2017-jee-advanced - neet ug coaching-jee main-iit jee...

2017-Jee-Advanced Question Paper-2_Key & Solutions

Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081

www. srichaitanya.net, [email protected]

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Paper-2 READ THE INSTRUCTIONS CAREFULLY

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PART-1:PHYSICS

SECTION -1 (Maximum Marks: 21)

This section Contains SEVEN questions.

Each question has FOUR options [A], [B], [C], and [D]. ONLY ONE of these four options is correct

For each question, darken the bubble corresponding to the correct option in the ORS.

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

Full Marks : +3 If only the bubble corresponding to the correct option is darkened.

Zero Marks : 0 If none of the bubbles is darkened

Negative Marks : -1 In all other cases.

*********************************************************************************************************

Q1. Consider regular polygons with number of sides 3,4,5.....n as shown in the figure. The center of mass of all

the polygons is at height h from the ground. They roll on a horizontal surface about the leading vertex without

slipping and sliding as depicted. The maximum increase in height of the locus of the center of mass for polygon is

. Then depends on n and h as

h h h

A) 2sinh

n

B)

2sinh

n

C)2tan

2h

n

D)

11

cos

h

n

Key: D

Sol:

h

d

d h

cos

hh

For regular polygon / n

( n is number of sides)

11

cos / cos /

hh h

n n

.




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Q2. Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is

such that the instantaneous density remains uniform throughout the volume. The rate of fractional change in

density1 d

dt

is constant. The velocity v of any point on the surface of the expanding sphere is proportional to

A) R B)1

R C)

3R D)2/3R

Key: A

Sol: 34

3R M

3 3 11

4

MR

2

2

3 13 2

4

dR M dR

dt dt

2 1 1gives constant

1 3

dR d

dt R dt

dR

vdt

v R

Q3. A Photoelectric material having work-function 0 is illuminated with light of wavelength0

hc

. The

fastest photoelectron has a de Broglie wavelength d . A change in wavelength of the incident light by results

in a change d in d . Then the ratio /d is proportional to

A)2 2/d B) /d C)

3 /d D)3 2/d

Key: D

Sol:

2

max

1

2 d

hc hK

m

2

2 3

2

2d

d

hc h

m

3

2

d dmcx

h

3

2

d d




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Q4. Three vectors ,P Q and R are shown in the figure. Let S be any point on the vector R . The distance between the

points P and S is b R . The general relation among vectors ,P Q and S is

O

Y

P SQ

b R

R Q P

Q

X

S

A) 21S b P bQ B) 1S b P bQ

C) 1S b P bQ D) 21S b P b Q

Key: C

Sol: 1S b Q P Q

1 1 1S Q b b P

1 b P bQ

Q5. A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the

sound of impact with the bottom of the well. The error in his measurement of time is 0.01T seconds and he

measures the depth of the well to be 20L meters. Take the acceleration due to gravity 210g ms and the

velocity of sound is 1300ms . Then the fractional error in the measurement, /L L , is closest to

A) 1% B) 5% C) 3% D) 0.2%

Key: A

Sol: 2L L

Tg V

2

02

dL dLdT

g VL

1

2

dLdT

g L

21

2L gT

2 2 20

210

LT s

g

1

2 22

dT dL g dL

T L LgL

2 2 0.01

0.012

dL dT

L T

1%




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6Q. A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the Sun and the

Earth. The sun is 53 10 times heavier than the Earth and is at a distance

42.5 10 times larger than the radius of

the Earth. The escape velocity from Earth’s gravitational field is 111.2ev kms . The maximum initial velocity

sv required for the rocket to be able to leave the sun-Earth system is closest to

(ignore the rotation and revolution of the Earth and the presence of any other planet)

A) 172Sv kms B)

122Sv kms

C)142Sv kms D)

162Sv kms

Key: C

Sol:

x

RsRE

2

02

SE

E E

GMMmV GMM

R x R

2

0 02

SE

E E

V GMGM

R x R

2

0

2

VG

0

22 SE

E E

GMGMV

R x R

52

4

2 3 10

2.5 10

Ee

E

GMV

R

22 2

0

3 10 2

5eV V

13 10eV

11.2 3.6

1

0 40.38V kms




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7Q. A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure. The

distance between the diametrically opposite vertices of the star is 4a . The magnitude of the magnetic field at the

center of the loop is

4a

I

A) 0 6 3 14

I

a

B) 0 6 3 14

I

a

C) 0 3 3 14

I

a

D) 0 3 2 34

I

a

Key: A

Sol:

a

d

2

1

a

012 sin sin

4

IB

d

60 30

0 3 1

4 2 2

I

d

012

3 1

4 2

IB

d

d a

0 1212B B

0 3 112

4 2d

0 6 3 14

I

a




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This Section Contains SEVEN questions.

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four options is(are)

Correct.

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


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

Partial Marks : +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is

darkened.

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


For example, if (A), (C)and (D) are all the correct options for a question, darkening all these three will get +4 marks;

darkening only (A) and (D) will get +2 marks; and darkening (A) and (B) will get -2 marks; as a wrong option is also

darkened.

***********************************************************************************************************

Q8. A Wheel of radius R and mass M is placed at the bottom of fixed step of a fixed step of height R as shown in the

figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without

slipping. Consider the torque about an axis normal to the plane of the paper passing through the pointQ . Which

of the following options is/are correct?

X

P

S

R

Q

A) If the force is applied normal to the circumference at point P then is zero

B) If the force is applied tangentially at point S then 0 but the wheel never climbs the step

C) If the force is applied at point P tangentially then decreases continuously as the wheel climbs

D) If the force is applied normal to the circumference at point X then is constant

Key: AD

Sol:

P

F

S

(A) net torque, when F acts at P is zero, as it intersects axis of rotation.

(B) The torque increases of the wheel can climb

Q9. Two coherent monochromatic point 1S and 2S of wavelength 600 nm are placed symmetrically on either side

of the center of the circle as shown. The sources are separate by a distance 1.8d mm. This arrangement

produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The

angular separation between two consecutive bright spots is . Which of the following options is/are correct?




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d1S

2S

2P

1P

A) The angular separation between two consecutive bright spots decreases as we move from 1p to 2p along the

first quadrant

B) A dark spot will be formed at the point 2p

C) The total number of fringes produced between 1p and 2p in the first quadrant is close to 3000

D) At 2p the order of the fringe will be maximum

Key: CD

Sol: At 2p the order of the fringe will be maximum

1S2S 2P

1P

x

0

3000n

1 maxP

2 maxP

cosn

d

sin

nd

d

increases, sin increases

d decreases

cosn

d

cosx

d

9

3

600 10cos

1.8 10n

6600 10cos

1.8n

36

1018

n

3cos 10

3

n




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Q10. A point charge Q is placed just outside an imaginary hemispherical surface of radius R as shown in the figure.

Which of the following statements is/are correct?

Q

R

A) The electric flux passing through the curved surface of the hemisphere is

0

11

2

Q

z

B) The component of the electric field normal to the flat surface is constant over the surface

C) Total flux through the curved and the flat surface is

0

Q

D) The circumference of the flat surface is an equipotential

Key: AD

Sol: A) 0

2 1 cos 454

q

0

11

2 2

q

45

B) as &x r are

Variable: E

E is not constant Over the surface

D) 2

kQ kQV

r R

R

r

x

E

Q

sinE E

2

kQ n

r r

y

KQx




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Q11. A rigid uniform bar AB of length L slipping from its vertical position on a frictionless floor (as shown in the

figure). At some instant of time, the angle made by the bar with the vertical is . Which of the following

statements about its motion is/are correct?

L

A

B

O

A) Instantaneous torque about the point in contact with the floor is proportional to sin

B) The trajectory of the point A is a parabola

C) The midpoint of the bar will fall vertically downward

D) When the bar makes an angle with the vertical, the displacement of its midpoint from the initial position is

proportional to 1 cos .

Key: ACD

Sol:

mg

A

CN

B

sin2

B mg A C

cos2 2

, ,

1 cos2

cX

D A C D

.

Q12. A source of constant voltage V is connected to a resistance R and two ideal inductors 1L and 2L through a switch

S as shown. There is no mutual inductance between the two inductors. The switch S is initially open. At t=0, the

switch is closed and current begins to flow. Which of the following options is/are correct?

V

R

1L 2L

S




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A) After a long time, the current through 1L will be 2

1 2

V L

R L L

B) After a long time, the current through 2L will be 1

1 2

V L

R L L

C) The ratio of the currents through 1L and 2L is fixed at all times 0t

D) At t=0, the current through the resistance R is V

R

Key: ABC

Sol:

1L

R

2L

i1i

2i

V

Left 1 2

1 2

L L

L L

t , V

iR

1 1 2 2 1 2;Li L i i i i 2 1L i i

0, 0t v 2 21

1 2 1 2

L V Li i

L L R L L

12

1 2

V Li

R L L

Q13. A uniform magnetic field B exists in the region between x=0 and 3

2

Rx (region 2 in the figure) pointing

normally into the plane of the paper. A particle with charge +Q and momentum p directed along x-axis enters

region 2 from region 1 at point 1P y R . Which of the following option(s) is/are correct?

O

Region 1 Region 2 Region 3

O2P x

y

1PO

+Q

y R

B

3 / 2R




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A) When the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the change

in its linear momentum between point 1P and the farthest point from y-axis is 2

p

B) For 8

,13

pB

QR the particle will enter region 3 through the point 2P on x-axis

C) For 2

,3

pB

QR the particle will re-enter region 1

D) For a fixed B, particle of same charge Q and same velocity , the distance between the point 1P and the point

of re-entry into region 1 is inversely proportional to the mass of the particle

Key: BC

Sol:

O

P

3

2

R2P

P

r

3

, 22

R mvr P P

qB

3

rsin2

R

P

rqB

3

sin2

P R

qB

2

sin3

PB

qR

290

3

PB

qR

d=2r 2 2p mv

qB qB

Q14. The instantaneous voltages at three terminals marked X,Y and Z are given by

0 sin ,XV V t

0

2sin

3YV V t and

0

4sin

3ZV V t




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An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is

connected between points X and Y and then between Y and Z. The reading(s) of the voltmeter will be

A) 0

1

2YZ

rmsV V

B) 0

3

2

rms

XYV V

C) Independent of the choice of the two terminals

D) 0

rms

XYV V

Key: BC

Sol: 0

2sin sin

3XY OV V t V t

0

2sin sin

3OV t V t

2

3 3

1

0 02 cos6

V V

03V

03 sinXYV V t

032

XY YZRms Rms

VV V


This section Contains TWO paragraphs.

Based on each table, there are TWO questions





Zero Marks : 0 If all other cases

***********************************************************************************************************

PARAGRAPH: 1

Consider a simple RC circuit as shown in Figure 1.

Process 1: In the circuit the switch S is closed at t=0 and the capacitor is fully charged to voltage 0V (i.e.,

charging continues for time T>>RC). In the process some dissipation DE occurs across the resistance R. The

amount of energy finally stored in the fully charged capacitor is CE .

Process 2: In a different process the voltage is first set to 0

3

V and maintained for a charging time T>>RC. Then

the voltage is raised to 02

3

V without discharging the capacitor and again maintained for a time T>>RC. The




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process is repeated one more time by raising the voltage to 0V and the capacitor is charged to the same final

voltage 0V as in Process 1.

These two processes are depicted in Figure 2.

V

R

S

C

Figure 1

0V

02

3

V

0

3

V

Process2

Process1

r 2T

T>>RC

t

Figure 2

Q15. In Process 1, the energy stored in the capacitor CE and heat dissipated across resistance DE are related by:

A) ln 2C DE E B) C DE E

C) 2C DE E D) 1

2C DE E

Key: B

Sol: 2

0

1

2cE CV

0W CV D CE W E 2

0

1

2CCV E

Q16. In Process 2, total energy dissipated across the resistance DE is:

A) 2

0

1 1

3 2DE CV

B)

2

0

13

2DE CV

C) 2

03DE CV D) 2

0

1

2DE CV

Key: A

Sol: Total energy suppliers in three stage s(Process 2

0

2

3CV

Energy storages in cap 2

0

1

2CV

Therefore, loss across R 2

0

1

6CV

2

0

1 1

3 2CV

Hence, ‘A’

PARAGRAPH: 2

One twirls a circular ring (of mass M and radius R) near the tip of one’s finger as shown in Figure 1. In the

process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone,

shown by the dotted line. The radius of the path traced out the point where the ring and the finger is in contact is

r. The finger rotates with an angular velocity 0 . The rotating ring rolls without slipping on the outside of a




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smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of

friction between the ring and the finger is and the acceleration due to gravity is g.

R

R

Figure 1 Figure 2

r

Q17. The total kinetic energy of the ring is

A) 22

0M R r B) 22

0

1

2M R r

C) 2 2

0M R D) 22

0

3

2M R r

Key: A

Sol: CMV of Ring=(R-r) 0w

For no supply

1

0 0Rw R r w rw

1

0 0Rw Rw

KE of ring 2 2 2 2

0 0

1 1

2 2M R r w mR w

Which is not matching with any answer given if r R

But we take 0 0Rw R r w

then2

KE 2

2 2

0M R r w

then A is nearest possible option

Q18. The minimum value of 0 below which the ring will drop down is

A) 2

g

R r B)

3

2

g

R r C)

g

R r D)

2g

R r

Key: C

Sol: If height of the cone h>>r

Then N mg

m 2

0R r m g

0

g

R r

PART-2:CHEMISTRY




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Q19. The order of basicity among the following compounds is

NH

3H C2NH

I II

N NH

III

NHN

NH

IV

2NH

2H N

A) IV > I > II > III B) IV > II> III > I

C) I > IV > III > II D) II > I > IV > III

Key: A

Sol: IV > I > II > III

HN N

2Sp least basic

2 2H N C NH strongest base

NH

Q20. The major product of the following reaction is

2) , ,0

) .

i NaNO HCl C

ii aq NaOH

OH

2NH

A)

O Na

2N Cl B) N N

OH

C)

OH

Cl D)

N N OH

Key: B




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Sol:

OH

.2NaNO HCl

oO C

2N

.aq NaOH

2NH

OH

OH

N N

Coupling reaction

N N

O

Q21. Which of the following combination will produce 2H gas?

A) Au metal and NaCN(aq) in the presence of air

B) Zn metal and NaOH(aq)

C) Fe metal and conc. 3HNO

D) Cu metal and conc. 3HNO

Key: B

Sol: 2 2 22 [ ]Zn NaOH Na ZnO H

Q22. The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are

1graphite 0f G C kJ mol

1diamonda 2.9f G C kJ mol

The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature.

The conversion of graphite [C(graphite) to diamond [C (diamond)] reduces its volume by 6 3 12 10 m mol . If

C(graphite) is converted to C(diamond) isothermally at 298T K , the pressure at which C(graphite) is in

equilibrium with C(diamond), is [Useful information: 2 21 1J kg m s ;

1 21 1Pa kg m s ; 51 10bar Pa ]

A) 14501 bar B) 29001 bar C) 1450 bar D) 58001 bar

Key: A

Sol: g DC C

( ) ( )

O O O

Rxn D gG G G

=2.9 – 0 = 2.9 KJ/mol

0G P V

62.9 / 2 10kj m P

1450 /P bar




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Q23. The order of the oxidation state of the phosphorus atom in 3 2H PO , 3 4H PO , 3 3H PO and 4 2 6H PO is

A) 3 4 4 2 6 3 3 3 2H PO H PO H PO H PO B) 3 2 3 3 4 2 6 3 4H PO H PO H PO H PO

C) 3 4 3 2 3 3 4 2 6H PO H PO H PO H PO D) 3 3 3 2 3 4 4 2 6H PO H PO H PO H PO

Key: A

Sol:

3 4 4 2 6 3 3 3 2H PO H P O H PO H PO

+5 +4 +3 +1

Q24. For the following cell,

4 4| || |Zn s ZnSO aq CuSO aq Cu s

When the concentration of 2Zn

is 10 times the concentration of 2Cu

, the expression for 1G in J mol is

[F is Faraday constant; R is gas constant; T is temperature; 0 1.1E cell V ]

A) 2.303 2.2RT F B) 1.1F

C) -2.2F D) 2.303 1.1RT F

Key: A

Sol: 0 2.303 logG G RT Q

0 0

2

2

0

2 1.1 2.2

[ ]2.303 log 2.303 log

[ ]

2.303 log10 2.303

2.303 2.2

G nFE

F F

ZnRT Q RT

Cu

RT RT

G RT F

Q.25. Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing

point of the solution. Use the freezing point depression constant of water as 12K kg mol . The figures shown

below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is

146 g mol ]

Among the following, the option representing change in the freezing point is

A)

water

water + Ethanol

Ice

270 273/T K

1

V.P

./bar

B)

water

water + EthanolIce

270 273/T K

1

V.P

./bar




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C)

water

water + EthanolIce

270 273/T K

1

V.P

./bar

D)

water

water + Ethanol

Ice

270 273/T K

1

V.P

./bar

Key: C

Sol:

.

34.5 1000273 2

46 500

273 3 270

f f

f

f

T K m

T

T K




Correct.





darkened.





darkened.

******** ***************************************************************************************************

Q26. In a bimolecular reaction, the steric factor P was experimentally determined to be 4.5. the correct option(s) among

the following is (are)

A) The activation energy of the reaction is unaffected by the value of the steric factor

B) Since P = 4.5, the reaction will not proceed unless an effective catalyst is used

C) The value of frequency factor predicted by Arrhenius equation is higher than that determined experimentally

D) Experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation

Key: AD

Sol: aE is independent of steric factor

/. Ea RTK Ae

A Z P

Steric factor

Exp. Z value is high

If “P” high reaction is very fast

So R n need not required a catalyst




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Q27. For a reaction taking place in a container in equilibrium with its surroundings , the effect of temperature on

its equilibrium constant K in terms of change in entropy is described by

A) with increase in temperature , the value of K for exothermic reaction decreases because favourable change

in entropy of the surroundings decreases.

B) with increase in temperature , the value of K for exothermic reaction decrease because the entropy change

of the system is positive

C) with increase in temperature , the value of K for endothermic reaction increases because the entropy

change of the system is negative

D) with increase in temperature, the value of K for endothermic reaction increases because unfavourable

change in entropy of the surroundings decreases

Key: AD

Sol: For Exothermic R n

H ve

K decreases with raise in Temp

sys

q HS H ve

T T

So sys

S ve

sorr

S ve

Q28. The correct statement (s) about surface properties is (are)

A) Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system

B) Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is dispersion medium

C) The critical temperatures of ethane and nitrogen are 563K and 126K ,respectively. The adsorption of ethane

will be more than that of nitrogen on same amount of activated charcoal at a given temperature

D) Brownian motion of colloidal particles does not depend on the size of the particles but depends on viscosity of

the solution

Key: AC

Sol: Adsorption degree of freedom

Decreases so S ve

tanG ve Spon eous

S ve

H ve

G H T S

ve T ve

At low temp. favours

Cloud is not emulsion

c

xT

m as cT (critical temp )

High liquification all reactions x

m also )




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Q29. Among the following , the correct statement(s) is (are)

A) 3AlCl has the three-centre two- electron bonds in its dimeric structure

B) 3BH has the three-centre two- electron bonds in its dimeric structure

C) 3 3Al CH has the three- centre two-electron bonds in its dimeric structure

D) The lewis acidity of 3BCl is greater than that of 3AlCl

Key: BCD

Sol: option ‘D’ is not given clearly

3 3BCl AlCl in case of hard lewis base

3 3BCl AlCl in case of soft lewis base. (refer. Inorganic chemistry atkins)

Q30. For the following compounds, the correct statement(s) with respect to nucleophilic substitution reactions

is(are)

Br

I

Br

II

C Br

3CH

3CH

3H C

III IV

Br

3CH

A) I and III follow 1NS mechanism

B) Compound IV undergoes inversion of configuration

C) The order of reactivity for I,III and IV is: IV>I>III

D) I and II follow 2NS mechanism

Key: ABCD

Sol: I)

Br

1NS

II)

Br

[Very slowly reactive to both the pahs]

III)

Br1NS only

IV)

Br

1NS

(A) I & III follow 1NS mechanism correct

(B) compound IV undergoes inversion of configuration

(C) the order of reactivity for I, III & IV is IV > I > III correct

(D) I & II follow 2NS Mechanism correct




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Q31. The option(s) with only amphoteric oxides is(are)

A) 2 3 2, , ,ZnO Al O PbO PbO B) 2 3, , ,Cr O CrO SnO PbO

C) 2 3, 2, ,Cr O BeO SnO SnO D) 2 3, 2, ,NO B O PbO SnO

Key: AC

Sol: 2 3 2 3 2, , , ,Cr O BeO B O PbO SnO are Amphoteric, 2PbO is not amphoteric

Q32. Compounds P and R upon ozonolysis produce Q and S, respectively. The molecular formula of Q and S is

8 8C H O . Q undergoes Cannizzaro reaction but not haloform reaction, whereas S undergoes haloform reaction but

not Cannizzaro reaction.

P3 2 2) /i O CH Cl

2) /ii Zn H OQ

8 8C H O

(i)

(ii)

2) /ii Zn H O 8 8C H O

R S3 2 2) /i O CH Cl

The option(s) with suitable combination of P and R, respectively, is(are)

A) 3CH

3CH

3H C

3CH

3CH

3CH

and

B)

and3H C

3CH

C)

and

3H C

3CH

3CH

3H C

3CH

D)

3H C and3H C

3CH

Key: BC




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Sol:

3H C

(P)

3 2 2

2

( ) /

( ) /

i O CH Cl

ii Zn H O

CHO

3CH

Undergoes cannizzaro's reaction , but not haloform reaction

(Q)

8 8( )C H O

Gives haloform reaction but not cannizzaro's reaction

( ) /3 2 2

( ) /2

i O CH Cl

ii Zn H O

(R)(S)

8 8( )C H O

O

Gives cannizzaro's reaction, but not haoform reaction

( ) /3 2 2

( ) /2

i O CH Cl

ii Zn H O

8 8( )C H O

O

3CH

3CH

3CH

3CH CHO

(Q)

Gives haoform reaction, but not cannizzaro's reaction,

( ) /3 2 2

( ) /2

i O CH Cl

ii Zn H O

O

+

O




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******** ***************************************************************************************************

PARAGRAPH 1

Upon heating 3KClO in the presence of catalytic amount of 2MnO , a gas W is formed. Excess amount of W

reacts with white phosphorus to give X. The reaction of X with pure 3HNO gives Y and Z

Q33. W and X are, respectively

A) 3O and 4 6PO B) 2O and 4 6PO

C) 2O and 4 10PO D) 3O and 4 10PO

Key: C

Sol:

3 2KClO KCl O4 10 2 5P O N O

2MnO3HNO4P

3 4H PO3HPO

(X)

Q34. Y and Z are, respectively

A) 2 5 3N O and HPO B) 2 3 3 4N O and H PO

C) 2 4 3N O and HPO D) 2 4 3 3N O and H PO

Key: A

Sol:

3 2KClO KCl O4 10 2 5P O N O

2MnO3HNO4P

3 4H PO3HPO

(X)

PARAGRAPH 2

The reaction of compound P with 3 ( )CH MgBr excess in 2 5 2C H O followed by addition of 2H O gives Q. The

compound Q on treatment with 2 4H SO at 00 C gives R. The reaction of R with 3CH COCl in the presence of

anhydrous 3AlCl in 2 2CH Cl followed by treatment with 2H O produces compound S. [Et in compound P is ethyl

group]

3 3H C C

2CO Et

Q R S

P




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Q35. The reactions, Q to R and R to S, are

A) Dehydration and Friedel-Crafts acylation

B) Friedel-Crafts alkylation, dehydration and Friedel-Crafts acylation

C) Friedel-Crafts alkylation and Friedel-Crafts acylation

D) Aromatic sulfonation and Friedel-Crafts acylation

Key: C

Sol:

2CO Et3

2

1. / .

2.

CH MgBr D E

excessH O

3HO C CH

3CH

2 4 ,0H SO C2OH

2H O

H

Freidel crafts alkylation

3 3

2

1. / .2.

Freidal-crafts acylation

CH COCl Anhy AlClH O

3CH R

P Q

S

Freidel-craft's alkylationQ R

Freidel-craft's acylationR S

Q36. The Product S is

A)

3 3H C C

3CHO

3COCH

3HO S

B)

3H COC

3 3H C C

3CH3H C

C)

3 3H C C

3H C3CH

3COCH D)

3 3H C C 3CH

3COCH

Key: C

Sol:




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2CO Et3

2

1. / .

2.

CH MgBr D E

excessH O

3HO C CH

3CH

2 4 ,0H SO C2OH

2H O

H


3 3

2

1. / .2.



3CH R

P Q

S

2CO Et3

2

1. / .

2.

CH MgBr D E

excessH O

3HO C CH

3CH

2 4 ,0H SO C2OH

2H O

H


3 3

2

1. / .2.



3CH R

P Q

S

Freidel-craft's alkylationQ R

Freidel-craft's acylationR S




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PART-3:MATHS









******************************************************************************************************

Q37. Let 1,2,3,...,9S For 1,2,...,5,k let kN be the number of subsets of S , each containing five elements

out of which exactly k are odd. Then 1 2 3 4 5N N N N N

A) 125 B) 210 C) 252 D) 126

Key: D

Sol: 5 4 5 4

1 4 2 3C C C C

5 4 5 5 5 2 4 4 4 2 4 3 4 4

0 1 2 0 1 2 3 41 1 ...x x C C x C x C C x C x C x C x

Coeff. Of 5x on LHS

9

4

9 8 7 6126

4 3 2C

Q38. The equation of the plane passing through the point 1,1,1 and perpendicular to the planes 2 2 5x y z

and 3 6 2 7x y z , is

A) 14 2 15 3x y z B) 14 2 15 27x y z

C) 14 2 15 1x y z D) 14 2 15 31x y z

Key: D

Sol: 1 1 1 0a x b y c z

2 2 0a b c

3 6 2 0a b c

14 2 15

a b c

: : 14 : 2 :15a b c

14 2 15 14 2 15 31x y z

Q39. Let O be the origin and Let PQR be an arbitrary triangles. The point S is such that

. . . . . .OP OQ OR OS OR OP OQOS OQOR OP OS Then the Triangle is PQR has S as it’s

A) in centre B) Circumcentre C) Orthocenter D) Centroid

Key: C

Sol: . .OP OQ OR OS OR OQ




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. . 0

. 0

. 0

OP RQ OS QR

RQ OP OS

SP RQ

SP RQ

Similarly SQ QR

& SR PQ

Orthocenter

Q40. How many 3 3 matrices M with entries from 0,1,2 are there, for which the sum of the diagonal entries of

TM M is 5?

A) 162 B) 135 C) 126 D) 198

Key: D

Sol: 2 2 2 2 2 2 2 2 2

11 21 31 12 22 32 13 23 33 5a a a a a a a a a

No of ways 9 9 8

4 1 1 198C C C

Q41. Three randomly chosen nonnegative integers ,x y and z are found to satisfy the equation 10x y z then

the probability that z is even, is

A) 6

11 B)

36

55 C)

1

2 D)

1

2

Key: A

Sol: Total Solutions = 10 3 1 12

3 1 2 66C C

10 2 1 11

2 1 10, 10z x y C C

=11

8 2 1 9

2 1 12, 8 9z x y C C

10, 0 1z x y

Sum = 1+3+5+7+9+11=36

36 6

66 11P

Q42. If :f R R is a twice differentiable function such that "( ) 0f x for all x R and 1

( )2

f = 1

2, 1 1f

then

A) 1

' 1 12

f B) 1

0 ' 12

f C) ' 1 0f D) ' 1 1f

Key: D

Sol: ' 1 1f




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1

2

1

1

21

C

As per LMVT ' 1f c for some

1,1

2c

And as " 0 'f x f x is increases

So ' 1f x for x C

As 1 , ' 1 ' 1c f f c

Q43. If y y x satisfies the differential equation 1

8 9 4 9 , 0x x dy x dx x

and

0 7y then 256y =

A) 80 B) 9 C) 16 D) 3

Key: D

Sol: 2 ,x t

dy

dx= .

dy dt

dt dx

2dx

tdt

1.

2

dy dy

dx t dt

1 1

.2 92 4 9

dy

dt tt

4 9y t

4 9y x c

0 7 0

256 4 9 16

4 5 3

y c

y




Pa

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Correct.





darkened.





darkened.

************************************************************************************************************

Q44. If the line x divides the area of region 2 3, : ,0 1R x y R x y x x into two equal parts

then,

A) 1

02

B) 4 22 4 1 0

C) 4 24 1 0 D)

11

2

Key: BD

Sol: 1

3 3

0 02 x x dx x x dx

0x

1 1

1

1

2 4 1 12

2 4 2 4

4 22 4 1 0

3 1

2

;

3 1

2

1 Discarded




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Q45. If

(2 ) 2 2

,

Cos x Cos x Sin x

f x Cosx Cosx Sinx

Sinx Sinx Cosx

then

A) f x attains its maximum at 0x

B) f x attains its minimum at 0x

C) ' 0f x at more than three points in ,

D) ' 0f x at exactly three points in ,

Key: AC

Sol:

0 cos2 sin2

2cos cos sin

0 sin cos

x x

f x x x x

x x

2cos cos3 cos4 cos2f x x x x x

' 4sin4 2sin2f x x x

2sin2 1 4cos2x x

24sin cos 8cos 3x x x

' '0 ; 0f ve f is ve f x has a max at 0x

' 20 sin2 0;cos 3 / 8 4f x x x solutions

' 0 sin2 0f x x ; 1

cos24

x

2x n

,2 2

x

Q46. If

98 1

1

1

1

k

kk

kI dx

x x

, then

A) 49

50I B)

49

50I C) log 99eI D) log 99eI

Key: AC

Sol:

198

1

1

1

k

k k

kI dx

x x

= 98

1

1k

k

log 1

1

k

k

x

x




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=

298

1

11 log

2K

kk

k k

=

2 22 32 42 992log 3.log 4.log ..... 99log

1.3 2.4 3.5 98.100

=

4 6 8 10 12 196 198

3 6 8 10 196 98 99

2 .3 .4 .5 .6 98 .99log

2 3 .4 .5 98 .99 .100

=

100

99

2.99log

100

log 99e

99100

99

2 99 9999 2 99 2 0.36 99 0.72 99

100 100

Clearly it is greater than 1

Q47. Let 1 1 1

1

x xf x

x

Cos

1

1 x

for 1x .then

A) 1

0x

Lim f x B) 1 0xLim f x

C) 2 0,xf x e in D) 1 0xLim f x does not exist

Key: BC

Sol: 1x 1 2 1

.cos1 1

x x

x x

1x 1 0x 1

1 cos 01

xx

1x 1 1

os1 1

x xc

x x

1x

1 0x 1

1 cos1

xx

does not exist

Q48. If :f R R is differentiable function such that ' 2 ,f x f x for all x R and 0f =1, then

A) f x is increasing in 0, B) 2' 0,xf x c in

C) 2 0,xf x e in D) f x is decreasing in 0,

Key: AC

Sol: ' 2 0f x f x

'

2 0xf x e

2 sinxf x e isincrea g 0x




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-34

2 00xf x e f e 2 1xf x e

2xf x e

Q49. If sin 2

1

sin

sin

x

x

g x t dt , then

A) ' 2

2g

B) ' 2

2g

C) ' 2

2g

D)

' 22

g

Key: Delete

Sol: by Leibnitz theorem

' 1 1sin sin2 2cos2 sin sin .cosg x x x x x 4 cos cosx x x x

' 2 . 1 .0 22 2

g

' 2 1 22

g

Q50. Let and be nonzero real numbers such that 2cos cos cos cos 1 . Then which of the

following is/are true?

A) tan 3 tan 02 2

B) 3 tan tan 0

2 2

C) tan 3 tan 02 2

D) 3 tan tan 0

2 2

Key: AC

Sol: 2 cos cos cos cos 1

1 2cos

cos2 cos

2

22

22

2

11 2

11tan / 2, tan / 2

112

1

x

xyx y

xy

x

2 2

2 2

1 3

1 3

y x

y x

2 23y x

3 tan tan 3 tan tan 02 2 2 2

Options A & C




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******** ***************************************************************************************************

PARAGRAPH-1

Let O be the origin, and , ,OX OY OZ be three unit vectors in the directions of the sides , ,QR RP PQ ,

respectively, of a triangle PQR.

Q51. OX OY

A) sin P R B) sin2R C) sin P Q D) sin Q R

Key: C

Sol: ;QR RP

OX OYQR RP

sin sin sinQR RP

OX OY QRP R P QQR RP

Q52. If the triangle PQR varies, then the minimum value of cos cosP Q Q R cos R P

A) 3

2 B)

3

2 C)

5

3 D)

5

3

Key: A

Sol: cos cos cosP Q Q R R P

cos cos cosP Q R

3

2 as

3cos cos cos

2P Q R P Q R

PARAGRAPH-2

Let p, q be integers and let , be the roots of the equation, 2 1 0x x , where .For 0,1,2,...,n

let n n

na p q .

FACT: If a and b are rational numbers and 5 0a b , then 0a b

Q53. If 4 28a , then 2p q =

A) 12 B) 21 C) 14 D) 7

Key: A

Sol: 1 1 5 1 5

,2 2




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1 2 1

4 43 2, 3 2

4 4

4 P q

28 3 2 3 2P q

1 3 5

82 2

p q p q

4p q 2 12p q

Q54. 12a

A) 11 102a a B) 11 10a a C) 11 10a a D) 11 102a a

Key: B

Sol: 5 5

6 12 128 5 144 89; 144 89

12 12

12a p q

12 144 89a p q p q

11 10 12a a a

****************

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