1

JEE-MAIN

MODEL GRAND TEST

IMPORTANT INSTRUCTIONS

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

and Mathematics having 30 questions in each part of equal weightage.

Each question is allotted 4 (four) marks for each correct response.

Candidates will be awarded marks as stated above in instruction for correct response

of each question. 1/4 (one fourth) marks will be deducted for indicating incorrect

response of each question.

No deduction from the total score will be made if no response is indicated for an item

in the answer sheet.

There is only one correct response for each question.

Filling up more than one response in each question will be treated as wrong response

and marks for wrong response will be deducted accordingly as per instruction

Maximum time is 3 hrs.

Maximum number of marks is 360

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PHYSICS

1. A square plate, of side a and mass M is placed in( )x z− plane as shown in figure and hinged at A to rotate about x-axis. The plate is allowed to fall upto( )x y− plane. Find the final angular velocity.

z

y

xA

1) 24

7 2ga

2) 1/224

7g

a

3) 1/212

7ga

4) 1/212

7 2ga

2. A string fixed at both ends is vibrating with 3 antinodes as the possible mode of vibration for which a point(p) at one sixth of its length from one end is a point of maximum displacement. The frequency of vibration is this mode is 100 Hz. What will be the frequency emitted when it vibrates in the next mode such that this point(p) is again a point of maximum displacement?

1) 200 Hz 2) 300 Hz 3) 400 Hz 4) 600 Hz

3. Unstretched length of a spring is2Rπ

,where R is the radius of smooth

vertical circular track in which spring is compressed. One end of the spring is fixed at lowest point A and compressed spring forms an angle of

045 at the centre. A ball of mass m placed at the free end of the spring just rises upto the highest point of the track, find the force constant.

045R

A

1) ( )

2

16 2 1mg

Rπ

− 2)

( )2

16 2 2 1mg

Rπ

+

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3) 2

16mgRπ

4) 2

16 2mgRπ

4. A particle of mass 4m which is at rest explodes into 3 fragments. Two of the fragments each of mass m are found to move with a speed V each in mutually perpendicular directions. The total energy released in the process of explosion is:

1) 232

mV 2) 22mV 3) 252

mV 4) 23mV

5. A particle is dropped from rest from point A and B is another point on the downward path of the particle. Time taken by the particle during its

journey from A to B is( )

th

17 1

−

of the time taken by particle during

its journey from B to the ground G. If AG h= ,find path length of AB.

G

B

A

1)7h

2)7

h 3)

( )7 1h−

4)( )( )

7 1

7 1

h −

+

6. A particle is projected vertically upwards from the surface of the earth (radius R) with a kinetic energy equal to half of the minimum value needed for it to escape. The height to which it rises above the surface of the earth is

1) 2.5h R= 2) 2.0h R= 3) 1.5h R= 4) h R= 7. A large container of negligible mass and uniform cross-sectional area A

has a small hole (of area a <<A) near its side wall at the bottom. The container is open at the top and kept on a smooth horizontal floor. It contains a liquid of density ρ and mass 0m when liquid starts flowing horizontally at time 0t = .Find the speed of container when 75% of the liquid has drained out (Assume the liquid surface remains horizontal throughout the motion)

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1)1/2

0m gAρ

2)1/2

04m gAρ

3)1/2

0

2m g

Aρ

4)1/2

02m gAρ

8. Acceleration-time graph of a particle is shown. Work done by all the forces acting on the particle of mass m in time interval t1 and t2, while a1 is the acceleration at time t1, is given by:

1) ( )2

3 312 1

14ma t t

t− 2) ( )

24 412 12

18ma t t

t−

3) ( )2

4 412 12

14ma t t

t− 4) ( )

22 212 12

12ma t t

t−

9. A cylinder A of a material of thermal conductivity 1K is surrounded by a cylindrical shell of inner radius equal to the cylinder. The shell is made of a material of thermal conductivity 2K .The two ends of the combined system are maintained at two different temperatures. If the effective

thermal conductivity of the system is 1 24 59

K K+,find the ratio of outer

radius of the shell to the radius of the cylinder. 1)2.0 2)5/2 3)5/3 4)3/2

10. In an adiabatic process,23 vR C= . Te pressure of the gas will be

proportional to:

1)5/3T 2)

5/2T 3)5/4T 4) 5/6T

11. The molar heat capacity for an ideal monoatomic gas is 3R.If dQ is the heat supplied to the gas and dU is the change in its internal energy, then for the process,

1)dQ dU= 2) 2dQ dU=

1a2a

1t 2t

acceleration

time

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3) 3dQ dU= − 4)32

dQ dU= −

12. In YDSE, bichromatic light of wavelengths 400 nm and 560 nm are used. The distance between slits is 0.1 mm and distance between the plane of the slits and the screen is 1.0 m. The minimum distance between two successive regions of complete darkness is 1) 4 mm 2) 5.6mm 3)14mm 4) 28mm

13. A convex lens is in contact with concave lens. Magnitude of the ratio of their focal lengths is 2/3.Their equivalent focal length is 30 cm. What are their individual focal lengths(in cm)?

1) –75,50 2) –10,15 3) 75,50 4) 10,-15

14. The charges appearing on inner surfaces of a parallel plate capacitor are q1 and q2. The charges appearing on outer surfaces of the capacitor are equal. Find the potential difference between plates of the capacitor, if capacitance is C.

1)1 2

2q q

C+

2)1 2

2q q

C−

3)1 2q qC+

4)1 2q qC−

15. Two parallel plate capacitors of capacitances C and 2C are connected in parallel and charged to a potential difference V. The battery is then disconnected and the region between the plates of capacitor C is completely filled with a material of dielectric constant K. The potential difference across the capacitors now becomes

1)

21

VK

+ 2)

31

VK

+ 3)

32

VK

+ 4)

33

VK

+

16. A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector?

1) 2 photon of energy 10.2 eV 2) 2 photon of energy of 1.4 eV

3) one photon of energy 10.2 eV and an electron of energy 1.4 eV 4) one photon of energy 10.2 eV and another photon of 1.4eV

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17. Two cylindrical bars of same material with radii r and 2r are kept in contact as shown. An electric current I is passed through the bars. Which one of the following is correct?

r2r

A B CL/2L/2

I

1) Heat produced in bar BC is 4 times the heat produced in bar AB 2) Electric field in both halves is equal 3) Current density across AB is double that across BC 4) Potential difference across AB is 4 times that across BC

18. Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper as shown. The variation of magnetic field B along the line 'XX is given by:

1) d d

x x'

2) d d

x x'

3) d d

x x'

4) d d

x x'

19. Photoelectric effect experiments are performed using three different metal

plates p,q and r having work functions 2.0 , 2.5 3.0p q reV eV and eVφ φ φ= = = respectively. A light beam

containing wavelengths of 550 nm,450nm and 350nm with equal intensities illuminates each of the plates. The correct I-V graph for the experiments istake hc=1240eV nm]

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1) 2)

3) 4) 20. The switch is closed at 0t = . The current through inductor and capacitor

are equal after time ( )ln 2CR . Then resistance R is equal to

v k

C R

L R

1) / 2L C 2) /L C 3)

2LC 4)

3/2LC

21. A nucleus with mass number 220 initially at rest emits an α -particle. If the Q value of the reaction is 5.5 MeV, calculate the kinetic energy of the α -particle

1) 4.4MeV 2) 5.4MeV 3) 5.6MeV 4) 6.5MeV

22. When 100 V dc is applied across a coil, a current of 1A flows through it. When 100V ac of 50 Hz is applied across the same coil, only 0.5 A flows.

The resistance and inductance of the coil are (take2 10π = )

1) 50 ,0.3HΩ 2) ( )50 , 0.3 HΩ

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3) 100 ,0.3HΩ 4) ( )100 , 0.3 HΩ 23. The near point of a person is 50 cm and the far point is 1.5 m. The

spectacles required for reading purpose and for seeing distant objects are respectively

1)

22 ,3

D D+ − 2)

2 , 23

D D+ − 3)

22 ,3

D D− + 4)

2 , 23

D D− +

24. A capacitor of 2 Fµ is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its energy dissipated is:

1 2

S

V2 Fµ 8 Fµ

1) 80% 2) 0% 3) 20% 4) 75%

25. The given network is equivalent to:

AB y

1) OR gate 2) NOR gate 3) NOT gate 4) AND gate

26. If 1θ and 2θ be the apparent angles of dip observed in two vertical planes at right angles to each other on either side of magnetic meridian, then the true angle of dip θ is given by

1) 2 2 21 2tan tan tanθ = θ + θ 2) 2 2 2

1 2cot cot cotθ = θ − θ

3) 2 2 21 2tan tan tanθ = θ − θ 4) 2 2 2

1 2cot cot cotθ = θ + θ

27. A message signal of frequency 10 kHz and peak voltage of 10 volts is used to module a carrier of frequency 1 MHz and peak voltage of 20 volts. The side bands are

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1)1010 kHz and 990 kHz 2)1000 kHz and 900 kHz 3)10kHz and 1 MHz 4)505 kHz and 445 kHz

28. The figure shows a charge q placed inside a cavity in an uncharged conductor. Now if an external electric field is switched on

1) Only induced charge on outer surface will redistribute 2) Only induced charge on inner surface will redistribute 3) Both induced charge on outer and inner surface will redistribute 4) Force on charge q placed inside the cavity will change.

29. Match the entries of column I with corresponding entries of column 2.where m is magnification produced by the mirror for real object]

1) A c→ and d ; B b→ and d: C b→ and c; D a→ and d

2) A b→ and c; B b→ and c; C b→ and d; D a→ and d

3) A a→ , and c; B a→ and d; C a→ and b; D c→ and d

4) A a→ and d; B b→ and d; C b→ and d; D a→ and d

30. A long solenoid of diameter 0.1 m has 42 10× turns per meter. At the

centre of the solenoid, a coil of 100 turns and radius 0.01 m is placed with

its axis coinciding with the solenoid axis. The current in the solenoid

reduces at a constant rate to 0A from 4A in 0.05 s. If the resistance of the

coil is 210π Ω . The total charge flown through the coil during this time is:

Column 1 Column 2 A) m=-2 a) Convex mirror B) 1

2m = −

b) Concave mirror

C) 2m = + c) Real image D) 1

2m = +

d) Virtual image

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1) 16 Cµ 2) 32 Cµ 3) 16 Cπµ 4) 32 Cπµ

MATHEMATICS

31. ( )f x is a quadratic function such that ( ) ( )0 1, 1 4f f= − = .If

( )( )22 1

f x dxx x +∫ is a rational function, then ( )10 ___f =

1) 512 2) 521 3) 432 4) 324

32. If cosx sin x cosy sin y 1cos sin cos sin

+ = + =θ θ θ θ

,where x y n ,n z− ≠ π ∈ , then

2 2cosxcosy sin xsin y

cos sin+ =

θ θ

1) 1/2 2) 1 3) -1 4) 3/2

33. For real numbers x and y, xRy 2x y⇔ − + is an irrational number. Then the relation R is

1) symmetric and transitive 2) symmetric but not transitive 3) transitive but not symmetric 4) neither symmetric nor transitive 34. With usual notation in triangle ABC,b,c,B are given such that b < c. If

1 2,a a are two positive values of third side satisfying 2 2 2 2

1 2 1 2(a a ) (a a ) tan B kb− + + = , then k = 1) 4 2) 2 3) 1 4) 3

35. The angle of elevation of a tower CD at a place A due south of it is 060and at a place B due west of A, the elevation is 030 . If AB=3KM, height of the tower is

1) 23 6

KM 2) 2 23 3

KM 3) 3 32 2

KM 4) 3 62

KM

36. If the roots of 2x a(x 1) b(x 1) 1 0+ − − + + = are real and unequal for all real values of b then a belongs to

1) ( )1,∞ 2) ( ),1−∞ 3) ( )1,− ∞ 4) ( ), 1−∞ −

37. If 2 2 2 2 2 21 1 1 1 1 1x 1 1 ...... 11 2 2 3 2019 2020

= + + + + + + + + then

4040(2020 x)− =

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

38. If 1 2p q 1 1

p q 2q 2p q 2q 2p q q(4 ) 3 (3 ) (2 )−

+ −− = − -such that p,q are positive in tangents and greatest common divisor of p,q is 1, then 3p-2q=

1) 0 2) -7 3) 7 4) 5 39. The number of natural numbers which are smaller than 82 10× and are

divisible by 3 using the digits 0,1,2 only is a four digited number abcd. Then ( ) ( )a c b d+ − + =

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

40. The digit in the 10’s place of 14147 is

1) 4 2) 0 3) 2 4) 5

41. Let ( )P E denotes probability of an event E, If 3P(B)4

= ,

1 1P(A B C) ,p(A B C)3 3

∩ ∩ = ∩ ∩ = , then p(B C)∩ =

1) 14

2) 512

3) 112

4) 16

42. If the length of the latus rectum of the parabola 2 2 2289(x 3) 289(y 1) (15x 8y 13)− + − = − + is k, then [k]=

1) 5 2) 2 3) 3 4) 11

43. Two ellipses 2 2 2 2

2 2 2 21, 1cos sin sin cos

x y x yα α α α+ = + = , 0,

4πα ∈

intersect at 4 points P,Q,R,S. If two statements given as

Statement-I: PQRS is a square of side sin 2α Statement-II: Eccentricities of two ellipse are equal, then 1) Both are true and II is correct explanation of I 2) Both are true and II is not correct explanation of I 3) I is true, II is false 4) I is false, II is true 44. Let 1 26 , 4 3z i z i= + = − and z be a complex number such that

1

2 2z zargz z

π −= −

.If two statements are given by

Statement-I: ( )( ) ( )( ) ( )( ) ( )( )ˆ6 4 3 4 3 6 0z i z i z i z i− + − + + − − − − =

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Statement-II:Z satisfies ( )5 5z i− − = ,Then

1) Both are true and II is correct explanation of I 2) Both are true and II is not correct explanation of I 3) I is true, II is false 4) I is false, II is true 45. A circle touches the line 2 3 1 0x y+ + = at the point(1,-1) and is

orthogonal to the circle which has the line segment having end points

( )( )0, 1 2,3− − as diameter is given by 2 2 2 2 0x y gx fy c+ + + + = .Then,

which of the following is not correct?

1) 52

g = − 2) 2 2 7g f c+ + = -

3) 54

f = − 4) 12

c = −

46. Let : 2P is a prime number as : 3q is rational be two propositions. Then the truth values of biconditional Statements

a) p q⇔ b) ⇔ p q c) q p⇔

d) ( )q q> in order are given by

1) FTFT 2) FTTF 3) FFTT 4) TFTF 47. If ( ) ( )2,5 , 5,11A B= = and a point P moves such that internal bisector

of APB passes through (4, 9). The maximum area of APB∆ is

1)45 .4

sq units 2)15 .2

sq units 3)15 .sq units 4)45 .2

sq units

48. Let ‘L’ be the projection of the line 1 1 32 1 4

x y z− + −= =

− on the plane

2 9x y z+ + = . Then, which of the following is not correct?

1) 1 17, ,02 4

is a point on L 2) 3 15, ,02 4

is a point on L

3) 13 4 25, ,6 3 6

is a point on L

4) Direction ratios on L are ( )4, 7,10−

49. The volume of triangular prism whose adjacent sides are , ,a b c each of

magnitude 4 units and each is inclined at an angle 3π

with other two is

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1) 16 2 2) 16 3 3) 8 3 4) 8 2

50. If , ,a p b q c r≠ ≠ ≠ and 0p b ca q ca b r

= , then ___p q ra p b q c r

+ + =− − −

1) 1 2) -1 3) 2 4) -2 51. If all the linear functions defined from[ ]1,1− onto [ ]0,3 intersect at ( ),p q

then 2019 2020p q+ = 1) 3030 2) 4038 3) -1 4) 4040

52. If ( ) ( ) ( )2 sin 2 cos tan , 0f x f x x x+ = > ,then ( )1

lim 1x

x f x→

− =

1) 2 2) 12

3) 2− 4) 12

−

53. The tangent at any point on the curve 2/3 2/3 2/3x y a+ = meets the axes at P and Q. The locus of midpoint of PQ is___

1) 2

2 2

2ax y+ = 2)

22 2

3ax y+ = 3)

22 2

4ax y+ = 4) 2 2 2x y a+ =

54. Let ( )p x be a polynomial of least degree having maximum value 6 at

x=1 and minimum value 2 at x=3.then ( ) ( )12 0p p+ = 1) 9 2) 11 3) 12 4) 13

55. The value of k for which the equation 3 3 0x x k− + = has two distinct roots is ( )0,1 is

1) in the interval ( )2, 1− − ( )1,2∪ 2) in the interval ( )1,1−

3) in the interval ( )0,1 4) not existing

56. In [ ]2 ,2− π π the number of solution of ln(sin x) x= is 1) 0 2) 2 3) 4 4) 8 57. The area of the region bounded by the curve xy e= and the lines

0,x y e= = is

1) 1 2) e-1 3) 1ee

− 4 ) ee-1

58. If ( )sin cos 1 cosdyy y x ydx

= − and ( )03

y π= , then ( )log 2y =

1) ( )1tan 2 log 2− + 2) ( )1sec 2 log 2− +

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3) ( )1sec 3 log 2− + 4) ( )1 3 log 2Tan− +

59. If σ is the standard deviation of the numbers 10, 15, 25, 30, x, then the value of x for which σ is least is

1) 5 2) 20 3) 30 4) 80

60. If 1 1x tan (tan( 6)),y cot (cot 4)− −= − = ,then 2x+3y = 1) 6 2) 6 − π 3) 6π − 4) π

CHEMISTRY

61. The enthalpy of combustion of H2(g) at 298 K to give 2H O is –1–298 kJ mol and bond enthalpies of H–H and O = O are 433 kJ mol–1

and –1492 kJ mol , respectively. The mean bond enthalpy of O–H is : ( vapH∆ of 2H O 40.65 kJ / mol= )

1) –1464 kJ mol 2) –1488.5 kJ mol

3) –1232 kJ mol 4) –1–232 kJ mol

62. In an fcc arrangement of A and B atoms, where A atoms are at the

corners of the unit cell, 13 atoms at the face centers, two atoms are missing from two corners in each unit cell, then the simplest formula of the compound is

1) 7 6A B 2) 6 7A B 3) 7 24A B 4) 4AB

63. By how much would the oxidising power of the 2 4MnO / Mn− + couple

change if the H+ ions concentration is decreased 100 times? 1) Increases by 189 mV 2) Decreases by 189 mV 3) Increases by 19 mV 4) Decreases by 19 mV

64.

1) A and B are identical mixtures of 3CH I and ( )3 3CH C — OH

2) A and B are identical mixtures of CH3OH and ( )3 3CH C — I

3) A is mixture of 3CH I and( )3 3CH C — OH ; B is a mixture of

CH3OH and( )3 3CH C — I

B (mix) Conc HI

2(CH3)3COCH3

anhyd. HI

1A (mix)

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4) None of these

65. Which of the following will react at faster rate with acetone to give a product containing ? 1) 6 5 2C H NH 2) ( )3 3

CH N

3) 6 5 6 5C H NHC H⋅ 4) 6 5 2C H NHNH

66. A negatively charged sol can be formed by peptizing a solution of

1) AgI with AgNO3 2) AgI with KI

3) Fe(OH)3 with FeCl3 4) Any of these

67. What is A in the following reaction?

Et2CO NaCN NH4Cl ANH4OH

1) 2) 3) 4) No reaction takes place 68. Among the following species identify the isostructural pairs:

–3 3 3NF , NO , BF , H3O+, HN3

1) –3 3[NF , NO ]and 3 3BF , H O+

2) [ ]3 3NF , HN and –3 3NO , BF

3) 3 3NF , H O+ and –3 3NO , BF

C N=3H C

3H C

CN

EtH2N

Et

CN

EtH2N

H

COOH

EtH2N

H

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4) 3 3NF , H O+ and [ ]3 3HN , BF 69.

1) (B) is BaSO4 and (D) is H2S 2) (A) is BaS and (D) is SO2

3) (C) is BaS and (D) is SO2 4) (A) is K2SO3 and (B) is BaS

70. The reaction; 3 22O 3O→ , is assigned the following mechanism.

(I) O3 O2 O (II) O3 2O2OSlow

The rate law of the reaction will therefore be 1) [ ] [ ]2

3 2r O Oα 2) [ ] [ ]2 –13 2r O Oα

3) [ ]3r Oα 4)

71.

OH

H+

→No.of possible products upon dehydration are (structuralonly)

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

72. How many chlorine atoms can you ionize in the process –Cl Cl e+→ + , by the energy liberated from the following process? – – Cl e Cl+ → for

236 10× atoms. Given electron affinity of Cl = 3.61 Ev/atom, and IP of Cl = 17.422 eV/atom. 1) 231.24 10 atoms× 2) 209.82 10 atoms× 3) 152.02 10 atoms× 4) None of these

73. 0.5 g of fuming 2 4H SO (oleum) is diluted with water. This solution is completely neutralised by 26.7 ml of 0.4 N NaOH. The percentage of free

3SO in the sample is 1) 30.6% 2) 40.6% 3) 20.6% 4) 50%

74. Arrange the elements with the following electronic configurations in the increasing order of electron affinity

(i) 1S2 2S2 2P3 (ii) 1S2 2S2 2P4 (iii) 1S2 2S2 2P3 (iv) 1S2 2S2 2P6 3S23P4 1) i < ii < iii < iv 2) iv < iii < ii < i

3) iii < ii < iv < i 4) ii < iii < i < iv

( ) ( )2BaCl Carbon HCl2

gaswhite ppt.KO S A (B) C D∆

∆+ → → → →

] [ 23 2r O O

α

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75. What is the maximum value of x ?

HC

OHHOHHOHHOH

CH2OH

H3COx HIO4

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

76. Sec. butyl chloride will undergo alkaline hydrolysis in the polar solvent by

1) 2NS 2) 1

NS

3) 1NS and 2

NS 4) None of these 77. A solution of sodium sulphate in water is electrolysed using inert

electrodes. The products at cathode and anode are, respectively, 1) H2, O2 2) O2, H2 3) O2, Na 4) O2, SO2 78. Which of the following compounds yields only one product on

monobromination? 1) Neopentane 2) Toluene 3) Phenol 4) Aniline 79. Which of the following reactions does not occur in Bessemer convertor in

the extraction of copper from chalcopyrites? 1) 2 2 2 22CuFeS O Cu S 2FeS SO+ → + +

2) 2 3FeO SiO FeSiO+ →

3) 2 22FeS 3O 2FeO 2SO+ → +

4) 2 2 2Cu S 2Cu O 6Cu SO+ → +

80. 2 moles of iso-pentylene on reaction with A gives compound B, which on treatment with 1-phenyl-2-butyne followed by H2O2 oxidation results in C. A is in-turn generated by treating 3BCl with LiAlH4. Compound C gives

1) Tollen’s reagent 2) Iodoform test 3) Fehling’s test 4) 1 & 3 81. How many enantiomers pairs are produced by the monochlorination of

2-methylbutane?

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

82. The correct order among the following is

1) ( )4 4 4ClO BrO IO basicity− − −> >

2) ( )2 3 4 delocalization of negative chargeClO ClO ClO ClO− − − −> > >

3) 2 3 4ClO ClO ClO ClO− − − −> > > (charge density)

4) 3 3 3NH PH AsH> > (delocalization of lone pair electron)

83. The number of isomers possible for C4H8 are

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

84. High concentration of fluoride is poisonous and harmful to bones and

teeth at levels over

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

85. Mole fraction of a non electrolyte in aqueous solution is 0.07. If Kf is

1.86° mol–1 kg, depression in f.p, ∆ Tf is,

1) 0.26° 2) 1.86° 3) 0.13° 4) 7.78°

86. Which of the following statements is incorrect?

1) The order of splitting energy is 2 2 24 4 4PtCl PdCl NiCl− − −> >

2) [Co(NH3)6]3+ is colourless whereas [Ni(H2O)6]2+ is coloured.

3) [M(en)(gly)]n+ will exhibit geometrical isomerism.

4) The magnetic moment of K3[Fe(CN)6] is 3 B.M.

87. The volume occupied by 2 mol of 2N at 200K and 10.1325 MPa pressure

is (Given that 3 2.218

C C r r

rC

P V PVandRT T

= = ), where , &r r rP V T are

reduced pressure , reduced volume & reduced temperature.

1) 320.7 cm 2) 3544 cm

3) 3272.0 cm 4) 3136 cm

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88. Choose the correct code to identify (X), (Y) and ‘Z’ in each case for the

changes indicated.

(i) ( ) ( ) ( )32 4 AgNOConc. H SOKOH2 2CrO Cl X Y Z→ → →

(ii) ( ) ( ) ( ) ( )2 2

2

Na Oexcess lead3 NaOH H O, boil acetateCrCl aq X Y Z→ → →

(iii) ( ) ( ) ( ) ( )2 3Na CO cobalt4 nitrate,ZnSO aq X Y Z∆

∆→ → →

(iv) ( ) ( ) ( ) ( )34

2

HNONH OH KCN2 H S excessCuCl aq X Y Z

∆→ → →

1) 2 4 2 2 7 2 4X K CrO ;Y K Cr O ;Z Ag CrO= = =

2) ( ) 2 4 43X Cr OH ;Y Na CrO ;Z PbCrO= = =

3) 3 2X ZnCO ;Y ZnO;Z CoZnO= = =

4) ( ) ( )3 3 42X CuS;Y Cu NO ;Z K Cu CN = = =

89. The pH of a buffer solution changes from 6.20 to 6.17 when 0.003 mole

of acid is added to 500 mL of the buffer. The buffer capacity of the

system is, therefore,

1) 0.1 2) 0.3 3) 0.2 4) 0.4

90. 2 2 2HOCH CH CH COOHon reaction with H+ forms

1) O

O

2)

O

3) O 4)

JEE MAIN MODEL GRAND TEST

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KEY SHEET PHYSICS

1 1 2 2 3 2 4 1 5 1

6 4 7 4 8 2 9 4 10 2

11 2 12 4 13 4 14 2 15 3

16 3 17 1 18 2 19 1 20 2

21 2 22 4 23 1 24 1 25 2

26 4 27 1 28 1 29 2 30 2

MATHEMATICS

31 2 32 3 33 4 34 1 35 3

36 1 37 2 38 4 39 1 40 2

41 3 42 1 43 2 44 3 45 4

46 2 47 3 48 2 49 1 50 4

51 1 52 2 53 3 54 4 55 4

56 2 57 1 58 3 59 2 60 4

CHEMISTRY

61 2 62 4 63 2 64 3 65 4

66 2 67 1 68 3 69 1 70 2

71 2 72 1 73 3 74 3 75 2

76 2 77 1 78 1 79 1 80 2

81 2 82 3 83 4 84 4 85 4

86 3 87 3 88 4 89 3 90 1

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JEE MAIN MODEL GRAND TEST SOLUTIONS

PHYSICS

1. 2 2 21 1 1 22 2 32

Mg Iw Ma ω = =

Which gives1/2 1/23 3

2 2g ga a

ω = =

2. For I mode

1

6 4L λ=

132

L λ=

For II mode

23

6 4L λ=

292

L λ=

1 1 2 2v f fλ λ= =

Or 1 12

2

9 33

ff f fλλ

= = × =

11

32 2n v Vf

L L= =

22

92 2n v Vf

L L= =

22

1

3. 300f f Hzf= =

3. ( )2 01 cos 452

Kx mg R R= +

112

mgR = +

But 2 4 4R R Rx π π π

= − =

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So, ( )2

16 2 1 2mgK

Rπ

+=

4. In figure 2p should be 2mV in opposite direction to RP

for conservation

of linear momentum

So, ( )2 ' 2m V mV=

'2

VV =

Total energy released ( )2 21 12 2 '2 2

mV m V= +

232

mV=

5. For AG, 21

12

h gt=

21

2htg

=

For AB,

22

12

h gtn=

22

2htng

=

( )1/2 1/2

1 22 2h ht tg ng

− = −

( ) ( )1/2

1/2 1/22

2 1 1h n t ng

= − = −

If ( ) ( )1/22 1 2, 1AB BG ABt t t t t t n= − = = −

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or( ) ( )1/2

1 11 7 1

AB

BG

tt n

= =− −

So n=7

So, 7

h hABg

= =

6. Kinetic energy required to escape eGM mR

=

Energy provided to the particle2

eGM mR

=

So, kinetic energy at the surface of the earth

( )2

e e eGM m GM m GM mR R h R

− − = − +

7. ( )1 22A

at H Hg

γ= −

02 112

mAa g Aρ

= −

22

omAa Agρ

=

2ma avρ=

0 2aa gA

=

2gadv dtA

=

022 22

Mga ga Av tA A a Agρ

= = ×

= 02mgAgρ

02m gAρ

=

8. ( )2 212f f f iW k k m v v= − = −

2 2

2 2 1 11 1 12 2 2

m a t a t = −

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4

2 2 221 1 1

1

18 2

tm a a tt

= −

22 1

1

, tSince a at

=

( )2

4 412 12

18ma t t

t= −

0F mg kxµ+ =

9. Radius of cylinder =R and outer radius of shell =nR

( )

( )

2 2 21 21 2

1 22 2 2

1 2

1 11

1 11

eq

K R K n RR RRR R

K R K n R

π π

π π

×−

= =+ +

−

On solving( )2

1 2

2

1K K nk

n

+ − =

1 24 59

K K+= (given)

1 25 / 49 / 4

k k+=

n=3/2

10. 23 vR C=

We know, p vC C R− =

or 1v

RC

γ − =

or ( )1vR C γ= −

Comparing 5 / 3γ =

1 yTP γ− =constant=K

Or ( )/ 1P T γ γ −=

Given 5 / 3γ =

( )1 5 / 2γ γ − =

So, 5/2p Tα

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11. Molar heat capacity

3 ( )dQC R givenndT

= =

0322vC As C R = =

2 2vnC TdQ dUCndT ndT ndT

= = =

So, 2dQ dU=

12. If nth minima of 400 nm is coinciding with mth minima of 560nm then

( ) ( )560 4002 1 2 12 2

m n− = −

( )( )

2 1 7 14 212 1 5 10 15

nm−

= = = =−

4th minima of 400 nm will coincide with 3rd minima of 560 nm. Its

location is given by

( )( )( )62 4 1 1000 400 102 0.4

−× − ×=

×

=14mm

Similarly,11th minima of 400 nm will coincide with 8th minima of

560nm.Its location is given by ( )( )( )62 11 1 1000 400 1042

2 0.1mm

−× − ×= =

×

Minima distance between two successive region of complete

darkness=42-14=28m

13. The total length of concave lens= 32

f

If focal length of concave lens is f

Now 1 1 2 130 3f f f

= − =

Or 10f cm= Focal length of concave lens=10 cm and that of concave lens=15cm

14. 0pE =

or 1 2q q q q+ + =

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1 2q q= −

1 122

q qVC C

= =

( )1 1 1 2

2 2q q q q

C C− − −

= =

15. 'Q Q=

( )2 2 'CV KC C V= +

3'2

V VK

= +

16. 13.6eV energy needed to liberate the electron form hydrogen atom.So

electron will liberate with kinetic energy=15-13.6=1.4eV

17. ( )2AB AB

BC BC

H R as H I RtH R

= =

( )( )

2

2 2

1/ 2 1 1 14 41/

ras R

rrα α = =

4BC ABH H=

18. Magnetic field in region be upwards as points in this region are to right of

the wires and in such condition, If current flows out of page, magnetic

field will be upwards. For the same logic, magnetic field in region AX

will be downwards. In region AC, points are closer to A than B and they

are to the right of A(and to the left of B),magnetic field will be upwards.

For the same logic, magnetic field in region BC will be downwards,

Hence, option (b) is correct.

19. 1

hcEλ λ=

1240 2.25550

eV eV= =

2

1240 2.8450

E eV eVλ = =

and3

1240 3.5250

E eV eVλ = =

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For metal, 3,r λ is able to generate photoelectron

For metal 2 2, ,q λ λ are able to generate electrtons.

For metal ,p all wavelength are able to generate electrons

Hence photoelectric current will be maximum for P and

minimum for r.

20. ( )1/1 tL

Vi eR

τ−= −

2/tC

Vi eR

τ−=

( )ln 2L Ci i at t CR= =

1 2/ /1 t te eτ τ− −− =

At ( ) ( )ln 2 ln 2t CR for CRτ τ= = =

So that ( )ln 2tτ=

/ 12

te τ− =

/2 1te τ− =

From eq(1), 1 2/ / 1t te eτ τ− −+ =

From eq(2) /2 1te τ− =

On camparing 1 2τ τ=

or L CRR=

/R L C=

21.

1 2 5.5K K MeV+ = ……(i)

and 1 2 1 22 216 2 4P P K K= = × = × …….ii)

⇒

P1P2

m1=216 m2=4M=220

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on solving equation(i)and (ii),we get

2 5.4K MeV=

22.

In dc circuit, resistance offered by inductor is zero in steady state

1001 100Vi RR R

= ⇒ = ⇒ = Ω

LR

100V,50Hz

1'2

i A=

2 2 2 2

1 100'2

L L

ViR X R X

= ⇒ =+ +

( ) ( ) ( )2 2 22 2 2200 200 100L LR X X+ = ⇒ = =

2 300 100LX = ×

( )22 300 100fLπ = ×

( ) ( )2 224 50 300 100Lπ × × = ×

2 0.3 0.3L L H= ⇒ =

23. ( )

1 1 150 25 f

− =− −

( )

1 1 100 100 250 50

p Df f cm= ⇒ = = =

I

LR

100V

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1 1 150 f

= =− ∞

100 2' 150150 3

f cm p D= − ⇒ = = −−

24. A capacitor of 2 Fµ is changed as shown in the diagram. When the switch

S is turned to position 2, the percentage of its stored energy dissipated is: 1 2

S

V2 Fµ 8 Fµ

V2 Fµ

2 Fµ8 FµV V

( ) ( )2

2

1 2 8. 02 2 8

0.8 . 80%1 .2.2

Vloss of energy i estored energy V

×−

+= =

25. The given electrical network is equivalent to:

AB y

A B A B A B+ → + → +

26. If 1θ and 2θ be the apparent angles of dip observed in two vertical planes

at right angles to each other, then the true angle of dip θ is given by:

( )1 2

tan tan tantan , tancos cos 90 sin

θ θ θθ θα α α

= = =−

2 2cos sin 1α α+ =

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2 2 21 2cot cot cotθ θ θ+ =

27. Signal frequency and peak voltage.

28. Applied electric field on the cavity of uncharged electrical conductor.

29. ;A band c B b and c→ →

,C b and a D a and d→ →

30. ( )NBAQ

R Rφ ∆∆

∆ = =

( )0 1 2N nA i iR

µ −=

( ) ( )27 4

2

100 4 10 2 10 0.05 4 010

π ππ

−× × × × × × −=

632 10 32C Cµ−= × =

MATHEMATICS

31. ( ) ( )2 21 4f x x x= + +

32. Multiplying both, get answer 2x y π θ+ = −

33. aRb bRa fails⇒

,aRb bRc aRc fails⇒

34. Cosine rule ,k=4

35. , 33

hAC BC h= =

2 2 2BC AC AB− = A

B

C

D h

36. ( )2 20 2 2 4 4 0,b a b a a b R∆ > ⇒ − − + + − > ∀ ∈ Again 0 1a∆ < ⇒ >

37. ( )

2

22

1 1 1 11 111n n nn

+ + = + − + +

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38. p xv= the 3

2n =

39. The number of natural numbers is 4373.

40. If is ‘0’ in 47

41. 3 1 1 14 3 3 12− − =

42. k=10017

43. Solving, the common points are 1 1sin 2 , sin 22 2

α α ± ±

,for both e=

21 tan α−

44. Arg= 02

z zπ=> + = => I is true z lies on ( )5z i− − = 5 => II is false

45. The circle is 2 2 5 15 02 2

x y x y+ − − + =

46. Take

and from the table.

47. ( )1 24,9 ,p p= = harmonic conjugate 1p max area = 1 21 12 2

AB PP×

48. L also lies in the plane 9x-2y-5z+4=0

49. 1 22

V abc = =

50. Expanding 0a b ra p b q c r

+ + =− − −

51. Two linear functions intersect at 30,2

52. ( )/2

1lim 1 sin sin2x

x f xπ→

− =

53. Comparing 1 1

, 2cos sin

x y x yax yθ θ

+ = + =

P Q

T F

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54. ( ) 3 26 9 2p x x x x= − + +

55. Since ( )' 1 0f ± = ,by rolle’s theorem no such k exists

56. Drawn both graph, two solutions is ( )2 ,π π− −

57. ln ydy

58. sec 1 xy x e= + +

59. ( )10 0 20x x− = ⇒ =

60. ( )( ) ( )1 1tan tan 6 2 6,cot cot 4 4π π− −− = − = −

CHEMISTRY 61. Bond enthalpies based on first law of thermodynamics.

62. Since there are six atoms (A) in the corner of the unit cell, the

contribution of atoms in 1 unit cell is 6/8. Since 3 face-centered atoms (B)

contribute to one unit cell, the formula is A6/8B3 or A6B24 or AB4.

63. MnO4- + 5e– + 8H+ → Mn2+ + 4H2O

Following Nernst equation, find out Ered(final) – Ered(initial). Thus Ered

decreases by 0.189 V. The tendency of the half cell to get reduced is its

oxidising power. Hence, the oxidising power decreases by 0.189 V.

64. Follows SN1 with conc.HI & SN

2 with anhyd.HI

65. More basic Nitrogen

66. When KI is added to AgI it would increase the concentration of I– which

is the negative part of the colloid and hence will give negatively charged

colloidal particle.

67. Cyanohydrin formation follows nucleophilic addition mechanism

68. NF3 and H3O+ have sp3 hybridisation; NO3– and BF3 have sp2

hybridisation.

69.

2BaCl Carbon2 2 4 4KO S K SO BaSO BaS∆

∆+ → → → (A)

(B) (C)

HCl

H2S (D)

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70. Step II, being r.d.s. Rate of overall reaction = Rate of Step II = KII

[O3][O]

substituting the value of [O] from the equilibrium of Step I, we get

Rate = KIIKC[O3]2/ [O2]

71.

72. Energy released in conversion of 6 × 1023 atoms of Cl– ions = 6 × 1023 ×

electron affinity

= 6 × 1023 × 3.61 = 2.166 × 1024 eV

Let x Cl atoms be converted to Cl+ ion.

Energy absorbed = x × ionization energy

⇒ x × 17.422 = 2.166 × 1024

⇒ x = 1.243 × 1023 atoms

73. Meq. of H2SO4 + Meq. of SO3 = Meq. of NaOH

⇒ [(0.5-x)/ (98/2) x 1000] + [x/(80/2) x 1000] = 26.7 × 0.4

⇒ x = 0.103

Percentage of SO3 = 0.103/ 0.5 x 100 = 20.6%

74. Influence of electronic configuration on electron gain enthalpies.

75. One mole HIO4 can oxidize 2 moles of alcoholic groups.

76. Polar medium sec. alkyl halide undergoes SN1 mechanism.

77. Na+ ions are not reduced at cathode and SO42- ions are not oxidized at

anode.

78. (CH3)4C has twelve equivalent 1°H. Hence, H forms only one product on

monobromination.

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79. It occurs during roasting in reverbetory furnace at moderate temperature

in presence of air.

80. A is Ipc2BH. C is a methyl ketone.

81. CH3CH(CH3)CH2CH3; it has four sets of equivalent hydrogens which

result in the formation of four different monochlorination product. When

chlorination takes place at first and third carbon, this results in the

formation of a racemic mixture.

82. Charge density = Charge / Volume

More the number of dx-px bonds, greater the delocalization of negative

charge, so lesser charge density

83. Chain, positional and stereo isomerism.

84. Optimum fluoride is below 3ppm. 10ppm fluoride will be very dangerous to human life.

85. Mole fraction = 0.07

Kf = 1.86

∆Tf = Kf m

solute

solute solvent

n 0.07n n

=+

solventn 0.93∴ =

of

0.07 1000T 1.86 7.780.93 18

×∆ = × =

×

86. (1) The magnitude of splitting energy increases down the group. Thus

Pt2+ – 5d8, Pd2+ – 4d8 and Ni2+ – 3d8.

(2) NH3 is a borderline ligand which forces pairing of electrons in Co3+,

whereas H2O is a weak-field ligand and it can not force the pairing in

Ni2+.

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Co3+ in [Co(NH3)6]3+ – 4s0, 3d6

Ni2+ in [Ni(H2O)6]2+ – 3d8, 4s0

So, there is no unpaired electron in [Co(NH3)63+] whereas there are

unpaired electron in [Ni(H2O)6]2+. Thus [Co(NH3)6]3+ is colourless and

[Ni(H2O)6]2+ is coloured due to d–d transition.

(3)

CH2

NH2

CH2

NH2

M(n – 1) +O C

O

CH2

NH2 As en is symmetrical ligand thus there is no geometrical isomerism.

(4) In K3[Fe(CN)6], the oxidation state of Fe is +3.

[Fe(CN)6]3– – 3d5, 4s0.

Number of unpaired electron = 1

∴µ = n(n 2)+ B.M. = 1(1 2)+ B.M. = 3 B.M.

Thus answer is (3).

87.

Volume of 2 mol of N2=272.0 cm3

.

88. Inorganic reactions are to be recollected properly.

89. Buffer capacity

= Number of moles of acid added per litre of buffer/Change in pH .

Buffer capacity = 0.0060.03

= 0.2

90. Lactone is formed

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