numerical If only the bubble In all II Value Type 6 corresponding — other — 18

(Up to second to correct answer casesdecimal place) is darkened

Ïi;k bu funsZ'kks a dks /;ku ls i<+ s a

lkekU; %

1. ;g eksgjcU/k iqfLrdk vkidk iz'ui= gSA bldh eqgj rc rd u rksM+s tc rd bldk funsZ'k u fn;k tk;sA

2. iz'uksa dk mÙkj nsus ds fy, vyx ls nh x;h vkWfIVdy fjLikal 'khV (vks- vkj- ,l-) (ORS) dk mi;ksx djsaA

3. dPps dk;Z ds fy, bl iqfLrdk esa [kkyh LFkku fn;s x;s gSaA

4. bl iqfLrdk ds fiNys i`"B ij fn, x, LFkku esa viuk uke o QkWeZ uEcj fyf[k, ,oa gLrk{kj cukb;sA

5. bl iqfLrdk dh eqgj rksM+us ds ckn Ïi;k tk¡p ysa fd blesa 28 i`"B gSa vkSj izR;sd fo"k; ds lHkh 18 iz'u vkSj muds mÙkj

fodYi Bhd ls i<+ s tk ldrs gSaA ;fn ugha] rks iz'ui= dks cnyus ds fy, fujh{kd ls lEidZ djsaA

6. ijh{kkFkhZ iz'ui= dks ijh{kk dh lekIrh ij ys tk ldrs gSaA

vkWfIVdy fjLikal 'khV (vks-vkj-,l-) %

7. vks- vkj- ,l- dks ijh{kk ds lekiu ij fujh{kd ds }kjk ,d= dj fy;k tk,xkA

8. vks- vkj- ,l- esa gsj&Qsj@foÏfr u djsaA vks-vkj-,l- dk dPps dke ds fy, iz;ksx u djs aA

9. viuk uke vkSj QkWeZ uEcj vks-vkj-,l- esa fn, x, [kkuksa esa dye ls fy[ksa vkSj vius gLrk{kj djsaA buesa ls dksbZ Hkh fooj.k

vks-vkj-,l- es a dgha vkSj u fy[ksaA QkWeZ uEcj ds gj vad ds uhps vuq:i cqycqys dks dkyk djsaA

vks-vkj-,l- ij cqycqyks a dks dkyk djus dh fof/k :

10. vks-vkj-,l- ds cqycqyksa dks dkys ckWy ikWbUV dye ls dkyk djsaA

11. cqycqys dks iw.k ± :i ls dkyk djsaA

12. cqycqys dks dkyk djus dk mi;qDr rjhdk gS :

13. vks-vkj-,l- e'khu tk¡P; gSA lqfuf'pr djsa dh cqycqys lgh fof/k ls dkys fd, x;sa gSaA

14. cqycqys dks rHkh dkyk djsa tc vki mÙkj ds ckjs esa fuf'pr gksA dkys fd, gq, cqycqys dks feVkus vFkok lkQ djus dk dksbZ

rjhdk ugha gSA

15. g = 10 m/s2 iz;qDr djsa] tc rd fd vU; dksbZ eku ugha fn;k x;k gksA

fujh

{kd

ds v

uqns'k

ksa d

s fcu

k eqgj

sa u r

ksM+s

Ïi;k 'ks"k funs Z'kks a ds fy, bl iqfLrdk ds vfUre i`"B dks i<+ s aA

Time : 3 Hours Maximum Marks : 186

)0000CJA102119027)(0000CJA102119027) Test Pattern

CLASSROOM CONTACT PROGRAMME(Academic Session : 2019 - 2020)

JEE(Main+Advanced) : LEADER & ENTHUSIAST COURSE

JEE(Advanced)AIOT

09-02-2020

PAPER-1

Hin

di

H-2/28

dPps dk;Z ds fy, LFkku

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 = 58Atomic 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,

0000CJA102119027

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

· Coulomb's law constant pe9

0

1 = 9×104

· 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 s = 5.67 × 10–8 Wm–2–K–4

· Wien's displacement law constant b = 2.89 × 10–3 m–K· Permeability of vacuum µ0 = 4p × 10–7 NA–2

· Permittivity of vacuum Î0 = 20

1

cm· Planck constant h = 6.63 × 10–34 J–s


H-3/280000CJA102119027


BEWARE OF NEGATIVE MARKINGHAVE CONTROL ¾® HAVE PATIENCE ¾® HAVE CONFIDENCE Þ 100% SUCCESS


Hkkx-1 : Hkk Sfrd foKku[k.M–I(i) : (vf/kdre vad : 12)

� bl [k.M esa pkj iz'u gSa

� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSaA ftuesa dsoy ,d gh lgh gSaA

� izR;sd iz'u ds fy , vks-vkj-,l ij lgh mÙkj fodYi ds vuq:i cqycqys dks dkyk djsaA

� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk,axs %

iw.kZ vad : +3 ;fn flQZ lgh fodYi ds vuq:i cqycqys dks dkyk fd;k gSA

'kwU; vad : 0 ;fn fdlh Hkh cqycqys dks dkyk ugha fd;k gSA

½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA

1. ,d æO;ekughu vforkU; jLlh fdlh 20 cm f=T;k o 2kg æO;eku okys csyu ij fyiVh gqbZ gSA ;g æO;eku 1 kg okys

,d yVdrs gq, CykWd ls tqM+h gSA ;fn lrg fpduh gks rks csyu t1 le; esa l nwjh rd igq¡prk gSA ;fn lrg i;kZIr :i ls

[kqjnjh gks rks csyu t2 le; esa l nwjh rd igq¡prk gSA rc 1

2

tt

dk eku gksxk%&

m

M

(A) 1 (B) 7

10 (C) 107 (D)

12

H-4/28 0000CJA102119027



2. ,d leksÙky ysal (µ = 1.5) dks fdlh leksory ysal (µ = 1.3) ds lkFk la;qDr fd;k x;k gSA lHkh lrgkas dh oØrk

f=T;k leku gSA (ySalksa dks iryk ekusa)

(A) ;g la;kstu 1.4 viorZukad okys ek/;e esa ,d vfHklkjh ysal dh rjg rFkk 1.6 viorZukad okys ek/;e esa ,d

vilkjh ysal dh rjg O;ogkj djrk gSA

(B) ;g la;kstu 1.2 rFkk 1.4 viorZukad okys ek/;eksa esa ,d vfHklkjh ysal dh rjg O;ogkj djrk gSA

(C) ;g la;kstu 1.2 rFkk 1.6 viorZukad okys ek/;eksa esa ,d vilkjh ysal dh rjg O;ogkj djrk gSA

(D) ;g la;kstu 1.4 viorZukad okys ek/;e esa ,d vilkjh ysal dh rjg rFkk 1.6 viorZukad okys ek/;e esa ,d vfHklkjh

ysal dh rjg O;ogkj djrk gSA

3. ;g fo'ks"k inkFkZ (,d vkn'kZ xSl ugha) dks tc :¼ks"e izØe ls xqtkjk tkrk gS rks bldk PV vkjs[k ,d ljy js[kk AB ds

:i esa izkIr gksrk gSA ;fn ;gh inkFkZ izØe A ® C ® B ls xqtjrk gS rks bls fdruh Å"ek nh x;h gS\

A C

B

2×10 Pa5

1×10 Pa5

1 litre 2 litreV

P

(A) 50 J (B) 200 J (C) 150 J (D) 100 J

H-5/280000CJA102119027



4. vkos'k +10µC o æO;eku 1gm okyk ,d d.k fp=kuqlkj x o y v{k ls 45° dks.k ij 106 m/s osx ls xfr'khy gSA bl ij

½.kkRed z-fn'kk ds vuqfn'k ,d pqEcdh; cy yxrk gSA tc bls /kukRed z-fn'kk ds vuqfn'k 106 m/s osx ls iz{ksfir

fd;k tkrk gS rks bl ij /kukRed x fn'kk esa 10–2 N dk ,d pqEcdh; cy yxrk gSA pqEcdh; {ks= Br

dk eku gS %&

x

y

45°

v

(A) ( )3 ˆ ˆ10 T i j-- + (B) ( )3 ˆˆ10 j k- + (C) 3 ˆ10 i- (D) 3 ˆ10 T j--

H-6/28 0000CJA102119027

All India Open Test/Leader & Enthusiast Course/JEE(Advanced)/09-02-2020/Paper-1ALLEN[kaM -I(ii) : (vf/kdre vad : 32)

� bl [kaM esa vkB iz'u gSaA� izR;sd iz'u ds lgh mÙkj (mÙkjksa) ds fy, pkj fodYi fn, x, gSaA bl pkj fodYiksa esa ls ,d ;k ,d ls vfèkd

fodYi lgh gS(gSa)A� izR;sd iz'u ds fy,] iz'u dk (ds) mÙkj nsus gsrq lgh fodYi (fodYiksa) dks pqusA� izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstuk ds vuqlkj gksxk %

iw.kZ vad : +4 ;fn dsoy (lkjs) lgh fodYi (fodYiksa) dks pquk x;k gSAvkaf'kd vad : +3 ;fn pkjksa fodYi lgh gSa ijUrq dsoy rhu fodYiksa dks pquk x;k gSAvkaf'kd vad : +2 ;fn rhu ;k rhu ls vf/kd fodYi lgh gSa ijUrq dsoy nks fodYiksa dks pquk x;k gS vkSj

pqus gq, nksuksa fodYi lgh fodYi gSaAvkaf'kd vad : +1 ;fn nks ;k nks ls vf/kd fodYi lgh gSa ijUrq dsoy ,d fodYi dks pquk x;k gS vkSj

pquk gqvk fodYi lgh fodYi gSaA'kwU; vad : 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS (vFkkZr~ iz'u vuqÙkfjr gS)A½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA

� mnkgj.k Lo:i % ;fn fdlh iz'u ds fy, dsoy igyk] rhljk ,oa pkSFkk lgh fodYi gSa vkSj nwljk fodYi xyrgS_ rks dsoy lHkh rhu lgh fodYiks a dk p;u djus ij gh +4 vad feys axs aA fcuk dksbZ xyr fodYi pqus(bl mnkgj.k esa nwljk fodYi) rhu lgh fodYiksa esa ls flQZ nks dks pquus ij (mnkgj.kr% igyk rFkk pkSFkk fodYi)+2 vad feysaxsA fcuk dksbZ xyr fodYi pqus (bl mnkgj.k esa nwljk fodYi)] rhu lgh fodYiksa esas ls flQZ ,ddks pquus ij (igyk ;k rhljk ;k pkSFkk fodYi) +1 vad feysaxsA dksbZ Hkh xyr fodYi pquus ij (bl mnkgj.kesa nwljk fodYi), –1 vad feysaxs] pkgs lgh fodYi (fodYiksa) dks pquk x;k gks ;k u pquk x;k gksA

5. iznf'kZr ifjiFk esa a o b ds e/; foHkokUrj dk eku 300 V gSA izR;sd izfrjks/kd 100 W dk gSA

M1

M2

R R R

ba

(A) ;fn Åijh ehVj M1 vkn'kZ oksYVehVj rFkk fupyk ehVj M2 vkn'kZ vehVj gks rks oksYVehVj dk ikB~;kad 300 V gksxkA

(B) ;fn nksuksa ehVj vkn'kZ oksYVehVj gks rks izR;sd 200 V n'kkZrk gSA

(C) ;fn nksuksa ehVj vkn'kZ vehVj gks rks izR;sd 6A n'kkZrk gSA

(D) ;fn Åijh ehVj M1 vkn'kZ oksYVehVj rFkk fupyk ehVj M2 vkn'kZ vehVj gks rks vehVj dk ikB~;kad 3A gksxkA


H-7/280000CJA102119027

All India Open Test/Leader & Enthusiast Course/JEE(Advanced)/09-02-2020/Paper-1ALLEN6. ?kuRo r okys ,d IykfLVd fi.M dks igys ra ?kuRo dh ok;q esa rFkk fQj rw ?kuRo ds ty esa fLFkj jksd dj j[kk tkrk

gSA bls fLFkj jksddj j[kus ds fy;s gesa bl ij ,d cy yxkuk gksxkA

(A) bls ty esa jksd dj j[kuk vklku gksxk ;fn r > ra rFkk w a

2r + r

r < gksA

(B) bls ty esa jksd dj j[kuk vklku gksxk ;fn r > rw gksA

(C) bls ok;q esa jksd dj j[kuk vklku gksxk ;fn r < rw rFkk w a

2r + r

r < gksA

(D) bls ty esa jksd dj j[kuk vklku gksxk ;fn r < rw rFkk w a

2r + r

r > gksA

7. æO;eku m okyh ,d yEch /kkfRod NM+ dks fu;r osx v ls nks fpduh {kSfrt /kkfRod iVfj;ksa ij xfr djk;h tkrh gSA iVfj;ksa

ds e/; dks.k dk eku a gSA ;gk¡ Å/okZ/kj fn'kk esa ,d le:i pqEcdh; {ks= (B) fp=kuqlkj fo|eku gSA la/kkfj= izkjEHk esa

vukosf'kr gSA ifjiFk dk izfrjks/k 'kwU; gSA lgh vkjs[k@vkjs[kksa dks pqfu;s%&

A

v

a

C

x0

(A)

time

icurrent in therail

(B)

time

External force to be

appliedon the rod

(C)

time

Distance ofpoint of

application of force

from lower end

(D)

time

Energystored incapacitor


H-8/28 0000CJA102119027


8. ,d jsfM;kslfØ; ukfHkd A nks a o rhu b– d.kksa ds mRltZu ds ckn C esa fo?kfVr gksrk gSA ,d vU; ukfHkd B ,d a

o ik¡p b– d.kksa ds mRltZu }kjk C esa fo?kfVr gksrk gSA A o B dh v/kZvk;q Øe'k% 1 ?k.Vk o 2 ?k.Vk gS rFkk e/;orhZ

mRiknksa dh vèkZvk;q ux.; gSA le; t = 0 ij A dh la[;k 4N0, B dh la[;k N0 rFkk C dh la[;k 'kwU; gSA

(A) A dh ijek.kq æO;eku la[;k – B dh ijek.kq æO;eku la[;k = 4(B) A dk ijek.kq Øekad – B dk ijek.kq Øekad = – 2

(C) tc A o B dh la[;k leku gksrh gS] rc C dh la[;k 09N2 gSA

(D) A o B dh fo?kVu nj t = 6 ?k.Vs ij leku gksrh gSA

9. tc ,d ladjs vuqukn LrEHk midj.k dks vkWtksu xSl (g = 4/3) ls Hkjk tkrk gS rks ;g ,d Lofj= ds lkFk izFke vuqukn esa

vuqukn djrk gSA tc bls ukbVªkstu xSl (N2) ls Hkjk tkrk gS rks ;g leku Lofj= ds lkFk izFke vuqukn esa vuqukn djrk gS

tcfd ty LrEHk dks 10 cm uhps dh vksj foLFkkfir fd;k x;k gSA ( 5 = 2.25 ysa)

(A) vkWtksu esa /ofu dh rjaxnS/; Z yxHkx 1m gSA

(B) ukbVªkstu esa /ofu dh rjaxnS/; Z yxHkx 1.6m gSA

(C) vkWtksu xSl esa f}Ùkh; vuqukn ds fy, ty LrEHk dks 60 cm uhps dh vksj foLFkkfir djuk gksxkA

(D) ukbVªkstu xSl esa f}Ùkh; vuqukn ds fy, ty LrEHk dks 80 cm uhps dh vksj foLFkkfir djuk gksxkA


H-9/280000CJA102119027

All India Open Test/Leader & Enthusiast Course/JEE(Advanced)/09-02-2020/Paper-1ALLEN10. ,d AC ifjiFk esa ckWDl esa ,d izsjd dq.Myh ;k ,d la/kkfj= gks ldrk gSA

Box

L=ksr ij oksYVrk iznf'kZr vkjs[k esa n'kkZ;h x;h gSA

V

t

;gk¡ Vbox o le; rFkk Vresistance o le; ds e/; vkjs[k [khaps x;s gSA laHkkfor lgh fodYi@fodYiksa dks pqfu;s%&

(A) ;fn ckWDl esa ,d iszjd dq.Myh gks rks

Vbox

t

(B) ;fn ckWDl esa ,d izsjd dq.Myh gks rks

Vresistance

t

(C) ;fn ckWDl esa ,d la/kkfj= gks rks

Vresistace

t

(D) ;fn ckWDl esa ,d la/kkfj= gks rks

Vbox

t


H-10/28 0000CJA102119027

All India Open Test/Leader & Enthusiast Course/JEE(Advanced)/09-02-2020/Paper-1ALLEN11. æO;eku m okys ,d fi.M ij dsUæh; cy F kr= -

r r yxrk gSA ;gk¡ rr cy ds dsUæ ls fLFkfr gS rFkk k ,d /kukRed

fu;rkad gSA ;g ,d can o`Ùkkdkj iFk esa bl izdkj xfr djrk gS fd cy ds dsUæ ds lkis{k bldk dks.kh; laosx DokUVhd`r

gksrk gS] vFkkZr~ nhL2

=p

gksrk gSA r = 0 ij fLFkfrt ÅtkZ = 0 ysaA

(A) bldh laHkkfor f=T;k bl izdkj DokUVhd`r gksrh gS fd r µ n1/2 gksrk gSA

(B) bldh laHkkfor ;kaf=d ÅtkZ bl izdkj DokUVhd`r gksrh gS fd E µ n gksrk gSA

(C) bldh pky bl izdkj DokUVhd`r gksrh gS fd v µ n1/2 gksrk gSA

(D) bldk dks.kh; osx bl izdkj DokUVhd`r gksrk gS fd w µ n gksrk gSA

12. leku æO;eku okyh 5 izR;kLFk xsans ; A,B, C, D o E ,d yEcs ?k"kZ.kjfgr {kSfrt rkj ij xfr'khy gSA budh fLFkfr x o le;

ds e/; ,d vkjs[k [khapk tkrk gSA bl vkjs[k ds vk/kkj ij lEHkkfor lgh dFku@ dFkuksa dks pqfu;sA

1m

2m

3m

4m

1 sec. 1.8 sec.0ED

C

B

A

1.5m

(A) ;gk¡ dqy 10 VDdjsa gksrh gSA (B) xsan A var esa +1.5 m/s osx ls xfr djrh gSA

(C) xsan C var esa 0.5 m/s osx ls xfr djrh gSA (D) xsan D var esa 10 m / s9

- osx ls xfr djrh gSA


H-11/280000CJA102119027

All India Open Test/Leader & Enthusiast Course/JEE(Advanced)/09-02-2020/Paper-1ALLEN[kaM-II : (vf/kdre vad : 18)

� bl [kaM esa N% iz'u gSaA� izR;sd iz'u dk mÙkj ,d la[;kRed eku (NUMERICAL VALUE) gSA� izR;sd iz'u ds mÙkj ds lgh la[;kRed eku (n'keyo vadu esa] n'keyo ds f}rh; LFkku rd :f.Mr@fudfVr_ mnkgj.k

6.25, 7.00, –0.33, –.30, 30.27, –127.30, ;fn mÙkj 11.36777..... gS] rks 11.36 vkSj 11.37 nksuksa lgh gksxsa)dks izfo"B djus ds fy, vks-vkj-,l- esa vuq:i cqycqys dks dkyk djsaA

mnkgj.k ds fy, : ;fn mÙkj –77.25, 5.2 gS] rks cqycqyksa dks fuEu izdkj ls dkyk djsaA

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

� izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstuk ds vuqlkj gksxk%&iw.kZ vad : +3 ;fn flQZ lgh la[;kRed eku (Numerical value) gh mÙkj Lo:i ntZ fd;k x;k gSA'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA

1. ,d YDSE midj.k ij l = 500 nm rjaxnS/; Z dh fo|qrpqEcdh; rjaxsa vkifrr gksrh gSA nksuksa fLyVsa leku pkSM+kbZ dh gSA

;gk¡ d = 3 mm, D = 2 m gSA ;fn dsUæh; mfPp"B ij rhozrk 1 W/m2 gks rks izFke Øe ds mfPp"B ds 1 mm Åij fLFkr

fdlh fcUnq ij rhozrk (W/m2 esa) D;k gksxh\

2. ,d fcUnq L=ksr rFkk ,d xzkgh fLFkj gSA ;g L=ksr vko`fÙk 256 Hz dh ,d rjax izsf"kr djrk gS tks lh/ks rFkk ,d nhokj ls

ijkofrZr gksus ds ckn] nksuksa çdkj ls xzkgh rd igq¡prh gSA nhokj 1 m/s pky ls xfr'khy gSA xzkgh }kjk lquh x;h foLiUn vko`fÙk

(Hz esa) Kkr dhft;sA /ofu dk osx = 321 m/s gSA

1m/s

S R


H-12/28 0000CJA102119027

All India Open Test/Leader & Enthusiast Course/JEE(Advanced)/09-02-2020/Paper-1ALLEN3. ds'kuyh esa p<+ s gq, æo dh Å¡pkbZ dks eki dj bl æo dk i`"B ruko Kkr fd;k tk ldrk gSA py lw{en'khZ ds ofuZ;j

iSekus dk ty Lrj B ds fy;s eq[; iSekus ij ikB~;kad 2.3 cm gS rFkk ofuZ;j iSekus dk prqFkZ Hkkx laikrh gSA uopUæd

ds fupys Lrj A ds fy;s eq[; iSekus ij ikB~;kad 4.7 cm gS rFkk ofuZ;j iSekus dk 9 ok¡ Hkkx laikrh gSA O;kl ds ekiu

ds fy;s ds'kuyh dk ck¡;k fljk C eq[; iSekus ij 3.1 cm ekirk gS rFkk ofuZ;j iSekus dk izFke Hkkx laikrh gSA ds'kuyh

dk nk¡;k fljk D eq[; iSekus ij 3.2 cm ekirk gS rFkk ofuZ;j iSekus dk prqFkZ Hkkx laikrh gSA ;gk¡ laidZ dks.k = 0°, ?kuRo

103 kg/m3 o xq:Roh; Roj.k = 10 m/s2 gS rFkk ofuZ;j iSekus ds 10 Hkkx = eq[; iSekus ds 9 Hkkx gSA i`"B ruko dk eku

(N/m esa) Kkr dhft;sA

A

B

C D

4. iznf'kZr ifjiFk esa fcUnq A dks 8µC vkos'k fn;k tkrk gS rFkk fcUnq B HkwlaifdZr fd;k x;k gSA ;fn izR;sd /kkfjrk 1 µFgks rks fudk; dh dqy ÅtkZ (µJ esa) D;k gksxh\

A

BdPps dk;Z ds fy, LFkku

H-13/280000CJA102119027


5. fdlh ukfHkd ds vanj vkos'k ?kuRo 2

0 2r1a

æ ör = r -ç ÷

è ø }kjk fn;k tkrk gS_ tgk¡ r £ a gS rFkk a = ukfHkd dh f=T;k = 3fm

o r0 = 5 × 1025 C/m3 gSA fo|qr {ks= r = r0 ij vf/kdre gSA r0 dk eku (fm esa) Kkr dhft;sA

6. ,d&nwljs ls 1m dh nwjh ij fLFkr nks n`<+ nhokjas 100 °C o 0°C okys ,d Å"ek ik= dh rjg O;ogkj djrh gSA

fp=kuqlkj bu nksuksa nhokjksa ds e/; 0°C ij 1m yEch dqpkyd le:i NM+ dks fLFkj j[kk tkrk gSA ;g ik;k x;k gS fd

tc fudk; LFkk;h voLFkk esa igq¡prk gS rks lgk;d cy dks gVk;k tk ldrk gS rFkk NM + ij nksuksa fljksa ij lhekUr ?k"kZ.k

yxrk gSA ;gk¡ NM+ ds fy;s a = 10–5/°C, Y = 2 × 1011 N/m2, µwalls = 0.3 rFkk {kS=Qy = 4mm2 gSA NM+ dk

æO;eku (Kg esa) Kkr dhft;sA

100°C 0°C


H-14/28 0000CJA102119027



Hkkx-2 : jlk;u foKku[k.M–I(i) : (vf/kdre vad : 12)

� bl [k.M esa pkj iz'u gSa

� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSaA ftuesa dsoy ,d gh lgh gSaA� izR;sd iz'u ds fy, vks-vkj-,l ij lgh mÙkj fodYi ds vuq:i cqycqys dks dkyk djsaA

� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk,axs %

iw.kZ vad : +3 ;fn flQZ lgh fodYi ds vuq:i cqycqys dks dkyk fd;k gSA

'kwU; vad : 0 ;fn fdlh Hkh cqycqys dks dkyk ugha fd;k gSA

½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA

1. ,slh vfHkfØ;k pqfu;s tks vxz fn'kk esa gksxh\

(A) O

O OH

OH+ NaHCO3

O

O ONa

ONa+ CO + H O2 2

(excess)

(B) Me –C º C – Me + NaNH2 Δ��

MeCH2 – C 3C Na NHÅº +1

(C) EtOH + NaOH � EtONa + H2O

(D) bues ls dksbZ ugha

2. xyr eq[; mRikn iznf'kZr dj jgh vfHkfØ;k pqfu;s\

(A) Me3C – Cl + MeONa MeOH¾¾¾¾® Me3C – OMe

(B)

Cl

+ Me2NH ¾¾®

NMe2

(C) OAc

Δ¾¾®

(D) 2 2 4Conc. H SOCatalytic Δ¾¾¾¾¾¾®

3. Cu, Cr rFkk Pd ds Øe'k% lgh cká bysDVªkWfu; foU;kl iznf'kZr djus okyk fodYi gS

(A) 3d104s1, 3d54s1, 4d105s0 (B) 3d94s2, 3d44s2, 4d85s2

(C) 3d104s1, 3d54s1, 4d85s2 (D) 3d94s1, 3d54s2, 4d105s0

4. ty esa Ag2CO3 dh foys;rk 5520 mg/L gS ] rks Ag2CO3 dk Ksp gS (Ag = 108)

(A) 4.0 × 10–10 (B) 8.0 × 10–15 (C) 3.2 × 10–14 (D) 1.6 × 10–11

H-15/280000CJA102119027



[kaM -I(ii) : (vf/kdre vad : 32)� bl [kaM esa vkB iz'u gSaA� izR;sd iz'u ds lgh mÙkj (mÙkjksa) ds fy, pkj fodYi fn, x, gSaA bl pkj fodYiksa esa ls ,d ;k ,d ls vfèkd



pqus gq, nksuks a fodYi lgh fodYi gSaAvkaf'kd vad : +1 ;fn nks ;k nks ls vf/kd fodYi lgh gSa ijUrq dsoy ,d fodYi dks pquk x;k gS vkSj



5. fuEu esa ls dkSu ls ijek.kqvksa dh vk;uu ÅtkZ] ukbVªkstu ijek.kq ls de gS.(A) Li (B) F (C) O (D) Ne

6. lgh dFku pqfu,

(A) C2 ls] 2CÅ cuus ij ca/k fo;kstu ÅtkZ ?kVrh gS rFkk 2C1 cuus ij ca/k fo;kstu c<+rh gS

(B) O2 ls 2OÅ ds fuekZ.k esa bysDVªkWu] ftjsM ca/kh vkf.od d{kd ls gVrk gS

(C) F2 dk HOMO, ftjsM gS tcfd Cl2 dk LUMO vuftjsM gS

(D) gsykstu esas a F2 ls I2 dh vksj c<+us ij jax dh rhozrk c<+rh gS

7. Ldanu ds ckjs esa dkSuls dFku lgh gS@gSa ?

(A) peM+s dh Vsfuax] ikjLifjd (mutual) Ldanu dk mnkgj.k gS

(B) MsYVk dk fuekZ.k oS|qr vi?kV~; }kjk Ldanu dk mnkgj.k gS

(C) dksVsªy vo{ksid oS|qr d.k lapyu }kjk LdfUnr djrk gS

(D) nzo Lusgh dksyksbM dks vU; mi;qDr foyk;d feykdj LdfUnr fd;k tk ldrk gS

H-16/28 0000CJA102119027



8. 298K rFkk vUkUr ru qrk ij H+ vk;u rFk k CH3COO– vk;u dk ek syj pkydRo Øe'k%3.474 × 10–2 ohm–1 m2 mol–1 rFkk 1.351 × 10–2 ohm–1 m2 mol–1 gSA 20 m–1 ds lSy fu;rkad ds pkydrklSy esa 298K ij 0.1 M CH3COOH foy;u dk izfrjks/k 2000 ohm gSA 298K ij fn;s x;s izkpyksa ds lkFklEcfU/kr lgh lwpuk,sa gS@gSa(1F = 96500 C/mol)(A) 0

m 3(CH COOH)L = 4.825 × 10–2 ohm–1 m2 mol–1

(B) k = 1.0 × 10–2 ohm–1m–1

(C) mL = 1.0 × 10–4 ohm–1 m2 mol–1

(D) [H+] = 2.07 × 10–3 M9. fuEu es ls dkSuls ;kSfxd] RMgCl ds lkFk vfHkfØ;k djkus ij mRikn ds :i es ,sYdsu nsxsa\

(A) PhOMe (B) Me3N (C) Me3C – Cl (D)

O

O10. fu;ksisUVsu dks] fuEu es ls fdlds }kjk vPNh yfC/k esa cuk;k tk ldrk gS\

(A) mRizsjdh; vip;u (B) oksYQ fd'uj vip;u

(C) oqV~tZ vfHkfØ;k (D) lksMkykbe dk iz;ksx djds fodkcksZfDlyhdj.k

H-17/280000CJA102119027


11.

Cl KO

CH3

CH3

CH3

D(X)eq[;mRikn

O3Zn / H2O

(A) + (B)

mRikn A rFkk B dks i`Fkd fd;k vksj fQj NaOH ds lkFk xeZ fd;k x;kA A, C nsrk gS tcfd B, D rFkk Ensrk gSAlgh dFku gS@gSa\

(A) mRikn C,

O

gSA (B) mRikn D, H – C – OO

gSA

(C) mRikn E, CH3 – OH gSA (D) eq[; mRikn (X), gSA

12. lgh dFku pqfu,(A) tc dkWij(II) lYQsV ds foy;u esa veksfu;k dk vkf/kD; feyk;k tkrk gS rks xgjs uhys jax dk foy;u cuus

dk dkj.k] izfrpqEcdh; [Cu(NH3)4]2Å vk;u fufeZr gksuk gS

(B) [Ni(H2O)6]2Å dh rqyuk esa [Cr(H2O)6]3Å esa vuqpqEcdRo vf/kd gS

(C) [Ni(CO)4] prq"Qydh; rFkk vuqpqEcdh; ladqy gS

(D) [Ni(DMG)2] fuEu pØ.k oxkZdkj leryh; ladqy gS


H-18/28 0000CJA102119027



[kaM-II : (vf/kdre vad : 18)� bl [kaM esa N% iz'u gSaA� izR;sd iz'u dk mÙkj ,d la[;kRed eku (NUMERICAL VALUE) gSA� izR;sd iz'u ds mÙkj ds lgh la[;kRed eku (n'keyo vadu esa] n'keyo ds f}rh; LFkku rd :f.Mr@fudfVr_

mnkgj.k 6.25, 7.00, –0.33, –.30, 30.27, –127.30, ;fn mÙkj 11.36777..... gS] rks 11.36 vkSj 11.37 nksuksa lghgksxsa) dks izfo"B djus ds fy, vks-vkj-,l- esa vuq:i cqycqys dks dkyk djsaA

mnkgj.k ds fy , : ;fn mÙkj –77.25, 5.2 gS ] rks cqycqyksa dks fuEu izdkj ls dkyk djsaA

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

� izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstuk ds vuqlkj gksxk %&iw.kZ vad : +3 ;fn flQZ lgh la[;kRed eku (Numerical value) gh mÙkj Lo:i ntZ fd;k x;k gSA'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA

1. 'kq¼ ty dk lkekU; DoFkukad 373.15 K gSA ,d tyh; foy;u ftlesa 260gm ty esa 11.7 gm NaCl rFkk

34.2 gm 'kdZjk , C12H22O11 mifLFkr gS] ds DoFkukad dh x.kuk (K esa) dhft;sA ty ds fy;s Kb = 0.52 K–Kg/mol.

[ijek.kq nzO;eku : Na = 23, Cl = 35.5]

2. vfHkfØ;k : A(g) ® 2B(g) + C(l) izFke dksfV cyxfrdh dk vuqlj.k djrh gSA vfHkfØ;k dsoy A(g) ds lkFk

izkjEHk dh x;h gS] rks fuEu vkadM+ks ls vfHkfØ;k ds nj fu;rkad dh x.kuk (min–1 esa) dhft;s:(ln2 = 0.6931)

le; (min) 0 10 ¥

ra=k dk dqy nkc (bar) 0.080 0.125 0.165

3. ,d okLrfod xSl dk ,d eksy (10 L, 2 atm, 300 K) ls (10L , 5 atm, 800K) rd ds izØe esa gSA bl izØe

esa xSl dh eksyj , sUFksYih esa ifjorZu (L-atm/mole esa) gSA

(fn;k gS : Cv,m= 3.2R, Cp,m= 4.1 R , R = 0.08 L-atm/K-mole).

H-19/280000CJA102119027



4. fuEu es ls fdrus bysDVªkWuLusgh izfrLFkkiu vfHkfØ;k ds fy;s csathu dh rqyuk es vf/kd fØ;k'khy gS\

(1) ,sfulkWy (2) ,sflVsfuykbM (3) dkcksZfyd vEy (4) TNT(5) esflfVyhu (6) csUtsfYMgkbM (7) ,sflVksQhuksu (8) Qsfuy csatks,sV

(9) esVk-tkbfyu (10) ,sfFky csatks,sV

5. fuEu esa ls , sls vkjs[kks a dh dqy la[;k fyf[k, ftuesa <ky] lgh :i ls iznf'kZr ugha fd;k x;k gS

0°C273K

400°C673K

800°C1073K

1200°C1473K

1600°C1873K

2000°C2273K

– 1200

– 1100

– 1000

– 900

– 800

– 700

– 600

– 500

– 400

– 300

– 200

– 100

0

4Cu + O 2Cu O2 2®

C + OCO

2

2

®

2C + O 2CO2 ®

4/3Al + O 2/3Al O2

2 3®

2Mg + O 2MgO2 ®

Temperature

DG/k

Jmol

orO

–12

A

2Zn + O 2ZnO2®

2CO+O 2CO2

2®

6. fuEu esa ls , sls ½.kk;uksa dh dqy la[;k crkb;s tks AgNO3 ds tyh; foy;u ds lkFk 'osr vo{ksi nsrs gSa

S2–, 3HSO1 , 23SO - , 2

2 3S O - , 3CH COO1 , Cl1 , I1 , 33BO -

H-20/28 0000CJA102119027



Hkkx-3 : xf.kr[k.M–I(i) : (vf/kdre vad : 12)

� bl [k.M esa pkj iz'u gSa� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSaA ftuesa dsoy ,d gh lgh gSaA� izR;sd iz'u ds fy , vks-vkj-,l ij lgh mÙkj fodYi ds vuq:i cqycqys dks dkyk djsaA� izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk,axs %

iw.kZ vad : +3 ;fn flQZ lgh fodYi ds vuq:i cqycqys dks dkyk fd;k gSA'kwU; vad : 0 ;fn fdlh Hkh cqycqys dks dkyk ugha fd;k gSA½.k vad : –1 vU; lHkh ifjfLFkfr;ksa esaA

1. Js<+h sinx sec3x + sin3x sec32x + sin32x sec33x+....ds n inksa rd dk ;ksxQy gksxk

(A) n1 cot3 x cot x2é ù-ë û (B) n1 tan3 x tan x

2é ù-ë û

(C) [tan3nx – tanx] (D) cot3nx – cotx

2. ;fn ,d leprqHkqZt dh nks vklUu Hkqtkvksa ax + y – 16 = 0 rFkk x + ay – 6 = 0 ls O;Dr djrs gS ftlds fod.kZfcUnq (3,5) ij izfrPNsn djrs gS] rks a ds laHko ekuksa dk ;ksxQy gksxk

(A) 4 (B) 2 (C) –2 (D) 94

-

3. ,d pj fcUnq P(x,y) dk fcUnqiFk] x3 + y3 + 3xy = 1 tgk¡ (x,y) ¹ (–1,–1) }kjk fn;k tkrk gSA oÙk dk lehdj.k] tksP ds fcUnqiFk dks fcUnq (–1,2) ij Li'kZ djrk gS rFkk fcUnq (1,–2) ls xqtjrk gS] fuEu }kjk fn;k trkk gS

(A) x2 + y2 + 2x + y – 5 = 0 (B) x2 + y2 + 12x + 6y – 5 = 0

(C) x2 + y2 + 4x + 2y – 5 = 0 (D) x2 + y2 – 4x – 2y – 5 = 0

4. ;fn ( ) ( ) ( )2`

sin x 1ƒ : R R,ƒ x 2x 1 x x 14x 2x 3

pé ùë û® = + - + - ++ +

(tgk¡ [x], x ls de ;k cjkcj egÙke iw.kk±d eku

dks n'kkZrk gS) ,d Qyu dks n'kkZrk gS] rks lehdj.k ƒ(x) = ƒ–1(x) ds okLrfod gyksa dh la[;k gksxh

(A) 0 (B) 1 (C) 2 (D) 3

H-21/280000CJA102119027



[kaM -I(ii) : (vf/kdre vad : 32)� bl [kaM esa vkB iz'u gSaA� izR;sd iz'u ds lgh mÙkj (mÙkjksa) ds fy, pkj fodYi fn, x, gSaA bl pkj fodYiksa esa ls ,d ;k ,d ls vfèkd



pqus gq, nksuks a fodYi lgh fodYi gSaAvkaf'kd vad : +1 ;fn nks ;k nks ls vf/kd fodYi lgh gSa ijUrq dsoy ,d fodYi dks pquk x;k gS vkSj



5. ekuk vUrjky [–4,4] esa ƒ nks ckj vodyuh; Qyu bl izdkj ifjHkkf"kr gS fd ƒ'(–4) = 10, ƒ'(4) = –14 rFkk

ƒ"(x) > – 3 " x Î [–4,4] gSA ;fn ( ) ( )x

0

g x ƒ t dt= ò rFkk ƒ(0) = 0 gks] rks

(A) vUrjky (–4,0) esa g(x) o/kZeku gksxkA

(B) vUrjky (0,4) esa g(x) ákleku gksxkA

(C) vUrjky (0,4) esa g(x) dk vkjs[k v/kkseq[kh gksxkA

(D) x = 2 ij ƒ(x) dk vf/kdre eku gksxkA

H-22/28 0000CJA102119027



6. ( )4

6x 1ƒ xx 1

+=

+ dk lekdyu gks ldrk gSA (tgk¡ 'C' lekdyu vpj gS)

(A) ( )1 1 31tan x tan x C3

- -+ + (B) ( )1 1 31 2tan x tan x Cx 3

- -æ ö- - +ç ÷è ø

(C) ( )1 1 31 tan x tan x C3

- -+ + (D) ( )1 1 32 1tan x tan x C3 x

- -æ ö- - +ç ÷è ø

7. ekuk 1 itz c1 it+æ ö- = rç ÷-è ø

tgk¡ c, lfEEJ la[;k] r rFkk t okLrfod izkpy gS] rks

(A) ;fn z = c rFkk r = 0 gks] rks t vifjHkkf"kr gksxkA

(B) ;fn z – c Î ¡ gks] rks r = z – c rFkk t = 0 gksxkA

(C) ;fn z ¹ c gks] rks r = ± |z – c|rFkk ( )( )

Im z ct

z c Re z c-

=± - + -

gksxkA

(D) ;fn z ¹ c gks] rks r = ± |z – c|rFkk ( )( )

Im z ct

z c Re z c-

=± - - -

gksxkA

H-23/280000CJA102119027


8. ekuk |r| < 1 rFkk n kn

k 1S k r

¥

=

=å gks] rks fuEu esa ls dkSulk fodYi lgh gksxk@gksxsa ?

(A) 01S

1 r=

-(B) ( )21 0rS S= (C) ( )n 1 n

dS r Sdr+ = (D) 2 3

2 0 0rS S 2S= +

9. ,d [ksy 'kks esa ,d izfr;ksxh rhu njoktksa esa ls ,d dk p;u djrk gSA njoktksa esa ls ,d njokts ds ihNs ,d iq:LdkjgS rFkk vU; nks njoktksa ds ihNs dksbZ iq:Ldkj ugha gSA izR;sd njokts esa ,d iq:Ldkj gksus dh leku izkf;drk gSA izfr;ksxhds ,d njokts dk p;u djus ckn] xse 'kks gksLV tks tkurk gS fd izR;sd njokts ds ihNs D;k gS] tkucw>dj 'ks"k njoktksa esals ,d dks [kksyrk gS rFkk irk pyrk gS fd blds ihNs dksbZ iq:Ldkj ugha gSA estcku rc izfr;ksxh ls iwNrk gS fd D;k ogviuh ialn ds vU; vukjf{kr njokts ls cnyuk (SWITCH) pkgrk gS ;k viuh ewy ialn dks j[kuk (STICK) pkgrk

gSA fuEu esa ls dkSulk fodYi lgh gksxk@gksxsa

(A) iq:Ldkj ds thrus dh izkf;drk leku gksxh ;fn izfr;ksxh j[krk (STICK) ;k cnyrk (SWITCH) gSA

(B) iq:Ldkj ds thrus dh izkf;drk] tc izfr;ksxh j[kuk (STICK) pkgrk gS, 13 gksxhA

(C) iq:Ldkj ds thrus dh izkf;drk] tc izfr;ksxh cnyuk (SWITCH) pkgrk gS, 23 gksxhA

(D) ;g j[kus (SWITCH) dh rqyuk esa cnyuk (STICK) izfr;ksxh ds fy;s mi;ksxh gksxkA


H-24/28 0000CJA102119027

All India Open Test/Leader & Enthusiast Course/JEE(Advanced)/09-02-2020/Paper-1ALLEN10. ekuk rhu pjksa esa js[kh; lehdj.k fudk;

–x + y + z = a

x – y + z = b

x + y – z = c

gks] rks

(A) a bx y c

2+

+ = + (B) b cy z a

2+

+ = + (C) a cx z b

2+

+ = + (D) x + y + z = a + b + c

11. vfrijoy; 3x2 – 2y2 = 6 ds fcUnq P(a,b) ls [khaph xbZ nks Li'kZ js[kk;sa x v{k dh /kukRed fn'kk ds lkFk dks.k A rFkk

B ij > qdh gqbZ gSA ekuk tanA.tanB = k gks] rks

(A) ;fn k = 2 gks] rks b2 = 2a2 + 7 gksxkA (B) ;fn k = 2 gks] rks b2 = 2a2 – 7 gksxkA

(C) ;fn k = 3 gks] rks b2 = 3a2 – 9 (D) ;fn k = 3 gks] rks b2 = 3a2 + 9 gksxkA

12. ;fn 2n–1Tn = 2nC0 + 3(2nC2) + 32 (2nC4) + ....+3n(2nC2n), n Î N gks] rks

(A) Tn, Tn+1, Tn+2 lekUrj Js<+h esa gksxsa sA (B) Tn, 2Tn+1, Tn+2 lekUrj Js<+h esa gksxsa sA

(C) T4 = 192 (D) T4 = 194


H-25/280000CJA102119027

All India Open Test/Leader & Enthusiast Course/JEE(Advanced)/09-02-2020/Paper-1ALLEN[kaM-II : (vf/kdre vad : 18)

� bl [kaM esa N% iz'u gSaA� izR;sd iz'u dk mÙkj ,d la[;kRed eku (NUMERICAL VALUE) gSA� izR;sd iz'u ds mÙkj ds lgh la[;kRed eku (n'keyo vadu esa] n'keyo ds f}rh; LFkku rd :f.Mr@fudfVr_

mnkgj.k 6.25, 7.00, –0.33, –.30, 30.27, –127.30, ;fn mÙkj 11.36777..... gS] rks 11.36 vkSj 11.37 nksuksa lghgksxsa) dks izfo"B djus ds fy, vks-vkj-,l- esa vuq:i cqycqys dks dkyk djsaA

mnkgj.k ds fy, : ;fn mÙkj –77.25, 5.2 gS] rks cqycqyksa dks fuEu izdkj ls dkyk djsaA

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

0 0 0

+

0

–

0 01 1 1 1 1 12 2 2 2 2 23 3 3 3 3 34 4 4 4 4 45 5 5 5 5 56 6 6 6 6 67 7 7 7 7 78 8 8 8 8 89 9 9 9 9 9

•

•

•

•

•

•

•

•

••

� izR;sd iz'u ds mÙkj dk ewY;kadu fuEu vadu ;kstuk ds vuqlkj gksxk%&iw.kZ vad : +3 ;fn flQZ lgh la[;kRed eku (Numerical value) gh mÙkj Lo:i ntZ fd;k x;k gSA'kwU; vad : 0 vU; lHkh ifjfLFkfr;ksa esaA

1.tanx

x tan x

x 0

tan xlimx

æ öç ÷-è ø

®

æ öç ÷è ø

dk eku gksxk (ekuk e = 2.7)

2. ;fn ( ) n 11ƒ x

2 -= ds fy;s n n 11 1x2 2 -< £ , x Î [0,1], n Î N rFkk ƒ(0) = 0 gSA ( )

1

0

ƒ x dxò dk eku gksxk

3. ;fn vody lehdj.k (tan–1y – x)dy = (1 + y2)dx }kjk ,d oØ dks n'kkZrs gS tks fcUnq (0,tan1) rFkk 1,b3pæ ö-ç ÷

è øls xqtjrk gS] rks b dk eku gksxk


H-26/28 0000CJA102119027



4. ;fn ar rFkk br

nks ijLij yEcor~ lfn'k gS ftlesa a 2, b 3= =rr rFkk c a b´ =

rr r gks] rks c a-r rdk eku gksxk

5. ;fn ,d lery lx + 13y + mz + n = 0 gS] tks leryksa 2x + 3y – z + 1 = 0 rFkk x + y – 2z + 3 = 0 ds

izfrPNsnu ls xqtjrk gS rFkk lery 3x – y – 2z = 4 ds yEcor~ gS] rks n

m-l

dk eku gksxk

6. leqPp; {1,2,3,.....,n} ds ifjfer Øep; Nk¡Vs x;s dgykrs gS] ;fn ,sls Øep;ksa esa ls ,d mi;qDr in gVk;sa tkus

ij 'ks"k n – 1 in c<+rs Øe gSA ;fn ,sls Nk¡Vs x;s Øep;ksa dh la[;k ƒ(n) gks] rks ( )( )

ƒ 5ƒ 4 dk eku gksxk

H-27/280000CJA102119027

All India Open Test/Leader & Enthusiast Course/JEE(Advanced)/09-02-2020/Paper-1ALLENdPps dk;Z ds fy, LFkku

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

+91-744-2757575 [email protected] www.allen.ac.in

iz'ui= dk izk:i vkSj vadu ;kstuk %16. bl iz'ui= esa rhu Hkkx gSa % HkkSfrd foKku] jlk;u foKku ,oa xf.krA17. izR;sd Hkkx esa nks [k.M gSa ftudk fooj.k fuEufyf[kr rkfydk esa fn;k x;k gSA

eSaus lHkh funsZ'kks a dks i<+ fy;k gS vkSj eS mudk

vo'; ikyu d:¡xk@d:¡xhA

____________________________

ijh{kkFkhZ ds gLrk{kj

eSaus ijh{kkFkhZ dk ifjp;] uke vkSj QkWeZ uEcj dks iwjh rjgtk¡p fy;k gS ,oa iz'u i= vkSj vks- vkj- ,l- dksM nksuks aleku gSaA

____________________________

fujh{kd ds gLrk{kj

ijh{kkFkh Z dk uke ................................................................................................

QkWeZ uEcj .............................................

Your Target is to secure Good Rank in JEE 2020 0000CJA102119027H-28/28


[k.M iz'u dk iz'uks a oxkZuqlkj izR;sd iz'u ds vad [k.M es aizdkj dh iw.kZ vad vkaf'kd vad 'kwU; vad ½.k vad vf/kdÙke

la[;k vad I(i) ,dy lgh +3 0 –1

fodYi 4 ;fn flQZ lgh fodYi ds — ;fn fdlh Hkh vU; lHkh 12vuq:Ik cqycqys dks cqycqys dks dkyk ifjfLFkfr;ksa

dkyk fd;k gS ugha fd;k gS esa,d ;k ,d ls +4 +1 0 –1

I(ii) vf/kd lgh ;fn flQZ lkjs lgh fodYi izR;sd lgh fodYi ds vuq:i ;fn fdlh Hkh vU; lHkhfodYi 8 (fodYiksa) ds vuq:Ik cqycqys dks dkyk djus ij] cqycqys dks ifjfLFkfr;ksa 32

cqycqys (cqycqyksa) dks ;fn dksbZ xyr fodYi dkyk dkyk ugha esa dkyk fd;k x;k gS ugha fd;k gS fd;k gS

la[;kRed eku +3 0 II izdkj ;fn flQZ lgh mÙkj ds vU; lHkh

(n'keyo ds nks 6 vuq:Ik cqycqys dks — ifjfLFkfr;ksa — 18

LFkku rd) dkyk fd;k gS esa

(acade ses : 2019 - 2020) jee(m+aance) : leader

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