practice set test rider advanced

PRACTICE SET 1 3

A to the Various IITs...Single Door Entry

Advanced

Paper 1

Duration : 3 Hours Max. Marks : 180

Pra

cti

ce

Se

t

Test RIDER

1

Question Booklet Code 0

Please read the instructions carefully. You are allotted 5 minutes speciallyfor this purpose.

Question Paper Format

Marking Scheme

}

}

}

}

}

}

}

}

}

}

This booklet is your question paper. Attempt all the questions.

Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers and

electronic gadgets are not allowed.

Write your name and roll number in the space provided on the bottom of this page.

The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists

of three sections.

contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d)

out of which only one is correct.

contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d)

out of which one or more than one is/are correct.

contains 5 questions. The answer to each question is a single–digit integer, ranging

from 0 to 9 (both inclusive)

For each question in Section 1, you will be awarded for correct answer and zero mark for

unattempted questions. No negative marks will be awarded for incorrect answers in this section.

For each question in Section 2, you will be awarded for correct answer(s) and zero mark for

unattempted questions. In all other cases, minus one (–1) mark will be awarded.

For each question in Section 3, you will be awarded for the correct answer and zero mark for

unattempted questions. In all other cases, minus one (–1) mark will be awarded.

Section 1

Section 2

Section 3

2 marks

4 marks

4 marks

Name of the Candidate

Roll Number

(in capital letters)

(in figures) (in words)

1. A conducting sphere is connected with earth and a concentric spherical shell is given a charge Q.Choose the correct option.

a. the charge on the sphere must be zero.

b. the charge on the sphere must be positive.

c. the magnitude of charge on the inner sphere isQr

R

d. None of the above

2. A half section of thin uniform pipe of massm and radius r is released from rest. Pipe rolls withoutslipping. The change in PE of pipe when it has rolled through 90°.

a.4mgr

pb.

2 mgr

pc. mgr d. 2 mgr

3. A potentiometer has an uncertainty of ± 0.0001 V. If it is used to measure current through astandard resistance of 0.1 ± 0.1% W (uncertainty). The voltage across the resistance is measured tobe 0.2514 V. What is the percentage uncertainty in the measurement of current?

a. 1.04% b. 2.01% c. 2.08% d. 0.08%

4. A ball rolls without slipping on horizontal surface and then moves upthe inclined shown if height attained is h1 and h2 with sufficientlyrough track and with smooth track then.

a. h h1 2= b. h h1 2< c. h h1 2> d. h h2 12=

5. The object O is placed in front of mirrors as given. If the velocity of image I 3 is v (magnitude),

then what will be speed of object?

a. v b. 2 v c. 3 v d. 5 v

Part I

Section 1 Single Correct Option Type

h

Q

r

R

O

I2 I3

I1

6. A uniform magnetic field exists in the space along the x-axis as shown in the figure.A rectangular loop of sides (a b´ ) is rotated in magnetic field about an axis passing through themid-point of side a and parallel to side b with angular speed w (this axis is parallel to direction of B).The induced current in the loop if the resistance of loop is R, is

a. B b Rw2 2/ b. B a Rw2 2/ c. zero d. None of these

7. Three point charges q, 2q and 8q are to be placed on a 9 cm long straight line. Find the positionwhere the charges should be placed such that the potential energy of this system is minimum. Inthis situation, what is the electric field at the position of the charge q due to the other twocharges?

a. 20 cm, 24N/C b. 12 cm, 2 N/C c. 3 cm, zero d. 9 cm, zero

8. A conical pendulum consists of a slender bar AB of length L and mass M. If it rotates withconstant angular velocity w, then value of cosa will be

a. cos aw

= g

L2b. cos a

w= 2

2

g

L

c. cos aw

= 3

2 2

g

Ld. cos a

w= 3

2

g

L

9. A solid sphere and a thin loop of equal masses m and radius R are harnessed together by riggingand free to roll without slipping down the incline plane. Neglect the mass of rigging, thencompressive force in rigging will be

a.3

17mgsin q b.

mgsin q17

c. 3mgsin q d.2

17

mgsin q

10. An electron and a proton are separated by a large distance. The electron starts approaching theproton with energy 2 eV. The proton captures the electron and forms a hydrogen atoms in firstexcited state. The resulting photon is incident on a photosensitive metal of threshold wavelength4600 Å. The maximum KE of the emitted photoelectron is

a. 2.4 eV b. 2.7 eV c. 2.9 eV d. 5.4 eV

PRACTICE SET 1 5

ω

θ

x

y B

11. A smooth track in form of a quarter circle of radius 6 m lies in verticalplane. A particle moves from P1 to P2 under action of force F F1 2, and F3 .Force F1 is always towards P2 and is always 20 N in magnitude. Force F2

is always act horizontally and is always 30 N in magnitude. F3 actstangentially and has magnitude 15 N. Select the correct options.

a. work done by F1 is 120 2 J b. work done by F2 is 180 J

c. work done by F3 45= pJ d. F3 is conservation in nature

12. A pendulum is constructed from two identical uniform thin rods A and B each of lengthL and mass m connected at right angle forming T shape. It suspended free end andswings in vertical plane.

a. moment of inertia of the T about axis of rotation is17

122ml

b. moment of inertia a of T about axis of rotation is 13 12 2/ ml

c. time period of small angular oscillation of T is 217

18p l

g

d. angular frequency of T for small oscillation is18

17

g

l

13. Variation of y-coordinate of two projectiles with time is given in figure.

If initial speed of both is same then

a. horizontal speed of 1 is less than 2 b. horizontal speed of 1 is greater than 2

c. horizontal range of 1 is greater than 2 d. Both has same horizontal range

14. A charged particle having a positive charge q approaches a groundedmetallic neutral sphere of radius R with a constant speed v as shown infigure. Now, choose the correct alternative (s).

a. As the charged particle draws nearer to the surface of the sphere, a current

flows into the ground

b. As the charged particle draws nearer to the surface of the sphere, a current flows out of the ground into

the sphere

c. As the charged particle draw nearer, the magnitude of current flowing in the connector joining the shell

to the ground increases

d. As the charged particle draws nearer, the magnitude of current flowing in the connector joining the

sphere shell to the ground decreases

AdvancedTest RIDER

y

t12

L

L

O P2

F3

F2

F1

P1

R v

q

Section 2 More Than One Correct Option

This section contains . Each question has four choices, (a), (b), (c)

and (d) out of which or is/are correct.

5 multiple choice questions

one more than one

15. A reflecting surface is represented by the equation yL

Lx x L= æ

èçöø÷

£ £20

pp

sin , . A ray of light

moving horizontally becomes vertical after reflecting. Find coordinates of points where ray isincident.

a.L L

4

2,

pæ

èç

ö

ø÷ b.

L L

3

3,

pæ

èç

ö

ø÷ c.

3

4

2L L,

pæ

èç

ö

ø÷ d.

2

3

3L L,

pæ

èç

ö

ø÷

16. A classroom is maintained at 20°C by a heater of resistance 20 W connected to 200 V mains. Thetemperature is uniform throughout the classroom and heat is transmitted through a glasswindow of area is A m2 and thickness 0.2 cm. So, outside temperature maintained is 15.24°C. Findthe value of A.

[Given Thermal conductivity of glass is 0.2 cal m s1 1- - (°C) -1 and mechanical equivalent of heatis 4.2 J/cal.]

17. Water is flowing in varying cross-section pipe. The areas of cross-sections 1, 2 and 3 are

1cm , 2 cm2 2 and A cm2 respectively. Water levels are shown in different vertical tubes. The

speed of water at cross-section 3 is1

xm/s. The value of x is

18. A block is moving with speed v towards a system of two blocks system. The first block hits thesecond block elastically. What will be the common velocity by which two blocks system willmove together after some time. If the maximum compression is x = 1 m.

19. If the K a radiation of Mo( )Z = 42 has a wavelength of 0.71 Å, calculate wavelength of the

corresponding radiation of Cu, i e. ., K a for Cu ( )Z = 29 . Find the value of nearest integer value(in Å) of wavelength.

PRACTICE SET 1 7

y

x

Section 3 Integer Answer Type

This section contains . The answer to each question is a single-digit

integer, ranging from to (both inclusive).


0 9

123

37.5 cm35 cm

200 cm

mmm

Plane surface

1 2 3v=2m/s k= 2N/m

20. The mean lives of a radioactive substance are 1000 and 400 years for a-emission and b-emissionrespectively. Find out the time during which three fourth of a sample will decay if it is decayingboth by a-emission and b-emission simultaneously and fill the number of significant digits ofyour answer in OMR.

Part II

P and Q are respectively

22. Boron differs from other members of group 13 elements on reaction with carbon it formscarbides, reaction of boron with metals, oxygen etc., are only due to its small size and highelectronegativity. Which of the following statement is/are incorrect?

a. Carbides of boron are abrasive

b. Boron show diagonal relationship with silicon due to smaller size and dissimilar e/r ratio

c. Boron on reaction with metals forms metal borides

d. Boron forms only covalent compounds

23. PbCl ( ) Pb + 2Cl ; =22 +

sp 1s K Ks-

If a constant concentration of [Cl ]= M- x is maintained in a solution in which the initial

[ ]Pb M2 + = y , what % of original [Pb ]2 + will remain unprecipitated?

a.K

x y

12

100´×

b.yK

x

12

100´

c.x y

K

2

1

100- ´ d. None of these

8 AdvancedTest RIDER

NO2

PCH Cl3 Sn/HClAlCl3

21. Q

a.

c.

b.

d.

NO2

NO2

NO2

CH3

NO2

CH3 CH3

NH2

CH3

NH2

, ,

, ,

NH2

CH3CH3

NH2


24. Me C2O

HCl KCN

2

== ¾¾® ¾¾®-

CH2 2P Q

D

Q will be

a. Me CH CH CN2 2¾ ¾ b. Me C (CN) Me2

c. Me C2 ==CH2 d. CH

CN

CH

Me

Me2½

¾½

¾

25. Consider the following case of competing 1st order reaction.

After the start of reaction at t = 0 with only A, the C is equal to the D at all times. The time inwhich all three concentrations will be equal is given by

a. tk

= 1

23

1

ln b. tk

= 1

23

2

ln c. tk

= 1

32

1

ln d. tk

= 1

32

2

ln

26. Which of the following order of acidic strength of oxoacids of phosphorous and oxoacids ofchlorine is/are correct? Choose the correct option.

I. H PO H PO H PO3 4 3 3 3 2> >II. H PO H PO H PO3 4 3 3 3 2< <III. HClO HClO HClO HClO4 3 2< < <IV. HClO HClO HClO HClO2 3 4< < <a. I and III are correct b. II and III are correct

c. I and IV are correct d. II and IV are correct

27. What is the reaction intermediate involved in Balz-Schiemann reaction?a. Carbocation b. Carbanionc. Free radical d. Carbene

28. Which of the following occur as sulphide ores?Ag, Hg, Pb, Au, Sn, Mg, Al, Fe

a. Ag, Hg, Pb and Fe b. Hg, Pb, Fe and Mg

c. Hg, Pb, Au and Fe d. Sn, Mg, Hg and Ag

29. There are some chemical reaction which produces only one product in large extent and are inminor amount. These reaction are known as stereoslective reaction and if only product isobtained as more than 98% reaction is said to be stereospecific reaction.

30. Reagent used during brown ring test of nitrate and the magnetic moment of brown ring complexformed during the reaction is

a. [Fe H O) ]SO2 6 4( and 3.87 BM b. FeSO 4 and 3 87. BM

c. FeSO 4 and 2.83 BM d. FeCO 3 and 3.87 BM

PRACTICE SET 1 9

A

C

D

k1

k2

NiMe

H+ D2 250°C

Me

HP, product isP

a. b. c. d.

Me

Me

H

HD

D

Me

Me

H

DD

H

Me

Me

H

DH

D

(±)

None of these

31. Which of the following pair of compounds cannot exist together in aqueous solution?

a. NaH PO2 4 and Na HPO2 4 b. Na CO2 3 and NaHCO 3

c. NaOH and NaH PO2 4 d. NaHCO 3 and NaOH

32. Identify the wrong statement.

a. Lower the packing fraction, more the stability

b. Lower the packing fraction, lesser the binding energy

c. Lower the packing fraction, more will be binding energy per nucleon, hence greater is stability

d. Higher the packing fraction, more the stability

a. Tollen’s reagent b. Fehling solution c. 2, 4 DNP d. KCN (alcohol)

34. Which of the following statements is/are incorrect?a. N O2 5 and N O2 4 contain one N—N bond each

b. Hybridisation of N(SiH )3 3 is sp3

c. Bond angle of H O2 is lower than O S2

d. H S O2 2 7 contain no S — S bond

35. In which of the following reaction(s), which product is more acidic than phenol?

36. Number of total possible stereoisomers of Pd[NH CH(CH ) CO ]2 3 2¾ ¾ - is

37. If pd versus p (where p in atm and d in g/L) is plotted for He gas at a particular temperature. If

d

dppd

p

( )é

ëê

ù

ûú =

= 8.21 atm

5, what is the temperature?

a.

OH

Ni

∆

OH

d.

OH

CH Cl3

OH

AlCl3

CH3

OH

CH3

+c.

OH

conc. H SO2 4

conc. HNO3

OH

NO2

NO2O N2

b.

OH OH

(i) CO , NaOH2COOH

(ii) H+

CHO

CH CHO and can be distinguished by333.





0 9





one more than one

38. The numbers of compounds which give NaHCO3 test are

39. If Q amount of heat is given to monoatomic ideal gas in a process in which the gas performs awork Q /2 on its surrounding. What is the molar heat capacity of the gas?

40. Dinitrogen is a diatomic molecule which contain s as well as p-bonds. How many electrons areremoved from oxygen that the bond order of both oxygen and nitrogen becomes equal?

Part III

41. a b c, , are sides of a triangle and a b c, , are in GP. If log log , log loga b b c- -2 2 3 and log log3c a-are in AP, then

a. angle A must be obtuse b. angle B must be obtuse

c. C = °90 d. None of these

42. If 0 1000< <x andx x x

x2 3 5

31

30

éëê

ùûú

+ éëê

ùûú

+ éëê

ùûú

= , where [ ]x is greatest integer function less than or

equal to x , then the number of possible values of x is

a. 32 b. 33 c. 34 d. None of these

43. In a certain test, there are n questions. In this test 2n i- , students gave wrong answers to atleast i

questions, where i n= ¼1 2 3, , , , . If the total numbers of wrong answer given is 2047, then n equals

a. 11 b. 10 c. 12 d. 13

44. One mapping is selected at random from all mappings of the set S n= ¼{ , , , , }1 2 3 into itself. If the

probability that the mapping is one-one is 3 32/ , then n equals.

a. 3 b. 2 c. 4 d. None of these

45. For any positive integers m, n with n m³ . Letn

mCn

m

æèç

öø÷ = . Then,

n

m

n

m

n

m

m

m

æèç

öø÷ +

-æèç

öø÷ +

-æèç

öø÷ +¼+ æ

èç

öø÷

1 2is equal to

PRACTICE SET 1 11


OH OH CH COOH2 COOH

SO H3

OH CH OH2 SO H3

NO2

CH3

NO2

C CH

NO2O N2

CH3

a.n

mC + 1 b.n

mC c.n

mC++

11 d. None of these

46. If Px

xdx Q

x dx

x=

+=

+

¥ ¥

ò ò0

2

4 0 41 1, and R

dx

x=

+

¥

ò0 4 1, then find the value of P Q R- +2 .

a.p2

b.p

2 2c.

p2

d.p

3 2

47. If f x ax bx cx a b c( ) , ,= + + -7 3 5 and are real constants and f ( )- =7 7, then the range of

f x( ) cos7 17+ is

a. [ , ]- 34 0 b. [ , ]0 34 c. [ , ]- 34 34 d. None of these

48. PQ is a chord of the parabola x x y2 2 12 11 0+ + - = through ( , )- -1 2 and a circle is drawn on PQ

as diameter, then the circle touches the

a. directrix of the parabola

b. tangent at the vertex of the parabola

c. latus rectum of the parabola

d. axis of the parabola

49. The solution of differential equation

xx

y

dy

dx

x

y

dy

dx

x

21

2 2

3 12

2

2

2

+ = + æèç

öø÷ +

æèç

öø÷

æèç

öø÷

+-

-

!

y

dy

dx

æèç

öø÷

æèç

öø÷

+¼

-3 3

3!, is

a. ( ) logxx

ey ce

22

33+ +æ

èç

ö

ø÷ = + b. ( ) logx

x

e

yce

22 2

33

2+ +æ

èç

ö

ø÷ = +

c. ( ) logxx

e

yce

22 2

33

2- -æ

èç

ö

ø÷ = + d. ( ) logx

x yce

22 2

33

2 2- +æ

èç

ö

ø÷ = +

50. If z a ib= + where a b> >0 0, , if b = 0, z is called purely real and if x = 0, z is called purely

imaginary, then

a. | | ( )z a b³ -1

2b. | | ( )z a b³ +1

2c. | | ( )z a b< +1

2d. None of these

51. a, b,c are three coplanar unit vectors such that a b c+ + = 0. If three vectors. p, q and r are parallel

to a b, and c respectively and have integral but different magnitudes, then | |p q r+ + can take avalue equal to

a. 3 b. 2 c. 1 d. 0

52. If f x y f x f y( ), ( ) ( )- and f x y( )+ are in AP for all. x y, and f ( ) ,0 0¹ then

a. ¢ = ¢ -f f( ) ( )2 2 b. f f( ) ( )2 2 0+ - =






one more than one

c. f f( ) ( )4 4= - d. ¢ + ¢ - =f f( ) ( )4 4 0

53. For any terms a a a1 2 3, , ¼ are in AP. Then1

1a,

1

2a,

1

3a… will be in HP. If a b c, , be three unequal

positive numbers in HP, then

a. a c b4 4 42+ > b. a c b5 5 52+ > c. a c b100 100 1002+ > d. a c b3 3 32+ >

54. If p q r, , be real, then f x

x p pq pr

pq x q qr

pr qr x r

( ) =+

++

2

2

2

, the value of determinants unchanged if

rows are changed into columns and columns are changed into rows.

a. increases in interval ( , )0 ¥

b. decreases in interval- + +æ

èç

ö

ø÷

2

30

2 2 2( ),

p q r

c. increases in interval - ¥ - + +æ

èç

ö

ø÷,

( )2

3

2 2 2p q r

d. None of the above

55. If cos a is a root of the equation 25 5 12 0 1 02x x x+ - = - < <, , then the value of sin 2a is

a.24

25b. - 12

25c. - 24

25d.

20

25

56. Let u be a vector on rectangular Cartesian system with angle 60° with x-axis. Suppose that | $ |u i-is geometric mean of | |u and | $ |u i- 2 where $i is the unit vector along x-axis. Then, the value of( ) | |2 1+ u is

57. The sum of the series 20130

20131

20132

20133C 8 C 15 C 22 C- + - +¼, upto 2014 terms is

58. If 0 4£ £arg / ,z p find the least value of 2 | |z i-

59. The coefficient of the quadratic equation ax a d x a d2 2 0+ + + + =( ) ( ) are consecutive terms of a

positively valued, increasing arithmetic progression. Then, the least integral value of d a/ , suchthat the equation has real solutions is

60. The line x c= cuts the triangle with vertices (0, 0), (1, 1) and (9, 1) into two regions. For the area

of two regions to be the same, c must be equal to

PRACTICE SET 1 13





0 9

1. (c) Idea The problem is based on conductingsphere when it is earthed it means that the netpotential on it is zero, it does not necessarilymeans that the charge on it must be zero.

⇒ Here net potential refers to the totalpotential due to all the charges present in thevicinity.

As the sphere is earthed the net potential on it mustbe zero. To make potential zero on the sphere someelectrons will flow from earth to sphere.

Q ( )V Anet = 0

⇒ ( ) ( )V VA r A R+ = 0

⇒ 14

14

00 0π ε π ε

q

r

Q

R+ =

[Potential due to Q on shell and on sphere will besame]

⇒ qQr

R= −

TEST Edge Different types of questions could beasked on the similar concept of earthing by justmaking different spheres or shells earthed or byincreasing the number of shells.

e g. . ,

For the above system of conducting sphere andshell if the outer sphere is earthed, then suppose ′qcharge flows from earth to the shell.

So, net charge on conducting sphere flow fromearth to the shell.

i e. ., ′ = − +q q Q( )

2. (b) Idea Here the problem is based on centre ofmass of a uniform semicircular pipe andconservation of total mechanical energy.

A half section of thin uniform pipe carryingmass m and radius r is released from rest.

∴ ∆PEmax = − −

mgR mg RR2π

= − +

mgR mgR mg

R2π

⇒ ∆PEmax =

mg

R2π

TEST Edge In every year, one question is asked inJEE Advanced/IIT. Students should concenrate onthis idea and relate with centre of mass, linearmomentum and collision. Suppose m be the massand R is the radius of a uniform semicircular wire.Take its centre as the origin, the line joining the endsas the x-axis and y-axis in the plane of wire as shownin figure.

As the wire is uniform, the mass per unit length of

the wire isM

Rπ.

The mass of the element is

dmM

RR d

Md= =

πθ

πθ( )

So, yM

ydmm

RM

dR= =

=∫∫

1 1 20

( sin )θπ

θπ

π

∴ The coordinates value of the centre of mass along y

- axis = 2R

π.

3. (a) Idea Here, the solution is based on percentageof fractional error.

i e. .,∆ ∆ ∆Z

Z

A

A

B

B× = ± +

×100 100

Fractional error in current is∆ ∆ ∆I

I

V

V

R

R= ± +

or = ± +

0.00010.2514

0.0010.1

= ± × +−[ . . ]3 977 10 0 014

= ± [ . ]0 0103977

AdvancedTest RIDER

Q

r

R

B

qA

+Q

r

R

+q

θ

y

dθ

R sinθx R cos θ

R R–2 /π

RCM

r

R

B+qQ+q′

A

Percentage of uncertainty in the measurement of

current i.e.,∆I

I× = ± ×100 0 0103977 100.

∆I

l× =100 1039. % ≈1.04%

TEST Edge This Idea is important according to JEEAdvanced. Students should focus on errors inmeasurement along with significant figures.

(i) In product If Z AB= , then maximum fractional

error is∆ ∆ ∆Z

Z

A

A

B

B= ± +

(ii) In division If Z A B= / , then maximum

fractional error∆ ∆ ∆Z

Z

A

A

B

B= ± +

≈ 104. %

4. (c) Idea This problem is based on the concept of arolling body trying to rolling up the inclineplane, ‘mg sin θ’ reduce the linear velocity and‘ fs’ will provide the torque to reduce theangular velocity. But for plane surface frictionis absent so the angular velocity of rollingbody will not change.

A ball rolls for rough surface, the linear and angularboth the velocities reduce and at the highest pointball will come at momentarily rest.

So,12

12

2 21mv T mgh+ =ω ...(i)

For plane surface, only the linear velocity of the ballwill become zero due to ‘mg sinθ’ at the highest point.

So,12

22mv mgh= ...(ii)

So, from Eqs. (i) and (ii), we geth h1 2>

TEST Edge This concept is important according toJEE Advanced. Students should focus on this ideaand relate with laws of motion and friction.

Note When the ball is placed on an inclined plane, then

Case I When a ball is rolling upward,

Case II When a ball is rolling downward,

5. (a) Idea This problem isbased on the concepton reflection of light.When magnitude ofspeed of three imagesis same then speed ofobject will be same.

Let say object moves with speed u image I1 moveswith same speed u towards mirror. Image I2 will movehorizontally with object with same speed in samedirection. As we know that during that imageformation all four (images and object) will lie on thecorners of a rectangle so like other three points, fourthpoint will have to move with same speed so speed ofimage I3 will be equal to object.

TEST Edge According to JEE Advanced, studentsshould focus on above idea and relate withrefraction of light and YDSE

Note

In the above case, the forming of locus point byjoining I I I O1 2 3, , and then it will be a square.

PRACTICE SET 1 15

Plane

surface

h2ω

v

ω O

I2 I3

I1v

fs

v

ω

mg

sin

θ

v v

v

I1 I3

I2O

rv

Rough

surface

h1ω

v

v

v

vv

I1

I2I3

O

fs

v

ω

mg

sin

θv

6. (c) Idea It is based on induced emf when magneticfield is perpendicular to the surface area ds,then,induced emf φ = °Bdscos 90

φ =zero

A rectangular loop of ( )a b× is rotated inmagnetic field about an axis passing throughthe mid-point of side a and parallel to side bwith angular speed ω, we get

Don’t be confused see clearly that the angle betweenthe area vector of loop and B will always remainconstant (90°).

⇒ Induced emf = Ba cos 90 (no change in flux)

⇒ Induced current = 0

TEST Edge Know exactly how emf is induced?

As the conducting rod is moving in the magneticfield, it’s free electrons are also moving in themagnetic field. There will be a magnetic force on theelectrons due to which they will drift towards theend ‘A’ and excess positive charge will remain atend ‘B’. Due to this an emf will induce in the

conducting rod.

7. (c) Idea The problem is based on potential energyand electric field between point charges. So,potential energy of the system i e. .,

U r Kq q

r( ) = 1 2 and electric field between

point charges E rK q

r( ) = 1

2

For potential energy to be minimum, the biggercharges should be farthest.

Suppose x be the distance of q to 2q. Then, potentialenergy of the system is shown in figure would be

U Kq q

x

q q

x

q q= +−

+

( )( ) ( )( )( )

( )( )2 89

8 29

Here, K = 14 0π ε

For U to be minimum2 8

9x x+

−should be minimum.

d

dx x x

2 89

0+−

=

∴ − +−

=2 89

02 2x x( )⇒ x = 3

i.e., distance of charge q from 2q should be 3 cm.

So, electric field at q

EK q K q=×

−×− −

( )( )

( )( )

23 10

86 102 2 2 2 = zero

TEST Edge This concept is important according toJEE Advanced. Every year one or two questions areasked in JEE Advanced/IIT. So, students shouldfocus on this concept and relate with equipotentialsurface and electric dipole moment.Such as

Suppose a point charge Q placed at a point A asshown in figure. We have to find the electricpotential at a point P where AP r= . Then, takeanother charge q is moved from r = ∞to the point P.

Then, electric potential energy of the system is

U UQq

rp − =∞

4 0π ε

8. (c) Idea The problem is based on concept ofequilibrium, net torque on the system must bezero. So, net torque due to mg and torque dueto centrifugal force must be zero.

For equilibrium of rigid body net torque acting onbody should be zero.Here only two forces, one is gravity force and otheris centrifugal force (if rod is considered as frame ofreference).Here notice that centrifugal force is variable becausedistance of mass from axis is variable that’s whyradius is variable and because of that centrifugalforce is variable.

Torque due to centrifugal force

= Torque due to weight


A B

Q Pr at r=∞

x 9–x2q q 8 q

v

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

F

e

F = e Bv

B

A

x

y

ωa

b

ω

α

r

( )dm g( = sin )r x α

dm rω2

x

( ) ( cos ) ( ) sindm r x dm g xω α α2 =ω α α α2 2( sin cos ) sinx gx=

cos ( )α ω20

20

L Lx dx g x dx∫ ∫=

cos α ω2 3 2

3 2L gL= ⇒ cos α

ω= 3

2 2g

L

TEST Edge Student should relate above concept withcentre of mass and rotational motion as a rod isfalling downwards or a plane surface as shown inthe figure. The position of end ‘A’ from point ‘O ’when it will completely fall down on the planesurface.

Here the simple point is that there is no force in thehorizontal direction, so horizontal position of CMwill not change.

9. (a) Idea The pure accelerated rolling is possibledue to friction (static) but as the sphere andring are attached then stress in the rod will beadjusted to keep the pure accelerated rolling inboth the rolling bodies.

In FBD arrangement, we have

F = force in agging.

So, force for ring,

mg f F masin θ − + =1 …(i)

f r mr12= α …(ii)

Force equation for spheremg f F masin θ − − =2 …(iii)

f r mr222

5= α …(iv)

[a r= α for pure rolling]Solve all equations

F mg= 317

sin θ

TEST Edge Rolling is a very complicated and vasttopic so different types of questions could be askedon it, let us consider more example.

e g. . ,When the particle hits the sphere and if it sticksto sphere, we can find the linear velocity ofsphere-particle system (as, particle will provide itonly linear velocity)

Then due to fk (backward direction), thesphere-particle system will start gaining theangular velocity. When the condition of pure rollingv R= ω is fullfilled then fk will stop acting.

10. (b) Idea The basis of the problem in frequencydependence of photoelectric emission. Theproblem based on emission of electron whenincident light with certain frequency (greaterthan on the threshold frequency) is focused ona metal surface then some electrons areemitted from the metal with substantial initialspeed.

Initial energy of electrons = 2 eV

Energy of electron in Ist excited state

( . .,i e n = 2) = − × = −13.6 3.4 eV12

2

2

Photon of energy = − − =2 ( 3.4) 5.4 eV will be emitted.

⇒ KE. 5.4Photonmax = − = −E W0124004600

= − =5.4 2.7 2.7 eV

TEST Edge The minimum energy is required toextract electron from the surface, they will have themaximum kinetic energy.

i e. . KEmax = −E W0

KEmax = −h Wν 0

This equation is known as Einstein’s photoelectricequation.

Note At a certain potential V0 having maximum kinetic

energy ( )Kmax also get stopped and current in the circuitbecomes zero. This is called stopping potential.

11. (a,b,c) Idea The work done due to variable force= ∫ Fdscosθi e. . , The work done due to a constant force= Fs [ ]∴ = °θ 0

Work done by a constant force is equal to= Force × (displacement in the direction of force)

So,W F RF1 1 2 20 2 6 120 2= × = × × = J

W F RF2 1 30 6 180= × = × = J

PRACTICE SET 1 17

N1

N2

aF

af1

f2

mg

mg

F

CM

O

A Bl

l

2

CM

O

hA

B

ω

l

θ

RM

CM

m → v

Sphere at rest

µ

F3 is a variable in direction only so work done in smalltangential displacement

W F ds F R d F R= = =∫ ∫3 3 3 2θ π( / )

= × × =15 6 3 45π π/ J

TEST Edge These type of problems are generallyasked in JEE Advanced/IIT. So, student shouldrelate this concept with potential energy and workenergy theorem.

For the process from A B A→ → , find thework done. AB s=

Don’t be confused with s = 0, here kinetic friction isvariable.

⇒ W f sA B k→ = −W f sB A k→ = −

W f s f sk knet = − − = −2 f sk

12. (a,c) Idea The basic idea is same as of simplependulum. It’s just there are two forces,one weight of vertical rod and otherweight of horizontal rod due to whichtorque will be produced.

Moment of inertia for AB rod about A is ml2 3/ and ofrod CD about A is

= + =Mlml

ml22

2

1213

12

Total moment of inertia is17

12

2ml

Restoring about A

τ θ θR mgl

mgl= +sin sin2

= 32mgl

sin θ

Small value of θ

τ θR

mgl=

32

TI

K= 2π

= ××

=217 212 3

21718

2

π πml

mgl

l

g

Tl

g= 2

1718

π

TEST Edge On SHM different types of questionscould be asked e g. . ,

A box is in SHM on a horizontal plane. Another boxis placed at the positive extreme. The oscillating boxhits the box which was at rest and sticks to it. Thenew amplitude of oscillation can be find out as.

When it hits the box B, then the mass at the systemwill become ‘2m’. So at the equilibrium position, theKE conservation could be applied.

1

2

1

220

2mv m v2 = ( ) ⇒ v v= 0 2/

But concentrate on the fact that it is the maximumspeed at the boxes that changed not the KE of thesystem. So, the amplitude will not change bymechanical energy conservation.

13. (b, c) Idea The idea is that by comparing the ‘usin θ’and ‘ucos θ’ of two projectile motions, we cansolve this question easily.

Slope at t = 0 gives initial vertical speed. So,dy

dt

dy

dtt t

<

= =1 0 2 0at at

u usin sinθ θ1 2<θ θ1 2< (angle of projection)

θ θ1 2<So, cos cosθ θ1 2>

u ucos cosθ θ1 2>∴ From graph we can see that

T T1 2>So, u T u Tcos cosθ θ1 1 2 2>

R R1 2> .


mg

mg

θ

m m

A Bv v0 max=

x = 0µ = 0

A DB

A

A B

m m

v v′ = 0

rough

A B

m m

vv′ = 0

rough

A B

m m

v

A B

m m

v

rough

fk

fk

s

s

TEST Edge These concepts are generally asked in JEEAdvanced/ IIT. In every year, one or two problemsare based on this idea. So, students shouldconcentrate on projectile motion along with circularmotion. Let us consider another example ofprojectile motion.

The average velocity of the projectile between thepoints A and B.

We know vr r

= =∆∆

∆t T

[Here, ∆t T= ]

For A B→ ∆ ∆ ∆r i + j= x y$ $

But here for A B→ , ∆ y = 0

So, there is no displacement along the y -axis.

So, v v= =( ) cosx u θ (as component of velocityalong x-axis is not changing.

14. (a,c) Idea It is based anelectric potential energyof a uniformly chargesphere. When thepotential on the surfaceof earth has to be zero, sosphere must get someelectrons from the earth to keep the potentialon it to be zero.

The potential of the groundedsphere has to be zero, due toparticle charge the potential ofthe sphere is positive so negativecharge must be possessed bysphere so that total potential ofsphere becomes zero. Thisnegative charge is acquired by sphere from theground, hence current flows into the ground. As thecharged particle comes nearer to sphere thepotential of sphere due to positive charge increasesand hence more electron per unit time will flow formground to sphere and thus magnitude of currentincreases.

TEST Edge According to JEE Advanced, this conceptis important and student should related withspherical charge distribution and earthing acondcutor. Let us consider an example based on thesame concept.

In the above system, if the shell is earthed, then thechange in the potential of point A is can be findout as

Here is the condition after earthing.

i e. . , ( )∆VKQ

RA =

15. (b,d) Idea The problem is based on the reflection oflight at smooth surface.

(i) The angle of incidence is equal to the angleof reflection.

(ii) The incident ray, the reflected ray and thenormal to the reflecting surface arecoplanar.

For reflection at general point A

2 90i = °So, i = °45

So, dy dx/ at A should be 45°

yL x

L=

2π

πsin

dy

dx

L x

L L

x

L=

× =

22

ππ π π

cos cos

1 2=

cos

x

L

π

π πx

L=

3⇒ x L y

L= =/ ,33π

and another point will be the just opposite point so, its

coordinates will be 2 33

LL

/ ,π

TEST Edge In every year, one question asked in JEEAdvanced/IIT. So, students should focus on thisidea and relate with snell’s law and refractiointhrough a glass slab. According to the basic law ofreflection are same for plane curve surface.A normal can be drawn from any point of thecurved by first drawing the tangent plane from thatpoint and then drawing the line perpendicular tothat plane.

PRACTICE SET 1 19

r

R

+qA

B+QConducting

sphere

Conducting

shell

R

e–

v q

e–

q

r

+q

–q

E

Eout = 0

Charge –( + ) is

flown to the shell

Q q

θA B

u

( , )x yi

i

i

y

x

(i) angle of incidence andreflection are defined fromthis normal as shown infigure.

(ii) the angle incident( )i is equalto the angle of reflection ( )r .The incident ray, the normaland the reflected ray are inthe same plane.

16. (1) Idea The problem is based on thermal

conductivity of the material.

i e. .,∆∆Q

tK

A T T

x

T T

x KA=

−=

−( )

/1 2 1 2

where K is constant for the material of the slab,and x KA/ is a thermal resistance. The idea isthat by comparing the power produced byheater and rate of heat flow through window.

Power produced by heater = rate of heat flownthrough window.

∴ V

R

Q2 20= = −Temperature differenceThermal resistance

( )/l KA

⇒ QV l

RKA= −20

2

Substituting the values, we get

QA

= − × ×× × ×

−20

200 0 2 100 2 4 2 20

2 2( ) ( . ). .

⇒ 15 24 20200 0 2 100 2 4 2 20

2 2

.( ) ( . )( . . )

= − × ×× × ×

−

A

⇒ A = ≈10004 12 2. m m

TEST Edge According to JEE Advanced, studentsshould relate this idea with heat transfer throughconduction, convection and radiation of a body. Ifthe area of cross-section is not uniform or if thesteady-state conditions are not reached, theequation can only be applied to an thin layer ofmaterial perpendicular to the heat flow.

If A be area of cross-section, dx be small thicknessand dT be temperature difference across the layer,then heat current through this cross-section is

∆∆Q

tKA

dT

dx= −

17. (2) Idea It is based on Bernoulli’s equation forflowing liquid for steady or streamline flow.

i e. ., p v gh+ + =1

22ρ ρ constant

where p is pressure energy,1

22ρv is kinetic

energy per unit volume andρgh is the potentialenergy per unit volume.

Apply Bernoulli’s equations at pipe (1), (2) and (3)

p v p v p v3 32

2 22

1 121

212

2+ = + = +ρ ρ ρ/

p gh v p gh v0 3 32

0 2 221

2 2+ + = + +ρ ρ ρ ρ

= + +p gh v0 1 12

2ρ ρ

ghv

ghv

ghv

332

222

112

2 2 2+ = + = + …(i)

Equations of continuity at (3) and (2) are

v A v A3 3 2 2=v A v3 2 2× = × …(ii)

v v2 12 1× = × …(iii)

From Eq. (i) , we get

ghv

ghv

222

112

2 2+ = +

1035

100 210

20100 2

22

12

× + = × +v v

1510 2

42

32

12

22

22

22

22

= − = − =v v v v v

v2 1= m/s, v1 2= m/s

From Eq. (i), we get

ghv

ghv

332

222

2 2+ = +

= × + = × +102

1035

10012

3237.5

100v

v32

212

3510

12

12

25100

= + − = − = −37.510

2.510

v32

214

= ⇒ v32 1

2= ⇒ v3

12

=

X = 2

TEST Edge According to the property of fluids, if thespeed of the fluid is zero everywhere, we get thesituation of hydrostatics. Putting v v1 2 0= = in theBernoulli’s equation, we get

p p g h h1 2 2 1− = −ρ ( )

18. (1) Idea The problem is based on the conservationof mechanical energy in elastic collision. Firstthe collision occured and then the spring startscompressing will reduce the velocity of 1st

block and increase the velocity of 2nd block

untill the velocities at both the blocks becomesame and it will be the condition of maximumstretch.

As the collision is elastic the first block will transfer itsspeed completly to second block. So, initial sceneriowill be like this


r

i

y

x

mmm

vv′=0

⇒ At maximum compression

Let the maximum compression is‘x’.

⇒ So, by mechanical energy conservation

⇒ 12

212

12

202 2mv mv kx=

+

⇒ mv mv kx202 22= +

⇒ mv kx mv2 2022− =

⇒ vmv kx

m0

2 2

2= − = × − ×

×1 4 2 1

2 1

v0 1= m/s

TEST Edge It is important according to JEEAdvanced. So, students should focus on collisionand try to relate with potential energy of a spring. Ifthe collision would be inelastic then you have tofind the velocity at 2nd block then apply the sameconcept of maximum compression.

19. (2) Idea This problem is based on Moseley’sobservation can be mathematically expressedas

ν = −a Z b( )

where ν is frequency, Z is atomic number anda b, are constants. This relation is known asMoseley’s law.

According to Moseley’s law Kα X-rays emission

ν = −a Z( )1

⇒ ( )Z − ∝1 2 ν or ( )Z − ∝112

λ

⇒ ( 1)( 1)

Mo2

Cu2

Cu

Mo

Z

Z

−−

= λλ

⇒ λ λCu MoMo

Cu

0.71 1.52 Å= −−

= ×

=( )

( )Z

Z

11

4128

2

2

2

[ ]λ = 2 Å

TEST Edge According to Moseley, that there must bea fundamental property of the atom whichincreased by regular steps as one moves from oneelement to the other, there is increase in the numberof proton in the nucleus and was referred as atomicnumber as shown in figure.

20. (4) Idea This problem is based on law ofradioactivity decay law i e. . , mean life. Theaverage or mean life ( )tav is the reciprocal ofthe decay constant ( )λ .

i e. ., tav = 1

λLet the element be X

X Y→ +λ α

α ⇒ X Z→ +λ β β

When a substance decays by α and β-emissionssimultaneously, the average rate disintegrationλ overall is given by

λ λ λα βoverall = +where λ α = disintegration constant for α-emissiononly and λ β = disintegration constant for β-emissiononly.

Mean life is given by τλ

= 1

⇒ λ λ λτ τ τα β

α βoverall

overall= + = = +1 1 1

( )

= + = × − −11600

1400

10 3 13.125 year

λ overall 102.303 logtN

Nt

= 0

⇒ (3.125 2.303× =−1010025

310) logt

⇒ t = ×× −2.303

3.1251

1043 10log = 443.5 year

Number of significant digit = 4

TEST Edge This problem is deal with activity of aradioactive susbtance and relate with equivalenceof mass, energy and binding energy per nucleon.According to the radioactive decay law, when thereis a nuclei in a sample, the rate of decay isproportional to the number of a nuclei.

i e. ., −

∝dN

dtN

where λ is the decay constant.

or lnN

Nt

0

= − λ

At t = 0, then N0 is the initial number of parentnuclei, then number of nuclei survive at time t ,

i e. ., N N e t= −0

λ

This function is plotted in figure.

PRACTICE SET 1 21

1

2

3

10 20 30 40 50

Atomic number ( )Z

ν(in

10

Hz

)9

1/2

mm

v0 v0 (The commonspeed at maximumcompression)

0.5 N0

N0

0.37 N0

t1/2 t1/4t

N


21. (d) Idea This problem is based on chemicalproperties of aromatic nitro compoundtowards Friedal Craft reaction and reductionof aromatic nitro compound. While solvingthis problem, students are advised to use boththe concepts notified and suggested to becareful while solving such problems.

As nitrobenzene does not give Friedal Craft reactionhence reaction medium contain mixture ofnitrobenzene and alkyl halide and when this mixtureundergo reduction with Sn/HCl, it produces aniline.The reaction sequence is as follows

Difficulty of this question is due to reactivity ofnitrobenzene towards Friedal Craft alkylation.

TEST Edge Similar problem based on chemicalreactivity of phenol, aromatic aldehyde etc., canalso be asked with conceptual mixing ofcomparative study of basicity or a acidity ofproduct, so students are advised to go throughthese topics as well.

22. (b) Idea This problem includes concept of propertiesof boron and compounds of boron.

To solve this problem, students must haveknowledge of chemical and physicalproperties of boron compounds as well as usesof them.

Boron show diagonal relation with aluminium due tosmall size and similar e/r ratio while all otherstatements are correct.

TEST Edge Similar types of problems based onchemical properties and uses of silicon and theircompounds such as silicones, silicates can be askedin JEE Advanced so students must have in depthknowledge of properties and nature bonding andstructure of silicones and silicates. Sometimes,questions from this segments are asked directly ongeneral formulae and net charge calculation ofvarious silicates hence students must be familiarwith these topics. A ready reference betweenformulae & charges of silicates is given below

NO2

+ CH Cl3

anhyd AlCl3

NO2

+ CH Cl3

Sn/HCl

Mixture

NH2

P

Q

Types of

Silicate

Structural

Formula

Number of

Oxygen Atom(s)

Shared

Net charge Example of Minerals

Island(Orthosilicate)

(SiO4)4− 0 Si = + 4

O = − 8

Net 4= −

Fenacite (Be SiO )2 4 Zircon (ZrSiO )4 ,Forestrite (Mg SiO )2 4

Island(Pyrosilicate)

(Si2O7)6− 1 Si = + 8

O = −14

Net 6= −

HemimorphiteZn Si O (OH)4 2 7 2Thorteveitite,Se (Si O )2 2 7

Single chain (SiO3)2− 2 Si 4= +

O 6= −

Net 2= −

Enstatite, MgSiO3 Wollastonite,Ca (Si O )3 3 9

Ring (SiO3)2− 2 Si 4= +

O 6= −

Net 2= −

Beryl, Be Al (SiO )3 2 3 6

Double chain (Si4O11)6−

2 12

Si 16= +

O 22= −

Net 6= −

Tremolite (Asbestos), Ca Mg (OH)2 5 2

(Si O )4 11 2

Sheet silicate (Si2O5)2− 3 Si 8= +

O 10= −

Net 2= −

Muscovite (Mica), KAl2(OH) Si AlO2 3 10

Three dimensional (SiO2)0 4 Si 4= +

O 4= −

Net 0=

Quartz, SiO2

23. (a) Idea This problem is based on concept ofsolubility product and equilibrium constant.To solve this problem, students are advised tocalculate Ksp and correlate this with totalconcentration of cation in order to calculate %of original cation.

K x12 2= +( ) ( )Pb ⇒ ( )Pb2 1

2+ =R

K

x

% ( )/

Pb2 +R

K x

y

K

x y= × = ×1

21

2100100

TEST Edge Similar problems based on conceptualmixing of solubility product and equilibriumconstant along with standard electrode potential ofelectrochemical shell can be asked in JEE Advancedsuch as for a spontaneous reaction occurring at 25°CAg|AgCl ; KCl(0.2M)|(s) KBR M( . )0 001 . AgBr(s)|Ag(s), calculate emf generated.

Given, Ksp AgCl 2.8( ) = × −8 10 10

Ksp AgBr( ) .= × −3 3 10 13 after solving this question,you will get answer E = 0 037. V.

Use formula

E En

= ° − 0 059.log

Products

Reactants

24. (c) Idea This problem is based on concept ofMarkonikov’s addition reaction of HCl toalkene and E2 elimination. Keep in mind theconcept of above two concepts while solvingthe problem.

Me

Me

Me

MeHCl ClC CH C CH== → →

+ −

2 3

Me

MeCl

MeKCN

2C

→ ==E

Here, presence of peroxide ion in step 1 will not affectthe product.

TEST Edge Similar problem including concept ofmarkonikov’s type addition of water to alkene canalso be asked in JEE Advanced so students areadvised to go through these topics problem such as

What will be the product of given reaction?

You can get answer as

Here, note that above addition is markonikov’s ofwhich looks like anti-markonikov’s due to thepresence of H− in the reagent.

25. (b) Idea This problem is based on concept ofkinetics of parallel reaction which can besolved by using the expression of rate constantobtained for parallel reaction. While solvingthis problem, students are suggested to becareful that reaction concentration isdistributed towards two pathway parallely.

k k1 2=

⇒ 23

rd of A has reaction for [ ] [ ] [ ]A C D= =

∴ k kt

A

A1 2

1+ = ln[ ]0

013

[ ]

⇒ tk k k

=+

=13

12

31 2 1

ln ln

= 12

32k

ln

TEST Edge In JEE Advanced, these types of problemare asked to judge the knowledge of studentsregarding kinetic study of chemical reaction andmathematics involved in calculation. Similarproblems based on half-life, quarter life, can also beasked.

26. (d) Idea The acidic character of oxy acids ofphosphorus and halogen can be explained onthe basis of polarity of O—H bond, which isdependent on number of P O== and otherelements P—OH bond. While in the case ofoxo acids of other elements, it depends onnumber of O atom.

Let us see the number of P O== and P—OH bond

Number of P O== bond1 1 1

Number of P—O H bond3 2 1

Polarity of OH bond is minimum when number ofP—OH bond is maximum because the effect ofP O== bond is distributed among all the threeP OHbond.

We may say as the number of P—OH bond increasesthe acidity of species decreases. The correct order ofacidic strength is

H PO < H PO < H PO3 4 3 3 3 2

In case of oxy acids of chlorine as number of oxygenatom increases, number of Cl = O bond increasesfinally acidic order increases as

PRACTICE SET 1 23

O

P

O O O

H H H

O O

P P

O OOH

H H

H HH

(i) BH(ii) H O

3

2 2

OHH

TEST Edge Similar trend related problems regardingoxidising capacity of oxyacids of chlorine, sulphuretc., can also be asked in JEE Advanced. So,students are advised to go through understandingof oxidising, reducing capacity of oxyacids ofchlorine and sulphur, questions including conceptof equivalent weight of oxoacids of sulphur,chlorine can also be asked.

27. (a) Idea This problem is based on concept ofmechanism involved in Balz Schiemannreaction. To solve the problem, students musthave a clear idea about mechanism ofconversion of diazonium salt tofluorobenzene. i.e., Balz Schiemann reaction.

Chemical conversion of benzene diazonium chlorideto fluorobenzene is known as Balz Schiemannprocess. This conversion can be made by heatingHBF4 with benzene diazonium chloride.

Mechanism

TEST Edge Similar problem based on conceptualmixing of chemical conversion, name of particularreaction and intermediate involved in givenconversion can also be asked in JEE Advanced.Students are advised to read any of chemicalreaction with proper mechanism of reaction.

What is the name of reaction which involves conversionof benzene diazonium salt to chlorobenzene and what isthe intermediate involved in the reaction?

After undergoing proper mechanism of aboveconversion one can get the answer as

Intermediate = free radical

Reagent = CuCl/HCl

Name = Sandmeyer process

28. (a) Idea This question is based on occurrence ofelement as sulphide ore so students areadvised to solve this problem by usingconcept of stability of sulphide of metals.

Ag occurs as Ag S2

Hg occurs as HgS (cinnabar)Pb occurs as PbS (galena)Fe occurs as FeS2 (iron pyrite)

TEST Edge Problems based on concept ofconcentration, reduction and roasting of ore canalso be asked so students are advised to go throughclear understanding of each term.

29. (a) Idea This problem is based on conceptualmixing of stereochemistry of reaction andaddition of D2 to alkene. While solving thesetypes of problem, students must be clear aboutstereochemistry of addition of dueterium toalkene.

Cis alkene + →°

D2(Syn adddition 250 C

Ni

)Meso compound.

TEST Edge The problem based on stereochemicalapproach of reagent and mechanism of productformation can also be asked in JEE Advanced.Students are advised to go through study ofstereospecific and stereoselective reaction. To solvethese types of problems such as

What will be the product when trans 2 butene undergoreaction with mCPBA followed by Br2?

By undergoing proper mechanistic step one can getanswer as trans 2-hydroxy 2-bromobutane useconcept of epoxide ring formation & ring openingfrom least hindered side.

Here, the concept of CAR can be very useful i e. . , forsteroselective reactions

cis alkene + anti addition → Recemic Mixture

trans alkene + anti addition → Meso compound

cis alkene + syn addition → Racemic mixture


Balz Schiemann process

N N.Cl≡≡ FHBF /4 ∆

r s

N NCl≡≡HBF4

∆

r s

+ BF + HCl + N3 2

F

+

F–

Syn additionMe

H+ D2

Me

Hcis

Me

Me

D

DHH

Meso

H C3

CH3mCPBA

CH3

CH3

OBr–Hδ– δ+

CH3

CH3

Br

Or s H+

H C3

CH3

Br

OH

Cl

OHO

Cl

O

OHO

Cl

O

O

OHO

Cl

HO

No. of O—H bond

No.of O atom

No. of Cl==O bond

1

1

0

1

2

1

1

3

2

1

4

3

Polarity of O—H increases

30. (b) Idea This problem involves reagent used inbrown ring test and magnetic moment ofbrown ring complex. While solving theproblem, students are advised to complete thechemical reaction involved in formation ofbrown ring complex then determination ofmagnetic moment of complex formed.

Brown ring test When the freshly prepared FeSO4solution is added to the solution containing nitrate ionalong with the conc. H SO2 4 added by the side of testtube, a brown ring of [Fe (H O) NO]SO2 5 4 is generatedbetween the junction of two liquids.

NO + FeSO + H SO3 4 2 4− →

3Fe (SO ) + 2NO + 4 H O4 3 2

[Fe (H O) ]SO + NO2 6 4 →[Fe (H O) NO] SO2 5

2 +4

Brown ring complex

2 −

Fe has +1 oxidation state due to addition of singleelectron from NO to Fe2 +.

[Fe(H O) ] SO [Fe (H O) NO]SO2 6 4

NO

2 5 4→

Fe + Fe2 + +es →

n = 3

µ = +n n( )2 BM = +3 3 2( ) BM

= =15 3.87 BM

TEST Edge Similar problems based on conceptualmixing of magnetic moment of coordinationcompounds and intensity of colour produces bydifferent coordination compound by absorption oflight can also be asked in JEE Advanced so studentsare advised to undergo study of properties andanalytical use of coordination compound, stabilityof coordination compound in detail to tacklequestion like

Which of the following has lowest negative value ofCFSE?

[Fe(OH)4]2−or [Fe(H O)622] +

Students can solve this question using aboverelations ∆ ∆t = 0.45 0

31. (c,d) Idea This problem involves chemical propertiesof sodium biphosphate and sodiumbicarbonate with NaOH.

The reaction in NaOH with NaHCO3 & NaH2PO4 is asfollowsNaHCO + NaOH Na CO + H O3 2 3 2→

NaH PO + NaOH Na PO + H O2 4 3 4 2→Since both of them form salt so they cannot existtogether.

TEST Edge Problems based on concept ofhydroxides, bicarbonates, carbonates, chloride,sulphates of sodium, potassium, magnesium and

calcium can also be asked. So, students are advisedto study the preparation, physical and chemicalproperties of these compounds which can be askedin JEE Advanced.

What is the difference between molecular mass ofplaster of Paris and dead burnt? After calculatingyou will get answer as 9.

32. (a,c) Idea While solving these types of problemswhich is based on packing fraction, studentsare advised to relate packing fraction withstability, binding energy and binding energyper nucleon.

Packing fraction

= − ×Actual isotopic mass mass number

Mass number104

As we know from characteristics of packing fractionvalue that lower the packing fraction, greater is thepacking fraction per nucleon, and hence greater isthe stability. Hence, options a and c are correct

TEST Edge Similar types of problems includingconcept of kinetics of radioactive decay, carbondating, nuclear fusion, nuclear fission can also beasked so students are advised to go through studyof these topics. One of such problems is

What will be the value of α activity in terms of dpm of

0.001 g sample of Pu239?

( /+ =1 2 24300 yrsand 1 107yr 3.15 seconds= × )

After solving the question by using formulae

λ = 0 693

1 2

.

/t, where λ = disintegration constant and

NdN

dtλ =

−one can get answer

= ×1.43 108 dpm

33. (b,d) Idea This problem is based on concept of test ofaldehyde and characteristics of benzoincondensation.

While solving this problem, students musthave knowledge of benzoin condensationand test to distinguish aldehyde andketones.

Aromatic aldehyde does not reduce Fehling solution.Aromatic aldehyde form benzoin with KCN.

TEST Edge In JEE Advanced, these types ofproblems are asked to judge the knowledge ofstudents regarding chemical characteristics ofaldehyde and ketone.

34. (a,b) Idea This problem includes concept of d pπ π−bonding, bent rule, structure of oxyacids ofsulphur, oxyacids of nitrogen. So students areadvised to go through structural characteristicof these compounds.

PRACTICE SET 1 25

Hybridisation of N(SiH )3 is sp2 due to back bonding.

TEST Edge Problems based on structure,hybridisation, bond angles, stability of inorganiccompounds usually asked in JEE Advanced

In [ ]ICN2− , how many atoms are in plane and how

many atoms are sphybridised ?

After drawing structure using concept of VSFPRone can get answer as

Atoms in plane = 5

atoms having sp hybridised = 4

35. (b,c) Idea This problem involves various chemicalproperties of phenoles. These types ofproblem can be solved by completing allchemical conversions and then comparingacidic strength of product with reactant.

(i)

Salicylic acid is a carboxylic acid which is a strongeracid than phenol.

(ii) In picric acid three strong –M group increases acidicstrength of phenol.

TEST Edge Problems based on concept of acidicstrength of substituted carboxylic acid, basicstrength of substituted amine can also be asked inJEE Advanced so students must be careful during

solving such problems. Students are advised to useconcept of ortho effect also like

Which of the following has highest value of basicity?

After comparing , one can get (d) as a answer.

36. (6) Idea This problem is based on structure andisomerism of coordination compound whilesolving the problem students are advised todraw the structure of coordination compoundthen draw the possible isomers ofcoordination compound. Students are advisedto keep in mind the concept of geometrical andoptical isomerism.

Every optically active isomer must have amirror image isomer. i.e., enantiomer ofoptically active isomer.

The two more optical isomers are mirror image ofoptically active isomer i.e., enantiomeric form ofoptical isomer.

Due to centre of symmetry. Total number of spaceisomers = 6.

TEST Edge Problems based on calculation ofnumbers of stereoisomers; geometrical isomers,optical isomers etc., can also be asked in JEE


NO

ONo N-N bond

NO

O

ON

O

Oone N-N bond

NO

O

S

Structure of H S O2 2 7

SOH

O

O

OO

HOO

No S-S bond

O

104°H H

O

92°S S

N C≡≡ Isp sp

C N≡≡sp sps

OH

COOH

NO2

NH2

(a)

(c)

(b)

(d)

NH2

NO2

NO2

O

N N

C

O C

N

–O

N

O–

C

CPd+II

MeH

O

Me

H

cis optically inactive

C

O C

N

–O

N

O–

C

CPd+II

MeH

O

Me

H

trans optically active

C

O C

N

–O

O–

N

C

CPd+II

MeH

OMe

H

cis optically inactive

trans optically inactivedue to centre of symmetry

C

O C

N

–O N–

C

CPd+II

Me

OMe

H

H2 H2

H2

H2

H2

H2

O–

H

H2

H2

N(CH ) Si3 3

Si(CH )3 3

Si (CH )3 3

sp2 hybridisation

Advanced so students are advised to go through indepth study of number of possible isomers. vantHoff factor of coordination compound,conductivity of coordination compound can also beasked hence students are suggested to go through ofthese types of problems.

What is the value of vant Hoff factor for brown ring

complex

One can get answer as 3.

37. (160) Idea This problem is based on concept of idealgas equation and their interconversion intopd vs p equation. To solve the problem,students must apply the creative idearegarding relation between p and d.

As we know

pV nRT=

⇒ pn

VRT=

⇒ p

w

MRT

V=

∴ nw

M=

⇒ pw RT

V M

w

V

RT

M= = ×.

.

⇒ p dRT

M= ⋅ Q

w

Vd=

⇒ dpM

RT=

Multiply both sides by p

⇒ pd ppM

RT

p M

RT= × =

2

…(i)

On differentiating Eq. (i) w.r.t. p, we will get

d p d

dp

d

dp

p M

RT

pM

RT

( . ) =

=

2 2

52= pM

RT

TpM

R= = × ×

×25

2 8 21 45 0 0821

..

[for He,M =4]

T =160 K

TEST Edge Problem based on various possiblerepresentation of ideal gas equation, gas laws etc;can be asked in JEE Advanced, Hence, it isadvisable to learn the equation and theirrepresentation in various form.

38. (5) Idea This problem can be solved by using theconcept of acidic strength of organiccompound and reactivity of the organiccompounds with base.

All those acids which are stronger acid than H CO2 3gives the NaHCO3 test.

These 5 organic compounds will give NaHCO3 test.

TEST Edge Similar problems based on relativeacidic/basic strength of organic compound can alsobe asked so students are advised to go throughstudy of factors affecting acidic and basic strengthof organic compound. Such as I effect, M effect, Reffect, hyperconjugation etc.

39. (3) Idea Problem is based on concept of first law ofthermodynamics. Students are advised towrite equation derived from first law ofthermodynamics then calculate value of molarheat capacity (c).

dQ dE dW= + −( )

Q nC dTQ

V= − −

2

n TQ

CV

∆ =2

CQ

n T=

∆

= =Q

Q CC

VV/ 2

2

C R R= × =232

3

TEST Edge Similar problem based on conceptualmixing of work done in various thermodynamicprocesses such as adiabatic, isochoric, isobaric,isothermal and internal energy can also be asked inJEE Advanced, student must go through study ofcriterion of spontaneity and entropy of system justlike

Calculate the enthalpy change of process when a sampleof argon gas at 1 atom pressure and 27°C expands

reversibly and adiabatically from 1.25dm 3 to 2.50 dm 3 .

After solving the question by using ideal gas

equation, and temperature volume relation for

adiabatic expansion.

i.e.,T

T

V

V1

2

2

1

1

=

−γ

and ∆ H C dTnp= . One can get

answer = −117.65

PRACTICE SET 1 27

NO2

NO2

OH

NO2 CH3

SO H3

OH

SO H3

CH3

COOH CH COOH2

40. (2) Idea This problem is based on concept ofmolecular orbital theory and molecular orbitalelectronic configuration of diatomicmolecules. Students are advised to writemolecular orbital electronic configuration ofN2 and NO so that they will able to comparetheir bond order after removing one electronfrom NO.

N2 7 7 14( )+ =Molecular orbital electronic configuration of N2

σ σ σ σ1 1 2 22 2 2 2s s s s∗ ∗ π π σ2 2 22 2 2p p px y z≡

Bond order = − =10 42

3

O2 ( )8 8 16+ =Molecular orbital electronic configuration of O2

σ σ σ σ σ1 1 2 2 22 2 2 2 2s s s s pz

∗ ∗ π π2 22 2p px y≡π π∗ ∗≡2 22 2p px y

Bond order =10 6

2− = 2

This is only possible if 2 electrons are removed from πantibonding molecular orbital.The Molecular orbital electronic configuration of O2

2+

becomes

σ σ σ σ σ1 1 2 2 22 2 2 2 2s s s s pz

∗ ∗ π π2 22 2p px y≡

Bond order = − =10 42

3

TEST Edge Similar problems based on conceptualmixing of molecular orbital theory, reducing,oxidising, nature of homonuclear as well asheteronuclear diatomic molecule can also be askedso students are advised to go through in depthstudy of information provided by molecular orbitalelectronic configuration of homonuclear andheteronuclear diatomic molecules.

41. (a) Idea Here, we use to the concepts of progressionAP and GP to find the value of a and b in termsof c which substitute in the trigonometric

relation cos Ab c a

bc= + −2 2 2

2to obtain the

result. If cos A > 0, then angle is acute and ifcos A < 0, then angle is obtuse.

We have b ac2 = …(i)

( , , )Qa b c are in GP

2 2 3 2(log log ) log logb c a b− = −+ −log log ( )3c a AP

⇒ 2 2 2 3 3 2log log log logb c c b− = −⇒ 3 2 3 3log logb c=⇒ 2 3b c= …(ii)From Eqs. (i) and (ii), we get

c a= 4 9/

Now, cos Ab c a

bc= + −2 2 2

2

=

+ −

× ×= − <

32

94

232

2948

0

22

2c

cc

c c

.

A is obtuse.

TEST Edge In JEE Advanced, the questions based onrelation of the terms of AP, GP and HP and the sumof n terms are asked. So, to solve such types ofquestions you should learn concepts of AP, GP andHP and also acquainted yourself with properties oftrigonometry, algebra, logarithm, etc. e.g., if x3, y3,z3 are in AP and log x y, log z x, log y z in GP alsoxyz = 64, then use the above concept to getx y z= = = 4.

42. (b) Idea Here, we use the concept of greatest integerfunction i e. . ,[ ]x n= if n x n≤ < + 1to get integerless than or equal to the number from thegiven equation and find the values of x whichsatisfies the equation.

Given equation isx x x

x2 3 5

3130

+

+

= …(i)

Now,x x x x x x

x2 3 5

15 10 630

3130

+ + = + + = …(ii)

Also LHS of given Eq. (i) is integer∴ RHS is an integer

As evident from Eq. (ii) sum ofx x x

2 3 5+ + is equal to

3130

x .

∴ x x x

2 3 5, , are all integers and

3130

x is also an integer

∴ x = multiple of 2, 3, 5 and 30

Thus, x = × × … ×30 1 30 2 30 33, , ,

TEST Edge In JEE Advanced, question based on

different type of functions such as smallest integer,

modulus, even function, etc., are asked in terms of

equation. So, to solve such types of question,

acquainted yourself with the definition of different

types of function and utilise it to solve the equation

e g. ., if f x x x( ) cos[ ] cos[ ]= + −π π2 2 , then to evaluate

fπ2

we utilise the concept of greatest integer

function to obtain [ ]π2 9= and [ ]− = −π2 10 and get

fπ2

1

= − .

43. (a) Idea Here, we use the concept of arrangements

to obtain different observations or results

whose sum can be obtained by the sum of n

terms of GP and equate it with the given sum

to get the value of n.


Given 2n i− students gave wrong answer to atleast i

questions.2 1n − students gave wrong answer to atleast 1questions.2 2n − students gave wrong answer to atleast 2questions.2 21 2n n− −− students gave wrong answer to exactlyone equation.

( ) ( )exactly 1 wrong exactly 2 wrong+ + …+ =( )exactly wrongn 2047

1 2 2 2 2 21 2 2 3( ) [ ]n n n n− − − −− + − + …+ − − + =( ) [ ] ( )n n1 2 2 1 20471 0

1 2 20471 2 3⋅ + + + … + ° =− − −n n n1.2 1.2 1.2

⇒ 22 1

2047 2 1 20471n

n−

−= ⇒ − =

⇒ 2 2048n =2 211n = ⇒ n =11

TEST Edge The question based on fundamentaltheorem of arrangements are asked. So, to solvesuch types of questions we need to deduce variousobservations whose sum of n terms are in AP, GPand HP. So, students acquainted themselves withthe concepts of arrangement and sum of series to getthe desired result e g. ., the value of the sum

∑ +=

+

n

n ni i1

131( ) , where i = −1can be obtain by the sum

of GP and answer will get i e. . , i − 1.

44. (c) IdeaThe number of mapping of a function

having n elements is nn and its one-one

mappings is n!. According to the definition of

probability P( )Favourable outomes

Total outcomesX = .

We get the probability that the mapping isone-one and equate it with given probabilityto get the value of n.

Total number of mappings = nn

We also know, number of one-one mappings= − − … =n n n n( ) ( ) , . . . !1 2 3 2 1

∴ Required probability = n

nn

!

⇒ 332

= n

nn

!

⇒ 325 = n

nn

! ⇒ 2428 = n

nn

!

⇒ 444

! != n

nn⇒ n = 4

TEST Edge There are various types of mappingone-one onto, into, etc., of functions are asked. So, tosolve such types of questions, students are advisedto learn various types of function mappings andformula to get number of mappings and basicsconcept of probability to get the desired result e g. ., IfA is non-empty set in which number of elements isn, then

(i) Number of functions from A to A nn=(ii) Number of one-one function = n!

(iii) Number of onto function = ∑ −=

−

r

nn r n

rnC r

11( )

45. (c) Idea To solve this question, we first write theseries in terms of binomial coefficients of( )1+ x n and using the formula of sum of n

terms of GP i.e., S ar

rn

n

= −−

1

1. [ ]∴ >r 1

Now,n

m

n

m

n

m

m

m

+−

+−

+ … +

1 2can be written

as nm

nm

nm

mmC C C C+ + + … +− −1 2

We know nrC = coefficient of x r in ( )1 + x n

∴ nm

nm

mmC C C+ + … +−1

= coefficient of xm in ( )1 + x n + coefficient of xm in

( )1 1+ + … +−x n coefficient of xm in ( )1 + x m

= coefficient of xm in

[( ) ( ) ( ) ]1 1 11+ + + + … + +−x x xn n m

= coefficient of xm in ( )1+ x m ( )1 11 1

1+ −+ −

− +x

x

n m

[using sum of n terms of GP]

= coefficient of xm+1 in [( ) ( ) ]1 11+ − ++x xn m

= coefficient of xm+1 in ( )1 1+ +x n

= ++

nmC1 1

TEST Edge In JEE Advanced, questions based onbinomial series and properties of binomialcoefficient and sum of nterms of AP, GP and HP areasked. So, students are advised to learn importantterms in binomial expansion such as general term,middle term, etc., and properties of binomialcoefficients such as 2

02

12

22 2C C C C Cn

nn+ + +…+ = ,

etc. and formula to find sum of n terms of AP, GPand HP. e.g., a, a d+ , a d+ 2 , … are in AP then

Sn

a n dn = + −2

2 1[ ( ) ].

46. (b) Idea We obtain the integral in the form ofpx q

ax bx cdx

++ +∫ 2 on substituting the value of P,

Q , R in P Q R− +2 , express ax bx c2 + + as

sum or difference of two squares and use

substitution to convert it to standard integral

form then apply limit to get desired result.

PRACTICE SET 1 29

123

n

123

n

P Q Rx

xdx

x dx

x− + =

+−

+∞ ∞

∫ ∫21

210

2

4 0 4

++

∞

∫0 4 1dx

x

⇒0

2

42 1

1

∞

∫− +

+x x

xdx

⇒0

2

4 0 411

21

∞ ∞

∫ ∫++

−+

x

xdx

x dx

x...(i)

Ix

xdx I

x dx

x1 0

2

4 2 0 411

21

= ++

=+

∞ ∞

∫ ∫,

⇒ I

xx

x xx

dx1 0

22

2 22

11

12 2

=+

+ −

+

∞

∫

⇒0

2

2

11

12

∞

∫+

−

+

x

xx

dx

⇒ xx

t− =1

Differentiate on both sides

11

2+

=

xdx dt

When x = 0 then t = − ∞and x = ∞ then t = ∞

−∞

∞

∫ +dt

t 2 22( )

⇒ 12 2

1tan−

−∞

∞

t ⇒ 12 2 2

π π− −

⇒ π2

Now, Ix dx

xx t2 0 4

221

=+

=∞

∫ ,

2x dx dt=

x dxdt=2

When x = 0 then t = 0

and x = ∞ then t = ∞

⇒ 12 10 2

∞

∫ +dt

t

⇒ 12

10[tan ]− ∞t

= −

12 2

0π = π

2 2Substituting values of I1 and I2 in Eq. (i) we get

⇒ 12 2

π π−

⇒ π2 2

TEST Edge In JEE Advanced, integration of varioustypes of functions such as trigonometric,exponential, etc based questions are asked. So, tosolve such types of questions, students are advisedto understand methods to convert different types of

integrals of formpx q

ax bx cdx

++ +∫ 2 ,

px q

ax bx c

+

+ +∫ 2,

etc., into standard form by substitution and usestandard formula of integration to get the result e g. . ,

1 12 2

1

x adx

a

x

a+=∫ −tan is one such standard result.

47. (a) Idea To get the range of f x( ) cos7 17+ which

involves different types of functions, we firstlyobtain value of f ( )7 from the equation of f x( )and f ( )−7 and use the principal value of cos xfunction i e. ., cos [ , ]x ∈ − 1 1 to calculate the rangeof the given function.

f a b c( ) ( ) ( ) ( )7 7 7 7 57 3= + + −f a b c( ) ( ) ( ) ( )− = − + − + − −7 7 7 7 57 3

f f( ) ( )7 7 10+ − = −and f ( )− =7 7

⇒ f ( )7 7 10+ = − ⇒ f ( )7 17= −∴ f x x( ) cos cos7 17 17 17+ = − +

= + −17 1(cos )x

Now, cos [ , ]x ∈ − 1 1

cos [ , ]x − ∈ −1 2 0

17 1 3 4 0(cos ) [ , ]x − ∈ −

TEST Edge The questions are asked to find domainand range of function which can be combination ofvarious types of functions such as trigonometricfunction, algebraic function, etc. So, to solve suchtypes of questions, students are advised tounderstand the concept of range and domain of ofthe functions.

48. (a) Idea Standard equation of the parabola is y ax2 4= .

Firstly, we convert given general equation intostandard form and find the focus and use theconcept of circle having focal chord as adiameter always touches the directrix ofparabola.

Given parabola is x x y2 2 12 11 0+ + − =⇒ ( ) ( )x y+ = − −1 12 12

Vertex ≡ −( , )1 1

Here, 4 12 3a a= ⇒ =∴ For focus, x y+ = − = −1 0 1 3,

x y= − = −1 2,

Focus ≡ − −( , )1 2Thus, the chord passing through the point ( , )− −1 2 isa focal chord of the parabola.∴ Circle having focal chord of a parabola as diameteralways touches the directrix of parabola.


TEST Edge In JEE Advanced, questions based onbasic properties of parabola and circle are asked. So,students are advised to understand all conceptsrelated to parabola and properties of circlerespective of parabola and also learn to deduceinformation from the general equation of parabolai e. ., focal chord, directrix, latus rectum, etc.

49. (b) Idea In this question, differential equation isgiven in exponential series form, so we firstreduce it to exponential function then find thesolution by the method of variable separationand concept of integration by substitution.

xx

y

dy

dx

x

y

dy

dx21

2 2

3 12

2

2+ = +

+

−

−

!

+

+ …

−2

3

3 3x

y

dy

dx

!

⇒ x

y

x

dy

dx

y

x

dy

dx2

2 2

3 1 21

2+ = +

+

! 2!

+

+ …

y

x

dy

dx23

3 3

!

⇒ x e

y

x

dy

dx2 23+ =

Qex xx = + + + …

1

1 2

2

! !

⇒ log ( )e xy

x

dy

dx

2 32

+ =

Separating the variables, we get⇒ 2 32x x dx y dyelog ( )+ =On integrating both sides, we get⇒ ∫ ∫+ =2 32x x dx y dyelog ( )

Put, x t2 3+ = ⇒ 2x dx dt=⇒ ∫ ∫=loge t dt y dy

⇒ t t ty

Ce. log ( ) − = +2

2

⇒ ( ) log ( ) ( )x x xy

Ce2 2 2

2

3 3 32

+ + − + = +

⇒ ( ) (log ( ) )x xy

Ce2 2

2

3 3 12

+ + − = +

⇒ ( ) logxx

e

yCe

22 2

33

2+ +

= +

TEST Edge In JEE Advanced, question based onsolution of differential equation in combinationwith its representation in various mathematicalseries form i e. ., logarithmic, sine function, etc., areasked. So, students are advised to learn varioustypes of mathematical series and concept ofintegration by substitution, partial method andmethod to find the solution of differential equation.

50. (b) Idea This question involves the concept ofmodulus of the complex number. To solve itapply the basic algebraic formula to deducethe desired result.

Given, z a i b= +

| |z a b= +2 2

| |z a b2 2 2= + …(i)Now, ( ) .a b a b ab− ≥ ⇒ + − ≥2 2 20 2 0⇒ a b ab2 2 2+ ≥ . …(ii)From Eqs. (i) and (ii), we get

| |z ab2 2≥Also, | |z a b a b ab2 2 2 2 2 2+ + ≥ + +

| | | | ( )z z a b2 2 2+ ≥ +2 2 2| | ( )z a b≥ +

| | ( )z a b≥ +12

TEST Edge The questions based on properties ofmodulus are asked to prove or find any algebraicrelations. So, students are advised to understandalgebra of complex number, its polar representationand properties of modulus and basic formula ofalgebra.

Such as| | | || |z z z z1 2 1 2=51. (a,b) Idea Three coplanar vetors a b, and c and

relation of its sum and with other vectors p q, ,rare given in the question. Firstly, we takevectors p, q, r values which satisfy the givenrelation and equate it with vectors p, q, r andtake modulus of sum of vectors. Then, we usethe concept of AM GM≥ to get the result.

Let a,b,c lie in the xy-plane.

Let a i j b i j= − + = − −1 23

212

32

/ $ $, $ $ and c i= $

Then, | | | |p q r a b c+ + = + +λ µ r

= − +

+ − −

+λ µ1

23

212

32

$ $ $ $ $i j i j ir

= − − +

+ −

$ $i j

12 2

32

32

λ µ λ µr

= − − +

+ −

λ µ λ µ2 2

32

32

2 2

r

= + + + −

−

λ µ λ µ2 22

4 42

2 2r

+ −

+ −

2

22

2λ µ

( )r r + + −34

22 2( )λ µ λµ

= + + − − −λ µ λµ λ µ2 2 2r r r

= − + − + − ≥ + + =12

12

1 1 4 32 2 2( ) ( ) ( )λ µ µ λr r

⇒ | |p q r+ + can take values equal to 3 and 2.

PRACTICE SET 1 31

TEST Edge In JEE Advanced, questions based ondifferent types of vectors and its relation andcombination with arithmetic and geometric meanare asked. So, students are advised to learn aboutdifferent types of vectors, calculate magnitude ofvectors and properties of arithmatic and geometricprogression.

52. (c,d) Idea As functions are in AP, so we use theformula such as 2b a c= + , first we find therelation between terms of AP and then itsderivative to get the desired result.

Since f x y f x f y f x y( ), ( ) ( ), ( )− + are in AP.

⇒ f x y f x y f x f y( ) ( ) ( ) ( )− + + = 2 ∀ x y, ...(i)Puting x y= =0We have

f f f( ) ( ) [ ( )]0 0 2 0 2+ =2 0 2 0 2f f( ) [ ( )]=

⇒ f ( )0 1= [ ( ) ]Q f 0 0≠Putting x y x= =0, in (i), we have

f x f x f f x( ) ( ) ( ) ( )− + = 2 0

f x f x f x( ) ( ) ( )− + = 2⇒ f x f x( ) ( )− = …(ii)

∴ f f( ) ( )− =4 4Differentiating of Eq. (ii) w.r.t x, we have

− ′ − = ′f x f x( ) ( )

⇒ ′ + ′ − =f x f x( ) ( ) 0∴ ′ + ′ − =f f( ) ( )4 4 0

TEST Edge In JEE Advanced, questions based ondifferent progression i e. ., AP, GP and HP and basicrule of derivatives are asked to prove certainresults. So, students should acquainted themselveswith the information and properties of progressionand basic formulae of derivatives of function tosolve these questions.

53. (a,b,c,d) Idea Given three numbers a, b, care in HP andwe need to prove the relations amongthem given in the options are true or false.So, we know the relationship betweenAM, GM and HM combining all relations,we deduce the correctness of the options.

b is the harmonic mean of a and c

We know GM HM>ac b>

also AM of an and cn > GM of an and cn .

a ca c

n nn n+ >

2

⇒ a c ac bn n n n+ > >2 2( ) .

Putting n = 4 5 100 3, , ,

We get the required options.

TEST Edge Questions based on different progressioni e. ., AP, GP and HP are asked on the basis of it weneed to find some relation. So, it is advisable to thestudents to understand these progressions andrelationships of its mean and various results whichcan be deduce by their combinations, e g. . if A, G, Hare AM, GM, HM of two positive numbers thenA G H≥ ≥ and G AH2 = .

54. (a,b,c) Idea This question involves the concept ofdeterminants and application ofderivatives. Differentiate the givenfunction f x( ) and equate f x′ =( ) 0 to findpoint of maxima or minima and interval inwhich f x( ) is increasing and decreasing.

f x

x p pq pr

pq x q qr

pr qr x r

( ) =+

++

2

2

2

′ = ++

f x pq x q qr

pr qr x r

( )1 0 0

2

2

++

+

x p pq pr

pr qr x r

2

20 1 0

++

+x p pq pr

pq x q qr

2

2

0 0 1

= + + +3 22 2 2 2x p q r x( )Putting ′ =f x( ) 0

⇒ x xp q r= = − + +

02

3

2 2 2

,( )

Clearly, ′ >f x( ) 0 for x ∈ ∞( , )0

′ <f x( ) 0 for x p q r∈ − + +

23

02 2 2( ),

′ >f x( ) 0 for x p q r∈ − ∞ − + +

, ( )

23

2 2 2


Interval value in the interval value of f x′ ( ) sign of f x′ ( )

( , )0 ∞ 1 3 2 2 2 2+ + +( )p q r positve

− + +

23

02 2 2( )

,p q r − + +1

32 2 2( )p q r − + +5

92 2 2( )p q r negative

− ∞ − + +

, ( )

23

2 2 2p q r− + +( )p q r2 2 2 ( )p q r2 2 2 2+ + positive

PRACTICE SET 1 33

TEST Edge In JEE Advanced, questions based onproperties of determinants and application ofderivatives are asked. To solve such types ofquestions, students are advised to learn propertiesof determinants and also aquainted yourself withthe concept of increasing and decreasing functionswith the use of derivatives and basic theoram suchas f x( ) is increasing in (a b, ), if f x′ ( ) > 0 ∀ ∈x (a b, ).

55. (a,c) Idea To solve this question, students are advisedto use the quadratic formula, If

ax bx c2 0+ + = , then xb b ac

a=

− −−+2 4

2to find

the root of the equation and use trigonometricrelation sin sin cos2 2x x x= .

It is given that cos α is root of equation25 5 12 02x x+ − = , − < <1 0x

Now, 25 5 12 02x x+ − =

x =− ± +5 25 1200

50

= − ±5 122550

x = − ±5 3550

⇒ x = −3050

4050

,

x = −1525

2025

,

− < <1 0x

cos α = − 2025

sin sin cos ,2 22425

α α α= = or− 2425

TEST Edge Questions based on quadratic formulaand basic trigonometric relations are asked.To solve such types of questions, students areadvised to understand the concept quadraticformula and its various roots i e. . , real, complex andalso learn basic formulae of trigonometry such assin cos2 2 1x x+ = , 1 2 2+ =tan secx x etc.

56. (1) Idea This question involves the concept of dotproduct of vectors i e. . , a b a b⋅ =| || |cosθ andGeometric Progression (GP) i e. . , b ac2 = if a b c, ,are in GP, Now applying both the concepts toget the magnitude of vector| |u .

Given, the angle between u and $i is 60°

∴ u. i u i$ | | | $|= cos 602

° = | |u

We are also given that

| $|u i− is the geometric mean of | |u and | |u 2i−

∴ | $|u i− 2 = −| | | $|u u i2

Squaring both sides, we have[| | | $ | . $] | | [| |u i u i u u2 2 2 2 22+ − = + −4 42| $ | . $]i u i

| || |

| | | || |

uu

u uu2

22 21

22

4 42

+ −

= + −

| | | | | | | | | |u u u u u4 2 2 31 2 2 2+ + + − −

= + −| | | | | |u u u4 2 34 2

| | | |u u2 2 1 0+ − =

⇒ | |u = − ±2 2 22

⇒ | |u = −2 1

∴ ( ) | |2 1+ u = + −( ) ( )2 1 2 1 = − =2 1 1

TEST Edge Dot and cross product based questions ofvectors and sequence and series, its properties areasked. So, students are advised to understand thebasic concept of vectors algebra, scalar tripleproducts etc., and also acquainted yourself with theproperties of sequence and series. For e g. . , if V bethe volume of parallelopiped formed by the vectorsa i j k= + +a a a1 2 3

$ $ $ , b i j k= + +b b b1 2 3$ $ $ and

c i j k= + +c c c1 2 3$ $ $ and ∑ + + =

=rr r ra b c L

1

3

3( ) , then to

prove V L≤ 3. We use the concept of scalar productand AM GM≥ .

57. (0) Idea Firstly, we apply the concept of AP toreduce the given series as an = + −a n d( )1 , isthe nth term of AP. Then, the concept ofbinomial expansion ( ) , ,x y C x Cn n n n+ = +0 1

x y C x yn n n− −+12

2 2,

+ … + + … +−nr

n r r nn

nC x y C y = ∑=

−

r

nn

rn r rC x y

0

and properties of binomial concept such asr C n Cn

rn

r= −−

11 to solve the question.

Consider the given series2013C C C C0

20131

20132

201338 +15 22 +− − …

upto 2014 terms

Clearly, sum of series = − +=∑ ( ) ( )1 7 1

0

20132013r

r

rr C

= − + −= =∑ ∑7 1 1

0

20132013

0

20132013( ) ( )r

r

rr

r

rC .r C

= + −

+ −

= =∑ ∑7 0 1 1

0

20132013

0

20132013( ) ( )r

r

rr

r

C .r Cr

= −

+ −

=−

=∑ ∑7 1 2013 1

0

20132012

10

2013201( ) ( )r

r

rr

r

C 3Cr

[ ]Q r c n Cnr

nr= −

−1

1

= × − −

=−∑7 2013 1 1

0

20132012

1( ) ( )r

r

rrC + −

=∑ ( )1

0

20132013r

r

rC

= × × + =7 2013 0 0 0

TEST Edge In JEE Advanced, questions based on

binomial expansion and properties of binomial

coefficient are asked. To solve such types of

questions, students are advised to understand

binomial theorem, important terms in the binomial

expansions and properties of binomial coefficients

such as C C C Cnn

0 1 2 2+ + + … + = , C C C0 2 4+ + + …= + + + … = −C C C n

1 3 512 , etc.


58. (1) Idea Represent the locus of | |z i− in the arg and

plane such that 04

≤ ≤arg zπ

and use the

formulaaz az b

z a1 2+ +

| |to find the length of

perpendicular from a point z1 to the lineaz az b+ + = 0

The shaded region represent

O z≤ ≤arg /π 4

The least value of | |z i− is the length of perpendicularfrom ( , )0 1 to the line y x= i e. ., x y− = 0

⊥ distance = −+

=0 11 1

12

∴ 2 212

1| |z i− =

=

TEST Edge The question based on geometricalrepresentation of complex numbers on the arg andplane are asked in JEE Advanced. To solve suchtypes of questions, students are advised to learngeometrical representation of complex number andits interpretation a line and circle. For e.g., equationof circle having centre z0 and radius r is| |z z r− =0 orzz z z z z r− + − =0 0 0

2 0

59. (6) Idea Nature of roots of the quadratic equationis found by the value of discriminant i e. . , D ≥ 0for real roots and concept of AP whose termsare a, a d+ , a d+ 2 … if series is increasing are

used to find ratio ofd

a.

ax a d x a d2 2 0+ + + + =( ) ( )Given, a a d a d, ,+ + 2 are in increasing AP ( ).d > 0∴ For real roots D ≥ 0.

⇒ ( ) ( ) .a d a a d+ − + ≥2 4 2 0

⇒ d ad a2 26 3 0− − ≥ .

⇒ ( )d a a− − ≥3 12 02 2

⇒ ( ) ( )d a a d a a− + − − ≥3 2 3 3 2 3 0

⇒ d

a

d

a− +

− −

≥( ) ( )3 2 3 3 2 3 0

Min value ofd

ad= + >3 2 3 0( )

⇒ least integeral value of d a/ = 6

TEST Edge In JEE Advanced, questions based on

nature of roots of quadratic equations and also

related to the terms of AP, GP and HP are asked. So,

students are advised to understand the different

roots of quadratic equation, i e. . , real, equal or

complex roots and also acquainted yourself with

the concept of progression such as sum of n terms,

nth term of AP, GP or HP etc.

60. (3) Idea Represent the given vertices of triangle onCartesian plane and use the formula, Area of

triangle = ×1

2Base × Altitude. To find the area

of the region formed by the line x C= cuts thistriangle.

ar OAB( ) ( ) ( ).∆ = 12

8 1

= 4 sq. units

Equation of OB,y x− = −0

19

0( )

y x= 19

Hence, point E is C C,19

Now, area ( )∆BDE = 2

⇒ 12

19

9 2−

− =C

C( )

⇒ ( )9 362− =C

⇒ 9 6− = ±C

⇒ 9 6m = C

⇒ C = 3

C =15

⇒ C = 3

TEST Edge In JEE Advanced, question based onequation of straight line in various forms and areaof the region formed on its representation onCartesian plane are asked.

So, to solve such types of questions, learn equationof straight lines in various forms such as parametricform, slope-intercept form, etc., and itsrepresentation on Cartesian plane. For e g. ., equationof a straight line in perpendicular form asx p pcos sin+ =α α .

3 – 2 3 3 + 2 3

A(1,1)

B(9,1)

D(C,1)

E (C,C/9)

y

xO

(0,0)

x=C

(0, 1)

O

yx=

x

y

practice set test rider advanced

Documents