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PHYSICS, CHEMISTRY & MATHEMATICS JEE Main 2019 Mock Tests-1 (Code:100365.1)

Time Allotted: 3 Hours Maximum Marks: 360

Do not open this Test Booklet until you are asked to do so. Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

Important Instructions:

1. Immediately fill in the particulars on this page of the Test Booklet with Blue / Black Ball Point Pen. Use of pencil is strictly prohibited.

2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the

Answer Sheet and fill in the particulars carefully. 3. The test is of 3 hours duration. 4. The Test Booklet consists of 90 questions. The maximum marks are 360. 5. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having

30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response. 6. Candidates will be awarded marks as stated above in instruction No.5 for correct response of each question. ¼

(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.

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

treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above. 8. Use Blue / Black Ball Point Pen only for writing particulars / marking responses on Side-1 and Side-2 of the

Answer Sheet. Use of pencil is strictly prohibited. 9. No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any

electronic device, etc. except the Admit Card inside the examination hall / room. 10. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room /

Hall. However, the candidates are allowed to take away this Test Booklet with them. 11. Do not fold or make any stray marks on the Answer Sheet.

Atomic No.: H=1, He=2, Li=3, Be=4, B=5, C=6, N=7, O=8, F=9, Na=11, Mg=12, Al = 13, Si = 14, P = 15, S = 16, Cl = 17, Ar =18, K=19, Ca=20,Cr=24, Mn=25, Fe=26, Co=27, Ni=28, Cu=29, Zn=30, As=33, Br = 35, Ag = 47, Si = 21, Sn = 50, Ti = 22,I = 53, Xe = 54, Ba = 56, Pb = 82, U = 92, V = 50. Atomic masses: H =1, He=4, Li=7, Be=9, B=11, C=12, N=14, O=16, F=19, Na=23, Mg=24, Al=27, Si=28, P=31, S=32, Cl=35.5, K=39, Ca=40, Cr=52, Mn=55, Fe=56, Co=59, Ni=58.7, Cu=63.5, Zn = 65.4, As = 75, Br = 80, Ag = 108, Sn = 118.7, I = 127, Xe = 131, Ba = 137, Pb = 207, U = 238.

Name of the Candidate (in Capital Letters) :_____________________________________

Enrolment Number :_________________________________________________________

Batch :________________________ Date of Examination : ________________________

FIITJEE - JEE (Main)

http://www.fiitjee.com.


IT2019JEE Main 2019 Mock Test-1-(100365.1)-PCM2

PPPAAARRRTTT ––– III ::: PPPHHHYYYSSSIIICCCSSS

(Single Correct Answer Type)

This part contains 30 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. One end of a massless rope, which passes over a massless

and frictionless pulley P is tied to a hook C while the other end is free. Maximum tension that the rope can bear is 360 N. With what value of minimum safe acceleration (in ms–2) can a monkey of 60 kg move down on the rope

(A) 16 (B) 6 (C) 4 (D) 8

P

C

2. A particle moves on a rough horizontal ground with some initial velocity say v0. If (3/4)th of its kinetic

energy is lost due to friction in time t0, then coefficient of friction between the particle and the ground is

(A) 0

0

v2gt

(B) 0

0

v4gt

(C) 0

0

3v4gt

(D) 0

0

vgt

3. A source is moving on a circle of radius 3 m with constant angular velocity

= 5 rad/s. If the observer is at a distance 5 m from the centre of circle, the time interval between maximum and minimum frequency received by the observer is

(A) 5 (B) 12 3cos

5 4

(C) 12 3cos5 5

(D) 12 3sin5 5

source

observer 3m

4. A spherical conductor A of radius r is placed concentrically inside a conducting shell B of radius

R(R > r). A charge Q is given to A, and then A is joined to B by a metal wire. The charge flowing from A to B will be

(A) Q RR r

(B) rQR r

(C) Q (D) zero

Space for rough work




5. A homogeneous rod of mass 3 kg is pushed along the smooth horizontal

surface by a horizontal force F equals to 40 N. The angle for which rod has pure translation motion is (g = 10 m/s2)

(A) 45 (B) 37 (C) 53 (D) 60

F=40 N

6. At a moment (t = 0), when the charge on capacitor C1 is zero, the switch is

closed. If I0 be the current through inductor at t = 0 , for t > 0 (initially C2 is uncharged)

(A) maximum current through inductor equals I0/2.

(B) maximum current through inductor equals 1 0

1 2

C IC C

.

C2

C1

L

S

(C) maximum charge on 1 0 21

1 2

C I LCC

C C

.

(D) maximum charge on 1 1 01 2

LC C IC C

.

7. A heated body emits radiation which has maximum intensity near frequency 0. The emissivity of

material is 0.5, if absolute temperature of body is doubled, the maximum intensity of radiation will be near the frequency.

(A) 0/2 (B) 20 (C) 80 (D) 160 8. The Boolean equation for the circuit shown in the figure is (A) Y A B C (B) Y A (B C)

(C) Y A (B C)

(D) Y A (B C)

A

B Y

C





9. Radius of a circular ring is changing with time and the ring is

placed in a uniform magnetic field perpendicular to its plane. The variation of r with time t is as shown in the figure. The magnitude of induced emf (e) is best represented by

r

t (A) e

t

(B) e

t (C) e

t

(D) e

t 10. For a telescope, the diameter of objective lens to be 6 cm and wavelength of light used to be 540 nm.

The resolving power of telescope is (A) 3.2 104 (B) 4 104 (C) 7.7 104 (D) 9.1 104 11. An atom of atomic number Z = 11 emits K wavelength which is . Find the atomic number for an

atom that emits K radiation with wavelength 4 (an integer). (A) 2 (B) 4 (C) 6 (D) 8 12. In the figure shown, a conducting wire PQ of length 1 m is

moving in a uniform magnetic field B = 3T with constant velocity v = 2m/sec towards right. R = 2, C = 1F and L = 2H. The currents through inductor at time t = 2 sec is

(A) 2A (B) 4A (C) 6A (D) 8A

R B C

Q

P

L





13. An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth. If the satellite is stopped suddenly in its orbit and allowed to fall freely, onto the earth, find the speed (in km/sec) with which it hits the surface of the earth (g = 10 m/s2 and R = 6400 km).

(A) 8 km/sec (B) 16 km/sec (C) 2 km/sec (D) 4 km/sec 14. A cube of mass 20 gm wettable by water floats on the surface of water. Each face of the cube is 3 cm

long. The distance between lower face of cube and surface of water is (surface tension of water = 70 dyne/cm, g = 10 m/s2, assume contact angle is 0)

(A) 1.012 cm (B) 1.631 cm (C) 2.315 cm (D) 2.52 cm 15. An amount Q of heat is added to a monoatomic ideal gas in a process in which the gas performs a

work Q/2 to its surrounding. The molar heat capacity for the process is (A) R (B) 2R (C) 3R (D) 4R 16. The figure shows a long straight wire of a circular cross-section (radius a) carrying

steady current I. The current I is uniformly distributed across this cross-section. The ratio of magnetic fields at a distance a/2 and 2a from the axis is

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

I

a

wire

17. A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a

particular point in space and time, E

= 6.3 j V/m. What is B

at this point?

(A) 2.1 × 108 T ˆ( k) (B) 2.1 × 108 T ˆ(k)

(C) 2.1 × 108 T ˆ( j) (D) 2.1 × 108 T ˆ( i)





18. Light with an energy flux of 18 W/cm2 falls on a nonreflecting surface at normal incidence. If the surface has an area of 20 cm2, the average force exerted on the surface during a half hour time span, will be

(A) 2.16 103 N (B) 6.48 105 N (C) 1.2 106 N (D) 360 N 19. A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations

of the rod are given to be 2.53 kHz. What is the speed of sound in steel? (A) 10.06 km s–1 (B) 5.06 km s–1

(C) 15.06 km s–1 (D) 0.06 km s–1

20. The ratio of free electron densities in copper and gold is given by Cu

Au

n1.4

n , and the ratio of the mean

free times of their free electrons is given by Cu

Au

5049

. The ratio of their conductivities is

(A) 107

(B) 710

(C) 250343

(D) 343250

21. A block of mass 1 kg is pulled along the curve path ACB by a

tangential force as shown in figure. The work done by the frictional force when the block moves from A to B is

(A) 5 J (B) 10 J (C) 20 J (D) none of these

=0.2 C

A B X=10m

22. A particle starts moving with velocity v i (m/s), and acceleration a i j

(m/s2). Find the radius

of curvature of the particle at t = 1sec. (A) 5 m (B) 5 5 m (C) 6 5 m (D) none





23. Energy stored in the capacitor in it’s steady state is

(A) CV (B) CV2

(C) 1QV2

(D) zero

C R

R E

24. A stone of mass m tied to one end of thread of length . The diameter of thread is ‘d’ and it is

suspended vertically. The stone is now rotated in horizontal plane and makes an angle ‘’ with vertical. The increase in length of wire is (young’s modulus of wire is y)

(A) 24mg

d ycos (B) 2

4mgd ysin

(C) 24mg

d y (D) 2

4mgd y sec

25. A rigid circular loop of radius 3.5 m and mass 11 kg lies at rest in x-y

horizontal plane, when a current is flowing through it. The magnetic field ˆ ˆB 5i 8k (T)

is present in space. Find the value of current. (g = 10 m/s2)

(A) 2A (B) 4A (C) 6A (D) 8A 26. In the Young’s double–slit experiment, when a glass plate ( = 1.5) of thickness t is introduced in the

path of one of the interfering beams (wavelength ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is

(A) 2 (B) 23

(C) 3 (D)





27. The potential deference across the resistor as measured by a

voltmeter is 60 1.5V and the current through the resistor as measured by an ammeter is 2 0.01 A. What is the maximum % error in the measured value of R?

(A) 3% (B) 1.5% (C) 2.5% (D) None of these

V

R

V

A 28. Two identical thermally insulated conducting rods AB and

CD are connected to a circular conducting and thermally insulated ring at two diametrically opposite points B and C. The radius of the ring is equal to the length of rods AB and CD each. The area of cross-section and thermal conductivity of the rod and ring are equal. Points A and D are maintained at temperatures of 100ºC and 0ºC. Temperature of point C

will be 227

A B C D100ºC 0ºC

(A) 62ºC (B) 37ºC (C) 28ºC (D) 45ºC 29. A conducting shell having inner radius 1R and outer radius

2R contains a charge + q which is placed at a distance x from its centre. Field at an exterior point P which is situated at a distance r from centre of shell 2r R

(A) depends on x (B) depends on 1R and 2R (C) depends only or r (D) None.

r R1

R2

+q

P

30. A glass hemisphere of radius R and of material having

refractive index 1.5 is silvered on its flat face as shown in fig. A small object of height h is located at a distance 2R from the surface of hemisphere as shown in the figure. The final image will form

2R

h

(A) at a distance of R from silvered surface, on the right side (B) on the object itself (C) at hemisphere surface (D) at a distance of 2R from the silvered surface, on left side.





PPPAAARRRTTT ––– IIIIII ::: CCCHHHEEEMMMIIISSSTTTRRRYYY

(Single Correct Answer Type)

This part contains 30 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. The wave number of the spectral line in the emission spectrum of hydrogen will be equal to 8/9 times

the Rydberg’s constant if the electron jumps from (A) n = 3, n = 1 (B) n = 10 to n = 1 (C) n = 9 to n = 1 (D) n = 2 to n = 1 2. If two compounds have same empirical formula but different molecular formula, they must have (A) same viscosity (B) same vapour density (VD) (C) different molecular weight (D) different percentage composition 3. For CO, isotherm is of the type as shown. Near the point A, compressibility factor Z is

A

pVCO

B Ideal

p

(A) b1V

(B) b1V

(C) a1RT V

(D) a1

RT V

4. Depression of freezing point of 0.01 molal aq. CH3COOH solution is 0.02046°. 1 molal urea solution

freezes at –1.86°C. Assuming molality equal to molarity, pH of CH3COOH is 23.0 Torr at 25°C, is (A) 0.25 (B) 0.75 (C) 0.50 (D) 0.66 5. The unstable nucleus 212

82 Pb decays with -particle emission, having a half-life of 10 h. From this it follows that the

I. mass number of the product is 212 II. atomic number of the product is 81 III. fraction of the original isotope remaining after 20 h is 1/4 IV. nucleus formed is stable (A) I, II and III (B) I and III (C) II and IV (D) IV





6. In the reaction, P Q R S the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with reaction time as shown in the figure. The overall order of the reaction is

Time

[Q]

0[Q]

(A) 2 (B) 3 (C) 0 (D) 1

7. 2 2 32SO O 2SO . Starting with 2 mole SO2 and 1 mole O2 in 1 L flask, mixture required 0.4

mole MnO4– in acidic medium. Hence, KC is

(A) 2 (B) 0.4 (C) 1.6 (D) 2.6 8. When con entreated HCl is added to a, solution of [Co(H2O)6]4– ion, an intense blue colour develops

due to the formation which one of the following ? (A) [CoCl6]4– (B) [CoCl4]2– (C) [CoCl2(H2O)4] (D) [CoCl(H2O)5]+ 9. A container of 1.0 L capacity filled with 1.0 mole of ideal gas is connected to an evaculated vessel of

9.0 L. Calculate the change in entropy. (R = 1.987 cal) (A) 0.188 cal K–1 mol–1 (B) 0.4576 cal K–1 mol–1 (C) 4.576 cal K–1 mol–1 (D) 4.366 cal K–1 mol–1 10. The standard enthalpy of formation of NH3 is –46.0 kJ mol–1. If the enthalpy of formation of H2 from its

atoms is –436 kJ mol–1 and that of N2 is –712 kJ mol–1, the average bond enthalpy of N–H bond in NH3 is

(A) –964 kJ mol–1 (B) +352 kJ mol–1 (C) +1056 kJ mol–1 (D) –1102 kJ mol–1 11. Mg2C3 reacts with water forming propyne. 4

3C has (A) two sigma an tow pi bonds (B) three sigma and one pi bonds (C) two sigma and one pi bonds (D) two sigma and three pi bonds 12. AgNO3 (aqueous) was added to an aqueous KCl solution gradually and the conductivity of the

solution was measured. The plot of conductance () versus the volume of AgNO3 is

volume(P) (Q)

volume(R)

volume volume(S)

(A) (P) (B) (Q) (C) (R) (D) (S)





13. Ferrimagnetism is in (A) (B) (C) (D) None of these 14. For the half-cell

OH

OH

O

O

2H 2e ; E 1.30 V

At pH = 2, electrode potential is (A) 1.36 V (B) 1.30 V (C) 1.42 V (D) 1.20 V 15. Select the incorrect order of size of ions/atoms. (A) I I I (B) Fe Co Ni (C) Ni < Cu < Zn (D) None is incorrect 16. In the following reaction,

OH

OH

O

O

acid or base

( = 13C) product formed is

(A)

OH

OH

O

O

(B)

OH

OH

O

O

(C)

OH

OH

O

O

(D)

OH

OH

O

O





17. CO

O

2NO Product of

This reaction by single SE reaction

(A) CO

ONO2

(B) CO

O

NO2

(C) CO

O

O2N

(D) CO

O

NO2

18. The first ionisation potential of Na is 5.1 eV. The value of electron gain enthalpy of Na+ will be (A) –2.55 eV (B) –5.1 eV (C) –10.2 eV (D) +2.55 eV

19. 3 2 3 2

3 2

3|

CH CH ONa CH CH OH3(CH CH OH) |(major ) (major )

3

CH

CHB CH C Br A

A and B are

(A) 3

|

3 2 3|

3

CH

CHCH C OCH CH in both case (B)

3|

3 2

CH

CH C CH in both case

(C)

2

3 3| |

3 2 3 3| ||

3

CH CH

CH CHCH C OCH CH and CH –C (D)

2

3 3| |

3 3 2 3|| |

3

CH CH

CH CHCH –C and CH C OCH CH





20.

O

3CH MgBr 3

3

H O HBr Mg/ether HCHOH O

A B C D, D is

(A) CH3 CH2OH

(B) CH3

CH2OH

(C)

CH3

CH3

OH

(D)

CH2OH 21. Which is the incorrect statement about silicones ? (A) They are repeating units (SiO4) in silicates (B) They are synthetic polymers containing repeated R2SiO2 units (C) They are formed by hydrolysis of R2SiCl2 (D) All the above are incorrect statements

22. 2 3||

NaNH /CH IHO OH H3 2

O

CH CCH C CH A, A is

(A) ||

3 2 2 2 3

O

CH CCH CH CH CH (B) ||

3 2 2 2 3

O

CH CH CCH CH CH

(C) ||

3 2 3

O

CH CCH C CCH (D) ||

3 2 3

O

CH CCH HC CHCH 23. Which sequence represents the best synthesis of hexanal ? 3 2 2 2 2CH CH CH CH CH CH O hexanal

(A)

3 2 2 21. CH CH CH CH Br NaC CH

2 2 4 42. H O, H SO , HgSO

(B) ||

3 2 2 2 3

O

1. CH CH CH CH CCH

||

3

O

2. CH COOH 43. LiAlH

24. H O 2 25. PCC, CH Cl

(C) ||

3 2 2 2 3

O

1. CH CH CH CH CH CH COOH 32. CH MgBr, diethyl ether

33. H O 2 24. PCC, CH Cl

(D)

3 2 2 21. CH CH CH CH MgBr H2C CH2

O 32. H O

2 23. PCC, CH Cl





24. The interhalogen compound not obtained is (A) ICI (B) IF5 (C) BrF5 (D) BrCl7

25.

OC CH

3 2

2 2

BH .THF I /NaOHH O , OH H

X,

X is

(A)

OCOOH

(B)

OCOCH3

(C)

OCH2CHO

(D)

O

26. AgCl and NaCl are colourless. NaBr and NaI are also colourless but AgBr and AgI are coloured. This

is due to (A) Ag+ polarises Br– and I– (B) Ag+ has unpaired d-orbital (C) Ag+ depolarises Br– and I– (D) None of the above is correct 27. (Gly-Ala) has structures

(A) H3NCH2CNHCHCO2

CH3

O

(B) H3NCHC NHCH2CO2

CH3

O

(C) O2CCHNH CCH2NH3

CH3

O

(D) O2CCHNH CCHNH3

CH3

O





28. Which one of the following is not correctly matched ?

(A) Neoprene [ 2 2 n||

Cl

CH C CHCH ] (B)

|| ||

2 6 2 4

n

O O

Nylon - 66 NH(CH ) NHC(CH ) C

(C) Terylene OCH2CH2C C

OO

n

(D) PMMAn

CCH2

CH3

COCH3

O 29. The process used for purification Al2O3.2H2O is (A) magnetic separation (B) froth floatation (C) leaching (D) liquation 30. Consider following pairs of organic compounds

CCH3

O

CH2C H

O

I :

II : CH3OH CH3CH2OH

III : CH3CCH3 HCHO

O

A test that can make distinction between each pair is (A) Lucas test (B) Silver-mirror test (C) Victor Meyrer’s test (D) Iodoform test





PPPAAARRRTTT ––– IIIIIIIII ::: MMMAAATTTHHHEEEMMMAAATTTIIICCCSSS

Straight Objective Type

This section contains 30 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which only ONE option is correct 1. If f(x) = x2 + bx + 3 is not injective for values of x in the interval, 0 x 1, then the interval in

which b lies (A) (–, ) (B) (–2, ) (C) (–2, 0) (D) (–, 2)

2. 1/3 1/2

2x 0

cosx cos xlim

sin x

is

(A) 112

(B) 16

(C) 13

(D) 12

3. Tangents are drawn from the origin to the curve y = sin x, then their point of contact lie on

the curve

(A) 2 21 1 1y x

(B) 2 21 1 1x y

(C) x2 – y2 = 1 (D) x2 + y2 = 1 4. 1 2cot x cot x cosec x dx is

(A) x2ln cos c2

(B) x2ln sin c2

(C) 1 xln cos c2 2

(D) ln(sin x) – ln(cosec x – cot x) + c





5. For 0,2

, the value of

0

ln 1 tan tanx dx

is equal to

(A) ln(sec ) (B) ln(cosec )

(C) ln22

(D) 2 ln(sec )

6. Area bounded by 21y

x 2x 2

and x-axis is (in square units)

(A) 2 (B) 2

(C) 2 (D) 7. The solution of differential equation ex(x + 1)dx + (yey – xex) dy = 0, if f(0) = 0 (A) xex + 2y2ey = 0 (B) 2xex + y2ey = 0 (C) xex – 2y2ey = 0 (D) 2xex – y2ey = 0 8. The point at which the origin be shifted so that the equation x2 + y2 – 5x + 2y – 5 = 0 has no

first degree terms,

(A) 5 , 12

(B) 5, 12

(C) 51,2

(D) 51,2

9. The circle x2 + y2 – 4x – 8y + 16 = 0 rolls up the tangent to it at 2 3, 3 by 2 units,

assuming the x-axis as horizontal, then the equation of circle in the new position is (A) x2 + y2 – 6x – 2 4 3 y + 24 + 8 3 = 0

(B) x2 + y2 – 6x + 2 4 3 y + 24 + 8 3 = 0

(C) x2 + y2 + 6x – 2 4 3 y + 24 + 8 3 = 0 (D) none of these





10. If the incident ray from the point (–3, 2) parallel to the axis of parabola y2 = 4x strike the parabola, then the equation of reflected ray is

(A) x + 1 = 0 (B) x – 1 = 0 (C) y = 0 (D) y – 1 = 0

11. If 2

2atan tanb

, then the chord joining two points and an the ellipse 2 2

2 2x y 1a b

will

subtend a right angle at (A) focus (B) centre (C) end of major axis (D) none of these 12. If 5x2 + y2 = 20 represents a rectangular hyperbola then equals (A) 5 (B) 4 (C) –5 (D) none of these 13. If a, b, c, d and p are distinct real numbers such that (a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) 0, then a, b, c, d are in (A) A.P. (B) G.P. (C) H.P. (D) none of these 14. If a > 0, b > 0, c > 0 and the minimum value of a(b2 + c2) + b(c2 + a2) + c(a2 + b2) is abc,

then is (A) 2 (B) 1 (C) 6 (D) 3 15. The number of real roots of (6 – x)4 + (8 – x)4 = 16 is (A) 0 (B) 2 (C) 4 (D) none of these 16. The coefficient of x65 in the expansion of (1 + x)131(x2 – x + 1)130 is (A) 130C65 + 129C66 (B) 130C65 + 129C55 (C) 130C66 + 129C65 (D) none of these





17. If 2001 i

1 i

= x + iy, the sin–1 x + cos–1 y is equal to

(A) 2 (B)

(C) 32 (D) none of these

18. A polygon has 9 diagonals. The number of its sides is (A) 3 (B) 6 (C) 9 (D) none of these

19. If maximum and minimum values of the determinant

2 2

2 2

2 2

1 sin x cos x sin2x

sin x 1 cos x sin2x

sin x cos x 1 sin2x

are and

, then (A) + 99 = 5 (B) 3 - 17 = 25 (C) (2n – 2n) is always an even integer for nN (D) a triangle can be constructed having it’s sides as , and –

20. Let A =

2

2

2

x 6 8

3 y 9

4 5 z

, B = 2x 3 52 2y 61 4 2z 3

. If trace A = trace B then x + y + z is equal to

(A) 0 (B) 6 (C) 5 (D) none of these





21. The odds against an event A are 5 : 2 and odds in favour of another independent event B are 6 : 5 the chance that neither A nor B occurs is

(A) 5277

(B) 2577

(C) 1077

(D) 1277

22. The vectors a

and b

are non-zero and non-collinear. The value of x for which vector c = (x – 2)a

+ b

and d

= (2x + 1)a – b

are collinear

(A) 1 (B) 1/2 (C) 1/3 (D) 2 23. If vectors ax ˆ ˆ ˆi 3 j 5k and x ˆ ˆ ˆi 2 j 2axk make an acute angle with each other, for all x R,

then a belongs to the interval

(A) 1,04

(B) (0, 1)

(C) 60,25

(D) 3 ,025

24. The distance between the parallel planes 2x – y + 2z + 3 = 0 and 4x – 2y + 4z + 5 = 0 is

(A) 18

(B) 116

(C) 16

(D) none of these

25. The number of solutions of the equation sin x2 3

= x2 – 2 3 x + 4 is

(A) 0 (B) 2 (C) 1 (D) none of these





26. If (tan–1x)2 + (cot–1x)2 = 25

8 , then x equals

(A) –1 (B) 1 (C) 0 (D) none of these 27. AB is a vertical pole. The end A is on the level ground. C is the middle point of AB, P is a

point on the level ground. The portion BC subtends an angle at P. If AP = nAB, then the tan is equal to

(A) 2n

2n 1 (B) 2

nn 1

(C) 2n

n 1 (D) none of these

28. If X = {4n 3n 1 : n N} and Y = {9(n 1) : n N}, then (A) X Y (B) X = Y (C) Y X (D) none of these 29. The statement (p ~q) p is logically equivalent to (A) p (B) ~p (C) q (D) ~q 30. S.D. of a data is 6. When each observation is increased by 1, then the S.D. of new data is (A) 5 (B) 7 (C) 6 (D) 8





PHYSICS, CHEMISTRY & MATHEMATICS JEE Main 2019 Mock Tests-1 (Code:100365.1)

ANSWER KEY PHYSICS (PART-I) 1. C 2. A 3. C 4. C

5. B 6. D 7. B 8. D 9. B 10. D 11. C 12. C 13. A 14. C 15. C 16. D 17. B 18. C 19. B 20. A 21. C 22. B 23. D 24. A 25. A 26. A 27. A 28. C

29. C 30. B

CHEMISTRY (PART-II) 1. A 2. C 3. D 4. B

5. B 6. D 7. A 8. A

9. D 10. B 11. A 12. D

13. C 14. C 15. D 16. A

17. C 18. B 19. C 20. D

21. A 22. C 23. D 24. D

25. D 26. A 27. A 28. C

29. C 30. D

MATHEMATICS (PART-III) 1. C 2. A 3. A 4. D

5. A 6. D 7. B 8. A

9. A 10. B 11. B 12. C

13. B 14. C 15. B 16. D

17. B 18. B 19. C 20. D

21. B 22. C 23. C 24. C

25. C 26. A 27. A 28. A

29. A 30. C

FIITJEE - JEE (Main)




HINTS AND SOLUTIONS

PHYSICS (PART-I) 1. 600 T = ma

min600 360a 4

60

m/s2

2. 00 0

mvmgt mv

2

0

0

v2gt

3. Maximum will be corresponding to source position A

and minimum to source friction B.

So, time interval =

1 32cos2 5

5

C

A

B

O

4. As A and B will be at same potential so

charge flown equal to Q.

A Q

r

R

B

5. N = Mg

L NLF sin cos2 2

mg 3tanF 4

F

N

mg

6. 2

20

1 2

1 q 1LI2 (C C ) 2

1 2 0q L C C I

1 1 1 0

1 2 1 2

q Lq C C IC C C C

7. 1T1 = 2T2

1 1

0

T 2T

= 20 8.

A

B Y

C

A

(B+C)

A .(B+C)

9. e = B(dA/dt) = B(2r)(dr/dt)

10. 4aR 9.1 101.22




11. 2

1 22

2 1

(z 1)(z 1)

2

22

(z 1)14 (11 1)

On solving z2 = 6 12. V = Bv = (3) (2) (1) = 6V VR = VC = VL = 6V (all are in parallel)

LdiV Ldt

6 = di2dt

di 3dt

A/sec, After 2 sec, current will be 6A

13. Given v0 = ve/2

1/2 1/2GM 1 2GMR h 2 R

On solving h = R From law of conservation of energy,

2GMm 1 GMmmv(R h) 2 R

or 21 GMm GMm GMmmv2 R 2R 2R

or GMv gRR

= [(10) (6.4 106)]1/2 = 8 km/sec 14. mg + T 4a = (a)2xg

x = 2

mg 4Taa g

15. U = Q/2 dU = dW CvdT = PdV

32

(PdV + VdP) = PdV

PdV + 3VdP = 0 PV1/3 = constant

C = Cv + R

1 1 3 = 3R.

16. 0 01 2

I Ia aB B(at r )2 2 4 a2 a

B2 = B(at r = 2a)

0 0I I2 (2a) 4 a

Ratio of the fields, B1/B2 = 1 : 1 17 The magnitude of B is

88

E 6.3 V / mB 2.1 10c 3 10 m / s

T




To find the direction, we note that E

is along y-direction and the wave propagates along x-axis. Therefore, B

should be in a direction perpendicular to both x- and y-axes. Using vector algebra,

E B

should be along x-direction. Since, ˆ ˆ ˆj k i , B is along the z-direction.

Thus, B = 2.1 × 108 kT . 18. The total energy falling on the surface is U = (18 W/cm2) × (20 cm2) × (30 × 60) J = 6.48 × 105 J Therefore, the total momentum delivered (for complete absorption) is

5

38

U 6.48 10 Jp 2.16 10c 3 10 m / s

kg m/s

The average force exerted on the surface is

3

64

p 2.16 10F 1.2 10 Nt 0.18 10

19. 1002 cm, = 200 cm

v = f = 2 2.53 103 = 5.06 103 m/s 20. we know

2ne

m

Cu Cu Cu

Au Au Au

501.449

= 10/7

21. Work done by friction x

0

dxF ds mgcoscos

= mg x = 20 J

N

f

mg dxcosds

dx

dy ds

22.

22

n

5VR 5 5m1a5

23. Potential across capacitor is zero, hence energy stored is zero.

24. TyA

, T mgLAy Aycos

25. In critical condition, torque on loop is equal to gravitational torque about an axis tangent to the loop, g = mgr xB is contributing a torque,

2xM B MBsin90 r iB

1/ms at t = 0

at t = 1sec

5

1/ms

2/ms

1m/s2

1m/s2




x

mg 7 11 10i 2ArB 22 3.5 5

26. Path difference due to slab should be integral multiple of x = n ( – 1)t = n, n = 1, 2, …..

nt1

For minimum value of t, n = 1

t = n 21 1.5 1

27. R V IR V I

R V I100 100 100R V I

3%

28. AB CDR RKA

eq BCR

2KA

we have, CT 0100 02KA 2KA KA

TC = 28C

29. Apply gaus theorem

2 in

0

qE 4 r

2 20

q 1E E4 r r

field depends only on r. 30. Here, three optical phenomena take place-first refraction, then reflection,

and finally refraction.

For 1st refraction, 1.5 1 1.5 1v 2R R

1.5 0 vv

i.e. rays after refraction get parallel to the axis and strike the mirror normally, get retraced back, and final image is formed at the same place where the object is kept.

2R

h =1.5

CHEMISTRY (Part-II)

1. A Wave number of spectral line in emission spectrum of hydrogen,

H 2 21 2

1 1Rn n

…(i)

Given, H8 R9

On putting the value of in eq. (i), we get




H H 2 21 2

8 1 1R R9 n n

2 22

8 1 19 (1) n

22

8 119 n

n2 = 3 Hence, electron jumps from n2 = 3 to n1 = 1 2. C Same empirical formula, it means ratio of atoms is identical. Hence, they differ in molecular weight. 3. D At point a near low pressure region volume is very high thur (V – b) V

2ap V RT

V

apV RTV

pV aZ 1RT RT V

4. B

3CH COOH f( T) K 0.01 i 0.02046

urea f( T) K 1 1.86 i 1.1 (1 x) x = 0.1 3

3[H ] in CH COOH Cx 1 10 M pH = 3 5. B 0212 212

82 83 1( )

Pb Bi e

y 2

00 0

N1 1N N N2 2 4

total time 20where, y 2half - life 10

Thus, I and III are true 6. D 7. A SO2 is oxidised by MnO4

– in acidic medium. 2 2

2 4 45SO 2MnO 2Mn 5SO 2mol MnO4

– 5 mol SO2 Hence, 0.4 mol MnO4

– 1 mol SO2

2 2 32SO O 2SO (2–2x) (1–x) 2x (2 – 2x) = 1 mol (as determined by MnO4

–) x = 0.5

32x[SO ] 1MV




22 2x[SO ] 0.5M

V

21 x[O ] 0.5 M

V

2

3c 2

2 2

[SO ]K 2

[SO ] [O ]

8. A 2 4

2 6 6 2[Co(H O) ] 6Cl [CoCl ] 6H O Cl– replace H2O ligands and intense blue colour is developed. 9. D

2

1

VS 2.303 Rlog

V

9S 2.303 1.987log1

1 1S 2.303 1.987 0.9542 4.366 cal K mol 10. B Given, 1

22H H H 436 kJmol

122N N H 712 kJmol

o 12 2 3 f

3 1H N NH , H 46 kJmol2 2

Thus, 1

(H H)(BE) 436 kJ mol

1(N N)(BE) 712 kJ mol

(N H)(BE) ?

Based on (BE) values of

H H N N N H3 1(BE) (BE) 3(BE)2 2

3 1436 712 3x 462 2

1N Hx(BE ) 352 kJ mol

11. A 2 3 2 3Mg C H O CH C CH

C C C

,

Two and two bond 12. D 13. C 14. C In this electrode [A] = [B] in quinhydrone electrode




OH

OH

O

O

2H 2e

(A) (B) Hence, Q = [H+]2

20.0591E E log[H ]2

E 0.0591pH = E° + 0.0591 pH = 1.30 + 0.0591 2 1.42 V 15. D A true, B true, C true 16. A There is enolisation of the keto ring

OH

OH

O

O

OH

OH

O

O

OH

OH

O

O

17. C

CO

O

A B

activation deactivation, thus, attacks ring (A)2NO

but at p-site due to avoid steric effect.

18. B 19. C 20. D

O

3

3

CH MgBrH O

CH3 OH

HBr

CH3 OH2

CH3

2H O

1, 2-methyl shiftring expansionBr

Br

MgBr

Mg / ether

3

(i) HCHO(ii) H O

CH2OH

3 carbocation

predominant 21. A 22. C




Glycol blocks carbonyl group

2 3NaNH /CH I3 2 3 2 3CH CCH C CH CH CCH C C – CH

OO OO

||H

3 2 3

O

CH CCH C CCH

23. D

3

(i)

3 2 2 2 3 2 2 2 2 2(ii) H OCH CH CH CH MgBr CH CH CH CH CH CH OH

H2C CH2O

oxid PCC, CH2Cl2

3 2 2 2 2CH CH CH CH CH CHO 24. D Due to steric effect 25. D

OC CH

OCCH3

OOO

COOH3

2 2

BH /THFH O /OH

2(i) I /NaOH(ii) H

26. A 27. A 28. C 29. C 30. D

I II III Lucas test No No No Silver mirror test Yes No Yes Victor Meryer’s Test No No Yes Iodoform test Yes Yes Yes

MATHEMATICS (PART-III) 1. f(x) is not one-one then atleast one horizontal line should intersect it at two points

b0 12

b (–2, 0)

2. Let cos x = y

1/3 1/2

2y 1

y ylim1 y

=

1/3 1/2

2y 1

y 1 y 1y 1 y 1

limy 1y 1

= 1 1

13 22 12

3. y = sin x

dy cosxdx

0 cos0

..... (1)

= sin ..... (2) Squaring and adding (1) and (2), we get

y

O

y = sin x

P(, )

x




2

2 2 2cos sin 2 2

1 11

2 21 1 1

2 21 1 1y x

4. 2I 1 2cosec x cot x 2cot x dx

= 2 2cosec x 2 cosec x cot x cot x dx

= cosec x cot x dx = ln(sin x) – ln(cosec x – cot x)

5. 0

I ln 1 tan tan x dx

I = ln sec

6. Area = 21

12 dxx 2x 2

= 11

2 tan x 1

= 2 02

sq. units

y

x n = 1 O

7. xex = t

x x dx dte xedy dy

; ydt y e t 0dy

ydt t yedy

I.F. = 1 dy ye e

2

y y y yt e ye e dy c2

xex – y = 2y c

2

x = y = 0 c = 0 2xex + y2ey = 0

8. 2

25 49x y 12 4

9. Equation of tangent at P x 2 3 y 3 2 x 2 3 4 y 3 16 0

3x y 2 3 0 B = (2 + 2 cos 60º, 4 + 2 sin 60º) = 3, 4 3

22 2x 3 y 4 3 2

A(2, 4)

B

P 2 3, 3 60º

Q

10. incident ray passes through focus 11. mOA·mOB = –1

A(a cos , b sin ) B(a cos , b sin )

x

y

O




12. Since the general equation of the second degree represents a rectangular hyperbola, if 0, h2 > ab and a + b = 0

Therefore the given equation represents a rectangular hyperbola, if + 5 = 0 = –5 13. The given inequality can be re-written as (ap – b)2 + (bp – c)2 + (cp – d)2 0 ap – b = bp – c = cp – d = 0

a b c 1b c d p a, b, c, d are in G.P.

14. Using AM G.M.

12 2 2 2 2 2

2 2 2 2 2 2 6ab ac bc ba ca cb ab .ac .bc .ba .ca .cb6

a(b2 + c2) + b(c2 + a2) + c(a2 + b2) 6abc Hence = 6 15. Let y = 7 – x. Then the given equation becomes (y + 1)4 + (y –1)4 = 16 y4 + 6y2 – 7 = 0 (y2 – 1)(y2 + 7) = 0 y2 – 1 = 0 [ y2 + 7 0] y = 1 7 – x = 1 x = 6, 8 16. (1 + x)131(x2 – x + 1)130 = (1 + x)(1 + x3)130 = (1 + x3)130 + x(1 + x3)130 Hence coefficient of x65 is zero

17. 2002i

2

= (i)200 = 1 x = 1, y = 0

18. nC2 – n = 9 n n 1n 9

2

n2 – 3n – 18 = 0 n = 6

19. Applying C1 C1 + C2, we get

2

2

2

2 cos x sin2x

2 1 cos x sin2x

1 cos x 1 sin2x

Applying R2 R2 – R1 and R3 R3 – R1, we get

22 cos x sin2x0 1 01 0 1

= 2 + sin 2x

Since the maximum value of sin 2x is 1, and min value of sin 2x is (–1) Therefore = 3, = 1

20. x2 + y2 + z2 = 2x + 2y + 2z – 3 x2 – 2x + 1 + y2 – 2y + 1 + z2 – 2z + 1 = 0 (x – 1)2 + (y – 1)2 + (z – 1)2 = 0 x = 1, y = 1, z = 1 x + y + z = 3

21. Given P (A) = 27

and P (B) = 611

Required probability = P (A B) = (1 – P(A)) · (1 – P(B)) 22. Since c

and d

are collinear We have pc qd

= 0 where p + q = 0 for some scalars p and q

p{(x – 2)a + b

} + q{(2a

+ 1)a – b

} = 0




p(x – 2) = –q(2x + 1) and pb

= qb

(since a, b are

non-collinear) p(x – 2) = –p(2x + 1) x = 1/3 23. Since vectors make an acute angle with each other so Wtheir dot product must be positive. i.e. ax2 – 10 ax + 6 > 0 x R –ax2 + 10ax – 6 < 0 x R –a < 0 and 100a2

< 24a 24. Let P(x1, y1, z1) be any point on 2x – y + 2z + 3 = 0, then 2x1 – y1 + 2z1 + 3 = 0 ..... (1) The length of perpendicular from P(x1, y1, z1) to 4x – 2y + 4z + 5 = 0 is

1 1 11 1 12 2 2

2 2x y 2z 54x 2y 4z 5 16364 ( 2) 4

(Using equation (1))

25. sin x2 3

= (x – 3 )2 + 1 which can hold only if x = 3

26. y2 + 2 25y

2 8

, where y = tan–1x y = tan–1x = –

4 , 3

4 x = tan 1

4

27. Let APC = . Then tan =

AC AC AB 1AP nAB 2nAB 2n

In APB, tan ( + ) = AB AB 1AP nAB n

Now, = + –

tan = tan( ) tan1 tan( )tan

= 2

1 1 nn 2n1 1 2n 11 n 2n

B

AP

C

Hence (A) is the correct answer 28. Here, 4n 3n 1 = (1 + 3)n 3n 1 = nC0 + nC1 3 + nC2 3

2 + + nCn 3n 3n 1

= 32 {nC2 + nC33 + + nCn 3n2} which is a multiple of 9. Hence, X Y

29.

p q ~ q p ~ q (p ~ q) p T T F F T T F T T T F T F F F F F T F F

Hence p is logically equal to statement p��~ q) ��statement (p 30. S.D. (and variance) of a data is not changed when each observation is increased (or decreased) by the same constant


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