sample

1. Several factors (steric, electronic, orbital interactions etc.) can affect the inversion barrier ofan amine. In the given pair which data is correctly placed ?

(a) (b)

(c) (d*) All of these

2. Circle represent most acidic hydrogens in these molecules. Which of the following is correctrepresentation ?

(a) (b)

(c) (d*) All of these

3. The pH at which maximum hydrate is present in an solution of oxaloacetic acid, is :

H O C||

C||

CH C||

O H

O O O

2-------

pK a =2 2. pK a =3 98.(a*) pH = 0 (b) pH = 12 (c) pH = 4 (d) pH = 6

Get Ready for IIT-JEE 1

vs

Me|N

MeMe

DG = 7.9 kcal/mol

i – Pr

i – Pr

|N

Me

DG = 0.2 kcal/mol+ DG = 20.5 kcal/mol

NMe

DG = 7.0 kcal/molMe

Nvs

Me|

NMeMe

DG = 7.9 kcal/mol++

vs

Cl|

NClCl

DG = 22.9 kcal/mol++

OH

CF3

OH

O

MeO C2

CO H2HO

S

F

By : M.S. Chouhan

Advanced Problems

Organic Chemistryfor

IIT-JEE

in

4. Which of the following isomeric hydrocarbons is most acidic ?

(a) (b*) (c) (d)

5. Compound A and B, both were treated with NaOH, producing a single compound C.

HO

heat

-

¾®¾¾ C. Compound (C) is :

(a*) (b) (c) (d)

6. ¾¾¾®RCO H2 ; Product of above reaction is :

(a) (b)

(c*) both a and b (d) None of these

7. 150° ==¾®¾¾+¾ ®¾¾¾¾¾¾

C H C CH — CN2( ) ( )A B

Compound (B) reacts acrylonitrile to give (C), structure of compound (A) is :

(a) (b*) (c) (d)

2 Get Ready for IIT-JEE

O

O

O

C — CH3O O

OTs

14C label

no label

OCOR OCOR

CH3O OCH3

CN

(C)CCH3CH3O

CH2

CH2

OCH3

O

CH3

OH

O

CH3

OH(A) (B)

+

8. Increasing order of rate of reaction with HNO H SO3 2 4 is :

(a) iii < ii< i (b) ii < iii < i(c) i < iii < ii (d*) i < ii < iii

9. Match the columns I, II and III (Matrix).

Col umn-I Col umn- II Column- IIII

Re ac tion Nature of product formed Num ber of chiral center pres - ent in product. (Consideronly one isomer in case of

racemic mixture orDiastereomer)

(a)Br

CCl2

4¾®¾¾

(p) Racemic mix ture (w) 0

(b) Br

CCl2

4¾®¾¾

(q) Meso (x) 1

(c) BrCCl

2

4¾®¾¾ (r) Diastereomer (y) 2

(d) (s) Vic i nal dihalide (z) 3


(i)

O

O O

OOO

O

OOO OO

(ii) (iii)

CH3

H

CH3

CH H3 C ==C H CH3

Br2

CCl4

CH3

H

10.

The synthetic steroid ethynylestradiol (1) is a compound used in the birth control pill.

A. How many sp3 hybridised carbon atoms are present in compound (1)?

(a) 8 (b) 9 (c) 10 (d) 11 (e) 12

B. How many sp2 hybridised carbon atoms are present in compound (1) ?

(a) 4 (b) 5 (c) 6 (d) 7 (e) 8

C. How many sp hybridised carbon atoms are present in compound (1) ?(a) 2 (b) 4 (c) 6 (d) 8 (e) 10

D. Which of the following functional group is contained in compound (1) ?(a) A ketone (b) An alcohol (c) A carboxylic acid (d) An ester

E. How many asymmetric (stereogenic) centres are present in compound (1) ?(a) 2 (b) 3 (c) 4 (d) 5

1. In the prep aration of iron from haematite (Fe O )2 3 by the reaction with carbonFe O + C Fe+ CO2 3 2¾®

How much 80% pure iron could be produced from 120 kg of 90% pure Fe O ?2 3(a*) 94.5 kg (b) 60.48 kg(c) 116.66 kg (d) 120 kg

2. van der Waals constant b of helium is 24 mL mol-1. Find molecular diameter of helium.(a) 1.335 cm´-10 10 (b) 1.335 cm´-10 8

(c*) 2.67 10 cm8́- (d) 4.34 cm´-10 8


Ethynylestradiol (1)HO

OH H

H H

H

H C3

3. There exist an equi librium between solid SrSO4 and Sr 2+ and SO42- ion in aque ous medium.

The possible equilibrium states are shown in figure as thick line. Now, if equilibrium isdist urbed b y a dd it ion of (a) Sr(NO )3 2 and (b) K SO2 4 and dotted line represent approachof system toward’s equilibrium. Match the columns given below :

(I) addition of Sr(NO )3 2 (II) addition of K SO2 4(a) (I) (iii), (II) (iv) (b*) (I)(iv),(II) (v) (c) (I) (vi), (II) (v) (d) (I) (iv), (II) (v)

4. Determine the potential of the following cell :

Pt|H bar H MnO2( , . )| ( , )|| ( , . ),– –g aq aq01 10 0134

+ M M Mn H Pt2 0 01 0 01+ +( , . ), ( , . )|aq aqM M

Given : EMnO4

–|Mn 2+1.51 V° =

(a) 1.54 V (b*) 1.48 V (c) 1.84 V (d) none of these

5. 0.1 M KI and 0.2 M AgNO3 are mixed in 3 1: volume ratio. The depression of freezing pointof the resulting solution will be [ ( )K f H O 1.862 = K kg mol-1] :

(a) 3.72 K (b) 1.86 K (c) 0.93 K (d*) 0.279 K

6. In an atomic bcc, what fraction of edge is not covered by atoms?(a) 0.32 (b) 0.16 (c*) 0.134 (d) 0.268

7. Select the correct statement(s) :(a*) A solution is prepared by addition of excess of AgNO3 solution in KI solution. The

charge likely to develop on colloidal particle is positive.(b*)The effects of pressure on physical adsorption is high if temperature is low(c) Ultracentrifugation process is used for preparation of lyophobic colloids.(d) Gold number is the index for extent of gold plating done

8. For a first order homogeneous gaseous reaction, A B C¾®+2

If the total pressure after time t was Pt and after long time ( )t ®¥ was P¥ then k in terms of P Pt , ¥ and t is :

(a) kt

PP Pt

=-

æ

èç

ö

ø÷

¥

¥

2.303log (b) k

tP

P Pt=

-

æ

èç

ö

ø÷

¥

¥

2.303log

2

(c*) k t

log P

(P P )t=

-

æ

èç

ö

ø÷

¥

¥

2.303 23

(d) none of these

9. Fixed mass of an ideal gas contained in a 24.63 L sealed rigid vessel at 1 atm is heated from -°73 C to 27°C. Calculate change in Gibb’s en ergy if entropy of gas is a function oftem per a ture as S =+-2 10 2 T (J/K): (Use 1 atm L 0.1= kJ)

(a) 1231.5 J (b) 1281.5 J (c*) 781.5 J (d) 0


2+Sr

2–SO4

2–SO4

Sr2+ 2+Sr

2–SO4

2–SO4

2+Sr(iii) (iv) (v) (vi)

10. Column-I Col umn-II

(A) Or bital an gu lar mo men tum of anelec tron

(P) s sh

( )+12p

(B) An gu lar mo men tum of an elec tronin an orbit

(Q) n n( )+2

(C) Spin an gu lar mo men tum of anelec tron

(R) nh2p

(D) Magnetic moment of atom (S) l lh

( )+12p

* Answer : A – S, B – R, C – P, D – Q

11. Which is the correct order of ionization energies?(a) F > F> Cl > Cl– – (b*) F > Cl > Cl > F– –

(c) F > Cl > Cl > F– – (d) F > Cl > F > Cl– – –

12. Ionization energy of an element is :(a*) Equal in magnitude but opposite in sign to the electron gain enthalpy of

the cation of the element(b) Same as electron affinity of the element(c*) Energy required to remove one valence electron from an isolated

gaseous atom in its ground state(d) Equal in magnitude but opposite in sign to the electron gain enthalpy of the anion of the

element

13. O F2 2 is an unstable yellow orange solid and H O2 2 is a colourless liquid, both have O—Obond and O—O bond length in H O2 2 and O F2 2 respectively is :

(a) 1.22Å, 1.48Å (b*) 1.48 Å, 1.22 Å

(c) 1.22Å, 1.22Å (d) 1.48Å, 1.48Å

14. Which of the following equilibria would have highest and lowest value of K p at a commontemperature.(a*) BeCO 3 � BeO CO 2+ (b) CaCO3 � CaO CO2+

(c) SrCO3 � SrO CO2+ (d*) BaCO3 � BaO CO2+

15. In the structure of H CSF2 4, which of the following statement is/are correct?

(a*) Two C—H bonds are in the same plane of axial S—F bonds(b) Two C—H bonds are in the same plane of equatorial S—F bonds(c*) Total six atoms are in the same plane(d) Equatorial S—F plane is perpendicular to plane of p-bond

16. In the isoelectronic series of metal carbonyl, the CO bond strength is expected to increase inthe order :(a) [Mn(CO) ] [Cr(CO) ] [V(CO) ]6 6 6

+ -< < (b*)[V(CO)6] [Cr(CO)6]<[Mn(CO)6]-< +

(c) [V(CO) ] <[Mn(CO) ] [Cr(CO) ]6 6 6- +< (d) [Cr(CO) ]<[Mn(CO) ] [V(CO) ]6 6 6

+ -<


17. [Fe(H O) ]2 62+ and [Fe(CN) ]6

4– differ in :

(a) geometry, magnetic moment (b) geometry, hybridization(c*) magnetic moment, colour (d) hybridization, number of d-electrons

18. Which of the following statements is true ?(a) In [PtCl (NH ) ]2 3 2

2+ the cis form is optically inactive while trans form is optically active

(b*)In [Fe(C O ) ]2 4 33– , geometrical isomerism does not exist while optical

isomerism exists(c) In Mabcd, square planar complexes show both optical as well as geometrical isomerism(d) In Mabcd tetrahedral complex, optical isomerism cannot be observed

19. The CFSE for [(CoCl)6] –4 complex is 18000cm –1. The D for [CoCl ]42– will be :

(a) 18000cm –1 (b) 16000 cm –1 (c*) 8000 cm–1 (d) 2000 cm –1

The sequential unknown reagents is/are:(a) H O2 2/neutral medium, H O /H , Cu2 2

+ 2+ salt solution(b*) H O / H , H O / OH , Fe2 2

+2 2

3+- salt solution(c) H O /OH , H O /H , Co2 2 2 2

2+- + salt solution(d) H O2 2/neutral medium, H O /OH Fe2 2

2+-, salt solution

21. Se lect cor rect state ment(s):(I) When excess FeCl3 solution is added to K [Fe(CN) ]4 6 solution, in addition to

Fe [Fe (CN) ]III II6

-, Fe [Fe (CN) ]II III6

- is also formed due to side redox reaction

(II) When FeCl2 is added to K [Fe(CN) ]3 6 solution, in addition to Fe [Fe (CN) ]II III6

-, Fe [Fe (CN) ]III II

6- is also formed due to side redox reaction

(III) Fe [Fe (CN) ]III II6

- is paramagnetic while Fe [Fe (CN) ]II III6

- is diamagnetic

(IV) Fe [Fe (CN) ]III II6

- is diamagnetic while Fe [Fe (CN) ]II III6

- is paramagnetic

(a*) I, II (b) III, IV (c) both (a) and (b) (d) None of these

22. Col umn-I(Pair of com plexes)

Col umn-II(Property which is similar in given pair)

(A) [Fe(CN) ]63– and [Co(NH ) ]3 6

2+ (P) Mag netic mo ment

(B) [Fe(H O) ]2 62+ and [Fe(CN) ]6

4– (Q) Ge om e try

(C) [Ni(CN) ]44– and [Ni(CO) ]4 (R) Hy bridi sa tion

(D) [Ni(H O) ]2 62+ and [NiCl ]4

2– (S) Num ber of d-electrons

* Answer : A – P, Q, R; B – Q, S; C – P, Q, R, S; D – P, S


A B A CX Y Z20.

Pale yellowsolution

Yellowsolution

Deep bluesolution

Deep bluesolution

Regenerated

Na (comH O CO SO Na2S I2 Ag+ salt2 / /¾®¾®¾®¾¾¾®¾¾¾®A B C E2 2

DD plex)

a. The compound B and C are :(a) Na CO , Na SO2 3 2 4 (b) NaHCO , Na SO3 2 4(c*) Na CO , Na SO2 3 2 3 (d) None of these

b. The compound D is :(a) Na SO2 4 (b) Na S O2 4 6 (c) Na S O2 2 5 (d*) Na S O2 2 3

c. Oxidation number of each ‘S’ atom in compound D :(a) +2, +2 (b) +4, 0 (c*) +6, –2 (d) +5, –1

The molecule in which an atom is associated with more than 8 electrons is known ashypervalent molecule and less than 8 electrons is known as hypovalent molecule. Allhypervalent molecules must have dp-pp bonding but the molecules having back bondingneed not to have always dp-pp-bonding.

a. Which of the molecule is not hypovalent but completes its octet :(a) AlCl3 (b) AlBr2 (c*) AlF3 (d) BF3

b. Which of the following molecule is having complete octet :(a) BeCl2 (dimer) (b) BeH2 (dimer) (c) BeH2(s) (d*) BeCl (s)2

c. Which of the following molecule is not having dp-pp-bonding :(a) SO2 (b) P O4 10 (c) PF3 (d*) B N H3 3 6

An unknown mixture contains one or two of the following : CaCO , BaCl , AgNO ,3 2 3Na SO , ZnSO2 4 4 and NaOH. The mixture is completely soluble in water and solutiongives pink colour with phenolphthalein. When dilute hydrochloric acid is graduallyadded to the solution, a precipitate is formed which dissolves with further addition of theacid.

a. Which of the following combination of compounds is soluble in water?(a) BaCl and AgNO2 3 (b) AgNO and NaOH3(c) BaCl and Na SO2 2 4 (d*) ZnSO 4 and excess NaOH

b. The aqueous solution of mixture gives white precipitate with dil. HCl which dissolves inexcess of dil. HCl. It confirms:(a) BaCl NaOH2 + (b) Na SO NaOH2 4 +

(c*) ZnSO NaOH4 + (d) AgNO NaOH3 +

c. The white pre cip i tate is:(a) ZnSO4 (b) Na ZnO2 2(c*) Zn(OH)2 (d) ZnCl2


P A S S A G E 2P A S S A G E



1. Initially blocks A and B are given impulse in opposite directions as shown in figure. Now wehave to calculate the :(i) Maximum stretch in spring(ii) Maximum velocity of block A and B in ground

frame.(iii)Minimum velocity of block A and B in ground

frame.

Sol. υυ υ

υCM =-

=( )( )2 2

30 0

0m m

m

At initial instant

v v v i®®®

=- =A A g gCM CM υ0$

v v v i®®®

=- =-B B g gCM CM 2 0υ$

From work energy equation in CM frameWe get wspring system CMKE=D

- =-+

12

012

22

320

2kXm mm mm

( )( )( )

( )υ

12

12

23

320

2kX mm =æèçö

ø÷( )υ

or Xmkm =

60υ


By : Er. Anurag Mishra

MechanicsIIT-JEE

for

Fig. 4.21

υ0m 2m

2υ0B A

(a)

2υ0m 2m

υ0

CM frame

Initial state

υ0

CM frame

At maximum stretch

υ0

υ= 0B/CM υ= 0A/CM

l + x0 m

Fig. 4.21

(b)

Vol. 1

When spring again regains its natural length in CM frame.

Most Important Concept

v v v® ®®

= +A Agroundframe

CMframe CM ground

frame

similarly v v v® ®®

= +B Bgroundframe

CMframe CM ground

frame

Note that velocity of any block in ground frame is superposition of two velocity vectors, velocity of block in CM frame and velocity of CM with respect to ground.

v®

A groundframe

is maximum when v®

A CMframe

has maximum magnitude and is in same direction as vector

v®

CM groundframe

.

Similarly v®

A groundframe

is minimum when v®

A CMframe

and v®

CM groundframe

vectors are in opposite direction.

| | | | | |v v v®® ®

= + =A AmaxCMframe CM ground

frame 2 0υ

Similarly v v v®® ®

= + =B Bmax CMframe CM ground

frame| | | | | 3 0υ

Minimum velocity of A is attained when block is at equilibrium

position (spring is relaxed) and v®

A CM and v®

CM g are

opposite to each other.

Thus | |min

v®

=A 0

Minimum velocity of B is attained at the instant B is moving toward left (opposite v®

CM ) and velocity

magnitude is υ0 (see figure)

Thus | |min

v®

=B 0


υ0

CM frame

|v = B/CM 0| υ0

|v = A/CM

υ|

AB

2

® ®

Fig. 4.22

Fig. 4.21

(c)

Blockdiagramsrepresentingsituation whenblock returnsto relaxedstate

What appears inCM frame

Blocks return to relaxed state

υ0

υ= 2υB/CM 0

m 2m

υ= υA/CM 0

What appears inground frame

υ= 3υA/g 0

m 2m

υ= 0B/g

ìíî

2. Two blocks A and B of masses 2m & 3m placed onsmooth horizontal surface are connected with a lightspring. The two blocks are given velocities as shownwhen spring is at natural length.

Col umn-I Col umn-II

(A) minimum magnitude of velocity of A ( )min

υA during motion (P) υ

(B) maximum magnitude of velocity of A ( )max

υA during motion (Q)υ

5

(C) maximum magnitude of velocity of B ( )max

υB during motion (R) 0

(D) velocity of centre of mass ( )υCM of the system comprised ofblocks A, B and spring

(S) 75υ

Sol. Step I:

υυ υυ

CM =-

=( )3 3

5 5m m

m

Step 2:

In COM frame

Initial velocity of A = --æèç ö

ø÷=-υ

υ υ

565

to left

Initial velocity of B =-==υυ

υυ5

45

45

to right

Blocks are executing SHM in CM frame with initial position as equilibrium position

Step III:

Velocity variation of B in ground frame, considering right as +ve

form 45 5υυ

υ+æèç ö

ø÷= to

45 5

35

υυυ+=-

so | |max

υ υB = & | |min

υB =0

Velocity variation A in ground frame

from 65 5

75

υυυ+æ

èç ö

ø÷= to

-+=-

65 5

υυυ

Thus minimum velocity of A is -υ

5 when spring is at maximum extension.


Fig. 4E.30

υ2m 3mK

υ

A B

3. The 10 kg block is resting on the horizontalsurface when the force 'F' is applied to it for7 second. The variation of 'F' with time isshown. Calculate the maximum velocityreached by the block and the total time 't'during which the block is in motion. Thecoefficient of static and kinetic friction areboth 0.50.

Sol. Block begins to move when

F =mN =́ =0 5 100 50. N

from t =0 to t =4sec F st=2 0 4££t =40N 4 7<£t

Block begins to move at t =2sec. after that

Fmddt

-=mNυ

25 50 10tddt

-=υ

or, d t dtυ0

υ

ò ò= -( . )2 5 52

4

υ= -2 5

25

2

2

4. t

t

=5m sAfter t = 4 sec.; block retards due to greater friction forceVelocity of block at t =7 sec(i.e. 3 sec. after force becomes 40 N)

υ=-́5 1 3 =2m s

At t =7 sec. force F is now only friction acts -=s aυ10

a =5m s2

0 2=-́( )s t t =4sec.

Total time interval for which block moves t =2sec. to t =7 9. sec

i.e., 5 4. sec.


(b)

N

FµN

mg

Fig. 2E.60

Fig. 2E.60

(c)

F = 40N

υ5m/s= initial

µ N = 5 0 Nk

Fig. 2E.60

(a)

100

40

F(N)

4 7t(s)

10 kg F

µ Nk

Fcos37°

Fsin37°

4. A force of 20 N is applied to a block at rest as shown in figure.After the block has moved a distance of 8m to the right thedirection of horizontal component of the force F is reversed indirection. Find the velocity with which block arrives at itsstarting point.

Sol. Stage 1: Motion till force reverses its direction

N =- °mg F sin 37

=-́20 2035

=8N F makcos37°-m=N

2045

0 25 8 2-́ =́́. a

a =7 m s2

Velocity of block after displacement of 8 m υ=́ =́2 7 8 12 m s

Concept

Kinetic friction is opposite to relative velocity it opposes relative motion. When horizontal componentforce is reversed, relative velocity is not changed therefore, direction of kinetic friction does notchange.

Stage 2: ¾®=υ112 m s

F cos37° is reversed, block continues along original direction, but due to retardetion createdby mkN and F cos37° block travels till it stops.

- °+=( cos )F N mak37 m

-́ + =́́( ( . )2045

0 25 7 8 2 a

or, a =-9m s2

Displacement of block in this phase

0 22=-υas;

sa

=υ

2

2

=´

=( )1122 9

569

m


Fig. 2E.61

(a)

m=2kg

37°

µ=0.25

Fig. 2E.61

(b)

µ Nk

N

Fcos37°

mg=20N

Fsin37°

Stage 3: Which block returns its a acceleration is: F N mgkcos37°-=m

a =7 m s2

Velocity of block when it returns to original position

υ2 2=as

=2 7 8569

´́ +æèç ö

ø÷

υ=16 7

3m s

5. A carriage of mass M and length l is joined to the end of aslope as shown in the figure. A block of mass m is releasedfrom the slope from height h. It slides till end of thecarriage (The friction between the body and the slope andalso friction between carriage and horizontal floor isnegligible) Coefficient of friction between block andcarriage is m. Find minimum h in the given terms.

(a) m1 +æèç ö

ø÷

Mm

l (b) 2 1m+æèç ö

ø÷

mM

l

(c) m2 +æèç ö

ø÷

mM

l (d) m1 +æèç ö

ø÷

mM

l

Sol.Concept:

Block slips relative to carriage, use relative motion equations of kinematics.

velocity of block, just before reaching carriage υ0 2=gh

Now acceleration of block

amgm

g1 =-=-m

m

acceleration of carriage amgM2 =

m

considering this moment as t =0, motion of block as seen from carriage

u ghrel ==υ0 2

a a a gmMrel =-=-+æ

èöø1 2 1m

Relative velocity of block when block moves through distance x with respect to carriage υυrel rel rel

2 2 2=+a x

when x l= =, υrel 0 Þ 2 2 1gh gmM

l= +æè

öø

m

Þ hmM

l=+æè

öø

m1


Fig. 2E.61

(c)

µ Nk

NFcos37°

mg

Fsin37°

m

h

Msmooth

Fig. 2E.102

N

A

µmg

mg

a1

N

A

µmg

mg

a2

µmg

Fig. 2E.102

(b)

Covering all the possible types of problems like numerical problems, objective problems, MCQ, matching type questions, integer answers type questions, previous year’s questions, comprehension based objective type problems & assertion reason type problems etc.

Covering all the possible types of problems like numerical problems, objective problems, MCQ, matching type questions, integer type questions, previous year’s questions, comprehension based objective type problems, etc.

Algebrafor

IIT-JEEBy : M. Uma Shankar & Dr. J.V. Rao

A Complete text book for IIT-JEE aspirants.

Objective Physics

Objective Physics

ConceptualInorganicChemistry

ConceptualPhysicalChemistry

By : Prabhat Kumar By : Prabhat Kumar & Adarsh Kumar

(IIT-JEE & AIEEE) (IIT-JEE & AIEEE)

Objective Physics

Entrance Exam.is

goal

sample

Documents

o o i o o o o o o o

compound c

c ch3 b o c o o d c

c c c ch o h pk

c ch2 d

c p d q

o o ch3

ocor b d