solution to iit jee 2019 (advanced) : paper - ii · so, e net should be to surface so 1/3 00 3 0 0...

(Pg.1)

Solution to IIT JEE 2019 (Advanced) : Paper - II

PAPER 2 : INSTRUCTIONS TO CANDIDATES Question Paper2 has three (03) parts : Physics, Chemistry and Mathematics. Each part has a total eighteen (18) questions divided into three (03) sections (Section1,

Section2 and Section3) Total number of questions in Question Paper-2 are Fifty Four (54) and Maximum

Marks are One Hundred Eighty Six (186)

Type of Questions and Marking Scheme

SECTION 1 (Maximum Marks:32)

This section contains EIGHT (08) questions. Each question has FOUR options ONE OR MORE THAN ONE of these four option(s)

is(are) the correct answer(s). For each question, choose the correct option(s) corresponding to (all) the correct

answer(s). 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 and both of which are correct. 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.


This section contains SIX (06) questions. The answer to each question is a NUMERICAL VALUE.

For each question, enter the correct numerical value of the answer using the mouse and the onscreen virtual numeric keypad in the place designated to enter the answer. If the numerical value has more than two decimal places, truncate/roundoff the value to TWO decimal places.

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

(2) Vidyalankar : IIT JEE 2019 Advanced : Question Paper & Solution

(Pg.2)


This section contains TWO (02) ListMatch sets. Each ListMatch set has TWO (02) Multiple Choice Questions. Each ListMatch set has two lists : ListI and ListII. ListI has Four entries (I), (II), (III) and (IV) ListII has Six entries (P), (Q), (R), (S),

(T) and (U). FOUR options are given in each Multiple Choice Question based on ListI and ListII

and ONLY ONE of these four options satisfies the condition asked in the Multiple Choice Question.

Answer to each question will be evaluated according to the following marking scheme : Full Marks : +3 If ONLY the option corresponding to the correct combination is

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

Answering Questions :

To select the option(s), use the mouse to click on the corresponding button(s) of the

option(s).

To deselect the chosen option for the questions of SECTION1, click on the button of the

chosen option again or click on the Clear Response button to clear the chosen option.

To deselect the chosen option(s) for the questions of SECTION3, click on the button(s)

of the chosen option(s) again or click on the Clear Response button to clear all the

chosen options.

To change the option(s) of a previously answered question of SECTION1 and

SECTION3 first deselect as given above and then select the new option(s).

To answer questions of SECTION2, use the mouse to click on numbers (and/or

symbols) on the on–screen virtual numeric keypad to enter the numerical value in the

space provided for answer.

To change the answer of a question of SECTION2, first click on the Clear Response

button to clear the entered answer and then enter the new numerical value.

To mark a question ONLY for review (i.e. without answering it), click on the Mark for

Review & Next button.

To mark is equation ONLY for review (i.e. without answering it), click on Mark for

Review & Next button

To mark is question for review (after answering it), click on mark for Review & Next

button– the answered question which is also marked for review will be evaluated.

To save the answer, click on the Save & Next button – the answered question will be

evaluated.

IIT JEE 2019 Advanced : Question Paper & Solution (Paper – II) (3)

(Pg.3)

PART I – PHYSICS








1. An electric dipole with dipole moment 0p ˆ î j2

is held

fixed at the origin O in the presence of an uniform electric field of magnitude E0. If the potential is constant on a circle of radius R centered at the origin as shown in figure, then the correct statement(s) is/are (0 is permittivity of free space. R >> dipole size)

(A) The magnitude of total electric field on any two points of the circle will be same.

(B) Total electric field at point A is A 0ˆ Ê 2E (i j)

(C)

1/3

0

0 0

pR

4 E

(D) Total electric field at point B is BE 0 1. (C) , (D) R >> Dipole size circle is equipotential

So, Enet should be to surface so 1/3

0 003

0

kp kpE r

Er

At point B net electric field will be zero. EB = 0

0A 0 0net 32kp

E E 3Er

Electric field at point A, A 03 Ê E i j2

(EB)net = 0 Because E0 is uniform and due to dipole electric field is different at different points, so

magnitude of total electric field will also be different at different points.


(Pg.4)

2. A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact point with the floor without slipping. Which of the following statement(s) is/are correct, when the rod makes an angle 60 with vertical? [g is the acceleration due to gravity]

(A) The angular acceleration of the rod will be 2g

L

(B) The normal reaction force from the floor on the rod will be Mg

16

(C) The radial acceleration of the rod’s center of mass will be 3g4

(D) The angular speed of the rod will be 3g

2L

2. (B), (C), (D) Wg = K.E.

2

21 mmg4 2 3

;

3g

2

Radial acceleration of C.M. of rod = 23g

2 4

Using = I about contact point

2mg m

sin602 3

3 3 g4

3 3 g4

Net vertical acceleration of C.M. of rod

av = ar cos60° + at cos30°

= 3g 1

cos304 2 2

= 3g 3 3g 3

8 4 2 2

=

3g 9g 15g

8 16 16

Applying Fnet = ma in vertical direction on rod as system

mg N = mav = m15

g16

mgN16

3. A free hydrogen atom after absorbing a photon of wavelength a gets excited from the state n = 1 to the state n = 4. Immediately after that the electron jumps to n = m state by emitting a photon of wavelength e. Let the change in momentum of atom due to the absorption and the emission are pa and pe respectively. If a/e = 1/5, which of the option(s) is/are correct?

[Use hc = 1242 eV nm; 1 nm = 109 m, h & c are Plank’s constant & speed of light, respectively] (A) The ratio of kinetic energy of the electron in the state n = m to the state n = 1 is 1/4 (B) m = 2 (C) pa/pe =1/2 (D) e = 418 nm 3. (A), (B)

It is given : a

e

= 1

5 or

2 2e

2a

11 1

1m 411 5

14

m = 2

e12400 4

364713.6


(Pg.5)

Clearly, a

e

p 1

p 2

The ratio of the kinetic energies is also the ratio of the corresponding total energies = 1

4.

4. In a Young’s double slit experiment, the slit separation d is 0.3 mm and the screen distance D is 1 m. A parallel beam of light of wavelength 600 nm is incident on the slits at angle as shown in figure. On the screen, the point O is equidistant from the slits and distance PO is 11.0 mm. which of the following statement(s) is/are correct?

(A) For = 0, there will be constructive interference at point P.

(B) For 0.36

degree, there will be destructive interference at point P.

(C) For 0.36

degree, there will be destructive interference at point O.

(D) Fringe spacing depends on . 4. (C)

(C) x = d sin = d (as is very small) 30.36 (2 10 ) rad180

4 3

7

x (3 10 )(2 10 )1

6 10

. So constructive interference

(D) D

d

(B) pdy

x dD

= 3 104 (2 103 + 11 103) = 39 107

7

p

7

x 39 106.5

6 10

so destructive

(A) 4 3pdy

x (3 10 ) 11 10D

= 33 107

7

p

7

x 33 105.5

6 10

destructive

5. Three glass cylinders of equal height H = 30 cm and same refractive index n = 1.5 are placed on a horizontal surface as shown in figure. Cylinder I has a flat top, cylinder II has a convex top and cylinder III has a concave top. The radii of curvature of the two curved tops are same (R = 3m). If H1, H2 and H3 are the apparent depth of A point X on the bottom of the three cylinders, respectively, the correct statement(s) is/are:

(A) H2 > H1 (B) H3 > H1 (C) 0.8 cm < (H2 H1) H3


(Pg.6)

5. (A), (D) For case I :

H = 30 cm , n = 3/2

1H 30 2

H 20 cmn 3

For case II :

R = 300 cm

2

31

1 3 2H 2 30 300

2600

H 20.684 cm29

For case III :

3

31

1 3 2H 2 30 300

3600

H 19.354 cm31

6. A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially at pressure P0, volume V0 and temperature T0. If the gas mixture is adiabatically compressed to a volume V0/4, then the correct statement(s) is/are,

(Given 21.2 = 2.3; 23.2 = 9.2; R is gas constant) (A) The final pressure of the gas mixture after compression is in between 9P0 and 10 P0.

(B) The average kinetic energy of the gas mixture after compression is in between 18 RT0 and 19 RT0

(C) Adiabatic constant of the gas mixture is 1.6 (D) The work |W| done during the process is 13 RT0 6. (A), (C), (D)

m

6 5 110

5 71 1 13 5

m = 1.6

m

m 00 0

VP V P

4

P = P0 23.2 = 9.2 P0

00 0 0

0 00 0

VP V 9.2P 1.3P V 134W 6RT 13RT

0.6 0.6 6

m

1m m

11 0.60

0 0 0 00

VT V TV T T T (4) 2.3T

V

Average kinetic energy of the gas mixture

= 03 5

5 RT 1 RT 10RT 23RT2 2

7. A block of mass 2M is attached to a massless spring with springconstant k. This block is connected to two other blocks of masses M and 2M using two massless pulleys and strings. The accelerations of the blocks are a1, a2 and a3 as shown in the figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is x0. Which of the following option(s) is/are correct?

[g is the acceleration due to gravity. Neglect friction]


(Pg.7)

(A) At an extension of 0x

4of the spring.

The magnitude of acceleration of the block connected to the spring as 3g

10

(B) 04Mg

xk

(C) When spring achieves an extension of 0x

2for the first time, the speed of the block

connected to the spring is M

3g5k

(D) a2 a1 = a1 a3

7. (D)

In the frame of pulley B, the hanging masses have accelerations : M (a2 a1), 2M (a3 a1) : downward. (a2 a1) = (a3 a1) [constant] Assuming that the extension of the spring is x We consider the FBD of A :

2

2

d x2M 2T kx

dt

where a1 = 2

2

d x

dt …(i)

and the FBD of the rest of the system in the frame of pulley B :

Upward acceleration of block M w.r.t. the pulley B = Downward acceleration of block

2M w.r.t. the pulley

1 1T M(g a ) 2M(g a ) T

M 2M

14M

T (g a )3

…(ii)

Substituting in equation (i), we get

1 18M

2M a (g a ) kx3

or 114M 8Mg

a kx3 3

…(iii)


(Pg.8)

This is the equation of SHM Maximum extension = 2 Amplitude

i.e. 08Mg

x 23k

At 0x

4, acceleration is easily found from equation (iii) ;

114M 8Mg 4Mg

a3 3 3

12g

a7

At 0x

,2

speed of the block (2M) = amplitude = 3k 8Mg14M 3k

8. A small particle of mass m moving inside a heavy, hollow and straight tube along the tube

axis undergoes elastic collision at two ends. The tube has no friction and it is closed at one end by a flat surface while the other end is fitted with a heavy movable flat piston as shown in figure. When the distance of the piston from closed end is L = L0 the particle

speed is v = v0. The piston is moved inward at a very low speed V such that 0dL

V vL

.

Where dL is the infinitesimal displacement of the piston. Which of the following statement(s) is/are correct?

(A) After each collision with the piston, the particle speed increases by 2V.

(B) If the piston moves inward by dL, the particle speed increases by dL

2VL

(C) The particle’s kinetic energy increases by a factor of 4 when the piston is moved

inward from L0 to 01

L2

(D) The rate at which the particle strikes the piston is v/L 8. (A), (C)

before collision

After collision Therefore change in speed = (2V + V0 V0) = 2V In every collision it acquires 2 V


(Pg.9)

V

f2x

dV

f 2Vdt

V

dV 2Vdt2x

V

dV 2( dx)2x

0

V x

V

dV dx

V x

0

V xn n

V 0

VV

x

where 0x , V 2V2

So, f = 0 02V 2V

22

fi

k4

k





1. An optical bench has 1.5 m long scale having four equal divisions in each cm. While measuring the focal length of a convex lens, the lens is kept at 75 cm mark of the scale and the object pin is kept at 45 cm mark. The image of the object pin on the other side of the lens overlaps with image pin that is kept at 135 cm mark. In this experiment, the percentage error in the measurement of the focal length of the lens is ______.

1. [1.38 & 1.39]

u = (x2 x1) = 75 45 = 30 cm

u = x2 + x1 = 1 1 1

cm4 4 2

v = (x3 x2) = 135 75 = 60 cm

v = x3 + x2 = 1 1 1

cm4 4 2

1 1 1v u f 1 1 1

60 30 f

f 20 cm


(Pg.10)

Also, 2 2 2

dv du df

v u f

2 2 2 2

df dv du 1 1 1f 20

f 2v u 60 30

df 1 1 50100 10 1.38 &1.39 (both)f 36 9 36

2. A perfectly reflecting mirror of mass M mounted on a spring constitutes a springmass system of angular frequency such

that 24 24 M

10 mh

with h as Planck’s constant. N photons of

wavelength = 8 106 m strike the mirror simultaneously at normal incidence such that the mirror gets displaced by 1m. If the value of N is x 1012, then the value of x is

[Consider the spring as massless] 2. [1] If v is speed of mirror just after absorption of photons

Mv N 2h 2Nh0 vM

If is the maximum compression in the spring

2 21 1

k mv2 2

kvm

62Nh 10M

N = Ω 106 6 6M M8 10 102h 2h

12 24 12 124 M

10 10 10 10h

x = 1

3. A monochromatic light is incident from air on a refracting surface of a prism of angle 75 and refractive index

0n 3 . The other refracting surface of the prism is coated by a thin film of material of refractive index n as shown in figure. The light suffers total internal reflection at the coated prism surface for an incidence angle of 600. The value of n2 is

3. [1.5]

At = 60° ray incidents at critical angle at second surface .

So, sin = 3sin r 1

13

3 sin r2

1r 30

r2 = 45° = C

3sin 45 = n sin90°

23 3

n n 1.502 2


(Pg.11)

4. Suppose a 22688 Ra nucleus at rest and in ground state undergoes –decay to a 22686 Rn nucleus

in its excited state. The kinetic energy of the emitted particle is found to be 4.44 MeV. 22286 Rn nucleus then goes to its ground state by –decay. The energy of the emitted photon is _____keV. [Given : atomic mass of 22688 Ra =226.005u, atomic mass of

22288 Rn = 222.000 u, atomic

mass of particle = 4.000 u, 1u = 931 MeV/c2, c is speed of light] 4. [135] mass defect = 0.005 amu

222decay2268688 Ra Rn

Total energy emitted = (m)C2 = 0.005 931.5 MeV = E0 (say) …(i) Also, E = 4.44 MeV …(ii)

ERn = 4.44 MeV 4

222 …(iii)

Er = E0 E ERn = 135 keV 5. A ball is thrown from ground at an angle with horizontal and with an initial speed u0.

For the resulting projectile motion, the magnitude of average velocity of the ball up to the point when it hits the ground for the first time is V1. After hitting the ground, the ball rebounds at the same angle but with a reduced speed of u0/. Its motion continues for a long time as shown in figure. If the magnitude of average velocity of the ball for entire duration of motion is 0.8 V1, the value of is __________

5. [4]

Let us assume 0 0 0 12u sin

T and u cos vg

(given)

Average velocity =

0 0 00 0 2 2

00 2

u u TTu cos T cos cos

T TT .......

= 0 0 2 4

1

0 2

1 1u cosT 1 ......

V 0.8 V1 1 1

T 1 ......

= 4

6. A 10 cm long perfectly conducting wire PQ is moving with a velocity 1 cm/s on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor L = 1 mH and a resistance R = 1 as shown in figure. The horizontal rails, L and R lie in the same plane with a uniform magnetic field B = 1T perpendicular to the plane. If the key S is closed at certain instant, the current in the circuit after 1 millisecond is x × 10–3A, where the value of x is _____________. [Assume the velocity of wire PQ remains constant (1 cm/s) after key S is closed. Given : e–1 = 0.37, where e is base of the natural logarithm]


(Pg.12)

6. [0.63] Since velocity of PQ is constant. So emf developed across it remains constant.

= B v where = length of wire PQ

Rt

Li 1 eR

Rt

LB v 1 eR

= 1

(1 0.37)1000

i = 0.63 103 A x = 0.63





Answer to each question will be evaluated according to the following marking scheme : Full Marks : +3 If ONLY the option corresponding to correct combination is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks: 1 In all other cases.

Answer the following by appropriately matching the lists based on the information given

in the paragraph.

A musical instruments is made using four different metal strings 1, 2, 3 and 4 with mass per unit length , 2, 3 and 4 respectively. The instrument is played by vibrating the strings by varying the free length in between the range L0 and 2L0. It is found that in string-1() at free length L0 and tension T0 the fundamental mode frequency is f0.

ListI ListII (I) Sting1 () (P) 1 (II) Sting2 (2) (Q) 1/2 (III) Sting3 (3) (R) 1

2

(IV) Sting4 (4) (S) 13

(T) 3/16 (U) 1/16

1. If the tension in each string is T0, the correct match for the highest fundamental frequency in f0 units will be:

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

1. (B)

Case I : 000

T1f

2L

; Case II : 0 01

0

T f1f

2L 2

Case III : 0 020

T f1f

2L 3

; Case IV : 0 04

0

T f1f

2L 2


(Pg.13)

2. The length of the strings 1, 2, 3 and 4 are kept fixed at L0, 0 03L 5L,

2 4 and 0

7L

4

respectively. Strings 1, 2, 3 and 4 are vibrated at their 1st, 3rd, 5th and 14th harmonics, respectively such that all the strings have same frequency. The correct match for the tension in the four strings in the units of T0 will be:

(A) I P, II R, III T, IV U (B) I P, II Q, III R, IV T (C) I P, II Q, III T, IV U (D) I T, II Q, III R, IV U 2. (C)

Case I : 000

T1f

2L

For string 2

Case II : 020 0

TT3 1 ;

3L 2 2L2

2

02T

T2

For string 3

Case III : 3 00 0

T T5 1 ;

5L 3 2L2

4

033T

T16

For string 4

Case IV : 040 0

TT14 1 ;

7L 4 2L2

4

04T

T16


in the paragraph.

In a thermodynamic process on an ideal monoatomic gas, the infinitesimal heat absorbed by the gas is given by TX, where T is temperature of the system and X is the infinitesimal change in a thermodynamic quantity X of the system. For a mole of monatomic ideal gas

A A

3 T VX R n R n

2 T V

. Here, R is gas constant, V is volume of gas. TA and VA are

constants. The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.

ListI ListII (I) Work done by the system in process 1 2 3 (P)

0

1RT n2

3

(II) Change in internal energy in process 1 2 3 (Q) 0

1RT

3

(III) Heat absorbed by the system in process 1 2 3 (R) RT0 (IV) Heat absorbed by the system in process 1 2 (S)

0

4RT

3

(T) 0

1RT (3 n2)

3

(U) 0

5RT

6


(Pg.14)

3. If the process on one mole of monatomic ideal gas is

as shown in the TV-diagram with 0 0 01

P V RT3

, the

correct match is, (A) I P, II R, III T, IV S

(B) I P, II T, III Q, IV T (C) I S, II T, III Q, IV U

(D) I P, II R, III T, IV P 3. (D)

(I) 01 2RT

W n23

2 3W 0

01 2 3 1 2 2 3RT

W W W n2 (1 P)3

(II) f if

U nR T T2

01 2 3 0T3

U 1 R T2 3

= RT0 (II R) (III) Q12 3 = U123 + W123

= RT0 + 0RT

n 23

= 0RT 3 n 23

(III T)

(IV) Q12 = U12 + W12

= 0 + 0RT

n 23

= 0RT

n 23

(IV P)

4. If the process carried out on one mole of

monoatomic ideal gas is as shown in figure in the

PV-diagram with 0 0 01

P V RT3

, the correct match is

(A) I S, II R, III Q, IV T (B) I Q, II R, III P, IV U

(C) I Q, II S, III R, IV U (D) I Q, II R, III S, IV U

4. (D)

1 2 3 0 0 01

W P V RT3

U123 = nCv (Tf Ti) = 1 0 0 0 03

(3P V P V )2

= RT0

1 2 3 04

Q U W RT3

1 2Q nCp T = 0 0 0 0 05 5

(2P V P V ) RT2 6


(Pg.15)

PART II : CHEMISTRY


This section contains EIGHT (08) 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.

1. Consider the following reaction (unbalanced)

2 4Zn hot conc. H SO G R X

Zn conc. NaOH T Q 2 4G H S NH OH Z(a precipitate) X Y Choose the correct option(s) (A) Bond order of Q is 1 in its ground state (B) The oxidation state of Zn in T is +1 (C) R is a V-shaped molecule (D) Z is dirty white in colour

1. (A), (C), (D)

2 4 4 2 2(Hot & conc.) (G) (R) (X)

Zn H SO ZnSO SO H O

2 2 2(T) (Q)

Zn conc. NaOH Na ZnO H

4 2 4 4 2 4 2(Z)(G) (Y) (X)

ZnSO H S NH OH(aq) ZnS NH ) SO H O

2. The cyanide process of gold extraction involves leaching out gold from its ore with CN in the presence of Q in water to form R. Subsequently, R is treated with T to obtain Au and Z. Choose the correct option (s)

(A) R is [Au(CN)4] (B) T is Zn (C) Q is O2 (D) Z is [Zn(CN)4]

2

2. (B), (C), (D)

2

CN , Q T

H OAu R Au Z

2 2 24Au(s) 8CN (aq) 2H O(aq) O (g) 4[Au(CN) ] (aq) 4OH (aq)

(Q)

2

242[Au(CN) ] (aq) Zn(s [Zn(CN) ] (aq) 2Au(s)

(Z)(R) (T)


(Pg.16)

3. Choose the correct option(s) that give(s) an aromatic compound as the major product

(A) (B)

(C) (D)

3. (A), (B)

(A)

(B)

(C)

(D)

4. Choose the correct option(s) for the following reaction sequence

Consider Q, R and S as major products (A)

(B)


(Pg.17)

(C) (D)

4. (A), (D)

5. With reference to aqua regia, choose the correct option(s) (A) Reaction of gold with aqua regia produces NO2 in the absence of air (B) The yellow colour of aqua regia is due to the presence of NOCl and Cl2. (C) Aqua regia is prepared by mixing conc. HCl and conc. HNO3 in 3 : 1 (v/v) ratio. (D) Reaction of gold with aqua regia produces an anion having Au in +3 oxidation state. 5. (B), (C), (D)

(A) Au + HNO3 + 4HCl 4 3 2AuCl H O NO H O

(B) Yellow colour of aqua regia is due to it’s decomposition into NOCl (orange yellow) and Cl2 (greenish yellow) (C) Aqua regia = 3HCl(conc.) + HNO3(conc.) (D) [AuCl4]

gets formed ; Au (+3). 6. The ground state energy of hydrogen atom is 13.6 eV. Consider an electronic state of

He+ whose energy, azimuthal quantum number and magnetic quantum number are 3.4 eV, 2 and 0, respectively. Which of the following statement(s) is (are) true for the state ?

(A) It is a 4d state (B) The nuclear charge experienced by the electron in this state is less than 2e, where e is the

magnitude of the electronic charge (C) It has 3 radial nodes (D) It has 2 angular nodes


(Pg.18)

6. (A), (D)

3.4 = 2

2

13.6 2

n

n = 4

= 2 Subshell = 4d

Angular nodes = = 2

Radial nodes = n = 4 2 1 = 1 7. Which of the following reactions produce(s) propane as a major product? (A) (B)

(C)

(D) 7. (A), (C)

electrolysis

3 2 2 2 2CH CH CH CO Na H O n hexane

NaOH CaO

3 2 2 2 3 2 3CH CH CH CO Na CH CH CH

dil.HCl

3 2 2 3 2 2 3 2 3CH CH CH Cl Zn CH CH CH ZnCl CH CH CH

8. Choose the correct option(s) from the following (A) Teflon is prepared by heating tetrafluoroethene in presence of a persulphate catalyst at

high pressure (B) Natural rubber is polyisoprene containing trans alkene units (C) Nylon-6 has amide linkages (D) Cellulose has only -D-glucose units that are joined by glycosidic linkages

8. (A),(C) (A) Teflon is prepared by heating tetrafluoroethene in presence of a persulphate catalyst at high pressure via radical addition mechanism.

Per sulphate2 2 2 2 nF C CF CF CF

(B) Natural rubber is polyisoprene containing cis alkene units.


(Pg.19)

(C) Nylon-6 has amide linkages.

(D) Cellulose has only -D-glucose units that are joined by glycosidic linkages





1. The decomposition reaction 2 5 2 4 22N O (g) 2N O (g) O (g) is started in a closed cylinder under isothermal isochoric condition at an initial pressure of 1 atm. After Y 103 s,

The pressure inside the cylinder is found to be 1.45 atm. If the rate constant of the reaction is 5 104 s1, assuming ideal gas behavior, the value of Y is ________.

1. [2.30] Unit of K represent it is first order reaction.

2N2O5 2N2O4 + O2 t = 0 1 0 0 t = t 1 P P P/2

P

1 P P2

= 1.45

P

2 = 0.45 , P = 0.9

t = 4

2.303 1log

1 P2

y 103 = 4

2.303 1log

12 =

4

2.303log10

2

y = 2.30 2. The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05.

If the density of the solution is 1.2 g/cm3, the molarity of urea solution is __________. (Given data : Molar masses of urea and water are 60 g mol1 and 18 g mol1,

respectively)


(Pg.20)

2. [2.98 or 2.99] Let mole of urea = x

x

900x

18

= 0.05 , x = 50

19

Mass of solution = 50

60 90019 =

20100

19

Volume of solution = 20100

mL19 1.2

= 20.1

litre19 1.2

Molarity =

50

1920.1

19 1.2

= 50 1.2

20.1

=

60

20.1 2.985

3. Total number of cis N-Mn-Cl bond angles (that is, Mn-N and Mn-Cl bonds in cis position) present in a molecule of cis [Mn(en)2Cl2] complex is _________ (en = NH2CH2CH2NH2)

3. [6]

Number of cis (Cl Mn N) = 6 4. The amount of water produced (in g) in the oxidation of 1 mole of rhombic sulphur by

conc. HNO3 to a compound with the highest oxidation state of sulphur is _______

(Given data : Molar mass of water = 18 g mol1)

4. [288]

2 8 2 432H O S 8H SO 48H 48e

3 2 2[e H HNO NO H O] 48

8 3 2 4 2 216 mol1mol

S 48HNO 8H SO 48NO 16H O

mass of H2O formed = (16) (18)g = 288 g

5. Total number of isomers considering both structural and stereoisomers, of cyclic ethers with the molecular formula C4H8O is _______________ .

5. [10]


(Pg.21)

6. Total number of hydroxyl groups present in a molecule of the major product P is____

6. [6]

Total 6 OH group present in a molecule of the major product.








in the paragraph.

Consider the Bohr’s model of a one-electron atom where the electron moves around the nucleus. In the following Column-I contains some quantities for the nth orbit of the atom and Column -II contains options showing how they depend on n.

Column-I Column-II (I) Radius of the nth orbit (P) n2 (II) Angular momentum of the electron in the nth orbit (Q) n1 (III) Kinetic energy of the electron in the nth orbit (R) n0 (IV) Potential energy of the electron in the nth orbit (S) n1 (T) n2 (U) n1/2


(Pg.22)

1. Which of the following has the correct combination considering List-I and List-II? (A) (III), (S) (B) (IV), (Q) (C) (III), (P) (D) (IV), (U) 1. (C) 2. Which of the following options has the correct combination considering List-I and List-II? (A) (II), (R) (B) (II), (Q) (C) (I), (P) (D) (I), (T) 2. (D)

r = 2n

0.529z

r n2

mvr = nh

2 (mvr) n

KE = +13.6 2

2

z

n KE n2

PE = 2 13.6 2

2

z

n PE n2


in the paragraph. Column-I includes starting materials and reagents of selected chemical reactions. Column-II gives structures of compounds that may be formed as intermediate products and/or final products from the reactions of Column-I

Column-I Column -II (I)

(P)

(II)

(Q)

(III)

(R)

(IV)

(S)

(T)

(U)


(Pg.23)

3. Which of the following options has correct combination considering Column-I and

Column -II

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

3. (A)

4. Which of the following options has correct combination considering Column -I and

Column-II

(A) (III), (S), (R) (B) (IV), (Q), (U) (C) (IV), (Q), (R) (D) (III), (T), (U)

4. (C) Solution 3 & 4


(Pg.24)

PART III – MATHEMATICS








1. Let f : R R be given by f(x) = (x 1) (x 2) (x 5). Define F(x) = x

0

f (t) dt, x 0 . Then

which of the following options is/are correct? (A) F(x) has a local maximum at x = 2 (B) F(x) has a local minimum at x = 1 (C) F(x) has two local maxima and one local minimum in (0, ) (D) F(x) 0, for all x (0, 5)

1. (A), (B), (D) f(x) = (x 1) (x 2) (x 5)

F(x) = x

0

f (t)dt as F(1) is negative and F(2) is also negative so F(x) cannot be zero for x

(0, 5). So option (A) is correct. F(x) = f(x) = (x 1) (x 2) (x 5)

F(x) has two local minima points at x = 1 and x = 5 F(x) has one local maxima point at x = 2 2. Three lines

L1 : ˆr i, R L2 : ˆ ˆr k j, R and L3 : ˆ ˆ ˆr i j k, R are given. For which point(s) Q on L2 can we find a point P on L1 and a point R on L3 so

that P, Q and R are collinear.

(A) 1ˆ ˆk j2

(B) ˆ ˆk j (C) k̂ (D) 1ˆ ˆk j2


(Pg.25)

2. (A), (D)

P.V. of point P, ˆp i P.V. of point Q, ˆ ˆq̂ j k P.V. of point R, ˆ ˆ ˆr i j rk PQR are collinear. Hence x(PQ) y (PR)

x 1 1

y

= r

1 ˆ ˆq j kr

or ˆ ˆq j k1

, where r 0, 0, 1

1

0, 1 Hence, ˆ ˆ ˆq k or j k 3. Let x R and let

1 1 1 2 x x

P 0 2 2 , Q 0 4 0

0 0 3 x x 6

and R = PQR1.

Then which of the following is/are correct. (A) there exists a real number x such that PQ = QP

(B) det R = det

2 x x

0 4 0

x x 5

+ 8 for all x R

(C) For x = 1 there exists a unit vector ˆ ˆ ˆi j k for which are 0

R 0

0

(D) For x = 0 if R

1 1

a 6 a

b b

then a + b = 5

3. (B), (D) det R = det P det Q det(P1) det R = det Q = 4(12 x2) = 48 4x2

Now

2 x x

0 4 0

x x 5

= 4(10 x2) = 40 4x2

det R = det 2 x x

0 4 0

x x 5

+ 8 x R

At x = 1, det Q = 48 4 = 44 = det R Because det R 0 so

R

0

0

0

= = = 0

So ˆ ˆ ˆi j k is not a unit vector


(Pg.26)

P =

1 1 1

0 2 2

0 0 3

, Q (x = 0) =

2 0 0

0 4 0

0 0 6

R = P Q P1

=

2 4 6 6 3 0 12 6 41 1

0 8 12 0 3 2 0 24 86 6

0 0 18 0 0 2 0 0 36

R =

2 1 2 / 3

0 4 4 / 3

0 0 6

(R 6I) 1 4 1 2 / 3 1

a 0 2 4 / 3 a

b 0 0 0 b

4 + a + 2b

3= 0

2a + 4

b3

= 0

8 + 8

3 = 0 b = 3

PQ = QP a12 for both are same x + 4 + x = 2 + 2x + 0 x No value exist

4. For nonnegative integer n, let f(n) =

n

k 0n

2

k 0

k 1 k 2sin sin

n 2 n 2

k 1sin

n 2

Assuming cos1x takes values in [0, ] which of the following options is/are correct?

(A) f(4) = 3

2

(B) If = tan (cos1 f(6)), then 2 + 2 1 = 0 (C) sin (7 cos1 f(5)) = 0

(D) n

1lim f (n)

2

4. (A), (B), (C)

f(x) =

n

k 0n

2

k 0

k 1 k 22 sin sin

n 2 n 2

k 12sin

n 2

n n

k 0 k 0n

k 0

2k 3cos cos

n 2 n 2

2k 21 cos

n 2


(Pg.27)

=

n 3 n 1cos .sin

n 2 n 2(n 1)cos

n 2sin

n 2

n 1cos .sin

n 2(n 1)

sinn 2

=

cos .sinn 2 n 2

(n 1)cosn 2

sinn 2

cos .sinn 2

(n 1)

sinn 2

=

(n 1) cos cosn 2 n 2

n 1 1

=

(n 2)cosn 2

cosn 2 n 2

(A) f(4) = cos 3

cos4 2 6 2

(B) = tan 1cos cos tan8 8

22

2 tan28tan 1

4 11 tan8

2 + 2 1 = 0

(C) sin 17 cos cos7

= sin 77

= sin = 0.

(D) n nlim f (n) lim cos

n 2

= cos (0) = 1

5. Let f : R R be a function we say that f has

PROPERTY 1 if h 0

f (h) f (0)lim

h

exist and is finite and

PROPERTY 2 if 2h 0

f (h) f (0)lim

h

exist and is finite.

Then which of the following options is/are correct? (A) f(x) = x2/3 has property 1 (B) f(x) = sin x has property 2 (C) f(x) = |x| has property 1 (D) f(x) = x|x| has property 2


(Pg.28)

5. (A), (C) P1 :

h 0

f (h) f (0)lim

h

= exist and finite

(B) f(x) = 2/3x , 2/3

h 0

h 0lim

h

=

2/3

h 0

hlim

h= 0

(D) f(x) = x , h 0

h 0lim

h

h 0lim h

= 0

P2 :

2h 0

f (h) f (0)lim

h

= exist and finite

(A) f(x) = x x , 2h 0

h h 0lim

h

=

2

2h 0

2

2h 0

hRHL lim 1

h

hLHL lim 1

h

(C) f(x) = sin x 2h 0

sinh 0lim

h

= DNE

6. For a R , |a| > 1, let

3 3

n 7/32 2 2

1 2 ....... nlim

1 1 1n ......

(na 1) (na 2) (na n)

= 54. Then possible value(s) a is/are

(A) 8 (B) 9 (C) 7 (D) 6 6. (A), (B)

1/3n1/3

r 1

nn7/3

2r 1

rn

nlim

1n

(an r)

= 54

1/3n

r 1nn

2r 1

1 r

n nlim

1 1

n (a r / n)

= 54

11/3

01

20

x dx

1dx

(a x)

= 54

3

41

a(a 1)

= 54

a(a + 1) = 72 a2 + a 72 = 0 a = 9, 8

7. Let

P1 = I = 2 3

1 0 0 1 0 0 0 1 0

0 1 0 , P 0 0 1 , P 1 0 0 ,

0 0 1 0 1 0 0 0 1

P4 = 5 6

0 1 0 0 0 1 0 0 1

0 0 1 , P 1 0 0 , P 0 1 0 ,

1 0 0 0 1 0 1 0 0

and X = 6

kk 1

2 1 3

P 1 0 2

3 2 1

PkT.


(Pg.29)

Where PkT denotes the transpose of matrix Pk. Then which of the following options is/are

correct? (A) X is a symmetric matrix (B) The sum of diagonal entries of X is 18

(C) If X

1 1

1 1

1 1

, then = 30

(D) X 30I is an invertible matrix 7. (A), (B), (C)

Clearly P1 = T 1

1 1P P

P2 = T 12 2P P

. . .

P6 = T 16 6P P

and AT = A, where A =

2 1 3

1 0 2

3 2 1

Using formula (A + B)T = AT + BT

XT = T T T T T T T1 1 6 6 1 1 6 6(P AP ..... P AP ) P A P ..... P A P X X is symmetric.

Let B =

1

1

1

XB = T T T1 1 2 2 6 6(P AP B P AP B ..... P AP B = P1 AB + PAB + …… + P6AB

XB = (P1 + P2 + ….. + P6) 6

3

6

=

6 2 3 2 6 2 30

6 2 3 2 6 2 30 30B

6 2 3 2 6 2 30

= 30

since X

1

1

1

= 30

1

1

1

(X 30I)B = 0 has a non trivial solution B = 1

1

1

|X 30I| = 0 X = T T1 1 6 6P AP ..... P AP

trace (x) = T Tr 1 1 r 6 6t (P AP ) ...... t (P AP ) = (2 + 0 + 1) + ….. + (2 + 0 + 1) = 3 6 = 18


(Pg.30)

8. Let

f(x) = 2

sin x

x

, x > 0.

Let x1 < x2 < x3 …… < xn < ……. be all points of local maximum of f and y1 < y2 < y3 < …… < yn < ……. be all the points of local minimum of f

Then which of the following options is/are correct? (A) |xn yn| > 1 for every n (B) x1 < y1 (C) xn+1 xn > 2 for every n

(D) xn 1

2n, 2n2

for every n

8. (A), (C), (D)

f (x) = 4

x2x cos x tan x

2

x





1. Suppose

det

n nn 2

kk 0 k 0

n nn n k

k kk 0 k 0

k C k

C k C 3

= 0, holds for some positive integer n. Then

nnk

k 0

C

k 1 equals.

1. [6.20]

nn 1

n 1 n 2

4n2

n(n 1)n2 n(n 1)2 0

2

2 2n(n 1) n n (n 1)

2 4 8

= 0

n = 0 or 4(n + 1) 2n n (n 1) = 0 4n + 4 2n n2 + n = 0 3n n2 + 4 = 0 n2 3n 4 = 0 (n 4) (n + 1) = 0 n = 4

4 5 54 4

r r 1

r 0 r 0

C C 2 1 31

r 1 5 5 5

= 6.20


(Pg.31)

2. The value of

sec1 10

k 0

1 7 k 7 (k 1)sec sec

4 12 2 12 2

in the interval 3,

4 4

equals

2. [0.00]

10

1

k 0

1 1sec

7 k 7 (k 1)4cos cos

12 12 12 2

= 10

1

k 0

7 (k 1) 7 ksin

12 2 12 21sec

7 k 7 (k 1)4cos cos

12 12 12 2

= 10

1

k 0

1 7 7 ksec tan (k 1) tan

4 12 2 12 2

= 11 11 7 7

sec tan tan4 2 12 12

= 1

1 7 7sec cot tan

4 12 12

= 11 1

sec7 74 sin cos12 12

= 11 1

sec72 sin6

= sec1(1) = 0.00

3. Let ˆ ˆ ˆa 2i j k and ˆ ˆ ˆb i 2 j k be two vectors. Consider a vector c a b , , R. If the projection of c on the vector (a b) is 2 , then the minimum value of

(c (a b)).c equal to 3. [18.00]

c.(a b)

3 2| a b |

a . a b.b ( )a.b

3 2

= 3 2

6 + 6 =( + )3 = 18 + = 2 2 2 2 2c (a b) . c | a | | b | 2 a .b (a b) . c = 62 + 62 + 6 (as c is linearly dependent on a b ) = 6[( + )2 ] , maximum = 1 min. (c (a b)) . c 18 4. Five persons A, B, C, D and E are seated in a circular arrangement. If each of them is

given a hat of one of the three colours red, blue and green, then the numbers of ways of distributing the hats such that the person seated in adjacent seats get different coloured hats is

4. [30.00] Maximum number of hats used of same colour are 2. They can not be 3 otherwise atleast 2 hats of same colour are consecutive.


(Pg.32)

Now, Let hats used are R, R, G, G, B (Which can be selected in 3 ways. It can be RGGBB or RRGBB also) Now, numbers of ways of distributing blue hat (single one) in 5 person equal to 5 Let blue hat goes to person A.

Now either position B & D are filled by green hats and C & E are filled by Reds hats Or B & D are filled by Red hats and C & E are filled by Green hats

2 ways are possible Hence total number of ways = 3 × 5 × 2 = 30 ways

5. The value of the integral

/2

50

3 cos

cos sin

d equals

5. [0.50]

I = /2

50

cos d3

( sin cos )

=

/2

50

3 sin d3

( cos sin )

/2

50

3 cos d

( sin cos )

=

/2

50

sin d3

( cos sin )

2I =

/2

40

d3

( sin cos )

2I

3=

/2 2

40

sec d

( tan 1)

Let tan = t2 sec2 d = 2tdt

2I

3=

40

2tdt

(t 1)

I

3 =

3 40

1 1dt

(t 1) (t 1)

I = 2 3

0

3 1

2(t 1) (t 1)

= 3

12 =

1

2

6. Let |X| denote the number of elements in a set X. Let S = {1, 2, 3, 4, 5, 6} be a sample

space, where each element is equally likely to occur. If A and B are independent events associated with S, then the number of ordered pairs (A, B) such that 1 |B| < |A| equals

6. [422.00]


(Pg.33)







Match Type

Let f(x) = sin ( cos x) and g(x) = cos (2 sin x) be two functions defined for x > 0. Define the following sets whose elements are written in increasing order

X = {x : f(x) = 0}, Y = {x : f (x) = 0} Z = {x : g(x) = 0}, W = {x : g(x) = 0} List I contains sets X, Y, Z and W ListII contains some information regarding these set.

List I List II (I) X (P) 3

, , 4 , 72 2

(II) Y (Q) an arithmetic progression

(III) Z (R) NOT an arithmetic progression

(IV) W (S) 7 13, ,

6 6 6

(T) 2, ,

3 3

(U) 3,

6 4

1. Which of the following is the only correct combination?

(A) IV (Q), (T) (B) III (R), (U) (C) III (P), (Q), (U) (D) IV (P), (R), (S) 2. Which is the following is only CORRECT combination?

(A) I (Q), (U) (B) I (P), (R) (C) II (Q), (T) (D) II (R), (S) 1. (D) & 2. (C)

(Solution Q. 1 & 2)

f(x) = 0 sin ( cos x) = 0 cos x = n cos x = n

cos x = 1, 0, 1 X = {n , (2n 1) }2

= {n , n I}

2


(Pg.34)

f (x) = 0 cos( cos x) ( sin x) = 0

cos x = (2n 1)2

or x = n

cos x = 1

n2

or x = n cos x = 1

2 or x = n

Y = 2

2n , 2n , n , n I3 3

= 2 2 4

..... , , 0, , , ,3 3 3 3 3

which is an arithmetic progression

g(x) = 0 cos (2 sin x) = 0 2 sin x = (2n 1)2

sin x = 2n 1

4

=

1 3,

4 4

Z = 1 11 3

n sin , n sin , n I4 4

g(x) = 0 sin(2 sin x) (2 cos x) = 0

2 sin x = n or x = (2n 1)2

sin x = n

2=

10, , 1

2 or x = (2n 1)

2

W = n , (2n 1) , n , n I2 6

Match Type

Let the circle C1 : x2 + y2 = 9 and C2 : (x 3)2 + (y 4)2 = 16 intersect at the points X and

Y. Suppose that another circle C3 : (x h)2 + (y k)2 = r2 satisfies the following conditions. (i) centre of C3 is collinear with the centres of C1 and C2

(ii) C1 and C2 both lie inside C3 and

(iii)C3 touches C1 at M and C2 at N

Let the line through X and Y intersect C3 at Z and W and let a common tangent of C1 and C3

be a tangent to the parabola x2 = 8y There are some expressions given in the list I, whose values are given in listII below :

List I List II (I) 2h + k (P) 6 (II) length of ZW

length of XY

(Q) 6

(III) Area of MZN

Area of ZMW

(R) 5

4

(IV) (S) 215

(T) 2 6

(U) 10

3


(Pg.35)

3. Which of the following is the only correct combination? (A) (II) (T) (B) (I) (S) (C) (II) (Q) (D) (I) (U) 4. Which of the following is the only Incorrect combination? (A) (IV) (S) (B) (I) (P) (C) (III) (R) (D)(IV) (U)

3. (C) & 4. (A) (Solution Q. 3 & 4)

Centre of C3 will lie on line on line MN : 3x = 4y at a distance of 3 units from origin in first quadrant

(h, k) = 9 12

,5 5

and r = 6

Equation of ZW : C1 C2 = 0 6x + 8y 18 = 0 or 3x + 4y = 9

Perpendicular distance from centre of C3, to ZW =

27 489

65 5

5 5

Length of ZW = 36 24

2 36 625 5

Similarly, length of XY = 81 24

2 925 5

ZW.MN

Area of MZN ZP.MN 2Area of ZMW MP.ZW MP.ZW

1.12

5224 45

Common tangent of C1C3 : C1 C3 = 0 (as they touch each other)

18 24

x5 5

y + 18 = 0

or 3x + 4y + 5 = 0

Tangent to x2 = 8y is x = Py + 2

P

Comparing 3 4 15

21 PP

P = 4 15P 10

33 2 3

I (P) II (Q) III (R) IV (U)

solution to iit jee 2019 (advanced) : paper - ii · so, e net should be to surface so 1/3 00 3 0 0...

Documents