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FIITJEE Solutions to JEE(Main)-2020

PHYSICS, CHEMISTRY & MATHEMATICS

Time Allotted: 3 Hours

Maximum Marks: 300

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

Important Instructions:

1. The test is of 3 hours duration. 2. This Test Paper 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 out of which 20 questions are MCQs and 5 questions are numerical value based. Each question is allotted 4 (four) marks for correct response.

4. (Q. No. 01 – 20, 26 – 45, 51 – 70) contains 60 multiple choice questions which have only one correct

answer. Each question carries +4 marks for correct answer and –1 mark for wrong answer. 5. (Q. No. 21 – 25, 46 – 50, 71 – 75) contains 15 Numerical based questions with answer as numerical

value. Each question carries +4 marks for correct answer. There is no negative marking. 6. Candidates will be awarded marks as stated above in instruction No.3 for correct response of each

question. One mark 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 box.

7. There is only one correct response for each question. Marked up more than one response in any

question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.

Paper - 1

Test Date: 4th September 2020 (Second Shift)

http://www.fiitjee.com.

JEE-MAIN-2020 (4th September-Second Shift)-PCM-2


PART –A (PHYSICS) 1. A capacitor C is fully charged with voltage V0. After disconnecting the voltage source, it

is connected in parallel with another uncharged capacitor of capacitance. C2

. The

energy loss in the process after the charge is distributed between the two capacitor is:

(A) 20

1 CV6

(B) 20

1 CV2

(C) 20

1 CV4

(D) 20

1 CV3

2. In photoelectric effect experiment, the graph of stopping

potential V versus reciprocal of wavelength obtained is shown in the figure. As the intensity of incident radiation is increased:

(A) Slope of the straight line get more steep (B) Straight line shifts to left (C) Straight line shifts to right (D) Graph does not change

3. Consider two uniform discs of the same thickness and different radii R1 = R and R2 = R

made of the same material. If the ratio of their moments of inertia I1 and I2, respectively, about their axes is I1 : I2 = 1 : 16 then the value of is:

(A) 4 (B) 2 (C) 2 (D) 2 2 4. A circular coil has moment of inertia 0.8 kg m2 around any diameter and is carrying

current to produce a magnetic moment of 20 Am2. The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of 4 T is applied along the vertical, it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by 60° will be:

(A) 10 rad s1 (B) 20 rad s1 (C) 20 rad s1 (D) 10 rad s1 5. Find the binding energy per neucleon for 120

50 Sn . Mass of proton mp = (A) 1.00783 U, mass of neutron mn = (A) 1.00867 U and mass of tin nucleus mSn = 119.902199 U. (take 1U = 931 MeV)

(A) 8.5 MeV (B) 9.0 MeV (C) 7.5 MeV (D) 8.0 MeV 6. A quantity X is given by (IF2/WL4) in terms of moment of inertia I, force F, velocity ,

work W and Length L. The dimensional formula for x is same as that of: (A) coefficient of viscosity (B) planck’s constant (C) energy density (D) force constant




7.

For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about

the axes perpendicular to the sheet and passing through O (the centre of mass) and O (corner point) is:

(A) 1/2 (B) 2/3 (C) 1/4 (D) 1/8 8. Identify the operation performed by the circuit given below:

(A) AND (B) NOT (C) NAND (D) OR 9. A particle of charge q and mass m is subjected to an electric field E = E0(1 – ax2) in the

x-direction, where a and E0 are constants. Initially the particle was at rest at x = 0. Other then the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is:

(A) 3a

(B) 1a

(C) a (D) 2a

10. The value of current i1 flowing from A to C in the circuit diagram is:

(A) 4 (B) 2 (C) 5 (D) 1




11. Two identical cylindrical vessels are kept on the ground and each contain the same

liquid of density d. The area of the base of both vessels is S but the height of liquid in one vessel is x1 and in the other, x2. When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is :

(A) 22 13 gdS x x4

(B) 22 1gdS x x

(C) 22 11 gdS x x4

(D) 2 22 1gdS x x

12. The driver of bus approaching a big wall notices that the frequency of his bus's horn

changes from 420 Hz to 490 Hz when he hears it after it gets reflected from the wall. Find the speed of the bus if speed of the sound is 330 ms1 :

(A) 90 kmh1 (B) 80 kmh1 (C) 61 kmh1 (D) 71 kmh1 13. A series L-R circuit is connected to a battery of emf V. If the circuit is switched on at t =

0, then the time at which the energy stored in the inductor reaches (1/n) times of its maximum value, is :

(A) L nlnR n 1

(B) L n 1lnR n 1

(C) L n 1lnR n

(D) L nlnR n 1

14. The electric field of a plane electromagnetic wave is given by 0 ˆ ˆE E x y sin kz t

Its magnetic field will be given by:

(A) 0E ˆ ˆx y sin kz tc

(B) 0E ˆ ˆx y cos kz tc

(C) 0E ˆ ˆx y sin kz tc

(D) 0E ˆ ˆx y sin kz tc

15. A person pushes a box on a rough horizontal platform surface. He applies a force of 200

N over a distance of 15m. Thereafter, he gets progressively tired and his applied force reduces linearly with distance to 100 N. The total distance through which the box has been moved is 30 m. What is the work done by the person during the total movement of the box?

(A) 5250 J (B) 5690 J (C) 3280 J (D) 2780 J 16. A paramagnetic sample shows a net magnetisation of 6A/m when it is placed in an

external magnetic field of 0.4 T at a temperature of 4K. When the sample is placed in an external magnetic field of 0.3T at a temperature of 24K, then the magnetisation will be :

(A) 1 A/m (B) 0.75 A/m (C) 2.25 A/m (D) 4 A/m




17. A body is moving in a low circular orbit about a planet of mass M and radius R. The

radius of the orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is :

(A) 2 (B) 1

(C) 2 (D) 12

18. A small ball of mass m is thrown upward with velocity u from the ground. The ball

experiences a resistive force mkv2 where v is it speed. The maximum height attained by the ball is:

(A) 21 kuln 1

k 2g

(B)

211 kutan

2k g

(C) 2

11 kutank 2g

(D) 21 kuln 1

2k g

19. Match the thermodynamics processes taking place in a system with the correct

conditions. In the table: Q is the heat supplied, W is the work done and U is change in internal energy of the system. Match the following:

Process Condition (I) Aidabatic (A) W = 0 (II) Isothermal (B) Q = 0 (III) Isochoric (C) U 0, W 0, Q 0 (IV) Isobaric (D) U = 0

(A) (I)(B); (II)(D); (III)(A); (IV)(C) (B) (I)(A); (II)(A); (III)(B); (IV)(C) (C) (I)(A); (II)(B); (III)(D); (IV)(D) (D) (I)(B); (II)(A); (III)(D); (IV)(C) 20. A cube of metal is subjected to a hydrostatic pressure 4GPa. The percentage change in

the length of the side of the cube is close to: (Given bulk modulus of metal, B = 8 × 1010 Pa)

(A) 0.6 (B) 5 (C) 1.67 (D) 20 21. The speed verses time graph for a particle is shown in the figure. The distance travelled

(in m) by the particle during the time interval t = 0 to t = 5 s will be …..

22. The distance between an object and a screen is 100 cm. A lens can produce real image

of the object on the screen for two different positions between the screen and the object.




The distance between these two positions is 40 cm. If the power of the lens is close to N D

100

where N is an integer, the value of N is ……….

23. The change in the magnitude of the volume of an ideal gas when a small additional

pressure P is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity T at constant pressure. The initial temperature and pressure of the gas were 300 K and 2 atm. respectively. If |T| = C|P| then value of C in (K/atm) is ….

24.

Four resistance 40 , 60 , 90 , and 110 make the arms of a quadrilateral ABCD.

Across AC is a battery of emf 40V and internal resistance negligible. The potential difference across BD in V is ……

25. Orange light of wavelength 6000 × 10–10 m illuminates a single slit of width 0.6 × 10–4 m.

the maximum possible number of diffraction minima produced on both sides of the central maximum is ……..




PART –B (CHEMISTRY) 26. The Crystal Field Stabilization Energy (CFSE) of [CoF3(H2O)3] (0 < P) is : (A) 0.8 0 (B) 0.4 0 + P (C) 0.8 0 + 2P (D) 0.4 0 27. The processes of calcination and roasting in metallurgical industries, respectively, can

lead to: (A) Photochemical smog and global warming (B) Global warming and photochemical smog (C) Global warming and acid rain (D) Photochemical smog and ozone layer depletion 28. 250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M

AgNO3 and 0.1 M AuCl. The solution was electrolyzed at 2 V by passing a current of 1 A for 15 minutes. The metal/metals electrodeposited will be :

o oAg /Ag Au /Au

E 0.8 V,E 1.69 V

(A) silver and gold in equal mass proportion (B) only silver (C) only gold (D) silver and gold in proportion to their atomic weights 29. Among the following compounds, which one has the shortest C – Cl bond?

(A)

(B)

(C) (D)

30. The process that is NOT endothermic in nature is: (A) (g) (g)H e H (B) 2

(g) (g)O e O

(C) (g) (g)Na Na e (D) (g) (g)Ar e Ar 31. If the equilibrium constant for A B C is (1)

(eq)K and that of B C P is (2)(eq)K , the

equilibrium constant for A P is:

(A) (1) (2)(eq) (eq)K K (B) (1) (1)

(eq) (eq)K / K

(C) (1) (2)(eq) (eq)K K (D) (2) (1)

(eq) (eq)K K




32. Which of the following compounds will form the precipitate with aq. AgNO3 solution most readily ?

(A) (B)

(C)

(D)

33. In the following reaction sequence, [C] is:

(A)

(B)

(C)

(D)

34. The major product [R] in the following sequence of reactions is:

(A)

(B)

(C)

(D)

35. The major product [C] in the following reaction sequence will be:

(A)

(B)

(C)

(D)




36. A sample of red ink (a colloidal suspension) is prepared by mixing eosine dye, egg white, HCHO and water. The component which ensures stability of the ink sample is :

(A) Egg white (B) HCHO (C) Water (D) Eosin dye 37. The one that can exhibit highest paramagnetic behaviour among the following is: gly = glycinato ; bpy = 2, 2'-bipyridine (A) [Fe(en)(bpy)(NH3)2]2+ (B) [Co(OX)2(OH)2] (0 > P) (C) [Pd(gly)2] (D) [Ti(NH3)6]3+ 38. The major product [B] in the following reactions is:

(A) (B)

(C)

(D)

39. The mechanism of action of "Terfenadine" (Seldane) is : (A) Inhibits the action of histamine receptor (B) Inhibits the secretion of histamine (C) Helps in the secretion of histamine (D) Activates the histamine receptor 40. An alkaline earth metal 'M' readily forms water soluble sulphate and water insoluble

hydroxide. Its oxide MO is very stable to heat and does not have rock-salt structure. M is:

(A) Sr (B) Be (C) Mg (D) Ca 41. The incorrect statement(s) among (a) - (c) is (are): (a) W(VI) is more stable than Cr(VI). (b) in the presence of HCl, permanganate titrations provide satisfactory results. (c) some lanthanoid oxides can be used as phosphors. (A) (b) and (c) only (B) (a) only (C) (b) only (D) (a) and (b) only 42. The molecule in which hybrid MOs involve only one d-orbital of the central atom is : (A) XeF4 (B) [CrF6]3- (C) [Ni(CN)4]2- (D) BrF5 43. The reaction in which hybridization of the underlined atom is affected is: (A) 4 5XeF SbF (B) 420 K

2 4H SO NaCl

(C) Disproportionation3 2H PO (D) H

3NH




44. The shortest wavelength of H atom in the Lyman series is 1. The longest wavelength in the Balmer Series of H+ is:

(A) 159 (B) 136

5

(C) 195 (D) 127

5

45. Five moles of an ideal gas at 1 bar and 298 K is expanded into vacuum to double the

volume. The work done is : (A) CV(T2 – T1) (B) RT ln V2/V1 (C) zero (D) RT V2/V1 46. The number of molecules with energy greater than the threshold energy for a reaction

increases five fold by a rise of temperature from 27º C to 42º C. Its energy of activation in J/mol is ………….. (Take in 5 = 1.6094; R = 8.314 J mol1K1)

47. Consider the following equations : 2Fe2+ + H2O2 x A + y B (in basic medium) (in acidic medium) 4 2 22MnO 6H 5H O x 'C y 'D z 'E (in acidic medium) The sum of the stoichiometric coefficients x, y, x’, y’ and z’ for products A, B, C, D and E,

respectively, is ……….. 48. The number of chiral centres present in threonine is _______. 49. A 100 mL solution was made by adding 1.43 g of Na2CO3.xH2O. The normality of the

solution is 0.1 N. The value of x is ……….. (The atomic mass of Na is 23 g/mol) 50. The osmotic pressure of a solution of NaCl is 0.10 atm and that of a glucose solution is

0.20 atm. The osmotic pressure of a solution formed by mixing 1 L of the sodium chloride solution with 2 L of the glucose solution is x × 10–3 atm. x is ………….. (nearest integer)




PART–C (MATHEMATICS)

51. If a and b are real numbers such that (2 + )4 = a + b, where 1 i 3 ,2

then a + b is

equal to: (A) 9 (B) 33 (C) 57 (D) 24 52. Contrapositive of the statement: 'If a function f is differentiable at a, then it is also continuous at a', is: (A) If a function f is not continuous at a, then it is not differentiable at a. (B) If a function f is continuous at a, then it is differentiable at a. (C) If a function f is not continuous at a. then it is differentiable at a. (D) If a function f is continuous at a, then it is not differentiable at a. 53. If for some positive integer n, the coefficients of three consecutive terms in the binomial

expansion of (1 + x)N+5 are in the ratio 5 : 10 : 14, then the largest coefficient in the expansion is:

(A) 252 (B) 330 (C) 792 (D) 462

54. The function

1tan x, x 14f(x)1 x 1 , x 12

is:

(A) both continuous and differentiable on R – {1}. (B) both continuous and differentiable on R – {–1}. (C) continuous on R – {1} and differentiable on R – {–1, 1}. (D) continuous on R – {–1} and differentiable on R – {–1, 1}. 55. If the system of equations x + y + z = 2 2x + 4y – z = 6 3x + 2y + z = has infinitely many solutions, then: (A) + 2 = 14 (B) 2 + = 14 (C) 2 = 5 (D) 2 = 5 56. Let 0 be in R. If and are the roots of the equation, x2 – x + 2 = 0 and and are

the roots of the equation, 3x2 – 10x + 27 = 0, then

is equal to:

(A) 18 (B) 36 (C) 9 (D) 27 57. The circle passing through the intersection of the circles, x2 + y2 – 6x = 0 and x2 + y2 – 4y

= 0, having its centre on the line, 2x – 3y + 12 = 0, also passes through the point: (A) (1, 3) (B) (3, 6) (C) (3, 1) (D) (1, 3)




58. The integral 3 3 2 2 2

6

tan x sin 3x 2sec x sin 3x 3 tan x sin6x dx

is

(A) 118

(B) 19

(C) 92

(D) 718

59. The area (in sq. units) of the largest rectangle ABCD whose vertices A and B lie on the

x-axis and vertices C and D lie on the parabola, y = x2 – 1 below the x-axis, is:

(A) 43 3

(B) 13 3

(C) 43

(D) 23 3

60. Suppose the vectors x1, x2 and x3 are the solutions of the system of linear equations, Ax

= b when the vector b on the right side is equal to b1, b2 and b3 respectively. If

1 2 3 1 2 3

1 0 0 1 0 0x 1 ,x 2 ,x 0 ,b 0 ,b 2 and b 0 ,

1 1 1 0 0 2

then the determinant of A is

equal of A is equal to :

(A) 12

(B) 4

(C) 2 (D) 32

61. The solution of differential equation e

dy y 3x 3 0dx log y 3x

is:

(where C is a constant of integration.)

(A) 2e1y 3x log x C2

(B) 2e1x log y 3x C2

(C) ex 2log y 3x C (D) ex log y 3x C 62. The minimum value of 2sinx + 2cosx is:

(A) 1122

(B) 1 22

(C) 1 22 (D) 1122

63. Let f : (0, ) (0, ) be a differentiable function such that f(1) = e and 2 2 2 2

t x

t f x x f (t)lim 0.

t x

If f(x) = 1, then x is equal to:

(A) e (B) 1e

(C) 12e

(D) 2e




64. The distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the

line x y z2 3 6

is:

(A) 17

(B) 7

(C) 75

(D) 1

65. In a game two players A and B take turns in throwing a pair of fair dice starting with

player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws a total of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six. The game stops as soon as either of the players wins. The probability of A winning the game is:

(A) 56

(B) 3161

(C) 3061

(D) 531

66. The angle of elevation of a cloud C from a point P, 200 m above a still take is 30°. If the

angle of depression of the image of C in the lake from the point P is 60°, then PC (in m) is equal to

(A) 200 3 (B) 100 (C) 400 (D) 400 3

67. Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 12

.

If P(1, ), > 0 is a point on this ellipse, then the equation of the normal to it at P is: (A) 4x – 2y = 1 (B) 4x – 3y = 2 (C) 7x – 4y = 1 (D) 8x – 2y = 5

68. Let 50 n

i ii 1 i 1

X Y T,

where each Xi contains 10 elements and each Yi contains 5

elements. If each element of the set T is an element of exactly 20 of sets Xi's and exactly 6 of sets Yi's then n is equal to:

(A) 15 (B) 30 (C) 45 (D) 50 69. If the perpendicular bisector of the line segment joining the points P(1, 4) and Q(k, 3)

has y-intercept equal to –4, then a value of k is: (A) 14 (B) 15 (C) 4 (D) 2 70. Let a1, a2, …. an be a given A.P. whose common difference is an integer and Sn = a1 + a2

+ … + an. If a1 = 1, a1 = 300 and 15 n 50, then the ordered pair (Sn–4, an–4) is equal to: (A) (2490, 248) (B) (2480, 248) (C) (2490, 249) (D) (2480, 249)




71. If ˆ ˆ ˆa 2i j 2k,

then the value of 2 2 2ˆ ˆ ˆ ˆ ˆ ˆi a i j a j k a k

is equal to:

72. A test consists of 6 multiple choice questions, each having 4 alternative answers of

which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is ……

73. Let PQ be a diameter of the circle x2 + y2 = 9. If and are the lengths of the

perpendiculars from P and Q on the straight line, x + y = 2 respectively, then the maximum value of is……

74. If the variance of the following frequency distribution : Class : 10 – 20 20 – 30 30 – 40 Frequency : 2 x 2 is 50, then x is equal to: 75. Let {x} and [x] denote the fractional part of x and the greatest integer x respectively of a

real number x. if n n

0 0

{x}dx, [x]dx and 10(n2 – n), (n N, n > 1) are three consecutive

terms of a G.P. then n is equal to ……




FIITJEE

Solutions to JEE (Main)-2020

PART –A (PHYSICS) 1. A

Sol. Qo = CVo

Ui = 20

1 CV2

o o1

q C 2qq C 3C2

C

+ + + + +

– – – – –

+qo –qo

C/2

oo

2

Cq q2q C 3C2

; 2 21 2

fq qU 2c2c

2

= 2 20 04q q

9 2c 9c

= 2 2 20 0 06q q cv

18c 3c 3

+ + + + +

– – – – –

+q2 –q2

+ + + + +

– – – – –

+q1 –q1

C/2

C

Energy loss in the process = Ui – Uf

= 2

2 00

CV1 CV2 3

= 20CV

6

2. D Sol.

slophceV

Slope of curve

tan = hc constante

as intensity of incident radiation is increased, there will be no effect on graph

V

1




3. A Sol. I1 2

1R & I2 22R

2

1 12

2 2

I R 1 1I R 16

= 4 4. None of the options Sol. I = 0.8 kg M2,

= 20 Am2

iU B 0

Uf = –B cos(30°) = –20 × 4 × 32

Ui – Uf = 2140 3 I2

= 0.4 2

2 = 100 3 ; = 1/410 (3 ) 5. A Sol. Binding energy = (50 Mp + 70 Mn – Msn)C2 = (50.3915 + 70.6069 – 119.902199)UC2 = (1.0962 U)C2 = 931 × 1.0962 MeV Binding energy per neucleon

= 931 1.0962 MeV120

= 8.5 MeV 6. C

Sol. 2

4IFVxWL

2I [ML ]

F = 2[MLT ] V2 = [L2T–2] W = [M L2T–2] Q4 = [L4] X = [M L–1 T–2]




7. C Sol. 1 = 0.6 M

2 = 0.8 M 2 21 2 1

MI about 0 = 2 21 2

M12

1MI12

O

O

1

2 1/2

MI about O’ = 2 21 22 2

1 2

MM12 4

I2 = M M 4m12 4 12

; 1

2

I 1I 4

8. A 9. A Sol. E = Eo (1 – ax2) F = qEo

2oqEF dvacceleration (1 ax ) vm m dx

x 0

2o

0 0

qE (1 ax )dx vdvm

; 3

0qE axx 0M 3

3axx 1 0

3

; x = 0 & 3x

a

10. D

Sol. i1 = 88

= 1 A

11. C Sol.

Ui = 1 21 2

x xsx dg sx dg2 2

Uf =

1 2 1 2S(x x )gd x x x22 4

= 2

1 2s(x x ) gd4

2 2 2i f 1 2 1 2

5gdU U 2x 2x (x x )4

= 21 2

5gd (x x )4

X2 X1




12. A Sol. f1 = 420 Hz

o2 1

o

V Vf f 490V V

330 V 420 490300 V

(330 + V) 6 = 7 (330 – V)

Wall

V

Vo

13 V = 330

33V (m / s)13

= 91 (km/hr) 13. 4 Sol. t

LVi 1 er

maxVir

21U Li2

r

V

L

V

= 2 2

rt / L2 2

1 V 1 L VL 1 e2 n 2r r

rt /L 11 en

; rt /L 1 n 1e 1

n n

ert/L = nInn 1

;

rt nInL n 1

T = L nInr n 1

14. C 15. A Sol.

F

F

x

200 N

100 N

x = 15 M 30 M Network done is equal to area under F-x curve




= 200 × 15 + 1 100 152 +100x15

= 4500 + 750 = 5250 16. B

Sol. From aerie’s law cxT

I = x H I1 = x1 H1 I2 = x2 H2

2 2 2

1 1 1

I HI H

2 2 2

1 1 1

I HI H

2 1 2

1 2 1

I THI T H

= 4 0.3 124 0.4 8

12

T 6I 0.758 8

17. D

Sol. Escape velocity = esf2GM V

R

Orbital speed = oGM VR

o

esp

V 1V 2

18. D Sol.

Mg + MkV2 = ma = –mv dVdx

Vdv = (–) (g + k 2V )dx

0 x

2a 0

Vdv dxg kv

02u

ln (g kV )x

2k

22In 2kx

g ku

X =

21 kuIn 12k g

V

u

a

Mg MKV2

19. A Sol. Q = heat supplied




W = work done U = change in internal energy (i) adiabatic (B) = 0 (ii) isothermal (D) U = 0 (iii) isochoric (A) W = 0 (iv) isobaric (C) U 0, W = 0, Q 0 20. C

Sol. VB PV

9

10V P 4 10 1

V B 208 10

V = 3

dV = 32 d

2

3dV 3 3ddV

V 3V

; 2V 1

3V 60

% 100 1.67%60

21. 20.00 Sol. Distance moved = Area under curve

= 1 8 5 202

22. 476 Sol. 1 NP D

f 100

2x + 40 = 100 x = 30 cm 100 – x = 70 cm 1 1 1v u f

1 1 170 30 f

A

Screen

x 40 cm

x

B

(100 – x)

100 cm

Lens

1 1 1 3 7 1

f 70 30 210 21

f = 21 cm = 0.21 M

Power = 1 1f 0.21 ; 100D D

21

N 100D D100 21




N = 1000021

= 476.19

N 476 23. 150 Sol. T = constant P = constant PV = nRT PdV = nRdT PdV + vdP = 0

vddV ( )P

nR TvP

PV VP

P nR TVP P

T = V PnR

V T 300C

nR P p = 150

24. 2.00 Sol.

140 2i

100 5

240 1i

200 5

VA – VB = 40 i1 = 2405

VA – BB = 16

VA – VD = 90I2 = 905

= 18

VB – VD = 18 – 16 = 2 volt

60 40

90 110

D

B

40 V

A C

25. 200.00 Sol. = 6 × 10–7 M

d = 6 × 10–5 M 2

osin (B)I I

; dsin

2

; d

= 5

76 10

6 10

= 100

d

So at also minima will form total number of minima = 2 × 100 = 200




PART –B (CHEMISTRY) 26. D Sol. [Co(H2O)3F3] Co3+ = 3d64s0 2,1,1 1,1

2g gt ,e CFSE = [-0.4nt2g + 0.6neg]0 + n(P) = [-0.4 4 + 0.6 2]0 + 0 = 0.4 0 27. C Sol. In Calcination and roasting CO2 and SO2 are released which are responsible for Global

warning and acid rain. 28. C

Sol. Charge(q) = it 1 15 60 900 9F F 0.0093F96500 96500 96500 965

No. of moles of Au+ = 0.025 & No. of moles of Ag+ = 0.025 Species with higher value of SRP will get deposited first at cathode. (i) Au aqs e Au s 0.025 0.0093 mole So only Au will get deposited 29. B Sol. Due to resonance C-Cl bond in option B is shortest. 30. A Sol. H can easily gain electron to form its anion. 31. A Sol. On adding reaction equilibrium constant will get multiplied. 32. C Sol. SN1 reaction depends on carbocation stability and cation form in 3 will be most stable. 33. C Sol.




34. D Sol.

35. A Sol.

36. A Sol. Egg white will stabilize blue ink easily. 37. B Sol. Cobalt has 2 unpaired electron




38. B Sol.

39. A Sol. Terfenadine act as antihistamine. 40. B Sol. BeO is hexagonal wurtzite type structure. 41. C Sol. KMnO4 oxidise HCl to Cl2. 42. C Sol. In [ Ni(CN)4]2- hybridization is dsp2 remaining are SP3d2 43. A Sol.

44. C Sol.




45. C Sol. Pext is zero so W = zero 46. 84297 J Sol.

47. 19 Sol.

48. 2 Sol.

49. 10 Sol.




50. 167 Sol.




PART–C (MATHEMATICS) 51. A

Sol. 4

4 3 32 i2 2

4i

63e

2 29 cos isin3 3

9 9 3i

2 2

1 i 30 9

2

a 0, b 9 Answer A 52. C Sol. Contrapositive of p q is ~ q ~ p . 53. D Sol. N 5 N 5 N 5

R 1 R R 1C : C : C

5 : 10 : 4

N 5 N 5R 1 R2 C C 3R N 6

N 5 N 5R R 17 C 5 C 12R 18 5N

Solving: N 6, R 4

Largest coefficient is N 5 11R 1 5C C 462

Answer is Option D.




54. C Sol. Graph of f(x) is

2

(–1, 0) (1, 0)

Option C is correct. 55. B

Sol. Here, 1 1 1

92 4 1 02

3 2

Also, 1 1 22 4 6 0 53 2

Option B is correct. 56. A Sol. Given: 23 10 27 0 ………..(i) 23 3 6 0 …………(ii) Subtract 7 21 0 3 0 By (ii) 29 3 2 0

10,9

1 2 1, , , 33 3 3

2 .33 1819

57. B Sol. By family of circle, passing through intersection of given circle will be member of

S S, 0 family 1




2 2 2 2x y 6x x y 4y 0

2 21 x 1 y 6x 4 y 0

2 2 6 4x y y 01 1

Centre 3 2,

1 1

Centre lies on i 2x 3y 12 0

3 22 3 12 0

1 1

6 18 0 3 Circle 2 2x y 3x 6y 0 58. A

Sol. 4 4/3

/6

d dtan x sin 3xdx dx.sin'3x tan' x.

2 2

/3

4 4

/6

1 d tan x.sin 3x dx2 dx

/34 4/6

1 tan x.sin 3x2

4

41 13 .O2 3

118

59. A Sol. A ,0 ,B ,0

2D , 1

Area (ABCD) = (AB) (AD) 2 2S 2 1 2 2

2ds 2 6d

20 a

13

B A

D C




13

Area = 2 2 22 23 3 3

4

3 3

60. C

Sol. Let 1 2 3

1 2 3

1 2 3

a a aA b b b

c c c

1 1Ax B 1 2 3a a a 1 1 2 3b b b 0 1 2 3c c c 0 Similar 3 32a a 0 and 3a 0 2 32b b 2 3b 0 2 32c c 0 3c 2 2 2 2a 0,b 1,c 1, 1 1 1a 1, b 1, c 1

1 0 0

A 1 1 0 A 21 1 2

61. B

Sol.

dy y 3x 3 0dx ln y 3x

dy y 3x3dx ln y 3x

d y 3xy 3x

dx ln y 3x

n y 3x

d y 3x dxy 3x

Let n y 3x t

1 d y 3x dty 3x

t dt dx




2t x c2

2n y 3x

x c2

62. D Sol. Since AM two positive quantities their G.M.

sinx cosx

sinx cosx2 2 2 .22

2 cos x

sinx cosx 42 2

2 sinx cosx 1// 2 1 1/ 22 2 2 2.2 2 63. B Sol Applying L – Hospital Rule

2 2

t x

2t f x x 2f t f ' tLim

1

2 22 f x x 2f x f ' x 0

f x x f ' x 0

f ' x 1 nf x n x C

f x x

At x 1,c 1 nf x nx 1

when f x 1 then nx 1

1xe

64. D Sol. Equation PQ

x 1 y 2 z 32 3 6

Let Q 2 1,3 2, 6 3 Q lies on x y z 5 2 1 3 2 6 3 5

17

5 17 15Q , ,7 7 7

Q

P (1, –2, 3)

x y z2 3 6




2 2 22 3 6PQ

7 7 7

PQ 1 65. C Sol. sum 6 (1, 5), (5, 1), (3, 3), (2, 4), (4, 2) sum 7 (1, 6), (6, 1), (5, 2), (2, 5), (3, 4), (4, 3) P A P A .P B .P A P A P B P A .P B .P A ...

This is infinite G.P. with common ratio P A P B

Probability of A wins

P A1 P A P B

53036

31 30 611 .36 36

66. C

Sol. o x 1tan30 y 3xy 3

……….(i)

and o x 400tan60 3yy

x 400 ……….(ii) Solving (i) and (ii), we get 2x 400, x 200

o x 200sin30 PC 400PC PC

C

P 60o

30o

y E

x

200

D

x + 200

Q

67. A

Sol. Given : a 4e and

2

21 b14 a

Solving : a 2,b 3 Parametric co – ordinates are 2cos , 3 sin 1,

o60




Equation of normal is a xsec – by 2 2cosec a b 4x 2y 1 68. B Sol. Let number of elements in T is R. 20R 500 R 25 and 6R 5N N 30 69. C Sol. Any point (x, y) on perpendicular bisector equidistant from p and q 2 2 2 2x 1 y 4 x k y 3 At x 0, y 4

21 64 k 49

2k 16 70. A Sol. Given, 300 1 N 1 d

N 1 d 299

N,d 24,13 is the only possible pair

20a 1 19 13 248 and, 201 248S 20

2

2490 71. 18 Sol. Let ˆ ˆ â xi yj zk

ˆ ˆ ˆ ˆ ˆ ˆ î a i i . i a a. i yj zk

Similarly ˆ ˆ ˆ ˆj a j xi zk

and ˆ ˆ ˆ ˆk a k xi yk

2 2 2ˆ ˆ ˆ ˆ ˆ î a i j a j k a k

2 2 2 2ˆ ˆ ˆ ˆ ˆ ˆyj zk xi zk xi yj 2 a 2 9 18

72. 135 Sol. Ways of selecting correct questions 6

4C 15 Ways of doing them correct = 1 Ways of doing remaining 2 questions incorrect 23 9 No. Of ways 15 1 9 135




73. 2 Sol. Let P 2cos ,2sin

Q 2cos , 2sin Given line x y 2 0

2cos 2sin 2

2

2cos 2sin 2

2

2 cos sin 1 . 2 cos sin 1

2 22 cos sin 2sin cos 1 2 sin2

Max sin 1 maximum 2 . 74. 4 Sol.

xi 5 15 25 fi 2 x 2

i i

i

f x 10 15x 50xf 4 x

60 15x 154 x

2 2i i2

i

f x50 x

f

250 225x 125050 154 x

1300 225x50 2254 x

275 4 x 1300 225x 50x 200x x 4

2225x 125050 154 x

1300 225x50 2254 x

275 4 x 1300 225x 50x 200 x 4




75. 21

Sol. Clearly, n

0

nx dx2

n

0

x dx 1 2 3.....n 1

n n 1

2

2n n 1 n 10n n 12 2

Solving, n = 21


FIITJEE...Sep 04, 2020 · FIITJEE Ltd., FIITJEE House, 29 -A, Kalu Sarai, Sarvapriya Vihar, New...

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Transcript of FIITJEE...Sep 04, 2020 · FIITJEE Ltd., FIITJEE House, 29 -A, Kalu Sarai, Sarvapriya Vihar, New...