Instructions

1. This test consists of 90 questions.

2. There are three parts in the question paper consisting of Physics, Chemistry & Mathematics having 30questions 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 no. 2 for correct response of eachquestion. 1 mark will be deducted for indicating incorrect response of each question. No deductionfrom the total score will be made if no response is indicated 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 questionwill be treated as wrong response and marks for wrong response will be deducted according as perinstructions.

Physics1. An observer is moving with half the

speed of light towards a stationarymicrowave source emitting waves atfrequency 10 GHz. What is thefrequency of the microwavemeasured by the observer? (speed oflight = ´ -3 108 ms 1)

(a) 12.1 GHz

(b) 17.3 GHz

(c) 15.3 GHz

(d) 10.1 GHz

2. The following observations weretaken for determining surfacetension T of water by capillarymethod. Diameter of capillary,d = ´ -125 10 2. m rise of water,

h = ´ -1 45 10 2. m. Using g = 9 80.

m/s2 and the simplified relation

Trhg

= ´2

103N m/ , the possible

error in surface tension is closest to

(a) 1.5% (b) 2.4%

(c) 10% (d) 0.15%

3. Some energy levels of a moleculeare shown in the figure. The ratio ofthe wavelengths r = l / l1 2 is givenby

(a) r = 23

(b) r = 34

(c) r = 13

(d) r = 43

4. A body of mass m = -10 2kg is

moving in a medium andexperiences a frictional forceF kv= - 2 . Its initial speed is v0 10=ms -1. If, after 10 s, its energy is1

802mv , the value of k will be

(a) 10 3 1- -kgs (b) 10 4 1- -kgm

(c) 10 1 1 1- - -kgm s (d) 10 3 1- -kgm

SOLVED PAPER 2017JEE Main

–E

– E4/3

– E2

– E3

λ2

λ1

5. Cp and CV are specific heats at

constant pressure and constantvolume, respectively. It is observedthat C C ap V- = for hydrogen gasC C bp V- = for nitrogen gas. Thecorrect relation between a and b is

(a) a b= (b) a b= 14

(c) a b= 28 (d) a b= 114

6. The moment of inertia of a uniformcylinder of length l and radius Rabout its perpendicular bisector is I.What is the ratio l/ R such that themoment of inertia is minimum?

(a) 32

(b) 1 (c) 32

(d) 32

7. A radioactive nucleus A with ahalf-life T , decays into a nucleus B.At t = 0, there is no nucleus B. Aftersometime t , the ratio of the number ofB to that of A is 0.3. Then, t is givenby

(a) t Te

= log .log

132

(b) t T= log .13

(c) tT=

log .13(d) t

T e= loglog .

22 13

8. Which of the following statements isfalse?

(a) In a balanced Wheatstone bridge, if thecell and the galvanometer areexchanged, the null point is disturbed

(b) A rheostat can be used as a potentialdivider

(c) Kirchhoff’s second law representsenergy conservation

(d) Wheatstone bridge is the most sensitivewhen all the four resistances are of thesame order of magnitude

9. A capacitance of 2 mF is required in

an electrical circuit across a potential

difference of 1kV. A large number of

1 mF capacitors are available which

can withstand a potential difference

of not more than 300 V. The

minimum number of capacitors

required to achieve this is

(a) 16 (b) 24 (c) 32 (d) 2

10. In the given circuit diagram, whenthe current reaches steady state inthe circuit, the charge on thecapacitor of capacitance C will be

(a) CEr

r r1

2( )+

(b) CEr

r r2

2( )+

(c) CEr

r r1

1( )+

(d) CE

11. In the below circuit, the current ineach resistance is

(a) 0.25 A (b) 0.5 A (c) 0 A (d) 1 A

12. In amplitude modulation, sinusoidalcarrier frequency used is denoted bywc and the signal frequency isdenoted by wm. The bandwidth( )Dwm of the signal is such that Dwm

<< wc . Which of the followingfrequencies is not contained in themodulated wave?

(a) wc (b) w wm c+

(c) w wc m- (d) wm

13. In a common emitter amplifiercircuit using an n-p-n transistor,the phase difference between theinput and the output voltages willbe

(a) 90° (b) 135°(c) 180° (d) 45°

E r

r1

r2

C

1Ω 1Ω 1Ω

2V 2V 2V

2V 2V 2V

14. A copper ball of mass 100 g is at atemperature T . It is dropped in acopper calorimeter of mass 100 g,filled with 170 g of water at roomtemperature. Subsequently, thetemperature of the system is found tobe 75°C. T is (Given, roomtemperature = °30 C, specific heat ofcopper = 0 1. cal/g°C)

(a) 885°C (b) 1250°C (c) 825°C (d) 800°C

15. In a Young’s double slit experiment,slits are separated by 0.5 mm and thescreen is placed 150 cm away. Abeam of light consisting of twowavelengths, 650 nm and 520 nm, isused to obtain interference fringeson the screen. The least distancefrom the common central maximumto the point where the bright fringesdue to both the wavelengthscoincide, is

(a) 7.8 mm (b) 9.75 mm

(c) 15.6 mm (d) 1.56 mm

16. An electric dipole has a fixed dipolemoment p, which makes angleq withrespect to X-axis. When subjected toan electric field E i1 = E$, it

experiences a torque T k1 = t $ . When

subjected to another electric field E2

= 3 1E $j, it experiences a torque

T T2 1= - . The angle q is

(a) 45° (b) 60° (c) 90° (d) 30°

17. A slender uniform rod of mass M andlength l is pivoted at one end so thatit can rotate in a vertical plane (seethe figure). There is negligiblefriction at the pivot. The free end isheld vertically above the pivot andthen released. The angularacceleration of the rod when itmakes an angleq with the vertical, is

(a) 23g

l

sin q (b) 32g

l

cos q

(c) 23g

l

cosq (d) 32

g

l

sinq

18. An external pressure P is applied ona cube at 0°C so that it is equallycompressed from all sides. K is thebulk modulus of the material of thecube and a is its coefficient of linearexpansion. Suppose we want tobring the cube to its original size byheating. The temperature should beraised by

(a) P

Ka(b) 3a

PK(c) 3PKa (d) P

K3a

19. A diverging lens with magnitude offocal length 25 cm is placed at adistance of 15 cm from a converginglens of magnitude of focal length 20cm. A beam of parallel light falls onthe diverging lens. The final imageformed is

(a) virtual and at a distance of 40 cm fromconvergent lens

(b) real and at a distance of 40 cm from thedivergent lens

(c) real and at a distance of 6 cm from theconvergent lens

(d) real and at a distance of 40 cm fromconvergent lens

20. An electron beam is accelerated by apotential difference V to hit ametallic target to produce X-rays. Itproduces continuous as well ascharacteristic X-rays. If lmin is thesmallest possible wavelength ofX-rays in the spectrum, the variationof log lmin with log V is correctlyrepresented in

z

x

θ

logλ

min

log V

(a)

logλ

min

log V

(b)

logλ

min

log V

logλ

min

log V

(c) (d)

21. The temperature of an open room ofvolume 30 m3 increases from 17°C to27°C due to the sunshine. Theatmospheric pressure in the roomremains 1 105´ Pa. If ni and n f are

the number of molecules in the roombefore and after heating, then n nf i-will be

(a) 138 1023. ´

(b) 2 5 1025. ´

(c) - ´2 5 1025.

(d) - ´161 1023.

22. In a coil of resistance 100 W, acurrent is induced by changing themagnetic flux through it as shown inthe figure. The magnitude of changein flux through the coil is

(a) 225 Wb (b) 250 Wb

(c) 275 Wb (d) 200 Wb

23. When a current of 5 mA is passedthrough a galvanometer having a coilof resistance 15 W, it shows full scaledeflection. The value of theresistance to be put in series with thegalvanometer to convert it into avoltmeter of range 0-10 V is

(a) 2 045 103. ´ W

(b) 2 535 103. ´ W

(c) 4 005 103. ´ W

(d) 1985 103. ´ W

24. A time dependent force F t= 6 acts

on a particle of mass 1 kg. If theparticle starts from rest, the workdone by the force during the first 1 swill be

(a) 22 J (b) 9 J

(c) 18 J (d) 4.5 J

25. A magnetic needle of magneticmoment 6 7 10 2 2. ´ - Am and

moment of inertia 7 5 10 6. ´ - kg m2

is performing simple harmonicoscillations in a magnetic field of0.01 T. Time taken for 10 completeoscillations is

(a) 8.89 s (b) 6.98 s

(c) 8.76 s (d) 6.65 s

26. The variation of acceleration due togravity g with distance d from centreof the Earth is best represented by(R = Earth’s radius)

27. A body is thrown verticallyupwards. Which one of thefollowing graphs correctly representthe velocity vs time?

28. A particle A of mass m and initialvelocity v collides with a particle B

of massm

2which is at rest. The

collision is head on, and elastic. Theratio of the de-Broglie wavelengthsl A to lB after the collision is

(a)ll

A

B

= 2 (b)ll

A

B

= 23

(c)ll

A

B

= 12

(d)ll

A

B

= 13

Time 0.5

10

Current(A)

(s)

(a) (b)

(c) (d)

g

O Rd

g

O R

d

g

O Rd

g

O Rd

v

t

v

t

v

t

v

t

(a) (b)

(c) (d)

Chemistry31. Which of the following compounds

will give significant amount ofmeta-product during mononitrationreaction?

32. DU is equal to

(a) isochoric work (b) isobaric work

(c) adiabatic work (d) isothermal work

33. The increasing order of reactivity ofthe following halides for the SN1reaction is

I. CH CH(Cl)CH CH3 2 3

II. CH CH CH Cl3 2 2

III. p -H CO C H CH Cl3 6 4 2¾ ¾

(a) (III) < (II) < (I) (b) (II) < (I) < (III)

(c) (I) < (III) < (II) (d) (II) < (III) < (I)

34. The radius of the second Bohr orbitfor hydrogen atom is (Planck’sconstant ( ) .h = ´ -6 6262 10 34 Js;

mass of electron = ´ -9 1091 10 31. kg

; charge of electron( ) .e = ´ -160210 10 19 C; permitivity

of vacuum ( ) .Î = ´ -0

128854185 10kg m A1 2- - 3 )

(a) 1.65 Å

(b) 4.76 Å

(c) 0.529 Å

(d) 2.12 Å

35. pKa of a weak acid (HA) and pKb of a

weak base (BOH) are 3.2 and 3.4,respectively. The pH of their salt(AB) solution is

(a) 7.2 (b) 6.9 (c) 7.0 (d) 1.0

36. The formation of which of thefollowing polymers involveshydrolysis reaction?

(a) Nylon-6 (b) Bakelite

(c) Nylon-6, 6 (d) Terylene

(a)

KE

tO T(b)

KE

tO TT/2

(c)

KE

tO T/4 TT/2

(d)

KE

tO TT/2 T

29. A particle is executing simple harmonic motion with a time period T . At timet = 0, it is at its position of equilibrium. The kinetic energy-time graph of theparticle will look, like

30. A man grows into a giant such that his linear dimensions increase by a factor of 9.Assuming that his density remains same, the stress in the leg will change by afactor of

(a) 19

(b) 81 (c) 181

(d) 9

(a)

OH

(b)

OCOCH3

(c)

NH2

(d)

NHCOCH3

37. The most abundant elements bymass in the body of a healthy humanadult are Oxygen (61.4%); Carbon(22.9%), Hydrogen (10.0 %); andNitrogen (2.6%). The weight which a75 kg person would gain if all 1H

atoms are replaced by 2H atoms is

(a) 15 kg (b) 37.5 kg (c) 7.5 kg (d) 10 kg

38. Which of the following, upontreatment with tert-BuONa followedby addition of bromine water, fails todecolourise the colour of bromine?

39. In the following reactions, ZnO isrespectively acting as a/an

(i) ZnO + Na O Na ZnO2 2 2¾®(ii) ZnO + CO ZnCO2 3¾®(a) base and acid (b) base and base

(c) acid and acid (d) acid and base

40. Both lithium and magnesiumdisplay several similar propertiesdue to the diagonal relationship;however, the one which is incorrectis

(a) Both form basic carbonates(b) Both form soluble bicarbonates(c) Both form nitrides(d) nitrates of both Li and Mg yield NO2

and O2 on heating41. 3-methyl-pent-2-ene on reaction

with HBr in presence of peroxideforms an addition product. Thenumber of possible stereoisomers forthe product is

(a) six (b) zero (c) two (d) four

42. A metal crystallises in a face centredcubic structure. If the edge length ofits unit cell is ‘a’, the closestapproach between two atoms inmetallic crystal will be

(a) 2 a (b) 2 2 a (c) 2 a (d) a

2

43. Two reactions R1 and R2 have

identical pre- exponential factors.Activation energy of R1 exceeds thatof R2 by 10 kJ mol- 1. If k 1 and k2 are

rate constants for reactions R1 andR2 , respectively at 300 K, then

lnk

k2

1

æ

èçç

ö

ø÷÷ is equal to

( . )R = - -8 314 J mol K1 1

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

44. The correct sequence of reagents forthe following conversion will be

(a) [Ag(NH ) ] OH , H / CH OH,3 2+ +

3-

CH MgBr3

(b) CH MgBr, H / CH OH,3+

3

[Ag(NH ) ] OH3 2+ -

(c) CH MgBr, [Ag(NH ) ] OH ,3 3 2+ -

H / CH OH+3

(d) [Ag(NH ) ] OH , CH MgBr,3 2+

3-

H / CH OH+3

45. The Tyndall effect is observed onlywhen following conditions aresatisfied

1. The diameter of the dispersedparticles is much smaller thanthe wavelength of the light used.

2. The diameter of the dispersedparticle is not much smallerthan the wavelength of the lightused.

3. The refractive indices of thedispersed phase and dispersionmedium are almost similar inmagnitude.

4. The refractive indices of thedispersed phase and dispersionmedium differ greatly inmagnitude.

(a) 1 and 4 (b) 2 and 4

(c) 1 and 3 (d) 2 and 3

O

(a)Br

(b)Br

C H6 5

O

(c)Br

(d)Br

HO CH3

OHCH3

CHO

O

H C3

46. Which of the following compoundswill behave as a reducing sugar in anaqueous KOH solution?

47. Given,C O CO(graphite) 2 2+ ¾®( ) ( )g g ;

D r H° = - 3935. kJ mol 1-

H ( )1

2O ( ) H O( )2 2 2g g l+ ¾® ;

D r H° = - 285 8. kJ mol 1-

CO ( ) 2 H O( ) CH ( )2 2 4g l g+ ¾®

+2O ( )2 g ;

D r H° = + 890 3. kJ mol 1-

Based on the above thermochemicalequations, the value of D r H° at298 Kfor the reaction,

C 2 H ( ) CH ( )(graphite) 2 4+ ¾®g g willbe

(a) + 78 8. kJ mol 1- (b) + 144 0. kJ mol 1-

(c) - 74 8. kJ mol 1- (d) - 144 0. kJ mol 1-

48. Which of the following reactions isan example of a redox reaction?

(a) XeF O F XeF O4 2 2 6 2+ ¾® +

(b) XeF PF [XeF] PF2 5++ ¾® -

6

(c) XeF H O XeOF 2HF6 2 4+ ¾® +

(d) XeF 2H O XeO F 4HF6 2 2 2+ ¾® +

49. The products obtained whenchlorine gas reacts with cold anddilute aqueous NaOH are

(a) ClO- and ClO3- (b) ClO2

- and ClO3-

(c) Cl- and ClO- (d) Cl- and ClO2-

50. The major product obtained in thefollowing reaction is

(a) ( ) C H CH(O Bu)CH C H6 5 2 6 5± t

(b) C H CH == CHC H6 5 6 5

(c) ( ) C H CH(O Bu)CH C H6 5t

2 6 5+

(d) ( )C H CH(O Bu)CH C H6 5t

2 6 5-

51. Sodium salt of an organic acid ‘X ’produces effervescence with conc.H SO .2 4 ‘X’ reacts with the acidifiedaqueous CaCl2 solution to give awhite precipitate whichdecolourises acidic solution ofKMnO .4 ‘X ’ is

(a) C H COONa6 5 (b) HCOONa

(c) CH COONa3 (d) Na C O2 2 4

52. Which of the following species is notparamagnetic?

(a) NO (b) CO

(c) O2 (d) B2

53. The freezing point of benzenedecreases by 0.45°C when 0.2 g ofacetic acid is added to 20 g ofbenzene. If acetic acid associates toform a dimer in benzene, percentageassociation of acetic acid in benzenewill be (K f for benzene = 512. K kgmol 1- )

(a) 64.6 % (b) 80.4 %

(c) 74.6 % (d) 94.6 %

OHOH C2 CH OH2

OCOCH3

HO

OH

(a)

OHOH C2 CH OH2

HO

OH

(b)

OHOH C2 CH OH2

OCH3

HO

OH

(c)

O

HOH C2

CH OCH2 3

OH

(d)OH

OH

BrH

C H6 5

C H6 5

tBuOK∆

54. Which of the following molecules isleast resonance stabilised?

55. On treatment of 100 mL of 0.1 Msolution ofCoCl3 .6H O2 with excess of AgNO3 ;1.2 1022´ ions are precipitated. Thecomplex is

(a) [Co(H O) Cl ] Cl .2H O2 4 2 2

(b) [Co(H O) Cl ]. 3H O2 3 3 2

(c) [Co(H O) ]Cl2 6 3

(d) [Co(H O) Cl] Cl .H O2 5 2 2

56. The major product obtained in thefollowing reaction is

57. A water sample has ppm levelconcentration of following anions

F 10; SO 100; NO 5042

3- - -= = =

the anion/anions that make/makesthe water sample unsuitable fordrinking is/are

(a) Only NO3- (b) Both SO4

2 - and NO3-

(c) Only F- (d) Only SO42 -

58. 1 g of a carbonate (M CO2 3) on

treatment with excess HCl produces0.01186 mole of CO2 . The molarmass of M CO2 3 in g mol 1- is

(a) 1186 (b) 84.3 (c) 118.6 (d) 11.86

59. Given, ECl / Cl2

-° = 136. V,

ECr / Cr3 +° = - 0 74. V

ECr O / Cr2 7

2 - +° =3 133. V,

EMnO / Mn4

2 +-° = 151. V

Among the following, the strongestreducing agent is

(a) Cr (b) Mn2 + (c) Cr3 + (d) Cl-

60. The group having isoelectronicspecies is

(a) O , F , Na , Mg2 + 2 +- -

(b) O , F , Na, Mg+- -

(c) O , F , Na, Mg2 2 +- -

(d) O , F , Na , Mg+ 2 +- -

Mathematics61. IfS is the set of distinct values ofb for

which the following system of linearequations

x y z+ + = 1, Þ x ay z+ + = 1and ax by z+ + = 0has no solution, then S is(a) an infinite set(b) a finite set containing two or more

elements(c) singleton set(d) a empty set

62. The statement( ) [(~ ) ]p q p q q® ® ® ® is

(a) a tautology(b) equivalent to ~ p q®(c) equivalent to p q® ~(d) a fallacy

63. If 5 2 2(tan cos )x x- = 2 2 9cos x + ,

then the value of cos 4x is

(a) - 35

(b) 13

(c) 29

(d) - 79

(a) (b)O

(c)N

(d)

O

O

DIBAL-H

COOH

O

(a)

COOH

CHO

OH

(b)

CHO

CHO

OH

(c)

COOH

CHO (d)

CHO

CHO

64. For three events A B, and C,

if P (exactly one of A or B occurs)= P(exactly one of B or C occurs) = P

(exactly one of C or A occurs) =1

4and P (all the three events occur

simultaneously) =1

16, then the

probability that atleast one of theevents occurs, is

(a) 732

(b) 716

(c) 764

(d) 316

65. Let w be a complex number suchthat 2 1w + = z, where z = - 3. If

1 1 1

1 1

1

32 2

2 7

- - =w ww w

k , then k is

equal to

(a) - z (b) z (c) - 1 (d) 1

66. Let k be an integer such that thetriangle with vertices ( , )k k- 3 , ( , )5 kand ( , )- k 2 has area28 sq units. Then, the orthocentre ofthis triangle is at the point

(a) 2 12

, -æèç

öø÷

(b) 1 34

,æèç

öø÷

(c) 1 34

, -æèç

öø÷

(d) 2 12

,æèç

öø÷

67. If 20 m of wire is available forfencing off a flower-bed in the formof a circular sector, then themaximum area (in sq m) of theflower-bed is

(a) 12.5 (b) 10 (c) 25 (d) 30

68. The area (in sq units) of the region( , ) : , ,x y x x y³ + £0 3 x y2 4£ and

y x£ +1 is

(a) 5912

(b) 32

(c) 73

(d) 52

69. If the image of the point P( , , )1 2 3- in

the plane 2 3 4 22 0x y z+ - + =measured parallel to the linex y z

1 4 5= = is Q, then PQ is equal to

(a) 3 5 (b) 2 42 (c) 42 (d) 6 5

70. For x Îæèç

öø÷0

1

4, , if the derivative of

tan-

-

æ

èçç

ö

ø÷÷

1

3

6

1 9

x x

xis x g x× ( ), then g x( )

equals

(a) 9

1 9 3+ x(b) 3

1 9 3

x x

x-

(c) 3

1 9 3

x

x-(d) 3

1 9 3+ x

71. If (2 + sin x)dy

dxy x+ + =( ) cos1 0

and y( )0 1= , then yp2

æèç

öø÷ is equal to

(a) 13

(b) - 23

(c) - 13

(d) 43

72. Let a vertical tower AB have its end Aon the level ground. Let C be themid-point of AB and P be a point onthe ground such that AP AB= 2 . IfÐ =BPC b, then tan b is equal to

(a) 67

(b) 14

(c) 29

(d) 49

73. If A =-

-

é

ëê

ù

ûú

2 3

4 1, then adj

( )3 122A A+ is equal to

(a)72 84

63 51

--

é

ëê

ù

ûú (b)

51 63

84 72

é

ëê

ù

ûú

(c)51 84

63 72

é

ëê

ù

ûú (d)

72 63

84 51

--

é

ëê

ù

ûú

74. For any three positive real numbersa b, and c, if9 25 25 32 2 2( ) ( )a b c ac+ + -= +15 3b a c( ), then

(a) b c, and a are in GP(b) b c, and a are in AP(c) a b, and c are in AP(d) a b, and c are in GP

75. The distance of the point ( , , )1 3 7-from the plane passing through thepoint ( , , )1 1 1- - having normalperpendicular to both the lines

x y z-=

+-

=-1

1

2

2

4

3and

x y-=

+-

=2

2

1

1

z +-

7

1, is

(a) 2074

units (b) 1083

units

(c) 583

units (d) 1074

units

76. Let I x dx nnn= >ò tan ( )1 . If

I I a4 6+ = tan5 5x bx C+ + , where C

is a constant of integration, then theordered pair ( , )a b is equal to

(a) -æèç

öø÷

15

1, (b) 15

0,æèç

öø÷

(c) 15

1, -æèç

öø÷

(d) -æèç

öø÷

15

0,

77. The eccentricity of an ellipse whosecentre is at the origin is 1/2. If one ofits directrices is x = - 4, then the

equation of the normal to it at 13

2,

æèç

öø÷

is

(a) 2 2y x- = (b) 4 2 1x y- =

(c) 4 2 7x y+ = (d) x y+ =2 4

78. If a hyperbola passes through thepoint P( , )2 3 and has foci at

( , )± 2 0 , then the tangent to thishyperbola at P also passes throughthe point

(a) ( , )3 2 2 3 (b) ( , )2 2 3 3

(c) ( , )3 2 (d) ( , )- -2 3

79. The function f R: ,® -é

ëêù

ûú1

2

1

2

defined as f xx

x( ) =

+1 2is

(a) invertible(b) injective but not surjective(c) surjective but not injective(d) neither injective nor surjective

80. limcot cos

( )/x

x x

x®

--p p2 32

equals

(a) 124

(b) 116

(c) 18

(d) 14

81. Let a = + -2 2$ $ $i j k, b i j= +$ $ and c be

a vector such that | |c a- = 3,|( ) |a b c´ ´ = 3 and the anglebetween c and a ´ b is 30°. Then,a c× is equal to

(a) 258

(b) 2 (c) 5 (d) 18

82. The normal to the curvey x x x( ) ( )- - = +2 3 6 at the point,where the curve intersects theY -axispasses through the point

(a) - -æèç

öø÷

12

12

, (b) 12

12

,æèç

öø÷

(c) 12

13

, -æèç

öø÷

(d) 12

13

,æèç

öø÷

83. If two different numbers are takenfrom the set 0, 1, 2, 3, …, 10, thenthe probability that their sum as wellas absolute difference are both

multiple of 4, is

(a) 655

(b) 1255

(c) 1445

(d) 755

84. A man X has 7 friends, 4 of them areladies and 3 are men. His wife Y alsohas 7 friends, 3 of them are ladiesand 4 are men. Assume X andY haveno common friends. Then, the totalnumber of ways in which X and Ytogether can throw a party inviting 3ladies and 3 men, so that 3 friends ofeach of X and Y are in this party, is

(a) 485 (b) 468 (c) 469 (d) 484

85. The value of( ) ( )21

110

121

210

2C C C C- + -+ - + - + +( ) ( ) ...21

310

321

410

4C C C C

( )2110

1010C C- is

(a) 2 221 11- (b) 2 221 10-(c) 2 220 9- (d) 2 220 10-

86. A box contains 15 green and 10yellow balls. If 10 balls are randomlydrawn one-by-one withreplacement, then the variance ofthe number of green balls drawn is

(a) 125

(b) 6 (c) 4 (d) 625

87. Let a b c R, , Î . If f x ax bx c( ) = + +2

be such that a b c+ + = 3 andf x y f x f y xy( ) ( ) ( ) ,+ = + +" Îx y R, , then

f nn

( )=å

1

10

is equal to

(a) 330 (b) 165 (c) 190 (d) 255

88. The radius of a circle havingminimum area, which touches the

curve y x= -4 2 and the lines

y x=| |, is

(a) 2 2 1( )+ (b) 2 2 1( )-(c) 4 2 1( )- (d) 4 2 1( )+

89. For a positive integer n, if thequadratic equation,x x x x( ) ( ) ( ) ...+ + + + +1 1 2

+ + - + =( ) ( )x n x n n1 10

has two consecutive integralsolutions, then n is equal to

(a) 12 (b) 9

(c) 10 (d) 11

90.p

p

/

/

cos4

3 4

1ò +dx

xis equal to

(a) - 2 (b) 2(c) 4 (d) - 1

Physics1. (b) As the observer is moving towards

the source, so frequency of wavesemitted by the source will be given by theformula

f fv c

v cobserved actual= × +

-

æ

èçç

ö

ø÷÷

1

1

1 2/

/

/

Here, frequency v

c= 1

2

So, f fobserved actual= æèç

öø÷

3 2

1 2

1 2/

/

/

\ fobserved = ´ =10 3 17 3. GHz

2. (a) By ascent formula, we have surfacetension,

Trhg= ´2

103 N

m

= ´dhg

4103 N

mQ r

d=æèç

öø÷2

Þ D D DT

T

d

d

h

h= + [given, g is constant]

So, percentage

= ´ = +æèç

öø÷

´D D DT

T

d

d

h

h100 100

= ´´

+ ´´

æ

èçç

ö

ø÷÷ ´

-

-

-

-0 01 10

1 25 10

0 01 10

145 10100

2

2

2

2

.

.

.

.= 15. %

\ DT

T´ =100 15. %

3. (c) We have, l =hc

EDSo, ratio of wave lengths

ll

1

2

= =hc E

hc E

E

E

/

/

DD

DD

1

2

2

1

=-æ

èçöø÷

-=

4

32

E E

E E

1

3

4. (b) Given, force, F k v= - 2

\ Acceleration, ak

mv= - 2

or dv

dt

k

mv= - 2 Þ dv

v

k

mdt

2= - .

Now, with limits, we havedv

v

k

mdt

tv

2 010= - òò

Þ -æèç

öø÷

= -1

10v

k

mt

v

Þ 1 0 1v

kt

m= +.

Þ vkt

m

k=

+=

+1

0 1

1

0 1 1000. .

Þ 1

2

1

82

02´ ´ =m v m v

Þ vv= =0

25

Þ 1

0 1 10005

. +=

kÞ 1 0 5 5000= +. k

Þ k = 0 5

5000

. Þ k = -10 4 kg/m

Answer with Explanation

5. (b) By Mayor’s relation, for 1 g mole of agas,

C C Rp V- =So, when n gram moles are given,

C CR

np V- =

As per given question,

a C CR

p V= - =2

; for H2 K (i)

b C Cp V= - = R

28; for N2 K (ii)

From Eqs. (i) and (ii), we geta b= 14

6. (d) MI of a solid cylinder about itsperpendicular bisector of length is

I ml R= +

æ

èçç

ö

ø÷÷

2 2

12 4

Þ ImR ml m

l

ml= + = +2 2 2 2

4 12 4 12pr

[ ]QrpR l m2 =

For I to be maximum,

dI

dl

m

l

ml= - æèç

öø÷

+ =2

24

1

60

pr

Þ m ml2 3

4 6pr=

Now, putting m R l= rp 2

\ l R l3 23

2= ×

prrp

l

R

2

2

3

2=

\ l

R= 3

2

7. (a) Decay scheme is ,

Given, N

N

B

A

= =0 3 3

10.

Þ N

N

B

A

= 30

100

So, N0 = + =100 30 130 atoms

By using N N et= -

0l

We have, 100 130= -e

tl

Þ 1

13.= -

etl Þ log .13 = lt

If T is half-life, then l = loge

T

2

Þ log . log13 2= ×e

Tt

\ tT

e

= × log ( . )

log

13

2

8. (a) In a balanced Wheatstone bridge,there is no effect on position of null point,if we exchange the battery andgalvanometer. So, option (a) is incorrect.

9. (c) As each capacitors cannot withstandmore than 300 V, so there should be fourcapacitors in each row became in thiscondition 1 kV i.e. 1000 V will be dividedby 4 (i.e. 250 not more than 300 V).

Now, equivalent capacitance of one row

= ´ =1

41 0 25m mF . F

[Q in series combination, Cc

neq = ù

ûú

Now, we need equivalent of 2mF, so letwe need n such rows

\ n ´ =0 25 2. mF

[Qin parallel combinationC nceq = ]

n = 2

0 25.

= 8\ Total number of capacitors

= number of rows´ number of capacitors in each row

= ´ =8 4 32

10. (b) In steady state, no current flowsthrough the capacitor. So, resistance rp

becomes ineffective.So, the current in circuit,

IE

r r=

+ 2 (Total Resistance)

QPotential drop across capacitor

= Potential dropacross r2 = =+

IrEr

r r2

2

2

\ Stored charge of capacitor, Q CV=

=+

CEr

r r

2

2

A

N

t=o

at 0

A, B

N N

Ao –

atoms of

N

B

atomsof

Let atoms decaysinto in timeN

B t

11. (c) Each resistance is converted withtwo cells combined in opposite direction,so potential drop across each resistor iszero. Hence the current through each ofresistor is zero.

12. (d) Frequency spectrum of modulatedwave is

Clearly, wm is not included in thespectrum.

13. (c)

In a CE npn transistor amplifier output is180° out of phase with input.

14. (a) Heat gained (water + calorimeter)= Heat lost by copper ball

Þ m s T m s T m s Tw w c c B BD D D+ =Þ 170 1 75 30 100 0 1´ ´ - + ´( ) .

´ -( )75 30= ´ ´ -100 0 1 75. ( )T

\ T = °885 C

15. (a) Let n1th fringe formed due to firstwavelength and n2th fringe formed dueto second wavelength coincide i.e. theirdistance from common central maximawill be samei.e. y yn n1 2

=

Þ n D

d

n D

d

1 1 2 2l l=

Þ n

n

1

2

1

2

520

650

4

5= = =l

l

Hence, distance of the point ofcoincidence from the central maxima is

yn D

d= 1 1l

= n D

d

2 2l

= ´ ´ ´´

-

-4 650 10 15

0 5 10

9

3

.

.= 7 8. mm

16. (b)

Torque applied on a dipole t = pE sin qwhere q = angle between axis of dipoleand electric field.For electric field E E1 = $i

it means field is directed along positive X

direction, so angle between dipole andfield will remain q, therefore torque inthis direction

E pE1 1= sin qIn electric field E E2 3= $j, it means fieldis directed along positive Y-axis, so anglebetween dipole and field will be 90 - q.Torque in this directionT pE2 90= -sin ( q)

= p E3 1 cos qAccording to questiont t t t2 1 2 1= - Þ =| | | |\ pE p E1 13sin cosq q=

tan q = 3Þ tan tanq = °60\ q = °60

17. (d) As the rod rotates in vertical plane soa torque is acting on it, which is due tothe vertical component of weight of rod.

Now, Torque t = force × perpendiculardistance of line of action of force fromaxis of rotation

= ´mglsin q2

Again, Torque t = Ia

Where, I = moment of inertia = ml2

3[Force and Torque frequency along axis

of rotation passing through in end]

ωm

USBLSB

ω ωc m– ωc ω ωc m+

Bandwidth

vi

t

(Input)

vo

t

(Output)

90 – qp

X

Y

q

mg sin q

mgqq

mg

cos q

l/2

l/2

Initial condition At any time t

a = angular acceleration

\ mgl mlsin q a´ =2 3

2

\ a = 3

2

g

l

sin q

18. (d) KP

V V=

-( / )D

Þ - =DV

V

P

K

Þ - =DVPV

K

Change in volume D = DV v TgWhere r = coefficient of volumeexpansionAgain, g a= 3a is coefficient of linear expansion

\ D = DV V T( )3a

\ PV

KV T= D( )3a

\ D =TP

K3a

19. (d) Focal length of diverging lens is25 cm.As the rays are coming parallel, so theimage ( )I1 will be formed at the focus ofdiverging lens i.e. at 25 cm towards left ofdiverging lens.

Now, the image (I1) will work as object forconverging lens.For convergying lens, distance of object u

(i.e. distance of I1) = - +( )25 15= - 40 cm

f = 20 cm

\ From len's formula 1 1 1

f v u= -

1

20

1 1

40= -

-vÞ 1 1

20

1

40v= -

1 1

90v= Þ v = 40 cm

v is positive so image will be real and willform at right side of converging lens at40 cm.

20. (d) lmin =hc

eV

log ( ) log logminl = æèç

öø÷

-hc

eV

y c mx= -So, the required graph is given in option(d).

21. (c) From pV nRTN

NRT

A

= =

We have, n npVN

RT

pVN

RTf i

A

f

A

i

- = -

Þ n nf i- = ´ ´ ´10 30

836 02 10

523

..

. 1

300

1

290-æ

èçöø÷

= - ´25 1025.

\ Dn = - ´2 5 1025.

22. (b) Induced constant, Ie

R=

Here, e = induced emf

= d

dt

f

Ie

R

d

dt R= = æ

èçöø÷

×f 1

d IRdtf =f = IRdtò

\ Here, R is constant\ f = òR Idt

I dt× =ò Area under I t- graph

= ´ ´ =1

210 0 5 25. .

\ f = ´R 25.= ´100 25.= 250 Wb.

23. (d) Suppose a resistance Rs is connectedin series with galvanometer to convert itinto voltmeter.

I G R Vg s( )+ =

Þ RV

IG

g

= -

Þ R = =1985 1985. kWor R = ´1985 103. W

15 cm

25 cm40 cm

I1

G

Rs

V

Ig

24. (d) From Newton’s second law,DD

p

tF=

Þ D Dp F t=

\ p dp F dt= =ò ò01

Þ p t dt= = æèç

öø÷ò 6 3

0

1kg m

s

Also, change in kinetic energy

D Dk

p

m=

2

2=

´3

2 1

2

= 4 5.From work-energy theorem, work done

= change in kinetic energy.So, work done = =Dk 4 5. J

25. (d) Time period of oscillation is

TI

MB= 2p

Þ T = ´´ ´

-

-2

7 5 10

6 7 10 0 01

6

2p .

. .

= 0 665. sHence, time for 10 oscillations is t = 6 65.s.

26. (c) Inside the earth surface gGM

Rr= 3

i.e. g rµ

Out the earth surface gGm

r=

2

i.e. gr

µ 12

So, till earth surface ‘g’ increases linearlywith distance r, shown only in graph (c).

27. (b) Initially velocity keeps on decreasingat a constant rate, then it increases innegative direction with same rate.

28. (a) For elastic collision,p pbefore collision after collision= .

mv mvm

vA B= +2

2 2v v vA B= + ...(i)Now, coefficient of restitution,

ev v

u v

B A

A B

= --

Here, uB = 0 (Particle at rest) and forelastic collisione = 1

\ 1 = - Þ = -v v

vv v vB A

B A K (ii)

From Eq. (i) and Eq. (ii)

vv

A =3

and vv

B = 4

3

Hence, ll

A

B

A

B

B

A

h

mV

h

mV

V

V=

æ

èçç

ö

ø÷÷

= = =

2

2

4 3

2 32

.

/

/

29. (c) KE is maximum at mean position and

minimum at extreme position at tT

=æèç

öø÷4.

30. (d) Stress =Weight

AreaVolume will become ( )93 times.

So weight = volume × density ´ g willalso become ( )9 3 times.

Area of cross section will become ( )9 2

times.

= ´´

=æ

èçç

ö

ø÷÷

9

99

30

20

0

0

W

A

W

A

Hence, the stress increases by a factorof 9.

Chemistry31. (c) Aniline in presence of nitrating

mixture (conc. HNO +3 conc. H SO2 4 )gives significance amount ( %)» 47 ofmeta-product because in presence ofH SO2 4 its protonation takes place andanilinium ion is formed.

Here, anilinium ion is stronglydeactivating group and meta-directingin nature. So, it gives meta-nitrationproduct.

32. (c) According to first law ofthermodynamics,

D DU q W q p V= + = -Aniline

NH2

H SO2 4

Anilinium ion

NH HSO3 4

NH2 NH HSO3 4

NO2

Conc.H SO2 4

+Conc.HNO3

In isochoric process ( )DV = 0 ,DU q=

In isobaric process ( )Dp = 0 ,DU q=

In adiabatic process ( )q = 0DU W=

In isothermal process (DT = 0) and DU = 0\ DU is equal to adiabatic work.

33. (b) (i) The rate of S N1 reaction dependsonly upon the concentration of thealkyl halide.

(ii) S N1 reaction proceeds through theformation of carbocation.

The reactivity is decided by ease ofdissociation of alkyl halide.

R X R X¾ +Åq

È

Higher the stability of R+ (carbocation),

higher would be the reactivitytowards SN1 reaction.

p - H CO C H CH3 6 4 2¾ ¾ Å is the most

stable carbocation due to resonance and

then CH C HCH CH3 2 3Å

(2° carbocation)

while CH CH C H3 2 2

Å(1°) is least stable.

Thus, the correct increasing order of thereactivity of the given halides towards theSN1 reaction isCH CH CH Cl < CH C

|Cl

HCH CH3 2 2 3 2 3

(II)

(I)

< -H CO C H CH Cl3 6 4 2p ¾ ¾(III)

34. (d) Bohr radius ( )rn = Î02 2

n h

rn h

me kZn =

2 2

2 24p

k =Î

1

4 0p

\ rn h

me Zn = Î2 2

02p

= na

Z

2 0

where, m = mass of electron

e = charge of electronh = Planck’s constantk = Coulomb constant

rn

Zn = ´2 0 53. Å

Radius of nth Bohr orbit for H-atom

= 0 53 2. n Å

[Z = 1 for H-atom]\ Radius of 2nd Bohr orbit for H-atom

= ´ =0 53 2 2122. ( ) . Å

35. (b) For a salt of weak acid and weakbase,

pH 7 1

2pK 1

2p= + -a bK

Given, p aK A( ) . ,H = 32 pK Ba( .OH) = 34

\ pH 7 + 1

2(3.2) 1

2(3.4)= -

= + - =7 16 17 6 9. . .

36. (a) Nylon-6 or perlon is prepared bypolymerisation of amino caproic acid athigh temperature. Caprolactam is firsthydrolysed with water to form aminoacid which on heating undergoespolymerisation to give nylon-6.

37. (c) Given, abundance of elements bymassoxygen = 614. %, carbon = 22 9. %,hydrogen = 10% and nitrogen = 2 6. %Total weight of person = 75 kg

Mass due to 1 H = ´ =75 10

1007 5. kg

1 H atoms are replaced by 2 H atoms,

Mass due to 2 H (7.5 2)= ´ kg

\ Mass gain by person = 7 5. kg

NH

Caprolactam

HydrolysisH N—(CH ) —C—O3 2 5

∆ Polymerisation

—[ HN— (CH ) —C ]—2 5 nNylon-6

OO

O

+ –

38. (a) To show decolourisation, compound

must be unsaturated.

39. (d) Zinc oxide (ZnO) when react with Na O2

it act as acid while with CO2 it act as base.

Therefore, it is an amphoteric oxide.

ZnO + Na O Na ZnOAcid

2Base

2 2Salt

¾®

ZnO +CO ZnCOBase

2Acid

3Salt

¾®

40. (a) Mg can form basic carbonate while Li

cannot.

5 Mg 6 CO 7H O2 +

32 –

2+ + ¾®

4MgCO Mg(OH) 5H O3 2 2× × + 2 HCO3-

41. (d) The number of stereoisomers in

molecules which are not divisible into two

equal halves and have n number of

asymmetric C-atoms = 2n.

3-methyl-pent-2-ene on reaction with HBr

in presence of peroxide forms an addition

product i.e. 2-bromo-3-methyl pentane. It

has two chiral centres. Therefore, 4

stereoisomers are possible

42. (d) For fcc arrangement, 4 2r a=where, r = radius and a = edge length

\ Closest distance = = =22

2 2r

a a

43. (d) According to Arrhenius equation

k AeE RTa= - /

where, A = collision number or

pre-exponential factor.

R = gas constant

T = absolute temperature

Ea = energy of activation

For reaction R1, k AeE RTa

11= - /

…(i)

For reaction R2 , k AeE RTa

22= - /

…(ii)

On dividing Eq. (ii) by Eq. (i), we get

k

ke

E E

RT

a a

2

1

2 1

=-

-( )

…(iii)

[Q Pre-exponential factor ‘A’ is same for

both reactions]

Taking ln on both the sides of Eq. (iii),

we get

lnk

k

E E

RT

a a2

1

1 2æ

èç

ö

ø÷ =

-

Given, E Ea a1 210= + kJ mol

1-

= +Ea210 000, J mol

1-

\ln,k

k

2

1

10 000=´

-

- -J mol

8.314 J mol K 300 K

1

1 1= 4

44. (a)

O

Br

O

O—tBu

tert-BuONa– +

(Saturated)

(cannot decolourise Br water)2

Br

tert–BuONa– +

(Unsaturated)

(decolourise Br water)2

C H6 5 C H6 5

O

Br

Otert–BuONa

– +

(Unsaturated)

(decolourise Br water)2

Br

tert–BuONa– +

(Unsaturated)

(decolourise Br water)2

O O

HBr1

2

3

4

5

Peroxide

1

2

3

4

5

Br

3-methylpent-2-ene Anti-Markownikoff's

addition Four stereoisomers

are possible (As molecule

has two chiral centres

and asymmetric).

CHO

[Ag (NH ) ]OH3 2

Tollen’s reagent

COOH

Esterification

CH OH, H3+

COOCH3

CH MgBr3

HO CH3

OH

O

OO

(Excess)

Before final product is formed, intermediate is

45. (b) Colloidal solutions show Tyndall effect due to scattering of light by colloidal particles in

all directions in space. It is observed only under the following conditions.

(i) The diameter of the colloids should not be much smaller than the wavelength of light

used.

(ii) The refractive indices of the dispersed phase and dispersion medium should differ

greatly in magnitude.

46. (a) Sugars that have an aldehyde, a ketone, a hemiacetal or a hemiketal group is able to

reduce an oxidising agent. These sugars are classified as reducing sugars.

Hemiacetal can be easily reduced by oxidising agent such as Tollen’s reagent.

47. (c) Based on given DrH°

Df H H° = ° = -CO2

3935. kJ mol1-

…(i)

D =f H H° ° = -H 2O 2858. kJ mol1-

…(ii)

D =f H H° ° =O 20.00 (elements) …(iii)

Required thermal reaction is for Df H° of CH4

Thus, from III

890.3 = ° °[ ( )]D Df fH H O(CH ) + 24 2

- ° + °[ ( )D Df fH HCO (H O)]2 22

= ° - - - ´Df H (CH ) + 0] 393.5 2 285.5]4 [

\ Df H° -(CH ) = 74.8 J / mol4 k

OHOH C2 CH OH2

OCOCH3

HO

OH

aq·KOH

–CH COOK3

– +

OHOH C2 CH OH2

O–

OH

OHHemiacetal

O

HOH C2

CH OH2

OH( -hydroxyketone)α

HO

O

HOH C2

CH OH2

OH

Tollen’s

reagentpositive silver

mirror test.

O

OH

(Reducing sugar)

HO CH3

O CH3

48. (a) The reaction in which oxidation and

reduction occur simultaneously are

termed as redox reaction.

( X ) eF O ( F ) X eF O+ 4

4

1

2 2

6

6

0

2+ ¾® ++ +

Since, Xe undergoes oxidation while O

undergoes reduction. So, it is an example

of redox reaction.

49. (c)Cl2, Br2 and I2 form a mixture of halide

and hypohalites when react with cold

dilute alkalies while a mixture of halides

and haloate when react with concentrated

cold alkalies.

Cl 2NaOH NaCl NaClO H O2 2+ ¾® + +Cold and dilute

\Cl-

and ClO-

are obtained as products

when chlorine gas reacts with cold and

dilute aqueous NaOH.

50. (b) An alkyl halide in presence of a

bulkier base removes a proton from a

carbon adjacent to the carbon bonded to

the halogen. This reaction is called E2

( b-elimination reaction).

51. (d) The reaction takes place as follows

Na C O + H SO Na SO + H O2 2 4( )

2 4(conc. )

2 4 2X

¾®

+ CO + COEffervescence

2­ ­

Na C O + CaCl CaC O + 2NaCl2 2 4( )

2 2 4White ppt.X

¾®

5CaC O 2KMnO 8H SO2 4 4Purple

2 4+ +

¾® + +K SO 5CaSO 2MnSO2 4 4 4Colourless

+ +10CO 8H O2 2

Hence, X is Na C O2 2 4 .

52. (b) To identify the magnetic nature we

need to check the molecular orbital

configuration. If all orbitals are fully

occupied, species is diamagnetic while

when one or more molecular orbitals

is/are singly occupied, species is

paramagnetic.

(a) NO ( ) , , ,*

7 8 15 1 1 22 2 2+ = - s s ss s s

s p p p*, , ,2 2 2 2

2 2 2 2s p p px y z=

p p* *2 2

1 0p px y=

One unpaired electron is present.

Hence, it is paramagnetic.

(b) CO ( ) , , ,*

6 8 14 1 1 22 2 2+ = - s s ss s s

s p p , s*,2 2 2 2

2 2 2 2s p p px y z=

No unpaired electron is present.

Hence, it is diamagnetic.

(c) O2 ( ) , , ,*

8 8 16 1 1 22 2 2+ = - s s ss s s

s s p p*, , ,2 2 2 2

2 2 2 2s p p pz x x=

p p* *2 2

1 1p px y=

Two unpaired electrons are present.

Hence, it is paramagnetic.

(d) B2( ) , , ,*

5 5 1 1 22 2 2+ - s s ss s s

s p p*,2 2 2

2 1 1s p px y=

Two unpaired electrons are present.

Hence, it is paramagnetic.

53. (d) Let the degree of association of

acetic acid (CH COOH)3 in benzene is a,

then

2CH COOH (CH COOH)3 3 2q

Initial moles 1 0

Moles at equilibrium 1 - aa2

\ Total moles = - + = -12

12

a a a

or i = -12

a

Now, depression in freezing point ( )DTf is

given as

DT i K mf f= K (i)

where, Kf = molal depression constant or

cryoscopic constant.

m = molality

Molality = number of moles of solute

weight of solvent (in kg)

= ´0 2

60

1000

20

.

Putting the values in Eq. (i)

\ 0 45 12

5120 2

60

1000

20. ( . )

.= -éëê

ùûú

´éëê

ùûú

a

12

0 45 60 20

512 0 2 1000- = ´ ´

´ ´a .

. .

Þ 12

0 527- =a. Þ a

21 0 527= - .

\ a = 0 946.

Thus, percentage of association = 94 6. %

HBr

C H6 5C H6 5

C==CH

C H6 5H

t-BuO K– +

+ -BuOH + Brtert –

HH

C H6 5

54. (d) Aromatic compounds are stable

due to resonance while non-aromatics

are not. According to Huckel’s rule (or

4 2n + rule), “For a planar, cyclic

compound to be aromatic, its p cloud

must contain ( )4 2n + p electrons,

where, n is any whole number.” Thus,

55. (d) Molarity (M)

= Number of moles of solute

Volume of solution (in L)

\ Number of moles of complex

= ´Molarity volume (in mL)

1000

= ´0 1 100

1000

. = 0 01. mole

Number of moles of ions precipitate

= ´´

12 10

6 02 10

22

23

.

.= 0 02. moles

\ Number of Cl-

present in ionisation

sphere

= Number of moles of ions precipitated

Number of moles of complex

= =0 02

0 012

.

.

\2 Cl-

are present outside the square

brackets, i.e. in ionisation sphere. Thus,

the formula of complex is

[Co(H O) Cl]Cl H O2 5 22 × .

56. (a) DIBAL-H (Di-isobutyl aluminium

hydride) is a reducing agent with

formula. This is generally used for the

preparation of aldehydes. Using DIBAL

¾H, Lactones are reduced directly to

aldehydes.

57. (c) NO3-

The maximum limit of nitrate

(NO )3-

in drinking water is 50 ppm and its

source is fertilisers. If the maximum limit

is increased in water it will cause

methemoglobinemia

(blue baby syndrome.)

SO42 -

The maximum limit of sulphate

(SO )42 -

according to WHO is 500 pm and its

sources are acid rain, industries. Excess

SO42 -

has laxative effect.

F-

The maximum limit of fluoride (F-) is

about 1.5 ppm. Its higher concentration

converts enamel to more harder

fluorapatite. Concentration (>2ppm)

causes brown mottling of teeth and high

concentration (>10 ppm) are harmful for

bones and teeth.

\SO42 -

(100 ppm) and NO3-

(50 ppm) in

water is suitable for drinking but the

concentration of F-

(10 ppm) makes water

unsuitable for drinking.

58. (b)

M CO 2HCl 2MCl H O CO2 31 g

2 20.01186mole

+ ¾® + +

Number of moles of M CO2 3 reacted =Number of moles of CO2 evolved

10 01186

M= . [M = molar mass of M CO2 3 ]

M = =1

0 0118684 3

.. g mol

1-

59. (a) The substances which have lower

reduction potentials are stronger reducing

agents. Therefore, Cr (E 0.74 V)Cr /Cr

3 +° = -

is the strongest reducing agent among all

the other given options.

60. (a) Isoelectronic species are those which

contains same number of electrons.

SpeciesAtomic

number

Number of

electrons

O2 - 8 10

F- 9 10

Na+ 11 10

Mg2 + 12 10

O- 8 9

Na 11 11

Mg+ 12 11

\ Option (a) is correct which contains

isoelectronic species O2 -

, F-, Na

+, Mg

2 +.

O

is non-aromatic, hence,

least stabilised by resonance.

O

and,

N

(6 system)πe– (4 + 2 system)πe e– – (6 system)πe–

are aromatic and stabilised by resonance.

They follow Huckel’s rule.

O

DIBAL–H CHO

COOHCOOH

OH

O

Mathematics

61. (d)Q D = = - - - + -1 1 1

1 1

1

1 1 1 12

a

a b

a b a b a( ) ( ) ( )

= - -( )a 12

D =1

1 1 1

1 1

0 1

a

b

= - - +1 1 1 1( ) ( ) ( )a b b

= - -( )a 1

D =2

1 1 1

1 1 1

0 1a

= - - + - =1 1 1 1 1 0 0( ) ( ) ( )a a and D =3

1 1 1

1 1

0

a

a b

= - - - + -1 1 12

( ) ( ) ( )b a b a

= - -a a( )1

For a = 1

D = D = D = D =1 2 3 0

D for b = 1 only

x y z+ + = 1,

x y z+ + = 1

and x y z+ + = 0

i.e. no solution (QRHS is not equal)

Hence, for no solution b = 1 only

62. (a) The truth table of the given expression is given below :

p q x p qº ® ~ p ~p q® y (~ p q) qº ® ® x y®

T T T F T T T

T F F F T F T

F T T T T T T

F F T T F T T

Hence, it is a tautology.

63. (d) Given, 5 2 2 92 2

(tan cos ) cosx x x- = +

Þ 51 2

1 2

1 2

22 2 9

-+

- +æ

èç

ö

ø÷ = +cos

cos

coscos

x

x

xx

Put cos2x y= , we have

51

1

1

22 9

-+

- +æ

èç

ö

ø÷ = +y

y

yy

Þ 5 2 2 1 22

( )- - - -y y y = + +2 1 2 9( )( )y y

Þ 5 1 4 2 2 9 2 92 2

( ) ( )- - = + + +y y y y y

Þ 5 20 52- -y y = + +22 18 4

2y y

Þ 9 42 13 02

y y+ + =Þ 9 3 39 13 0

2y y y+ + + =

Þ 3 3 1 13 3 1 0y y y( ) ( )+ + + =Þ ( )( )3 1 3 13 0y y+ + =

Þ y = - -1

3

13

3,

\ cos ,21

3

13

3x = - -

Þ cos21

3x = - Qcos2

13

3x ¹ -é

ëêùûú

Now, cos cos4 2 2 12x x= -

= -æèç

öø÷

-21

31

2

= - = -2

91

7

9

64. (b) We have, P (exactly one of A or B

occurs)= È - ÇP A B P A B( ) ( )

= + - ÇP A P B P A B( ) ( ) ( )2

According to the question,

P A P B P A B( ) ( ) ( )+ - Ç =21

4…(i)

P B P C P B C( ) ( ) ( )+ - Ç =21

4…(ii)

and P C P A P C A( ) ( ) ( )+ - Ç =21

4…(iii)

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

2 [ ( ) ( ) ( ) ( )P A P B P C P A B+ + - Ç

- Ç - Ç =P B C P C A( ) ( )]3

4

Þ P A P B P C P A B( ) ( ) ( ) ( )+ + - Ç

- Ç - Ç =P B C P C A( ) ( )3

8

\ P (atleast one event occurs)

= È ÈP A B C( )

= + + - ÇP A P B P C P A B( ) ( ) ( ) ( )

- Ç - ÇP B C P C A( ) ( )

+ Ç ÇP A B C( )

= + =3

8

1

16

7

16

Q P A B C( )Ç Ç =éëê

ùûú

1

16

65. (a) Given, 2 1w + = z

Þ 2 1 3w + = -[Q z = - 3]

Þ w = - +1 3

2

i

Since, w is cube root of unity.

\ w2 1 3

2= - - i

and w3 1n =

Now,

1 1 1

1 1

1

32 2

2 7

- - =w ww w

k

Þ1 1 1

1

1

32

2

w ww w

= k

[Q1 02+ + =w w and w w w w7 3 2= × =( ) ]

On applying R R R R1 1 2 3® + + , we get

3 1 1

1

1

3

2 2

2

2

+ + + +=

w w w ww ww w

k

Þ =3 0 0

1

1

32

2

w ww w

k

Þ 3 32 4( )w w- = k

Þ ( )w w2 - = k

\ ki i= - -æ

èç

ö

ø÷ - - +æ

èç

ö

ø÷

1 3

2

1 3

2

= - = -3i z

66. (d) Given, vertices of triangle are( , ), ( , )k k k- 3 5 and ( , )- k 2 .

\ 1

2

3 1

5 1

2 1

28

k k

k

k

-

-= ±

Þk k

k

k

-

-= ±

3 1

5 1

2 1

56

Þ k k k k k( ) ( ) ( )- + + + +2 3 5 1 10 2

= ± 56Þ k k k k k

2 2 22 15 3 10 56- + + + + = ±Þ 5 13 10 562

k k+ + = ±Þ 5 13 66 02

k k+ - =or 5 13 46 02

k k+ - =Þ k = 2 [Qk IÎ ]Thus, the coordinates of vertices oftriangle are A( , )2 6- , B( , )5 2 and C( , )- 2 2 .

Y

C (–2, 2)B (5, 2)

D

(2, 1/2)

E

A (2, –6)

X′

Y′

XO

Now, equation of altitude from vertex A is

y x- - = --

- -æ

èç

ö

ø÷

-( ) ( )61

2 2

2 5

2 Þ x = 2 …(i)

Equation of altitude from vertex C is

y x- = -- -

-é

ëê

ù

ûú

- -21

2 6

5 2

2( )

[ ( )]

Þ 3 8 10 0x y+ - = …(ii)On solving Eqs. (i) and (ii), we get x = 2

and y = 1

2.

\ Orthocentre = æèç

öø÷

21

2,

67. (c) Total length = + =2 20r r q

Þ q = -20 2r

r

Now, area of flower-bed,

A r= 1

2

2q Þ A rr

r= -æ

èçöø÷

1

2

20 22

Þ A r r= -10 2

\ dA

drr= -10 2

For maxima or minima, putdA

dr= 0.

Þ 10 2 0- =r Þ r = 5

\ Amax ( )( )= -é

ëêùûú

1

25

20 2 5

5

2

= ´ ´1

225 2 = 25 sq m

68. (d) Required area

= + + -ò ò0

1

1

21 3( ) ( )x dx x dx

- ò02 2

4

xdx

= +é

ëê

ù

ûú + -

é

ëê

ù

ûú -

é

ëê

ù

ûúx

xx

x x3 2

0

1 2

1

2 3

0

2

3 23

2 12

/

/

= +æèç

öø÷

+ - - +æèç

öø÷

- æèç

öø÷

12

36 2 3

1

2

8

12

= + -5

3

3

2

2

3= +1

3

2= 5

2sq units

69. (b) Any line parallel tox y z

1 4 5= = and

passing through P( , , )1 2 3- is

x y z-=

+=

-=

1

1

2

4

3

5l (say)

Any point on above line can be written as

( , , )l l l+ - +1 4 2 5 3 .

\ Coordinates of R are

( , , )l l l+ - +1 4 2 5 3 .

Since, point R lies on the above plane.

\2 1 3 4 2 4 5 3( ) ( ) ( )l l l+ + - - + + =22 0

Þ l = 1

So, point R is (2, 2, 8).

Now, PR = - + + + -( ) ( ) ( )2 1 2 2 8 32 2 2

= 42

\ PQ PR= =2 2 42

70. (a) Let yx x

x=

-

æ

èç

ö

ø÷-tan 1

3

6

1 9

=×

-

é

ëê

ù

ûú-tan

( )

( )

/

/1

3 2

3 2 2

2 3

1 3

x

x

= -2 31 3 2tan ( )/x

Q22

1

1 1

2tan tan- -=

-

é

ëê

ù

ûúx

x

x

\ dy

dx xx= ×

+× ´2

1

1 33

3

23 2 2

1 2

( )( )

/

/

=+

×9

1 9 3x

x

\ g xx

( ) =+

9

1 9 3

(0, 3)

(0, 1)

(0, 0) (1, 0)(2, 0) (3, 0)

(2, 1)

x y+ =3

(1, 2)4y=x 2

y= + x1 √Y

Y′

XX′

θ

r r

r θ

2 + 3 – 4

+ 22 = 0

x y zR

P (1, –2, 3)

<1, 4, 5>x1

y4

z5

= =

Q

71. (a) We have,

( sin ) ( )cos2 1 0+ + + =xdy

dxy x

Þ dy

dx

x

xy

x

x+

+= -

+cos

sin

cos

sin2 2

which is a linear differential equation.

\ IF = +ò

e

x

xdx

cos

sin2 = +e

xlog ( sin )2

= +2 sin x

\ Required solution is given by

y xx

x× + = -

+ò( sin )cos

sin2

2

× + +( sin )2 x dx C

Þ y x x C( sin ) sin2 + = - +

Also, y( )0 1=

\ 1 2 0 0( sin ) sin+ = - + C

Þ C = 2

\ yx

x= -

+2

2

sin

sin

Þ yp

p

p2

22

22

1

3

æèç

öø÷

=-

+=

sin

sin

72. (c) Let AB h= , then AD h= 2 and

AC BCh

= =2

Again, let Ð =CPA a

Now, in DABP,

tan ( )a b+ = = =AB

AP

h

h2

1

2

Also, in DACP, tan a = = =AC

AP

h

h

22

1

4

Now, tan tan[( ) ]b a b a= + -

= + -+ +

tan( ) tan

tan( )tan

a b aa b a1

=-

+ ´

1

2

1

4

11

2

1

4

=

1

49

8

= 2

9

73. (b) We have, A =-

-é

ëê

ù

ûú

2 3

4 1

\ A A A2 = × =

--

é

ëêù

ûú-

-é

ëêù

ûú2 3

4 1

2 3

4 1

=+ - -

- - +é

ëêù

ûú4 12 6 3

8 4 12 1

=-

-é

ëêù

ûú16 9

12 13

Now,

3 12 316 9

12 1312

2 3

4 12

A A+ =-

-é

ëêù

ûú+

--

é

ëêù

ûú

=-

-é

ëêù

ûú+

--

é

ëêù

ûú48 27

36 39

24 36

48 12

=-

-é

ëêù

ûú72 63

84 51

\ adj ( )3 1251 63

84 722

A A+ = é

ëêù

ûú

74. (c) We have,

225 9 25 75 452 2 2a b c ac ab+ + - -

- =15 0bc

Þ ( ) ( ) ( ) ( )( )15 3 5 15 52 2 2a b c a c+ + -

- - =( )( ) ( )( )15 3 3 5 0a b b c

Þ 1

215 3 3 52 2[( ) ( )a b b c- + -

+ - =( ) ]5 15 02c a

Þ 15 3a b= , 3 5b c= and 5 15c a=

\ 15 3 5a b c= =

Þ a b c

1 5 3= = = l (say)

Þ a b c= = =l l l, ,5 3

Hence, a b, and c are in AP.

75. (b) Given, equations of lines arex y z- = +

-= -1

1

2

2

4

3

andx y z- = +

-= +

-2

2

1

1

7

1

Let n i j k1 2 3= - +$ $ $ and n i j k2 2= - -$ $ $

\ Any vector n perpendicular to both

n n1 2, is given by

n n n= ´1 2

24 JEE Main Solved Paper 2017

A P

B

C

h/2

h/2

h

2h

αβ

Þ n

i j k

= -- -

$ $ $

1 2 3

2 1 1

= + +5 7 3$ $ $i j k

\ Equation of a plane passing through( , , )1 1 1- - and perpendicular to n is givenby

5 1 7 1 3 1 0( ) ( ) ( )x y z- + + + + =

Þ 5 7 3 5 0x y z+ + + =

\ Required distance = + - +

+ +

5 21 21 5

5 7 32 2 2

= 10

83units

76. (b) We have, I x dxnn= ò tan

\ I I x dxn nn+ =+ ò2 tan

+ +ò tannx dx

2

= +ò tan ( tan )nx x dx1 2

= ò tan secnx x dx

2

=+

++tann

x

nC

1

1

Put n = 4, we get I Ix

C4 6

5

5+ = +tan

\ a = 1

5and b = 0

77. (b) We have, e =1

2and

a

e= 4

\ a = 2

Now, b a e2 2 21= -( ) = - æ

èçöø÷

é

ëê

ù

ûú( )2 1

1

2

22

= -æèç

öø÷

4 11

4= 3 Þ b = 3

\ Equation of the ellipse isx y

2

2

2

22 31

( ) ( )+ =

Þ x y2 2

4 31+ =

Now, the equation of normal at 13

2,æ

èçöø÷

is

a x

x

b y

ya b

2

1

2

1

2 2- = -

Þ 4

1

3

3 24 3

x y- = -( / )

Þ 4 2 1x y- =

78. (b) Let the equation of hyperbola bex

a

y

b

2

2

2

21- = .

\ ae = 2 Þ a e2 2 4=

Þ a b2 2 4+ =

Þ b a2 24= -

\ x

a

y

a

2

2

2

241-

-=

Since, ( , )2 3 lie on hyperbola.

\ 2 3

41

2 2a a

--

=

Þ 8 2 3 42 2 2 2- - = -a a a a( )

Þ 8 5 42 2 4- = -a a a

Þ a a4 29 8 0- + =

Þ ( )( )a a4 48 1 0- - =

Þ a4 8= , a

4 1=

\ a = 1

Now, equation of hyperbola isx y

2 2

1 31- = .

\ Equation of tangent at ( , )2 3 is givenby

23

31x

y- =

Þ 23

1xy- =

which passes through the point( , )2 2 3 3 .

79. (c) We have, f xx

x( ) =

+1 2

\ fx

x

x

x

x

11

11 12

2

æèç

öø÷

=+

=+

= f x( )

\ f f1

22æ

èçöø÷

= ( ) or f f1

33æ

èçöø÷

= ( ) and so on.

So, f x( ) is many-one function.Again, let y f x= ( )

Þ yx

x=

+1 2

Þ y x y x+ =2

Þ yx x y2 0- + =

As, x RÎ\ ( ) ( )( )- - ³1 4 02

y y

Þ 1 4 02- ³y

Þ y Î -éëê

ùûú

1

2

1

2,

\ Range = Codomain = -éëê

ùûú

1

2

1

2,

So, f x( ) is surjective.

Hence, f x( ) is surjective but not injective.

80. (b) limcot cos

( )/x

x x

x®

--p p2 32

= × -

-æèç

öø÷

®lim

cos ( sin )

sin/x

x x

x xp p2 3

1

8

1

2

= ×-æ

èçöø÷

- -æèç

öø÷

éëê

ùûú

-®

lim

cos sin

sinh

h h

0

1

8

21

2

2

p p

ph h

æèç

öø÷

- +æèç

öø÷

p p2 2

3

= -×®

1

8

1

0 3lim

sin ( cos )

cosh

h h

h h

=

æèç

öø÷

×®

1

8

22

0

2

3lim

sin sin

cosh

hh

h h

=× æ

èçöø÷

®

1

4

20

2

3lim

sin sin

cosh

hh

h h

= æèç

öø÷

æ

è

ççç

ö

ø

÷÷÷

× ×®

1

42

2

1 1

40

2

limsin

sin

cosh

h

h

h

h h

= ´1

4

1

4= 1

16

81. (b) We have, a i j k= + -2 2$ $ $

Þ | |a = + + =4 1 4 3

and b i j= +$ $ Þ| |b = + =1 1 2

Now, | |c a- = 3 Þ| |c a- =2 9

Þ ( ) ( )c a c a- × - = 9

Þ | | | |c a c a2 2 2 9+ - × = …(i)

Again, |( ) |a b c´ ´ = 3

Þ | || |sina b c´ ° =30 3 Þ| || |

ca b

=´6

But a b

i j k

i j k´ = - = - +

$ $ $

$ $ $2 1 2

1 1 0

2 2

\ | |c =+ +

=6

4 4 12 …(ii)

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

( ) ( )2 3 2 92 2+ - × =c a

Þ 4 9 2 9+ - × =c a

Þ c a× = 2

82. (b) Given curve is

y x x x( )( )- - = +2 3 6 …(i)Put x = 0 in Eq. (i), we get

y( ) ( )- - =2 3 6 Þ y = 1

So, point of intersection is (0, 1).

Now, yx

x x= +

- -6

2 3( )( )

Þ dy

dx=

1 2 3 63 2

2 32 2

( )( ) ( )( )

( ) ( )

x x x

x x

x x

- - - +- + -

- -

Þ dy

dx

æèç

öø÷

= +´( , )0 1

6 30

4 9= =36

361

\ Equation of normal at (0, 1) is given by

y x- = - -11

10( )

Þ x y+ - =1 0

which passes through the point1

2

1

2,æ

èçöø÷.

83. (a) Total number of ways of selecting 2different numbers from 0, 1, 2, ..., 10= =11

2 55C

Let two numbers selected be x and y.

Then, x y m+ = 4 …(i)

and x y n- = 4 …(ii)

Þ 2 4x m n= +( ) and 2 4y m n= -( )

Þ x m n= +2( )

and y m n= -2( )

\ x and y both are even numbers.

x y

0 4, 8

2 6, 10

4 0, 8

6 2, 10

8 0, 4

10 2, 6

\ Required probability = 6

55

84. (a) Given, X has 7 friends, 4 of them areladies and 3 are men while Y has 7friends, 3 of them are ladies and 4 aremen.\ Total number of required ways

= ´ ´ ´ + ´33

40

40

33

32

41C C C C C C

´ ´ + ´ ´ ´41

32

31

42

42

31C C C C C C

+ ´ ´ ´30

43

43

30C C C C

= + + +1 144 324 16 = 485

26 JEE Main Solved Paper 2017

85. (d) ( ) ( )211

101

212

102C C C C- + -

+ - + + -( ) ... ( )213

103

2110

1010C C C C

= + + + -( ... ) (211

212

2110

101C C C C

+ + +102

1010C C... )

= + + +1

2

211

212

2120( ... )C C C - -( )2 110

= + + + -1

2121

121

221

21( ... )C C C

- -( )2 110

= -1

22 221( ) = - - +2 1 2 120 10

- -( )2 110

= -2 220 10

86. (a) Given box contains 15 green and 10yellow balls.

\ Total number of balls = + =15 10 25

P(green balls) = = = =15

25

3

5p Probability

of success

P(yellow balls) = = = =10

25

2

5q Probability

of unsuccess and n = =10 Number oftrials.

\ Variance = npq = ´ ´103

5

2

5= 12

5

87. (a) We have, f x ax bx c( ) = + +2

Now, f x y f x f y xy( ) ( ) ( )+ = + +Put y = 0 Þ f x f x f( ) ( ) ( )= + +0 0

Þ f( )0 0= Þ c = 0

Again, put y x= -\ f f x f x x( ) ( ) ( )0 2= + - -

Þ 0 2 2 2= + + - -ax bx ax bx x

Þ 2 02 2ax x- =

Þ a = 1

2

Also, a b c+ + = 3

Þ 1

20 3+ + =b

Þ b = 5

2

\ f xx x

( ) = +2 5

2

Now, f nn n

n n( ) = + = +2

25

2

1

2

5

2

\ f n n n

n n n

( )= = =

å å å= +1

102

1

10

1

101

2

5

2

= × ´ ´ + ´ ´1

2

10 11 21

6

5

2

10 11

2

= + = =385

2

275

2

660

2330

88. (c) Let the radius of circle with leastarea be r .Then, then coordinates of centre= -( , )0 4 r .

Since, circle touches the line y x= in firstquadrant.

\ 0 4

2

- - =( )rr

Þ r r- = ±4 2

Þ r =+

4

2 1or

4

1 2-

But r ¹-4

1 2Q

4

1 20

-<é

ëêùûú

\ r =+

= -4

2 14 2 1( )

89. (d) Given quadratic equation isx x x x x n( ) ( )( ) ... ( )+ + + + + + + -1 1 2 1

( )x n n+ = 10

Þ ( ... ) [( ...x x x2 2 2 1 3 5+ + + + + + +

+ -( )]2 1n x

+ × + × + + - =[( ... ( ) ]1 2 2 3 1 10n n n

Þ nx n xn n

n2 2

2 1

310 0+ + - - =( )

Þ x nxn2

2 1

310 0+ + - - =

Þ 3 3 31 02 2x nx n+ + - =

Let a and b be the roots.

Since, a and b are consecutive.

\ | |a b- = 1

Þ ( )a b- =2 1

rr

(0, 4)

–r

y=|x|

y= x4 – 2

X′

Y′

XO

Y

Again, ( ) ( )a b a b ab- = + -2 2 4

Þ 13

34

31

3

2 2

= -æèç

öø÷

- -æ

èç

ö

ø÷

n n

Þ 14

3312 2= - -n n( )

Þ 3 3 4 1242 2= - +n n

Þ n2 121=

Þ n = ± 11

\ n = 11 [Qn > 0]

90. (b) Let Idx

x=

+òp

p

/

/

cos4

3 4

1

= --ò

1

1 24

3 4 cos

cos/

/ x

xdx

p

p

= -ò

124

3 4 cos

sin/

/ x

xdx

p

p

= -òp

p

/

/

4

3 4(cosec cosec2

x x

cot )x dx

= - +[ cot ] //

x xcosec pp

43 4

= + - - +[( ) ( )]1 2 1 2 = 2

28 JEE Main Solved Paper 2017