jee advanced offline exam 2017 (paper 2) code: 0 · pdf file1 300 2 2 25 gm gm v rr v 13 2 gm...

JEE Advanced Offline Exam 2017 (Paper – 2) CODE: 0

Copyright © Ace Creative Learning Pvt Ltd., www.deekshalearning.com

1

PART 1: PHYSICS

1. Consider regular polygons with number of sides 3,4,5......n as shown in the figure. The centre 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 each polygon is . Then depends on n and h as

(A) 2sinhn

(B)

2sinh

n

(C) 2tan2

hn

(D)

11

cos

h

n

Sol:

cos2

h

n

11

cos cos

hh h

n n

Ans: (D)

2. 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

Sol: 33

3

4 43

M M

RR

h

h h

cos

h

n



2

33

4

d M dR

dt dt

433

4

d M dRR

dt dt

4

9

4

d Mv

dt R

4

3

9

1 343

4

Mv

d vRMdt R

R

v R

Ans: (A)

3. A photoelectric material having work-function 0 is illuminated with light of wavelength

0

hc

. The fastest photoelectron has a de Broglie wavelength d . A change in wavelength

of the incident light by results in a change ind d . Then the ratio d

is proportional to

(A) 2

2

d

(B) d

(C)

3d

(D)

3

2

d

Sol: 2

022

h hc

m d

2

22

2

d

a

dh hcd

m

3

2 2

dd hc m

d h

3

2

d dd

d

Ans: (D)

4. Three vectors , andP Q R shown in the figure. Let S be any

point on the vector R . The distance between the points

andP S is b R . The general relation among vectors

, andP Q S is

(A) 21S b P bQ

(B) 1S b P bQ

(C) 1S b P bQ

(D) 21S b P b Q

X

Y

O

Q

SP Q

P

S

b R

R Q P



3

Sol: T b Q P

P T S b Q P P

1bQ b P S

Ans: (C)

5. 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 210msg 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%

Sol:

20mL

1m as 1mL LC

1

20

l

l

1100

20

5%

Ans: (B)

6. 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

111.2kmsev . The minimum 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) 172kmssv (B) 122kmssv (C) 142kmssv (D) 162kmssv

P

T

Q

Q P R



4

Sol: Only for earth escape speed is

12112 kmse

e

GMV

R

For both earth and sun

21

2

e s

e s

GM m GM mmv

R R

5

4

3 10

2.5 10

ee

e e

G M mGM m

R R

21 300

2 25

GM GMV

R R

113 213 11.2 42 kms

GMV

R

Ans: (C)

7. 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 centre of the loop is

(A) 6 3 14

oI

a

(B) 6 3 14

oI

a

(C) 3 3 14

oI

a

(D) 3 2 34

oI

a

Sol:

2 sin30d a

a

0each segment cos30 cos120

4

IB

a

0 3 1

4 2 2

I

a

net 12B B due to each segment

0 3 112

4 2

I

a

0 6 3 14

I

a

Ans: (A)

I

4a

Sun

42.5 10 R

eM

R sM

120

30

2a

90

30

d

i

cos cos4

oiBd

d



5

8. A wheel of radius R and M is placed at the bottom 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 point

Q . Which of the following options is/are correct?

(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 the 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

Sol:

(a) Force passing thorough the axis of rotation the 0

(b) Force at s in tangential direction produces a constant clockwise torque that is opposed by the

torque due to gravity about centre of mass. If the clockwise torque is greater than torque due to

gravity then the wheel will rotate.

(c) Tangential force about p will rotate it and since the perpendicular between Q and centre of

mass decreases, torque increases as the objet rotates.

(d) Since the perpendicular distance charges torque is not constant when applied form X .

Ans: (A,B)

9. Two coherent monochromatic point sources 1S and 2S of wavelength 600 nm are placed

symmetrically on either side of the center of the as shown. The sources are separated by a

distance 1.8 mmd . 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?

P

S

RX

Q

1S2S

d

1P

2P

p



6

(A) The angular separation between two consecutive bright spots decrease 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 quadratn is close to

3000

(D) At 2P the order of the fringe will be maximum

Sol: For bright fringe, path difference should be integral multiple of .

At 2p , path difference 1.8 mm

3

91.8 10600 10 3000

x

n

At 1p path difference 0 ,

Hence 3000 maximas are possible.

Ans: (ACD)

10. 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 ?

(A) The electric flux passing through the curved surface of the

hemisphere is 0

11

2 2

Q

(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

Sol: The total electric flux is zero.

The flux through flat face is 0

1 cos2

Q

0

11

2 2

Q

flux through curved area is 0

11

2 2

Q

The circumference is equipotential.

Ans: (A, D)

Q

R

1p

1.6 mm2p

R

R



7

11. A rigid uniform bar AB of length L is slipping from its vertical position on the frictionless floor

(as shown in the figure). At the some instant of time, the

angle made by the bar with the vertical is . Which of the

following statement about its motion is/are correct?

(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 mid-point of the bar 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

Sol: sin2

Lmg

sin

The displacement of the midpoint falls through a height

of cos 1 cos2 2 2

L L L

The midpoint falls vertically downward and ' 'A will have

parabolic both.

Ans: (A, B, C, D)

12. 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 0t the switch is closed and

current beings to flow. Which of the following options is/are

correct ?

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

1 2

Lv

R L L

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

1 2

Lv

R L L

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

(D) At 0t the current through the resistance R is v

R

B

O

A

L

V 1L

2L

R

S

B

mg

A

2

L

sin2

L



8

Sol: At 0t 0i

Ratio of currents 1 21 2

di diL L

dt dt

1 1 2 2L i L i

1 2

2 4

i L

i

11 1

2

Li i i

L

21

2

4LVi

R L

21

1 2

L Vi

L L R

12

1 2

L Vi

L L R

Ans: (ABC)

13. A uniform magnetic field B exists in the region between

0x 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?

(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 change Q and same velocity v , the distance between the

point 1P and the point of re-entry in to region 1 is inversely proportional to the mass of the

particle

2PxO

B

Q 1P

y R

3 / 2R

yRegion -1 Region -2 Region -3



9

Sol: 3

32 2

mv p Rr QR

QB Q p

For 2

3

PB

QR , the particle will re-enter region 1

To re-enter the region ‘1’ the particle will cover semi-circle,

2

2mv

rqB

2r m

For 8

13

pB

QR , the particle will come to region 3

‘B’ and C are the correct answer

Ans: (B,C)

14. The instantaneous voltage at three terminal marked ,X Y and Z are given by

0 sinXV V t

02

sin3

YV V t

and 04

sin3

ZV V t

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

terminal. It is connected between points X and Y and then between Y and Z . The reading (s) of

the voltmeter will be

(A) rms0

1

2YZV V

(B) rms0

3

2XYV V

(C) independent of the choice of the two terminals

(D) rms0XYV V

Sol: 02

sin sin3

X YV V V t t

03

2V

0 02 4

sin sin3 3

YZV V t V t

01

2V

Ans: (A,B)



10

Paragraph-1

Consider a simple RC circuit as shown in figure1

Process 1: in the circuit the switch S is closed at 0t 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 stored in the fully charged capacitor is

CE

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

3

Vand maintained for a charging time

T RC . Then the voltage is raised to 02

3

Vwithout discharging the capacitor and again maintained

for the capacitor is charged to the same final voltage 0V as in process 1

These two processes are depicted in figure 2.

15. In process 1 , the energy stored in the capacitor CE and heat dissipated across resistance DE are

related by

(A) 0 ln 2CE E (B) C DE E (C) 2C DE E (D) 1

2C DE E

Sol: / /0 01 t RC t RCq q e i i e where 0

0

qi

RC

2

2 20

0 0

t

RCDE i Rdt i R e dt

220

01

2 2

qCV

C and 2

01

2CE CV

D CE E

Ans: (B)

V

S

R

C

Figure -1

T 2T

Processs 2

Processs1

T RC0 / 3V

02 / 3V

0V

t

Figure -2



11

02

3

V

R

0q

At t T

0

3

V

02

3

V

R

0q

At between and 2t t T T

V

i

16. In process 2 , total energy dissipated across the resistance DE is

(A) 20

1 1

3 2DE CV

(B) 2

01

32

DE CV

(C) 203DE CV (D) 2

01

2DE CV

Sol: Between 0t to t T RC

2 2 2 22 0 00

0

1

2 2 9

t

RCD

q c vE i R e dt

C C

20

1 1

9 2CV

From t T to 2t T

02

3

V qiR

C

02q dqR q

C dt C

By solving, we get

/0

t RCi i e by assuming t T as 0t

heat dissipated in the 2nd interval also

22 0

1 1

9 2Q CV

From 2t T to t

P.D. across capacitor changes from 2

3

Vto 0V

Heat dissipated in this interval also

23 0

1 1

9 2Q CV

Total heat dissipated from 0t to t is

1 2 3Q Q Q Q

20

1 13

9 2CV

20

1 1

3 2CV

Ans: (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 by 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 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 .



12

17. The total kinetic energy of the ring is

(A) 22

0M R r (B) 22

01

2M R r

(C) 2 20M R (D)

220

3

2M R r

Sol: From the centre the velocity of the figure and ring should be same

'0

Kinetic energy about the point of rotation,

i.e., the tangent is 2

2 112

2MR

Ans: (C)

18. 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

Sol: The normal reaction force is provided by centrifugal force 2

0M R r

20M R r Mg

0

g

R r

Ans: (C)

Figure -1 Figure -2

R

R

r

R r

R



13

PART II: CHEMISTRY

19. The order of basicity among the following compounds is

NH

CH3 NH2

I

NHN NNH

NH2

NH2 NH

II III IV

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

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

Sol:

NH

CH3 NH2

NHN NNH

NH2

NH2 NH

> > >

Ans: (A)

20. The major product of the following reaction is

NH2

OH

i) NaNO2, HCl, 0oC

ii) aq. NaOH

(A) (B)

(C) (D)

N2Cl

ONa+

N N

OH

Cl

OH OHN N



14

Sol:

OH

NH2

NaNO2, HCl

0 C

OH

N N+

NaOH (aq)

OH

N N(Major product)

(Intramolecular coupling)

Ans: (B)

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

(A) Au metal and aqNaCN in the presence of air

(B) Zn metal and aqNaOH

(C) Fe metal and conc. 3HNO

(D) Cu metal and conc. 3HNO

Sol: 2 2 22Zn NaOH Na ZnO H

Ans: (B)

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

are

1graphite 0kJ molf G C

1diamond 2.9kJ molf G C

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) and 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 1 1 2 51J = 1 kg m s ; 1 Pa 1kg m s ; 1 bar = 10 Pa ]

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



15

Sol: 1

Graphite diamand 2.9kJ molC C G

G P V

3 62.9 10 1 2 10P

92.910

2P Pa

91.45 10P Pa 51Bar 10 Pa

41.45 10 BarP

14500BarP

Ans: (A)

23. The order of the oxidation state of the phosphorous atom in 3 2 3 4 3 3 4 2 6, , andH PO H PO H PO H P O

is

(A) 3 4 4 2 6 3 3 3 2H PO H P O H PO H PO (B) 3 2 3 3 4 2 6 3 4H PO H PO H P O H PO

(C) 3 4 4 2 3 3 4 2 6H PO H PO H PO H P O (D) 3 3 3 2 3 4 4 2 6H PO H PO H PO H P O

Sol: Oxidation state of P:

3 4 5H PO

4 2 6 4H P O

3 3 3H PO

3 2 1H PO

Ans: (A)

24. For the following cell,

4 4aq aqZn s ZnSO CuSO Cu s

When the concentration of 2Zn is 10 times the concentration of 2Cu , the expression for G

1in J mol as

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

(A) 2.303RT 2.2F (B) 1.1 F

(C) 2.2F (D) 2.303RT +1.1 F

Sol: 2.303 logG G RT K

Cell reaction: 2 2Zn Cu Zn Cu

2

2

ZnK

Cu



16

2 210Zn Cu

10K

2 1.1 2.2G nFE F F

2.2 2.303 log10G F RT

Or 2.303 2.2G RT F

Ans: (A)

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

12 K 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) (B)

(C) (D)

Sol: 1000f A

A B

K WT

M W

2 34.5 1000

46 500

3.0

Freezing point of solution 270K

Graph (C) will be correct match.

Ans: (C)

270 273

T/K

V.P./bar

1

271 273

T/K

V.P./bar

1

V.P./bar

1

270 273

T/K

V.P./bar

1

271 273

T/K



17

26. 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 4.5P , 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

Sol: The activation energy is not dependent on steric factor

Steric factor Experimental value

Theoritical value

Hence, the value of Arrhenius factor (theoretical) is less than the experimental values.

Ans: (A, D)

27. 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 decreases 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 increase because

unfavourable change in entropy of the surroundings decreases

Sol: With increase in temperature, K for exothermic reaction decreases due to decrease in the

magnitude of positive value of S

With increase in temperature, K for endothermic reaction increases, due to decrease in the

unfavourable value of S i.e., negative value for S

Ans: (A, D)



18

28. 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 563 K and 126 K, 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

Sol: Adsorption of gases is exothermic, H is –ve and S is negative

The gas with higher critical temperature will have higher vander waal’s forces and hence can be

adsorbed on charcoal to a higher extent.

Cloud is not an emulsion

Brownian movement depends on size of particles

Ans: (A, C)

29. Among for 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

Sol: 2 6 2 3 6,B H Al CH have 2 3e c bond known as Banana bond. 3BCl is more acidic than 3AlCl

due to 3 3p p overlap for back bonding in 3AlCl . 2 6Al Cl does not have Banana bond, it is

having co-ordinate bond

Ans: (B, C, D)

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

reactions is (are)

Br BrBr

CH3

CH3 C Br

CH3

CH3

I II III IV

(A) I and III follow 1NS mechanism

(B) Compound IV undergoes inversion of configuration



19

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

(D) I and II follow 2NS mechanism

Sol:

Br Br

CH3

CH3

CH3 BrBr

(I) (II) (III) (IV)

Order of 1:NS IV III I

Order of 2 :NS I IV III

(I) And (II) being primary will follow 2NS

(I) and (III) form stable benzyl carbocation and follow 1NS

Compound (IV) can undergoes inversion in 2NS

Ans: (A, B, D)

31. 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

Sol: 2 3 2, , ,ZnO Al O PbO PbO amphoteric oxides. 2 3 2, , ,Cr O BeO SnO SnO amphoteric oxides.

Ans: (A, C)

32. Compounds P and Q 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.

P i) O3/CH2Cl2

Q

(C8H8O)ii) Zn/H2O

(i)

R (ii)i) O3/CH2Cl2

(C8H8O)ii) Zn/H2O

S

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

(A)

CH3

CH3

CH3

and

CH3

CH3

CH3



20

(B)

(C)

(d)

Sol:

CH3

(1) O3

(2) Zn / H2O

CHO

CH3

+ CH3 C CH3

O

(P) (Q)

CH3

CH3

CH3

(1) O3

(2) Zn / H2O

(R)

CH3

CH3

O

+ CH3CHO

(S)

(A)

(B) CHCH3 CH2

(P)

(1) O3

(2) Zn / H2O

CHO CH3 HCHO +

(R)

(1) O3

(2) Zn / H2O

O

(S)

HCHO +

(Q)

CH3and

CH3

CH3

CH3

CH3

and

CH3

CH3

CH3

CH3 and

CH3



21

(C)

(P)

(1) O3

(2) Zn / H2O

H

O

+ CH3CHO

(Q)

(R)

(1) O3

(2) Zn / H2O

O(S)

+ CH3 C CH3

O

(D) CH3

(1) O3

(2) Zn / H2O

CHO CH3 + HCHO

CH3

(1) O3

(2) Zn / H2O

CH3

O

CH3

+ HCHO

Ans: (B, C)

PARAGRAPH 1

Upon heating 3KClO in the presence of catalytic amount of 2 ,MnO 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 .

33. W and X are, respectively

(a) 3 4 6andO P O (b) 2 4 6andO P O (c) 2 4 10andO P O (d) 3 4 10andO P O

Ans: (C)

34. Y and Z are, respectively

(a) 2 5 3andN O HPO (b) 2 3 3 4andN O H PO

(c) 2 4 3andN O HPO (d) 2 4 3 3andN O H PO

Sol: (Q. 33 to Q. 34 Solution)

23 2

MnO

w

KClO KCl O

2 4 10X

P O P O

4 10 3 2 5 3Y Z

P O HNO N O HPO

Ans: (A)



22

PARAGAPH 2

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

gives Q . The compound Q on treatment with 2 4H SO at 0 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]

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

(a) Dehydration and Friedel-Crafts acylation

(b) Friedel-Crafts alkylatin, dehydration and Friedel-Crafts acylation

(c) Friedel-Crafts alkylation and Friedel-Crafts acylation

(d) Aromatic sulfonation and Friedel-Crafts acylation

Ans: (C)

36. The product S is

(a) (b)

(c) (d)

Sol: (Q. 35 to Q. 36 Solution)

(H3C)3C

COCH3

CH3O

HO3S

(H3C)3C CH3CH3

H3COC

(H3C)3C

CH3 CH3

COCH3

(H3C)3C

COCH3

CH3

(H3C)3CCO2Et

P

Q R S



23

CO2Et C(CH3)3

CH3MgBr

H3O

C(CH3)3 OH

H2SO4

H2O

C(CH3)3C

P

Q R

CH3COCl

AlCl3

C(CH3)3C

COCH3

S

Ans: (C)



24

PART III: MATHS

37. Let 1,2,3,...,9 .S 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

Sol: 1,2,3,...9 , 1,2,....5S K

kN number of subsets of S each containing 5 elements out of which exactly k are odd.

1N number of subsets of S each containing

5 elements out of which one is odd

5

1 5C

5 4

2 2 3 10 4 40N C C

5 4

3 3 2 10 6 60N C C

5 4

4 4 1 5 4 20N C C

5

5 5 1N C

1 2 3 4 5 5 40 60 20 1N N N N N

126

Ans: (D)

38. The equation of the plane passing through the point 1,1,1 and Perpendicular to planes

2 2 5x y z and 3 6 2 7x y z , is

(A) 14 12 15 3x y z (B)14 2 15 27x y z

(C) 14 12 15 1x y z (D) 14 12 15 31x y z

Sol: Pass the 1,1,1 and perpendicular to planes

2 2 5x y z

3 6 2 7x y z

The equation is

1 1 1

2 1 2 0

3 6 2

x y z

1 2 12 1 4 6 1 12 3 0x y z

14 1 2 1 15 1 0x y z



25

14 14 2 2 15 15 0x y z

14 2 15 31 0x y z

14 2 15 31 0x y z

Ans: (C)

39. Let O be the origin and let PQR be an arbitrary triangle. The point S is such that

OP OQ OR OS OR OP OQ OS OQ OR OP OS

Then then triangle PQR has S as its

(A) incentre (B) circumcentre (C) orthocentre (D) centroid

Sol: OP OQ OR OS OR OP OQ OS OQ OR OP OS

OP OQ OR OP OQ OS OR OS

OP OQ OR OS OQ OR

0OP OS OQ OR

0SP RQ S in orthocentre

Ans: (C)

40. 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

Sol: If

1 1 1

2 2 2

3 3 3

a b c

M a b c

a b c

Sum of diagonal elements of TM M

2 2 2 2 2 2 2 2 2

1 2 3 1 2 3 1 2 3 5a a a b b b c c c

One possible case is 5 ones’s and rest zeros

Possibilities 9

5 1 126C

Other possibilities cases is 4 in one place

One in one place & zero in other places:

9 8 1 72 ways

Total possibilities 126 72 198

Ans: (B)

R r

P Q

Q S



26

41. 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)

5

11

Sol: 10x y z

3 1

10 3 1 12

2

12 1166

2n s C C

z is even 2 , 0,1,2,3,4,5z k k

10 2 , 0,1,2,3,4,5x y k k

11 9 7 5 3

1 1 1 1 1 1n E C C C C C

36

36 6

66 11P E

Ans: (A)

42. If :f is a twice differentiable function such that " 0f x for all ,x and

1 1

, 1 1,2 2

f f

then

(A) 1

1 12

f (B) 1

0 12

f (C) 1 0f (D) 1 1f

Sol: ' 0f x

1 1

1 12 2

f f

f x is increasing function

f b f a

f cb a

11

21

2

f c

1

1 12

f c

y

x

1

2 0,0



27

1

12

f f

1 1f

Ans: (D)

43. If y y x satisfies the differentiable equation 1

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

and

0 7,y then 256y

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

Sol: 8 9

4 9

dxx x dy

x

29 16x t

132

2dx t dt

x

32

4 4 4 4

t dtdy

t t

2

2 1

dt

t

2 1dy t

92 1

16

xdy c

3

0 2 14

y 7

24

7 7 c

9 256

256 2 116

y

9 162 1

16

52 1

4

3

Ans: (D)



28

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

then

(A) 1

02

(B) 4 22 4 1 0

(C) 4 24 1 0 (D) 1

12

Sol: 1

3 3

0

x x dx x x dx

2 4 2 41 1

2 4 2 4 2 4

2 4 12

2 4 4

42 1

2 4

2 44 2 1

4 22 4 1 0

2 4 16 8

4

.

2 4 2 2

4

2 11

2

2 11

2

2 11

2

10

2

Ans: (B,C)

0

y

x 1



29

45. If

cos 2 cos 2 sin 2

cos cos sin ,

sin sin cos

x x x

f x x x x

x x x

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 ,

Sol:

cos 2 cos 2 sin 2

cos cos sin

sin sin cos

x x x

f x x x x

x x x

0 cos 2 sin 2

2cos cos sin 2cos cos cos 2 sin sin 2

0 sin cos

x x

f x x x x x x x x x

x x

2cos cos3f x x x

' 2sin cos3 3sin3 cosf x x x x x

'' 2cos cos3 sin sin3 9cos3 cos 3cos sin3f x x x x x x x x x

'' 0 2 9f

'' 0 0f

A is correct

' 0f x 2sin cos3 3sin3 cosx x x x

2 22sin cos 4cos 3 3 3 4sin sin cosx x x x x x

2 2sin cos 8cos 6 9 12sin 0x x x x

sin 0x at 0x , cos 0x at 2 2

2 28 8sin 3 12sin 0x x

211 20sin x

2 11sin

20x

So (C)

Total solution is (A), (C)



30

46. If

198

1

1

1

k

k k

kI dx

x x

, then

(A) 49

50I (B)

49

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

Sol:

198

1

1

1

k

k k

kI dx

x x

198

1

k

k k

dxI

x

98

1

1log

k

kI

k

2 3 99log ...........

1 4 98I

log99I

1.99I

49

50I

Ans: (A, C)

47. Let 1 1 1 1

cos1 1

x xf x

x x

for 1x . Then

(A) 1

lim 0x

f x

(B) 1

lim 0x

f x

(C) 1

limx

f x

does not exist (D) 1

limx

f x

does not exist

Sol:

2

2

1 1cos , 1

1 1

1 1cos , 1

1 1

xx

x xf x

xx

x x

1 1

11 lim lim 1 cos 0

1x xf f x x

x

1 1

11 lim lim 1 cos

1x xf f x x

x

does not exist.

Ans: (B, C)

48. If :f is a differentiable function such that ' 2f x f x for all x , and 0 1f , then

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

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



31

Sol: ' 2f x f x

' 2f x f x , where 0

2d

f x f xdx

2 2 22x x xde f x e f x e

dx

2 2x xde f x e

dx

2 2x xe f x e dx

2 2

2

x xe f x e

2

2

xf x e

… (1)

0 1 1 12 2

f

… (2)

21 xf x e

21 1xf x e … (3)

For 20, 2 0, 1, 0xx x e from (2)

0 ' 2 0f x f x f x

f is increasing

(2) 1 1 12

2 21 1x xe e

2xf x e

Ans: (A, C)

49. If sin 2 1

sinsin

x

xg x t dt , then

(A) ' 22

g

(B) ' 22

g

(C) ' 22

g

(D) ' 22

g



32

Sol: sin 2

1

sin

sin

x

x

g x t dt

sin 2

1

2

sin

sin1

x

x

tt t dt

t

sin 21 2

sin

sin 1x

x

t t t

2 sin 2 cos2 sin cosg x x x x x x x

cos2 cos sin 4cos 1x x x x x

2sin 2 sin sin 4sin 4cos 1 cos sing x x x x x x x x x x

0 1 4 1 12 2

g

1 2 1

2

0 1 4 1 1 1 2 1 22 2

g

Ans: (C)

50. Let and be non-zero real numbers such that 2 cos 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

Sol:

2

2

1 tan12cos1

1 tan2

x

x

2tan2

x

1cos

1

y

y

2tan

2y

2 cos cos 1 cos cos

1 11 12 1

1 1 1 1

x yy x

y x x y

3x y



33

2 2tan 3tan2 2

tan 3 tan2 2

3 tan tan 02 2

or 3 tan tan 0

2 2

Ans: (A, C)

51. OX OY

(A) sin P R (B) sin 2R (C) sin P Q (D) sin Q R

Sol: ,QR RP

OX OYQR RP

1QR RPOX OY QR RP

QR RP QR RP

1

sin 180QR RP RQR RP

sin P Q

Ans: (C)

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

(A) 3

2 (B)

3

2 (C)

5

3 (D)

5

3

Sol: cos cos cosP Q Q R R P

cos cos cosR P Q

3 3

2 2

Ans: (B)

53. If 4 28a , then 2p q

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

Sol: 4 28a

4 41 5 1 5

282 2

p q

Q

P

R



34

14 2 5 14 2 5 4 28p q

14 14 2 2 5 4 28p q p q

14 14 4 28p q and 2 2 0p q

8p q and p q

4p q

2 12p q

Ans: (A)

54. 12a

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

Sol:

11 11 10 10

11 101 5 1 5 1 5 1 5

2 2 2 2a a p q p q

10 101 5 3 5 1 5 3 5

2 2 2 2p q

12 121 5 1 5

2 2p q

12a

Ans: (B)

jee advanced offline exam 2017 (paper 2) code: 0 · pdf file1 300 2 2 25 gm gm v rr v 13 2 gm...

Documents