sri chaitanya · 2020. 4. 10. · sri chaitanya iit academy # 304, kasetty heights, ayyappa...

2019-Jee-Advanced Question Paper-2_Key & Solutions

Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]

Page

-1



Page

-2

PHYSICS Max Marks: 62 SECTION-1 (Maximum Marks : 32)

This section contains EIGHT (08) questions Each question has FOUR options. ONE MORE THAN ONE of these four option(s) is(are) correct answer(s) For each question, choose the 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. For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct answers, then Choosing ONLY (A), (B), and (D) will get +4 marks ; Choosing ONLY (A) and (B) will get +2 marks ; Choosing ONLY (A) and (D) will get +2 marks ; Choosing ONLY (B) and (D) will get +2 marks ; Choosing ONLY (A) will get +1 marks; Choosing ONLY (B) will get +1 marks; Choosing ONLY (D) will get +1 marks; Choosing no option (i.e. the question is unanswered) will get 0 marks ; and Choosing any other combination of options will get -1 mark. Question ID: 337911170

1. 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 ap and ep , respectively. If a e1/5

, which of the

option (s) is /are correct ?

[Use hc = 1242 eV nm; 1 nm = 10-9 m, h and c are Planck’s constant and speed of light,

respectively]

1) a e1p / p2

2) The ratio of kinetic energy of the electron in the state n = m to the state n = 1 is 14

3) m = 2

4) e = 418 nm

Key: 2,3



Page

-3

Sol: 1 1116a

R

2 2

2 2

2 2

2

1 1 14

1 114

15 516

1 1 3164

1 1 24

e

a

e

Rm

m

m

mm

Kinetic energy 21n

21

1 1142

1 1 124213.64 163 124213.6

16

m

e

e

KK

Question ID: 337911165

2. 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 V << dLL

v0, where dL is the infinitesimal displacement of the piston. Which of the

following statement(s) is/are correct ?



Page

-4

1) If the piston moves inward by dL, the particle speed increases by 2v dLL

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

3) The rate at which the particle strikes the piston is v/L

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

inward from Lo to 12

L0

Key: 2,4

Sol:

Before the collision speed is v.

After the collision speed is v + 2V.

Change in speed is 2V.

If V1 is the instantaneous speed and ‘x’ remaining distance then V1x = constant.

V1 = 2v.


3. 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)

1) The work W done during the process is 13RT0

2) The final pressure of the gas mixture after compression is in between 9P0 and 10P0

3) Adiabatic constant of the gas mixture is 1.6

4) The average kinetic energy of the gas mixture after compression is in between 18RT0

and 19RT0

Key: 1,2,3

Sol: 15 o

o

RTPV



Page

-5

2o

o

RTPV

1 2

1 11

mixture 1 11

6

5 15 1

oo

o

P P

V V

RTP P PV

C CC C

5 75 12 2

3 55 12 2

32 1.620

R R

R R

1.61.6

1.6 3.2

constant

P / 4

4 2

9.2

o o o

o o

o

PV

V P V

P P P

P P

2 2 1 19.2

41 1 1.6

2.3 1 1.3 60.6 0.6

13

oo o o

o o o

o

VP PVPV PVW

PV RT

W RT


4. 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 600 with vertical ?

[g is the acceleration due to gravity]

1) The normal reaction force from the floor on the rod will be Mg16

2) The angular acceleration of the rod will be 2gL

3) The angular speed of the rod will be 3g2L

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

Key: 1,3,4

Sol: 2

sin602 3l mlmg



Page

-6

34 3

g l 3 3

4g

l

2

2

2

2

11 cos602 2 3

4 632

l mlmg

g l

gl

radial acceleration 2

2la

34ga

3 1 3 3 34 2 4 2

3 98 16

1516 16

g gMg N M ll

Mg N Mg

Mg MgN Mg


5. An electric dipole with dipole moment 0p i 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

statements(s) is/are :

( 0 is permittivity of free space. R >>dipole size)



Page

-7

1) Total electric field at point B is BE 0

2) Total electric field at point A is A 0E 2E i j

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

4) 1/3

0

0 0

pR4 E

Key: 1,4

Sol:

B A

At ‘B’ electric field due to dipole is

3

3

13

42 3

oB

oo

o

o o

A o o o

KPER

KP ER

PRE

E E E E



Page

-8


6. 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 = 3 m). If H1, H2, and H3 are the apparent depths of a point X

on the bottom of the three cylinders, respectively, the correct statement(s) is/are :

1) 0.8 cm < (H2 – H1) < 0.9 cm

2) H3 > H1

3) H2 > H1

4) H2 > H3

Key: 3,4

Sol: (I)

12 2 30 20

3 3HH cm

(II)



Page

-9

1 1.5 1 1.5

1 1 1.52 3 0.3

V H R

V

1 1 56

1 2966 0.193529

V

V

V

(III)

1 1 56631

V

V


7. A block of mass 2M is attached to a massless spring with spring-constant k. This block

is connected to two other blocks of masses M and 2M using two massless pulleys and

strings. The acceleration 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]



Page

-10

1) At an extension of 0x4

of the spring, the magnitude of acceleration of the block

connected to the spring is 3g10

2) When spring achieves an extension of 0x2

for the first time, the speed of the block

connected to the spring is 3g M5k

3) 04Mgx

k

4) a2 – a1 = a1 – a3 Key: 4

Sol:

2 1 1 3a a a a

Effective mass of pulley-block system is 4 2 8

3 3M M M

M .

163oMgxK

22

oo

xx A A

Acceleration of block is 2 4 3 22 3 14 7A Mg k g

k M

Speed = 3 8 32 14 3 14

ox k Mg KAM K M

Question ID: 337911169 8. 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 ?



Page

-11

1) Fringe spacing depends on

2) For 0.36

degree, there will be destructive interference at point O.

3) For 0 , there will be constructive interference at point P.

4) For 0.36

degree, there will be destructive interference at point P.

Key: 4

Sol: sind x

2

3 70.36 36 100.3 10 6 105 180 180

d

Fringe width 7

46 10 1 2

3 10D mmd



Page

-12

SECTION-2 (Maximum Marks : 18) 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 on-screen virtual numeric

keypad in the place designated to enter the answer. If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.

Answer to each question will be evaluated according to the following marking scheme : Full Marks : + 3 If ONLY the correct numerical value is entered. Zero Marks : +0 In all other cases Question ID: 337911172

1. 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 = 1 T 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]

Key: 0.63 (0.62 – 0.64)

Sol: EMF = BVl = 2 1 31 1 10 10 10 VV

Time constant = 31 10 1 .sec

1L mR

/5

31

3

1

10 11

10 0.63 0.63

toi i e

i e

mA



Page

-13


2. A ball is thrown from ground at 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 0u / . 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 _____

Key: 4

Sol: 20

1sin 2uRg

20 1

2 2 21 sin 2u RR

g

01

12

11 2 1

2 sin

........

1..... 1 11

utg

tt

tT t t t

Average speed = 1

1

Rt

Average speed = RT

4 .



Page

-14

Question ID: 337911176 3. 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 _____

Key: 1.39 (1.35 - 1.45)

Sol: 1 1 1v u f

2 2 2

1 1 1 2060 30

f cmf

df dv duf v u

Question ID: 337911174 4. A perfectly reflecting mirror of mass M mounted on a spring constitutes a spring-mass

system of angular frequency such that 24 24 M 10 mh

with h as Planck’s

constant. N photons of wavelength 68 10 m strike the mirror simultaneously at normal incidence such that the mirror gets displaced by 1 m . If the value of N is x 1012, then the value of x is ______

[Consider the spring as massless]

Key: 1



Page

-15

Sol: Km

KM

Momentum change for single photon = 2h

Momentum change = 2Nh

12

12

2

2

452 10 2

10

NhA MV

NhMA

M Nh

N

Question ID: 337911173 5. A monochromatic light is incident from air on a refracting surface of a prism of angle

750 and refractive index n0 = 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 060 . The value of n2 is

Key: 1.5 Sol: 11 sin60 3 sin r

1 1

2

2

1sin 302

451 3 sin90 1.52

o

o

o

r r

r

n n



Page

-16


6. Suppose a 22688 Ra nucleus at rest and in ground state undergoes -decay to a 222

86 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 = 226.005 u, atomic mass of 222

86 Rn = 222.000 u,

atomic mass of particle = 4.000 u, 1 u = 931 MeV/c2, c is speed of the light]

Key: 135 (130 – 140)

Sol: 226 222 488 86 2Ra Rn He

226.005 222 4 0.005Rnm mRa m m

u

Q-Value = 35 10 931 4.655MeV MeV ,K.E. of 22286

4.44 4 0.08222

Rn

Energy of 4.655 4.52 135MeV KeV

SECTION-3 (Maximum Marks : 12) This section contains TWO (02) List-Match sets. Each List-Match set has TWO (02) Multiple Choice Questions. Each List-Match set has two lists : List-I and List-II. List-I has Four entries (I), (II), (III) and (IV) and List-II has Six entries (P), (Q), (R), (S), (T) and (U). FOUR options are given in each Multiple Choice Questions based on List-I and List-II 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 Question ID: 337911177

1. Answer the following by appropriately matching the lists based on the information given in the paragraph.

A musical instrument 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.

List-I gives the above four strings while list-II lists the magnitude of some quantity.



Page

-17

List-I List-II

I) String-1 ( ) P) 1

II) String-2 (2 ) Q) 1/2

III) String-3 (3 ) R) 1 / 2

IV) String-4 (4 ) S) 1 / 3

T) 3/16

U) 1/16

If the tension in each string is T0, the correct match for the highest fundamental

frequency in f0 units will be,

1) IP, IIQ, IIIT, IVS 2) IQ, IIP, IIIR, IVT

3) IP, IIR, IIIS, IVQ 4) IQ, IIS, IIIR, IVP

Key: 3

Sol: 12

oo

o

TfL

1

2

3

4

2

3

2

o

o

o

o

f fff

ff

ff


2. Answer the following by appropriately matching the lists based on the information

given in the paragraph.

A musical instrument 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.

List-I gives the above four strings while list-II lists the magnitude of some quantity.



Page

-18

List-I List-II

I) String-1 ( ) P) 1

II) String-2 (2 ) Q) 1/2

III) String-3 (3 ) R) 1 / 2

IV) String-4 (4 ) S) 1 / 3

T) 3/16

U) 1/16

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

, and 07L4

,

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

1) IP, IIQ, IIIT, IVU

2) IT, IIQ, IIIR, IVU

3) IP, IIQ, IIIR, IVT

4) IP, IIR, IIIT, IVU

Key: 1

Sol: 12

oo

o

TfL

1

22

33

12

232 2 3

453 2 5

o

o

o

o

TfL

TfL

TfL



Page

-19




In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed

by the gas is given by T X , 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 ln R ln2 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.

List-I List-II

I) Work done by the system in process1 2 3 P) 01 RT ln 23

II) Change in internal energy in process 1 2 3 Q) 01 RT3

III) Heat absorbed by the system in process 1 2 3 R) 0RT

IV) Heat absorbed by the system in process 1 2 S) 04 RT3

T) 01 RT 3 ln 23

U) 05 RT6

If the process carried out on one mole of monatomic ideal gas is as shown in figure in

the PV-diagram with P0V0 = 01 RT3

, the correct match is,



Page

-20

1) IQ, IIR, IIIP, IVU 2) IQ, IIR, IIIS, IVU

3) IQ, IIS, IIIR, IVU 4) IS, IIR, IIIQ, IVT

Key: 2

Sol: Work done in 1 2 = PoVo 2 3 = 0 Total work in 1 2 3 = P0V0

Q in 1 2 3 = 43 oRT

U in 1 2 3 = RTo Question ID: 337911180



In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed

by the gas is given by T X , 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 ln R ln2 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.



Page

-21

List-I List-II

I) Work done by the system in process1 2 3 P) 01 RT ln 23

II) Change in internal energy in process 1 2 3 Q) 01 RT3

III) Heat absorbed by the system in process 1 2 3 R) 0RT

IV) Heat absorbed by the system in process 1 2 S) 04 RT3

T) 01 RT 3 ln 23

U) 05 RT6

If the process on one mole of monatomic ideal gas is as shown in the TV-diagram with

P0V0 = 01 RT3

, the correct match is,

1) IP, IIR, IIIT, IVS 2) IP, IIT, IIIQ, IVT 3) IS, IIT, IIIQ, IVU 4) IP, IIR, IIIT, IVP

Key: 4

Sol: Work done in 12 = ln 23

oRT

2 3 0 U in 1 2 = 0

O in 1 2 = ln 23

oRTW



Page

-22

CHEMISTRY Max Marks: 62 SECTION-1 (Maximum Marks : 32)

This section contains EIGHT (08) questions Each question has FOUR options. ONE MORE THAN ONE of these four option(s) is(are) correct answer(s) For each question, choose the 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. For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct answers, then Choosing ONLY (A), (B), and (D) will get +4 marks ; Choosing ONLY (A) and (B) will get +2 marks ; Choosing ONLY (A) and (D) will get +2 marks ; Choosing ONLY (B) and (D) will get +2 marks ; Choosing ONLY (A) will get +1 marks; Choosing ONLY (B) will get +1 marks; Choosing ONLY (D) will get +1 marks; Choosing no option (i.e. the question is unanswered) will get 0 marks ; and Choosing any other combination of options will get -1 mark.


1. With reference to aqua regia, choose the correct options(s)

1) Reaction of gold with aqua regia produces NO2 in the absence of air

2) Reaction of gold with aqua regia produces an anion having Au in +3 oxidation state

3) Aqua regia is prepared by mixing conc. HCl and conc. HNO3 in 3 : 1 (v/v) ratio

4) The yellow colour of aqua regia is due to the presence of NOCl and Cl2

Key: 2,3,4

Sol: Aquaregia is a mixture of conc.HCl and conc.HNO3 in 3:1 ratio.

When gold dissolves in aquaregia, the species formed is 4AuCl in which gold is in +3

oxidation state.

In the absence of air the reaction between gold and aquaregia

3 4 24 2Au HNO HCl H AuCl H O NO

NO2 will not be produced. Yellow colour is due to the presence of NOCl and Cl2.



Page

-23

Question ID: 337911181 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)

1) Q is O2 2) T is Zn 3) Z is [Zn(CN)4]2- 4) R is [Au(CN)4]- Key: 1,2,3 Sol: Gold extraction:

2 2 2

22 4

4 8 2 4 4

4 2 2 4

R

TR Z

QAu NaCN H O O Na Au CN NaOH

Na Au CN Zn Na Zn CN Au

Question ID: 337911185 3. Which of the following reactions produce(s) propane as a major product ?

1)

2)

3)

4) Key: 1,3 Sol:

CH3

O

ONa NaOHCaO,Δ

CH3 CH3

CH3

O

ONa H2Oelectrolysis CH3

CH3

Cl

CH3Zn

dil.HCl CH3CH3

Br

CH3

Br

ZnCH2CH3



Page

-24


4. Choose the correct option(s) from the following.

1) Nylon-6 has amide linkages

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

3) Teflon is prepared by heating tetrafluoroethene in presence of a persulphate catalyst

at high pressure

4) Natural rubber is polyisoprene containing trans alkene units

Key: 1,3

Sol: Nylon-6 has amide linkages.

Cellulose has only β-D-glucose units that are joined by glycosidic linkages.

Teflon is prepared by heating tetrafluoromethane in presence of a persulphate catalyst

at high pressure.

Natural rubber is polyisoprene containing cis alkene units


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

1)

2)

3)

4)

Key: 2,3

Sol:

BrNaOEt

Anti aromatic



Page

-25

CH3

Br

Br

1.alcKOH2.NaNH2

3.redhot iron tube,873K

CH3

CH3 CH3Aromatic

NaOMe- Na+

Aromatic

+ Cl2 (excess) UV,500K

Cl

Cl

Cl

Cl

Cl

Cl

Non aromatic Question ID: 337911184

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 from the

state ?

1) It is a 4d state

2) It has 3 radial nodes

3) It has 2 angular nodes

4) The nuclear charge experienced by the electron in this state is less than 2e, where e

is the magnitude of the electronic charge

Key: 1,3

Sol: 2

2213.6 3.4En

2

2

2 2

213.6 3.4

4 4n

n n

Wave function corresponds to 4,2,0 represents 24z

d – orbital which has only one radial

nodes and two angular nodes. It experiences nuclear change of 2e units.



Page

-26


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

Consider Q, R and S as major products.

1)

2)

3)

4) Key: 1,4 Sol:

MeO C C CH2

CHO i)Hg2+,dilH2SO4 MeO C CH2 CH2

CHOOii)AgNO3 NH4OH

MeO C CH2 CH2

COOHO

MeO CH2 CH2 CH2

COOH

MeO CH2 CH2 CH2

COCl MeO

OMeO

iii)Zn-Hg.conc.HCl

i)SOCl2 pyridine

ii)AlCl3 Zn-Hgconc.HCl

'Q'

'R' 'S'



Page

-27


8. Consider the following reactions (unbalanced)

Zn + hot conc. H2SO4 G + R + X

Zn + conc. NaOH T + Q

G + H2S + NH4OH Z (a precipitate) + X + Y

Choose the correct option(s)

1) The oxidation state of Zn in T is + 1

2) Bond order of Q is 1 in its ground state

3) Z is dirty white in colour

4) R is a V-shaped molecule

Key: 2,3,4

Sol: 2 4Hot & Conc.

Zn H SO G R X

2 4 4 2 2

2 2 2

4 2 4 4 4 22

2

Conc.G R X

T Q

ZG XY

Zn H SO ZnSO SO H O

Zn NaOH Na ZnO H

ZnSO H S NH OH ZnS NH SO H O

1) Oxidation state of Zn in Na2ZnO2 is +2 2) Bond order of Q is one for H2(Q) 3) ZnS is white in colour 4) SO2 is angular in shape

SECTION-2(Maximum Marks : 18) 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 on-screen virtual numeric


Answer to each question will be evaluated according to the following marking scheme : Full Marks : + 3 If ONLY the correct numerical value is entered. Zero Marks : +0 In all other cases Question ID: 337911190

1. Total number of cis N – Mn – Cl bond angles (that is Mn – N and Mn – Cl bonds in cis

positions) present in a molecule of cis –[Mn(en)2Cl2] complex is ___

(en = NH2CH2CH2NH2)

Key: 6



Page

-28

Sol:

Mn

2H N Cl

Cl2NH

2CH 2CH

2NH

2 2CH CH

2NH

3 1

24

Similarly from bottom N-there will be another (2).

Total N – Mn – Cl angle are 6.


2. 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 = 18g mol-1) s

Key: 288 (288.00 – 288.30)

Sol: S8 + HNO3 H2SO4 + NO2 + H2O

Rhombic sulphur

Balancing the equation

S8 + 48HNO3 8H2SO4 + 48NO2 + 16H2O

Rhombic sulphur

1 mole S8 produces 16 x 18 gm of H2O i.e. 288 gms of H2O.


3. The decomposition reaction 2 5 2 4 22 2N O g N O g O g is started in a closed

cylinder under isothermal isochoric condition at an initial pressure of 1 atm. After 310Y s the pressure inside the cylinder is found to be 1.45 atm. If the rate constant of

the reaction is 4 15 10 s , assuming ideal gas behavior, the value of Y is __

Key: 2.30

Sol: 2 2 4 252 2gN O N O g O g

Given rate constant of the reaction = 4 15 10 sec



Page

-29

2 5

2 5

2 5

2 5

overall

overall

12

2

N ON O

N ON O

dPK P

dtdp

K Pdt

2 5 2 4 21 /2

2 2xx x

N O g N O g O g

3

4

1 1.4520.90 atm

2.303 110 log0.12 5 10

23.03 2.3010

x

x

y

y


4. Total number of hydroxyl groups present in a molecule of major product P is ___

Key: 6

Sol:

i)H2,Pd-BaSO4,quinoline

ii)dil.KMnO4(excess),273K

OH

OH

OH

OH

OHOH


5. Total number of isomers, considering both structural and stereoisomers of cyclic ethers

with the molecular formula 4 8C H O is____

Key: 10

Sol:

OO

CH3

O

CH3

O

CH3CH3

O

CH3 CH3

O

CH3

* * * *



Page

-30


6. The mole fraction of urea in an aqueous urea solution containing 900g of water is 0.05.

If the density of the solution is 1.2g cm-3, the molarity of urea solution is ___

(Given data: Molar masses of urea and water are 60g mol-1 and 18g mol-1, respectively)

Key: 2.98 (2.80 – 3.05)

Sol: Xurea = 0.05 Mass of water = 900 gm

Number of moles of water = 2

900 5018H On Let moles of urea = Un

0.05

505019

U

U

U

nn

n

Mass of urea = 50 6019

Given density (d) of solution = 1.2 g/ml.

50 60900massof solution 19density / 1.2solutionV

g cc

50191000 2.98

50 60900 1000191.2

U

ml

nMV

SECTION-3(Maximum Marks : 12) This section contains TWO (02) List-Match sets. Each List-Match set has TWO (02) Multiple Choice Questions. Each List-Match set has two lists : List-I and List-II. List-I has Four entries (I), (II), (III) and (IV) and List-II has Six entries (P), (Q), (R), (S), (T) and (U). FOUR options are given in each Multiple Choice Questions based on List-I and List-II 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



Page

-31




Consider the Bohr’s model of a one – electron atom where the electron moves around

the nucleus. In the following List – I contains some quantities for the nth orbit of the

atom and List – II contains options showing how they depend on n

List – I List – II

(I) Radius of the nth orbit (P) 2n

(II) Angular momentum of the electron in the nth orbit (Q) 1n

(III) Kinetic energy of the electron in the nth orbit (R) 0n

(IV)Potential energy of the electron in the nth orbit (S) 1n

(T) 2n

(U) 1/2n

Which of the following options has the correct combination considering List – I and List – II?

1. (II), (R) 2. (II), (Q) 3. (I), (P) 4. (I), (T) Key: 4

Sol: 2nr

Z

2

2

2

2

L n

ZKEn

ZPEn




Consider the Bohr’s model of a one – electron atom where the electron moves around

the nucleus. In the following List – I contains some quantities for the nth orbit of the

atom and List – II contains options showing how they depend on n



Page

-32


(I) Radius of the nth orbit (P) 2n

(II) Angular momentum of the electron in the nth orbit (Q) 1n

(III) Kinetic energy of the electron in the nth orbit (R) 0n

(IV) Potential energy of the electron in the nth orbit (S) 1n

(T) 2n

(U) 1/2n

Which of the following options has the correct combination considering List – I and

List – II?

1. (III), (S) 2. (IV), (Q) 3. (III), (P) 4. (IV), (U)

Key: 3

Sol: 2nr

Z

2

2

2

2

L n

ZKEn

ZPEn



given in the paragraph List – I includes starting materials and reagents of selected

chemical reactions. List – II gives structures of compounds that may be formed as

intermediate products and/ or final products from the reactions of List – I



Page

-33


(I) (P)

(II) (Q)

(III) (R)

(IV) (S)

(T)

(U)

Which of the following options has correct combination considering List – I and List –

II?

1. (I), (S), (Q), (R) 2. (II), (P), (S), (U)

3. (I), (Q), (T), (U) 4. (II), (P), (S), (T)

Key: 2



Page

-34

Sol:

CN

O

O

i)DIBAL-Hii)dilHCl

CHO

CHO

NaBH4

iii)conc.H2SO4

OH

OH

O

i)O3

ii)Zn,H2O

iv)conc.H2SO4

iii)NaBH4CH2

COOH

O

COOH

OH

COOH

O

O

Cl

COOCH3

i)KCN ii)H3O+,Δ iii)LiAlH4CN

COOCH3

COOH

COOH OH

OH

O

iv)conc.H2SO4

COOMe

COOMe

i)LiAlH4 iI)conc.H2SO4OH

OH

O=

O



given in the paragraph List – I includes starting materials and reagents of selected

chemical reactions. List – II gives structures of compounds that may be formed as

intermediate products and/ or final products from the reactions of List – I



Page

-35


(I) (P)

(II) (Q)

(III) (R)

(IV) (S)

(T)

(U)

Which of the following options has correct combination considering List – I and List –

II?

1. (IV), (Q), (R) 2. (IV), (Q), (U) 3. (III), (S), (R) 4. (III), (T), (U)

Key: 1



Page

-36

Sol:

CN

O

O

i)DIBAL-Hii)dilHCl

CHO

CHO

NaBH4

iii)conc.H2SO4

OH

OH

O

i)O3

ii)Zn,H2O

iv)conc.H2SO4

iii)NaBH4CH2

COOH

O

COOH

OH

COOH

O

O Cl

COOCH3

i)KCN ii)H3O+,Δ iii)LiAlH4CN

COOCH3

COOH

COOH OH

OH

O

iv)conc.H2SO4

COOMe

COOMe

i)LiAlH4 iI)conc.H2SO4OH

OH

O=

O



Page

-37

MATHEMATICS Max Marks: 62 SECTION-1 (Maximum Marks : 32)

This section contains EIGHT (08) questions Each question has FOUR options. ONE MORE THAN ONE of these four option(s) is(are) correct answer(s) For each question, choose the 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. For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct answers, then Choosing ONLY (A), (B), and (D) will get +4 marks ; Choosing ONLY (A) and (B) will get +2 marks ; Choosing ONLY (A) and (D) will get +2 marks ; Choosing ONLY (B) and (D) will get +2 marks ; Choosing ONLY (A) will get +1 marks; Choosing ONLY (B) will get +1 marks; Choosing ONLY (D) will get +1 marks; Choosing no option (i.e. the question is unanswered) will get 0 marks ; and Choosing any other combination of options will get -1 mark. Question ID: 337911204

1. For , 1a a , let

3 3

7/32 2 2

1 2 ......lim 541 1 1....

1 2

n

n

nan an an n

Then the possible value(s) of a is/are

1. – 9 2. – 6 3. 7 4. 8

Key: 1,4

Sol:

1/33

11

7/32 2

1 1

1

lim lim1 1 1

nn

rrn nn n

r r

rr n n

nn ran r a

n

=

11/3 4/3 1

00

1102

0

3 34 4 541 11

1

x dx x l

ldx a x a aa x

1 8 9 9 8a a or 8 9a or



Page

-38


2. Let :f be a function. We say that f has

PROPERTY 1 if

0

0limh

f h fh

exists and is finite, and

PROPERTY 2 if 20

0limh

f h fh

exists and is finite

Then which of the following options is/are correct ?

1. f x x has PROPERTY 1

2. sinf x x has PROPERTY 2

3. f x x x has PROPERTY 2

4. 2/3f x x has PROPERTY 1

Key: 1,4

Sol: If 0 0

0lim lim 0h h

hf x x h

h

If 20

sinh 0sin lim .h

f x x D Eh

If 20

0lim .h

h hf x x x D E

h

If

2/31/32/3

1/30 0

0lim lim 0h h

hf x x hh


3. For non – negative integers n, let

0

2

0

1 2sin sin2 2

1sin2

n

kn

k

k kn nf n

kn

Assuming cos-1x takes values in 0, which of the following options is/are correct?

1. If 1tan cos 6f ,then 2 2 1 0

2. 1sin 7cos 5 0f

3. 342

f

4. 1lim2n

f n



Page

-39

Key: 1,2,3

Sol: 0

0

2 3cos cos2 2

2 21 cos2

n

k

n

k

kn nf n

kn

1sin

2 31 cos cos2 2 2sin

21

sin2 21 cos

2 2sin2

nn nn

n n nn

nn nn

n nn

2 cos

2 cos2 2

nn

n n

.

26 cos 2 1 2 1 08 8

f Tan

1sin 7cos 5 sin 7. 07

f

34 cos

6 2f

lim 1n

f n


4. Let :f be given by 1 2 5f x x x x . Define

0

, 0x

F x f t dt x

Then when of the following options is/are correct?

1. F has a local minimum at x = 1

2. 0F x for all 0,5x

3. F has a local maximum at x = 2

4. F has two local maxima and one local minimum in 0,



Page

-40

Key: 1,2,3

Sol: 1 1 2 5F x f x x x x

At 1x and 5x F x has local minima

At 2x , F x has local maxima and 0 0,5F x x

1 2 5


5. Let

1 2 3

1 0 0 1 0 0 0 1 00 1 0 , 0 0 1 , 1 0 00 0 1 0 1 0 0 0 1

P I P P

4 5 6

0 1 0 0 0 1 0 0 10 0 1 , 1 0 0 , 0 1 01 0 0 0 1 0 1 0 0

P P P

and 6

1

2 1 31 0 23 2 1

Tk k

k

X P P

Where TkP denotes the transpose of the matrix kP . Then which of the following options

is/ are correct?

1. X is a symmetric matrix

2. The sum of diagonal entries of X is 18

3. X – 30I is an invertible matrix

4. If 1 11 1 ,1 1

X

then 30

Key: 1,2,4



Page

-41

Sol: Let 6

1

2 1 31 0 2 ,3 2 1

Tk k

k

A X P AP

Here T Tk k k kP P P P I for all k =1,2,3,4,5,6.

1 &T Tk k kP P P A A

6 6

1 1

TT T Tk k k k

k k

X P AP P AP X

Let 6 6 6

1 1 1

1 2 2 2 30 11 2 2 2 30 30 1 301 2 2 2 30 1

TK K K K

K K KB XB P AP B P AB P AB AB B

30 and 30X I is singular

6

1

18Tk k

k

Tr P AP


6. Let 2sin , 0xf x x

x

Let x1 < x2 < x3 < …….. < xn <……. be all the 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 ?

1. 1 1x y

2. 1 2n nx x

3. 12 ,22nx n n

for every n

4. 1n nx y for every n

Key: 2,3,4

Sol: 2

14

cos 2 sin , 0x x x xf x xx

4

2 cos tan2 0

xx x xcos x

x



Page

-42

y

0 1/23/2 3

1y 1x2y


7. Let x and let

1 1 1 20 2 2 , 0 4 00 0 3 6

x xP Q

x x

and 1R PQP

Then which of the following options is/are correct?

1. det 2

det 0 4 0 85

x xR

x x

, for all x

2. For x = 1, there exists a unit vector i j k for which 000

R

3. There exists a real number x such that PQ = QP

4. For x = 0, if 1 1

6R a ab b

, then a + b = 5

Key: 1,4

Sol: 1 2 2

24 12 4 10 8 0 4 0 8

5

x xR PQP Q x x

x x



Page

-43

If 1 44 0 0x R

For 1 1 1 2 0 0 6 3 0

10 0 2 2 0 4 0 0 3 26

0 0 3 0 0 6 0 0 2x R

12 6 4

10 24 8

60 0 36

24 134

6 0 23

0 0 0

R I

2 44 0, 2 03 3b ba a

3, 2 5b a a b

If PQ = QP no value of x .

Question ID: 337911206 8. Three lines

1 : ,L r i

2 : ,L r k j

3 : ,L r i j vk v

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?

1. k 2. 12

k j 3. k j 4. 12

k j

Key: 2,4 Sol: , 0,0 , 0, ,1 , 1,1,P Q R

If P,Q,R are collinear 11 1

1 11 1,01



Page

-44

SECTION-2 (Maximum Marks : 18) 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 on-screen virtual numeric


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


1. Suppose

2

0 0

0 0

det 03

n nn

kk k

n nn n k

k kk k

k C k

C k C

Holds for some positive integer n. The 0 1

nnk

k

Ck equals____

Key: 6.20

Sol: 2 1

1

11 2 2 02

2 4

n n

n n

n nn n n

n

2 1 2 3 21 2 1 2 2 0n n nn n n n

4 1 1 2 0n n n n

2 3 4 0 4 0,n n n n n N

4 54

0

2 1 31 6.201 5 5k

k

Ck


2. The value of

10

1

0

11 7 7sec sec sec4 12 2 12 2k

kk

In the interval 3,4 4

equals ___

Key: 0



Page

-45

Sol: 10

0

7 7sec sec12 2 12 2 2k

k kS

10

0

72cos6k

ec k

10

1

0

2 1 cos6

k

k

ec

4 1 4

1 15sec sec 1 04


3. The value of the integral

/2

50

3 cos

cos sind

equals ____

Key: 0.5

Sol:

/2 /2

5 50 0

3 cos 3 sin

cos sin sin cosI d d

/2 /2 2

4 40 0

3 3sec2

sin cos 1I d d

Tan

Put, 2 2sec 2Tan t d tdt

4 3 4

0 0

6 1 16

1 1 1t dt dtt t t

3 2

0

1 1 1 1 13 3 0 1 0 1 0.503 2 3 2 2t t

I l


4. 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 ____



Page

-46

Key: 422

Sol: If n(B) = 1, n(A) = 6 no. of ordered pairs (A, B) = 6 If n(B) = 2, n(A) = 3 no. of ordered pairs (A, B) = 180 If n(B) = 2, n(A) = 6 no. of ordered pairs (A, B) = 15 If n(B) = 3, n(A) = 4 no. of ordered pairs (A, B) = 180 If n(B) = 3, n(A) = 6 no. of ordered pairs (A, B) = 20 If n(B) = 4, n(A) = 6 no. of ordered pairs (A, B) = 15 If n(B) = 5, n(A) = 6 no. of ordered pairs (A, B) = 6 Total number of ordered pairs (A, B) = 422.

Question ID: 337911208 5. 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 number of ways of distributing the hats such that the persons seated in adjacent seats get different coloured hats is ___

Key: 30 Sol: Total number of ways = 2 3 5 30 Question ID: 337911212

6. Let 2a i j k and 2b i j k

be two vectors. Consider a vector

, ,c a b

. If the projection of c

on the vector a b

is 3 2 , then the

minimum value of .c a b c

equals_____

Key: 18 Sol: 3 3a b i j

.3 2 2

c a b

a b

2

18 . 18c c c a b

SECTION-3 (Maximum Marks: 12) This section contains TWO (02) List-Match sets. Each List-Match set has TWO (02) Multiple Choice Questions. Each List-Match set has two lists : List-I and List-II. List-I has Four entries (I), (II), (III) and (IV) and List-II has Six entries (P), (Q), (R), (S), (T) and (U). FOUR options are given in each Multiple Choice Questions based on List-I and List-II 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



Page

-47


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

in the paragraph. Let sin cosf x x and cos 2 sing x x be two functions

defined for x > 0. Define the following sets whose elements are written in the increasing

order :

: 0 ,X x f x 1: 0Y x f x

: 0 ,Z x g x 1: 0W x g x

List – I contains the sets X, Y, Z and W. List – II contains some information regarding

these sets.


(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

Which of the following is the only CORRECT combination?

1. (IV), (P), (R), (S) 2. (III), (P), (Q), (U)

3. (III), (R), (U) 4. (IV), (Q), (T)

Key: 1

Sol: For x , 0 sin cos 0 cosf x x x n

cos 0 ( )1( ) 1x or or

,2

nx n N



Page

-48

For Y, 1 0 cos sin 0f x cos x x

sin 0 ( ) cos cos 0x or x

, ( ) cos 2 1 ,2

x n n N or x n n N

1 1cos ,2 2

x

22 ,2 , , 03 3

x n n n I x

For Z, 0 cos 2 sin 0 2 sin 2 1 ,2

g x x n n N

2 1 1 1 3 3sin , , ,4 4 4 4 4

nx

For W, 1 0 2 sin 2 sin cos 0g x x

sin 2 sin 0 cos 0x or x

2 1 , 2 sin2

x n n N or x n

1sin 1, ,02 2nx

2 1 , ,2

x n n n N

, 1 , 16 6

n nn n n



given in the paragraph

Let sin cosf x x and cos 2 sing x x be two functions defined for x > 0.

Define the following sets whose elements are written in the increasing order :

: 0 ,X x f x 1: 0Y x f x

: 0 ,Z x g x 1: 0W x g x

List – I contains the sets X, Y, Z and W. List – II contains some information regarding

these sets.



Page

-49


(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. (I), (P), (R) 2. (I), (Q), (U) 3. (II), (R), (S) 4. (II), (Q), (T)

Key: 4

Sol: For x , 0 sin cos 0 cosf x x x n

cos 0 ( )1( ) 1x or or

,2

nx n N

For Y, 1 0 cos sin 0f x cos x x

sin 0 ( ) cos cos 0x or x

, ( ) cos 2 1 ,2

x n n N or x n n N

1 1cos ,2 2

x

22 ,2 , , 03 3

x n n n I x

For Z, 0 cos 2 sin 0 2 sin 2 1 ,2

g x x n n N

2 1 1 1 3 3sin , , ,4 4 4 4 4

nx

For W, 1 0 2 sin 2 sin cos 0g x x



Page

-50

sin 2 sin 0 cos 0x or x 2 1 , 2 sin2

x n n N or x n

1sin 1, ,02 2nx 2 1 , ,

2x n n n N

, 1 , 16 6

n nn n n

Question ID: 337911215 3. Answer the following by appropriately matching the lists based on the information

given in the paragraph. Let the circles 2 2

1 : 9C x y and 2 22 : 3 4 16C x y , intersect at the points

X and Y. Suppose that another circle 2 2 23 :C x h y k r satisfies the

following conditions :

(i) Centre of 3C is collinear with the centres of 1C and 2C

(ii) 1C and 2C both lie inside 3C , and

(iii) 3C touches 1C at M and 2C 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 2 8x y .

There are some expressions given in the List – I whose values are given in List – II

below


(I) 2h k (P) 6

(II) Length of ZWLength of XY

(Q) 6

(III) Area of triangle MZNArea of triangle ZMW

(R) 54

(IV) (S) 215

(T) 2 6

(U) 103


1. (I), (U) 2. (II), (T) 3. (II), (Q) 4. (I), (S)



Page

-51

Key: 3

Sol:

WN

M ZO(0,0)

Y

X

(3,4)

2 21

2 2 2 22 2

2 2 23

9 0,0

3 4 16 6 8 9 0 3, 4

C x y O

C x y x y x y C

C x h y k r

Let the centre of C3 be C.

29 12: 3: 2 ,5 5

OC CC C

2 6, 2 3 5 4 12 6h k r r

Eq. of common chord of C1 and C2 is 3x+4y-9=0. Let P be the point of intersection of

OC1 and common chord.

6 12 6 24, ,5 5 5

81 242 925 5

6

54

CP ZP WP MP

XY

ZWXYMZNZMW

Equation of the common chord of C1 and C3 is 103 4 15 03

x y .



Page

-52

Question ID: 337911216 4. Answer the following by appropriately matching the lists based on the information

given in the paragraph

Let the circles 2 21 : 9C x y and 2 2

2 : 3 4 16C x y , intersect at the points

X and Y. Suppose that another circle 2 2 23 :C x h y k r satisfies the

following conditions :

(i) Centre of 3C is collinear with the centres of 1C and 2C

(ii) 1C and 2C both lie inside 3C , and (iii) 3C touches 1C at M and 2C 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 2 8x y .

There are some expressions given in the List – I whose values are given in List – II

below


(I) 2h k (P) 6

(II) Length of ZWLength of XY

(Q) 6

(III) Area of triangle MZNArea of triangle ZMW

(R) 54

(IV) (S) 215

(T) 2 6

(U) 103

Which of the following is the only INCORRECT combination?

1. (III), (R) 2. (IV), (U) 3. (IV), (S) 4. (I), (P)

Key: 3



Page

-53

Sol:

WN

M ZO(0,0)

Y

X

(3,4)

2 21

2 2 2 22 2

2 2 23

9 0,0

3 4 16 6 8 9 0 3, 4

C x y O

C x y x y x y C

C x h y k r

Let the centre of C3 be C.

29 12: 3: 2 ,5 5

OC CC C

2 6, 2 3 5 4 12 6h k r r

Eq. of common chord of C1 and C2 is 3x+4y-9=0. Let P be the point of intersection of

OC1 and common chord.

6 12 6 24, ,5 5 5

81 242 925 5

6

54

CP ZP WP MP

XY

ZWXYMZNZMW

Equation of the common chord of C1 and C3 is 103 4 15 03

x y .

sri chaitanya · 2020. 4. 10. · sri chaitanya iit academy # 304, kasetty heights, ayyappa...

Documents