Time : MM : 360

Instructions

1. This test consists of 90 questions.

2. There are three parts in the question paper A,B,C consisting of Physics, Chemistry and Mathematics

having 30 questions in each part of equal weightage. Each question is allotted 4 marks for correct

response.

3. Candidates will be awarded marks as stated above in instruction 2 for correct response of each

question. 1/4 mark will be deducted for indicating incorrect response for an item in the answer sheet.

4. There is only one correct response for each question. Filling up more than one response in any

question will be treated as wrong response and marks for wrong response will be deducted

accordingly as per instructions.

Physics1. The current voltage relation of diode is

given by I e T= −( )/1000 1V mA, where the

applied voltage V is in volt and thetemperature T is in kelvin. If a studentmakes an error measuring ± 0 01. V whilemeasuring the current of 5 mA at 300K,what will be the error in the value ofcurrent in mA?

(a) 0.2 mA (b) 0.02 mA

(c) 0.5 mA (d) 0.05 mA

2. From a tower of height H, a particle isthrown vertically upwards with a speed u.The time taken by the particle to hit theground, is n times that taken by it to reachthe highest point of its path. The relationbetween H, u and n is

(a) 2 2 2gH n u= (b) gH n u= −( )2 2 2

(c) 2 22gH nu n= −( ) (d) gH n u= −( )2 2 2

3. A mass m supported by amassless string woundaround a uniform hollowcylinder of mass m andradius R. If the string doesnot slip on the cylinder,with what acceleration willthe mass fall on release?

(a)23g

(b)g2

(c)56g

(d) g

4. A block of mass m is placed on a surface

with a vertical cross-section given by

y x= 3 6/ . If the coefficient of friction is

0.5, the maximum height above the

ground at which the block can be placed

without slipping is

(a)16

m (b)23

m (c)13

m (d)12

m

Solved Paper 2014

Joint Entrance ExaminationJoint Entrance Examination

Time : 3 hrs MM : 360

R

m

m

5. When a rubber band is stretched by adistance x, it exerts a restoring force ofmagnitude F ax bx= + 2 , where a and b are

constants. The work done in stretching theunstretched rubber-band by L is

(a) aL bL2 3+ (b)12

2 3( )aL bL+

(c)aL bL2 3

2 3+ (d)

12 2 3

2 3aL bL+

6. A bob of mass m attached to aninextensible string of length l is suspendedfrom a vertical support. The bob rotates in ahorizontal circle with an angular speed ωrad/s about the vertical support. About thepoint of suspension

(a) angular momentum is conserved

(b) angular momentum changes in magnitude butnot in direction

(c) angular momentum changes in direction but notin magnitude

(d) angular momentum changes both in directionand magnitude

7. Four particles, each of mass M and

equidistant from each other, move along a

circle of radius R under the action of their

mutual gravitational attraction, the speed

of each particle is

(a)GMR

(b) 2 2GMR

(c)GMR

( )1 2 2+

(d)12

1 2 2GMR

( )+

8. The pressure that has to be applied to the

ends of a steel wire of length 10 cm to keep

its length constant when its temperature is

raised by 100°C is (For steel, Young’s

modulus is 2 1011× Nm−2 and coefficient of

thermal expansion is1.1 × − −10 5 1K )

(a) 2 2 108. × Pa (b) 2 2 109. × Pa

(c) 2 2 107. × Pa (d) 2 2 106. × Pa

9. There is a circulartube in a verticalplane. Two liquidswhich do not mix andof densities d1 and d2

are filled in the tube.Each liquid subtends90° angle at centre. Radius joining theirinterface makes an angle α with vertical.Ratio d d1 2/ is

(a)11

+−

sinsin

αα

(b)11

+−

coscos

αα

(c)11

+−

tantan

αα

(d)11

+−

sincos

αα

10. On heating water, bubbles beingsformed at the bottom of the vessel detachand rise. Take the bubbles to be spheresof radius R and making a circular contactof radius r with the bottomof the vessel. If r <<R andthe surface tension of wateris T, value of r just beforebubbles detach is (densityof water is ρ)

(a) Rg

Tw2

3ρ

(b) Rg

Tw2

6ρ

(c) Rg

Tw2 ρ

(d) Rg

Tw2 3ρ

11. Three rods of copper, brass and steel arewelded together to form a Y-shapedstructure. Area of cross-section of eachrod is 4 cm2 . End of copper rod is

maintained at 100°C whereas ends ofbrass and steel are kept at 0°C. Lengthsof the copper, brass and steel rods are 46,13 and 12 cm respectively. The rods arethermally insulated from surroundingsexcept at ends. Thermal conductivities ofcopper, brass and steel are 0.92, 0.26 and0.12 in CGS units, respectively. Rate ofheat flow through copper rod is

(a) 1.2 cal/s (b) 2.4 cal/s

(c) 4.8 cal/s (d) 6.0 cal/s

2 JEE Main Solved Papers

α

d1

d2

R

2 r

12. One mole of diatomicideal gas undergoes acyclic process ABC asshown in figure. Theprocess BC is adiabatic.The temperatures at A,B and C are 400 K, 800 K and 600 K,respectively. Choose the correctstatement.

(a) The change in internal energy in whole cyclicprocess is 250 R

(b) The change in internal energy in the process CAis 700 R

(c) The change in internal energy in the process ABis −350 R

(d) The change in internal energy in the process BCis −500 R

13. An open glass tube is immersed in mercuryin such a way that a length of 8 cm extendsabove the mercury level. The open end ofthe tube is then closed and sealed and thetube is raised vertically up by additional 46cm. What will be length of the air columnabove mercury in the tube now?(Atmospheric pressure = 76 cm of Hg)

(a) 16 cm (b) 22 cm (c) 38 cm (d) 6 cm

14. A particle moves with simple harmonicmotion in a straight line. In first τ sec, afterstarting from rest it travels a distance a andin next τ sec, it travels 2a, in samedirection, then

(a) amplitude of motion is 3a

(b) time period of oscillations is 8π(c) amplitude of motion is 4a

(d) time period of oscillations is 6π

15. A pipe of length 85 cm is closed from oneend. Find the number of possible naturaloscillations of air column in the pipe whosefrequencies lie below 1250 Hz. Thevelocity of sound in air is 340 m/s.

(a) 12 (b) 8 (c) 6 (d) 4

16. Assume that an electric field E i= 30 2x $

exists in space. Then, the potentialdifference V VA O− , where VO is thepotential at the origin and VA the potentialat x = 2 m, is

(a) 120 J (b) −120 J (c) −80 J (d) 80 J

17. A parallel plate capacitor is made of twocircular plates separated by a distance of5 mm and with a dielectric of dielectricconstant 2.2 between them. When theelectric field in the dielectric is3 104× V/m, the charge density of the

positive plate will be close to

(a) 6 10 7× − C/m2 (b) 3 10 7× − C/m2

(c) 3 104× C/m2 (d) 6 104× C/m2

18. In a large building, there are 15 bulbs of40 W, 5 bulbs of 100 W, 5 fans of 80 Wand 1 heater of 1 kW. The voltage of theelectric mains is 220 V. The minimumcapacity of the main fuse of the buildingwill be

(a) 8 A (b) 10 A (c) 12 A (d) 14 A

19. A conductor lies along the z-axis at− ≤ <1.5 z 1.5 m and carries a fixedcurrent of 10.0 A in −az direction (seefigure). For a fieldB = × − −3 0 10 4 0 2. ,.e ax

y T find the power

required to move the conductor atconstant speed to x = 2 0. m, y = 0 in5 10 3× − s. Assume parallel motion along

the x-axis.

(a) 1.57 W (b) 2.97 W

(c) 14.85 W (d) 29.7 W

20. The coercivity of a small magnet wherethe ferromagnet gets demagnetised is3 103× Am−1. The current required to bepassed in a solenoid of length 10 cm andnumber of turns 100, so that the magnetgets demagnetised when inside thesolenoid is

(a) 30 mA (b) 60 mA (c) 3 A (d) 6 A

JEE Main Solved Paper 2014 3

–1.5

1.5

2.0

x

yB

z

l

P

V

A

B

C

800 K

600 K

400 K

21. In the circuit shown here, the point C

is kept connected to point A till thecurrent flowing through the circuitbecomes constant. Afterward,suddenly point C is disconnectedfrom point A and connected to pointB at time t = 0. Ratio of the voltageacross resistance and the inductor att L R= / will be equal to

(a)e

e1 −(b) 1

(c) −1 (d)1 − e

e

22. During the propagation ofelectromagnetic waves in a medium,

(a) electric energy density is double of themagnetic energy density

(b) electric energy density is half of themagnetic energy density

(c) electric energy density is equal to themagnetic energy density

(d) Both electric and magnetic energydensities are zero

23. A thin convex lens made from crown

glass µ =

32

has focal length f.

When

it is measured in two different liquids

having refractive indices43

and53

. It

has the focal lengths f1 and f2 ,respectively. The correct relationbetween the focal length is

(a) f f f1 2= <(b) f f1 > and f2 becomes negative

(c) f f2 > and f1 becomes negative

(d) f1 and f2 both become negative

24. A green light is incident from the water to theair-water interface at the critical angle (θ).Select the correct statement.

(a) The entire spectrum of visible light will come out ofthe water at an angle of 90° to the normal

(b) The spectrum of visible light whose frequency is lessthan that of green light will come out of the airmedium.

(c) The spectrum of visible light whose frequency ismore than that of green light will come out to the airmedium.

(d) The entire spectrum of visible light will come out ofthe water at various angles to the normal.

25. Two beams, A and B, of plane polarised lightwith mutually perpendicular planes ofpolarisation are seen through a polaroid. Fromthe position when the beam A has maximumintensity (and beam B has zero intensity), arotation of polaroid through 30° makes the twobeams appear equally bright. If the initialintensities of the two beams are IA and IB

respectively, then I IA B/ equals

(a) 3 (b)32

(c) 1 (d)13

26. The radiation corresponding to 3→ 2 transitionof hydrogen atom falls on a metal surface toproduce photoelectrons. These electrons aremade to enter a magnetic field of 3 10 4× − T. If

the radius of the largest circular path followedby these electrons is 10.0 mm, the workfunction of the metal is close to

(a) 1.8 eV (b) 1.1 eV (c) 0.8 eV (d) 1.6 eV

27. Hydrogen( )11H , deuterium( )1

2H , singly ionised

helium ( He )24 + and doubly ionised lithium

( Li )38 ++ all have one electron around the

nucleus. Consider an electron transition fromn = 2 to n =1. If the wavelengths of emittedradiation are λ λ λ 31 2, , and λ 4 , respectively forfour elements, then approximately which oneof the following is correct?

(a) 4λ λ λ λ1 = = =2 22 3 4

(b) λ λ λ λ1 = = =2 22 3 4

(c) λ λ λ λ1 = = =2 3 44 9

(d) λ λ λ λ1 = = =2 3 42 3 4


A C R

B

L

28. The forward biased diode connection is

(a)

(b)

(c)

(d)

29. A student measured the length of a rod andwrote it as 3.50 cm. Which instrument didhe use to measure it?

(a) A meter scale

(b) A vernier calliper where the 10 divisions in vernierscale matches with 9 divisions in main scale andmain scale has 10 divisions in 1 cm

(c) A screw gauge having 100 divisions in thecircular scale and pitch as 1 mm

(d) A screw gauge having 50 divisions in the circularscale and pitch as 1 mm

30. Match List I (Electromagnetic wavetype) with List II (Itsassociation/application) and select thecorrect option from the choices givenbelow the lists.

List I List II

A. Infrared waves 1. To treat muscular strain

B. Radio waves 2. For broadcasting

C. X-rays 3. To detect fracture of bones

D. Ultraviolet 4. Absorbed by the ozone layerof the atmosphere

Codes

A B C D

(a) 4 3 2 1

(b) 1 2 4 3

(c) 3 2 1 4

(d) 1 2 3 4


Chemistry31. The correct set of four quantum numbers

for the valence electrons of rubidium atom( )Z = 37 is

(a) 5 0 012

, , , + (b) 5 1 012

, , , +

(c) 5 1 112

, , , , + (d) 5 0 112

, , , +

32. If Z is a compressibility factor, van derWaals’ equation at low pressure can bewritten as

(a) ZRTpb

= +1 (b) Za

VRT= −1

(c) ZpbRT

= −1 (d) ZpbRT

= +1

33. CsCl crystallises in body centred cubiclattice. If ‘a’ its edge length, then which ofthe following expressions is correct?

(a) r r aCs Cl+ −+ = 3

(b) r ra

Cs Cl+ −+ = 32

(c) r r aCs Cl+ −+ = 3

2

(d) r r aCs Cl+ −+ = 3

34. For the estimation of nitrogen, 1.4 g of anorganic compound was digested byKjeldahl's method and the evolved ammonia

was absorbed in 60 mL ofM

10sulphuric acid.

The unreacted acid required 20 mL of M /10sodium hydroxide for completeneutralisation. The percentage of nitrogenin the compound is

(a) 6% (b) 10%

(c) 3% (d) 5%

35. Resistance of 0.2 M solution of an

electrolyte is 50 Ω . The specific

conductance of the solution of 0.5 M

solution of same electrolyte is 1.4 S m 1−

and resistance of same solution of the

same electrolyte is 280 Ω . The molar

conductivity of 0.5 M solution of the

electrolyte in Sm mol2 1− is

(a) 5 10 4× − (b) 5 10 3× −

(c) 5 103× (d) 5 102×

–2V +2V

2V 4V

–3V –3V

+2V –2V

36. For the complete combustion of ethanol,

C H OH( )+3O ( ) 2CO ( )+3H O( ),2 5 2 2 2l g g l→the amount of heat produced as measured in

bomb calorimeter, is 1364.47 kJ mol 1− at

25°C. Assuming ideality the enthalpy of

combustion, ∆CH, for the reaction will be

( = 8.314 J mol )–1 –1R K

(a) − 1366 95. kJ mol 1−

(b) − 1361 95. kJ mol 1−

(c) − 1460 50. kJ mol 1−

(d) − 1350 50. kJ mol 1−

37. The equivalent conductance of NaCl atconcentration C and at infinite dilution areλ C and λ ∞, respectively. The correctrelationship between λ C and λ ∞ is given as(where, the constant B is positive)

(a) λ λC (B)C= +∞(b) λ λC B C= −∞ ( )

(c) λ λC B C= −∞ ( )

(d) λ λC C= +∞ ( )B

38. Consider separate solution of 0.500 M

C H OH2 5 (aq), 0.100 M Mg3(PO )4 2(aq),

0.250 M KBr(aq) and 0.125 M Na PO3 4 (aq)

at 25°C. Which statement is true about these

solution, assuming all salts to be strong

electrolytes?

(a) They all have the same osmotic pressure

(b) 0.100 M Mg (PO )3 4 2(aq) has the highest osmoticpressure

(c) 0.125 M Na PO3 4(aq) has the highest osmoticpressure

(d) 0.500 M C H OH2 5 (aq) has the highest osmoticpressure

39. For the reaction,

SO O SO2 2 3( ) ( ) ( )g g g+ 12

q

if K K RTp Cx= ( )

where, the symbols have usual meaning,then the value of x is (assuming ideality)

(a) − 1 (b) − 12

(c)12

(d) 1

40. For the non-stoichiometric reaction2 A B C D+ +→ , the following kineticdata were obtained in three separateexperiments, all at 298 K.

Initial

concen tration

( )A

Initial

concentration

( )B

Initial rate of

formation of C

(mol L−1S−1)

(i) 0.1 M 0.1 M 1 2 10 3. × −

(ii) 0.1 M 0.2 M 1 2 10 3. × −

(iii) 0.2 M 0.1 M 2 4 10 3. × −

The rate law for the formation of C is

(a)dCdt

= k A[ [B]] (b)dCdt

= k A B[ ] [ ]2

(c)dCdt

= k A B[ ][ ]2 (d)dCdt

= k A[ ]

41. Among the following oxoacids, thecorrect decreasing order of acidstrength is

(a) HOCl > HClO > HClO > HClO2 3 4

(b) HClO > HOCl > HClO > HClO4 2 3

(c) HClO > HClO > HClO > HOCl4 3 2

(d) HClO > HClO > HClO > HOCl2 4 3

42. The metal that cannot be obtained by

electrolysis of an aqueous solution of its

salts is

(a) Ag (b) Ca(c) Cu (d) Cr

43. The octahedral complex of a metal ion

M3+ with four monodentate ligands L1,

L2 , L3 and L4 absorb wavelengths in the

region of red, green, yellow and blue,

respectively. The increasing order of

ligand strength of the four ligands is

(a) L < L , L < L4 3 2 1

(b) L < L < L < L1 3 2 4

(c) L < L < L < L3 2 4 1

(d) L < L < L < L1 2 4 3

44. Which of the following properties is notshown by NO?

(a) It is diamagnetic in gaseous state

(b) It is a neutral oxide

(c) It combines with oxygen to form nitrogendioxide

(d) Its bond order is 2.5


45. In which of the following reactions H O2 2

acts as a reducing agent?

I. H O H H O2 2 22 2 2+ + →+ −e

II. H O O H2 2 22− →− ++ 2e

III. H O OH2 2 2 2+ →− −e

IV. H O OH O H O2 2 2 22 2 2+ +− −− →e

(a) I and II (b) III and IV

(c) I and III (d) II and IV

46. The correct statement for the molecule,CsI3 is

(a) it is a covalent molecule

(b) it contains Cs+ and I3−

(c) it contains Cs3+ and I− ions

(d) it contains Cs+ , I− and lattice I2 molecule

47. The ratio of masses of oxygen and nitrogenof a particular gaseous mixture is 1 : 4.The ratio of number of their molecule is

(a) 1 : 4 (b) 7 : 32 (c) 1 : 8 (d) 3 : 16

48. Given below are the half-cell reactions

Mn Mn2 2+ −+ →e ; E° = −1.18 eV

2 2( );Mn Mn3+ − ++ →e E° = + 1.51eV

The E° for 3Mn Mn + 2Mn2 3++ → will

be(a) − 2 69. V; the reaction will not occur(b) − 2 69. V; the reaction will occur

(c) − 0 33. V; the reaction will not occur(d) − 0 33. V; the reaction will occur

49. Which series of reactions correctlyrepresents chemical relations related toiron and its compound?

(a) Fe FeSO4

Dil. H SO H SO ,O2 4 2 4→ → 2 Fe (SO )

2 4 3

→Heat Fe

(b) Fe Fe2 2 4, Heat Dil. H SO→ →O

O FeSO4

→Heat Fe

(c) Fe FeClCl , Heat Heat, air2

3→ → FeCl2

→Zn Fe

(d) Fe Fe OO , Heat CO, 600 C2

3 4→ → °

FeO

→ °CO, 700 C Fe

50. The equation which is balanced andrepresents the correct product(s) is

(a) Li O + 2KCl 2LiCl + K O2 2→(b) [CoCl(NH ) ] + 5H Co3 5

+ 2++ →

+5NH + Cl4+ −

(c) [Mg(H O) + (EDTA)2 62+ 4 Excess NaOH] − →

[Mg(EDTA)] + 6H O2+2

(d) CuSO + 4KCN K [Cu(CN) ]4 2 4→+ K SO2 4

51. In S 2N reactions, the correct order of

reactivity for the following compoundsCH Cl,CH CH Cl,3 3 2 (CH ) CHCl3 2 and(CH ) CCl3 3 is

(a) CH CI > (CH ) CHCI > CH CH CI3 3 2 3 2

> (CH ) CCI3 3

(b) CH CI> CH CH CI> (CH ) CHCI3 3 2 3 2

> (CH ) CCI3 3

(c) CH CH Cl > CH Cl > (CH ) CHCl3 2 3 3 2

> (CH ) CCI3 3

(d) (CH ) CHCI > CH CH CI > CH CI3 2 3 2 3

> (CH )3 3CCI

52. On heating an aliphatic primary aminewith chloroform and ethanolic potassiumhydroxide, the organic compound formedis

(a) an alkanol

(b) an alkanediol

(c) an alkyl cyanide

(d) an alkyl isocyanide

53. The most suitable reagent for the

conversion of

R R → CH2 OH CHO is(a) KMnO4

(b) K Cr O2 2 7

(c) CrO3

(d) PCC (pyridinium chlorochromate)

54. The major organic compound formed bythe reaction of 1,1,1-trichloroethane withsilver powder is

(a) acetylene (b) ethene

(c) 2-butyne (d) 2-butene


55. Sodium phenoxide when heated withCO2

under pressure at 125°C yields a productwhich on acetylation produces C.

+ → →° +CO2

125

5 atm

H

Ac O2

B C

The major product C would be

56. Considering the basic strength of aminesin aqueous solution, which one has thesmallest pKb value?

(a) (CH ) NH3 2 (b) CH NH3 2

(c) (CH ) N3 3 (d) C H NH6 5 2

57. For which of the following moleculesignificant µ ≠ 0 ?

(a) Only I (b) I and II

(c) Only III (d) III and IV

58. Which one is classified as a condensationpolymer?

(a) Dacron (b) Neoprene(c) Teflon (d) Acrylonitrile

59. Which one of the following bases is notpresent in DNA?

(a) Quinoline (b) Adenine(c) Cytosine (d) Thymine

60. In the reaction,

CH COOH3 → → →LiAlH PCl Alc. KOH4 5

A B C

The product C is

(a) acetaldehyde (b) acetylene(c) ethylene (d) acetyl chloride


Cl

ClI

CN

CNII

OH

OHIII

SH

SHIV

COOH

OCOCH3

COCH3

OH

COCH3

COOCH3

OH

COOH

OCOCH3

(a) (b)

(d)(c)

Mathematics61. If X n n Nn= − − ∈ : 4 3 1 and

Y n n N= − ∈ ( ) : 9 1 ; where N is the set ofnatural numbers, then X Y∪ is equal to

(a) N (b) Y X−(c) X (d) Y

62. If z is a complex number such that| |z ≥ 2,

then the minimum value of z + 12

(a) is equal to52

(b) lies in the interval (1, 2)

(c) is strictly greater than52

(d) is strictly greater than32

but less than52

63. If a R∈ and the equation

− − + −3 22( [ ]) ( [ ])x x x x + =a2 0 (where,[ ]x

denotes the greatest integer ≤ x) has nointegral solution, then all possible valuesof a lie in the interval

(a) ( , ) ( , )− ∪1 0 0 1 (b) ( , )1 2

(c) ( , )− −2 1 (d) ( , ) ( , )−∞ − ∪ ∞2 2

64. Let α and β be the roots of equation

px qx r2 0+ + = , p ≠ 0. If p q, and r are in

AP and1 1

4α β

+ = , then the value of| |α β−

is

(a)619

(b)2 17

9(c)

349

(d)2 13

9

ONa

65. If α, β ≠ 0 and f n n n( ) = +α β and

3 1 1 1 21 1 1 2 1 31 2 1 3 1 4

+ ++ + ++ + +

f f

f f f

f f f

( ) ( )( ) ( ) ( )( ) ( ) ( )

= − −K( ) ( )1 12 2α β ( )α β− 2 ,

then K is equal to

(a) αβ (b)1

αβ(c) 1 (d) −1

66. If A is a 3 3× non-singular matrix such

that AA A AT T= and B A AT= −1 , then BBT

is equal to

(a) I B+ (b) I (c) B−1 (d) ( )B T−1

67. If the coefficients of x3 and x4 in theexpansion of ( )( )1 1 22 18+ + −ax bx x in

powers of x are both zero, then ( , )a b isequal to

(a) 162513

,

(b) 14

2513

,

(c) 14272

3,

(d) 16

2723

,

68. The angle between the lines whosedirection cosines satisfy the equationsl m n+ + =0 and l m n2 2 2= + is

(a)π3

(b)π4

(c)π6

(d)π2

69. If( ) ( ) ( ) ( ) ( ) ... ( )10 2 11 10 3 11 10 10 119 1 8 2 7 9+ + + += k( )10 9 , then k is equal to

(a)12110

(b)441100

(c) 100 (d) 110

70. Three positive numbers form anincreasing GP. If the middle term in thisGP is doubled, then new numbers are inAP. Then, the common ratio of the GP is

(a) 2 3+ (b) 3 2+(c) 2 3− (d) 2 3+

71. limsin( cos )

x

x

x→ 0

2

2

πis equal to

(a)π2

(b) 1

(c) − π (d) π

72. If g is the inverse of a function f and

′ =+

f xx

( )1

1 5, then ′g x( ) is equal to

(a) 1 5+ x (b) 5 4x

(c)1

1 5+ ( )g x(d) 1 5+ ( )g x

73. If f and g are differentiable functions in

(0, 1) satisfying f g( ) ( )0 2 1= = , g( )0 0= andf( )1 6= , then for some c ∈ ] , [0 1

(a) 2/ ′ = ′f c g c( ) ( ) (b) 2 3′ = ′f c g c( ) ( )

(c) ′ = ′f c g c( ) ( ) (d) ′ = ′f c g c( ) ( )2

74. If x = −1 and x = 2 are extreme points of

f x x x x( ) log| |= α + β +2 , then

(a) α = − 6, β = 12

(b) α = − 6, β = − 12

(c) α = 2 , β = − 12

(d) α = 2 , β = 12

75. The integral 11

1

+ −

+

∫ xx

e dxx

x is equal

to

(a) ( )x e Cx

x− ++

11

(b) xe Cx

x+

+1

(c) ( )x e Cx

x+ ++

11

(d) − ++

xe Cx

x

1

76. The integral 1 42

42

2

0

+ −∫ sin sinx x

dx

π

is

equal to

(a) π − 4 (b)23

4 4 3π − −

(c) 4 3 4− (d) 4 3 43

− − π

77. The area of the region described byA x y x y= + ≤( , ) : 2 2 1and y x2 1≤ − is

(a)π2

43

+ (b)π2

43

− (c)π2

23

− (d)π2

23

+

78. Let the population of rabbits surviving at

a time t be governed by the differential

equationdp t

dtp t

( )( )= −1

2200. If p( )0 100= ,

then p t( ) is equal to

(a) 400 300 2− et

(b) 300 200 2−−

et

(c) 600 500 2− et

(d) 400 300 2−−

et


79. If PS is the median of the triangle withvertices P ( , )2 2 , Q ( , )6 1− and R( , )7 3 , thenequation of the line passing through( , )1 1−and parallel to PS is

(a) 4 7 11 0x y− − = (b) 2 9 7 0x y+ + =(c) 4 7 3 0x y+ + = (d) 2 9 11 0x y− − =

80. Let a b c, , and d be non-zero numbers. If

the point of intersection of the lines4 2 0ax ay c+ + = and 5 2 0bx by d+ + =lies in the fourth quadrant and isequidistant from the two axes, then

(a) 2 3 0bc ad− = (b) 2 3 0bc ad+ =(c) 2 3 0ad bc− = (d) 3 2 0bc ad+ =

81. The locus of the foot of perpendiculardrawn from the centre of the ellipsex y2 23 6+ = on any tangent to it is

(a) ( )x y x y2 2 2 2 26 2− = +(b) ( )x y x y2 2 2 2 26 2− = −(c) ( )x y x y2 2 2 2 26 2+ = +(d) ( )x y x y2 2 2 2 26 2+ = −

82. Let C be the circle with centre at (1, 1) andradius 1. If T is the circle centred at ( , )0 y

passing through origin and touching thecircle C externally, then the radius of T isequal to

(a)3

2(b)

32

(c)12

(d)14

83. The slope of the line touching both theparabolas y x2 4= and x y2 32= − is

(a)12

(b)32

(c)18

(d)23

84. The image of the linex y z− = − = −

−1

3

3

1

4

5

in the plane 2 3 0x y z− + + = is the line

(a)x y z+ = − = −

−3

35

12

5

(b)x y z+−

= −−

= +33

51

25

(c)x y z− = + = −

−3

35

12

5

(d)x y z−−

= +−

= −33

51

25

85. If [ ]a b b c c a× × × = λ [ ]a bc2 , then λ is

equal to

(a) 0 (b) 1

(c) 2 (d) 3

86. Let A and B be two events such that

P A B( )∪ = 16

, P A B( )∩ = 14

and P A( ) = 14

,

where A stands for the complement of the

event A. Then , the events A and B are

(a) independent but not equally likely(b) independent and equally likely

(c) mutually exclusive and independent(d) equally likely but not independent

87. The variance of first 50 even naturalnumbers is

(a)8334

(b) 833

(c) 437 (d)437

4

88. If f x k x xkk k( ) / (sin cos )= +1 , where x R∈

and k ≥ 1, then f x f x4 6( ) ( )− is equal to

(a)16

(b)13

(c)14

(d)1

12

89. A bird is sitting on the top of a verticalpole 20 m high and its elevation from apoint O on the ground is 45°. It flies offhorizontally straight away from the pointO. After 1s, the elevation of the bird fromO

is reduced to 30°. Then, the speed (in m/s)of the bird is

(a) 40 2 1( )−(b) 40 3 2( )−(c) 20 2

(d) 20 3 1( )−

90. The statement ~( ~ )p q↔ is

(a) equivalent to p q↔(b) equivalent to ~ p q↔(c) a tautology

(d) a fallacy


Solutions

Physics

1. Given, I e V T= −( )/1000 1 mA,

dV = ± 0 0. 1 V

T = 300 K

I = 5 mA

So, I e V T= −1000 1/

I e V T+ =1 1000 /

Taking log on both sides, we get

log( )IV

T+ =1

1000

On differentiating,dI

I T+=

11000

dV

dIT

I= × +10001( )dV

⇒ dI = × + ×1000300

5 1 0 01( ) . = 0.2 mA

So, error in the value of current is 0.2 mA.

2. Time taken to reach the maximum height tug1 =

If t 2 is the time taken to hit the ground,

i.e., − = −H ut gt2 221

2

But t nt2 1= [Given]

So, − = −H unug

gn u

g

12

2 2

2

− = −Hnug

n ug

2 2 212

⇒ 2 22gH nu n= −( )

3. For the mass m,

mg T ma− =

As we know, a R= αSo, mg T mR− = α …(i)

Torque about centre of pully

T R mR× = 2α ...(ii)

From Eqs. (i) and (ii), we get

ag=2

Hence, the acceleration with the mass of a body fall isg2

.

4. A block of mass m is placed on a surface with a

vertical cross-section, then

tanθ =

=dydx

dx

dxx

3

26

2

H

t1

t2

u

Answers

1. (a) 2. (c) 3. (b) 4. (a) 5. (c) 6. (c) 7. (d) 8. (a) 9. (c) 10. (*)

11. (c) 12. (d) 13. (a) 14. (d) 15. (c) 16. (c) 17. (a) 18. (c) 19. (b) 20. (c)

21. (c) 22. (c) 23. (b) 24. (d) 25. (d) 26. (b) 27. (c) 28. (a) 29. (b) 30. (d)

31. (a) 32. (b) 33. (c) 34. (b) 35. (a) 36. (a) 37. (c) 38. (a) 39. (b) 40. (d)

41. (c) 42. (b) 43. (b) 44. (a) 45. (d) 46. (b) 47. (b) 48. (a) 49. (d) 50. (b)

51. (b) 52. (d) 53. (d) 54. (c) 55. (a) 56. (a) 57. (d) 58. (a) 59. (a) 60. (c)

61. (d) 62. (b) 63. (a) 64. (d) 65. (c) 66. (b) 67. (d) 68. (a) 69. (c) 70. (d)

71. (d) 72. (d) 73. (d) 74. (c) 75. (b) 76. (d) 77. (a) 78. (a) 79. (b) 80. (c)

81. (c) 82. (d) 83. (a) 84. (a) 85. (b) 86. (a) 87. (b) 88. (d) 89. (d) 90. (a)

NOTE

R

m a

m

mg

T

T

Y

θy

X

m

At limiting equilibrium, we get

µ θ= tan

0 52

2

. = x ⇒ x2 1= ⇒ x = ± 1

Now, putting the value of x in yx=

3

6, we get

When x = 1

y = =( )16

16

3

When x = −1

y = − = −( )16

16

3

So, the maximum height above the ground atwhich the block can be placed without slipping is16

m .

5. / We know that change in potential energy of asystem corresponding to a conservativeinternal force as

U U W df i

f

i− = − = −∫ F r.

Given, F ax bx= + 2

We know that work done in stretching the rubberband by L is

| | | |dW Fdx=

| | ( )W ax bx dx

O

L

= +∫ 2

=

+

ax bx

O

L

O

L2 3

2 3

= − ×

+ × − ×

aL a b L b2 2 3 3

20

2 30

3( ) ( )

| |WaL bL= +

2 3

2 3

6. Angular momentum of the pendulum about thesuspension point O is

L r v= ( )m ×

Then, v can be resolved into two components,

radial component rrad and tangential component

rtan. Due to vrad, L will be tangential and due to v tan,

L will be radially outwards as shown.

So, net angular momentum will be as shown in

figure whose magnitude will be constant

(| | )L mvl= . But its direction will change as shown in

the figure.

L r v= ( )m ×where, r = radius of circle

7. Net force acting on any one particle M,

= + °GM

R

GM

R

2

2

2

22 245

( ) ( )cos + °GM

R

2

2245

( )cos

= +

GM

R

2

2

14

1

2

This force will equal to centripetal force.

So,Mv

RGM

R

2 2

2

14

1

2= +

vGM

R= +

41 2 2( ) = +1

22 2 1

GMR

( )

Hence, speed of each particle in a circular motionis

12

2 2 1GMR

( )+ .

8. / If the deformation is small, then the stress in abody is directly proportional to the corres-ponding strain.

According to Hooke's law i.e.,

Young’s modulus ( )Tensile stressTensile strain

Y =

So, YF AL L

FLA L

= =//∆ ∆

If the rod is compressed, then compressive stressand strain appear. Their ratio Y is same as that fortensile case.

Given, length of a steel wire (L) = 10cm,

Temperature ( )θ = °100 C

As length is constant.

∴ Strain = =∆ ∆L

Lα θ


O

L L

O

R

M

m

v

vv

m

M

RR

45°

45°

R

v

Now, pressure = stress = Y × strain

[Given, Y = ×2 1011N / m2 and α = × − −1.1 K. 10 5 1]

= × × × ×−2 10 11 10 10011 5.

= ×2 2 108. Pa

9. Equating pressure at A, we getR d R d R dsin cos ( cos )α α α2 2 11+ + −

= −R d( sin )1 1α

(sin cos ) (cos sin )α α α α+ = −d d2 1

⇒ d

d1

2

11

= +−

tantan

αα

10. The bubble will detach if,

Buoyant force ≥ Surface tension force43

3π ρ θR g T dlw ≥ ∫ × sin

( ) ( ) ( )sinρ π π θw R g T r43

23

≥

sinθ = rR

Solving, rR g

Tw= 23

4ρ = Rg

Tw2 2

3ρ

No option matches with the correct answer.

11. / In thermal conduction, it is found that insteady state the heat current is directlyproportional to the area of cross-section Awhich is proportional to the change intemperature ( )T T1 2− .

Then,∆∆Qt

KA T T

x= −( )1 2

According to thermal conductivity, we getdQ

dt

dQ

dt

dQ

dt1 2 3= +

0 92 10046

026 013

. ( ) . ( )− = −T T + −012 012

. ( )T

⇒ T = °40 C

∴ dQ

dt1 0 92 4 100 40

404 8= × − =. ( ). cal/s

12. According to first law of thermodynamics, we get

(i) Change in internal energy from A to B i.e., ∆UAB

∆U nC T TAB V B A= −( )

= × − =152

800 400 1000R

( ) R

(ii) Change in internal energy from B to C

∆U nC T TBC V C B= −( )

= × − = −152

600 800 500R

( ) R

(iii) ∆Uisothermal = 0

(iv) Change in internal energy from C to A i.e., ∆UCA

∆U nC T TCA V A C= −( )

= × − = −152

400 600 500R

( ) R

13. / In this question, the system is acceleratinghorizontally i.e., no component of accelerationin vertical direction. Hence, the pressure in thevertical direction will remain unaffected.i.e., p p gh1 0= + ρ

Again, we have to use the concept that thepressure in the same level will be same.

For air trapped in tube, p V p V1 1 2 2=p p g1 76= =atm ρV A1 8= ⋅ [A = area of cross-section]

p p g x g x2 54 22= − − = +atm ρ ρ( ) ( )

V A x2 = ⋅ρ ρg A g x A x76 8 22× = +( )

x x2 22 78 8 0+ − × =

⇒ x = 16 cm


K2

l2

K3

l3

K1

l1

100°C

0°C0°C

dQ3dQ2dtdt

dQ1

dt

8 cm 54x

54−x

p2

α

d1

RR α R sin α

d2

A

90 − α R

θ

θr

∫ θ = θsin × (2 ) sinT dl T pr

R

14. In SHM, a particle starts from rest, we have

i.e., x A t= cosω , at t = 0, x A=When t = τ, then

x A a= − ....(i)

When t = 2τ, then

x A a= − 3 ....(ii)

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

A a A− = cosωτA a A− =3 2cos ωτ

As cos cos2 2 12ωτ ωτ= −

⇒ A a

A

A a

A

− = −

−32 1

2

⇒ A a

A

A a Aa A

A

− = + − −3 2 2 42 2 2

2

A aA A a Aa2 2 23 2 4− = + −

a aA2 2=

A a= 2

Now, A a A− = cosωτ⇒ cos /ωτ = 1 2

23

πτ πT

=

T = 6 π

15. For closed organ pipe

= +( )2 14

n v

l[ , , , ...]n = 0 1 2

( )2 14

1250n v

l

+ <

( ).

2 1 12504 0 85

340n + < × ×

( ) .2 1 12 5n + < 2 1150n < .

n < 525.

So, n = 0 1 2 3 5, , , ,...,

So, we have 6 possibilities.

Alternate method

In closed organ pipe, fundamental node

i.e.,λ4

0 85= . ⇒ λ = ×4 0 85.

As we know, νλ

= c ⇒ 3404 0 85

100×

=.

Hz

∴ Possible frequencies = 100 Hz, 300 Hz, 500 Hz,700 Hz, 900 Hz, 1100 Hz below 1250 Hz.

16. As we know, potential difference V VA O− is

dV Edx= −

dV x dx

V

V

o

A

= − ∫∫ 30 2

0

2

V Vx

A O− = − ×

30

3

3

0

2

= − × −10 2 03 3[ ( ) ]

= − × = −10 8 80 J

17. When free space between parallel plates of

capacitor, E = σε0

When dielectric is introduced between parallel

plates of capacitor, EK

′= σε0

Electric field inside dielectricσεK

0

3 104= ×

where, K = dielectric constant of medium = 2.2

ε0 = permitivity of free space = × −8 85 10 12.

⇒ σ = × × × ×−2 2 8 85 10 3 1012 4. .

= × × −6 6 8 85 10 8. . = × −5 841 10 7.

= × −6 10 7 C/m2

18. Total power (P) consumed

= × + × + ×( ) ( ) ( )15 40 5 100 5 80

+ × =( )1 1000 2500 W

As we know,

Power, P VI=

⇒ I = 2500220

A = 12511

= 113. A

Minimum capacity should be 12 A.

19. When force exerted on a current carryingconductor

F Bext = IL

Average power = Work done

Time taken

Pt

F dx= ⋅∫1

0

2

ext. = ∫1

0

2

tB x IL dx( )

=×

× × ×−

− −∫1

5 103 10 10 3

3

4

0

20 2e dxx.

= − −9 1 0 4[ ].e

= −

9 110 4e .

= 2 967.

≈ 2 97. W


l 0.85= = λ4

85

cm

20. For solenoid, the magnetic field needed to bemagnetised the magnet.

B nI= µ0

[Where, n cm m m= = = =100 1010100

01, .l ]

⇒ 3 1010001

3× = ×.

I

I = 3A

21. After connecting C to B hanging the switch, thecircuit will act like an L-R discharging circuit.

Applying Kirchhoff’sloop equation,

V VR L+ = 0

⇒ V VR L= −V

VR

L

= −1

22. Both the energy densities are equal i.e., energy isequally divided between electric and magneticfield.

23. / It is based on lens maker's formula and itsmagnification

i.e.,1

11 1

1 2f R R= − −

( )µ .

According to lens maker’s formula, when the lensin the air.

1 32

11 1

1 2f R R= −

−

1 12f x

= ⇒ f x= 2

Here,1 1 1

1 2x R R= −

In case of liquid, where refractive index is43

and53,

we get

Focal length in first liquid

11

1 1

1 1 21f R R

s

l

= −

−

µµ

13243

11

1f x= −

⇒ f1 is positive.1 1

81

4 2141f x x f

= = =( )

⇒ f f1 4=

Focal length in second liquid

11

1 1

2 1 22f R R

s

l

= −

−

µµ

⇒ 13253

11

2f x= −

⇒ f2 is negative.

24. / For total internal reflection of light take place,following conditions must be obeyed.

(i) The ray must travel from denser to rarermedium.

(ii) Angle of incidence (θ) must be greater than or

equal to critical angle ( )C i.e.,

C =

−sin 1 µµ

rarer

denser

Here, sinC = 1nwater

and nwater = +ab

λ2

If frequency is less ⇒ λ is greater and hence, RIn( )water is less and therefore, critical angleincreases. So, they do not suffer reflection andcome out at angle less than 90°.

25. By law of Malus i.e., I I= 02cos θ

Now, I IA A′ = °cos2 30

I IB B′ = °cos2 60

As, ′ = ′I IA B

I IA Bcos cos2 230 60° = °

⇒ I IA B34

14

=

I

IA

B

= 13

26. / The problem is based on frequency dependenceof photoelectric emission. When incident lightwith certain frequency (greater than on thethreshold frequency is focus on a metal surface)then some electrons are emitted from the metalwith substantial initial speed.

When an electron moves in a circular path, then

rmveB

= ⇒ r e B m v2 2 2 2 2

2 2=

KEmax = ( )mvm

2

2⇒ r e B

m

2 2 2

2=( )maxKE


Green

C

Air

Water

IA IA

IB IBPolaroid Polaroid

Transmission axisTransmission axis

Initially Finally

R

L

Work function of the metal (W),

i.e., W = hν − KEmax

1892

12 2

2 2 2 2 2

. − = =φ r e Bm

r eBm

eV eV

[hv → 1.89 eV, for the transition on from third tosecond orbit of H-atom]

= × × × × ×× ×

− − −

−100 10 16 10 9 10

2 91 10

6 19 8

31

.

.

φ= − ××

18916 92 91

..

.

= −189 0 79. .

= 11. eV

27. For hydrogen atom, we get

1 1

1

1

2

2

2 2λ= −

R z

11

341

2

λ=

R( )

1 =

λ 2

2134

R( )

12

343

2

λ=

R( )

13

344

2

λ=

R( )

1 = = =λ λ λ λ1 3 4 2

14

19

1

28. For forward bias, p-side must be a higher potential

than n-side.

So, is forward biased.

29. If student measures 3.50 cm, it means that there inan uncertainly of order 0.01 cm.

For vernier scale with 1 MSD = 1mm

and 9 MSD = 10 VSD

LC of Vernier calliper

= 1MSD − 1VSD

= −

110

19

10

= 1100

cm

30. (a) Infrared rays are used to treat muscular strain.

(b) Radiowaves are used for broadcastingpurposes.

(c) X-rays are used to detect fracture of bones.

(d) Ultraviolet rays are absorbed by ozone.

Chemistry

31. Given, atomic number of Rb, Z = 37

Thus, its electronic configuration is [ ]Kr 5 1s . Since

the last electron or valence electron enter in 5ssubshell.

So, the quantum numbers are n = 5, l = 0, (for s

orbital) m = 0( )Qm l l= + −to , s = + 12

or − 12

32. / To solve this problem, the stepwise approachrequired i.e.,

(i) Write the van der Waals’ equation, thenapply the condition that at low pressure,volume become high,

i.e., V b V− −~

(ii) Now calculate the value of compressibilityfactor ( ).Z

[ / ]Z pV RT=According to van der Waals’ equation,

pa

VV b RT+

− =2

( )

At low pressure, pa

VV RT+

=2

pVaV

RT+ =

pV RTaV

= −

Divide both side by RT,pVRT

aRTV

= −1

33. In CsCl, Cl− lie at corners of simple cube and Cs+

at the body centre, Hence, along the body

diagonal, Cs+ and Cl− touch each other so

r r rCs Cl

+ −+ = 2

Calculation of r

In ∆EDF,

Body centred cubic unit cell

FD b a a a= = + =2 2 2

In ∆AFD,

c a b2 2 2= + = +a a2 22( ) = +a a2 22

c a2 23=

c a= 3


+2 V –2 V

BG

H

a

F C

DE a

b

c

A

a

r

2r

r

As ∆AFD is an equilateral triangle.

∴ 3 4a r= ⇒ ra= 3

4

Hence, r r r a aCs Cl+ + = = × =2 2

34

32

34. / This problem is based on the estimation ofpercentage of N in organic compound usingKjeldahl’s method. Use the concept ofstoichiometry and follow the steps given belowto solve the problem.

( )i Write the balanced chemical reaction forthe conversion of N present in organiccompound to ammonia, ammonia toammonium sulphate and ammoniumsulphate to sodium sulphate.

( )ii Calculate millimoles (m moles) of N

present in organic compound followed by

mass of N present in organic compound

using the concept of stoichiometry.

( )iii At last, calculate % of N present in organic

compound using formula

% of N = ×Mass of N 100

Mass of organic compound

Mass of organic compound = 1 4. g

Let it contain x mmole of N atom.

Organic compound → NH3

x mmole

2NH + H SO (NH ) SO3 2 4 4 2 4

6 m moleinitially taken

→ …(i)

H SO NaOH Na SO2 4 2 42+ → + 2 H O2 ...(ii)

2 mmole NaOH reacted.

Hence, mmoles of H SO2 4 reacted in Eq. (ii) = 1

⇒ mmoles of H SO2 4 reacted from Eq. (i)

= −6 1 = 5mmoles

⇒ mmoles of NH3 in Eq. (i) = ×2 5

= 10 mmoles

⇒ mmoles of N atom in the organic compound

= 10 mmoles

⇒ Mass of N = × ×−10 10 143 = 0.14 g

% of N =Mass of N present in

organic compound

Mass of organic compound× 100

⇒ % of N = ×0 141 4

100..

= 10%

35. / In order to solve the problem, calculate thevalue of cell constant of the first solution andthen use this value of cell constant to calculatethe value of k of second solution. Afterwards,finally calculate molar conductivity usingvalues of k and m.

For first solution,

k = 1.4 Sm 1− , R = 50 Ω , M = 0 2.

Specific conductance ( )κ = ×1R

lA

1.4 Sm− = ×1 150

lA

lA

= × −50 11.4 m

For second solution,

R = 280,lA

= × −50 1 4 1. m

κ = × × =1280

1 4 5014

.

Now, molar conductivity

λ κm m

=×1000

=×

1 41000 0 5

/.

= 12000

= × − −5 10 4 2 1Sm mol

36. C H OH( ) + 3O ( ) 2CO ( )2 5 2 2l g g→ + 3H O()2 l

∆U = − 1364 47. kJ/mol

∆ ∆ ∆H U n RTg= +∆ng = −1

∆H = 1364.47 +1 8.314 298

1000− − × ×

[Here, value of R in unit of J must beconverted into kJ]

= − −1364 47 2 4776. .

= − 1366 94. kJ/mol

During solving such problem, students areadvised to keep much importance in unitconversion. As here, value of R (8.314 J K mol1 1− − )

in JK mol1 1− − must be converted into kJ by dividing

the unit by 1000.

37. According to Debye Huckel Onsager equation,

λ λC B C= −∞

where, λC = limiting equivalent conductivity atconcentration C

λ∞ = limiting equivalent conductivity at

infinite dilution

C = concentration

38. / This problem includes concept of colligativeproperties (osmotic pressure here) and van’tHoff factor. Calculate the effective molarity ofeach solution.i.e., effective molarity

= van’t Hoff factor × molarity

0.5 M C H OH2 5 (aq) i = 1

Effective molarity = 0 5.

0.25 M KBr (aq) i = 2

Effective molarity = 0 5. M

0.1 M Mg (PO )3 4 2(aq) i = 5



0.125 M Na PO3 4(aq) i = 4


Hence, all colligative properties are same.

NOTE

Mg (PO )3 4 2

39. For the given reaction, ∆n = n ng p R−

where, np = number of moles of products

nR = number of moles of reactants

K K RTp cng= ( )

∆

∆ng = − 12

40. / This problem can be solved by determining theorder of reaction w.r.t. each reactant and thenwriting rate law equation of the given equationaccordingly as

r =dCdt

= k A Bx y[ ] [ ]

where, x = order of reaction w.r.t. A

y = order of reaction w.r.t. B

1 2 10 0 1 0 13. ( . ) ( . )× =− k x y

1 2 10 01 023. ( . ) ( . )× =− k x y

2 4 10 02 013. ( . ) ( . )× =− k x y

R = k A B[ ] [ ]1 0

As shown above, rate of reaction remains constant

as the concentration of reactant (B) changes from

0.1 M to 0.2 M and becomes double when

concentration of A change from 0.1 to 0.2

(i.e., doubled).

41. Decreasing order of strength of oxoacids

HClO > HClO > HClO > HOCl4 3 2

Reason Consider the structures of conjugatebases of each oxyacids of chlorine.

Negative charge is more delocalised on ClO4−

due

to resonance, hence ClO4− is more stable (and less

basic).

Hence, we can say as the number of oxygenatom(s) around Cl-atom increases as oxidationnumber of Cl-atom increases and thus, the abilityof loose the H+ increases.

42. Higher the position of element in theelectro-chemical series more difficult is thereduction of its cations.

If Ca2+(aq) is electrolysed, water is reduced inpreference to it. Hence, it cannot be reducedelectrolytically from their aqueous solution.

Ca ( ) + H O Ca + OH + H2+2

2+2aq → ↑−

43. / Arrange the complex formed by differentligands L , L , L1 2 3 and L4, according towavelength of their absorbed light, then usethe following relation to answer the question.

Ligand field strength

∝ Energy of light absorbed

∝ 1

Wavelength of light absorbed

λ L1 L2 L3 L4

Absorbed light Red Green Yellow Blue

Wavelength of absorbed light decreases

∴ Increasing order of energy of wavelengthsabsorbed reflect greater extent of crystal-fieldsplitting, hence higher field strength of the ligand.

Energy Blue (L4) > green (L2) > yellow ( L3)> red ( )L 1

∴ L > L > L > L4 2 3 1 in field strength of ligands.

44. NO is paramagnetic in gaseous state because ingaseous state, it has one unpaired electron.

Total number of electron present = +7 8

= 15 e−

Hence, there must be the presence of unpairedelectron in gaseous state while in liquid state, itdimerises due to unpaired electron.

45. Release of electron is known as reduction. So,H O2 2 acts as reducing agent when it releaseselectrons.

Here, in reactions (II) and (IV), H O2 2 releases twoelectron, hence reactions (II) and (IV) is known asreduction.

In reactions (I) and (III), two electrons are beingadded, so (I) and (III) represents oxidation.

46. I3− is an ion made up of I2 and I− which has linear

shape.

while Cs+ is an alkali metal cation.


O

ClO

O

ClO

OO

O

––

ClO

O –Cl—O

–

CH Cl > CH CH Cl > CH3 3 2 3 CH

C

CHCH

> CH3

C Cl

3

H3

1°(Less

crowded)

(More crowded)2° 3°

Cl

3

47.n

n

m

M

m

M

O

N

O

O

N

N

2

2

2

2

2

2

( )

( )

( )

( )

=

where, m O2= given mass of O2

mN2= given mass of N2

MO2= molecular mass of O2

M N2= molecular mass of N2

nO2= number of moles of O2

nN2= number of moles of N2

=

= ×

m

mO

N

2

2

2832

14

2832

= 732

48. Standard electrode potential of reaction [E0 ] can

be calculated as

Ecell0 = −E ER P

where, ER = SRP of reactantE =P SRP of product

If Ecell0 = +ve, then reaction is spontaneous

otherwise non-spontaneous.

Mn Mn3+ 2+1 = 1.51VE ° →

Mn Mn2+ VE = .2 118° − →

∴ For Mn2+ disproportionation,

E° = − −1.51 V 1.18 V = − <2 69 0. V

49. / Analyse each reaction given in the questionand choose the correct answer on the basis ofoxidation state and stability of iron compounds.Use the concept of Ellingham diagram to solvethis problem.

The correct reactions are as follows

(a) Fe + dil H SO FeSO + H2 4 4 2→

H SO + 2FeSO +12

O2 4 4 2 →

Fe (SO ) + H O2 4 3 2

Fe (SO ) Fe O ( ) + 3SO2 4 3 2 3 3∆ → ↑s

The given reaction is incorrect in question

(b) Fe FeOO2 →∆

[It could also be Fe O2 3 or Fe O3 4]

FeO + H SO FeSO + H O2 4 4 2→

2FeSO Fe O + SO + SO4 2 3 2 3→∆

(c) Fe FeClCl 3

2 Air

∆ ∆ → → No reaction

[It cannot give FeCl2]

(d) Fe Fe O FeOO

3 4CO

600 C

2 → →°∆

CO

700 CFe →

°

This is correct reaction.

50. / This problem is based on conceptual mixing ofproperties of lithium oxide and preparation,properties of coordination compounds. Toanswer this question, keep in mind that onadding acid ammine complexes get destroyed.

(a) Li O +KCl 2LiCl + K O2 2→This is wrong equation, since a stronger base

K O2 cannot be generated by a weaker base

Li O2 .

(b) [CoCl(NH ) ] + 5H3 5++

→ + + −Co ( ) 5NH + Cl2+

4aq

This is correct. All ammine complexes can bedestroyed by addingH⊕ . Hence, on adding acid to

[CoCl(NH ) ]3 5 , It gets converted to Co ( )2+ aq

NH4+

and Cl– .

(c) [Mg(H O) EDTA2 6 ]2 4+ −+ →−

+

excess

OHMg EDTA[ ( )]2

+ 6 2H O

This is wrong, since the formula of complexmust be [Mg(EDTA)]2+ as EDT.

(d) The 4th reaction is incorrect. It can be correctlyrepresented as

2CuSO + 10KCN 2K [Cu(CN) ]4 3 4→+ 2K SO (CN)2 4 2+ ↑

51. / Steric hindrance (crowding) is the basis of SN 2

reaction, by using which we can arrange thereactant in correct order of their reactivitytowards SN 2 reaction.

Rate of SSteric crowding of 'C'N2

1∝

As steric hinderance (crowding) increases, rate ofSN2 reaction decreases.

NOTE SN 2

52. This reaction is an example of carbylaminereaction which includes conversion of amine toisocyanide.

R + →••

°NH CHCl2 3

a e

C H OH

KOH

2 5

1 min

R—N C+ –

Alkyl isocyanide

≡≡ :

NOTE


CH Cl > CH CH Cl > CH3 3 2 3 CH

C

CHCH

> CH3C Cl

3

H3

1°(Less

crowded)

(More crowded)2° 3°

Cl

3

53. R R— —CH OH CH2PCC → == O

Pyridinium chlorochromate is the mild oxidisingagent which causes conversion of alcohol toaldehyde stage. While others causes conversionof alcohol to acid.

54. The reaction is

2CH CCl3 3

6Ag →

∆CH C C CH3 3

But -2 -yne

≡≡ + 6AgCl

55. It is a Kolbe Schmidt reaction.

The second step of the reaction is an example ofacetylation reaction.

56. / This problem can be solved by using the conceptof effect of steric hindrance, hydration andH-bonding in basic strength of amines.

Order of basic strength of aliphatic amine inaqueous solution is as follows (order of Kb)

(CH ) NH > CH NH > (CH ) N > C H NH3 2 3 2 3 3 6 5 2

•• •• •• ••

As we know, p = logK Kb b−

So, (CH ) NH3 2

••

will have smallest pKb value.

In case of phenyl amine, N is attached to sp2

hybridised carbon, hence it has highest pKb andleast basic strength.

57. / Draw the structure of organic compoundsindicating net dipole moment which includeslone pair and bond angle also.

OH groups and SH groups do not cancel theirdipole moment as they exist in differentconfirmation.

58. Dacron is a condensation polymer of ethyleneglycol and methyl terepthalate. Formation ofdacron can be shown as

Here, elimination of MeOHoccurs as a by product.So, this reaction is known as condensationpolymerisation.

59. Quinoline is an alkaloid, it is not present in DNA.DNA has four nitrogen bases in adenine, guanine,cytosine and thymine.

60. / This problem is based on successive reduction,chlorination and elimination reaction. To solvesuch problem, use the function of the givenreagents.

(i) LiAlH4 causes reduction

(ii) PCl5 causes chlorination

(iii) Alc. KOH causes elimination reaction

CH COOH CH CH OH3

LiAlH

3 2

( )

4 →A

→PCl

3 2( )

5CH CH Cl

B

→ ==−HCl

Alc.KOH2

( )2CH CH

Ethylene

C

Mathematics

61. We have, X n n Nn= − − ∈ : 4 3 1

X = , , , , ...0 9 54 243 [Put n = 1 2 3, , , ...]

Y n n N= − ∈ ( ) : 9 1

Y = , , , , ...0 9 18 27 [Put n = 12 3, , ,...]

It is clear that X Y⊂ .

∴ X Y Y∪ =

62. | |z ≥ 2 is the region on or outside circle whose

centre is (0, 0) and radius is 2.

Minimum z + 12

is distance of z, which lie on circle

| |z = 2 from −

12

0, .

∴ Minimum z + =12

Distance of −

12

0, from

( , )−2 0


ONa

125°C

5 atm

COONa

OHH+/Ac O2

COOH

OC—CH3

O

Aspirin

(pain killer)

+ CO2

= − +

+ =2

12

032

2

= − +

+ =1

22 0

32

2

Geometrically Min z AD+ =12

Hence, minimum value of z +

12

lies in the interval

(1, 2).

63. / Put t = x x X− =[ ] , which is a fractional part

function and lie between 0 1≤ < X and thensolve it.

Given, a R∈ and equation is

− − + − + =3 2 02 2 [ ] [ ]x x x x a

Let t x x= − [ ], then equation is

− + + =3 2 02 2t t a ⇒ ta= ± +1 1 3

3

2

Q t x x X= − =[ ] [Fractional part]

∴ 0 1≤ ≤t

01 1 3

31

2

≤ ± + ≤a

Taking positive sign, we get

01 1 3

31

2

≤ + + <a[Q x > 0]

⇒ 1 3 22+ <a ⇒ 1 3 42+ <a ⇒ a2 1 0− <

⇒ ( )( )a a+ − <1 1 0

a ∈ −( , )1 1

For no integral solution of a, we consider the interval( , ) ( , )− ∪1 0 0 1.

NOTE a = 0

64. / If ax bx c2 0+ + = has roots α and β, then

α β+ = −ba

and αβ = ca

. Find the values of α β+

and αβ and then put in ( ) ( )α β α β αβ− = + −2 2 4

to get required value.

Given, α and β are roots of px qx r2 0+ + = , p ≠ 0.

∴ α β+ = −qp

, αβ = rp

...(i)

Since, p q, and r are in AP.

∴ 2q p r= + ...(ii)

Also,1 1

4α β

+ =

⇒ α βαβ+ = 4

⇒ α β αβ+ = 4

⇒ − =qp

rp4

[From Eq. (i)]

⇒ q r= − 4

On putting the value of q in Eq. (ii), we get

⇒ 2 4( )− = +r p r ⇒ p r= − 9

Now, α β+ = − = =−

= −qp

rp

rr

4 49

49

and αβ = =−

=−

rp

rr9

19

∴ ( ) ( )α β α β αβ− = + −2 2 4 = +1681

49

= +16 3681

⇒ ( )α β− =2 5281

⇒ | |α β− = 29

13

65. / Use the property that, two determinants can bemultiplied row-to-row or row-to-column, towrite the given determinant as the product oftwo determinants and then expand.

Given, f n n n( ) = +α βf( )1 = +α β, f( )2 2 2= +α β ,

f( )3 3 3= +α β , f( )4 4 4= +α β

Let ∆ =+ +

+ + ++ + +

3 1 1 1 21 1 1 2 1 31 2 1 3 1 4

f ff f ff f f

( ) ( )( ) ( ) ( )( ) ( ) ( )

⇒ ∆ =+ + + +

+ + + + + ++ + + + +

3 1 1

1 1 1

1 1 1

2 2

2 2 3 3

2 2 3 3

α β α βα β α β α β

α β α β α β4 4+

=⋅ + ⋅ + ⋅⋅ + ⋅ + ⋅

⋅ + ⋅ + ⋅

⋅ + ⋅ + ⋅⋅

1 1 1 1 1 11 1 1 1

1 1 1 1

1 1 1 11 1

2 2

α βα β

α β+ ⋅ + ⋅

⋅ + ⋅ + ⋅α α β β

α α β β1 1 2 2

1 1 1 1

1 1

1 1

2

2 2

2 2 2 2

⋅ + ⋅ + ⋅⋅ + ⋅ + ⋅

⋅ + ⋅ + ⋅

α βα α β β

α α β β

2

=1 1 11

1

1 1 11

12 2 2 2

α βα β

α βα β

=1 1 11

1 2 2

2

α βα β

On expanding, we get

∆ = − − −( ) ( ) ( )1 12 2 2α β α β

But given, ∆ = − − −K( ) ( ) ( )1 12 2 2α β α βHence, K( ) ( ) ( )1 12 2 2− − −α β α β

= − − −( ) ( ) ( )1 12 2 2α β α β

∴ K = 1


+ +

–1 1

(2,0)(0,0)D A

( 2,0)−

Y

XX′

Y′

1—,2 0–

66. / Use the following properties of transpore

( )AB B AT T T= , ( )A AT T = and A A I− =1 and

simplify.

If A is non-singular matrix, then| |A ≠ 0

AA A AT T= and B A AT= −1

BB A A A AT T T T= − −( )( )1 1

= − −A A A AT T1 1( ) [ ( ) ]Q A B AT T TB =

= − −A AA AT T1 1( ) [Q AA A AT T= ]

= −IA AT T( )1 [ ]Q A A I− =1

= −A AT T( )1

= −( )A A T1 [Q( )AB B AT T T= ]

= IT = I

67. / To find the coefficient of x 3 and x 4, use the

formula of coefficient of x r in ( )1 − x n is

( )−1 r nrC and then simplify.

In expansion of ( )( )1 1 22 18+ + −ax bx x ,

Coefficient of x3 = Coefficient of x3 in ( )1 2 18− x

+ Coefficient of x2 in a( )1 2 18− x

+ Coefficient of x in b x( )1 2 18−

= − ⋅ + ⋅ − ⋅183

3 182

2 1812 2 2C a C b C

Given, coefficient of x3 0== ⋅ + ⋅ − ⋅ =18

33 18

22 18

12 2 2 0C a C b C

⇒ − × ××

⋅ + ⋅ × ⋅18 17 163 2

818 17

222a

− ⋅ ⋅ =b 18 2 0

⇒ 1734 16

3a b− = ×

...(i)

Similarly, coefficient of x4

184

4 183

3 182

22 2 2 0C a C b C⋅ − ⋅ + ⋅ ⋅ =

∴ 32 3 240a b− = …(ii)

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

a = 16, b = 2723

68. We know that angle between two lines is

cosθ = + +

+ + + +

a a b b c c

a b c a b c

1 2 1 2 1 2

12

12

12

22

22

22

l + + =m n 0

⇒ l = − +( )m n

⇒ ( )m n+ =2 2l

⇒ m n mn m n2 2 2 22+ + = +

[Ql2 2 2= +m n , given]

⇒ 2 0mn =When m= 0 ⇒ l = − n

Hence, ( , , )l m n is ( , , )1 0 1− .

When n = 0, then l = − m

Hence, ( , , )l m n is (1, 0, –1).

∴ cosθ = + +×

=1 0 0

2 2

12

⇒ θ π=3

69. Given,

k ⋅ = +10 10 2 11 109 9 1 8( ) ( )

+ + +3 11 10 10 112 7 9( ) ( ) ... ( )

k = +

+

+ +

1 2

1110

31110

101110

2 9

... ...(i)

1110

11110

21110

2

=

+

k

+ +

+

... 9

1110

101110

9 10

...(ii)

On subtracting Eq. (ii) from Eq. (i), we get

k 11110

11110

1110

2

−

= + +

+ +

−

...

1110

101110

9 10

⇒ k10 11

10

11110

1

1110

11

10

−

=

−

−

− 01110

10

Q In G ,sum of terms whenP na r

rr

n

= −−

>

( )

,1

11

⇒ − k =

− −

10 10

1110

10 101110

10 10

∴ k = 100

70. Let a ar ar, , 2 be in GP ( )r > 1.

On multiplying middle term by 2, a ar, ,2 ar2 are in

AP.

⇒ 4 2ar a ar= +

⇒ r r2 4 1 0− + =

⇒ r = ± − = ±4 16 4

22 3

⇒ r = +2 3 [QAP is increasing]

71. limsin( cos )

x

x

x→ 0

2

2

π

= limsin ( sin )

x

x

x→

−0

2

2

1π = −→lim

sin( sin )

x

x

x0

2

2

π π

=→lim

sin( sin )x

x

x0

2

2

π[ sin ( ) sin ]Q π θ θ− =

= ×

→lim

sin sin

sin( )

sinx 0

2

2

2

2

ππ

πx

x

x

x

= π Q limsin

θ

θθ→

=

0

1


72. Here, ‘g’ is the inverse of f x( ).

⇒ fog x x( ) =On differentiating w.r.t. x, we get

′ × ′ =f g x g x ( ) ( ) 1

′ =′

g xf g x

( )( ( ))

1 =

+

11

1 5 ( )g x

Q ′ =+

f x

x( )

1

1 5

′ = +g x g x( ) ( )1 5

73. Given, f g( ) ( ),0 2 1= = g( )0 0= and f( )1 6=

f and g are differentiable in (0, 1).

Let h x f x g x( ) ( ) ( )= − 2 …(i)

h f g( ) ( ) ( )0 0 2 0= −h( )0 2 0= −h( )0 2=

and h f g( ) ( ) ( )1 1 2 1= − = −6 2 2( )

h( )1 2= , h h( ) ( )0 1 2= =Hence, using Rolle’s theorem,

′ =h c( ) 0, such that c ∈ ( , )0 1

Differentiating Eq. (i) at c, we get

⇒ ′ − ′ =f c g c( ) ( )2 0

⇒ ′ = ′f c g c( ) ( )2

74. Here, x = −1 and x = 2 are extreme points of

f x x x x( ) log| | ,= + β +α 2 then

′ = α + β +f xx

x( ) 2 1

′ − = − − β + =f ( )1 2 1 0α …(i)

[At extreme point, ′ =f x( ) 0]

′ = + β + =f ( )22

4 1 0α

…(ii)

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

α = 2, β = − 12

75. 11

1

+ −

+

∫ xx

e dxx

x

= + −

+ +

∫∫e dx xx

e dxx

xx

x

1

2

1

11

= + −+ + +

∫∫e dx xeddx

x e dxx

xx

xx

x

1 1 1

( )

= + −+ + +

∫∫e dx xe e dxx

xx

xx

x

1 1 1

Q 112

1 1

−

=

+ +

∫x

e dx ex

xx

x

= + −+ + +∫ ∫e dx xe ex dxx x x x x1 1 1/ / /

= ++

xe Cx

x

1

76. / Use the formula,|x a|− = − ≥− − <

x a x ax a x a

,( ) ,

to break given integral in two parts and thenintegrate separately.

1 22

2

0−

∫ sin

xdx

π= | sin |1 2

20−∫ x

dxπ

= −

− −

∫ ∫1 2

21 2

203

3

sin sinx

dxx

dxπ

ππ

= +

− +

x

xx

x4

24

20

3

3

cos cos

π

π

π

= − −4 3 43π

77. Given, A x y x y= + ≤( , ) : 2 2 1and y x2 1≤ −

Required area = + −∫12

2 12 2

0

1πr y dy( )

= + −

1

21 2

32

3

0

1

π( ) yy

= +π2

43

78. Given differential equationdpdt

pt− = −12

200( ) is a

linear differential equation.

Here, p t Q t( ) , ( )= − = −12

200

IF =∫

=−

−

e edt t1

2 2

Hence, solution is

p t Q t dt( ). ( )IF F= ∫ I

p t e e dtt t

( )⋅ = − ⋅− −

∫2 2200

p t e et t

( )⋅ =− −

2 2400 + K

⇒ p t ke( ) /= + −400 1 2

If p( )0 100= , then k = − 300

⇒ p t et

( ) = −400 300 2


Y

XX′

Y′

(0,1)(–1,0)

79. Coordinate of S = + −

7 62

3 12

, =

132

1,

[QS is mid-point of line QR]

Slope of the line PS is−29

.

Required equation passesthrough ( , )1 1− and parallelto PS is

y x+ = − −129

1( )

⇒ 2 9 7 0x y+ + =

80. Let coordinate of the intersection point in fourthquadrant be (α, − α).Since, (α, − α) lies on both lines 4 2 0ax ay c+ + =and 5 2 0bx by d+ + = .

∴ 4 2 0a a cα α− + = ⇒ α = −ca2

…(i)

and 5 2 0b b dα α− + = ⇒ α = −db3

…(ii)

From Eqs. (i) and (ii), we get− = −c

a

d

b2 3

⇒ 3 2bc ad= ⇒ 2 3 0ad bc− =

81. Equation of ellipse is x y2 23 6+ = orx y2 2

6 21+ = .

Equation of the tangent isx

ay

bcos sinθ θ+ = 1

Let ( , )h k be any point on the locus.

∴ ha

kb

cos sinθ θ+ = 1 …(i)

Slope of the tangent line is−ba

cotθ.

Slope of perpendicular drawn from centre (0,0) to

( , )h k iskh.

Since, both the lines are perpendicular.

∴ kh

ba

× −

= −cotθ 1

⇒ cos sinθ θha kb

= = α [Say]

⇒ cosθ = αha, sinθ = αkb

From Eq. (i), we getha

hakb

kb( ) ( )α + α = 1

⇒ h k2 2 1α + =α

⇒ α =+1

2 2h k

Also, sin cos2 2 1θ θ+ =⇒ ( ) ( )α + α =kb ha2 2 1

⇒ α + α =2 2 2 2 2 2 1k b h a

⇒ k b

h k

h a

h k

2 2

2 2 2

2 2

2 2 21

( ) ( )++

+=

⇒ 2 61

2

2 2 2

2

2 2 2

k

h k

h

h k( ) ( )++

+=

[ , ]Q a b2 6 2= =2

⇒ 6 22 2 2 2 2x y x y+ = +( )

[Replacing k by y and h by x]

82. / Use the property, when two circles touch eachother externally, then distance between thecentre is equal to sum of their radii, to getrequired radius.

Let the coordinate of the centre of T be ( , )0 k .

Distance between their centre

k k+ = + −1 1 1 2( ) [ ]Q C C k1 2 1= +

⇒ k k k+ = + + −1 1 1 22

⇒ k k k+ = + −1 2 22

⇒ k k k k2 21 2 2 2+ + = + − ⇒ k = 14

So, the radius of circle T is k i e. .,14.

83. / Let the tangent to parabola be y mx= + a/m, if

it touches the other curve, then D = 0, to get the

value of m.

For parabola, y x2 4=

Let y mxm

= + 1be tangent line and it touches the

parabola x y2 32= − .

∴ x mxm

2 321= − +

⇒ x mxm

2 3232

0+ + =

Q D = 0

∴ ( )32 432

02mm

− ⋅ = ⇒ m3 1

8=

∴ m = 12


XX′

Y′

(1,1)

(1– )kT(0, )k

O (1,0)

C

C1

C2

1

Y

84. / Here, plane, line and its image are parallel toeach other. So, find any point on the normal tothe plane from which the image line will be

passed and then find equation of image line.

Here, plane and line are parallel to each other.Equation of normal to the plane through the point(1, 3, 4) is

x y zk

− = −−

= − =12

31

41

[Say]

Any point in this normal is ( , , ).2 1 3 4k k k+ − + +

Then,2 1 1

23 3

24 4

2k k k+ + − + + +

, , lies on

plane.

⇒ 2 ( )kk k+ − −

+ +

+ =16

28

23 0

⇒ k = −2

Hence, point through which this image pass is

( , , )2 1 3 4k k k+ − +i.e., ( ( ) , , )2 2 1 3 2 4 2− + + − = ( , , )−3 5 2

Hence, equation of image line isx y z+ = − = −

−3

35

12

5

85. / Use the following formula to simplify

a b c a c b a b c× × = ⋅ − ⋅( ) ( ) ( )

[ ] [ ] [ ]a b c b c a c a b= =and [ ] [ ] [ ] .a a b a b b a c c= = = 0

Now, [ ]a b b c c a× × ×= × ⋅ × × ×a b b c c a[( ) ( )]

= × ⋅ × ×a b k c a[ ( )] [Let k b c= × ]

= × ⋅ ⋅ − ⋅a b k a c k c a[( ) ( ) ]

= × ⋅ × ⋅ − × ⋅( ) [( ) ( ) ]a b b c a c b c c a

= × ⋅ −( ) ([ ] )a b b c a c 0 Q[ ]b c c× ⋅ = 0

= × ⋅( [ ]a b) c b c a = [ ][ ]a b c b c a

= [ ]a b c 2 Q [ ] [ ]a b c b c a=

But given, [ ] [ ]a b b c c a× × × = λ a b c 2

So, [ ] [ ]a bc a bc2 2= λ ⇒ λ = 1

86. Given, P A B( )∪ = 16

, P A B( )∩ = 14, P A( ) = 1

4

∴ P A B P A B( ) ( )∪ = − ∪1 = − =116

56

and P A P A( ) ( )= −1 = − =114

34

P A B P A P B P A B( ) ( ) ( ) ( )∪ = + − ∩⇒ 5

634

14

= + −P B( )

P B( ) = 13

⇒ A and B are not equally likely.

P A B P A P B( ) ( ) ( )∩ = ⋅ = 14

So, events are independent.

87. Here, XX

n

i= ∑ = + + + + +2 4 6 8 100

50

K

= ×50 51

50= 51

[Q∑ = +2 1n n n( ), here n = 50]

Variance, σ2 2 21= ∑ − −n

Xi x( )

σ = + + + −2 2 2 2 2150

2 4 100 51( ) ( )K

= 833

88. Given, f xk

x xkk k( ) (sin cos )= +1

,

where x R∈ and k > 1

f x f x4 6( ) ( )− = +14

4 4(sin cos )x x

− +16

6 6(sin cos )x x

= − ⋅14

1 2 2 2( sin cos )x x

− − ⋅16

1 3 2 2( sin cos )x x = − =14

16

112

89. In ∆OA B1 1,

tan45 1 1

1

° = A B

OB⇒ 20

11OB

=

⇒ OB1 20=

In ∆OA B2 2,

tan3020

2

° =OB

⇒ OB2 20 3=

⇒ B B OB1 2 1 20 3+ =⇒ B B1 2 20 3 20= −⇒ B B1 2 20 3 1= −( )m

Now, speed = DistanceTime

= −20 3 11

( )

= −20 3 1( )m/s

90.

p q ~p ~q p q↔ p q↔ ~ ~ p q↔ ~ ~( )p q↔

T F F T F T T F

F T T F F T T F

T T F F T F F T

F F T T T F F T

~( ~ )p q↔ is equivalent to ( )p q↔ .


20

45°

30°O

20 m

A2

A1

B1B2

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Time : MM : 360