AIEEE Solved Paper 2007

1. The displacement of an object attached to a

spring and executing simple harmonic motion

is given by x = × −2 10 2 cos πt metre. The time

at which the maximum speed first occurs is

(a) 0.5 s (b) 0.75 s

(c) 0.125 s (d) 0.25 s

2. In an AC circuit the voltage applied is

E E t= 0 sin ω . The resulting current in the

circuit is I I t= −

0

2sin ω π

. The power

consumption in the circuit is given by

(a) PE I= 0 0

2(b) P = zero

(c) PE I= 0 0

2(d) P E I= 2 0 0

3. An electric charge 10 3− µC is placed at the origin

(0, 0) of X-Y coordinate system. Two points A

and B are situated at ( , )2 2 and (2, 0)

respectively. The potential difference between

the points A and B will be

(a) 9 V (b) zero

(c) 2 V (d) 4.5 V

4. A battery is used to charge a parallel plate

capacitor till the potential difference between

the plates becomes equal to the electromotive

force of the battery. The ratio of the energy

stored in the capacitor and the work done by

the battery will be

(a) 1 (b) 2

(c) 1

4(d)

1

2

5. An ideal coil of 10 H is connected in series with

a resistance of 5Ω and a battery of 5 V. 2 s after

the connection is made, the current flowing (in

ampere) in the circuit is

(a) ( )1 − e (b) e

(c) e−1 (d) ( )1 1− −e

6. A long straight wire of radius a carries a steady

current i. The current is uniformly distributed

across its cross-section. The ratio of the

magnetic field at a

2 and 2a is

(a) 1

4(b) 4

(c) 1 (d) 1

2

7. A current I flows along the length of an

infinitely long, straight, thin walled pipe. Then

(a) the magnetic field is zero only on the axis

of the pipe(b) the magnetic field is different at different

points inside the pipe(c) the magnetic field at any point inside the

pipe is zero(d) the magnetic field at all points inside the

pipe is the same, but not zero

8. If M O is the mass of an oxygen isotope 817O ,

M p and M n are the masses of a proton and a

neutron, respectively, the nuclear binding

energy of the isotope is

(a) ( )M MO p− 8 2c

(b) ( )M M MO p n− −8 9 2c

(c) M Oc2

(d) ( )M MO n− 17 2c

9. In gamma ray emission from a nucleus

(a) both the neutron number and the proton

number change(b) there is no change in the proton number

and the neutron number(c) only the neutron number changes

(d) only the proton number changes

10. If in a p-n junction diode, a square input signal

of 10 V is applied as shown

1

Solved Paper 2007

AIEEE(All India Engineering Entrance Examination)

Physics

AIEEE Solved Paper 2007

Then the output signal across R L will be

11. Pho ton of fre quency ν has a mo men tumas so ci ated with it. If c is the ve loc ity of light, the mo men tum is

(a) ν/c (b) hνc(c) h cν/ 2 (d) h cν/

12. The velocity of a particle is v v gt ft= + +02. If

its position is x = 0 at t = 0, then itsdisplacement after unit time ( )t = 1 is

(a) v g f0 2 3+ + (b) v g f0 2 3+ +/ /(c) v g f0 + + (d) v g f0 2+ +/

13. For the given uniform square lamina ABCD,whose centre is O

(a) 2 I IAC EF= (b) I IAD EF= 3

(c) I IAC EF= (d) I IAC EF= 2

14. A point mass oscillates along the x-axisaccording to the law x x= 0 cos ( / )ω πt − 4 . Ifthe acceleration of the particle is written asa A t= +cos( ),ω δ then(a) A x= = −0 4, /δ π(b) A x= =0

2 4ω δ π, /

(c) A x= = −02 4ω δ π, /

(d) A x= =02 3 4ω δ π, /

15. Charges are placed on

the vertices of a square

as shown. Let E→

be the

electric field and V the

potential at the centre.

If the charges on A and

B are interchanged

with those on D and C

respectively, then

(a) E→

remains unchanged, V changes

(b) both E→

and V change

(c) E→

and V remain unchanged

(d) E→

changes, V remains unchanged

16. The half-life period of a radioactive element X

is same as the mean life time of another

radioactive element Y. Initially they have the

same number of atoms. Then

(a) X will decay faster than Y

(b) Y will decay faster than X

(c) Y and X have same decay rate initially

(d) X and Y decay at same rate always

17. A Carnot engine, having an efficiency of

η = 1 10/ as heat engine, is used as a

refrigerator. If the work done on the system is

10 J, the amount of energy absorbed from the

reservoir at lower temperature is

(a) 99 J (b) 90 J

(c) 1 J (d) 100 J

18. Carbon, silicon and germanium have four

valence electrons each. At room temperature

which one of the following statements is most

appropriate ?

(a) The number of free conduction electrons is

significant in C but small in Si and Ge(b) The number of free conduction electrons is

negligibly small in all the three(c) The number of free electrons for

conduction is significant in all the three(d) The number of free electrons for

conduction is significant only in Si and Ge

but small in C

19. A charged particle with charge q enters a

region of constant, uniform and mutually

orthogonal fields E→

and B→

with a velocity v→

perpendicular to both E→

and B→

, and comes out

without any change in magnitude or direction

of v→

. Then

2

D C

O

BAE

F

–5 V

(a) (b)

(c) (d)

+5 V

10 V

–10 V

A B

CD

q q

–q –q

5 V

– 5 V

RL

(a) v E B→ → →

= × / B2 (b) v B E→ → →

= × / B2

(c) v E B→ → →

= × / E 2 (d) v B E→ → →

= × / E 2

20. The potential at a point x (measured in µm)

due to some charges situated on the x-axis isgiven by : V x x( ) / ( )= −20 42 volt

The electric field E at x = 4 µm is given by

(a) 5

3 V/µm and in the –ve x direction

(b) 5

3V/µm and in the +ve x direction

(c) 10

9 V/µm and in the –ve x direction

(d) 10

9 V/µm and in the +ve x direction

21. Which of the following transitions in hydrogenatoms emit photons of highest frequency ?

(a) n = 2 to n = 6 (b) n = 6 to n = 2(c) n = 2 to n = 1 (d) n = 1 to n = 2

22. A block of mass m is connected to another block of mass M by a spring (massless) of springconstant k. The blocks are kept on a smoothhorizontal plane. Initially the blocks are at restand the spring is unstretched. Then a constantforce F starts acting on the block of mass M topull it. Find the force on the block of mass m.

(a) mF

M(b)

( )M m F

m

+

(c) mF

m M( )+(d)

MF

m M( )+

23. Two lenses of power –15D and +5D are incontact with each other. The focal length of thecombination is :

(a) –20 cm (b) –10 cm(c) +20 cm (d) +10 cm

24. One end of a thermally insulated rod is kept ata temperature T1 and the other at T2. The rod iscomposed of two sections of lengths l1 and l2and thermal conductivities K1 and K2

respectively. The temperature at the interfaceof the two sections is

(a) ( )/( )K l T K l T K l K l2 2 1 1 1 2 1 1 2 2+ +(b) ( )/( )K l T K l T K l K l2 1 1 1 2 2 2 1 1 2+ +(c) ( )/( )K l T K l T K l K l1 2 1 2 1 2 1 2 2 1+ +(d) ( )/( )K l T K l T K l K l1 1 1 2 2 2 1 1 2 2+ +

25. A sound absorber attenuates the sound level by20 dB. The intensity decreases by a factor of

(a) 1000 (b) 10000

(c) 10 (d) 100

26. If C p and CV denote the specific heats ofnitrogen per unit mass at constant pressure and constant volume respectively, then

(a) C C Rp V− = /28 (b) C C Rp V− = /14

(c) C C Rp V− = (d) C C Rp V− = 28

27. A charged particle moves through a magneticfield perpendicular to its direction. Then

(a) the momentum changes but the kineticenergy is constant

(b) both momentum and kinetic energy of theparticle are not constant

(c) both momentum and kinetic energy of theparticle are constant

(d) kinetic energy changes but the momentumis constant

28. Two iden ti cal con duct ing wires AOB and CODare placed at right an gles to each other. Thewire AOB car ries an elec tric cur rent I1 and CODcar ries a cur rent I2. The mag netic field on apoint ly ing at a dis tance d from O, in a di rec tion per pen dic u lar to the plane of the wires AOBand COD, will be given by

(a) µ

π0 1 2

1 2

2

I I

d

+

/

(b) µπ0

12

22 1 2

2 dI I( ) /+

(c) µπ0

1 22 d

I I( )+ (d) µπ0

12

22

2 dI I( )+

29. The resistance of a wire is 5 Ω at 50°C and 6 Ωat 100°C. The resistance of the wire at 0°C willbe(a) 2Ω (b) 1 Ω(c) 4 Ω (d) 3Ω

30. A par al lel plate con denser with a di elec tric ofdi elec tric con stant K be tween the plates has aca pac ity C and is charged to a po ten tial V volts.The di elec tric slab is slowly re moved frombe tween the plates and then re in serted. Thenet work done by the sys tem in this process is

(a) 1

21 2( )K CV− (b) CV K K2 1( )/−

(c) ( )K CV− 1 2 (d) zero

31. If gE and gM are the ac cel er a tions due tograv ity on the sur faces of the earth and themoon re spec tively and if Millikan’s oil dropex per i ment could be per formed on the twosur faces, one will find the ra tio electronic charge on the moon

electronic charge on the earth to be

3

T1 T2

K1 K2

l1 l2

(a) 1 (b) zero(c) g gE M/ (d) g gM E/

32. A cir cu lar disc of ra dius R is re moved from abig ger cir cu lar disc of ra dius 2R, such that thecir cum fer ence of the discs co in cide. The cen tre

of mass of the new disc is αR

from the cen tre of

the big ger disc. The value of α is

(a) 1

3(b)

1

2

(c) 1

6(d)

1

4

33. A round uni form body of ra dius R, mass M andmo ment of in er tia I, rolls down (with outslip ping) an in clined plane mak ing an an gle θwith the hor i zon tal. Then its ac cel er a tion is

(a) g

I MR

sin

/

θ1 2+

(b) g

MR I

sin

/

θ1 2+

(c) g

I MR

sin

/

θ1 2−

(d) g

MR I

sin

/

θ1 2−

34. Angular momentum of the particle rotatingwith a central force is constant due to

(a) constant force(b) constant linear momentum(c) zero torque(d) constant torque

35. A 2 kg block slides on a hor i zon tal floor with aspeed of 4 m/s. It strikes a un com pressedspring, and com presses it till the block ismo tion less. The ki netic fric tion force is 15 Nand spring con stant is 10,000 N/m. The springcom presses by

(a) 5.5 cm (b) 2.5 cm

(c) 11.0 cm (d) 8.5 cm

36. A par ti cle is pro jected at 60° to the hor i zon talwith a ki netic en ergy K. The ki netic en ergy atthe high est point is

(a) K (b) zero(c) K/4 (d) K/2

37. In a Young’s dou ble slit ex per i ment thein ten sity at a point where the path dif fer ence is

λ6

(λ be ing the wave length of the light used) is

I. If I0 de notes the max i mum in ten sity, I I/ 0 isequal to

(a) 1

2(b)

3

2

(c) 1

2(d)

3

4

38. Two springs, of force con stants k1 and k2, arecon nected to a mass m as shown. Thefre quency of os cil la tion of the mass is f. If both k1 and k2 are made four times their orig i nalval ues, the fre quency of os cil la tion be comes

(a) f

2(b)

f

4(c) 4 f (d) 2 f

39. When a sys tem is taken from state i to state falong the path iaf , it is found that Q = 50 caland W = 20 cal. Along the path ibf Q = 36 cal.W along the path ibf is

(a) 6 cal (b) 16 cal(c) 66 cal (d) 14 cal

40. A par ti cle of mass m ex e cutes sim ple har monicmo tion with am pli tude a and fre quency ν. Theav er age ki netic en ergy dur ing its mo tion fromthe po si tion of equi lib rium to the end is

(a) π ν2 2 2ma (b) 1

42 2ma ν

(c) 4 2 2 2π νma (d) 2 2 2 2π νma

41. The en er gies of ac ti va tion for for ward andre verse re ac tions for A B AB2 2 2+ a are

180 kJ mol−1 and 200 kJ mol−1 re spec tively.The pres ence of a cat a lyst low ers the ac ti va tionen ergy of both (for ward and re verse) re ac tionsby 100 kJ mol−1. The enthalpy change of the

re ac tion ( )A B AB2 2 2+ → in the pres ence ofcat a lyst will be (in kJ mol−1)

(a) 300 (b) 120(c) 280 (d) 20

42. The cell, Zn|Zn (1M)||Cu (1M)|Cu2+ 2+

( .E° =cell 1 10 V), was allowed to be completely

4

k1 k2m

i

a f

b

Chemistry

discharged at 298 K. The relative concentration

of Zn2+ to Cu2+ [ ]

[ ]

Zn

Cu

2

2

+

+

is

(a) antilog (24.08) (b) 37.3

(c) 1037 3. (d) 9 65 104. ×

43. The pKa of a weak acid ( )H A is 4.5. The pOH of

an aqueous buffered solution of HA in which

50% of the acid ionised is

(a) 4.5 (b) 2.5

(c) 9.5 (d) 7.0

44. Consider the reaction,

2 A B+ → products

When concentration of B alone was doubled,

the half-life did not change. When the

concentration of A alone was doubled, the rate

increased by two times. The unit of rate

constant for this reaction is

(a) L mol s− −1 1 (b) no unit

(c) mol L s− −1 1 (d) s−1

45. Identify the incorrect statement among the

following

(a) d-block elements show irregular and erratic

chemical properties among themselves

(b) La and Lu have partially filled d orbitals

and no other partially filled orbitals

(c) The chemistry of various lanthanoids is

very similar(d) 4 f and 5 f orbitals are equally shielded

46. Which one of the following has a square planar

geometry ?

(a) [CoCl ]42– (b) [FeCl ]4

2–

(c) [NiCl ]42– (d) [PtCl ]4

2–

(At. no. Co = 27, Ni = 28, Fe = 26, Pt = 78)

47. Which of the following molecules is expected

to rotate the plane of plane-polarised light ?

48. The secondary structure of a protein refers to

(a) α-helical backbone

(b) hydrophobic interactions

(c) sequence of α-amino acids

(d) fixed configuration of the polypeptide

backbone

49. Which of the following reactions will yield,

2, 2-dibromopropane ?

(a) CH C CH+ 2HBr3 ≡≡ →(b) CH CH ==CHBr+ HBr3 →(c) CH CH+ 2HBr≡≡ →(d) CH CH ==CH + HBr3 2 →

50. In the chemical reaction,

CH CH NH + CHCl 3KOH3 2 2 3 + →

( ) ( )A B+ + 3H O2 , the compounds (A) and

(B) are respectively

(a) C H CN2 5 and 3KCl

(b) CH CH CONH3 2 2 and 3KCl

(c) C H NC2 5 and K CO2 3

(d) C H NC2 5 and 3KCl

51. The reaction of toluene with Cl2 in presence of

FeCl3 gives predominantly

(a) benzoyl chloride

(b) benzyl chloride

(c) o-and p-chlorotoluene(d) m-chlorotoluene

52. Presence of a nitro group in a benzene ring

(a) activates the ring towards electrophilicsubstitution

(b) renders the ring basic(c) deactivates the ring towards nucleophilic

substitution(d) deactivates the ring towards electrophilic

substitution

53. In which of the fol low ing ion is ation pro cesses,the bond or der has in creased and the mag neticbe hav iour has changed ?

(a) C C2 2+→ (b) NO NO+→

(c) O O2 2+→ (d) N N2 2

+→

54. The actinoids ex hibit more num ber ofox i da tion states in gen eral than thelanthanoids. This is be cause

(a) the 5 f orbitals are more buried than the 4 forbitals

(b) there is a similarity between 4 f and 5 forbitals in their angular part of the wavefunction

(c) the actinoids are more reactive than thelanthanoids

(d) the 5 f orbitals extend farther from thenucleus than the 4 f orbitals

5

CHO

H

CH OH2

HO(a)

SH

(b)

COOH

H

H

H N2

H N2

Ph

H

Ph

NH2

H(c) (d)

55. Equal masses of methane and oxygen aremixed in an empty container at 25°C. Thefraction of the total pressure exerted by oxygenis

(a) 2

3(b)

1

3

273

298×

(c) 1

3(d)

1

2

56. A 5.25% so lu tion of a sub stance is iso tonic witha 1.5% so lu tion of urea (mo lar mass = 60 g mol−1) in the same sol vent. If the den si ties of

both the so lu tions are as sumed to be equal to1.0 g cm−3, mo lar mass of the sub stance will be

(a) 90.0 g mol–1 (b) 115.0 g mol–1

(c) 105.0 g mol–1 (d) 210.0 g mol–1

57. As sum ing that wa ter vapour is an ideal gas, thein ter nal en ergy change ( )∆U when 1 mole ofwa ter is vapourised at 1 bar pres sure and100°C, (Given : mo lar enthalpy of vapori sa tionof wa ter at 1 bar and 373 K = 41 kJ mol−1 and

R = 8 3. J mol−1K−1) will be

(a) 4.100 kJ mol−1 (b) 3.7904 kJ mol−1

(c) 37.904 kJ mol−1 (d) 41.00 kJ mol−1

58. In a saturated solution of the sparingly soluble

strong electrolyte AgIO3 (Molecular mass

= 283) the equilibrium which sets in is

AgIO Ag IO3 3( ) ( ) ( )s aq aqa+ −+

if the solubility product constant K sp of AgIO3

at a given temperature is 1 0 10 8. × − , what is the

mass of AgIO3 contained in 100 mL of its

saturated solution ?

(a) 28 3 10 2. × − g (b) 2 83 10 3. × − g

(c) 1 0 10 7. × − g (d) 1 0 10 4. × − g

59. A radioactive element gets spilled over the

floor of a room. Its half-life period is 30 days. If

the initial activity is ten times the permissible

value, after how many days will it be safe to

enter the room ?

(a) 1000 days (b) 300 days

(c) 10 days (d) 100 days

60. Which one of the following conformations of

cyclohexane is chiral ?

(a) Twist boat (b) Rigid

(c) Chair (d) Boat

61. Which of the following is the correct order of

decreasing S 2N reactivity ?

(a) R X R X R XCH C CH22 3> >

(b) R X R X R XCH CH C2 2 3> >(c) R X R X R X3 2 2C CH CH> >(d) R X R X R X2 3 2CH C CH> >

(X a= halogen)

62. In the following sequence of reactions,

CH CH OH3 2

P + I Mg

ether

HCHO2→ → →A B

C D→H O2

the compound ‘D’ is

(a) butanal (b) n-butyl alcohol

(c) n-propyl alcohol (d) propanal

63. Which of the following sets of quantum

numbers represents the highest energy of an

atom ?

(a) n l m s= = = = +3 1 1 1 2, , , /

(b) n l m s= = = = +3 2 1 1 2, , , /

(c) n l m s= = = = +4 0 0 1 2, , , /

(d) n l m s= = = = +3 0 0 1 2, , , /

64. Which of the following hydrogen bonds is the

strongest ?

(a) O H N K (b) F H F K

(c) O H O K (d) O H F K

65. In the reaction,

2Al 6HCl 2Al3+( ) ( ) ( )s aq aq+ →

+ +−6Cl 3H( ) ( )aq g2

(a) 6 L HCl( )aq is consumed for every 3L H2( )g

produced(b) 33.6 L H2 (g) is produced regardless of

temperature and pressure for every mole

Al that reacts(c) 67.2 L H2( )g at STP is produced for every

mole Al that reacts(d) 11.2 L H2( )g at STP is produced for every

mole HCl(aq) consumed

66. Regular use of which of the following fertilisers

increases the acidity of soil ?

(a) Potassium nitrate

(b) Urea

(c) Superphosphate of lime

(d) Ammonium sulphate

67. Identify the correct statement regarding a

spontaneous process

(a) For a spontaneous process in an isolated

system, the change in entropy is positive

(b) Endothermic processes are never

spontaneous

6

(c) Exothermic processes are always

spontaneous(d) Lowering of energy in the reaction process

is the only criterion for spontaneity

68. Which of the following nuclear reactions will

generate an isotope ?

(a) Neutron particle emission

(b) Positron emission

(c) α-particle emission

(d) β-particle emission

69. The equivalent conductances of two strong

electrolytes at infinite dilution in H O2 (where

ions move freely through a solution) at 25°C

are given below

Λ°CH COONa2

3S cm /equiv= 91 0.

Λ°HCl S cm equiv= 426 2 2. /

What additional information/quantity one

needs to calculate Λ° of an aqueous solution of

acetic acid ?

(a) Λ° of NaCl

(b) Λ° of CH COOK3

(c) The limiting equivalent conductance of

HH

+ ° +( )λ(d) Λ° of chloroacetic acid (ClCH COOH)2

70. Which one of the following is the strongestbase in aqueous solution ?(a) Trimethylamine (b) Aniline(c) Dimethylamine (d) Methylamine

71. The compound formed as a result of oxidationof ethyl benzene by KMnO4 is

(a) benzophenone (b) acetophenone(c) benzoic acid (d) benzyl alcohol

72. The IUPAC name of is

(a) 1, 1-diethyl-2, 2-dimethylpentane(b) 4, 4-dimethyl-5, 5-diethylpentane(c) 5, 5-diethyl-4, 4-dimethylpentane(d) 3-ethyl-4, 4-dimethylheptane

73. Which of the fol low ing spe cies ex hib its thedia mag netic be hav iour ?

(a) O22− (b) O2

+

(c) O2 (d) NO

74. The stability of dihalides of Si, Ge, Sn and Pbincreases steadily in the sequence

(a) Ge Si Sn Pb2 2 2 2X X X X< < <(b) Si Ge Pb Sn2 2 2 2X X X X< < <

(c) Si Ge Sn Pb2 2 2 2X X X X< < <(d) Pb Sn Ge Si2 2 2 2X X X X< < <

75. Identify the incorrect statement among thefollowing

(a) Ozone reacts with SO2 to give SO3

(b) Silicon reacts with NaOH( )aq in thepresence of air to give Na SiO2 3 and H O2

(c) Cl2 reacts with excess of NH3 to give N2

and HCl (d) Br2 reacts with hot and strong NaOH

solution to give NaBr, NaBrO3 and H O2

76. The charge/size ra tio of a cat ion de ter mines itspo lar iz ing power. Which one of the fol low ingse quences rep re sents the in creas ing or der ofthe po lar iz ing power of the cationic spe cies,

K+ , Ca2+ , Mg2+ , Be2+ ?

(a) Mg < Be < K < Ca2+ 2+ + 2+

(b) Be < K < Ca < Mg2+ + 2+ 2+

(c) K < Ca < Mg < Be+ 2+ 2+ 2+

(d) Ca < Mg < Be < K2+ 2+ 2+ +

77. The den sity (in g mL−1) of a 3.60 M sulphuric

acid so lu tion that is 29% H SO2 4 (mo lar mass = 98 g mol−1) by mass will be

(a) 1.64 (b) 1.88(c) 1.22 (d) 1.45

78. The first and sec ond dis so ci a tion con stants ofan acid H2A are 1.0 10–5× and 5 0 10 10. × −

re spec tively. The over all dis so ci a tion con stantof the acid will be

(a) 5 0 10 5. × − (b) 5 0 1015. ×(c) 5 0 10 15. × − (d) 0 2 105. ×

79. A mix ture of ethyl al co hol and propyl al co holhas a vapour pres sure of 290 mm at 300 K. Thevapour pres sure of propyl al co hol is 200 mm. Ifthe mole frac tion of ethyl al co hol is 0.6, itsvapour pres sure (in mm) at the sametem per a ture will be

(a) 350 (b) 300(c) 700 (d) 360

80. In conversion of limestone to lime,

CaCO CaO CO3 2( ) ( ) ( )s s g→ +the values of ∆H° and ∆S° are +179.1 kJ mol−1

and 160.2 J/K respectively at 298 K and 1 bar.Assuming that ∆H° and ∆S° do not change withtemperature, temperature above whichconversion of limestone to lime will bespontaneous is

(a) 1008 K (b) 1200 K(c) 845 K (d) 1118 K

7

81. In a geo met ric pro gres sion con sist ing ofpos i tive terms, each term equals the sum of thenext two terms. Then the com mon ra tio of thispro gres sion equals

(a) 1

21 5( )− (b)

1

25

(c) 5 (d) 1

25 1( )−

82. If sin− −

+

=1 1

5

5

4 2

xcosec

π, then a value of

x is

(a) 1 (b) 3(c) 4 (d) 5

83. In the bi no mial ex pan sion of ( )a b n− , n ≥ 5, the

sum of 5th and 6th terms is zero, then a

b equals

(a) 5

4n −(b)

6

5n −

(c) n − 5

6(d)

n − 4

5

84. The set S : , , , . . . , = 1 2 3 12 is to bepartitioned into three sets A B C, , of equal size.Thus, A B C S∪ ∪ = ,

A B B C A C∩ = ∩ = ∩ = φ.

The number of ways to partition S is

(a) 12!/3! ( !)4 3 (b) 12!/3!( !)3 4

(c) 12!/( !)4 3 (d) 12!/( !)3 4

85. The larg est in ter val ly ing in −

π π2 2

, for which

the func tion

f xx

xx( ) cos log (cos )= + −

+

− −42

12 1

is defined, is

(a) [ , ]0 π (b) −

π π2 2

,

(c) −

π π4 2

, (d) 02

,π

86. A body weighing 13 kg is suspended by twostrings 5 m and 12 m long, their other endsbeing fastened to the extremities of a rod 13 mlong. If the rod be so held that the body hangsimmediately below the middle point. Thetensions in the strings are

(a) 12 kg and 13 kg (b) 5 kg and 5 kg(c) 5 kg and 12 kg (d) 5 kg and 13 kg

87. A pair of fair dice is thrown independently

three times. The probability of getting a score

of exactly 9 twice is

(a) 1/729 (b) 8/9

(c) 8/729 (d) 8/243

88. Con sider a fam ily of cir cles which are pass ing

through the point (–1, 1) and are tan gent to

x-axis. If ( , )h k are the co or di nates of the cen tre

of the cir cles, then the set of val ues of k is given

by the in ter val

(a) 0 1 2< <k / (b) k ≥ 1 2/

(c) − ≤ ≤1 2 1 2/ /k (d) k ≤ 1 2/

89. Let L be the line of in ter sec tion of the planes

2 3 1x y z+ + = and x y z+ + =3 2 2. If L

makes an an gle α with the pos i tive x-axis, then

cos α equals

(a) 1 3/ (b) 1 2/

(c) 1 (d) 1 2/

90. The dif fer en tial equa tion of all cir cles pass ing

through the or i gin and hav ing their cen tres on

the x-axis is

(a) x y xydy

dx2 2= +

(b) x y xydy

dx2 2 3= +

(c) y x xydy

dx2 2 2= +

(d) y x xydy

dx2 2 2= −

91. If p and q are pos i tive real num bers such that

p q2 2 1+ = , then the max i mum value of ( )p q+is

(a) 2 (b) 1

2

(c) 1

2(d) 2

92. A tower stands at the cen tre of a cir cu lar park.

A and B are two points on the bound ary of the

park such that A B a( )= sub tends an an gle of

60° at the foot of the tower and the an gle of

el e va tion of the top of the tower from A or B is

30°. The height of the tower is

(a) 2

3

a(b) 2 3a

(c) a

3(d) a 3

8

Mathematics

93. The sum of the series 20

020

120

220

320

10C C C C C− + − + − +K K is

(a) − 2010C (b)

1

220

10C

(c) 0 (d) 2010C

94. The nor mal to a curve at P x y( , ) meets the

x-axis at G. If the dis tance of G from the or i gin

is twice the ab scissa of P, then the curve is a

(a) ellipse (b) parabola

(c) circle (d) hyperbola

95. If | |z + ≤4 3, then the maximum value of

| |z + 1 is

(a) 4 (b) 10

(c) 6 (d) 0

96. The re sul tant of two forces P N and 3 N is a

force of 7 N. If the di rec tion of the 3 N force

were re versed, the re sul tant would be 19 N.

The value of P is

(a) 5 N (b) 6 N

(c) 3 N (d) 4 N

97. Two aeroplanes I and II bomb a target in

succession. The probabilities of I and II scoring

a hit correctly are 0.3 and 0.2, respectively. The

second plane will bomb only if the first misses

the target. The probability that the target is hit

by the second plane is

(a) 0.06 (b) 0.14

(c) 0.2 (d) 0.7

98. If D xy

= ++

111

11

1

11

1 for x y≠ ≠0 0, , then

D is

(a) divisible by neither x nor y

(b) divisible by both x and y

(c) divisible by x but not y

(d) divisible by y but not x

99. For the hy per bola x y2

2

2

21

cos sinα α− = , which of

the fol low ing re mains con stant when α var ies ?

(a) Eccentricity

(b) Directrix

(c) Abscissae of vertices

(d) Abscissae of foci

100. If a line makes an an gle of π4

with the pos i tive

di rec tions of each of x-axis and y-axis, then the

an gle that the line makes with the pos i tive

di rec tion of the z-axis is

(a) π6

(b) π3

(c) π4

(d) π2

101. A value of C for which the con clu sion of Mean

Value The o rem holds for the func tion

f x xe( ) log= on the in ter val [1, 3] is

(a) 2 3log e (b) 1

23loge

(c) log3 e (d) loge 3

102. The func tion f x x x( ) tan (sin cos )= +−1 is an

in creas ing func tion in

(a) ( / , / )π π4 2 (b) ( / , / )−π π2 4

(c) ( , / )0 2π (d) ( / , / )−π π2 2

103. Let A =

500

5

055

αα

αα

If | |A2 25= , then | |α equals

(a) 52 (b) 1

(c) 1

5(d) 5

104. The sum of the se ries 1

2

1

3

1

4! ! !− + − K upto

in fin ity is

(a) e−2 (b) e−1

(c) e−

1

2 (d) e+

1

2

105. If $u and $v are unit vec tors and θ is the acute

an gle be tween them, then 2 3$ $u v× is a unit

vec tor for

(a) exactly two values of θ(b) more than two values of θ(c) no value of θ(d) exactly one value of θ

106. A par ti cle just clears a wall of height b at a

dis tance a and strikes the ground at a dis tance c

from the point of pro jec tion. The an gle of

pro jec tion is

(a) tan−1 b

ac(b) 45°

(c) tan( )

−

−1 bc

a c a(d) tan−1 bc

a

107. The average marks of boys in a class is 52 and

that of girls is 42. The average marks of boys

and girls combined is 50. The percentage of

boys in the class is

(a) 40 (b) 20

(c) 80 (d) 60

9

108. The equation of a tangent to the parabola

y x2 8= is y x= + 2 . The point on this line

from which the other tangent to the parabola is

perpendicular to the given tangent is

(a) ( , )−1 1 (b) (0, 2)

(c) (2, 4) (d) ( , )−2 0

109. If (2, 3, 5) is one end of a diameter of the

sphere x y z x y z2 2 2 6 12 2+ + − − − + =20 0,

then the coordinates of the other end of the

diameter are

(a) ( , , )4 9 3− (b) (4, –3, 3)

(c) (4, 3, 5) (d) (4, 3, –3)

110. Let a i j k b i j k→ →

= + + = − +$ $ $ , $ $ $2 and

c i j k→

= + − −x x$ ( )$ $2 . If the vector c→

lies in the

plane of a→

and b→

, then x equals

(a) 0 (b) 1

(c) – 4 (d) – 2

111. Let A h k( , ), B( , )1 1 and C( , )2 1 be the vertices of

a right angled triangle with AC as its

hypotenuse. If the area of the triangle is 1, then

the set of values which ‘k’ can take is given by

(a) , 1 3 (b) 0, 2

(c) –1, 3 (d) –3, –2

112. Let P Q= − =( , ), ( , )1 0 0 0 and R = ( , )3 3 3 be

three points. The equation of the bisector of the

angle PQR is

(a) 3 0x y+ = (b) x y+ =3

20

(c) 3

20x y+ = (d) x y+ =3 0

113. If one of the lines of

my m xy mx2 2 21 0+ − − =( ) is a bisector of the

angle between the lines xy = 0, then m is

(a) − 1

2(b) −2

(c) ± 1 (d) 2

114. Let F x f x fx

( ) ( )= +

1, where

f xt

tdt

x( )

log.=

+∫1 1 Then F e( ) equals

(a) 1

2(b) 0

(c) 1 (d) 2

115. Let f R R: → be a function defined by f x x x( ) min ,| | = + +1 1 . Then which of thefollowing is true ?

(a) f x( ) ≥ 1 for all x R∈(b) f x( ) is not differentiable at x = 1(c) f x( ) is differentiable everywhere(d) f x( ) is not differentiable at x = 0

116. The function f R R: / 0 → given by

f xx e x

( ) = −−

1 2

12

can be made continuous at x = 0 by defining f( )0 as

(a) 2 (b) –1(c) 0 (d) 1

117. The solution for x of the equation

2 2 1 2

x dt

t t∫

−= π

is

(a) 2 (b) π(c) 3 2/ (d) 2 2

118. ∫ +dx

x xcos sin3 equals

(a) 1

2 2 12log tan

xc+

+π

(b) 1

2 2 12log tan

xc−

+π

(c) log tanx

c2 12

+

+π

(d) log tanx

c2 12

−

+π

119. The area enclosed between the curves y x2 =and y x=| | is

(a) 2

3 sq unit (b) 1 sq unit

(c) 1

6 sq unit (d)

1

3 sq unit

120. If the difference between the roots of theequation x ax2 1 0+ + = is less than 5, then

the set of possible values of a is

(a) ( , )−3 3 (b) ( , )− ∞3(c) ( , )3 ∞ (d) ( , )−∞ − 3

10

11

Answers

à PHYSICS

1. (a) 2. (b) 3. (b) 4. (d) 5. (d) 6. (c) 7. (c) 8. (b)9. (b) 10. (d) 11. (d) 12. (b) 13. (c) 14. (d) 15. (d) 16. (b)

17. (b) 18. (d) 19. (a) 20. (d) 21. (c) 22. (c) 23. (b) 24. (c)25. (d) 26. (a) 27. (a) 28. (b) 29. (c) 30. (d) 31. (a) 32. (a)33. (a) 34. (c) 35. (a) 36. (c) 37. (d) 38. (d) 39. (a) 40. (a)

à CHEMISTRY

41. (d) 42. (c) 43. (c) 44. (a) 45. (d) 46. (d) 47. (a) 48. (a)49. (a) 50. (d) 51. (c) 52. (d) 53. (b) 54. (d) 55. (c) 56. (d)57. (c) 58. (b) 59. (d) 60. (c) 61. (b) 62. (c) 63. (b) 64. (b)65. (d) 66. (d) 67. (a) 68. (a) 69. (a) 70. (c) 71. (c) 72. (d)73. (a) 74. (c) 75. (d) 76. (c) 77. (c) 78. (c) 79. (a) 80. (d)

à MATHEMATICS

81. (d) 82. (b) 83. (d) 84. (c) 85. (d) 86. (c) 87. (d) 88. (b)89. (a) 90. (c) 91. (d) 92. (c) 93. (b) 94. (a,d) 95. (c) 96. (a)97. (*) 98. (b) 99. (d) 100. (d) 101. (a) 102. (b) 103. (c) 104. (b)

105. (d) 106. (c) 107. (c) 108. (d) 109. (a) 110. (d) 111. (c) 112. (a)113. (c) 114. (a) 115. (c) 116. (d) 117. (*) 118. (a) 119. (c) 120. (a)

Note. None of the option is match (*).

1. x t= × −( )cos2 10 2 π

Here, a = × −2 10 2m = 2 cm

At t = 0, x = 2 cm, i.e., the object is at positive

extreme, so to acquire maximum speed (ie, to

reach mean position) it takes 1

4th of time

period.

∴ Required time = T

4

where ω π π= =2

T

⇒ T = 2s

So, required time = =T

4

2

4 = 0 5. s

2. For given circuit current is lagging the voltageby π/2, so circuit is purely inductive and thereis no power consumption in the circuit. Thework done by battery is stored as magneticenergy in the inductor.

3. Potential at A due to charge at O

VOA

A =4 ε

−1 10

0

3

π( )

=ε

⋅+

−1

4

10

2 20

3

2 2π( )

( ) ( )

Potential at B due to charge at O

VOB

B =ε

⋅−1

4

10

0

3

π( )

= ⋅−1

4

10

20

3

πε( )

So, V VA B− = 0

4. Ratio of energy stored in the capacitor and thework done by the battery

= =

1

2 1

2

qV

qV

5. Rise of current in L-R circuit is given by

I I e t= − −0 1( )/ τ

where IE

R0 = = =5

51 A

Now, τ = = =L

R

10

52s

After 2s, ie, at t = 2s

Rise of current I e= − −( )1 1 A

6. Current density Ji

a=

π 2

From Ampere’s circuital law

B dl i⋅ = ⋅∫ µ0 enclosed

For r a<B r J r× = × ×2 0

2π µ π

⇒ Bi

a

r= ×µπ

02 2

At r a= /2 Bi

a1

0

4= µ

π

For r a>

B r i Bi

r× = ⇒ =2

20

0π µ µπ

At r a= 2 , Bi

a2

0

4= µ

π

So, B

B1

2

1=

7. Using Ampere’s circuital law the magnetic field

at any point inside the pipe is zero.

8. Binding energy

BE nucleus nucleons= −( )M M c2

= − −( )M M M cO p n8 9 2

12

r

a

10H 5Ω

5V

Y

B

A

X(0,0) (2,0)

–310 µc

O

2, 2 ))

Hints & Solutions

Physics

9. In gamma ray emis sion the en ergy is re leasedfrom nu cleus, so that nu cleus get sta bi lised.

10. For Vi < 0

the diode is reverse biased and hence offerinfinite resistance, so circuit would be like asshown in Fig. (2) and Vo = 0.For Vi > 0, the diode is forward biased andcircuit would be as shown in Fig. (3) andV Vo i= .

Hence, the opitcal (d) is correct.

11. The momentum of the photon

ph h

c= =

λν

12. ν = + +v gt ft02

or dx

dtv gt ft= + +0

2

⇒ dx v gt ft dt= + +( )02

So, 0 0

1

02x

dx v gt ft dt∫ ∫= + +( )

⇒ x v g f= + +0 2 3/ /

13. Let the each side of square lamina is d.

So, I IEF GH= (due to symmetry)

and I IAC BD= (due to symmetry)

Now, according to theorem of perpendicularaxis,

I I IAC BD+ = 0

⇒ 2 0I IAC = …(i)and I I IEF GH+ = 0

⇒ 2 0I IEF = …(ii)

From Eqs. (i) and (ii), we get I IAC EF=

∴ I Imd

AD EF= +2

4

= +md md2 2

12 4 as I

mdEF =

2

12

So, Imd

IAD EF= =2

34

14. x x t= −0 4cos( / )ω π

Acceleration, ad x

dt=

2

2

= − −

ω ω π20

3

4x tcos

= +

ω ω π20

3

4x tcos

So, A x= ω20

and δ π= 3

4

15. Direction of E→

reverses while magnituderemains same and V remains unchanged.

16. T X Y1 2/ ( ) ( )= τ

⇒ 0 693 1.

λ λX Y

=

⇒ λ λY

X=0 693.

⇒ λ λY X>

So, Y will decay faster than X.

17. For Carnot engine using as refrigerator

W QT

T= −

2

1

2

1

It is given η = 1

10

⇒ η = −1 2

1

T

T

⇒ T

T2

1

9

10=

So, Q2 90= J (as W = 10 J)

18. The number of free electrons for conduction issignificant only in Si and Ge but small in C, as Cis an impurity.

13

Vi Vo

Fig. (2)

Vi Vo

Fig. (3)

5V

–5V

RLVi Vo

Fig. (1)

D

G

AE

B

H

CF

d

O

19. As v→

of charged particle is remaining constant,it means force acting on charged particle iszero.

So, q q( )v B E→ → →

× =

⇒ v B E→ → →

× =

⇒ vE B→→ →

= ×B2

20. E i j k→

= − ∂∂

− ∂∂

− ∂∂

V

x

V

y

V

z$ $ $

⇒ EV

xx = − ∂

∂ = −

−

d

dx x

20

42

=−

40

42 2

x

x( )

⇒ Ex at x = =410

9µm V/µm

and is along positive x direction.

21. Emission spectrum would be when electronmakes a jump from higher energy level tolower energy level.

Frequency of emitted photon is proportional tochange in energy of two energy levels, i.e.,

ν = −

RcZ

n n

2

12

22

1 1

22. Acceleration of system, aF

m M=

+

So, force acting on mass m ma=

=+

mF

m M

23. Power of a lens is reciprocal of its focal length.

Power of combined lens is

P P P= +1 2

= − + = −15 5 10 D

∴ fP

= =−

1 100

10cm

f = −10 cm

24. Let temperature at the interface is T.

For part AB,

Q

t

T T K

l1 1 1

1

∝ −( )

For part BC,Q

t

T T K

l2 2 2

2

∝ −( )

At equilibrium, Q

t

Q

t1 2−

∴ ( ) ( )T T K

l

T T K

l1 1

1

2 2

2

− = −

⇒ TT K l T K l

K l K l= +

+1 1 2 2 2 1

1 2 2 1

25. Let intensity of sound be I and I′.Loudness of sound initially

β10

10=

log

I

I

Later, β 20

10= ′

log

I

I

Given, β β2 1 20− =

∴ 20 10= ′

logI

I

∴ I I′ = 100

26. According to Mayer’s relation,

C CR

m

Rp V− = =

28

27. In case of motion of a charged particleperpendicular to the motion, i.e.,displacement, the work done

W Fds= ⋅ = =→ →

∫ ∫F ds cosθ 0 (as θ = °90 )

and by work-energy theorem, W = ∆KE, thekinetic energy and hence speed v remains

constant. But v→

changes, so, momentum changes.

28. The magnetic field induction at a point P, at adistance d from O in a direction perpendicularto the plane ABCD due to currents through AOBand COD are perpendicular to each other, is

14

m Mk

F

T1l1 l2 T2

K1 K2A B C

T C

A

P

D

B

d

I1

I2O

Hence, B B B= +12

22

=

+

µπ

µπ

0 12

0 22 1 2

4

2

4

2I

d

I

d

/

= +µπ0

12

22

2 dI I( )

29. From R R tt = +0 1( )α 5 1 500= +R ( )α …(i)

and 6 1 1000= +R ( )α …(ii)

∴ 5

6

1 50

1 100

1

200= +

+⇒ =α

αα

Putting value of α in Eq (i), we get

5 1 50 1 2000= + ×R ( / )

∴ R0 4= Ω30. On in tro duc tion and re moval and again on

in tro duc tion, the ca pac ity and po ten tial re main same. So, net work done by the sys tem in thisprocess

W U Uf i= −

= −1

2

1

22 2C V C V

= 0

31. According to Millikan’s oil drop experiment,electronic charge is given by,

qr v v

E= +6 1 2πη ( )

which is independent of g.

So, electronic charge on the moon

electronic charge on the earth= 1

32. In this question distanceof centre of mass of newdisc from the centre ofmass of remaining disc is αR.

Mass of remaining disc

= − =MM M

4

3

4

∴ − + =3

4 40

MR

MRα

⇒ α = 1

3

NOTE In Q.No. 32, the given distance must be αR for real

approach to the solution.

33. Assuming that no energy is used up againstfriction, the loss in potential energy is equal tothe total gain in the kinetic energy.

Thus, Mgh I v R Mv= +1

2

1

22 2 2( / )

or 1

22 2v M I R Mgh( / )+ =

or vMgh

M I R

gh

I MR

22 2

2 2

1=

+=

+/ /

If s be the distance covered along the plane,

then

h s= sin θ

vgs

I MR

22

2

1=

+sin

/

θ

Now, v as2 2=

∴ 22

1 2as

gs

I MR=

+sin

/

θ

or ag

I MR=

+sin

/

θ1 2

34. Ac cord ing to the prin ci ple of con ser va tion of

an gu lar mo men tum, in the ab sence of ex ter nal

torque, the to tal an gu lar mo men tum of the

sys tem is constant.

35. aF

mk= = =15

27 5. m/s2

Now, ma kx= 1

22

or 2 7 51

210000 2× = × ×. x

or x2 33 10= × −

or x = 0 055. m

⇒ x = 5 5. cm

36. Kinetic energy at highest point

(KE)H mv= 1

22 2cos θ

= K cos2 θ

= °K(cos )60 2

= K

4

15

h

θ

C

s

C

3

v

RO

O1O2

αR

37. Phase difference = ×2πλ

path difference

i e, φ = ×2

6

πλ

λ

= π3

As, I I= φmax cos ( / )2 2

or I

Imax

= φcos ( / )2 2

or I

I0

2

6=

cosπ

= 3

4

38. fk k

m= +1

21 2

π

and fk k

mf′ = ⋅ + =1

22 21 2

π

39. From first law of thermodynamics,

Q U W= +∆For path iaf ,

50 20= +∆U

∴ ∆U U Uf i= − = 30 cal

For path ibf ,

Q U W= +∆or W Q U= − ∆

= −36 30 = 6 cal

40. Average kinetic energy of particle

= 1

42 2ma ω

= 1

422 2ma ( )πν

= π ν2 2 2ma

41. A B AB2 2+ 2a

Ea( )forward kJ mol–1= 180

Ea (backward) = 200 kJ mol−1

In the presence of catalyst :

Ea (forward) = − =180 100 80 kJ mol−1

Ea (backward) = − =200 100 100 kJ mol−1

∆H Ea= (forward) − Ea (backward)

= −80 100

= − 20 kJ mol−1

42. Cell is completely discharged, it meansequilibrium gets established, Ecell = 0

Zn/Zn (1M)||Cu (1M)|Cu2+ 2+

cell reaction : Zn + Cu Zn + Cu2+ 2+a

Keq = [Zn ]

[Cu

2+

2+]

We know,

E En

Keqcell cell= ° − 0 0591.log

En

Keq° =cell

0 0591.log

or 1 100 0591

2.

.log= Keq

Keq = [Zn ]

[Cu ]= antilog

2+

2+

2 20

0 0591

.

.

= antilog 37.3

43. Aqueous buffered solution of H A

50% H A is ionised

⇒ [H ] [A A= −]

Buffer solution of weak acid H A → acidicbuffer

pH = p + log[ ]

[H ]

–

KA

Aa

or pH p= =Ka 4 5.

pOH p pH= Kw –

pOH = − =14 4 5 9 5. .

44. 2 A B+ → products

[B] is doubled, half-life didn’t change

Half-life is independent of change in conc. of reactant ⇒ First order

First order w.r.t. to B

[A] is doubled, rate increased by two times

⇒ First order w.r.t A

Hence, net order of reaction = + =1 1 2

Unit for the rate constant = conc.( )1 1− −n t

= ⋅ ⋅ −( )–mol L s–1 1 1

= ⋅L mol s–1 –1

45. 4 f and 5 f belongs to different shell, experience different amount of shielding.

46. Cl− is a weak ligand but Cl− cause the pairing ofelectron with large Pt2+ and consequently give dsp2 hybridisation and square planar geometry.

16

Chemistry

47. The molecule, which is optically active, has

chiral centre, is expected to rotate the plane of

polarised light.

One chiral centre ⇒ optically active

Two chiral centres, but plane of symmetry

within molecule ⇒ optically inactive

48. Primary structure involves sequence of α-amino

acids polypeptide chain.

Secondary structure involves α-helical and

β-pleated sheet like structures.

49. CH C CH+ HBr CH C

Br

==CH3 3 2 ≡≡ →

↓

H Br+

Rearrangement3 2H C CH

Br

CH←

+

CH C

Br

Br

CH CH C

Br

CH3 3Br

3

+

dibromopropane

←

−

−

2 2,

3

more stablecarbocation

CH CH==CHBr3 → HBr3 2 2CH CH CHBr

CH CH CH CHBr2HBr

3 2≡≡ →

CH CH ==CH CH CHBr CH3 2HBr

3 3 →

50. CH CH NH + CHCl + KOH 3H O3 2 2 3 2→ + +A B

CHCl + KOH CCl + KCl+ H O3 2 2→ ••

CH CH NH + CCl CH CH NC + 2HCl3 2 2 2 3 2carbylamine

•• →

51.

52.

NO2 group withdraw electron from the ring, shows −M effect makes ring electron deficient,thus, deactivates ring for electrophilicsubstitution.

53. NO NO+→

(NO)+

Total e− = 14

σ σ σ σ π1 1 2 2 22 2 2 2 1 1s s s s px* * + = +2 21 1 2p py zσ

Diamagnetic

Bond order = − =10 4

23

(NO) Total e− = 15

σ σ σ σ1 1 2 22 2 2 2s s s s* * , σ π π2 2 22 1 1 1 1p p pz x y+ +

π π* 2 21p px y=

Paramagnetic

Bond order = − =10 5

22 5.

Electron is taken away from non-bondingmolecular orbital, that’s why bond orderincreases.

54. The actinoid (5f-elements) exhibits morenumber of oxidation states in general than thelanthanoid because 5f-orbitals extend fartherfrom the nucleus than the 4 f -orbitals.

55. Suppose the equal mass of methane andoxygen = w = 1 g

Mole fraction of oxygen =+

w

w w

/

/ /

32

32 16

= =

1

323 32

1

3/

Let the total pressure = p

17

H N2

H

Ph Ph

H

NH2

HO

CHO

* H

CH OH2

N+

O

O–

Cl FeCl2, 3

+(Cl )

CH3

Cl

CH3

Cl

CH3

toluene p-chloro toluene

o-chloro toluene

+

chlorinationelectrophilicsubstitution

Pressure exerted by oxygen (partial pressure)

= ×X PO total2 = ×P

1

3

56. Solution is isotonic.

⇒ C RT C RT1 2= C C1 2=Density of both the solutions are assumed to beequal to 1.0 g cm−3 ⇒ molality = molarity

5.25%

in 100 g 5.25 g of substance therefore, in1000 g 52.5 g of substance

Hence, 52 5 15

60

.

M= ,

M = molecular mass of the substance

M = × =52 5 60

15210

.

57. H O H O ( ) 2 2( )l g→∆n = − =1 0 1

∆ ∆ ∆E H nRT= − = − × × × = ×− −41 1 8 3 373 10 8 3 103 3. ( . )R

= 37 9. kJ mol−1

58. AgIO Ag IO3 3( ) ( ) ( )s aq aqa+ −+

Let solubility of AgIO3 be S

K sp =[Ag ][IO ]+3–

1 0 10 8 2. × =− S or S = × −1 10 4 mol/L

In 1000 mL mol of AgIO3 dissolved

= × −1 10 4 mol

In 100 mL of mole of AgIO3 dissolved

= × −1 10 5 mol

Mass of AgIO3 in 100 mL = × ×−1 10 2835

= × −2 83 10 3.

59. Activity ∝ N

N

N

n

0

1

2=

or 1

10

1

2=

n

or 10 2= n

Taking log on both sides

log log10 2= n

n = =1

0 3013 32

..

time = ×n half-life

= ×3 32 30. = 99 6. days

60.

Chair form is unsymmetrical and absence ofany element of symmetry.

61. S 2N reactions are greatly controlled by stericfactor.

R X °

CH21

> R X22

CH°

> R X33

C °

S 2N reactivity decreases as bulkyness of alkylgroup increases.

62.

63. n l+ = 5 maximum.

64. Hydrogen bond is strongest in HF due to higher electronegativity of F.

65. 2Al 6HCl 2Al Clmol

( ) ( ) ( ) ( )s aq aq aq+ → ++ −

6

3 6

+×

3H22.4 L2

3

( )g

3 2× 22.4 L H ( )g at STP is produced for 6 mole

HCl

hence, 11.2 L H2( )g at STP is produced for

1 mole HCl

66. (NH ) SO 2NH SO4 2 4

+

4 42–

acidic

→ +

67. In an isolated system where either mass and

energy are not exchanged with surrounding for

the spontaneous process, the change in entropy

is positive.

68. ZA

ZAX X n→ +− 1

01

Isotopes are species having same number of

proton, but different number of neutron.

69. We know from Kohlrausch’s law

λ° = + λ° − λ°CH COOH CH COONa HCl NaCl3 3λ°

70. In aqueous solution, basicity order :

Dimethylamine> methylamine2 1° °

> trimethylamine> aniline3°

18

CH CH OH3 2 CH CH3 2

__ __CH CH CH3 2 2

CH CH CH + Mg(OH)I3 2 2

OMgI

OH

O

CH CH MgI3 2

__I

P + I2

‘A’

‘B’‘C’

‘D’n-propyl alcohol

H O2

Mg, ether

__ __H C H

71. Any aliphatic carbon with hydrogen attachedto it, in combination with benzene ring, will beoxidised to benzoic acid by KMnO /H4

+.

72.

73. The correct option is O22−. This species has 18

electrons, which are filled in such a way that all molecular orbitals are fully filled, sodiamagnetic.

σ σ σ σ1 1 2 22 2 2 2s s s s* *, σ2 2pz , π π2 22 2p px y= ,

π π* *2 22 2p px y=

74. Due to inert pair effect, the stability of +2oxidation state increases as we move down this group.

∴ Si < Ge < Sn < Pb2 2 2 2X X X X

75. Br2 reacts with hot and strong NaOH to giveNaBr, NaBrO3 and H O2 .

6NaOH + 3Br 5NaBr+ NaBrOconc. and hot

2 3→

+ 3H O2

76. Higher the charge/size ratio, more is thepolarising power.

∴ K < Ca < Mg < Be+ 2+ 2+ 2+

77. Molarity =10 density wt. of solute

mol. wt. of the solute

× ×

density = ××

=3 60 98

10 291 21

..

78. K k k= ×1 2

= × × ×− −1 0 10 5 0 105 10. .

= × −5 10 15

79. According to Raoult’s law :

P P X P XA A B B= ° + ° 290 200 0 4 0 6= × + ° ×. .P B

P B° = 350

80. ∆ ∆S

H

T=

∆S = 160 2. J/ K

∆H J Mol= ×179 1 103. /

T = ×179 1 10

160 2

3.

.

J/ Mol

J/K

= =1117 97 1118. K K

= 1118 K.

81. Since, each term is equal to the sum of the next

two terms.

∴ ar ar arn n n− += +1 1

⇒ 1 2= +r r ⇒ r r2 1 0+ − =

⇒ r = −5 1

2Q r ≠ − −

5 1

2

82. Since, sin− −

+

=1 1

5

5

4 2

xcosec

π

⇒ sin sin− −

+

=1 1

5

4

5 2

x π

⇒ sin sin− −

= −

1 1

5 2

4

5

x π

⇒ sin cos− −

=

1 1

5

4

5

x

⇒ sin sin− −

=

1 1

5

3

5

x

⇒ x = 3.

83. Since, in binomial expansion of ( )a b n− , n ≥ 5,

the sum of fifth and sixth term is equal to zero.

∴ n n n nC a b C a44 4

55 55 0− −− + − =( ) ( )

⇒ n

na b

n

na bn n!

( )! !

!

( )! !−⋅ −

−=− −

4 4 5 504 4 5 5

⇒ n

na b

a

n

bn!

( )! !−⋅

−−

=−

5 4 4 505 4

⇒ a

b

n= − 4

5 .

84. Required number of ways = × ×124

84

44C C C

=×

××

×12

8 4

8

4 41

!

! !

!

! !

= 12

4 3

!

( !).

85. 42−x is defined for −

π π2 2

, .

cos− −

1

21

x is defined if − ≤ − ≤1

21 1

x.

19

7

6

54

3

2

1

3-ethyl-4, 4-dimethylheptane

Mathematics

⇒ 02

2≤ ≤x ⇒ 0 4≤ ≤x

and log (cos )x is defined, if cos x > 0

⇒ − < <π π2 2

x

Hence,

f xx

xx( ) cos log (cos )= + −

+− −42

12 1

is defined, if x ∈

02

, .π

86. Since, OC CA CB= =⇒ ∠ = ∠AOC OAC and ∠ = ∠COB OBC

∴ sin sinθ = =A5

13

and cos θ = 12

13

Now by Lami’s theoremT T1 2

180 90sin ( ) sin ( )° −=

° +θ θ

⇒ T T1 2

sin cosθ θ=

⇒ T T1 2cos sinθ θ=

⇒ T T1 2

12

13

5

13

=

⇒ T T1 2

5

12=

.

Also, T T1 2 13sin cosθ θ+ =

⇒ T2

5

12

5

13

12

1313⋅ +

=

⇒ T2

169

12 1313

⋅

=

⇒ T T2 112 5= =kg and kg.

87. Prob a bil ity of get ting score 9 in a sin gle throw

= =4

36

1

9.

Probability of getting score 9 exactly twice

= ×

× =32

21

9

8

9

8

243C .

88. Equa tion of cir cle which touches x-axis andcoordinates of centre are ( , ),h k is

( ) ( )x h y k k− + − =2 2 2

Q It is passing through (– 1, 1), then

( ) ( )− − + − =1 12 2 2h k k

⇒ h h k2 2 2 2 0+ − + =Since, h is real, therefore

D ≥ 0

⇒ 2 1 01

2k k− ≥ ⇒ ≥ .

89. If di rec tion co sines of L be l, m, n, then

2 3 0l m n+ + = K(i)

and l m n+ + =3 2 0 K(ii)

On solving Eqs. (i) and (ii), we getl m n

3 3 3=

−=

∴ l m n: : : := −

1

3

1

3

1

3

⇒ l = 1

3 ⇒ cos .α = 1

3

90. Gen eral equa tion of all such cir cles which passthrough the origin and whose centre lie onx-axis, is

x y gx2 2 2 0+ + = . K(i)

On differentiating w.r.t. x, we get

2 2 2 0x ydy

dxg+ + =

2 2 2g x ydy

dx= − +

On putting the value of 2g in Eq. (i), we get

x y x ydy

dxx2 2 2 2 0+ + − −

=

⇒ x y x xydy

dx2 2 22 2 0+ − − =

⇒ y x xydy

dx2 2 2= +

Which is required equation.

91. Using AM ≥ GM,

p q

pq2 2

2

+ ≥

⇒ pq ≤ 1

2(Q p q2 2 1+ = )

20

π/2 − θθπ/2 − θ θ

T2

T15 m

12 m

T sin θ2 O T cos θ1

13 kg

A

C

B

13 m

Now ( )p q p q pq+ = + +2 2 2 2

⇒ ( )p q pq+ = +2 1 2

⇒ ( )p q+ ≤ +2 1 1

⇒ p q+ ≤ 2 .

92. Let h be the height of a tower,

Q ∠ = °AOB 60

∴ ∆ OAB is equi lat eral.

∴ OA OB AB a= = =Now in ∆OAC,

tan 30° = h

a

⇒ 1

3= h

a

⇒ ha=3

.

93. We know that

( ) . . .1 20 200

201

2010

10+ = + + + +x C C x C x

. . . + 2020

20C x

On putting x = −1, we get

0 200

201

209

2010

2011= − + − + − +C C C C C. .

. . .+ 2020C

⇒ 0 200

201

209

2010= − + − +C C C C. . .

− + +209

200C C. . .

⇒ 0 2 200

201

209

2010= − + − +( . . . . . )C C C C

⇒ 2010

200

201

20102C C C C= − + +( . . . )

⇒ 200

201

2010

2010

1

2C C C C− + + =. . . . .

94. Equa tion of nor mal is Y ydx

dyX x− = − −( )

⇒ G x ydy

dx= +

, 0

According to question

x ydy

dxx+ = 2

⇒ ydy

dxx= or y

dy

dxx= − 3

⇒ y dy x dx= or y dy x dx= − 3

⇒ y x

c2 2

2 2= + or

y xc

2 2

2

3

2= − +

⇒ x y c2 2 2− = − or 3 22 2x y c+ = .

95. From the Ar gand di a gram max i mum value of | |z + 1 is 6.

Alternative

| | | | | | | |z z z+ = + − ≤ + + − ≤1 4 3 4 3 6

Thus maximum value of | |z + 1 is 6.

96. 7 3 2 32 2 2= + + × ×P P cosθ ... (i)

and ( ) ( ) ( ) cos19 3 2 32 2 2= + − + × − ×P P θ

...(ii)

On adding Eqs. (i) and (ii), we get

68 2 18 52= + ⇒ =P P N

97. The de sired prob a bil ity

= (0.7) (0.2) + (0.7) (0.8) (0.7) (0.2) +

(0.7) (0.8) (0.7) (0.8) (0.7) (0.2) + .....

= + + +0 14 1 0 56 0 56 2. [ ( . ) ( . ) . . . . ]

=−

= = =0 14

1

1 0 56

0 14

0 44

7

220 32.

.

.

.. .

From above it is clear that no option is correct.

98. Given, D x

y

= ++

1 1 1

1 1 1

1 1 1

Applying C C C C C C2 2 1 3 3 1→ − → −and

= =1 0 0

1 0

1 0

x

y

xy .

Hence, D is divisible by both x and y.

99. The given equation of hyperbola is x y2

2

2

21

cos sin.

α α− = Here, a2 2= cos α and

b2 2= sin α

21

P

7

3O

θ

−3

19

(−7, 0) (−4, 0) (−1, 0)

y

x

O

B

A

C

30°

30°a a

a

Coordinates of foci are (± ae, 0).

Q eb

a= +1

2

2

⇒ e = + = +1 12

22sin

costan

αα

α

⇒ e = sec α.

Hence, abscissae of foci remain constant when

α varies.

100. Since, a line makes an angle of π4

with positive

direction of each of x-axis and y-axis, therefore

α π β π= =4 4

,

We know that cos cos cos2 2 2 1α β γ+ + =

⇒ cos cos cos2 2 2

4 41

π π γ+ + =

⇒ 1

2

1

212+ + =cos γ

⇒ cos2 0γ = ⇒ γ = °90 .

101. Using Mean Value theorem

′ = −−

f cf f

( )( ) ( )3 1

3 1

⇒ 1 3 1

2ce e= −log log

⇒ c ee

= =2

32 3

loglog .

102. Since, f x( ) tan= −1 (sin cos )x x+

∴ ′ =+ +

−f xx x

x x( )(sin cos )

(cos sin )1

1 2

=+

+ +

24

1 2

cos

(sin cos )

x

x x

π

f (x) is increasing, if − < + <π π π2 4 2

x

⇒ − < <3

4 4

π πx

Hence, f (x) is increasing when x ∈ −

π π2 4

, .

103. Since, A =

5 5

0

0 0

α αα α5

5

∴ A2

5 5

0 5

0 0 5

5 5

0 5

0 0 5

=

α αα α

α αα α

=+ +

+

25 25 5 10 25

0 5 25

0 0 25

2 2

2 2

α α α αα α α

⇒ | |A2

2 2

2 2

25 25 5 10 25

0 5 25

0 0 25

=+ +

+α α α α

α α α

= +25

25 25 5

0

2

2

α αα

= 625 2α

But | |A2 25=

∴ 625 252α = ⇒ | | .α = 1

5

104.1

2

1

3

1

4! ! !− + − K

= − + − + −1 11

2

1

3

1

4! ! !K = −e 1.

105. Let ( $ $ )2 3u v× is a unit vector. Then

| $ $|2 3 1u v× =

⇒ 6| $ $u | v| | | sin | = 1θ

⇒ sin θ = 1

6

Hence, there is exactly one value of θ for which

(2 3$ $u v× ) is a unit vector.

106. a u t= ( cos )α and b u t gt= −( sin )α 1

22

b aga

u= −tan

cosα

α1

2

2

2 2

Also, cu

g=

2 2sin α

⇒ b aa g

cg= −

tan

sinsecα α α

22

2

2

⇒ b aa

c= −tan tanα α

2

22

⇒ aa

cb−

=

2

tan α

⇒ tan( )

.α =−

bc

a c a

⇒ α =−

−tan( )

1 bc

a c a

107. Let the number of boys and girls are x yand .

∴ 52 42 50x y x y+ = +( )

22

uB

DA a

C

b

c

⇒ 52 42 50 50x y x y+ = +⇒ 2 8x y= ⇒ x y= 4

∴ Total number of students in the class

= + = + =x y y y y4 5

∴ Required % of boys = × =4

5100 80

y

y

108. The point of intersection of two perpendicular

tangents to the parabola lies on direct of the

parabola.

Since, equation of directrix is y x= + 2

So, point is ( , ).−2 0

109. Equation of sphere is

x y z x y z2 2 2 6 12 2 20 0+ + − − − + =

whose co or di nates of cen tre are (3, 6, 1).

Let the coordinates of the other end of

diameter are (α, β, γ),

then α β γ+ = + = + =2

23

3

26

5

21, ,

Hence α β= =4 9, and γ = − 3.

Thus the coordinates of other point are

(4, 9, – 3).

110. Since, a i j k→

= + +$ $ $ , b i j k→

= − +$ $ $2

and c i j k→

= + − −x x$ ( )$ $2 are coplanar

∴x x − −

−=

2 1

1 1 1

1 1 2

0

⇒ 3 2 2 0x x+ − + =

⇒ 2 4x = −⇒ x = − 2.

111. Since, A h k B( , ), ( , )1 1 and C( , )2 1 are the

vertices of a right angled triangle ABC.

Now, AB h k= − + −( ) ( )1 12 2

BC = − + −( ) ( )2 1 1 12 2

and CA h k= − + −( ) ( )2 12 2

Using, Pythagorus theorem

AC AB BC2 2 2= +⇒ h h k k2 24 4 1 2+ − + + −

= + − + + − +h h k k2 21 2 1 2 1

⇒ 5 4 3 2− = −h h

⇒ h = 1 K(i)

Now, given that area of the triangle is 1.

Then, area ( )∆ABC AB BC= × ×1

2

⇒ 11

21 1 12 2= × − + − ×( ) ( )h k

⇒ 2 1 12 2= − + −( ) ( )h k K(ii)

On putting h = 1 from Eq. (i), we get

2 1 2= −( )k

On squaring both the sides, we get

4 1 22= + −k k

⇒ k k2 2 3 0− − =⇒ ( )( )k k− + =3 1 0⇒ k = −1 3,Thus, the set of values of k is −1, 3.

112. Slope of QR = 3,

So, ∠ = °PQR 120

Slope of the line QM = tan2

33

π = −

Hence equation of line QM is y x= − 3

or 3 0x y+ = .

113. Equa tion of bi sec tors of lines xy = 0 are y x= ± .

Put y x= ± in my m xy mx2 2 21 0+ − − =( ) , we

get

mx m x mx2 2 2 21 0+ − − =( )

⇒ ( )1 02 2− =m x ⇒ m = ± 1.

114. Since, f xt

tdt

x

( )log=

+∫ 11

and F e f e fe

( ) ( )= +

1

23

A(h, k)Y

O X

C(2, 1)B(1, 1)

π/3π/3P

M

Q (0, 0)x

y

R (3, 3 3)

(−1, 0)x'

y'

2π/3

⇒ F et

tdt

t

tdt

ee

( )log log

/

=+

++∫∫ 1 1

1

1

1

=+

++∫∫

log log

( )

t

tdt

t

t tdt

ee

1 111

= =

∫

log (log )t

tdt

tee 2

112

= −1

212 2[(log ) (log ) ]e = 1

2

115. f x x x( ) min , | | = + +1 1

f x x x R( ) .= + ∀ ∈1

∴ f x( ) is differentiable every where.

116. limx xx e→

−−

0 2

1 2

1

= − −−→

lim( )x

x

x

e x

x e0

2

2

1 2

1

= −− +→

lim( )x

x

x x

e

e xe0

2

2 2

2 2

1 2

=+

=→

limx

x

x x

e

e xe0

2

2 2

4

4 41

∴ f x( ) is continuous at x = 0, then

lim ( ) ( )x

f x f→

=0

0

⇒ 1 0= f ( ).

117. Given question is wrong.

118.dx

x xcos sin+∫ 3

= −

∫

1

2 3sec x dx

π

= − +

+1

2 2 6 4log tan

xc

π π

= +

+1

2 2 12log tan .

xc

π

119. Required area, A x x dx= −∫ ( )0

1

= −

2

3 23 2

2

0

1

xx/

= − =2

3

1

2

1

6 sq unit

120. Let α and β be the roots of equation

x ax2 1 0+ + = , then

α β+ = − a and αβ = 1

Now | | ( )α β α β αβ− = + −2 4

⇒ | |α β− = −a2 4

According to question,

a2 4 5− <

⇒ a2 4 5− <

⇒ a2 9 0− <

⇒ a ∈ −( , ).3 3

24

(1, 0)Ox

y'

x'

y

y = –x y = x

2y = x