[Page 1]

Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005, Ph.011-47623456

Mock Test

for JEE (Main) - 2020

PART – A : PHYSICS

SECTION - I

Multiple Choice Questions: This section contains

20 multiple choice questions. Each question has 4

choices (1), (2), (3) and (4), out of which ONLY

ONE is correct.

Choose the correct answer :

1. Linear momentum of a particle moving along

a straight line as a function of time is given as 3–

0tp p e ; where p0 and are constants.

Time is measured with a stop watch of least

count 10–2 s and value of is

1 s–3. The percentage error in the

measurement of p at t = 1 s is

(1) 2.5% (2) 3.0%

(3) 1.5% (4) 1%

2. The graph of velocity (v) versus time (t) for a

particle of mass 1 kg moving along a straight

line is as shown. The slope of kinetic energy

versus position graph of the particle at t = 3 s

in SI units is

(1) 12

(2) 6

(3) 3

(4) 4

3. A uniform disc is given a linear velocity v0 on a rough surface as shown. Regarding angular momentum of the disc, mark the correct option

Time : 3 hrs MM : 300

General Instructions:

1. Duration of Test is 3 hrs.

2. The Test booklet consists of 75 questions. The maximum marks are 300.

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

having 25 questions in each part of equal weightage. Each part has two sections.

(i) Section-I : This section contains 20 multiple choice questions which have only one correct answer. Each

question carries 4 marks for correct answer and –1 mark for wrong answer.

(ii) Section-II : This section contains 5 questions. The answer to each of the questions is a

numerical value. Each question carries 4 marks for correct answer and there is no negative

marking for wrong answer.

A CODE

Mock Test (Code-A)

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(1) Angular momentum of disc remains

conserved about point C

(2) Angular momentum of disc is conserved about point O

(3) Angular momentum of disc is conserved about any point lying on a vertical line passing through point O

(4) Angular momentum is not conserved about any point

4. A cylindrical vessel of radius R and height H is half filled with liquid. It is rotated with

angular speed about the central axis as shown. The free surface of the liquid takes

paraboloidal shape. The value of for which

vertex of the parabola is at a distance 2

H

below P is

(1) 2gH

R (2)

3gH

R

(3) gH

R (4)

3

gH

R

5. Consider three travelling waves

1 sin –3

y A kx t

2 sin – –3

y A kx t

3 2 cos –y A kx t

The ratio of intensity of first wave and the

wave formed due to superposition of all three

waves is

(1) 1

5 (2)

1

2

(3) 1

9 (4)

1

3

6. Two short electric dipoles having dipole moment p are kept on two corners of a rectangle as shown in figure. The direction of electric field at the centre C of rectangle, is

(1) Along –Y-axis (2) Along +Y-axis

(3) Along +X-axis (4) Along –X-axis

7. Acceleration-time graph of a train moving in

straight line is as shown in graph. If its initial

speed is zero then its maximum speed is

(1) 80 m/s (2) 40 m/s

(3) 25 m/s (4) 55 m/s

8. A particle is projected along an inclined plane

as shown in figure. After what time the

particle will hit the inclined plane again?

(g = 10 m/s2)

(1) 1 s (2) 2 s

(3) 3 s (4) 1

s3

9. Time constant of the given L-R circuit is

(1) 7

4

L

R (2)

L

R

(3) 7

11

L

R (4)

3

11

L

R

10. If potential at the surface of earth is taken

to be zero, then potential at the centre of

earth will be (Mass of earth = m and radius of

earth = R)

Mock Test (Code-A)

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(1) 2

Gm

R (2) Zero

(3) Gm

R (4)

2Gm

R

11. 2 moles of a monoatomic ideal gas is mixed

with 3 moles of a diatomic ideal gas. Molar

specific heat of mixture at constant volume is

(1) 3

10

R (2)

11

10

R

(3) 13

10

R (4)

21

10

R

12. A gas undergoes the cyclic process as shown

in figure. The cycle is repeated 25 times per

second. Determine the power generated

(1) 600 W (2) 300 W

(3) 450 W (4) 750 W

13. A convex lens of focal length 20 cm is cut into

two equal parts as shown in figure. An object

is placed at 30 cm from point C. What should

be the distance between the images formed

by the two halves of the lens?

(1) 6 cm

(2) 12 cm

(3) 8 cm

(4) 9 cm

14. A monochromatic light source of frequency f

illuminates a metallic surface and ejects

photoelectrons. The most energetic

photoelectrons are just able to ionise the

hydrogen atom in ground state. When

frequency of incident radiation is made 5

,6

f

the most energetic photoelectrons excites the

H-atom in ground state to first excited state.

The work function of the metal is

(1) 1.13 eV (2) 6.8 eV

(3) 3.4 eV (4) 2 eV

15. A person uses +1.5 D glasses to have normal

vision from 25 cm onwards. He uses a 20 D

lens as a simple microscope to see an object.

The maximum magnifying power if he uses

the microscope together with his glasses and

without the glasses are respectively

(1) 6, 3 (2) 5, 8

(3) 4, 6 (4) 6, 9

16. A force of 10 N is required to keep a ball of

weight 70 N, completely immersed in a liquid.

What fraction of volume of the ball is outside

the liquid, if it floats by itself?

(1) 7

8

(2) 1

8

(3) 1

7

(4) 6

7

17. Let t1 be the time interval in which a particle

executing SHM moves from a position,

mid-way between the mean and extreme

position, to the extreme position. Let t2 be the

time interval in which the particle executing

SHM moves from extreme position to a

position mid-way between extreme and mean

position. Then 1

2

t

t

(1) 1 (2) 2

(3) 1

2 (4)

1

3

18. The equation of state for a gas is given by

PV = nRT + V, where n is the number of

mole and is constant. Initial pressure and

temperature of 1 mole of the gas contained in

Mock Test (Code-A)

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a cylinder is P0 and T0 respectively. The work

done by the gas when its temperature

doubles isobarically will be

(1) 0 0

0 –

P T R

P (2) 0 0

0

P T R

P

(3) P0T0R ln2 (4) 0 0

0

ln 2P T R

P

19. When a charge q is kept at (–a, 0, 0), the

electric field vector at (0, b, 0) is ˆ .̂ai bj Now,

another charge 2q is also kept at (a, 0, 0).

The net electric field vector at (0, b, 0) is

(1) ˆbj (2) ˆ ˆ2ai bj

(3) ˆ ˆ3ai bj (4) Zero

20. A rod of length l and cross-sectional area A

has a variable thermal conductivity given by

K = aT, where a is constant and T is

temperature in kelvin. Two ends of rod are

maintained at temperature T1 and T2.

(T1 > T2). Heat current (Q) flowing through rod

will be

(1) 2 21 2–

aAQ T T

l

(2) 2 21 2–

2

aAQ T T

l

(3) 21 2–2

aAQ T T

l

(4) 21 2–aA

Q T Tl

SECTION - II

Numerical Value Type Questions: This section

contains 5 questions. The answer to each question

is a NUMERICAL VALUE. For each question,

enter the correct numerical value (in decimal

notation, truncated/rounded‐off to the second

decimal place; e.g. 06.25, 07.00, 00.33, 00.30,

30.27, 27.30) using the mouse and the on-screen

virtual numeric keypad in the place designated to

enter the answer.

21. A small ball of mass m = 100 gm and radius

r = 5

R rolls without slipping along the track

shown in figure. Radius of circular part of the

track is R. If the ball starts from rest at a

height of 8R above the bottom, the normal

force on the ball at point P is ________N.

(g = 9.80 m/s2)

22. A tuning fork produces 5 beats/s with another

tuning fork of frequency 80 Hz. The first fork is now loaded with a little wax and the beat frequency is again found to be 5 beats/s. The frequency of first tuning fork before waxing was ______Hz.

23. In the modified YDSE arrangement shown in

the figure, y0 = 3

D

d

, where is wavelength

of light used. The ratio of maximum to minimum intensity observed on screen is n. The value of 5n is_______.

24. The spherical balls A, B and C of equal size

are joined and placed inside a liquid as

shown in figure. Density of ball A and B are

equal to and that of liquid is 2If the balls

stay at equilibrium, then density of ball C is

n. The value of 50

n is______.

25. A particle starts moving in straight line with

constant acceleration. After 10 seconds,

acceleration changes its sign (Opposite to the

initial direction) remaining the same in

magnitude. The time (in seconds) from the

beginning of motion in which the particle

returns to the initial positions is _______.

2 1.414

Mock Test (Code-A)

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PART – B : CHEMISTRY

SECTION - I

Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct.

26. If radii of B3+ ions and O2– ions are 27 pm and 140 pm respectively, then which of the following should be the correct crystal structure of boric anhydride? [‘o’ represents O2– ions while ‘ ’ represents B3+ ions)

(1)

(2)

(3)

(4) Hexagonal close packed structure of oxide

ions in which two-thirds of the octahedral voids are occupied by Boron atoms

27. S3O9 is a trimer of SO3. Number of S – S bonds

in S3O9 is

(1) Three (2) Two

(3) One (4) Zero

28. Correct order of bond strength is

(1) 3s – 3s > 3p– 3p> 3p– 3d

(2) 3s – 3s > 3p– 3d> 3p– 3p

(3) 3p – 3p > 3s– 3s> 3p– 3d

(4) 2s – 2s > 2p– 3s> 2p– 2s

29. In an exothermic reaction A B, the Ea of

reaction is 120 kJ/mole of A. The enthalpy of reaction is –140 kJ/mole. The activation energy of the reaction B A is

(1) 20 kJ/mole (2) 260 kJ/mole

(3) 360 kJ/mole (4) 120 kJ/mole

30. Ionic hydride among following is

(1) BeH2 (2) NH3

(3) CaH2 (4) CuH2

31. In which type of binary solution formation,

Hmixing is positive?

(1) Ideal solution

(2) Non-ideal solution showing positive deviation

(3) Non-ideal solution showing negative deviation

(4) Ideal as well as non-ideal solution

32. Solution of which of the following species is most basic?

(1) (2) NH4Cl

(3) (4) NaCl

33. Consider the following reactions involving Williamson’s Ether Synthesis

(i)

(ii)

(iii)

(iv)

Which of the above reactions constitute good synthesis?

(1) (i), (ii) & (iv) (2) (ii) & (iii) only

(3) (iii) & (iv) only (4) (i) & (ii) only

CH3 C

O

ONa

CH3 C

O

ONH4

– +

NaONa

CH CH Br3 2OH

O–CH CH2 3

OHNaOH

ONaCH – Br3

OCH3

CH3

CH3

OHH C3

Na

CH3

CH3

ONaH C3

Br

CH3

CH3

OH C3

CH OH3

NaCH ONa3

CH3

CH3

BrCH3

CH – O 3

CH3

CH3

CH3

Mock Test (Code-A)

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34. Match column-I with column-II and select the correct answer using the codes given below the lists

Column-I Column-II

a. (i) DIBAL-H

b. (ii) (CH3)2Cd

c. (iii) 2 4N H /OH

d. (iv) LiAlH4

(v) NaOH/Br2

(1) a(i), b(ii), c(iii), d(iv)

(2) a(ii), b(iii), c(i), d(iv)

(3) a(i), b(ii), c(iii), d(v)

(4) a(ii), b(i), c(iii), d(v)

35. A layered sequence for an fcc arrangement of metal is shown below : (where 'a' is the edge length of cube)

A face diagonal that passes through the centre

of atom 'd' will not pass through the centre of which other atoms?

(1) b, e (2) h, l

(3) i, j (4) g, k

36. Which of the following reaction is accompanied with emission of proton?

(1) 10 4 135 2 7B He N

(2) 14 4 177 2 8N He O

(3) 241 4 24495 2 97Am He Bk

(4) 2 3 41 1 2H H He

37. According to Lewis theory, which compound has expanded octet?

(1) BF3 (2) H2O

(3) AlCl3 (4) PCl5

38. Ethylbenzene can be prepared from benzene by the following two methods.

Method (II) is preferable because

(1) Acylation is slower than alkylation

(2) Alkylation is accompanied with polysubstitution

(3) The yield is greater in method (I)

(4) In this alkylation, rearrangement is possible

39. The magnetic moment (spin only value) of Co3+ ion is

(1) 1.732 BM (2) 4.90 BM

(3) 2.828 BM (4) 3.873 BM

40. Which of the following species does not give positive Fehling's reagent test?

(1) Formaldehyde

(2) Acetaldehyde

(3) Benzaldehyde

(4) All of these give positive test

41. Select the incorrect statement regarding the blue vitriol.

(1) Colour of anhydrous blue vitriol is black

(2) It cannot be prepared by dissolving copper in HCl

(3) It contains ionic, covalent, coordinate and hydrogen bonds

(4) Blue colour is due to crystal field splitting due to H2O ligand

42. How much quick lime is obtained on heating 200 kg of 95% pure lime stone?

(1) 106.4 g (2) 106.4 kg

(3) 10.64 g (4) 10.64 kg

43. A decimolar solution of potassium ferrocyanide is 50% dissociated at 300 K. Calculate osmotic pressure of solution.

(1) 2.46 atm (2) 7.38 atm

(3) 1.84 atm (4) 0.61 atm

Cl

O

C H

OC

Cl

O

C CH3

O

CH3

O

C

NH2

O

C NH2

(I) + C H Cl2 5

Anh. AlCl3

CH CH2 3

(II) + CH COCl3

Anh. AlCl3

COCH3

Zn(Hg)

HCl

CH CH2 3

Mock Test (Code-A)

[Page 7]

44. Compressibility factor for CO2 at 400 K and 5

bars is 0.9. Molar volume of CO2 under these

conditions is (approx)

(Take R = 0.08 atm L K–1mol–1)

(1) 22.4 L (2) 2.24 L

(3) 5.8 L (4) 19.5 L

45. For the reaction 2SO3(g) 2SO2(g) + O2(g)

initial concentration of SO3 is 1 M. At equilibrium the vapour density is 30. Calculate the equilibrium constant in closed container.

(1) 3/4 (2) 1/4

(3) 2/3 (4) 4/3

SECTION - II

Numerical Value Type Questions: This section

contains 5 questions. The answer to each question

is a NUMERICAL VALUE. For each question,

enter the correct numerical value (in decimal

notation, truncated/rounded‐off to the second

decimal place; e.g. 06.25, 07.00, 00.33, 00.30,

30.27, 27.30) using the mouse and the on-screen

virtual numeric keypad in the place designated to

enter the answer.

46. The volume of O2 liberated at S.T.P from 1.5 ml

of 11.2 volume H2O2 (in mL) is

47. For the following graph, it is given that

= 45º

OP = 10

If the value of H° is X cal/mol then the value of

10X is (to the nearest integer)

48. pH of 1 ml of a strong monobasic acid

increases by 1 unit on mixing x ml H2O. The

value of x is

49. 100 ml solution of NaOH (containing 4 g NaOH

per litre) and 50 ml of HCl (containing 7.3 g HCl

per litre) react as

NaOH (aq) + HCl (aq) NaCl (aq) + H2O (l)

at any instant 0.5265 g of NaCl is formed. Thus

unreacted HCl is (in mg) (Given atomic mass of

Na = 23 u, O = 16 u, Cl = 35.5 u)

50. Consider the reaction :

Trans-2-butene 2 2Br /H O A (major product).

Number of chiral carbon in the product A is

PART – C : MATHEMATICS

SECTION - I

Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct.

51. A matrix is chosen at random from the set of

all matrices of order 2 with elements 0, 1 or 2.

The probability that the matrix chosen is

singular, is

(1) 35

81

(2) 31

81

(3) 32

81

(4) 50

81

52. If –1 –1cot ( cos ) – tan ( cos ) u , then

sinu is equal to

(1) 2cot2

(2) 2tan

2

(3) 2sin2

(4) 2cos

2

53. The number of value(s) of x satisfying the equation

2 3 2log (log (log )) 8 8x x x is

(1) Zero (2) 1

(3) 3 (4) 4

54. 2 5x nyy ex e

, then x e

dy

dx

is

(1) 3 12

e

ee

(2) 3 1 1

3 6

e

ee

Mock Test (Code-A)

[Page 8]

(3) 3

e

e (4) 1

2 eee

55. If 2 2 1( ) ( ) ( ) ,k kf x x m x n m n , then

(where k Z ) which of the following must be

true for function f (x)?

(1) x = m is point of maxima

(2) x = n is point of maxima

(3) x = m is point of neither maxima nor

minima

(4) x = n is point of neither maxima nor

minima

56. The number of point(s) of intersection of the

curves y = cosec–1 x and y = x is

(1) 0 (2) 1

(3) 2 (4) 3

57. The area of the quadrilateral formed by

the tangents at the vertices of pair of conics

x2y2 = 4, is

(1) 4 sq. units (2) 8 sq. units

(3) 16 sq. units (4) 32 sq. units

58. (p ~q) (~p q) is

(1) A tautology

(2) A contradiction

(3) Neither a tautology nor a contradiction

(4) Equivalent to p q

59. If 1, log9(31–x + 2), log3(4·3x – 1) are in A.P,

then x is equal to

(1) log34 (2) 1 – log34

(3) 1 – log43 (4) log43

60. If 0 0

1, then

n n

n n nr rr r

ra

C C

equals

(1) (n – 1)an (2) nan

(3) 1

2 nna (4) 1

4 nna

61. The mean and standard deviation of 1, 2, 3,

4, 5, 6 are respectively

(1) 7 35

,2 12

(2) 3, 3

(3) 7

, 32

(4) 35

3,12

62. Let F denotes the set of all onto functions from A = {a1, a2, a3, a4} to B = {x, y, z}.

A function f is chosen at random from F. The probability that f –1(x) exists

(1) 2

3 (2)

1

3

(3) 1

6 (4) Zero

63. The value of ln ln lnx a a x a ae e e dx is

equal to (where a > 0, x > 0)

(1) 1

ln1

ax ax

a a a x ca

(2) 1

ln 1

x aaa x

a x ca a

(3) axlna + xalnx + axx + c

(4) lnln

xa aa

x x a x ca

64. The angle between the two lines represented by 2x2 + 5xy + 3y2 + 2y – 8 = 0 is tan–1m. Then m is equal to

(1) 1

5 (2)

7

5

(3) 5 (4) 7

65. The set of values of for which exactly one

root of the equation x2 – x + 1 = 0 lies

between 1 and 2, is

(1) (1, 2) (2) (–2, –1)

(3) (–5, –2) (4) 5

2,2

66. The number of complex numbers z, satisfying the equations | z + 5 | + | z – 5 | = 10 and |z + 1| = 2, is

(1) Zero (2) 1

(3) 2 (4) Infinite

67. If 2 3 , – 2 , 2 ,a i j k b i j k c j k

then

a a a b a c

b a b b b c

c a c b c c

equals

Mock Test (Code-A)

[Page 9]

(1) 2 (2) 4

(3) 8 (4) 16

68. Equation of a plane passing through the

points

(2, 1, –1) and (1, 1, –2) and perpendicular to

the plane x + 2y + 3z = 4, is

(1) x + 2y + z – 3 = 0

(2) x + y – z – 4 = 0

(3) 2x + y + z – 4 = 0

(4) 2x + y – z – 5 = 0

69. The integrating factor of the differential

equation sin2x dy

dx + y = cot x, is

(1) ex (2) e

(3) ecotx (4) e–cotx

70. Total number of arrangements of the letters

of the word 'IITKANPUR' taking all at a time

so that all vowels come together is

(1) 8640

(2) 4320

(3) 17280

(4) 8760

SECTION - II

Numerical Value Type Questions: This section contains 5 questions. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal

notation, truncated/rounded‐off to the second

decimal place; e.g. 06.25, 07.00, 00.33, 00.30,

30.27, 27.30) using the mouse and the on-screen

virtual numeric keypad in the place designated to enter the answer.

71. The value of c2 for which the lines joining the origin to the points of intersection of the line

y = 3x c and the circle x2 + y2 = 2 are

perpendicular to each other, is _________.

72. The value of 2

0

{3 }

4

xdx , where {.} denotes the

fractional part function, is_________.

73. 1 cos(7( ))

lim( )x

x

x

is equal to _______.

74. The number of solution(s) of the system of equations x + 4y – z = 0, 3x – 4y – z = 0 and x – 3y + z = 0 is_______

75. The area of the region bounded by the curves

1 – y2 = |x| and |x| + |y| = 1 is A, then 12A

equals ________.

[Page 1]

Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005, Ph.011-47623456

Mock Test

for JEE (Main) - 2020

ANSWERS

PHYSICS CHEMISTRY MATHEMATICS

1. (2)

2. (1)

3. (1)

4. (2)

5. (1)

6. (1)

7. (1)

8. (3)

9. (3)

10. (1)

11. (4)

12. (1)

13. (2)

14. (2)

15. (4)

16. (2)

17. (1)

18. (1)

19. (3)

20. (2)

21. (12.25)

22. (85.00)

23. (45.00)

24. (12.50)

25. (34.14)

26. (2)

27. (4)

28. (2)

29. (2)

30. (3)

31. (2)

32. (1)

33. (4)

34. (3)

35. (4)

36. (2)

37. (4)

38. (2)

39. (2)

40. (3)

41. (1)

42. (2)

43. (2)

44. (3)

45. (4)

46. (16.80)

47. (46.00)

48. (09.00)

49. (36.50)

50. (02.00)

51. (2)

52. (2)

53. (2)

54. (2)

55. (3)

56. (3)

57. (3)

58. (2)

59. (2)

60. (3)

61. (1)

62. (4)

63. (2)

64. (1)

65. (4)

66. (3)

67. (4)

68. (2)

69. (4)

70. (1)

71. (04.00)

72. (00.25)

73. (00.00)

74. (01.00)

75. (08.00)

Time : 3 hrs MM : 300

A CODE

Mock Test(Code-A)

[Page 2]

Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005, Ph.011-47623456

Mock Test

for JEE (Main) - 2020 HINTS & SOLUTIONS

PART – A : PHYSICS 1. Answer (2)

3–

0tp p e

3– 2

0 (–3 )tdpp e t

dt

2– 3dp

t dtp

= –3 × 1 × 1 × 10–2 = –0.03

% error = 3%

2. Answer (1)

21

2K mv

dK dV

mvdS dS

3. Answer (1)

Angular momentum is conserved only about C since torque of friction is zero only about that point.

4. Answer (2)

2 2

2

xy

g

,

2 23

2 2

H R

g

, =

3gH

R

5. Answer (1)

I1 A2

I2 = Iresultant 25A

Ratio = 1

5

6. Answer (1)

When –1tan 2

E p

7. Answer (1)

V adt

8. Answer (3)

2 sin30

cos30

UT

g

9. Answer (3)

Time constant of the circuit is

= Equivalent resistance across

after shorting the battery

L

L =

7

11

L

R

10. Answer (1)

A CODE

Mock Test(Code-A)

[Page 3]

V(centre) – V(surface) = 2

Gm

r

11. Answer (4)

1 1 2 2mix

1 2v

n C n CC

n n

12. Answer (1)

W = (P2 – P1) (V2 – V1)

= 24 J

P = W × 25 = 600 W

13. Answer (2)

1 1 1

30 20V

V = 60 cm

distance = 2(4) + 4

= 12 m

14. Answer (2)

hf = W + KEmax

For first case KEmax = 13.6 eV

5

6

hf = W + KEmax

For second case KEmax = 10.2 eV

5 5 5

13.66 6 6

hf W

5

10.26

hf W

0 – 1.1336

W ,

W = 6.8 eV

15. Answer (4)

The focal length of the glass (lens) used 100

m,1.5

f if y is the distance of near point,

then

1 1 1.5

––25 100y

{y = –40 cm}

The focal length of lens of microscope

fe = 1 100

20P = 5 cm

Magnifying power of microscope together with glass

M = 1e

D

f =

251

5 = 6

In case of without glass, D = 40 cm = y

M = 1e

D

f

=

401

5 = 9

16. Answer (2)

Vg = 70 N

Vg = 80 N

7

8

1

18

n

17. Answer (1)

Time taken does not change.

18. Answer (1)

P0V0 – V0 = nRT0

V0 = 0

0( – )

nRT

P , n = 1

w = P0V = 0 0

0( – )

P RT

P

19. Answer (3)

1 2E E E

1E ai b j

2 2 2E ai b j

3E ai b j

20. Answer (2)

Q = –dT

TAdx

0

lQdx

A = 2

1

–T

T

T dT

Q = 2 21 2( – )

2

AT T

l

21. Answer (12.25)

7mgR = 2

22

1 7

2 5

vmr

r

v = 10gR

N = 2mv

R r =

10 50mg

4 45

mgRR

22. Answer (85.00)

|n – 80| = 5

n = 85 or 75

23. Answer (45.00)

0 3

DY

d

04PI I

Mock Test(Code-A)

[Page 4]

24. Answer (12.50)

Use weight = Buoyant force

25. Answer (34.14)

t = 2 2 t0

PART – B : CHEMISTRY 26. Answer (2)

Since radius ratio r

0.193r

which shows

that B3+ ions are present in triangular voids. Also in option (2), each corner of triangle is shared by two unit cells. So, no. of boron atoms per unit cell = 1, Number of O2– ions per unit

cell = 1 3

32 2

So, formula of the solid is B1O3/2 or B2O3.

27. Answer (4)

28. Answer (2)

3s-3s overlap results in the formation of sigma

bond. Relative strength of 3p-3p and 3p -

3d is based on the extent of overlapping which

is higher in 3p -3d bond.

Correct order of bond strength is :

3s – 3s > 3p – 3d > 3p – 3p

29. Answer (2)

H = Ea(F) – Ea(R)

Ea(R) = 120 + 140 = 260 kJ

30. Answer (3)

BeH2 is covalent hydride.

31. Answer (2)

In case of non-ideal solution showing +ve

deviation: Hmix > O

32. Answer (1)

Salt of weak acid and strong base are most basic in nature.

33. Answer (4)

34. Answer (3)

35. Answer (4) If the layers are superimposed, the unit cell will

appear as shown below and the face diagonals originating from the centre of atom d are the lines connecting the centres of atoms.

d e b

d h l

d i j

36. Answer (2)

14 4 17 17 2 8 1N He O H

37. Answer (4)

PCl5 has expanded octet.

O

SO

S OO

O O

OS

O O

Pot

en

tial e

nerg

y

Progress of reaction

R

P

E (R)aE (F)a

H

CH3

CH — C — Br + CH ONa3 3

CH3

C CH2

Me

Me

C — Cl CHODiBAL-H

O

C — CH 3

NH – NH2 2

O

–OH

C — NH 2 NH2

Br /aq NaOH2

O

Mock Test(Code-A)

[Page 5]

38. Answer (2)

is more preferable as alkylation may lead to polysubstitution.

39. Answer (2)

Co+3 [Ar] 4s0 3s6

Number of unpaired electrons = 4

Magnetic moment = n n 2

= 4 4 2

= 24

= 4.9 BM

40. Answer (3)

Aromatic aldehydes do not give Fehling's test.

41. Answer (1)

Anhydrous CuSO4 is white.

42. Answer (2)

Amount of quick lime obtained = 200

56 0.95100

= 106.4 kg

43. Answer (2)

Osmotic pressure () = iCRT

2

4 6 6K Fe CN 4K Fe CN

1 4 CRT

= 1

3 0.0821 30010

= 7.38 atm

44. Answer (3)

0.9 0.08 400 1.01325

Molar volume5

5.8 L

45. Answer (4)

2SO3 2SO2 + O2

1 – 2

e i

i e

n M=

n M

80

12 30 2

4

12 3

1

2 3

2

3

[SO3] = 1

3

[SO2] = 2

3

[O2] = 1

3

2

C 2

2 143 3K =31

3

46. Answer (16.80)

Volume of O2 liberated = 1.5 × 11.2 = 16.8 mL

47. Answer (46.00)

G° = H° – TS° = –2.303 RT log Kc

cH 1 S

logK .2.303R T 2.303R

Slope = H

tan1352.303R

H

12.303R

H° = 2.303 × 2 = 4.606 cal/mol

48. Answer (09.00)

pH = 1

Change in pH by 1 unit when solution is diluted 10 times.

Volume of H2O added = 9 mL

49. Answer (36.50)

Milli moles of NaOH in sol = 10

Milli moles of HCl in sol = 10

At any instant millimoles of NaCl

formed = 30.526510

58.5 = 9

Amount of HCl left = 36.5 mg

50. Answer (02.00)

Number of optical centre = 2

+ CH3 C

O

Cl/AlCl3 C

O

CH3

Zn – HgHCl

CH2 CH3

C

C

H CH3

CH3 H

Br /H O2 2

CH Br

CH OH

Me

Me

Mock Test(Code-A)

[Page 6]

PART – C : MATHEMATICS

51. Answer (2)

n(S) = 3 × 3 × 3 × 3 = 81

n(E) = all four elements are 0

+ exactly three elements are 0

+ exactly two elements are 0

+ no element is 0

n(E) = 1 + (4C3 × 2C1) + (4 × 2 × 2) + (1 + 1 + 4)

= 1 + 8 + 16 + 6 = 31

31

81

n EP E

n S

52. Answer (2)

1 1cot cos tan cos u

12 tan cos2

u

1sin cos 2tan cos cos2 ,u

Let 1tan cos

tan cos

tan2 = cos

cos2 = 2

2

1 tan 1 cos

1 cos1 tan

cos2 =

2

2

2

1 1 2sin2 tan

21 2cos 12

sinu = 2tan2

53. Answer (2)

x – 8 0, 8 – x 0

x = 8

when x = 8. L.H.S = 0, R.H.S = 0

x = 8 is only solution

54. Answer (2)

2 5lnx yy ex e

5

22 5ln

lnln

xy x y

y

ex e

e

y6lnx = e2x ...(i)

6

5 2ln 6 2 xy dyx y e

x dx

...(ii)

From (i), put x = e, y6 = e2e

3

e

y e

From (ii), 52

226 2e

ee dye e

e dx

31

26

e

dy e

dx e

55. Answer (3)

f (m) = 0, f (m–) < 0, f (m+) < 0

x = m is point of inflection for y = f (x)

f (n) = 0, f (n–) < 0, f (n+) > 0

x = n is point of local minima for y = f (x)

56. Answer (3)

Number of intersection points of y = cosec–1x and y = x is two.

57. Answer (3)

x2y2 = 4

xy = ±2

Vertices are 2, 2

Equation of tangent to xy = 2 at 2, 2 is

12 2 2

2x y

Mock Test(Code-A)

[Page 7]

2 2x y

2 2, 0A and 0, 2 2B

Area of quadrilateral

14 2 2 2 2 16

2ABC sq. units

58. Answer (2)

(p q) (p q)

p p = F

q q = F

So, p q p q = F (contradiction)

59. Answer (2)

2log9(31–x + 2) = log33 + log3(4.3x – 1)

1

3

3

2log (3 2)

log 9

x = log33 + log3(4.3x – 1)

log3(31–x + 2) = log3 3(4.3x – 1)

31–x + 2 = 12·3x – 3

3 + 2·3x = 12.(3x)2 – 3.3x

12.(3x)2 – 5·3x – 3 = 0

3x = 3 1

,4 3

(Impossible)

x = 33

log4

= 1 – log34

60. Answer (3)

If 0

1,

n

n nr r

aC

let 0

n

nr r

ry

C

0 1 2

1 1 1 1...n n n n n

n

aC C C C

Now 0 1 2 3

0 1 2 3...

n n n n nn

ny

C C C C C

1 2 0

1 2 0...

n n n nn n n

n n ny

C C C C

nCr = nCn – r

0 1 2

02 ...

n n n nn

n n ny

C C C C

2y = n[an]

2 nn

y a

61. Answer (1)

Mean = 1 2 3 4 5 6 6 7 7

3.56 2 6 5

Variance =

2 2 2 22 5 3 1

6 2 2 2ix x

n

2 1 2579 1 35

3 4 12

Now, standard deviation = 2 35

12

62. Answer (4)

a1

a2 x

a3 y

a4 z

For f–1(x) to have exactly one element, exactly one element from set A should be present but in that case it will not be a function

Hence probability = 0

63. Answer (2)

ln ln lnx a a x a ae e e dx

= 1

.ln 1

x ax a a aa x

a x a dx a x ca a

64. Answer (1)

22

tanh ab

a b

5

, 2, 32

h a b

252 6

2 14tan5 10 5

1

5m

65. Answer (4)

Let f(x) = x2 – x + 1

Exactly one root in (1, 2), f(1) f(2) < 0

2 < < 5

2

66. Answer (3)

| z1 – z2 | = 2a

Locus of z is a line segment joining the

points (5, 0) and (–5, 0)

and |z + 1| = 2 represents a circle with centre

(–1, 0) and radius 2 units.

Total number of common points = 2.

Mock Test(Code-A)

[Page 8]

67. Answer (4)

1 2 3

–1 2 1 4

0 1 2

a b c

2

16a bc

68. Answer (2)

The plane is passing through the points A(2, 1, –1) and B(1, 1, –2).

2 1 1

1 1 2 0

1 2 3

x y z

x y z

x + y – z – 4 = 0

69. Answer (4)

dy

dx + cosec2x y = cotx cosec2x

2cosec cotIFx dx xe e

70. Answer (1)

71. Answer (04.00)

x2 + y2 – 2 = 0; 3y x

c

= 1

x2 + y2 – 2·12 = 0

2

2 2 32

y xx y

c

= 0 for perpendicular

Coeff. of x2 + coeff. of y2 = 0

i.e. 2 2

2 61 1

c c = 0 2 =

2

8

c c2 = 4

72. Answer (00.25)

12 3

0 0

3 6 3I x dx x dx

{3x} is periodic with 1

3T

1 1 123 3 3

00 0

6 3 3 6 3 0 182

xI x x dx x dx

= 1 1

18 12 9

73. Answer (00.00)

1 cos 7

limx

xl

x

Form : 0

0, using L.H. rule

= 7sin 7

lim 01x

x

74. Answer (01.00)

1 4 1

3 4 1

1 3 1

= 1(–4 – 3) –4(3 + 1) – 1(–9 + 4)

= –7 – 16 + 5 = –18

System of equations has trivial solution

75. Answer (08.00)

Required area

= 1

2

1

12 1 4 1 1

2y dy

=

13

02 2 2

3

yy

= 2 2

4 23 3

sq. units