2019 Jee-Main Q P K S...2019/10/01 · 2019_Jee-Main Question Paper_Key & Solutions Sri Chaitanya...

2019_Jee-Main Question Paper_Key & Solutions


Sri Chaitanya IIT Academy # 304, Kasetty Hegihts, Ayappa Society, Madhapur, Hyderabad – 500081

www. srichaitanya.net, [email protected]

PHYSICS 1. The Wheatstone bridge shown in figure here, gets balanced when the carbon resistor used

as R1 has the colour code (Orange, Red, Brown). The resistor R2 and R4 are 80 and

40 , respectively. Assuming that the colour code for the carbon resistors gives their

accurate values, the colour code for the carbon resistor, used as R3, would be

1) Brown, Blue, Brown 2) Brown, Blue, Brown

3) Red, Green, Brown 4) Grey, Black, Brown

Key: 1

Sol: 1 23 4

R RR R

13 2RR

1 320R

2. Consider the nuclear fission 20 4 122Ne He C .

Given that the binding energy / nucleon of Ne20, He4 and C12 are, respectively, 8.03 MeV,

7.07 MeV and 7.86 MeV, identify the correct statement:

1) Energy of 12.4 MeV will be supplied

2) 8.3 MeV energy will be released

3) Energy of 3.6 MeV will be released

4) Energy of 11.9 MeV has to be supplied

Key: 4

Sol: 20 20 8.03Ne Mev 42 8 7.07He Mev

12 12 7.86C Mev

Q value = BEProducts - BEReactants




3. A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic

material with their magnetic moment parallel to their respective axes. But the magnetic

moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in

such a manner that their magnetic moments make a small angle with the field. If the

oscillation periods of hoop and cylinder are Th and TC respectively, then

1) Th = TC 2) Th = 2TC 3) Th = 1.5 TC 4) Th = 0.5 TC

Key: 1

Sol: Time period ITB

4. An unknown metal of mass 192 g heated to a temperature of 100°C was immersed into a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4°C. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5°C. (Specific heat of brass is 394 J kg-1 K-1)

1) 458 J kg-1 K-1 2) 1232 J kg-1 K-1 3) 916 J kg-1 K-1 4) 654 J kg-1 K-1 Key: 3

Sol: 1 2 2u u w w b bm s t m s t m s t = u u w w b bm s t m s t m s t

5. Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod

of length 2R and mass M (see figure). The moment of inertia of the system about the axis

passing perpendicularly through the centre of the rod is

1) 2137 MR5

2) 217 MR15

3) 2209 MR15

4) 2152 MR15

Key: 1

Sol: 2 2

22 22 212 5

M R MRI M R

6. The self induced emf of a coil is 25 volts. When the current in it is changed at uniform

rate from 10A to 25A in 1s, the change in the energy of the inductance is

1) 740 J 2) 437.5 J 3) 540 J 4) 637.5 J

Key: 2




Sol: 25diLdt

53

L H , Energy charge 2 21 25 102

E L

7. The actual value of resistance R, shown in the figure is 30 . This is measured in an

experiment as shown using the standard formula VR =I

, where V and I are the readings

of the voltmeter and ammeter, respectively. If the measured value of R is 5% less, then then internal resistance of the voltmeter is

1) 600 2) 570 3) 35 4) 350

Key: 2

Sol: 00

0.95 RRRR R

8. At some location on earth the horizontal component of earth’s magnetic field is 18 × 10-6

T. At this location, magnetic needle of length 0.12 m and pole strength 1.8 Am is

suspended from its mid-point using a thread, it makes 45° angle with horizontal in

equilibrium. To keep this needle horizontal, the vertical force that should be applied at

one of its ends is

1) 3.6 × 10-5 N 2) 1.8 × 10-5 N 3) 1.3 × 10-5 N 4) 6.5 × 10-5 N Key: 4

Sol: sin 45 sin 452lMB F

56.5 10F N 9. Two vectors A

and B

have equal magnitudes. The magnitude of A B

is ‘n’ times the

magnitude of A B

. The angle between A

and B

is

1) 2

12

1cos1

nn

2) 1 1cos1

nn

3) 2

12

1sin1

nn

4) 1 1sin1

nn

Key: 1

Sol: 2 2

2A B n A B

2 222 21 cos 1 cosa n a




10. A metal plate of area 1 × 10-4 m2 is illuminated by a radiation of intensity 16 mW/m2. The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be : [1 eV = 1.6 × 10-19 J]

1) 1014 and 10 eV 2) 1012 and 5 eV 3) 1011 and 5 eV 4) 1010 and 5 eV Key: 3

Sol: ,nE NI nA t

0KE hv hv

11. A particle which is experiencing a force, given by 3 12F i j

, undergoes a displacement

of 4d i

. If the particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement ?

1) 9 J 2) 12 J 3) 10 J 4) 15 J Key: 4

Sol: Work = 12 J, Net work done = KS , Final kinetic energy = 15 J 12. Consider a Young’s double slit experiment as shown in figure. What should be the slit

separation d in terms of wavelength such that the first minima occurs directly in front

of the slit (S1)?

1) 2 5 2

2)

5 2

3)

2 5 2

4)

5 2

Key: 1

Sol: 2 1 / 2S P S P , 5 2 / 2d d

13. The eye can be regarded as a single refracting surface. The radius of curvature of this

surface is equal to that of cornea (7.8 mm). This surface separates two media of refractive

indices 1 and 1.34. Calculate the distance from the refracting, surface at which a parallel

beam of light will come to focus.

1) 1 cm 2) 2 cm 3) 4.0 cm 4) 3.1 cm

Key: 4




Sol: f f iiv u R

,

1.34 1.34 17.8f

f =30.7 mm

14. A current of 2mA was passed through an unknown resistor which dissipated a power of

4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is

1) 11 × 10-5 W 2) 11 × 10-3 W 3) 11 × 10-4 W 4) 11 × 105 W

Key: 1

Sol: Power = 2i R R = 61.1 10

New power dissipation 2

511 10v WR

15. The diameter and height of a cylinder are measured by a meter scale to be 12.6 ± 0.1 cm

and 34.2 ± 0.1 cm, respectively. What will be the value of its volume in appropriate

significant figures ?

1) 4264 ± 81 cm3 2) 4264.4 ± 81.0 cm33) 4260 ± 80 cm3 4) 4300 ± 80 cm3

Key: 3

Sol: 2

4d hV V=4260 (after rounded off to 3 significant figures)

2V d hV d h

V =80

16. Four equal point charges Q each are placed in xy plane at (0, 2), (4, 2), (4, – 2) and (0, –

2). The work required to put a fifth charge Q at the origin of the coordinate system will

be

1) 2 11

4 3Q

2)

2 114 5Q

3)

2

02 2Q

4) 2

04Q

Key: 2

Sol: Potential at origin 0

14 2 2 20 20

Q Q Q Q

Work required = final initialV V q 0

11 04 5

Q Q




17. The modulation frequency of an AM radio station is 250 kHz, which is 10 % of the

carrier wave. If another AM station approaches you for license what broadcast frequency

will you allot ?

1) 2750 kHz 2) 2900 kHz 3) 2250 kHz 4) 2000 kHz

Key: 4

Sol: 2500cf

250mf

Side bonds are 2250, 2750

18. A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones

that can be distinctly heard by a person with this organ pipe will be (Assume that the

highest frequency a person can hear is 20,000 Hz)

1) 6 2) 4 3) 7 4) 5

Key: 3

Sol: 02 1 20000n f 0 1.5f kHZ n=7 19. For the circuit shown n below, the current through the zener diode is

1) 9 mA 2) 5 mA 3) Zero 4) 14 mA

Key: 1

Sol: Potential drop across series resistance is 70V

Current through load = 5010

mA

=5mA

Current through series resistance = 705

mA

= 14mA




20. The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is

given by an expression ( , ) 10 cos (6 8 )E x y j x z . The magnetic field ( , , )B x z t

is given

by :

(c is the velocity of light)

1) 1 6 8 cos 6 8 10k i x z ctc 2) 1 6 8 cos 6 8 10k i x z ctc

3) 1 6 8 cos 6 8 10k i x z ctc 4) 1 6 8 cos 6 8 10k i x z ctc

Key: 2

Sol: 0

1S E B

0 0E CB

21. Charges – q and + q located at A and B, respectively, constitute an electric dipole.

Distance AB = 2a, O is the midpoint of the dipole and OP is perpendicular to AB. A

charge Q is placed at P where OP = y and y > > 2a. The charge Q experience an

electrostatic force F. If Q is now moved along the equatorial line to P ' such that

OP3

' y

, the force on Q will be close to : 23y a

1) 3F 2) 3F 3) 9 F 4) 27 F

Key: 4

Sol: Electric field due to dipole is 3 30

1 2 cos sin4

rP Pe e

r r

At equatorial point 090 . According to new position r becomes r/3




22. Two starts of masses 3 × 1031 kg each, and at distance 2 × 1011 m rotate in a plane about

their common centre of mass O. A meteorite passes through O moving perpendicular to

the stars rotation plane. In order to escape from the gravitational field of this double star,

the minimum speed that meteorite should have at O is (Take gravitational constant G =

6.67 × 10-11 Nm2 kg-2)

1) 2.4 × 104 m/s 2) 1.4 × 105 m/s 3) 3.8 × 104 m/s 4) 2.8 × 105 m/s

Key: 4

Sol: By energy conservation

21 0 02

GMm GMm mVr r

4GMVr

23. Half mole of an ideal monoatomic gas is heated at constant pressure of 1 atm from 20°C

to 90°C. Work done by gas is close to : (Gas constant R = 8.31 J/mol.K)

1) 581 J 2) 291 J 3) 146 J 4) 73 J

Key: 2

Sol: Work = Pdv nR T 24. A parallel plate capacitor having capacitance 12 pF charged by a battery to a potential

difference of 10 V between its plates. The charging battery is now disconnected and a

porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by

the capacitor on the slab is

1) 692 pJ 2) 508 pJ 3) 560 pJ 4) 600 pJ

Key: 2

Sol: Initial energy of capacitor = 212

CV

Charge on capacitor = CV

Final capacitance C’ = 6.5 C

` Final energy stored = 21 / '2

Q C

Work = i fU U




25. A particle starts from the origin at time t = 0 and moves along the positive x-axis. The

graph of velocity with respect to time is shown in figure. What is the position of the

particle at time t = 5s ?

1) 10 m 2) 6 m 3) 3 m 4) 9 m

Key: 4

Sol: Area under the graphs

1 2 2 2 2 3 12

26. A rigid massless rod of length 3l has two masses attached at each end as shown in the

figure. The rod is pivoted at point P on the horizontal axis (see figure). When released

from initial horizontal position, its instantaneous angular acceleration will be

1) 13g

l 2)

3gl

3) 2gl

4) 73gl

Key: 1

Sol: Torque about P 0 02 2 5M g l M g l oI M gl oM gl

I

27. Two forces P and Q, of magnitude 2F and 3F, respectively are at an angle with each

other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle

is

1) 120° 2) 60° 3) 90° 4) 30°

Key: 1




Sol: 2 2 12 3 2 2 3 cosF F F F F

2 2 12 6 2 2 6 cos 2F F F F F

28. A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left

floating in a pond with still water. If pressed downward slightly and released, it starts

performing simple harmonic motion at angular frequency . If the radius of the bottle is

2.5 cm then is close to: (density of water = 103 Kg / m3)

1) 3.75 rad s-1 2) 1.25 rad s-1 3) 2.50 rad s-1 4) 5.00 rad s-1

Key: No Ans

Sol: AgWm

7.9 / secW rad 29. A particle executes simple harmonic motion with an amplitude of 5 cm. When the

particle is at 4cm from the mean position, the magnitude of its velocity in SI units is

equal to that of its acceleration. Then, its periodic time in seconds is :

1) 43 2) 3

8 3) 8

3 4) 7

3

Key: 3

Sol: Velocity of particle 2 2V A Y

Acceleration of particle 2a A |V|= |a|

30. Two kg of a monoatomic gas is at a pressure of 4 × 104 N/m2. The density of the gas is 8

kg/m3. What is the order of energy of the gas due to its thermal motion?

1) 103 J 2) 105 J 3) 104 J 4) 106 J

Key: 3

Sol: Energy of gas 32

nRT 32

PV

Volume = 32 18 4

m m




CHEMISTRY 31. The ground state energy of hydrogen atom is – 13.6 eV. The energy of second excited

state of He+ ion in eV is

1) – 54.4 2) – 3.4 3) – 6.04 4) – 27.2

Key: 3

Sol:

2

1 2n H

zE En

2

3 2

213.6 6.043

E evHe

32. Haemoglobin and gold sol are examples of

1) positively and negatively charged

2) positively charged sols

3) negatively charged sols

4) negatively and positively charged sols, respectively

Key: 1

Sol: Hemoglobin is positive solution

Gold solution is negative solution

33. The major product of the following reaction is

1) 2)

3) 4)




Key: 3

Sol: H3CO OH

O

O

CH3(i)dil.HCl,Δ

COOHCOOH

OH OH

OH

OH OH

O

OOH

O

OH OH

O

OOH

O

Polymerisation

OH

O OO

O

n 34. The amount of sugar (C12H22O11) required to prepare 2 L of its 0.1 M aqueous solution is

1) 136.8 g 2) 17.1 g 3) 68.4 g 4) 34.2 g

Key: 3

Sol: cos

cos

0.1glu eglu e in L

MMolarity

M V

, cos 0.1 2 342 68.4glu eM gm

35. Among the following reactions of hydrogen with halogens the one that requires a catalyst

is

1) H2 + I2 2HI 2) H2 + Cl2 2HCl

3) H2 + Br2 2HBr 4) H2 + F2 2HF

Key: 1

Sol: The reaction between 2H and 2I ions takes place in the presence of platinum sponze

catalyst at about 0440 . Even then it will not take place completely and it is a reversible reaction.

36. 5.1 g NH4SH is introduced in 3.0 L evacuated flask at 327°C. 30% of the solid NH4SH

decomposed to NH3 and H2S as gases. The Kp of the reaction at 327°C is (R = 0.082 L

atm mol-1K-1, Molar mass of S = 32 g mol-1, molar mass of N = 14 g mol-1)

1) 0.242 × 10-4 atm2 2) 1 × 10-4 atm2

3) 4.9 × 10-3 atm2 4) 0.242 atm2




Key: 4

Sol: 4 3 21 0 0

g gSNH HS NH H S

mol

0.1 0.1 0.11

0.3 0.6 0.082 600 0.984P atm

2 3

0.984 0.4922H S NH

P P

3 2

2. 0.492 0.492 0.242P NH H SK P P atm

0.242PK 37. The reaction that is NOT involved in the ozone layer depletion mechanism in the

stratosphere is

1) 2 2 2( ) ( ) ( )uvCF Cl g C l g C F Cl g

2) 2( ) ( ) ( ) ( )Cl O g O g C l g O g

3) 4 3 2 22 3 3CH O CH O H O 4) ( ) ( ) ( )hvHOCl g O H g C l g

Key: 3 Sol: In the depletion of ozone layer, the following reactions takes place.

1) 2 2 2UVCF Cl C l C F Clg g g

2) 2Cl O O C l Og g g g

3) hvHOCl O H C lg g g

But 4 3 2 22 3 3CH O CH O H O do not take place. 38. In the cell Pt(s)|H2(g, 1bar)|HCl(aq)|AgCl(s)|Pt(s) the cell potential is 0.92 V when a 10-6

molal HCl solution is used. The standard electrode potential of (AgCl / Ag.Cl–) electrode

is

2.303 RTGiven 0.06 V at 298 KF

1) 0.94 V 2) 0.76 V 3) 0.40 V 4) 0.20 V




Key: 4

Sol: Given cell

6

2

10/ / / /,1s S S

mpt H HCl AgCl Agg bar

2 2 2aqg

H H e

2 2 2 aqS SAgCl e Ag Cl

CELL REACTION: 22 2 2aq aqS SAgCl H H Cl Ag

0 2 20.06 log[ ] [ ]

2E E H Cl

0 240.92 0 0.031 log 10

/ /ECl AgCl Ag

0/ /

0.2Cl AgCl Ag

E V

39. The 71st electron of an element X with an atomic number of 71 enters into the orbital

1) 6p 2) 4f 3) 5d 4) 6s

Key: 3

Sol: The 71st element is lutetium. At ytterbium (z=70) the 4f sub shell is completely filled.

Hence the last election enters into 5d.

40. The correct match between Column – I and Column – II is

Column – I Column – II

(Compound) (Reagent)

A) Lysine P) 1-naphthol

B) Furfural Q) ninhydrin

C) Benzyl alcohol R) KMnO4

D) Styrene S) Ceric ammonium nitrate

1) A – Q, B – P , C – S, D – R 2) A – Q, B – P, C – R, D – S

3) A – R, B – P, C – Q, D – S 4) A – Q, B – R, C – S, D – P




Key: 1

Sol: Lysine can be tested by ninhydrin.

Furfural can be tested by 1-napthol.

Benzyl alcohol can be tested by Ceric ammonium nitrate.

Styrene can be tested by KMnO4.

41. An aromatic compound ‘A’ having molecular formula C7H6O2 on treating with aqueous

ammonia and heating forms compound ‘B’. The compound ‘B’ on reaction with

molecular bromine and potassium hydroxide provides compounds ‘C’ having molecular

formula C6H7N. The structure of ‘A’ is

1) 2)

3) 4)

Key: 1

Sol:

COOH CONH2 NH2COO-NH4

+

aq.NH3 Δ Br2KOH

42. The process with negative entropy change is

1) Dissociation of CaSO4(s) to CaO(s) and SO3(g)

2) Sublimation of dry ice

3) Dissolution of iodine in water

4) Synthesis of ammonia from N2 and H2

Key: 4

Sol: S ve

0f i

f i

S SS S

2 2 33 2N H NH S ve .




43. An ideal gas undergoes isothermal compression from 5 m3 to 1 m3 against a constant

external pressure of 4 Nm-2. Heat released in this process is used to increase the

temperature of 1 mole of Al. If molar heat capacity of Al is 24 J mol-1 K-1, the

temperature of Al increases by

1) 3 K2

2) 2 K 3) 2 K3

4) 1 K

Key: 3

Sol: , f iiso irr ext V VW P =16 N.m Being isothermal 0,V q W = -16 Nm = -16J

mq nc T 216 1 243

T T K

44. Elevation in the boiling point for 1 molal solution glucose is 2K. The depression in the

freezing point for 2 molal solution of glucose in the same solvent is 2K. The relation

between Kb and Kf is

1) Kb = 1.5 Kf 2) Kb = Kf 3) Kb = 0.5 Kf 4) Kb = 2Kf

Key: 4

Sol: b b

f f

T KT K

2b fK K

45. The major product of the following reaction:

1) 2)

3) 4)

Key: 4




Sol: Solution: Imine and Ketone will be reduced by NaBH4, but not alkene.

NCH3 CH3

ONaBH4 NH

CH3 CH3

OH

imine alekene ketone

46. Sodium metal on dissolution in liquid ammonia gives a deep blue solution due to the

formation of

1) Sodium – ammonia complex 2) Sodamide

3) Sodium ion-ammonia complex 4) ammoniated electrons

Key: 4

Sol: The deep blue colour of sodium metal in liquid ammonia solution is due to the formation

of ammoniated electrons.

47. For an elementary chemical reaction,

2 11 2 ,k

kA A

the expression for

[ ]d Adt

is

1) 21 2 1[ ] [ ]k A k A 2) 21 2 12 [ ] [ ]k A k A

3) 21 2 1[ ] [ ]k A k A 4) 21 2 12 [ ] 2 [ ]k A k A

Key: 4

Sol: 11

2 2KK

A A

1K & 1K are rate constants with regard to reaction

21 2 1[ ] 2 2[ [ ] [ ] ]d A r K A K Adt

48. Which of the following tests cannot be used for identifying amino acids ?

1) Biuret test 2) Barfoed test 3) Ninhydrin test 4) Xanthoproteic test

Key: 2




Sol: Biuret Test: The Biuret Test is a general test for proteins. When a protein reacts with

copper(II) sulfate (blue), the positive test is the formation of a violet colored complex.

Biuret test can be used for identification of peptide bond formed by amino acids.

Barfoed’s test : Barfoed’s test is a chemical test used for detecting the presence of

monosaccharides. It is based on the reduction of copper(II) acetate to copper(I) oxide

(Cu2O), which forms a brick-red precipitate.

Ninhydrin Test: The Ninhydrin Test is a test for amino acids and proteins with a free –

NH2 group. When such an –NH2 group reacts with ninhydrin, a purple-blue complex is

formed.

Xanthproteic Test: Phenyl rings containing an activating group of amino acids can be

nitrated producing a yellow product.

49. The difference in the number of unpaired electrons a metal ion in its high-spin and low-

spin octahedral complexes is two. The metal ion is

1) Ni2+ 2) Fe2+ 3) Co2+ 4) Mn2+

Key: 3

Sol: Cobalt has 7d configuration in +2 oxidation state. In low spin complete its electronic

config. is 62gt and 2ge

Low spin complex 1 unpaired electron

High spin complex 3 unpaired electrons.

Difference 2

50. The major product obtained in the following reaction is




1) 2)

3) 4)

Key: 4

Sol: Aldol condensation: Keto is more acidic than ester.

O

COOEt

ONaOEt

O

COOEt

O O

O–

COOEt

O

OHCOOEt

O

OHCOOEt

O

COOEt

Δ

acidic

51. The pair that contains two P – H bonds in each of the oxoacids is

1) H2P2O5 and H4P2O6 2) H3PO2 and H4P2O5

3) H3PO3 and H3PO2 4) H4P2O5 and H3PO3

Key: 2

Sol:

||

|

3 2

HO H

H

OP

HPO

|| ||

| |

4 2 5

HO O

H

O OP P OH

H HP O

52. Which is the most suitable reagent for the following transformation?

1) Tollen’s reagent 2) I2/NaOH

2) CrO2Cl2/CS2 3) alkaline KMnO4




Key: 2

Sol: CH CH CH2CH3 CH CH3

OH(i)I2,NaOH(ii) dil.H+ CH CH CH2

CH3 COOH + CHI3

53. What is the IUPAC name of the following compound ?

1) 3-Bromo-1, 2-dimethylbut-1-ene 2) 3-Bromo-3-mehtyl-1, 2-dimethylprop-1-ene

3) 2-Bromo-3-methylpent-3-ene 4) 4-Bromo-3-mehtylpent-2-ene

Key: 4

Sol: Longest Chain and alkene with minimum locant will be preferred.

CH3

H

CH3

CH3

H

Br

: 4-bromo-3-methylpent-2-ene

1

2 34

5 54. The number of 2-centre-2-eleectron and 3-centre-2-electron bonds in B2H6, respectively

are

1) 2 and 1 2) 4 and 2 3) 2 and 2 4) 2 and 4

Key: 2

Sol: Structure of 2 6B H is

It contains four 2c-2e and two 3c-2e bonds.

55. In the reaction of oxalate with permanganate in acidic medium, the number of electrons

involved in producing one molecule of CO2 is

1) 1 2) 10 3) 2 4) 5




Key: 1

Sol: 4 2 4 2 2 4 2 4 4 2 22 3 5 2 8 10KMnO H SO H C O K SO MnSO H O CO Each KMnO4 molecule takes up 5 electrons. So 10 electrons for two moles of KMnO4

from 5 H2C2O4 molecules to produce 10 molecules of CO2. So one electron is involved in

producing one molecule of CO2.

56. A reaction of cobalt (III) chloride and ethylenediamine in a 1 : 2 mole ratio generates two

isomeric products A (violet coloured) and B(green coloured). A can show optical activity,

but, B is optically inactive. What type of isomers does A and B represent?

1) Geometrical isomers 2) Coordination isomers

3) Linkage isomers 4) Ionisation isomers

Key: 1 Sol: They are geometrical isomers. Since trans isomer cannot exhibit optical isomerism but

Cis isomer can exhibit optical isomerism. 57. The electrolytes usually used in the electroplating of gold and silver, respectively are 1) [Au(CN)2]– and [Ag(CN)2] – 2) [Au(CN)2] – and [AgCl2] – 3) [Au(OH)4– and [Ag(OH)2] – 4) [Au(NH3)2]+ and [Ag(CN)2] –

Key: 1 Sol: In electroplating with gold and silver, their cyno complexes are used as electrolytes. 58. A compound of formula A2B3 has the hcp lattice. Which atom forms the hcp lattice and

what fraction of tetrahedral voids is occupied by the other atoms

1) hcp lattice – A, 23

Tetrahedral voids – B

2) hcp lattice – A, 13

Tetrahedral voids – B

3) hcp lattice – B, 23

Tetrahedral voids – A

4) hcp lattice – B, 13

, Tetrahedral voids – A

Key: 4

Sol: 2 3A B being Hcp , If ‘A’ forms Hcp, 34

th

of TVs must occupied by B

If ‘B’ forms Hcp, 13

rd

of TVs must occupied by A




59. The major product of the following reaction is :

1) 2) 3) 4)

Key: 2

Sol:

CH3OH

CH3O

–CH3

OCH3aq.NaOH CH3I

SN2

60. What will be the major product in the following mononitration reaction?

1) 2)

3) 4)

Key: 4

Sol: The ring with high electron density will be nitrated. -NH group has +R effect whereas –

CO- group has –R effect.




MATHEMATICS 61. The value of such that sum of the squares of the roots of the quadratic equation,

2 (3 ) 2x x has the least value is

1) 158

2) 1 3) 49

4) 2

Key: 4

Sol: 22 23, 2 1

If 2 2 is least, then 2

62. The value of 2 3 10 10cos .cos ......cos .sin2 2 2 2 is

1) 1512

2) 11024

3) 1256

4) 12

Key: 1

Sol: 9

10

2 3 4 109

10

sin 22

cos cos cos ...........cos2 2 2 2 2 .sin

2

2 3 10 10 91 1cos cos ........cos sin

2 2 2 2 2 512

63. The curve amongst the family of curves represented by the differential equation, 2 2( ) 2 0x y dx xy dy which passes through (1, 1) is

1) a circle with centre on the x-axis 2) an ellipse with major axis along the y-axis

3) a circle with centre on the y-axis 4) a hyperbola with transverse axis along the x-

axis

Key: 1

Sol: 2

22 4 0 ydx xydy dx x kx

It passes through (1,1)

K=2equation of curve is 2 2 2 0x y x




64. Let :( 1, 1)f R be a function defined by 2( ) max | |, 1f x x x . If K be the set of all points at which f is not differentiable, then K has exactly

1) Five elements 2) one element 3) three elements 4) two elements Key: 3

Sol: f(x) is not differentiable at 1 , 02

x x

65. The positive value of for which the co-efficient of x2 in the expression 10

22x x x

is 720, is 1) 4 2) 2 2 3) 5 4) 3 Key: 1

Sol: Constant in 10

xx

is 10 2 720 4rC

66. The tangent to the curve, 2xy xe passing through the point (1, e) also passes though the

point

1) (2, 3e) 2) 4 , 23

e

3) 5 , 23

e

4) (3, 6 )e

Key: 2 Sol: Equation of tangent of (1,e) to xy xe is 3ex-y=2e

It also passes through 4 , 23

e

67. Let N be the set of natural numbers and two functions f and g be defined as , :f g N N such that

1 if is odd2( )if is even

2

n nf n

n n

and ( ) ( 1)ng n n . Then fog is

1) onto but not one-one 2) one-one but not onto

3) both one-one and onto 4) neither one-one nor onto




Key: 1

Sol: 1, 12 ,

1,2

n n is odd n n is oddf x g x

n n n is evenn is even

,

21,

2

n n is evenf x

n n is odd

It is many one and onto

68. The number of values of (0, ) for which the system of linear equations 3 7 0x y z

4 7 0x y z (sin 3 ) (cos 2 ) 2 0x y z has a non-trivial solution, is

1) three 2) two 3) four 4) one

Key: 2

Sol: 1 3 7

1 11 4 7 0 14 7sin 3 14cos 2 0 sin 02 2

sin 3 cos 2 2or or

Number of solutions in 0, is 2

69. Let 2 a b

and 4 2 3a b

be two given vectors where vectors a

and b

are non-collinear. The value of for which vectors

and

are collinear, is 1) – 4 2) – 3 3) 4 4) 3 Key: 1

Sol: 2 1 44 2 3

70. Two sides of a parallelogram are along the lines, x + y = 3 and x – y + 3 = 0. If its

diagonals intersect at (2, 4), then one of its vertex is

1) (3, 5) 2) (2, 1) 3) (2, 6) 4) (3, 6)

Key: 4 Sol: P (2,4)




71. If 1

2 2

0

( ) ( )x

x

f t dt x t f t dt , then 1'2

f

is

1) 2425

2) 1825

3) 45

4) 625

Key: 1

Sol: 22f x x x f x 1 2 12 2f x xf x x f x 1 1 242 25f

72. Let 5 5

3 32 2 2 2

i iz

. If R(z) and I(z) respectively denote the real and

imaginary parts of z, then

1) I(z) = 0 2) R(z) > 0 and I(z) > 0

3) R(z) < 0 and I(z) > 0 4) R(z) = – 3

Key: 1

Sol: 5 5 5cos cos 2cos 0, 06 6 6

z IM Z

73. If the probability of hitting a target by a shooter, in any shot, is 13

, then the minimum

number of independent shots at the target required by him so that the probability of

hitting the target at least once is greater than 56

, is

1) 3 2) 6 3) 5 4) 4

Key: 3

Sol: 2 5 2 11 53 6 3 6

n n

n

74. If 3 35 4 41 ( )

48x xx e dx e f x C where C is a constant of integration, then f (x) is equal

to

1) 32 1x 2) 34 1x 3) 32 1x 4) 34 1x

Key: 2




Sol: 35 ,axx e dx Put, 3x t

32 53 ax axx dx dt x e dx te dt

34 1 1 448

axe t e f x x

75. If the area of an equilateral triangle inscribed in the circle, 2 2 10 12 0x y x y c

is 27 3 sq. units then c is equal to

1) 13 2) 20 3) – 25 4) 25

Key: 4

Sol: 23 27 3 6 3 6 25 36 25

4o a R c c

76. Consider the following three statements:

P : 5 is a prime number

Q : 7 is a factor of 192

R : L.C.M of 5 and 7 is 35.

Then the truth value of which one of the following statements is true?

1) (~ ) ( )P Q R 2) ( ) (~ )P Q R

3) (~ ) (~ )P Q R 4) (~ )P Q R

Key: 4

Sol: Conceptual

77. The length of the chord of the parabola 2 4x y having equation 2 4 2 0x y is

1) 3 2 2) 2 11 3) 8 2 4) 6 3

Key: 4

Sol: Obviously, 2 4 2 0x y is a normal chord to 2x ay

Length of the chord is 6 3

78. Let 22 1

11 2

bA b b b

b

where b > 0. Then the minimum value of det( )Ab

is

1) 2 3 2) 2 3 3) 3 4) 3

Key: 1




Sol: 2 33 2 3 0AA b b bb b

79. Let 2 2

2( , ) : 11 1

y xS x y Rr r

, where 1r . Then S represents

1) a hyperbola whose eccentricity is 21 r

, when 0 < r < 1

2) an ellipse whose eccentricity is 2 ,1r

when r > 1

3) a hyperbola whose eccentricity is 2 ,1r

when 0 < r < 1

4) an ellipse whose eccentricity is 1 ,1r

where r > 1

Key: 2

Sol: If (0,1)r then s is a hyperbole whose 21

er

If r>1, then s is an ellipse whose 21

er

80. If 25

50 50 5025 25

0. rr r

rC C K C

then K is equal to

1) (25)2 2) 225 – 1 3) 224 4) 225

Key: 4

Sol: 25 25 25

50 50 50 2525 25

0 0 0

50 2525 25 25

rr r r

r r r

C C C Cr r

= 50 2525 2C

81. The plane which bisects the line segment joining the points ( 3, 3, 4) and (3, 7, 6) at

right angles, passes through which one of the following points?

1) (– 2, 3, 5) 2) (4, – 1, 7) 3) (2, 1, 3) 4) (4, 1, – 2)

Key: 4

Sol: Equation of required plane is 2 6 2 10 2 2 60x y Z 3x+5y+z=15




82. The value of 19

1

1 1cot cot 1 2

n

n pp

is

1) 2119

2) 1921

3) 2223

4) 2322

Key: 1

Sol:

1 1 1 1

1

1cot 1 2 tan tan 1 tan

1 1

n

p

n nP n n

n n

19 19

1 1 1

1 1 1 1

21cot 1 2 tan 20 cot cot 1 214 19

n n

n p n p

P

83. If mean and standard deviation of 5 observations 1 2 3 4 5, , , ,x x x x x are 10 and 3,

respectively, then the variance of 6 observations 1 2 5, ,.....,x x x and – 50 is equal to

1) 509.5 2) 586.5 3) 582.5 4) 507.5

Key: 4

Sol: 15 5

2

1 1

50, 109i ii i

x x

Required vawana=

25 522

1 1

50 50

6 6

i ii i

x x

=507.5

84. Let f be a differentiable function such that 3 ( )'( ) 7 , ( 0)4

f xf x xx

and (1) 4f .

Then 0

1limx

x fx

1) Exists and equals 47

2) exists and equals 4

3) Does not exist 4) exists and equals 0 Key: 2

Sol: I.F=

3 73/4 3/4 3/44 4. 7 4

dxxe x y x x dx x d

3

1/440

1 14 4 lim 4x

xy cx f x f cx xfx x




85. Two vertices of a triangle are (0, 2) and (4, 3). If its orthocenter is at the origin, then its third vertex lies in which quadrant?

1) Third 2) second 3) first 4) fourth Key: 2 Sol: Third vertex is the orthocenter of the triangle formed by (0, 2)(4,3)(0,0) which is

3,3

4

86. The value of 2

2

sin 4dx

x x

, where [t] denotes the greatest integer less than or equal

to t, is

1) 1 (7 5)

12 2)

1 (7 5)12

3) 3 (4 3)

20 4)

3 (4 3)10

Key: 3

Sol: 3 0 1 2

1 0 12

3 92 1 4 1 1 4 0 0 4 1 0 4 5 20

dx dx dx dx

87. On which of the following lines lies the point of intersection of the line, 4 5 3

2 2 1x y z

and the plane 2x y z ?

1) 3 4 1

3 3 2x y z

2) 4 5 5

1 1 1x y z

3) 1 3 4

1 2 5x y z

4) 2 3 3

2 2 3x y z

Key: 3

Sol: Let a port on the given line is 4 2 ,5 2 ,3 , if it is on x+y+z=2 2

88. Let 1 2 3 10, , , .....,a a a a be in G.P with 0ia for 1, 2, ...., 10i and S be the set of pairs

(r, k), r, k N (the set of natural numbers) for which

1 2 2 3 3 4

4 5 5 6 6 7

7 8 8 9 9 10

log log loglog log log 0log log log

r k r k r ke e e

r k r k r ke e e

r k r k r ke e e

a a a a a aa a a a a aa a a a a a

. Then the number of elements in S, is

1) 4 2) infinitely many 3) 2 4) 10




Key: 2

Sol: 3 3 2 2 2 1,C C C C C C Given det values D=0

89. With the usual notation, in ABC , if 120 , 3 1A B a and 3 1b , then the

ratio :A B is

1) 7 : 1 2) 5 : 3 3) 9 : 7 4) 3 : 1

Key: 1

Sol: A+B=120, 2 2A B a b A BTan Tan

a b

=1 090A B 0 0105 , 15A B

90. A helicopter is flying along the curve given by3

2 7, ( 0)y x x . A soldier positioned

at the point 1 , 72

wants to shoot down the helicopter when it is nearest to him. Then

this nearest distance is

1) 56

2) 1 73 3

3) 1 76 3

4) 12

Key: 3

Sol: 1/2 1/2 1/23 3 77 112 2

2

dy yy x x x xdx x

3/2

2 1 13 1 2 0 73 3

x x x x y

Shortest distance = 7 1 7

108 6 3

Date post:	21-Oct-2020
Category:	Documents
Upload:	others
View:	1 times
Download:	0 times

2019 Jee-Main Q P K S...2019/10/01 · 2019_Jee-Main Question Paper_Key & Solutions Sri Chaitanya...

Documents