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SAMPLE TEST - 2

Time Allotted: 3 Hours

Maximum Marks: 360

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having 30 questions in each section of equal weightage. Each question is allotted 4 (four) marks for correct response.

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FIITJEE - JEE (Main)

JEE-MAIN -SAMPLE TEST -2-PCM-2


Useful Data Chemistry:

Gas Constant R = 8.314 J K1 mol1 = 0.0821 Lit atm K1 mol1 = 1.987 2 Cal K1 mol1 Avogadro's Number Na = 6.023 1023 Planck’s Constant h = 6.626 10–34 Js = 6.25 x 10-27 erg.s 1 Faraday = 96500 Coulomb 1 calorie = 4.2 Joule 1 amu = 1.66 x 10-27 kg 1 eV = 1.6 x 10-19 J

Atomic No : H=1, D=1, Li=3, Na=11, K=19, Rb=37, Cs=55, F=9, Ca=20, He=2, O=8, Au=79. Atomic Masses: He=4, Mg=24, C=12, O=16, N=14, P=31, Br=80, Cu=63.5, Fe=56, Mn=55, Si = 28 Pb=207, u=197, Ag=108, F=19, H=2, Cl=35.5

Useful Data Physics:

Acceleration due to gravity g = 10 2m / s



Section – I (Physics) 1.

If A B B A then the angle between A and B is

(A) (B) / 3 (C) / 2 (D) / 4. 1. A Sol.

A B B A or A B – B A 0

or A B A B 0 or 2 A B 0

ˆor 2ABsin n 0. As A 0 nor B 0, So sin =0. Hence =0° or 2. A ball is projected from the top of a tower of height 40 m with a velocity of 20 m/s at an angle of

300 with horizontal. The ratio of time of flight to the time taken to reach the ground is (A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) 2 : 3 2. B

Sol. t1 = 2 20 sin 3010

t1 = 2 sec

40 = - 20 sin 30 t +21

10 t2

t2 = 4 sec t1 : t2 = 1 : 2 3. A particle starting from the origin (0, 0) moves in a straight line in (x, y) plane. Its coordinates at a

later time are (3, 3). The path of the particles makes with the x-axis an angle of. (A) 45° (B) 60° (C) 0° (D) 30°. 3. B Sol. Refer fig. let at a given time the particle be (A) and be the angle between OA and OX. Then

3tan 3 tan60 or 603

4. A body of mass 1 kg is projected with a velocity of 10m/s at an angle 600 w.r.t. horizontal ground. The maximum value of gravitational potential energy in its motion is

(A) 50J (B) 25J (C) 35J (D) 37.5J 4. D

Sol . Hmax. = 4

15 m

PE = mg Hmaximum

PE = 4

15101 = 37.5



5. A block is kept on a frictionless inclined surface with angle of inclination ‘’ as shown in fig. the incline is given an acceleration ‘a’ to keep the block stationary. Then a is equal to

(A) gtan (B) g (C) gcosec (D) g /tan . 5. A Sol. As is clear from fig, the block is provided by stationary, when

gsinmacos mgsin a gtan .cos

6. A stationary particle explodes into two particles of masses m1 and m2, which move in opposite

directions, with velocities v1 and v2. The ratio of their kinetic energies (E1/E2) is (A) m2 / m1 (B) m1 / m2 (C) 1 (D) m1 v2/m2v1 6. A Sol. According to the principle of conservation of linear momentum, m1v1 – m2v2 = 0

1 2

2 1

m vm v

2 21 1

1 1 2 2

22 2 1 12 2

1 m vE m m m21E m m mm v2

7. Two spheres of masses m and M are situated in air and the gravitational force between them is F. the space around the masses is now filled with a liquid of specific gravity 3. The gravitational force will now be.

(A) 3 F (B) F (C) F / 3 (D) F / 9. 7. B Sol. The gravitational force between two masses is independent of the presence of other masses. 8. In the fig. a common emitter configuration on NPN transistor with current gain = 100 is used.

The output voltage of the amplifier will be

(A) 10 mV (B) 0.1 V (C) 1.0 V (D) 10 V

8. (C)



Sol. Voltage gain, 0 LV ac

V RAV R

i i

L0 ac

Ror V VR

ii

–3 10010 V 100 1V.

1

9. A particle is moving in a X-Y plane under the action of a force such that its instantaneous momentum jtsin3itcos3p . The instantaneous angle between the force and momentum is

(A) 2

(B) 4

(C) (D) zero 9. A

Sol. F = dtpd jtcos3itsin3

p.F = 0 = 900 10. A cylinder contains 10 kg of gas at pressure of 107 Nm–2, the quantity of gas taken out of the

cylinder, if final pressure is 2.5 × 106 Nm–2 (A) 7.5 kg (B) 10.5 kg (C) 5.2 kg (D) none of these. 10. (A)

Sol. 21 MP C ;so,P M3 V

1 1

2 2

P MP M

62

2 1 71

P 2.5 10or M M 10 2.5kgP 10

Quantity of gas taken out = 7.5Kg. 11. A particle of mass M and charge Q moving with velocity describes a circular path of radius R

when subjected to a uniform transverse magnetic field of induction B. The work done by the field when the particle completes one full circle is

2MA 2 RR

(B) zero (C) BQ 2 R (D) BQ 2 R

11. (B) Sol. In one complete circle, the displacement s = 0. As work done

W F.S Fscos 0.

12. Two masses A of 0.5 kg and B of 0.3 kg having specific heat capacities of 0.85 J/kg K and 0.9

J/kg K respectively are at temperatures 600C and 900C respectively. When connected with each other with a conducting rod, heat will flow from

(A) A to B (B) B to A (C) Initially from A to B and then from B to A (D) Heat can’t flow 12. B Sol. Irrespective of the heat content and nature of surfaces, heat will flow always from high

temperature eto low temperature object. 13. In an oscillating LC circuit, the maximum charge on the capacitor is Q. The charge on the

capacitor, when the energy is stored equally between the electric and magnetic field is (A) Q/2 (B) Q / 2 (C) Q / 3 (D) Q/3



13. (B)

Sol. Max. Energy stored in a capacitor 2

1QE2C

When energy is stored equally between the electric and magnetic fields, then energy in the

capacitor is 2 11E E2

If Q’ is the charge on the capacitor in this case, then 2

2QE2C

2 2Q 1 Q

2C 2 2C

QQ' .2

14. In YDSE, two slits are made 5 mm apart and the screen is placed 2 m away. What is the fringe separation when light of wavelength 500 nm is used?

(A) 0.002 mm (B) 0.02 mm (B) 0.2 mm (D) 2 mm 14. (C) Sol. Here, d = 5 mm = 5 × 10–3 m, D = 2m = 500 nm = 5 × 10–7 m

–7–4

–3D 5 10 2 2 10 md 5 10

= 0.2 mm. 15. A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the

lens. Then its focal length will (A) Become zero (B) become infinite (C) Reduce (D) increase. 15. (B) Sol. when 1 = 2, then from

2

1 1 2

u1 1 1– 1 –f u R R

1 0 or ff

i.e. focal length becomes infinite. 16. Momentum of a photon of energy 1 MeV in kg ms–1 will be (A) 5 × 10–22 (B) 0.33 × 106 (C) 7 × 10–24 (D) 10–22 16. (A) Sol. Energy of photon, E = 1 MeV = 1 × 106 × 1.6 × 10–19 J

Momentum of photon, Epc

6 –19–22 –1

810 1.6 10 5.3 10 kg ms .

3 10

17. If a star converts all helium in its core to oxygen then energy released per oxygen nuclei is [Mass of He – 4.0026 a.m.u., mass of O – 15.9994 a.m.u.] (A) 10.24 MeV (B) 0 (C) 7.56 MeV (D) 5 MeV 17. (A)

Sol. 4 1684 2He O Q

4 4.0026 15.9994m 0.011m 0.011 931.25E 10.24MeV 18. In the middle of the depletion layer of reverse biased p-n junction, the



(A) Electric Field is zero (B) Potential is maximum (C) Electric field is maximum (D) Potential is zero 18. (D) Sol. In case of a reserve biased p-n junction, the voltage applied supports the barrier voltage. Due to

which the electric potential at the middle of depletion layer becomes zero. 19. Consider telecommunication through optical fibres. Which of the following statements is not

correct? (A) Optical fibres have extremely low transmission loss (B) Optical fibres may have homogeneous core with a suitable cladding (C) Optical fibres can be graded refractive index (D) Optical fibers are subject to electromagnetic interference from outside 19. (D) Sol. In optical fibers, the electromagnetic waves from outside cannot interfere with the

telecommunication through optical fibers. 20. A tuning fork of frequency 256 Hz produces a faint sound when kept near the mouth of a one end

closed cylindrical tube. When water is poured, such that 31 cm of air column exists, the sound becomes loud. The speed of sound in air is

(A) 1317 ms (B) 1371ms (C) 1340 ms (D) 1332ms 20. A

Sol. vf4l

as the cylinder will behave as a closed pipe.

So, ms 1v 4lf 4 0.31 256 317.44 21. A Carnot engine takes 3 × 106 cals of heat from a reservoir at 627°C and gives it to a sink at

27°C. the work done by the engine is (A) 4.2 × 106J (B) 8.4 × 106 J (C) 16.8 × 106J (D) 3 × 106J 21. (B) Sol. Here, T1 = 627º = 627 + 273 = 900 K T2 = 27ºC = 27 + 273 = 300 K Q1 = 3 × 106 cals, W = ?

2 2

1 1

Q T 300 1Q T 900 3

1 1

2 1 2 1Q QQ W Q – Q Q –3 3

61

2 2Q 3 10 cals3 3

W = 2 × 106 cals = 8.4 × 106 J. 22. A car moving with a speed of 50 km/h, can be stopped by brakes after at least 6 m. if the same

car is moving at a speed of 100 km/h, the minimum stopping distance is (A) 12 m (B) 18 m (C) 24 m (D) 6 m. 22. (C)

Sol. –1500When u 50km / h ms ;36

v 0,s 6m

22 2–20 – 500 / 36v – uThen a 16ms

2s 2 6

500when u 100km / h 2 27.78m / s.36

22 2 2 27.78v – u us – 24m

2s 2a 2 16



23. Three charge are placed at the vertices of an equilateral triangle of side a as shown in Fig. The force experienced by the charge placed at the vertex A in a direction parallel to BC is

(A) Q2/4 0 a2 (B) – Q2/4 0 a2 (C) Zero (D) Q2/2 0 a2 23. (A)

Sol. 2

AC 20

QF f, along CA.4 a

2

AB 20

QF f, along AB.4 a

Angle between the two forces, = 120º. Net force on the charge at A. 2 2F f f 2ff cos120º

2 2 2 2f f – f f The direction of F is parallel to CB. 24. 2kg of ice at –20°C is mixed with 5 kg of water at 20°C in an insulating vessel having a negligible

heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water and ice are 1 kcal/kg./°C and 0.5 kcal/kg/°C. while the latent heat of fusion of ice = 80 kcal/kg

(A) 7 kg (B) 6 kg (C) 4 kg (D) 2 kg. 24. (B) Sol. heat released by 5 kg of water when its temperature falls from 20°C to 0°C Q1 = cm (T) = 103 × 5 × 20 = 0.2 × 105 cals This heat is used in raising the temp. of 2 kg of ice at –20° C to 0°C and then melting it

subsequently. Heat energy taken by 2kg of ice at – 20°C in coming to 0°C is Q2 = cmT = 500 × 2 × 20 = 0.2 × 105 cal The remaining heat Q = Q1 – Q2 = 0.8 × 105 cal.

5

3

Q 0.8 10Mass of ice melted,m 1kgL 80 10

Temperature of mixture will become 0°C Mass of water in it = 5 + 1 = 6 kg Mass of ice left in it = 2 – 1 = 1 kg. 25. The rms value of the electric field of the light coming from the sun is 720 NC–1. The average total

energy density of the electromagnetic wave is (A) 3.3 × 10–3 Jm–3 (B) 4.58 × 10–6 Jm–3 (C) 6.37 × 10–9 Jm–3 (D) 81.35 × 10–12 Jm–3 25. (B) Sol. Total average energy density of electromagnetic waves is

2 2 2

0 0 0 rms 0 rms1 1u E E 2 E2 2

= (8.85 × 10–12) × (720)2 = 4.58 × 10–6Jm–3



26. The displacement of the body in metres varies with time as x = 2t3 + 5. The mass of the body is 2

kg. What is the increase in its kinetic energy one second after the start of the motion (A) 20 J (B) 30 J (C) 36 J (D) 48 J 26. C

Sol. dtdx

= 6t2 = v

21

mv2 = 21 2. (6)2 = 36 J

27. A closed organ pipe of length L and an open organ pipe contain gases of densities 1 and 2

respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open organ pipe is

(A) 3L

(B) 43L

(C) 1

2

43L

(D) 2

1

43L

27. (C)

Sol. 1 21 2 1

1 2 2

4'4 2 ' 33 2

B Bv v LLL L

28. The velocity of the small ball of mass M and density d1 when dropped in a container filled with

glycerin becomes constant after sometime. If the density of glycerin is d2, the viscous force acting on the ball is.

(A) Mg (1 – d2/d1) (B) Mg d1/d2 (C) Mg (d1 – d2) (D) Mg d1d2. 28. (A) Sol. when ball is moving through a medium with a terminal velocity, then viscous force Fv = W – FB

2v 2

1 1

dMSo F Mg – d g Mg 1–d d

29. The current flowing through wire depends on time as, I = 3t2 + 2t + 5. The charge flowing through the cross-section of the wire in time t = 0 to t = 2 sec is

(A) 22 C (B) 20 C (C) 18 C (D) 5 C 29. (A)

Sol. 2dqI 3t 2t 5dt

or dq = (3t2 + 2t + 5)dt

t 2x2

t 0

q 3t 2t 5 dt

22 2

023 2

0

3r 2t 5t3 2

t t t 22 C.

30. If the maximum speed of a particle in SHM is 5 m/s. The average speed of the particle is SHM is equal to

(A) 5

m/s (B)

10 m/s (C)

25

m/s (D) zero

30. B



Sol. vavg. =

10A2

dt

tsinA

T

0

T

0 m/s

space for rough work



Section – II (Chemistry) 1. What is the equivalent weight of 4HClO ? (A) 100.5 (B) 50.3 (C) 60.1 (D) 90.5 1. A

Sol. Equivalent weight = molecular mass 1Basicity

Basicity

100.5 100.5

1gm

2. 1.82 gm of a metal requires 32.5 ml of 1N HCl to dissolve it. What is the equivalent weight of metal? (A) 46 (B) 65 (C) 56 (D) 42 2. C Sol. mille equivalents of metal = N mlV

1000 1 32.5weightE

1.82 18201000 32.5 56

32.5E

E

3. For a p-electron, the orbital angular momentum is

(A) 32h

(B) 2h

(C) 22h

(D) h

3. C Sol. For p – electron 1l

. . . 1 22 2h hO A M l l

4. Calculate the frequency of the spectral line emitted when the electron in n = 3 in H-atom de-

excites to ground state. (A) 15 12.92 10 sec (B) 15 14.52 10 sec (C) 10 13.2 10 sec (D) 10 11.9 10 sec 4. A

Sol. 2 22 2 2 21 2 1 2

1 1 1 1 1.H H

cR Z v R Z cn n n n

15 12.92 10 secv 5. Which one of the following transition metal ions is diamagnetic? (A) 2Co (B) 2Ni (C) 2Cu (D) 2Zn 5. D Sol. 2Zn ions have all paired electrons, so it is diamagnetic. 6. Which of the following is expected not to exist? (A) 3 3H BO (B) 2NaBO (C) 2B N (D) 2 6B H 6. C Sol. Boron nitrite is BN



7. The hydrogen bond is strongest in (A) O H S (B) S H O (C) F H F (D) F H O 7. C Sol. F H F hydrogen bond is strongest, because as the electronegativity difference increase, H-bond strength increases 8. What is the increasing order of lattice energies of , ,MgO CaO SrO and BaO (A) MgO CaO SrO BaO (B) CaO MgO BaO SrO (C) BaO SrO CaO MgO (D) BaO MgO CaO SrO 8. C Sol. BaO SrO CaO MgO

Lattice energy 1size of cation

9. Which is the correct order for the graph below?

(A) 1 2 3

P P P (B) 2 1 3

P P P

(C) 3 2 1

P P P (D) 3 1 2

P P P

9. C Sol.

Since 1 acc. to Boyle's lawVP

We can see from the graph

1 2 3V V V

3 2 1P P P

10. Which is the incorrect graph?

(A)

(B)

(C)

(D)

10. A Sol.

1 acc. to Boyle's lawPV

11. HI was heated in sealed tube at 4000C till the equilibrium was reached. HI was found to be 22%

decomposed. The equilibrium constant for dissociation is (A) 1.99 (B) 0.0199 (C) 0.0796 (D) 0.282 11. B

V

T

P1 P2

P3

V

P

V

PV

T

V

V

P

V

P

V

T

P1 P2

P3



Sol. 2 2

2

1 0 01 .22 .11 .11

HI H I

2 22 2

0.11 0.11 0.01990.78C

H IK

HI

12. When sulphur in the form of S8 is heated at 900K, the initial pressure of 1 atm falls by 30% at

equilibrium. This is because of conversion of same S8 to S2. Find the value of equilibrium constant for this reaction?

(A) 2.55 atm3 (B) 2.96 atm3 (C) 0.71 atm3 (D) 3.4 atm3 12. B

Sol.

8 24

Initially 1 0At equilibrium 1 .30 4 0.30 = 0.70 atm = 1.2 atm

S g S g

2

8

4' 43

'

1.22.96 atm

0.70S

P

S

PK

P

13. Catalyst is a substance which (A) Supplies energy to the reaction (B) Increases the equilibrium concentration of the product (C) Changes the equilibrium constant of the reaction (D) Lowers the activation energy barrier 13. D Sol. Catalyst lowers the activation energy barrier 14. What is the pH of solution which have 0.1M NH3 and 0.05 M 4NH Cl Given that

53 10bK NH

(A) 9.3 (B) 5 (C) 4.74 (D) 8.26 14. A Sol. For basic buffer

logb

SaltpOH pK

Base

4.7pOH 14 4.7pH = 9.3 15. DDT is an example of (A) Fungicide (B) Herbicide (C) Insecticide (D) Analgesic 15. C Sol. DDT is an example of Insecticide 16. The catalyst decrease the Ea from 100 kJ mol-1 to 80 kJmol-1. At what temperature the rate of

reaction in the absence of catalyst at 500 K will be equal to rate of reaction in presence of catalyst.

(A) 200 K (B) 400 K (C) 625 K (D) None of these



16. B Sol. Using relation,

log log2.303

aEk ART

Rate constant will not change in the presence of catalyst. 1 2log logk k

1 2

1 2

a aE ET T

2

100 80500 T

2 400T K 17. The increasing order of stability of the following free radicals is

(A) 3 3 6 5 6 52 3 2 3CH C H CH C C H C H C H C

(B) 6 5 6 5 3 33 2 3 2C H C C H C H CH C CH C H

(C) 6 5 6 5 3 32 3 3 2C H C H C H C CH C CH C H

(D) 3 3 6 5 6 52 3 3 2CH C H CH C C H C C H C H

17. A Sol. Free radicals stability

H5C6 C C6H5

C6H5

H5C6 CH

C6H5

CH3 C CH3

CH3

H3CC H

CH3Highly stable by delocalisation 9-hyperconjugative hydrogens

and +I effect

> > >

18. Among the following structures which is not a correct resonance form:

(A) CH2+

N O

CH3

(B) CH2 N O

CH3 (C) CH2 N

+O

CH3

(D) H2C N+

O

CH3 18. B Sol. In (B), resonating structure is not proper Lewis structure. 19. The reaction

CH3

CH3

OHH

2SOCl CH3

CH3

ClH

2SO HCl

Proceeds by the ……….mechanism. (A) SNi (B) SN2 (C) SE2 (D) SE1 19. A Sol. Since the given reaction proceeds through retention of configuration the mechanism of the

reaction is SNi



20. Which of the following statement about the sulphates of alkali metal is correct? (A) Except 2 4Li SO , all sulphates of other alkali metals are soluble in water

(B) All sulphates of alkali metals except 2 4Li SO forms alum. (C) The sulphates of alkali metals cannot be hydrolysed (D) All of these 20. D Sol. All are facts 21. The compound which undergoes SN1 reaction most rapidly is

(A) Br

(B) Br

(C) CH2Br

(D) Br

21. B Sol. In case of the given compound the carbocation intermediate is resonance stabilized. Hence the

reaction is most favoured by SN1. 22. A solution of (+) 1-chloro-1-phenylethane in toluene racemises slowly in the presence of small

amount of SbCl5, due to the formation of (A) carbanion (B) carbene (C) free-radical (D) carbocation 22. D Sol.

H Cl

Ph

CH3

65

SbClSbCl Ph C

CH3

H6

5

SbClSbCl

H Cl

Ph

CH3

HCl

Ph

CH3

23. A mixture of benzoic acid and phenol may be differentiated by treatment with (A) NaHCO3 (B) NaOH (C) NH3 solution (D) KOH 23. A Sol. 3 2 2Ph COOH NaHCO PhCOO Na CO H O

3PhOH NaHCO No reaction

24. The compound which reacts fastest with Lucas’ reagent at room temperature is (A) 1-butanol (B) 2-butanol (C) 2-methyl propan-1-ol (D) 2-methyl propan-2-ol 24. D Sol. 2 –methylpropan-2-ol forms the most stable 30 carbocation hence (D) is most reactive.

25. The ether O

when treated with HI produces

(A)

IOH+

(B)

OHI+



(C)

OHI +

(D) OH + I

25. A Sol.

O+ H

+ Ph O+

H Ph2 2

IPhOH CH Ph Ph CH I

26. 1 g of a monobasic acid HB (having pKa = 5) in 100g water lower the freezing point by 0.155K. If

0.45 g of same acid required 15 ml 5M

NaOH solution for complete neutralization

-1

2 2, 1.86 kg mol , density of H O is 1 g/ml

fk H O K , then which of the following option is

incorrect regarding above question (A) degree of ionization of acid is 0.25 (B) the pH of the resultant solution at the end point of neutralization is greater than 7 (C) normal molecular mass of acid is 150 (D) 10 g of acid HB is hypotonic with 0.625 g urea at the same temperature and same volume of

H2O. 26. D Sol.

f fT K m

120acid observed

M

meq of acid = meq of base 150

acid normalM

150 1.25 1120

i

0.25 pH > 7 as the solution contains salt of strong base and weak acid. 27. Which of the following electrolyte is most effective in the coagulation of gold sol? (A)

3NaNO (B) 64

K Fe CN (C) 3 4

Na PO (D) 2

MgCl

27. B Sol. According to Hardy Schulze Rule. 28. Which of the following is false? (A) When NaCl is heated in the atmosphere of Na, metal excess defect arise due to the migration

of Na from vapour to NaCl lattice. (B) Both Schottky and Frenkel defects can effect electrical conductivity and this conduction is

known as intrinsic semiconduction. (C) Density decreases in Frenkel defect but remains same in Schottky defect (D) In compounds having metal excess defect F-centres are present which makes them

paramagnetic, coloured and help in n-type semiconduction. 28. C Sol. C is false. 29. Ammonia gas can be dried by (A) conc. H2SO4 (B) P2O5 (C) quick lime (D) None of these 29. C Sol. 3 2 4 4 42NH H SO NH SO



3 2 5 4 43NH P O NH PO 30. What is the relation between the following compounds?

NH

NH

HN

O

O O

N

N

N

OH

OH OH

NH+

NH

+

HN+

O-

O-

O-

(a) (b) (c) (I) ‘b’ and ‘c’ are resonating structures. (II) ‘a’ and ‘b’ are resonating structures. (III) ‘a’ and ‘b’ are tautomers. (IV) ‘a’ and ‘c’ are tautomers. Which of the above statement/s is/are correct? (A) (I), (III) only (B) (II), (IV) only (C) (III) only (D) (IV) only 30. C Sol. Tautomers are functional group isomers in equilibrium

pace for rough work



Section – III (Mathematics) 1. A man standing on a horizontal plane, observes the angle of elevation of the top of a tower to be

. After walking a distance equal to double the height of the tower, the angle of the elevation becomes 2 , then is equal to

(A) 2 (B)

6 (C)

12 (D)

18

1. B Sol.

tanhx

2

3 2 tantan 21 tan

hx

22

2 tan3tan 3 3 tan 21 tan

tan 0

2 2 13tan 1 tan3

1tan .63

A

B

2h

O x

2

C

h

2. If 1 1cos cos ,2yx then

2 24 4 cosx xy y is equal to: (A) 24sin (B) 24sin (C) 4 (D) 2sin 2 2. B

Sol. 1 1cos cos2

yx

2

1 2cos 1 12 4xy yx

2

21 1 cos2 4xy yx

2

22 1 1 2 cos4yx xy

2

2 2 2 24 1 1 4cos 4 cos4yx x y xy

2 2 2 2 21 4 4cos 4 cosx y x y xy

2 2 2 2 2 2 24 4 4cos 4 cosx y x y xy x y 2 2 24 4 4cos 4 cosx y xy

2 2 2 24 4 cos 4 1 cos 4sinx xy y

3. 1 3sin 2 cos5

is equal to

(A) 625

(B) 2425

(C) 45

(D) 2425

3. D



Sol. Put 1 3 3cos cos5 5

given expression sin 2

2sin cos 4sin5

4 3 242.5 5 25

4. If 12 3 63cos ,cos ,cos ,13 5 65

then cos is:

(A) 1 (B) 2 (C) 3 (D) 0 4. D Sol. cos cos cos cos sin sin cos sin cos sin cos sin sin

12 3 63 5 4 63 5 3 16 12 4 16. . . . . . . .13 5 65 13 5 65 13 5 65 13 5 65

2268 1260 240 768 013.5.65

5. India plays two matches each with West Indies and Australia. In any match probabilities of India

getting points 0, 1 and 2 are 0.45, 0.05 and 0.5 respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points is

(A) 0.8750 (B) 0.0875 (C) 0.0625 (D) 0.0250 5. B Sol. Since there are just four matches to be played, India can get a maximum of 8 points. P(India gets at least 7 points) = P(getting exactly 7 points) + P(getting exactly 8 points) = P(getting 2 in each of the 3 matches and 1 in one match) + P(getting 2 in each of the four matches) 3 44 4

3 40.5 0.05 0.5c c

30.5 0.2 0.5 0.125 0.7 0.0875 6. The equation of the plane through intersection of planes 2 3 4x y z and 2 5x y z and

perpendicular to the plane 5 3 6 8 0,x y z is (A) 7 2 3 81 0x y z (B) 23 14 9 48 0x y z (C) 23 14 9 48 0x y z (D) 51 15 50 173 0x y z 6. D Sol. Any plane through the intersection of the given planes is 2 3 4 2 5 0x y z K x y z

i.e. 1 2 2 3 5 4 0K x K y K z K This is to 5 3 6 8 0x y z 5 1 2 3 2 6 3 0K K K 29 7 0K

297

K

reqd. plane is

58 29 29 1451 2 3 4 07 7 7 7

x y z

51 15 50 173 0x y z 51 15 50 173 0x y z



7. Image of the point 1,6,3A in the line 1 21 2 3x y z is

(A) 7,1,0 (B) 2,5,7 (C) 1,0,7 (D) None of these 7. C Sol. Any point P on the given line is , 2 1,3 2r r r direction ratios of AP are 1,2 5,3 1 r r r Now AP to the given line if 1 1 2 2 5 3 3 1 0r r r 14 14 0 1r r P is 1,3,5 .

Thus the foot of the from A on the line is 1,3,5 . Let , ,B a b c be the image of A in the given line. Then P is the mid point of AB.

1 1 12

a a and 3 5 7

2c c

6 3 02

b b

B is (1, 0, 7) 8. If 2 4 ,

a b c d a c d b c d then

(A) 6 (B) -6 (C) 10 (D) 8 8. A Sol. 2 4 2 4

a b c d a c d b c d

a c d b c d

2, 4 6 9. If ,

x y a x y b and 1, x a then

(A) 2

2 2

1,

a a a ba a bx ya a

(B) 2

2 2,a a b a a b

x ya a

(C)

2 ,b a b

x ya

can have any value (D)

2

2

1b b a by

a

,y can have any value

9. A Sol. x y a y a x

…1

x y b …2

. 1x a …3

By (1) and (2), x a x b

x a b

x a b

a x a a b

. 1.a a x a a b

2 1.a x a a b

2

a a bx

a



2

a a by a x a

a

2

2

1a a a ba

10. The product of the perpendicular and drawn from any point on a hyperbola to its asymptotes is

(A) aba b

(B) 2 2

aba b

(C) 2 2

2 2

a ba b

(D) 2 2

2 2

a ba b

10. C

Sol. Product of perpendicular and drawn from any point on a hyperbola to its asymptotes = 2 2

2 2

a ba b

11. The number of values of c such that the straight line 4y x c touches the curve 2

2 14x y is

(A) 0 (B) 1 (C) 2 (D) infinite 11. C

Sol. 4y x c touches 2 2

14 1x y

If 2 2 2 2 2 2Here 4, 1, 4c a m b a b m

. .i e If 2 4 16 1 65c This gives two values of c.

12. If 2x ya b touches the ellipse

2 2

2 2 1,x ya b

then its eccentric angle is equal to

(A) 00 (B) 090 (C) 045 (D) 060 12. C

Sol. Let be the eccentric angle, then tangent at ' ' is cos sin 1x ya b

Also, 12 2x y

a b is the tangent

cos sin 11 12 2

01 1cos ;sin 452 2

13. The latus rectum of a parabola whose focal chord is PSQ such that SP = 3 and SQ = 2 is given by

(A) 245

(B) 125

(C) 65

(D) None of these

13. A

Sol. Since 2 1 1l SP SQ (Where l is the semi-latus rectum of the parabola)

2 1 1 5 12 2423 2 6 5 5

l ll

latus rectum = 245



14. Let PQ and RS be tangent at the extremities of the diameter PR of a circle of radius r. If PS and RQ intersect at a point X on the circumference of the circle, then 2r equals

(A) .PQ RS (B) 2

PQ RS (C) 2 .PQ RSPQ RS

(D) 2 2

2PQ RS

14. A Sol.

tan2

PQ PQPR r

Also tan2 2

RSr

. . cot2RSi e

r

2

.tan .cot4

PQ RSr

24 . 2r PQ RS r PQ RS

S Q

P R r r

/ 2

2

15. The line of the system 2 2 3 4 0a x y b x y situated farthest from the point (1, 1) is (A) 4 0x y (B) 2 6 0x y (C) 2 6 0x y (D) None of these 15. A Sol. Point of intersection of 2 2 0 and 3 4 0x y x y is (2, 2).

The line of the system 2 2 3 4 0a x y b x y situated farthest from the point (1, 1) is the line passing through (2, 2) and perpendicular to the line passing through points (1,1) and (2,2).

16. Solution of the equation

tandy yx y xdx x

is

(A) sin x Cxy (B) sin y Cx

x (C) sin x Cy

y (D) sin y Cy

x

16. B Sol. Put y vx

dv dyv xdx dx

cot log sin logdxvdv v Cxx

sin sin yv Cx Cxx

17. The solution of the equation 2 2x xy dy xy y dx is

(A) /y xxy Ce (B) /x yxy Ce (C) 2 1/ xyx Ce (D) None of these 17. B

Sol. 2

2 .dy xy ydx x xy

Put y vx

2 2

1 1dv v v dv v vv x x vdx v dx v



2 2 221 1

v v v v vv v

2

1 2v dxdvxv

2

1 1 2 dxdvv xv

1 log 2log .v x constv

21log log logx y x C

y x

21 1log . . logx y x C C xy

y x

/1

x yC xy e

/ /

1

1 .x y x yxy e C eC

(Take 11CC

)

18. / 2

0

sinsin cos

x dxx x

(A) 4 (B)

2 (C) 0 (D) 1

18. A

Sol. / 2

0

sin4sin cos

x dxx x

19. 10

0

sin x dx

is

(A) 20 (B) 8 (C) 10 (D) 18 19. A

Sol. 10 / 2

0 0 / 2

sin 10 sin sinx dx xdx xdx

= / 2

0 / 210 cos 10 cos

x x

10 1 1 10 2 20

20. 31

dxx x is equal to

(A) 3

3

1 1 1log3 1 1

x cx

(B) 3

3

1 1 3log3 1 3

x cx

(C) 3

2 1log3 1

cx

(D) 3

1 1log3 1

cx

20. A Sol. Put 3 2 21 3 2x y x dx ydy

23 3 3

2 23 11 3 1

dx ydy ydyy yx x x x



2

2 2 1 1. log3 3 2 11

dy y cyy

3

3

1 1 1log3 1 1

x cx

21. If 21 log , 0,8

f x x bx x x where 0b is a constant, then

(A) f x has local minimum at 14

x for b = 1

(B) f x has no extremum for 0 1b

(C) f x has a local minimum at 2 1 , 1

4b bx b

(D) f x has a local maximum at 2 1 , 1

4

b bx b

21. B

Sol. f’(x) = 2

1 12 ; " 28 8

b x f xx x

For max or min, f’(x) = 0 1 – 8bx + 16x2 =0

2 28 64 64 1

32 4

b b b bx

Case I. When b = 0, f’(x) = 1 2 08

xx

Hence there is no extreme value for b = 0 Case II. When 0 < b < 1, then 2 1 0b x is not real. there is no extreme value for 0 1b

Case III. When 1,b then 14

x

" 2 2 0f x

3

1 1''' 16 044

f x for xx

14

x is a point of inflexion.

Case IV. When b > 1

Let 2 21 1,

4 4

b b b b

Clearly We have

2

2 21 2 1 1 2 1' 2 18 2 16 4 16

bf x b x x bx x bx x x

2 22 1 11 14 4 4 4b bx b x b

x

2 2x x x xx x

' 0, 0f x for x

f ' x < 0, for x and ' 0f x for x

f x has a local Max. at



21 14

x b b and local min. at

21 14

x b b

22. 1 22 cot log 1 ,y x x x x then y

(A) increases in 0, only (B) decreases in 0

(C) neither increases nor decreases in 0 (D) increases in , 22. D

Sol. 2 2

1 121 1

dydx x x x 2

1 .2 12 1

xx

2

2 2 2

1 1 121 1 1

x xx x x x

= 2 2

1 121 1x x

2

2 2

2 1 11 1

xx x

2 2

2

2 1 11

x xx

Now 2 20 2 1 1 0 dy x xdx

22 22 1 1x x

4 24 3 0x x Which is true for all real value of x. Hence y increases in , .

23.

2

2 2 2

sin1 1 1sin sin sin

0lim 1 2 ...

x

x x xx

n

=

(A) (B) 0 (C) 12

n (D) n

23. D Sol. We have

2

2 2 2 sin1/ sin 1/ sin 1/ sin

0lim 1 2 ...

xx x x

xn

1/

20

1lim 1' 2 ' .... where 1sin

tt

tn t

x

Since 1 2 3 ... 1 2 3 ... tt t t tn nn n

1

1 2 ...2

tt t t n n

n nn

12

tnn

1/ 1/ 11 2 ... .2

tt t t t nn n

1/ 1/ 1 1lim 1 2 ... lim2 2

tt t t t

t t

n nn n



11 2 12

nn n n n

1/lim 1 2 ...

tt t t

tn n

24. 2 2 2 2

0 1 2 .... 1 ,nC C C C where n is an even integer is

(A) 2nnC (B) 21 n n

nC (C) 211 n n

nC (D) None of these 24. D Sol. We have 2

0 1 21 ... ... 1n nnx C C x C x C x

1 20 2

11 ... 1 ... 2n

n nn

CC CCx x x x

2 2 2 20 1 2 ... 1 n

nC C C C = co-eff. Of the term independent of x in Product of R.H.S. (1) and (2) = co-eff. Of term independent of x in

11 1n

nxx

= co-eff. Of 2/ 21 1

nnn nnx in x C nis even

25. The digit at unit’s place in the number 1225 1915 122513 11 23 is equal to: (A) 0 (B) 1 (C) 2 (D) None of these 25. B Sol. 1225 1915 122510 3 10 1 20 3

I II III

Last term of Ist and IIIrd expansion gets cancelled and rest are divisible by 10 and in IInd expansion last term is 1 and rest are divisible by 10 so at unit place this number has digit 1.

26. The number of parallelograms that can be formed from a set of four parallel lines intersecting

another set of three parallel lines is (A) 6 (B) 10 (C) 12 (D) None of these 26. D Sol. Since each parallelogram is formed by choosing two parallel straight lines from the first and two

from the second set. total no. of parallelograms formed 4 3

2 2 6 3 18C C 27. The number of triangles which can be formed from 12 points out of which 7 are collinear is (A) 105 (B) 210 (C) 175 (D) 185 27. D Sol. Reqd. no. of triangles

12 73 3

12 11 10 7 6 51 2 3 1 2 3

C C

220 35 185



28. One root of the equation

3 8 3 3

3 3 8 3 03 3 3 8

xx

x

is which of the following?

(A) 83

(B) 23

(C) 13

(D) 16 .3

28. B Sol. Operate 1 2 3 ,R R R

3 8 3 3

3 3 8 3 03 3 3 8

xx

x

2 / 3x as one of the roots. 29. If is a complex number such that 2 1 0 , then 31 is (A) (B) 2 (C) 0 (D) 1 29. A Sol. Since 2 1 0

1 1 4 1 32 2

i 2 .or

Take

1031 31 30 3. .

101 .

Take 312 31 2

62 60 2 2 2. 1

Hence 31 . 30. The domain of

1

2 2cot ,xf x x R

x x

is

(A) R (B) 0R (C) ,R n n N (D) None of these.

30. D

Sol. Domain of 1cot x is R and 2 2

x

x x is defined if 2 2x x .

2. .i e x is not integer 2 2x x

Hence 2x non negative integer . .i e 0 or +ve integer. Hence domain = 0,R n n Z

Thus correct answer is (D)

space for rough work

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