(A S : 2018 - 2019) JEE (Main + Advanced)

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 19 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 : 195

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)1001CJA103118014)Paper Code

(1001CJA103118014)

PAPER – 2

EN

GLI

SH

CLASSROOM CONTACT PROGRAMME(Academic Session : 2018 - 2019)

Test Type : FULL SYLLABUS Test Pattern : JEE-Advanced

TEST DATE : 21 - 04 - 2019

JEE (Main + Advanced)LEADER & ENTHUSIAST COURSE SCORE(ADVANCED)

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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,

1001CJA103118014

· 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

Target : JEE (Main + Advanced) 2019/21-04-2019/Paper-2

Leader & Enthusiast Course/Score(Advanced)/21-04-2019/Paper-2

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

PART-1 : PHYSICSSECTION-I(i) : (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 : –2 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 correctoption(s) will result in –2 marks.

1. A siphon has a uniform circular base of diameter 8

p cm with its crest A, 1.8 m above water

level as in figure. Then [Use P0 = 105 N/m2 & g = 10 m/s2]

3.6m

1.8mA

(A) Velocity of flow = 6 2 m/sec.

(B) Discharge rate of flow is = 39.6 2 10-´ m3/sec.(C) Absolute pressure at the crest level A is = 4.6 × 104 N/m2.(D) Speed of flow will descrease with time.


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2. A particle of unit mass, strikes a horizontal smooth floor with a velocity u = 10 m/sec, makingan angle θ= 600 with the floor and rebounds with velocity v making an anglef with the floor.If the coefficient of restitution between the particle and the floor is e = 0.6 then :(A) the impulse delivered by the floor to the body is = 13.86 Kg m/sec.

(B) f = tan–1 ( )0.6 3

(C) v 7m / sec.»

(D) the ratio of the final kinetic energy to the intial kinetic energy is 0.523. A vessel containing water is kept on an inclined plane of coefficient of friction µ. The angle

made by the free surface of liquid with horizontal as shown in steady state is q. If surface isas shown in figure (b), mark the angle q as negative. Then

q

a = 45°

Figure (a)

a = 45°

Figure (b)

q

(A)µ = 1.2, q = 0° (B) µ = 0, q = 45° (C) µ = 34 , q = 8° (D) µ =

34 , q = –8°

4. Mx and My denote the atomic masses of the parent and the daughter nuclei respectively in a

radioactive decay. The Q-value for a b decay is Q1 and that for a +b decay is Q2. If me denotesthe mass of an electron, then which of the following statements are incorrect ?(A) Q1 = (Mx – My) c2 and Q2 = (Mx – My – 2me)c2

(B) Q1 = (Mx – My) c2 and Q2 = (Mx – My)c2

(C) Q1 = (Mx – My – 2me) c2 and Q2 = (Mx – My + 2me)c2

(D) Q1 = (Mx – My + 2me) c2 and Q2 = (Mx – My + 2me)c2


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5. A particle 'A' of mass 103

kg is moving in the positive x–direction. Its initial position is x = 0& initial speed is unknown. The velocity at x = 10m can be : (use the graph given)

4

2

10

Power (in watts)

(in m)x

(A) 4 m/s (B) 2 m/s (C) 3 m/s (D) 100/3 m/s6. Consider the situation shown in figure. The two sits S1 and S2 placed symmetrically around

the central line are illuminated by a monochromatic light of wavelength l . The separationbetween the slits is d. The light transmitted by the slits falls on Screen-1 placed at a distanceD from the slits. The slit S3 is at the central line and the slit S4 is at a distance z from S3.Another Screen-2 is placed a further distance D away from S1.Then choose the CORRECToptions - (All slits are narrow)

d

S1

S2

S4

S3

z

Screen–1 Screen–2D D

(A) If z = D

2dl

then the ratio of the maximum to minimum intensity observed on S2 is 1

(B) If z = Dd

l then the ratio of the maximum to minimum intensity observed on S2 tends to ¥

(C) If z = Dd

l then the ratio of the maximum to minimum intensity observed on S2 is

approximately 12

(D) If z = D

4dl

then the ratio of the maximum to minimum intensity observed on S2 isapproximately 34


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1001CJA103118014

7. A cylinder of height h, diameter h/2 and mass M and with a homogeneous mass distributionis placed on a horizontal table. One end of a string running over a pulley is fastened to thetop of the cylinder, a body of mass m is hung from the other end and the system is released.Friction is negligible everywhere. Then choose the correct options.

M

h/2

h

m

(A) If m = M, m Ma a=r r (B) If m = 2M, cylinder will not tilt

(C) If m = m MM , | a | | a |2

=r r (D) If

Mm2

= , cylinder will not tilt

8. Two particles A and B move anticlockwise with the same speed v in a circle of radius R andare diametrically opposite to each other. At t=0, A is given a constant acceleration (tangential)

at = 272v

25 Rp. Then select the correct options :-

(A) the time in which A collides with B is 5 R6vp

(B) the angle traced by A at the time of collision is 116

p

(C) angular velocity of A at the time of collision is 17v5R

(D) radial acceleration of A at the time of collision is 2289v

25R



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SECTION–I(ii) : (Maximum Marks : 12)� This section contains TWO paragraphs.� Based on each paragraph, there are TWO questions.� Each question has FOUR options (A), (B), (C) and (D) ONLY ONE of these four options

is correct.� 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 answer is darkened.Zero Marks : 0 In all other cases.

Paragraph for Questions 9 and 10An Ideal gas sample is taken through process A ® B ®C ®A. It absorb 360J of Heat inA ® B & rejects 56 J of Heat in C ® A.

V

BA P

AC

V0 3V0 4V0

4P0

P0

9. What is degree of freedom of gas ?(A) 3 (B) 5 (C) 6 (D) None of these

10. What is heat rejected by Gas in process B ® C ?

(A) 220 J (B) 260 J (C) 280 J (D) 300 JSpace for Rough Work

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1001CJA103118014

Paragraph for Questions 11 and 12Consider an Earthed solid metal sphere of radius 3R with a spherical cavity of radius R,center of sphere is O2 & that of cavity is O1.

R

O2O1

3R

A point charge Q is placed at 4R distance from center of this metal sphere on the linejoining O1 and O2 and rotated in plane of this paper in circular path.

11. If Q is rotated about O2 as center then ratio of maximum & minimum induced charge onmetal sphere is :

(A) 1 (B) 2 (C) 3 (D) 13

12. If Q is rotated about O1 as centre then ratio of maximum & minimum induced charge onmetal sphere is :

(A) 56 (B) 1 (C)

32 (D)

43



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SECTION-II : (Maximum Marks: 21)� This section contains SEVEN 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 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.

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� 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. A particle of mass m is projected from a celestial body. At point A, its distance from thecentre of the celestial body is 400 km and it has a velocity v1 = 20km / sec . Determine thevelocity v2 (in km/sec) of the satellite as it reaches point B, a distance 800 km from the centreof the celestial body. Take the radius of the celestial body = 400 km. The acceleration due togravity on the surface of the celestial body is 10 m/s2.

2. A 4 kg block is placed on top of a long 12 kg block which is accelerating along a smoothhorizontal table at a = 5.2 m/s2 under application of an external constant force. Let minimumcoefficient of friction between the two blocks which will prevent the 4 kg block from slidingis m and coefficient of friction between blocks is only half of this minimum value of (i.e., m/2).Find the amount of heat (in joules) generated due to sliding between the two blocks during

the time in which 12 kg block moves 10 m starting from rest. Give answer in terms of loss10 .

4Kg

12Kg a = 5.2 m/s2

smooth


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1001CJA103118014

3. Two containers A and B are connected by a conducting solid cylindrical rod of length242

7 cm and radius 8.3 cm. Thermal conductivity of the rod is 693 watt/mK. The containerA contains two mole of oxygen gas and the container B contains four mole of helium gas. Attime t = 0 temperature difference of the containers is 50ºC, after what time (in seconds)temperature difference between them will be 25ºC. Transfer of heat takes place through the

rod only. Neglect radiation loss. Take R = 8.3 J/mole-K and p = 227 .

A

O2

B

He

4. The emissive power of a black body of surface area A = 10 m2 at T = 300 K is 10 watt/m2.Consider a body B of area A = 10m2, coefficient of reflectivity r = 0.3 and coefficient oftransmission t = 0.5. Its temperature is 300K. Then emissive power of B in W/m2 is ..........

5. Two blocks of masses 1 kg and 2 kg are connected by a metal wire going over a smooth pulleyas shown in figure. The breaking stress of the metal is 2 × 109 N/m2. If minimum radius of thewire used is r × 10–5 m so that it is not to break. Find the value of r. Take g = 10 m/s2.

1 kg 2 kg



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6. A cubical block of coefficient of linear expansion as is submerged partially inside a liquid of

coefficient of volume expansion gl. On increasing the temperature of the system by DT, the

height of the cube inside the liquid remains unchanged. Find gl/as = ?

7. A series LCR circuit containing a resistance of 120W has an angular resonance frequency4 × 105 rad/sec. At resonance the voltage across resistance and inductance are 60 V and 40 Vrespectively. At frequency w = X × 105 rad/s the current in the circuit lags the voltage by 45º.Find the value of X.


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1001CJA103118014

PART-2 : CHEMISTRYSECTION-I(i) : (Maximum Marks: 32)







1. 3 moles of triatomic non linear ideal gas is adiabatically compressed by using constant externalpressure of 10 atm from initial pressure and temperature of one atmosphere and 227°C.(considering vibrational degree of freedom to be active). Choose the correct statement(s) forthe given process. [Given : ln 2.28 = 0.824 ln 10 = 2.303](A) In the given process final temperature of gas is approx 1143 K.(B) Entropy change of system is – 0.65 R(C) Entropy change of surrounding is – 10.4 R(D) Work done on the system is approx 11574 R

2. Choose incorrect statement(s) from the following :

(A) 4 6 2 3K [Fe(CN) ] K Fe CO NO+ -+++¾¾® + + + , equivalent weight of K4[Fe(CN)6] is M61 .

(B) 2 2 4 3 2Fe (C O ) Fe CO+++¾¾® + , equivalent weight of Fe2(C2O4)3 is M7

(C) 2 2 3 2 4Na S O Na SO¾¾® , equivalent weight of Na2S2O3 is M2 .

(D) 2 3 4 2SnCl HCl O SnCl H O+ + ¾¾® + , equivalent weight of O3 is M2



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3. 50 mL, 0.5 M H3PO4 solution is titrated against 0.5 M NaOH solution. Choose the correctstatement(s) about the given process.

[Given H3PO4 1

3aK 10-= ,

28

aK 10-= , 3

13aK 10-= log 2 = 0.30]

(A) On adding 25 mL NaOH, pH of resulting solution will be 3.(B) At first equivalence point, pH of resulting solution will be 5.5(C) On adding 125 mL NaOH, pH of resulting solution will be 4(D) At last equivalent point pH of resulting solution is 13.05

4.

CHCl3/ KOH

CO2/ KOH

C6H5OH

H AÅ¾¾¾®

H BÅ

¾¾¾®Which of the following statements are incorrect.(A) A and B are functional isomers of each other.(B) A and B can be distinguish by fehling solution.(C) A evolves CO2 when heated with sodalime.(D) B gives positive test with tollen's reagent.

5. 3 3CH CH(OH) CH(OH) CH(OH) CH- - - -

Which of the following statements are true for the given compound.(A) It have 4 stereoisomers. (B) It is always optically inactive.(C) It may present in meso form (D) It have 3 stereoisomers.

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6. In a qualitative inorganic analysis, consider the following sequence-

O.S. + dil HCl

Precipitate (1)

Filtrate (2)

Precipitate (3)

Filtrate (4)

H S2

Which of the following is/are correct?(A) Filtrate(2) can contain Pb2+ ion (B) Precipitate (3) can not be Ag2S(C) Filtrate (4) can contain Zn2+ (D) Precipitate (1) can contain Hg2+

7. Which of the following disproportionation results into change in magnetic property ofproduct/s compared to that of reactant?(A) 2 2Cu Cu Cu+ +¾¾® +

(B) 2 22Hg Hg Hg+ +¾¾® +

(C) 24 2 4 23 4 2 2- + -+ ¾¾® + +MnO H MnO MnO H O

(D) 33 2Au Au Au+ +¾¾® +

8. Which of the following gives orthoboric acid as one of the products?

(A) Hydrolysis of B2H6 (B) Acidic hydrolysis of ( )2 5 3C H B

(C) Reaction of 2 6B H with ethanol (D) Hydrolysis of 3BCl


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Paragraph for Questions 9 and 10Cleavage of unsymmetrical ether (R–O–R') by HI (1 mole) follows SN2 mech. But if any ofthe alkyl group forms stable carbocation, then it follows SN-1 mechanism.

9.O

HI(1mole)

A¾¾® , product A is :

(A) OH

I (B) I

OH (C) OH

OH (D)

I

I

10. Ph O Me HI(1mole)¾¾® product

Select the correct statement for the product.(A) Product formed by SN1 mechanism.(B) Product is an aromatic alcohol.(C) DBE of product alcohol is 4.(D) All of these


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Paragraph for Questions 11 and 12Analysis of salt can involve some interfering radicals and typical combinations special measuresand precautions are taken in such cases during the analysis.

11. Which of the following is the interfering acid radical?(A) 2NO- (B) 3 2CH CO- (C) 3

4PO - (D) Br–

12. For the detection of Cd2+ in presence of 2Cu + , which reagent must be added in excess before

passing 2H S ?

(A) HCl (B) 4NH Cl (C) KCN (D) 2H O



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1. 100 mL of 1 M CuSO4 solution is electrolysed using Pt - electrode by passing a currentof 9.65 Ampere for 50 seconds, then the pH of resulting solution after electrolysis is :[Given : log 2 = 0.3]

2. 100 gm solution of NH3 in H2O having mole fraction =3NHX 0.2 . What will be the molality

of solution.3. An ionic compound AB crystallizes in rock salt structure. The ionic radius of A+ and B- are

60 pm and 100 pm respectively and atomic masses of A and B are 200 gm/mol and120 gm/mole respectively, the density of unit cell in gm/ cm3 is : (Given NA = 6×1023)


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1001CJA103118014

4.Cl

D

¾¾¾¾®alc.KOH A

Cl

CH3

D¾¾¾¾®alc.KOH B

Cl

CH3

D¾¾¾¾®alc.KOH C

A,B,C are major product of the reactions and let x, y, z are the total number of allylic hydrogenpresent in A,B,C respectively then find x + y – z.



E-19/281001CJA103118014

5. How many of the following are more stable than

CH2

OCH3

(i)

CH2

OCH3

(ii)

CH2

NO2

(iii)

CH2

OCH3H3CO

(iv)

CH2

CN

(v)

CH2

NO2

(vi)

CH2OCH3 (vii)

CH2

CH3

(viii)

CH2

6. What is the spin only magnetic moment (in Bohr Magneton) of the complex

( )2 45Fe H O NO SOé ùë û ?

7. For the molecule of 2 2XeO F consider no distortion due to lone pair, atoms of differentelectronegativity and p bonds according to VBT, what will be the percentage 's' character oforbital that contains lone pair of electrons ?


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1001CJA103118014

PART-3 : MATHEMATICSSECTION-I(i) : (Maximum Marks: 32)







1. Let a, b, g, d be the roots of 4 3 2x x x 1 0- - - = . Also consider ( ) 6 5 3 2p x x x x x x= - - - - .

Then the value of ( ) ( ) ( ) ( )p p p pa + b + g + d cannot be

(A) 4 (B) 5 (C) 6 (D) 72. Let all the letters of the word ‘MATHEMATICS’ are arranged in all possible order.

Three events A, B and C are defined asA : Both M are together. ; B : Both T are together. ; C : Both A are together.Which of the following hold (s) good ?

(A) 2P(A) P(B)

11= = (B)

2P(A B) P(B C) P(C A)55

Ç = Ç = Ç =

(C) 4P(A B C)495

Ç Ç = (D) 58P((A B) / C)405

Ç =



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3. Let ƒ is a differentiable function satisfying 3ƒ(x + 2y) = ƒ(2x) + ƒ(y) + 2x + 10y + 1 "x,yÎRsuch that ƒ'(0) = 2ƒ(0), then-

(A) ƒ(x) is an odd function (B) ( ) ( )= +ò ldx

n ƒ(x) Cƒ x

(C) ƒ(0.5) = ƒ'(5.0) (D) sin(ƒ(x)) is a periodic function with period p.(where C is constant of integration)

4. The equations of the common tangents to the parabola y2 = 8x and x2 + y2 –12x + 4 = 0 are -(A) y = – x + 2 (B) y = x – 2 (C) y = x + 2 (D) y = – x – 2

5. Sum of the series ( )r 4 n n 4 rn 3

r rr 4 n 4

r 0 r 04 4

3 C C 3SC C

+ +

+ += =

= +å å is

(A) n 4

n 3 n 3n n 1

4C C

+

+ +-+ (B) ( )n 4 2n

44 C+ (C) n 4

n 44

4C

+

+ (D) n 4 n 4

n 44

3 2C

+ +

+

+

6. The triangle ABC has ÐA = 60°, ÐB = 45°. The bisector of ÐA meets the side BC at T whereAT = 24. Then which of the following is correct?(A) length of BC = 18 2 (B) length of AC < AT

(C) radius of the circumcircle is 12 2 (D) area of the triangle is 72(3 3)+ sq. units.7. A solution of the differential equation (x2y2 – 1)dy + 2xy3dx = 0 is

(A) 1 + x2y2 = cx (B) 1 + x2y2 = cy (C) y = 0 (D) 2

1y

x= -

8. The number of arrangements of the letters of the word 'GOOGLE' so that no two 'O' and notwo 'G' are together, is-(A) 180 (B) 216 (C) 84 (D) 60

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1001CJA103118014





Paragraph for Questions 9 and 10Set A consist all Natural number formed by using digit 1, 3, 5, 8 not more than 4 digit.

9. If a number is selected from set A then the probability that the number will not exceed 8531is –

(A) 229340 (B)

230340 (C)

312340 (D)

313340

10. A 4 digit number (all digits distint) is selected from set A then the probability that in number8 is not at unit place, 5 is not at tens place, 3 is not at hundreds place and 1 is not at thousandsplace

(A) 58 (B)

38 (C)

18 (D)

68


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Paragraph for Questions 11 and 12Consider a system of equations :

f - f - l =xsin 2ycos z 0x + 2y + z = 0– x + y + z = 0

11. The number of integer values of l for which system have nontrivial solution –(A) 2 (B) 3 (C) 1 (D) 0

12. Consider the 3 statements -Statement 1 : For l = 1 there exist f for which the system of equation will have infinitesolution.Statement 2 : For l = 3 there exist f for which the system of equation will have uniquesolution.Statement 3 : Number of integer value of l is 3 for the system of equation to have nontrivialsolution..(A) All statement are incorrect (B) Exactly 2 statement are correct(C) at most 2 statement are correct (D) at most 2 statement are incorrect


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1. Tangents are drawn from any point on the hyperbola 4x2 – 9y2 = 36 to the circle

x2 + y2 – 9 = 0. The locus of mid-point of chord of contact is ÷÷ø

öççè

æ-

4y

9x 22

= 222

kyx

÷÷ø

öççè

æ +,

where k Î N. Find k.2. In DABC, if sin 2A = sin 2B but ÐA ¹ ÐB and 3 tan A – 4 = 0 then find the value of the

expression 21

1 tan B+ +

sin(B A)2

-+ cot C ÷

øöç

èæ +-+ Bsin1AcosAsin1Bcos 22 .

3. If z1 and z2 be two distinct complex numbers satisfying 22

21 zz - = 21

22

21 zz2zz -+ and if

(arg z1 – arg z2) = bap

, then find the least possible value of |a b|

2+

(a, b Î integers).



E-25/281001CJA103118014

4. The roots of the equation z4 + az3 + (12 + 9i) z2 + bz = 0 (where a and b are complex numbers)are the vertices of a square, then find the value of |a|

5. If f : [0, ¥) ® [1, ¥) is a quadratic function which is invertible and satisfying the equation

( ) ( )22 )x("fx4)x('f l-+- = f "'(x) " x ³ 0, where l is constant then find the value

of ò -

l

3

1

1 dx)x(xf15.


E-26/28


1001CJA103118014

6. Find the number of common solution(s) of the trigonometric equations

cos 2x + ( )31- = ( )32 - cos x and sin 3x = 2 sin x which satisfy the inequality

3 tan x – 1 ³ 0 in [0, 5p].

7. Straight line L1 is parallel to the bisector of first and third quadrant, forms a triangleof area 2 square units with coordinate axes in second quadrant. Line L2 passes throughM(1, 1) and has positive x and y intercepts. L2 makes a triangle of minimum area withcoordinates axes. Find the area of triangle formed by L1, L2 and x-axis



E-27/281001CJA103118014


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 section 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 2019 1001CJA103118014E-28/28

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

Que. Marks Marks Marks Marks section+4 +1 0 –2

One or more If only the bubble(s) For darkening a bubble If none In all I(i) correct 8 corresponding corresponding to each of the other 32option(s) to all the correct correct option, provided bubbles is cases

option(s) is(are) NO incorrect option darkeneddarkened darkened

Paragraph +3 0Based If only the bubble In all

I(ii) (Single 4 corresponding to — other — 12correct the correct option casesoption) is darkened

+3 0Numerical If only the bubble In all

II Value Type 7 corresponding — other — 21(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 19 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[ks aA 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

Time : 3 Hours Maximum Marks : 195

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

)1001CJA103118014)Paper Code

(1001CJA103118014)

PAPER – 2

HIN

DI

CLASSROOM CONTACT PROGRAMME(Academic Session : 2018 - 2019)

Test Type : FULL SYLLABUS Test Pattern : JEE-AdvancedTEST DATE : 21 - 04 - 2019

JEE (Main + Advanced)LEADER & ENTHUSIAST COURSE SCORE(ADVANCED)

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,

1001CJA103118014

· 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-3/281001CJA103118014

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS

-1 :

-I(i) : ( : 32)

() ()

: +4 : +3 : +2

: +1

: 0 : –2

+4 +2 +1 , –2

1. 8

A1.8

P0 = 105 N/m2 g = 10 m/s2 )

3.6m

1.8mA

(A) = 6 2 m/sec

(B) = 39.6 2 10 m3/sec

(C) A= 4.6 × 104 N/m2

(D)

H-4/28


1001CJA103118014


2. bdkbZ æO;eku okyk ,d d.k ,d {kSfrt fpdus Q'kZ ls u = 10 m/sec osx ds lkFk Q'kZ ls θ= 600 dks.k cukrs gq, Q'kZ

ls Vdjkrk gS rFkk Q'kZ ls f dks.k cukrs gq, v osx ls okil ykSV tkrk gSA ;fn d.k o Q'kZ ds e/; izR;koLFkku xq.kkad

e = 0.6 gS rks :(A) Q'kZ }kjk oLrq dks fn;k vkosx 13.86 Kg m/sec. gksxkA

(B) f = tan–1 ( )0.6 3

(C) v 7m / sec.»

(D) vfUre xfrt ÅtkZ rFkk izkjfEHkd xfrt ÅtkZ dk vuqikr 0.52 gSA

3. ty ls Hkjk ,d ik= ?k"kZ.k xq.kkad µ okys urry ij j[kk gSA fp=kuqlkj LFkk;h voLFkk eas æo dh eqDr lrg }kjk {kSfrt ds

lkFk cuk;k x;k dks.k q gSA ;fn lrg fp= (b) ds vuqlkj gks rks dks.k q dks ½.kkRed ekusaA rc

q

a = 45°

Figure (a)

a = 45°

Figure (b)

q

(A)µ = 1.2, q = 0° (B) µ = 0, q = 45° (C) µ = 34 , q = 8° (D) µ =

34 , q = –8°

4. ekuk Mx o My fdlh jsfM;kslfØ; fo?kVu esa larfr o iq=h ukfHkdksa ds ijekf.od æO;eku gSA b –fo?kVu dk Q eku Q1

rFkk +b fo?kVu ds fy;s Q2 gSA ;fn me bysDVªkWu dk æO;eku gks rks fuEu esa ls xyr dFku pqfu;sA

(A) Q1 = (Mx – My) c2 rFkk Q2 = (Mx – My – 2me)c2

(B) Q1 = (Mx – My) c2 rFkk Q2 = (Mx – My)c2

(C) Q1 = (Mx – My – 2me) c2 rFkk Q2 = (Mx – My + 2me)c2

(D) Q1 = (Mx – My + 2me) c2 rFkk Q2 = (Mx – My + 2me)c2


H-5/281001CJA103118014


5. æO;eku 103

fdxzk dk d.k A /kukRed x fn'kk esa xfr dj jgk gSA bldh izkjfEHkd fLFkfr x =0 gS rFkk izkjfEHkd pkYk

vKkr gS rks x=10 ehVj ij osx gks ldrk gS & (fn;s x;s xzkQ dk mi;ksx djsa)

4

2

10

Power (in watts)

(in m)x

(A) 4 m/s (B) 2 m/s (C) 3 m/s (D) 100/3 m/s

6. fuEu fp= esa çnf'kZr fLFkfr ij fopkj dhft;sA dsUæh; js[kk ds lkis{k lefer :i ls j[kh gqbZ nks fLyVksa S1 o S2 dks

rjaxnS/;Z l okys ,do.khZ; çdk'k ls çdkf'kr fd;k tkrk gSA fLyVksa ds e/; nwjh d gSA fLyVksa ls ikjxfer çdk'k fLyVksa

ls D nwjh ij j[ks insZ-1 ij fxjrk gSA fLyV S3 dsUæh; js[kk ij gS rFkk fLyV S4, S3 ls z nwjh ij gSA ,d vU; inkZ-2, S1

ls vfrfjDr nwjh D ij j[kk gqvk gSA lHkh fLyVsa ladjh gSA lgh dFku pqfu;sA

d

S1

S2

S4

S3

z

Screen–1 Screen–2D D

(A) ;fn z = D

2dl

gks rks S2 ij çsf{kr vf/kdre rFkk U;wure rhozrk dk vuqikr 1 gksxkA

(B) ;fn z = Dd

l gks rks S2 ij çsf{kr vf/kdre rFkk U;wure rhozrk dk vuqikr ¥ dh vksj vxzlj gksxkA

(C) ;fn z = Dd

l gks rks S2 ij çsf{kr vf/kdre rFkk U;wure rhozrk dk vuqikr yxHkx 12 gksxkA

(D) ;fn z = D

4dl

gks rks S2 ij çsf{kr vf/kdre rFkk U;wure rhozrk dk vuqikr yxHkx 34 gksxkA

H-6/28


1001CJA103118014

7. ,d h Å¡¡pkbZ] M æO;eku rFkk h/2 O;kl dk ,dleku æO;eku forj.k okyk csYku fp«kkuqlkj {kSfrt VscYk ij j[kk gSA

f?kjuh ds Åij ls xqtjrh gqbZ Mksjh dk ,d fljk csYku ds 'kh"kZ ls tqM+k gqvk gS rFkk ,d m æO;eku dk fi.M nwljs fljs ls

yVdk gqvk gS rFkk fudk; dks fojkekoLFkk ls NksM+k tkrk gSA ?k"kZ.k dks loZ= ux.; ekusaA rc lgh fodYi pqusA

M

h/2

h

m

(A) ;fn m = M gS, rc m Ma a=r r gksxkA (B) ;fn m = 2M gS, rc csYku ugha > qdsxkA

(C) ;fn m = M2

gS, rc m M| a | | a |=r r gksxkA (D) ;fn

Mm2

= gS] rc csYku ugha > qdsxkA

8. nks d.k A rFkk B] R f=T;k ds o`Ùk esa leku pky v ls okekorZ fn'kk esa xfr djrs gSa rFkk nksuksa O;klh; :i ls ,d nwljs ds

foijhr fLFkr gSA t=0 ij d.k A dks fu;r Roj.k (Li'kZjs[kh;) at = 272v

25 Rp fn;k tkrk gS] rc lgh fodYi pqfu,A

(A) le; 5 R6vp

ij d.k A, B ls Vdjkrk gSA (B) VDdj ds le; A }kjk cuk;k x;k dks.k 116

pgSA

(C) VDdj ds le; A dk dks.kh; osx 17v5R

gSA (D) VDdj ds le; A dk f=T;h; Roj.k 2289v

25RgSA



H-7/281001CJA103118014

[k.M–I(ii) : (vf/kdre vad : 12)� bl [k.M esa nks vuqPNsn gSa

� izR;sd vuqPNsn ij nks iz'u fn, x;sa gSaA

� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa dsoy ,d 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 vU; lHkh ifjfLFkfr;ksa esaA

iz'u 9 ,oa 10 ds fy;s vuqPNsn

,d vkn'kZ xSl çfrn'kZ dks izØe A ® B ®C ®A ls xqtkjrs gSA ;g A ® B esa 360J Å"ek vo'kksf"kr djrh gS ,oa

C ® A esa 56 J Å"ek fu"dkflr djrh gSA

V

BA P

AC

V0 3V0 4V0

4P0

P0

9. xSl dh Lora=rk dh dksfV gS :(A) 3 (B) 5 (C) 6 (D) buesa ls dksbZ ugha

10. izØe B ® C esa xSl }kjk fu"dkflr Å"ek gS %

(A) 220 J (B) 260 J (C) 280 J (D) 300 J


H-8/28


1001CJA103118014


,d HkwlaEifdZr Bksl /kkfRod xksys (f=T;k 3R) esa f=T;k R okyh ,d xksykdkj xqfgdk cuh gqbZ gSA xksys dk dsUæ O2 ,oa

xqfgdk dk dsUæ O1 gSA

R

O2O1

3R

,d fcUnq vkos'k Q dks bl /kkfRod xksys ds dsUæ ls 4R nwjh ij O1 rFkk O2 dks tksM+us okyh js[kk ij j[kdj bls ,d

o`Ùkkdkj iFk esa bl dkxt ds ry esa ?kqek;k tkrk gSA

11. ;fn Q dks O2 (dsUæ) ds lkis{k xfr djk;h tk;s rks /kkfRod xksys ij vf/kdre ,oa U;wure izsfjr vkos'k dk vuqikr

gksxk :

(A) 1 (B) 2 (C) 3 (D) 13

12. ;fn Q dks O1 (dsUæ) ds lkis{k xfr djk;h tk;s rks /kkfRod xksys ij vf/kdre ,oa U;wure izsfjr vkos'k dk vuqikr

gksxk :

(A) 56 (B) 1 (C)

32 (D)

43



H-9/281001CJA103118014

-II : (: 21)

(NUMERICAL VALUE)

6.25, 7.00, –0.33, –.30, 30.27, –127.30, 11.36777..... 11.36 11.37 )

: –77.25, 5.2

0 0 0

+

0

–

0 0

1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

4 4 4 4 4 4

5 5 5 5 5 5

6 6 6 6 6 6

7 7 7 7 7 7

8 8 8 8 8 8

9 9 9 9 9 9

•

•

•

•

•

•

•

•

•

•

0 0 0

+

0

–

0 0

1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

4 4 4 4 4 4

5 5 5 5 5 5

6 6 6 6 6 6

7 7 7 7 7 7

8 8 8 8 8 8

9 9 9 9 9 9

•

•

•

•

•

•

•

•

•

•

: +3 (Numerical value) : 0

1. mA

400 km v1 = 20km / sec 800 km

Bv2 (km/sec) = 400 km

10 m/s2

2. 4 kg 12 kga = 5.2 m/s2

4 kg

/212 kg

10 m

10

4Kg

12Kga = 5.2 m/s2

smooth

H-10/28


1001CJA103118014

3. nks ik= A rFkk B ,d pkyd Bksl csyukdkj NM+ }kjk tqM+s gq;s gS ftldh yEckbZ 242

7 cm gS rFkk f=T;k 8.3 cm gSA

NM+ dh Å"eh; pkydrk 693 watt/mK gSA ik= A esa 2 eksy vkWDlhtu xSl gS rFkk ik= B esa 4 eksy ghfy;e xSl gSA

le; t = 0 ij ik=ksa dk rkikarj 50ºC gS rks fdrus le; (lSd.M esa) i'pkr buds e/; rkikUrj 25ºC jg tk;sxk\ Å"ek

LFkkukUrj.k dsoy NM+ }kjk laiUu gksrk gSA fofdj.k âkl dks ux.; ekusaA R = 8.3 J/mole-K rFkk p = 227 yasA

A

O2

B

He

4. i`"Bh; {kS=Qy A = 10 m2 okys d`f".kdk fi.M dh mRltZu {kerk T = 300 K ij 10 watt/m2 gSA {kS=Qy

A = 10m2] ijkorZu xq.kkad r = 0.3 o ikjxeu xq.kkad t = 0.5 okys ,d fi.M B ij fopkj dhft;sA bldk rkieku

300K gSA rc B dh mRltZu {kerk W/m2 esa Kkr dhft;sA

5. æO;eku 1 kg o 2 kg okys nks CykWd fp=kuqlkj fpduh f?kjuh ij ls gksdj xqtj jgs ,d /kkfRod rkj }kjk vkil esa tqM+s

gq, gSA /kkrq dk Hkatu çfrcy 2 × 109 N/m2 gSA ;fn ç;qDr rkj dh U;wure f=T;k r × 10–5 m gks rkfd ;g VwVs ugha rks

r dk eku Kkr dhft;sA g = 10 m/s2 ysaA

1 kg 2 kg



H-11/281001CJA103118014

6. js[kh; çlkj xq.kkad as okyk ,d ?kuh; CykWd vk;ru çlkj xq.kkad gl okys æo es vkaf'kd :i ls Mwck gqvk gSA fudk; ds

rkieku esa DT dh o`f¼ djus ij æo ds vanj ?ku dh Å¡pkbZ vifjofrZr cuh jgrh gSA gl/as dk eku Kkr dhft;sA

7. ,d Js.kh LCR ifjiFk esa 120W çfrjks/k yxk gqvk gS rFkk bldh dks.kh; vuquknh vkofÙk 4 × 105 rad/sec gSA vuqukn

ij çfrjks/k rFkk çsjd dq.Myh ij oksYVrk Øe'k% 60 V o 40 V gSA vkofÙk w = X × 105 rad/s ij ifjiFk esa /kkjk]

oksYVrk ls 45º ihNs gks tkrh gSA X dk eku Kkr dhft;sA


H-12/28


1001CJA103118014

-2 : -I(i) : ( : 32)

() ()

: +4 : +3 : +2

: +1

: 0 : –2

+4 +2 +1 , –2

1. 3 227°C 10

(

[ : n 2.28 = 0.824 n 10 = 2.303]

(A) 1143 K

(B) – 0.65 R

(C) – 10.4 R

(D) 11574 R


H-13/281001CJA103118014


2. fuEufyf[kr esa ls vlR; dFku pqfu, :

(A) 4 6 2 3K [Fe(CN) ] K Fe CO NO+ -+++¾¾® + + + , K4[Fe(CN)6] dk rqY;kadh Hkkj M61 gSA

(B) 2 2 4 3 2Fe (C O ) Fe CO+++¾¾® + , Fe2(C2O4)3 dk rqY;kadh Hkkj M7 gSA

(C) 2 2 3 2 4Na S O Na SO¾¾® , Na2S2O3 dk rqY;kadh Hkkj M2 gSA

(D) 2 3 4 2SnCl HCl O SnCl H O+ + ¾¾® + , O3 dk rqY;kadh Hkkj M2 gSA

3. 50 mL, 0.5 M H3PO4 foy;u dks 0.5 M NaOH foy;u ds lkFk vuqekiu fd;k x;kA fn;s x;s izØe ds fy,

lR; dFku pqfu,&

[fn;k x;k gS& H3PO4 1

3aK 10-= ,

28

aK 10-= , 3

13aK 10-= log 2 = 0.30]

(A) 25 mL NaOH foy;u feykus ij izkIr foy;u dk pH eku 3 gksxkA

(B) izFke rqY;kad fcanq ij ifj.kkeh foy;u ds pH dk eku 5.5 gksxkA

(C) 125 mL NaOH, foy;u feykus ij ifj.kkeh foy;u dk pH eku 4 gksxkA

(D) vafre rqY;kad fcanq ij ifj.kkeh foy;u dk pH eku 13.05 gksxkA

4.

CHCl3/ KOH

CO2/ KOH

C6H5OH

H AÅ¾¾¾®

H BÅ

¾¾¾®

fuEu esa ls dkSu ls dFku vlR; gS&

(A) A o B ,d nqljs ds fØ;kRed leko;oh gSA

(B) A o B dks Qsgfyax foy;u }kjk foHksfnr fd;k tk ldrk gSA

(C) A dks lksMkykbZe ds lkFk xeZ djus ij CO2 eqDr gksrh gSA

(D) VkWysu vfHkdeZd ds lkFk B /kukRed ijh{k.k nsrk gSA

H-14/28


1001CJA103118014


5. 3 3CH CH(OH) CH(OH) CH(OH) CH- - - -

fn;s x, ;kSfxd ds fy, dkSu ls dFku lR; gS&

(A) blds 4 f=foe leko;oh gSA (B) ;g lnSo izdk'k vfØ; gksxkA

(C) ;g ehlks :i esa mifLFkr gks ldrk gSA (D) blds 3 f=foe leko;oh gSA

6. ,d xq.kkRed vdkcZfud fo'ys"k.k esa fuEUk Øe ij fopkj dhft,A

H S2

vo{ksi(1)

vo{ksi(3)

Nfur(2)

Nfur(4)

O.S. + dil HCl

fuEu esa lgh dks pqfu;sA

(A) Nfur (2) esa Pb2+ vk;u mifLFkr gks ldrs gSa (B) vo{ksi (3) esa Ag2S ugha gks ldrk gS

(C) Nfur (4) esa Zn2+ vk;u mifLFkr gks ldrs gaS (D) vo{ksi (1) esa Hg2+ gks ldrs gSa

7. fØ;kdkjd dh rqyuk esa mRikn esa pqEcdh; xq.k dk ifjorZu dkSulh fo"kekuqikrhdj.k vfHkfØ;kvksa esa gksxk\

(A) 2 2Cu Cu Cu+ +¾¾® +

(B) 2 22Hg Hg Hg+ +¾¾® +

(C) 24 2 4 23 4 2 2- + -+ ¾¾® + +MnO H MnO MnO H O

(D) 33 2Au Au Au+ +¾¾® +

8. fuEu esa dkSUklh vfHkfØ;kvksa esa] ,d mRikn ds :i esa vkFkksZcksfjd vEy izkIr gksrk gS\

(A) B2H6 dk ty vi?kVu (B) ( )2 5 3C H B dk vEyh; ty vi?kVu

(C) 2 6B H dh bFksuky ds lkFk fØ;k (D) 3BCl dk ty vi?kVu


H-15/281001CJA103118014


� izR;sd vuqPNsn ij nks iz'u fn, x;sa gSaA� izR;sd iz'u esa pkj mÙkj fodYi (A), (B), (C) vkSj (D) gSa ftuesa dsoy ,d lgh gSaA

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





HI (1 eksy) }kjk vlefer bZFkj (R–O–R') dk fonyu SN2 fØ;kfof/k dk ikyu djrk gS ijarq ;fn dksbZ Hkh

,Ydkby lewg] LFkkbZ dkcZ/kuk;u cukrk gS rks ;g SN-1 fØ;kfof/k dk ikyu djrk gSA

9.O

HI(1mole)

A¾¾® , mRikn A gS :

(A) OH

I (B) I

OH (C) OH

OH (D)

I

I

10. Ph O Me HI(1mole)¾¾® mRikn

mRikn ds fy, lgh dFku dk p;u dhft ,&

(A) mRikn dk fuekZ.k SN1 fØ;kfof/k ls gksrk gSA

(B) mRikn ,d , sjkseSfVd ,Ydksgy gSA

(C) mRikn ,Ydksgy dk DBE 4 gSA

(D) mijksDr lHkh


H-16/28


1001CJA103118014


fdlh yo.k ds xq.kkRed fo'ys"k.k esa dqN ewyd ck/kk Mkyus okys gksrs gS rFkk dqN fof'k"V la;ksx Hkh gksrs gS , sls

fo'ys"k.k esa fo'ks"k lko/kkfu;kWa j[kh tkrh gSA

11. fuEu esa ls dkSu] ck/kkdkjh (interfering) vEyh; ewyd gS ?

(A) 2NO- (B) 3 2CH CO- (C) 34PO - (D) Br–

12. Cd2+ dks 2Cu + dh mifLFkfr esa Kkr djus ds fy, 2H S izokfgr djus ds igys dkSuls vfHkdeZd dh vf/kd

ek=k feykuk gksxh ?

(A) HCl (B) 4NH Cl (C) KCN (D) 2H O



H-17/281001CJA103118014

-II : (: 21)

(NUMERICAL VALUE)

6.25, 7.00, –0.33, –.30, 30.27, –127.30, 11.36777..... 11.36 11.37 ) : –77.25, 5.2

0 0 0

+

0

–

0 0

1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

4 4 4 4 4 4

5 5 5 5 5 5

6 6 6 6 6 6

7 7 7 7 7 7

8 8 8 8 8 8

9 9 9 9 9 9

•

•

•

•

•

•

•

•

•

•

0 0 0

+

0

–

0 0

1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

4 4 4 4 4 4

5 5 5 5 5 5

6 6 6 6 6 6

7 7 7 7 7 7

8 8 8 8 8 8

9 9 9 9 9 9

•

•

•

•

•

•

•

•

•

•

: +3 (Numerical value) : 0

1. 100 mL , 1 M CuSO4 9.65 50

pH [ : log 2 = 0.3]

2. NH3 100 gm

3NHX 0.2

H-18/28


1001CJA103118014

3. ,d vk;fud ;kSfxd AB jkWd lkYV lajpuk esa fØLVyhd`r gksrk gSA ;fn A+ rFkk B- dh vk;fud f=T;k Øe'k%

60 pm rFkk 100 pm gS rFkk ijek.kq Hkkj Øe'k% 200 gm/mol rFkk 120 gm/mole gS rks bdkbZ lsy dk ?kuRo

(gm/ cm3 esa) dh x.kuk dhft,A (fn;k x;k gS] NA = 6×1023)

4.Cl

D

¾¾¾¾®alc.KOH A

Cl

CH3

D¾¾¾¾®alc.KOH B

Cl

CH3

D¾¾¾¾®alc.KOH

C

A,B,C vfHkfØ;kvksa ds eq[; mRikn gaS vkSj x, y, z Øe'k% A,B,C esa mifLFkr ,yk;fyd gkbMªkstuksa dh la[;kgSA Kkr dhft, x + y – z



H-19/281001CJA103118014

5. fuEu esa ls fdruh lajpuk,¡]

CH2

OCH3

ls vf/kd LFkkbZ gaS\

(i)

CH2

OCH3

(ii)

CH2

NO2

(iii)

CH2

OCH3H3CO

(iv)

CH2

CN

(v)

CH2

NO2

(vi)

CH2OCH3

(vii)

CH2

CH3

(viii)

CH2

6. ladqy ( )2 45Fe H O NO SOé ùë û dk dsoy pØ.k pqEcdh; vk?kw.kZ Bohr Magneton bdkbZ esa D;k gksxk ?

7. VBT ds vuqlkj 2 2XeO F v.kq dh lajpuk esa ml d{kd ds s y{k.k dh izfr'krrk D;k gksxh ftlesa ,dkadh

bysDVªkWu ; qXe mifLFkr gS\ ekusa fd ,dkadh bysDVªk Wu ; qXe] vyx&vyx fo|qr½.krk ds ijek.kqvksa rFkk p ca/kksa

ds dkj.k dksbZ fod`fr mRiUu ugha gks jgh gS\


H-20/28


1001CJA103118014


Hkkx-3 : xf.kr[kaM -I(i) : (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, 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

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 : –2 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), –2 vad feysaxs] pkgs lgh fodYi (fodYiksa) dks pquk x;k gks ;k u pquk x;k gksA

1. ;fn lehdj.k 4 3 2x x x 1 0- - - = ds ewy a, b, g rFk k d gk s ,oa ( ) 6 5 3 2p x x x x x x= - - - - , rc

( ) ( ) ( ) ( )p p p pa + b + g + d dk eku gksxk &

(A) 4 (B) 5 (C) 6 (D) 72. 'kCn ‘MATHEMATICS’ ds lHkh v{kjksa ls fofHkUu 'kCn cuk;s tkrs gS ,oa fdlh 'kCn dks pquus ij rhu ?kVuk,¡

A, B rFkk C fuEukuqlkj ifjHkkf"kr dh tkrh gSA : 'kCn esa nksuksa M ,d lkFk gS ; B : 'kCn esa nksuksa T ,d lkFk gS; C : 'kCn esa nksuksa A ,d lkFk gS]

rc

(A) 2P(A) P(B)

11= = (B)

2P(A B) P(B C) P(C A)55

Ç = Ç = Ç =

(C) 4P(A B C)

495Ç Ç = (D)

58P((A B) / C)405

Ç =


H-21/281001CJA103118014


3. ;fn ƒ ,d vodyuh; Qyu bl izdkj gS fd 3ƒ(x + 2y) = ƒ(2x) + ƒ(y) + 2x + 10y + 1 " x , y Î R ,oa

ƒ'(0) = 2ƒ(0), rc -

(A) ƒ(x) ,d fo"ke Qyu gSA (B) ( ) ( )= +ò ldx

n ƒ(x) Cƒ x

(C) ƒ(0.5) = ƒ'(5.0) (D) sin(ƒ(x)) ,d vkorhZ Qyu gS] ftldk vkoÙkZ p gSA

(tgk¡ C lekdyu fu;rkad gS )4. ijoy; y2 = 8x rFkk oØ x2 + y2 –12x + 4 = 0 dh mHk;fu"B Li'kZ js[kkvksa dk lehdj.k -

(A) y = – x + 2 (B) y = x – 2 (C) y = x + 2 (D) y = – x – 2

5. Js.kh ( )r 4 n n 4 rn 3

r rr 4 n 4

r 0 r 04 4

3 C C 3SC C

+ +

+ += =

= +å å dk ;ksxQy gksxk

(A) n 4

n 3 n 3n n 1

4C C

+

+ +-+ (B) ( )n 4 2n

44 C+ (C) n 4

n 44

4C

+

+ (D) n 4 n 4

n 44

3 2C

+ +

+

+

6. ;fn f=Hkqt ABC esa ÐA = 60°, ÐB = 45° ,oa ÐA dk v/kZd Hkqtk BC dks fcanq T ij feyrk gS tgk¡ AT = 24.rc fuEufyf[kr esa ls lR; dFku gS ?

(A) BC = 18 2 (B) AC < AT

(C) ifjo`Ùk dh f=T;k 12 2 (D) f=Hkqt dk {ks=Qy = 72(3 3)+ oxZ bdkbZ

7. vody lehdj.k (x2y2 – 1)dy + 2xy3dx = 0 dk gy gS&

(A) 1 + x2y2 = cx (B) 1 + x2y2 = cy (C) y = 0 (D) 2

1y

x= -

8. 'kCn 'GOOGLE' ds v{kjksa dh O;oLFkkvksa dh la[;k rkfd uk rks nks 'O' rFkk uk gh nks 'G' ,dlkFk gks] gksxh&(A) 180 (B) 216 (C) 84 (D) 60

H-22/28


1001CJA103118014



� izR;sd vuqPNsn ij nks iz'u fn, x;sa gSaA

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





leqPp; A mu lHkh izkd`r la[;kvks a dk leqPp; gS tk s vad 1, 3, 5, 8 ls cuh gS rFkk la[;k 4 vadksa ls

vf/kd dh ugha gSA

9. ,d la[;k ;kn`fPNd :i ls leqPp; A esa ls pquh x;h gS rks pquh xbZ la[;k ds 8531 ls vf/kd ugha gksus dh

izkf;drk gksxh

(A) 229340 (B)

230340 (C)

312340 (D)

313340

10. ,d la[;k ;kn`fPNd #i ls leqPp; A esa ls pquh x;h gSA pquh x;h la[;k 4 fofHkék vadksa dh gS rks izkf;drk

Kkr dhft, fd pquh x;h la[;k esa 8 bdkbZ okYks LFkku ij u gks] 5 ngkbZ okYks LFkku ij u gks] 3 lSadMs okYks

LFkku ij u gks o 1 gtkj okYks LFkku ij u gksA

(A) 58 (B)

38 (C)

18 (D)

68


H-23/281001CJA103118014


,d lehdj.k fudk; fuEu izdkj gS

f - f - l =xsin 2ycos z 0x + 2y + z = 0– x + y + z = 0

11. l ds iw.kk±d ekuka s dh la[;k ftlds fYk, lehdj.k fudk; ds v'kwU; gYk gks axs &(A) 2 (B) 3 (C) 1 (D) 0

12. fuEu 3 dFku fn, x, gS %

dFku 1 : l =1 ds fYk, f dk eku dk vfLrRo gksxk ftlds fYk, lehdj.k fudk; ds vuUr gYk gksaxsA

dFku 2 % l = 3 ds fYk, f dk eku dk vfLrRo gksxk ftlds fYk, lehdj.k fudk; dk vf}rh; gYk gksxkA

dFku 3 % l ds 3 iw.kk±d eku gks axs ftlds fYk, lehdj.k fudk; ds v'kwU; gYk gksaxsA

(A) lHkh dFku vlR; gS (B) Bhd 2 dFku lR; gSa

(C) vf/kdre 2 dFku lR; gSa (D) vf/kdre 2 dFku vlR; gSa


H-24/28


1001CJA103118014

[kaM-II : (vf/kdre vad : 21)� bl [kaM esa lkr 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. vfrijoy; 4x2 – 9y2 = 36 ij fLFkr fdlh fcUnq ls o`Ùk x2 + y2 – 9 = 0 ij Li'kZ js[kk,¡ [khaph tkrh gS] ,oa Li'kZ

t hok d s e/ ; fcUnq d k fcUnqi Fk ÷÷ø

öççè

æ-

4y

9x 22

= 222

kyx

÷÷ø

öççè

æ +, tgk¡ k Î N gS] rks k =

2. ;fn DABC ds fy, sin 2A = sin 2B ysfdu ÐA ¹ ÐB ,oa 3 tan A – 4 = 0,

rks 21

1 tan B+ +

sin(B A)2

- + cot C ÷

øöç

èæ +-+ Bsin1AcosAsin1Bcos 22 dk eku Kkr dhft,

3. ;fn z1 ,oa z2 nks fHkUu lfEeJ la[;k,¡ gS tks 22

21 zz - = 21

22

21 zz2zz -+ dks larq"V djrh gSA

,oa (arg z1 – arg z2) = bap

, rks |a b|

2+

dk U;qure laHko eku Kkr dhft, (a, b Î iw.kk±d)



H-25/281001CJA103118014

4. lehdj.k z4 + az3 + (12 + 9i) z2 + bz = 0 (tgk¡ a ,oa b lfEeJ la[;k,¡ gS ) ds ewy ,d oxZ ds 'kh"kZ gS] rks |a|dk eku Kkr dhft,

5. ;fn f : [0, ¥) ® [1, ¥) f}?kkr Qyu gS] tks fd O;qRØe.kh; gS ,oa

( ) ( )22 )x("fx4)x('f l-+- = f "'(x) " x ³ 0, tgk¡ l vpj gS] rks ò -

l

3

1

1 dx)x(xf15 dk eku Kkr dhft,


H-26/28


1001CJA103118014

6. f=dks.kfefr lehdj.k cos 2x + ( )31- = ( )32 - cos x ,oa sin 3x = 2 sin x ds mHk;fu"V gyksa dh fdruh

la[;k gksxh] tks fd vlfedk 3 tan x – 1 ³ 0, vUrjky [0, 5p] esa larq"V djrs gSA

7. izFke ,oa r`rh; prqFkk±'k dks lef}Hkkftr djus okyh js[kk ds lekukUrj js[kk L1] f}rh; prqFkk±'k es funsZ'kkad v{kksa ds

lkFk ,d f=Hkqt cukrh gS] ftldk {ks=Qy nks oxZ bdkbZ gSA js[kk L2 fcUnq M(1, 1) ls xqtjrh gS ,oa /kukRed x-v{k

,oa y-v{k ds lkFk vUr%[k.M cukrh gS ,oa js[kk L2 funsZ'kkad v{kksa ds lkFk U;qure {ks=Qy dk f=Hkqt cukrh gS] rks

js[kkvksa L1, L2 ,oa x-v{k ls cus f=Hkqt ds {ks=Qy dk eku Kkr dhft,



H-27/281001CJA103118014



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 2019 1001CJA103118014H-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,dy ;k ,d ls +4 +1 0 –2

I(i) 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

vuqPNsn ij +3 0 I(ii) vk/kkfjr ;fn flQZ lgh fodYi ds vU; lHkh

(,dy lgh 4 vuq:Ik cqycqys dks — ifjfLFkfr;ksa — 12

fodYi) dkyk fd;k gS esala[;kRed eku +3 0

II izdkj ;fn flQZ lgh mÙkj ds vU; lHkh(n'keyo ds nks 7 vuq:Ik cqycqys dks — ifjfLFkfr;ksa — 21

LFkku rd) dkyk fd;k gS esa

(A S : 2018 - 2019) JEE (Main + Advanced)

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